tag:blogger.com,1999:blog-42740384326889519632024-03-19T05:03:18.709-07:00A Level H1 and H2 BiologyUnited JC Unionhttp://www.blogger.com/profile/06131630985986252085noreply@blogger.comBlogger27125tag:blogger.com,1999:blog-4274038432688951963.post-14353702063739495742011-02-03T02:14:00.002-08:002011-02-03T02:14:56.166-08:00ISC Animations<div class="MsoNormal" style="text-align: justify;"><b style="mso-bidi-font-weight: normal;"><u>Sequencing<o:p></o:p></u></b></div><div class="MsoNormal"><b style="mso-bidi-font-weight: normal;"><u><span style="mso-spacerun: yes;"> </span><a href="http://www.dnalc.org/ddnalc/resources/cycseq.html"><span style="mso-fareast-font-family: +mn-ea;">http://www.dnalc.org/ddnalc/resources/cycseq.html</span></a><o:p></o:p></u></b></div><div class="MsoNormal"><b style="mso-bidi-font-weight: normal;"><u><span style="mso-fareast-font-family: +mn-ea;"><a href="http://www.dnai.org/text/mediashowcase/index2.html?id=552">http://www.dnai.org/text/mediashowcase/index2.html?id=552</a><br />
<br />
</span><a href="http://smcg.cifn.unam.mx/enp-unam/03-EstructuraDelGenoma/animaciones/secuencia.swf">http://smcg.cifn.unam.mx/enp-unam/03-EstructuraDelGenoma/animaciones/secuencia.swf</a><br />
<br />
Linkage mapping<br />
<br />
<a href="http://courses.bio.psu.edu/fall2005/biol110/tutorials/tutorial8.htm">http://courses.bio.psu.edu/fall2005/biol110/tutorials/tutorial8.htm</a> <o:p></o:p></u></b></div>United JC Unionhttp://www.blogger.com/profile/06131630985986252085noreply@blogger.com1tag:blogger.com,1999:blog-4274038432688951963.post-10606737532547665582011-02-03T02:14:00.000-08:002011-02-03T02:14:03.739-08:00Diversity and Evolution Animations<div class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0cm;"><b style="mso-bidi-font-weight: normal;"><u><span style="font-family: "Arial","sans-serif"; font-size: 12.0pt;">Animations - Evolution<o:p></o:p></span></u></b></div><div class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0cm;"><br />
</div><div class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0cm;"><span style="font-family: "Arial","sans-serif"; font-size: 12.0pt;"><a href="http://www.echalk.co.uk/Science/Biology/PepperedMoth/Peppered_MothWEB.swf">http://www.echalk.co.uk/Science/Biology/PepperedMoth/Peppered_MothWEB.swf</a><o:p></o:p></span></div><div class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0cm;"><span style="font-family: "Arial","sans-serif"; font-size: 12.0pt;">(Peppered Moth Game)<o:p></o:p></span></div><div class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0cm;"><br />
</div><div class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0cm;"><span style="font-family: "Arial","sans-serif"; font-size: 12.0pt;"><a href="http://www.johnkyrk.com/evolution.html">http://www.johnkyrk.com/evolution.html</a><o:p></o:p></span></div><div class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0cm;"><span style="font-family: "Arial","sans-serif"; font-size: 12.0pt;">(Interesting overview of evolution)<o:p></o:p></span></div><div class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0cm;"><br />
</div><div class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0cm;"><span style="font-family: "Arial","sans-serif"; font-size: 12.0pt;"><a href="http://www.wwnorton.com/college/biology/discoverbio3/core/content/index/animations.asp">http://www.wwnorton.com/college/biology/discoverbio3/core/content/index/animations.asp</a><o:p></o:p></span></div><div class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0cm;"><span style="font-family: "Arial","sans-serif"; font-size: 12.0pt;">(Various Animation Links)<o:p></o:p></span></div><div class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0cm;"><br />
</div><div class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0cm;"><span style="font-family: "Arial","sans-serif"; font-size: 12.0pt;"><a href="http://trc.ucdavis.edu/biosci10v/bis10v/media/ch13/archipelago.html">http://trc.ucdavis.edu/biosci10v/bis10v/media/ch13/archipelago.html</a><o:p></o:p></span></div><div class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0cm;"><span style="font-family: "Arial","sans-serif"; font-size: 12.0pt;">(Allopatric Speciation)<o:p></o:p></span></div><div class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0cm;"><br />
</div><div class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0cm;"><span style="font-family: "Arial","sans-serif"; font-size: 12.0pt;"><a href="http://www.pbs.org/wgbh/evolution/darwin/origin/index.html">http://www.pbs.org/wgbh/evolution/darwin/origin/index.html</a><o:p></o:p></span></div><div class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0cm;"><span style="font-family: "Arial","sans-serif"; font-size: 12.0pt;">(Adapative Radiation)<o:p></o:p></span></div><div class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0cm;"><br />
</div><div class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0cm;"><span style="font-family: "Arial","sans-serif"; font-size: 12.0pt;"><a href="http://trc.ucdavis.edu/biosci10v/bis10v/media/ch12/finches.html">http://trc.ucdavis.edu/biosci10v/bis10v/media/ch12/finches.html</a><o:p></o:p></span></div><div class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0cm;"><span style="font-family: "Arial","sans-serif"; font-size: 12.0pt;">(Adaptive Radiation)<o:p></o:p></span></div><div class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0cm;"><br />
</div><div class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0cm;"><span style="font-family: "Arial","sans-serif"; font-size: 12.0pt;"><a href="http://bcs.whfreeman.com/thelifewire/content/chp24/2402001.html">http://bcs.whfreeman.com/thelifewire/content/chp24/2402001.html</a><o:p></o:p></span></div><div class="MsoNormal" style="line-height: normal; margin-bottom: .0001pt; margin-bottom: 0cm;"><span style="font-family: "Arial","sans-serif"; font-size: 12.0pt;">(Speciation)<o:p></o:p></span></div>United JC Unionhttp://www.blogger.com/profile/06131630985986252085noreply@blogger.com2tag:blogger.com,1999:blog-4274038432688951963.post-1039626458117456362011-02-03T02:13:00.001-08:002011-02-03T02:13:23.239-08:00Homeostasis and Cell Signalling Animations<h1><span lang="EN-GB">Animations - Homeostasis<o:p></o:p></span></h1><div class="MsoNormal"><br />
</div><div class="MsoNormal"><span lang="EN-GB" style="font-family: "Arial","sans-serif";">http://pennhealth.com/health_info/animationplayer/homeostasis.html<o:p></o:p></span></div><div class="MsoNormal"><br />
</div><div class="MsoNormal"><span lang="EN-GB" style="font-family: "Arial","sans-serif";">http://trc.ucdavis.edu/biosci10v/bis10v/week10/07homeostasis.html<o:p></o:p></span></div><div class="MsoNormal"><br />
</div><div class="MsoNormal"><br />
</div><div class="MsoNormal"><br />
</div><h1><span lang="EN-GB">Animations – Cell Signalling<o:p></o:p></span></h1><div class="MsoNormal"><br />
</div><div class="MsoNormal"><span lang="EN-GB" style="font-family: "Arial","sans-serif";">http://entochem.tamu.edu/G-Protein/index.html<o:p></o:p></span></div><div class="MsoNormal"><br />
</div><div class="MsoNormal"><span lang="EN-GB" style="font-family: "Arial","sans-serif";">http://www.wiley.com/college/pratt/0471393878/student/animations/signal_transduction/index.html<o:p></o:p></span></div><div class="MsoNormal"><br />
</div><div class="MsoNormal"><span lang="EN-GB" style="font-family: "Arial","sans-serif";">http://www.wiley.com/legacy/college/boyer/0470003790/animations/signal_transduction/signal_transduction.htm<o:p></o:p></span></div><div class="MsoNormal"><br />
</div><div class="MsoNormal"><span lang="EN-GB" style="font-family: "Arial","sans-serif";">http://www.vivo.colostate.edu/hbooks/pathphys/endocrine/moaction/surface.html<o:p></o:p></span></div><div class="MsoNormal"><br />
</div>United JC Unionhttp://www.blogger.com/profile/06131630985986252085noreply@blogger.com1tag:blogger.com,1999:blog-4274038432688951963.post-30966719050438387362011-02-03T02:12:00.001-08:002011-02-03T02:12:39.629-08:00Cell Membrane Animations<div class="MsoNormal"><br />
</div><div class="MsoNormal"><a href="http://home.earthlink.net/~shalpine/anim/Life/memb.htm">http://home.earthlink.net/~shalpine/anim/Life/memb.htm</a><o:p></o:p></div><div class="MsoNormal"><br />
</div><div class="MsoNormal"><a href="http://telstar.ote.cmu.edu/Hughes/tutorial/cellmembranes/orient2.swf">http://telstar.ote.cmu.edu/Hughes/tutorial/cellmembranes/orient2.swf</a><o:p></o:p></div><div class="MsoNormal"><br />
</div><div class="MsoNormal"><a href="http://telstar.ote.cmu.edu/biology/downloads/membranes/index.html">http://telstar.ote.cmu.edu/biology/downloads/membranes/index.html</a><o:p></o:p></div><div class="MsoNormal"><br />
</div><div class="MsoNormal"><a href="http://www.stolaf.edu/people/giannini/flashanimat/lipids/membrane%20fluidity.swf">http://www.stolaf.edu/people/giannini/flashanimat/lipids/membrane%20fluidity.swf</a><o:p></o:p></div><div class="MsoNormal"><br />
</div><div class="MsoNormal"><a href="http://www.microscopy-uk.org.uk/index.html?http://www.microscopy-uk.org.uk/amateurs/movigifs.html">http://www.microscopy-uk.org.uk/index.html?http://www.microscopy-uk.org.uk/amateurs/movigifs.html</a><o:p></o:p></div><div class="MsoNormal"><br />
</div><div class="MsoNormal"><a href="http://microscope.mbl.edu/baypaul/microscope/animations/rarbur/rarbur_01.htm">http://microscope.mbl.edu/baypaul/microscope/animations/rarbur/rarbur_01.htm</a><o:p></o:p></div><div class="MsoNormal"><br />
</div><div class="MsoNormal"><a href="http://www.bioscope.org/taste/cd1/a0255.htm">http://www.bioscope.org/taste/cd1/a0255.htm</a><o:p></o:p></div><div class="MsoNormal"><br />
</div><div class="MsoNormal"><br />
</div>United JC Unionhttp://www.blogger.com/profile/06131630985986252085noreply@blogger.com2tag:blogger.com,1999:blog-4274038432688951963.post-31665610403752380662011-02-03T02:11:00.000-08:002011-02-03T02:11:53.574-08:00Biological Molecules Animation Links<table border="1" cellpadding="0" cellspacing="0" class="MsoTableGrid" style="border-bottom-style: none; border-collapse: collapse; border-color: initial; border-left-style: none; border-right-style: none; border-top-style: none; border-width: initial; width: 669px;"><tbody>
<tr style="mso-yfti-firstrow: yes; mso-yfti-irow: 0;"> <td colspan="2" style="border: solid windowtext 1.0pt; mso-border-alt: solid windowtext .5pt; padding: 0cm 5.4pt 0cm 5.4pt; width: 501.4pt;" valign="top" width="669"> <div class="MsoNormal"><b style="mso-bidi-font-weight: normal;">Topic: Proteins<o:p></o:p></b></div><div class="MsoNormal"><br />
</div></td> </tr>
<tr style="mso-yfti-irow: 1;"> <td style="border-top: none; border: solid windowtext 1.0pt; mso-border-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0cm 5.4pt 0cm 5.4pt; width: 28.5pt;" valign="top" width="38"> <div align="center" class="MsoNormal" style="text-align: center;"><b style="mso-bidi-font-weight: normal;">No.<o:p></o:p></b></div></td> <td style="border-bottom: solid windowtext 1.0pt; border-left: none; border-right: solid windowtext 1.0pt; border-top: none; mso-border-alt: solid windowtext .5pt; mso-border-left-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0cm 5.4pt 0cm 5.4pt; width: 472.9pt;" valign="top" width="631"> <div align="center" class="MsoNormal" style="text-align: center;"><b style="mso-bidi-font-weight: normal;">Web-links<o:p></o:p></b></div></td> </tr>
<tr style="mso-yfti-irow: 2; mso-yfti-lastrow: yes;"> <td style="border-top: none; border: solid windowtext 1.0pt; mso-border-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0cm 5.4pt 0cm 5.4pt; width: 28.5pt;" valign="top" width="38"> <div align="center" class="MsoNormal" style="text-align: center;">1<o:p></o:p></div></td> <td style="border-bottom: solid windowtext 1.0pt; border-left: none; border-right: solid windowtext 1.0pt; border-top: none; mso-border-alt: solid windowtext .5pt; mso-border-left-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0cm 5.4pt 0cm 5.4pt; width: 472.9pt;" valign="top" width="631"> <div class="MsoNormal"><span style="font-size: 10.0pt;"><a href="http://www.bbc.co.uk/education/asguru/biology/02biologicalmolecules/01proteins/12polymers/index.shtml">http://www.bbc.co.uk/education/asguru/biology/02biologicalmolecules/01proteins/12polymers/index.shtml</a><o:p></o:p></span></div><div class="MsoNormal" style="margin-left: -46.5pt; text-indent: 46.5pt;"><br />
</div></td> </tr>
</tbody></table><br />
<div class="MsoNormal"><br />
</div><div class="MsoNormal"><br />
</div><table border="1" cellpadding="0" cellspacing="0" class="MsoTableGrid" style="border-bottom-style: none; border-collapse: collapse; border-color: initial; border-left-style: none; border-right-style: none; border-top-style: none; border-width: initial; width: 669px;"><tbody>
<tr style="mso-yfti-firstrow: yes; mso-yfti-irow: 0;"> <td colspan="2" style="border: solid windowtext 1.0pt; mso-border-alt: solid windowtext .5pt; padding: 0cm 5.4pt 0cm 5.4pt; width: 501.4pt;" valign="top" width="669"> <div class="MsoNormal"><b style="mso-bidi-font-weight: normal;">Topic: Enzymes<o:p></o:p></b></div><div class="MsoNormal"><br />
</div></td> </tr>
<tr style="mso-yfti-irow: 1;"> <td style="border-top: none; border: solid windowtext 1.0pt; mso-border-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0cm 5.4pt 0cm 5.4pt; width: 28.5pt;" valign="top" width="38"> <div align="center" class="MsoNormal" style="text-align: center;"><b style="mso-bidi-font-weight: normal;">No.<o:p></o:p></b></div></td> <td style="border-bottom: solid windowtext 1.0pt; border-left: none; border-right: solid windowtext 1.0pt; border-top: none; mso-border-alt: solid windowtext .5pt; mso-border-left-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0cm 5.4pt 0cm 5.4pt; width: 472.9pt;" valign="top" width="631"> <div align="center" class="MsoNormal" style="text-align: center;"><b style="mso-bidi-font-weight: normal;">Web-links<o:p></o:p></b></div></td> </tr>
<tr style="mso-yfti-irow: 2;"> <td style="border-top: none; border: solid windowtext 1.0pt; mso-border-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0cm 5.4pt 0cm 5.4pt; width: 28.5pt;" valign="top" width="38"> <div align="center" class="MsoNormal" style="text-align: center;">1<o:p></o:p></div></td> <td style="border-bottom: solid windowtext 1.0pt; border-left: none; border-right: solid windowtext 1.0pt; border-top: none; mso-border-alt: solid windowtext .5pt; mso-border-left-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0cm 5.4pt 0cm 5.4pt; width: 472.9pt;" valign="top" width="631"> <div class="MsoNormal"><a href="http://www.stolaf.edu/people/giannini/flashanimat/enzymes/prox-orien.swf">http://www.stolaf.edu/people/giannini/flashanimat/enzymes/prox-orien.swf</a><o:p></o:p></div><div class="MsoNormal"><br />
</div></td> </tr>
<tr style="mso-yfti-irow: 3;"> <td style="border-top: none; border: solid windowtext 1.0pt; mso-border-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0cm 5.4pt 0cm 5.4pt; width: 28.5pt;" valign="top" width="38"> <div align="center" class="MsoNormal" style="text-align: center;">2<o:p></o:p></div></td> <td style="border-bottom: solid windowtext 1.0pt; border-left: none; border-right: solid windowtext 1.0pt; border-top: none; mso-border-alt: solid windowtext .5pt; mso-border-left-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0cm 5.4pt 0cm 5.4pt; width: 472.9pt;" valign="top" width="631"><pre style="line-height: 14.4pt;"><span class="HTMLTypewriter3"><span style="color: black; font-size: 10.0pt;"><a href="http://resources.emb.gov.hk/biology/english/virtual_lab/flash/enzyme_lab.html" target="_blank">http://resources.emb.gov.hk/biology/english/virtual_lab/flash/enzyme_lab.html</a><o:p></o:p></span></span></pre><div class="MsoNormal"><br />
</div></td> </tr>
<tr style="mso-yfti-irow: 4;"> <td style="border-top: none; border: solid windowtext 1.0pt; mso-border-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0cm 5.4pt 0cm 5.4pt; width: 28.5pt;" valign="top" width="38"> <div align="center" class="MsoNormal" style="text-align: center;">3<o:p></o:p></div></td> <td style="border-bottom: solid windowtext 1.0pt; border-left: none; border-right: solid windowtext 1.0pt; border-top: none; mso-border-alt: solid windowtext .5pt; mso-border-left-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0cm 5.4pt 0cm 5.4pt; width: 472.9pt;" valign="top" width="631"><pre style="line-height: 14.4pt;"><span class="HTMLTypewriter3"><span style="color: black; font-size: 10.0pt;"><a href="http://www.stolaf.edu/people/giannini/flashanimat/enzymes/allosteric.swf" target="_blank">http://www.stolaf.edu/people/giannini/flashanimat/enzymes/allosteric.swf</a><o:p></o:p></span></span></pre><div class="MsoNormal"><br />
</div></td> </tr>
<tr style="mso-yfti-irow: 5; mso-yfti-lastrow: yes;"> <td style="border-top: none; border: solid windowtext 1.0pt; mso-border-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0cm 5.4pt 0cm 5.4pt; width: 28.5pt;" valign="top" width="38"> <div align="center" class="MsoNormal" style="text-align: center;">4<o:p></o:p></div></td> <td style="border-bottom: solid windowtext 1.0pt; border-left: none; border-right: solid windowtext 1.0pt; border-top: none; mso-border-alt: solid windowtext .5pt; mso-border-left-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0cm 5.4pt 0cm 5.4pt; width: 472.9pt;" valign="top" width="631"> <div class="MsoNormal"><a href="http://www.northland.cc.mn.us/biology/Biology1111/animations/enzyme.swf">http://www.northland.cc.mn.us/biology/Biology1111/animations/enzyme.swf</a><o:p></o:p></div><pre style="line-height: 14.4pt;"><span class="HTMLTypewriter3"><span style="color: black; font-size: 10.0pt;"><o:p> </o:p></span></span></pre></td> </tr>
</tbody></table><div class="MsoNormal"><br />
</div>United JC Unionhttp://www.blogger.com/profile/06131630985986252085noreply@blogger.com0tag:blogger.com,1999:blog-4274038432688951963.post-86618685607470529612011-02-03T01:44:00.000-08:002011-02-03T01:44:22.512-08:00Common Mistakes & Misconceptions in Biology<div align="center" class="MsoNormal" style="text-align: center;"><b style="mso-bidi-font-weight: normal;"><span style="font-family: "Arial","sans-serif"; font-size: 14.0pt;">Common Mistakes & Misconceptions in Biology <span style="mso-spacerun: yes;"> </span><o:p></o:p></span></b></div><div class="MsoNormal"><br />
</div><div class="MsoNormal"><b style="mso-bidi-font-weight: normal;"><u><span style="font-family: "Arial","sans-serif";">Cell Structure and Cell Membrane <o:p></o:p></span></u></b></div><div class="MsoNormal" style="margin-left: 21.0pt; mso-list: l0 level1 lfo1; tab-stops: list 21.0pt; text-indent: -21.0pt;"><!--[if !supportLists]--><span style="font-family: Wingdings; font-size: 8.0pt; mso-bidi-font-family: Wingdings; mso-fareast-font-family: Wingdings;"><span style="mso-list: Ignore;">l<span style="font: 7.0pt "Times New Roman";"> </span></span></span><!--[endif]--><span style="font-family: "Arial","sans-serif";">Cisterna </span><span lang="ZH-CN" style="font-family: 宋体; mso-bidi-font-family: Arial;">≠</span><span style="font-family: "Arial","sans-serif";"> crista <o:p></o:p></span></div><div class="MsoNormal" style="margin-left: 21.0pt;"><span style="font-family: "Arial","sans-serif";">Cisterna = lumen of RER or Golgi Apparatus <o:p></o:p></span></div><div class="MsoNormal" style="margin-left: 21.0pt;"><span style="font-family: "Arial","sans-serif";">Crista (cristae= <i style="mso-bidi-font-style: normal;">plural</i>) = infolding of the inner mitochondrial membrane <o:p></o:p></span></div><div class="MsoNormal" style="margin-left: 21.0pt;"><br />
</div><table border="1" cellpadding="0" cellspacing="0" class="MsoTableGrid" style="border-collapse: collapse; border: none; mso-border-alt: solid windowtext .5pt; mso-padding-alt: 0cm 5.4pt 0cm 5.4pt; mso-yfti-tbllook: 480;"><tbody>
<tr style="mso-yfti-firstrow: yes; mso-yfti-irow: 0;"> <td style="border: solid windowtext 1.0pt; mso-border-alt: solid windowtext .5pt; padding: 0cm 5.4pt 0cm 5.4pt; width: 149.4pt;" valign="top" width="199"> <div align="center" class="MsoNormal" style="text-align: center;"><b style="mso-bidi-font-weight: normal;"><span style="font-family: "Arial","sans-serif";">Wrong<o:p></o:p></span></b></div></td> <td style="border-left: none; border: solid windowtext 1.0pt; mso-border-alt: solid windowtext .5pt; mso-border-left-alt: solid windowtext .5pt; padding: 0cm 5.4pt 0cm 5.4pt; width: 276.7pt;" valign="top" width="369"> <div align="center" class="MsoNormal" style="text-align: center;"><b style="mso-bidi-font-weight: normal;"><span style="font-family: "Arial","sans-serif";">Correct<o:p></o:p></span></b></div></td> </tr>
<tr style="mso-yfti-irow: 1;"> <td style="border-top: none; border: solid windowtext 1.0pt; mso-border-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0cm 5.4pt 0cm 5.4pt; width: 149.4pt;" valign="top" width="199"> <div class="MsoNormal"><span style="font-family: "Arial","sans-serif";">Ions are polar…..<o:p></o:p></span></div><div class="MsoNormal"><br />
</div></td> <td style="border-bottom: solid windowtext 1.0pt; border-left: none; border-right: solid windowtext 1.0pt; border-top: none; mso-border-alt: solid windowtext .5pt; mso-border-left-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0cm 5.4pt 0cm 5.4pt; width: 276.7pt;" valign="top" width="369"> <div class="MsoNormal"><b style="mso-bidi-font-weight: normal;"><span style="font-family: "Arial","sans-serif";">Ions</span></b><span style="font-family: "Arial","sans-serif";"> are <b style="mso-bidi-font-weight: normal;">charged</b> hence they cannot pass through the hydrophobic core of the Cell Surface Membrane. <o:p></o:p></span></div></td> </tr>
<tr style="mso-yfti-irow: 2;"> <td style="border-top: none; border: solid windowtext 1.0pt; mso-border-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0cm 5.4pt 0cm 5.4pt; width: 149.4pt;" valign="top" width="199"> <div class="MsoNormal"><span style="font-family: "Arial","sans-serif";">Nucleolus contains rRNA<o:p></o:p></span></div></td> <td style="border-bottom: solid windowtext 1.0pt; border-left: none; border-right: solid windowtext 1.0pt; border-top: none; mso-border-alt: solid windowtext .5pt; mso-border-left-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0cm 5.4pt 0cm 5.4pt; width: 276.7pt;" valign="top" width="369"> <div class="MsoNormal"><span style="font-family: "Arial","sans-serif";">Nucleolus <b style="mso-bidi-font-weight: normal;">synthesizes</b> rRNA (which forms part of ribosome), but it itself is made of DNA (coiled around histones)<o:p></o:p></span></div></td> </tr>
<tr style="mso-yfti-irow: 3;"> <td style="border-top: none; border: solid windowtext 1.0pt; mso-border-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0cm 5.4pt 0cm 5.4pt; width: 149.4pt;" valign="top" width="199"> <div class="MsoNormal"><span style="font-family: "Arial","sans-serif";">Fatty acid/ hydrocarbon tails interact with hydrophobic bonds<o:p></o:p></span></div></td> <td style="border-bottom: solid windowtext 1.0pt; border-left: none; border-right: solid windowtext 1.0pt; border-top: none; mso-border-alt: solid windowtext .5pt; mso-border-left-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0cm 5.4pt 0cm 5.4pt; width: 276.7pt;" valign="top" width="369"> <div class="MsoNormal"><span style="font-family: "Arial","sans-serif";">Fatty acid/ hydrocarbon tails interact with hydrophobic <b style="mso-bidi-font-weight: normal;">interactions</b><o:p></o:p></span></div></td> </tr>
<tr style="mso-yfti-irow: 4;"> <td style="border-top: none; border: solid windowtext 1.0pt; mso-border-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0cm 5.4pt 0cm 5.4pt; width: 149.4pt;" valign="top" width="199"> <div class="MsoNormal"><span style="font-family: "Arial","sans-serif";">Ribosomes have one membrane<o:p></o:p></span></div></td> <td style="border-bottom: solid windowtext 1.0pt; border-left: none; border-right: solid windowtext 1.0pt; border-top: none; mso-border-alt: solid windowtext .5pt; mso-border-left-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0cm 5.4pt 0cm 5.4pt; width: 276.7pt;" valign="top" width="369"> <div class="MsoNormal"><span style="font-family: "Arial","sans-serif";">Ribosomes do not have membranes<o:p></o:p></span></div></td> </tr>
<tr style="mso-yfti-irow: 5;"> <td style="border-top: none; border: solid windowtext 1.0pt; mso-border-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0cm 5.4pt 0cm 5.4pt; width: 149.4pt;" valign="top" width="199"> <div class="MsoNormal"><span style="font-family: "Arial","sans-serif";">Facilitated diffusion uses only channel proteins. Active transport uses only carrier proteins.<o:p></o:p></span></div></td> <td style="border-bottom: solid windowtext 1.0pt; border-left: none; border-right: solid windowtext 1.0pt; border-top: none; mso-border-alt: solid windowtext .5pt; mso-border-left-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0cm 5.4pt 0cm 5.4pt; width: 276.7pt;" valign="top" width="369"> <div class="MsoNormal"><span style="font-family: "Arial","sans-serif";">Facilitated diffusion uses <b style="mso-bidi-font-weight: normal;">both</b> channel protein and carrier proteins. The carrier protein can change conformation without ATP hydrolysis. <o:p></o:p></span></div><div class="MsoNormal"><span style="font-family: "Arial","sans-serif";">Active transport uses only carrier proteins which can change conformation only upon ATP hydrolysis. (Campbell 8<sup>th</sup> edition pg 135). </span><i style="mso-bidi-font-style: normal;"><span style="font-family: "Arial","sans-serif"; font-size: 8.0pt;">Once, Cambridge question goes against this rule (see 2006/P2/Q1) but you still have to answer the question based on diagram.</span></i><span style="font-family: "Arial","sans-serif";"><o:p></o:p></span></div></td> </tr>
<tr style="mso-yfti-irow: 6;"> <td style="border-top: none; border: solid windowtext 1.0pt; mso-border-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0cm 5.4pt 0cm 5.4pt; width: 149.4pt;" valign="top" width="199"> <div class="MsoNormal"><span style="font-family: "Arial","sans-serif";">Passage of small, hydrophobic molecules through the phospholipid bilayer is called “diffusion”<o:p></o:p></span></div></td> <td style="border-bottom: solid windowtext 1.0pt; border-left: none; border-right: solid windowtext 1.0pt; border-top: none; mso-border-alt: solid windowtext .5pt; mso-border-left-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0cm 5.4pt 0cm 5.4pt; width: 276.7pt;" valign="top" width="369"> <div class="MsoNormal"><span style="font-family: "Arial","sans-serif";">Passage of small, hydrophobic molecules through the phospholipid bilayer is called “simple diffusion”<o:p></o:p></span></div></td> </tr>
<tr style="mso-yfti-irow: 7; mso-yfti-lastrow: yes;"> <td style="border-top: none; border: solid windowtext 1.0pt; mso-border-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0cm 5.4pt 0cm 5.4pt; width: 149.4pt;" valign="top" width="199"> <div class="MsoNormal"><span style="font-family: "Arial","sans-serif";">Exocytosis is the same as active transport<o:p></o:p></span></div></td> <td style="border-bottom: solid windowtext 1.0pt; border-left: none; border-right: solid windowtext 1.0pt; border-top: none; mso-border-alt: solid windowtext .5pt; mso-border-left-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0cm 5.4pt 0cm 5.4pt; width: 276.7pt;" valign="top" width="369"> <div class="MsoNormal"><span style="font-family: "Arial","sans-serif";">Exocytosis is a type of bulk transport requiring vesicles, while active transport uses carrier proteins.<o:p></o:p></span></div></td> </tr>
</tbody></table><div class="MsoNormal"><br />
</div><div class="MsoNormal"><b style="mso-bidi-font-weight: normal;"><u><span style="font-family: "Arial","sans-serif";">Biological Molecules<o:p></o:p></span></u></b></div><table border="1" cellpadding="0" cellspacing="0" class="MsoTableGrid" style="border-collapse: collapse; border: none; mso-border-alt: solid windowtext .5pt; mso-padding-alt: 0cm 5.4pt 0cm 5.4pt; mso-yfti-tbllook: 480;"><tbody>
<tr style="mso-yfti-firstrow: yes; mso-yfti-irow: 0;"> <td style="border: solid windowtext 1.0pt; mso-border-alt: solid windowtext .5pt; padding: 0cm 5.4pt 0cm 5.4pt; width: 149.4pt;" valign="top" width="199"> <div align="center" class="MsoNormal" style="text-align: center;"><b style="mso-bidi-font-weight: normal;"><span style="font-family: "Arial","sans-serif";">Wrong<o:p></o:p></span></b></div></td> <td style="border-left: none; border: solid windowtext 1.0pt; mso-border-alt: solid windowtext .5pt; mso-border-left-alt: solid windowtext .5pt; padding: 0cm 5.4pt 0cm 5.4pt; width: 276.7pt;" valign="top" width="369"> <div align="center" class="MsoNormal" style="text-align: center;"><b style="mso-bidi-font-weight: normal;"><span style="font-family: "Arial","sans-serif";">Correct<o:p></o:p></span></b></div></td> </tr>
<tr style="mso-yfti-irow: 1;"> <td style="border-top: none; border: solid windowtext 1.0pt; mso-border-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0cm 5.4pt 0cm 5.4pt; width: 149.4pt;" valign="top" width="199"> <div class="MsoNormal"><span style="font-family: "Arial","sans-serif";">Cellulose is a protein OR collagen is a polysaccharide<o:p></o:p></span></div><div class="MsoNormal"><br />
</div></td> <td style="border-bottom: solid windowtext 1.0pt; border-left: none; border-right: solid windowtext 1.0pt; border-top: none; mso-border-alt: solid windowtext .5pt; mso-border-left-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0cm 5.4pt 0cm 5.4pt; width: 276.7pt;" valign="top" width="369"> <div class="MsoNormal"><span style="font-family: "Arial","sans-serif";">Cellulose is a polysaccharide AND collagen is a (fibrous) protein<o:p></o:p></span></div><div class="MsoNormal"><br />
</div></td> </tr>
<tr style="mso-yfti-irow: 2;"> <td style="border-top: none; border: solid windowtext 1.0pt; mso-border-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0cm 5.4pt 0cm 5.4pt; width: 149.4pt;" valign="top" width="199"> <div class="MsoNormal"><span style="font-family: "Arial","sans-serif";">Collagen is made up of triple helix<o:p></o:p></span></div></td> <td style="border-bottom: solid windowtext 1.0pt; border-left: none; border-right: solid windowtext 1.0pt; border-top: none; mso-border-alt: solid windowtext .5pt; mso-border-left-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0cm 5.4pt 0cm 5.4pt; width: 276.7pt;" valign="top" width="369"> <div class="MsoNormal"><b style="mso-bidi-font-weight: normal;"><span style="font-family: "Arial","sans-serif";">Tropocollagen</span></b><span style="font-family: "Arial","sans-serif";"> is made up of triple helix.<o:p></o:p></span></div></td> </tr>
<tr style="mso-yfti-irow: 3; mso-yfti-lastrow: yes;"> <td style="border-top: none; border: solid windowtext 1.0pt; mso-border-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0cm 5.4pt 0cm 5.4pt; width: 149.4pt;" valign="top" width="199"> <div class="MsoNormal"><span style="font-family: "Arial","sans-serif";">Globular protein is globular in shape<o:p></o:p></span></div></td> <td style="border-bottom: solid windowtext 1.0pt; border-left: none; border-right: solid windowtext 1.0pt; border-top: none; mso-border-alt: solid windowtext .5pt; mso-border-left-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0cm 5.4pt 0cm 5.4pt; width: 276.7pt;" valign="top" width="369"> <div class="MsoNormal"><span style="font-family: "Arial","sans-serif";">Globular protein is<b style="mso-bidi-font-weight: normal;"> spherical </b>in shape. <i style="mso-bidi-font-style: normal;">Use synonyms!</i><b style="mso-bidi-font-weight: normal;"><o:p></o:p></b></span></div></td> </tr>
</tbody></table><div class="MsoNormal"><br />
</div><div class="MsoNormal"><br />
</div><div class="MsoNormal"><b style="mso-bidi-font-weight: normal;"><u><span style="font-family: "Arial","sans-serif";">Enzymes<o:p></o:p></span></u></b></div><div class="MsoNormal" style="margin-left: 21.0pt; mso-list: l0 level1 lfo1; tab-stops: list 21.0pt; text-indent: -21.0pt;"><!--[if !supportLists]--><span style="font-family: Wingdings; font-size: 8.0pt; mso-bidi-font-family: Wingdings; mso-fareast-font-family: Wingdings;"><span style="mso-list: Ignore;">l<span style="font: 7.0pt "Times New Roman";"> </span></span></span><!--[endif]--><b style="mso-bidi-font-weight: normal;"><span style="font-family: "Arial","sans-serif";">Active site</span></b><span style="font-family: "Arial","sans-serif";"> can only be used when describing <b style="mso-bidi-font-weight: normal;">enzymes.</b> <u><o:p></o:p></u></span></div><div class="MsoNormal" style="margin-left: 21.0pt;"><span style="font-family: "Arial","sans-serif";">For other proteins eg. Receptors, pumps, transcription factors, use <b style="mso-bidi-font-weight: normal;">allosteric sites / binding sites.<o:p></o:p></b></span></div><div class="MsoNormal" style="margin-left: 21.0pt; mso-list: l0 level1 lfo1; tab-stops: list 21.0pt 84.0pt; text-indent: -21.0pt;"><!--[if !supportLists]--><span style="font-family: Wingdings; font-size: 8.0pt; mso-bidi-font-family: Wingdings; mso-fareast-font-family: Wingdings;"><span style="mso-list: Ignore;">l<span style="font: 7.0pt "Times New Roman";"> </span></span></span><!--[endif]--><span style="font-family: "Arial","sans-serif";">Failure to name the <i style="mso-bidi-font-style: normal;">types</i> of bonds affected when subjected to denaturation agents/ factors. E.g. Candidates must mention <b style="mso-bidi-font-weight: normal;">hydrogen/ionic bonds</b> when high temperature is applied, <b style="mso-bidi-font-weight: normal;">ionic bonds</b> when pH is changed or heavy metals added, hydrophobic interactions when organic solvents added and disulphide bonds when a reducing agent is added.<b style="mso-bidi-font-weight: normal;"><o:p></o:p></b></span></div><div class="MsoNormal" style="margin-left: 21.0pt; mso-list: l0 level1 lfo1; tab-stops: list 21.0pt; text-indent: -21.0pt;"><!--[if !supportLists]--><span style="font-family: Wingdings; font-size: 8.0pt; mso-bidi-font-family: Wingdings; mso-fareast-font-family: Wingdings;"><span style="mso-list: Ignore;">l<span style="font: 7.0pt "Times New Roman";"> </span></span></span><!--[endif]--><span style="font-family: "Arial","sans-serif";">Denaturation has two meanings: <u><o:p></o:p></u></span></div><div class="MsoNormal" style="margin-left: 39.0pt; mso-list: l0 level2 lfo1; tab-stops: list 39.0pt; text-indent: -18.0pt;"><!--[if !supportLists]--><span style="font-family: "Arial","sans-serif"; font-size: 8.0pt; mso-fareast-font-family: Arial;"><span style="mso-list: Ignore;">-<span style="font: 7.0pt "Times New Roman";"> </span></span></span><!--[endif]--><span style="font-family: "Arial","sans-serif";">For proteins: disruption of tertiary structure, causing the protein to lose its 3D conformation <o:p></o:p></span></div><div class="MsoNormal" style="margin-left: 39.0pt; mso-list: l0 level2 lfo1; tab-stops: list 39.0pt; text-indent: -18.0pt;"><!--[if !supportLists]--><span style="font-family: "Arial","sans-serif"; font-size: 8.0pt; mso-fareast-font-family: Arial;"><span style="mso-list: Ignore;">-<span style="font: 7.0pt "Times New Roman";"> </span></span></span><!--[endif]--><span style="font-family: "Arial","sans-serif";">For DNA : breaking of hydrogen bonds between complementary base pairs causing the two DNA strands to separate <u><o:p></o:p></u></span></div><div class="MsoNormal"><br />
</div><div class="MsoNormal"><b style="mso-bidi-font-weight: normal;"><u><span style="font-family: "Arial","sans-serif";">Cell and Nuclear Division <o:p></o:p></span></u></b></div><div class="MsoNormal" style="margin-left: 21.0pt; mso-list: l0 level1 lfo1; tab-stops: list 21.0pt; text-indent: -21.0pt;"><!--[if !supportLists]--><span style="font-family: Wingdings; font-size: 8.0pt; mso-bidi-font-family: Wingdings; mso-fareast-font-family: Wingdings;"><span style="mso-list: Ignore;">l<span style="font: 7.0pt "Times New Roman";"> </span></span></span><!--[endif]--><span style="font-family: "Arial","sans-serif";">Chromatids separate to become chromosomes <o:p></o:p></span></div><div class="MsoNormal" style="text-indent: 21.0pt;"><i style="mso-bidi-font-style: normal;"><span style="font-family: "Arial","sans-serif";">Chromatids are no longer called chromatids after they separate from the centromere <o:p></o:p></span></i></div><div class="MsoNormal"><br />
</div><table border="1" cellpadding="0" cellspacing="0" class="MsoTableGrid" style="border-collapse: collapse; border: none; mso-border-alt: solid windowtext .5pt; mso-padding-alt: 0cm 5.4pt 0cm 5.4pt; mso-yfti-tbllook: 480;"><tbody>
<tr style="mso-yfti-firstrow: yes; mso-yfti-irow: 0;"> <td style="border: solid windowtext 1.0pt; mso-border-alt: solid windowtext .5pt; padding: 0cm 5.4pt 0cm 5.4pt; width: 194.4pt;" valign="top" width="259"> <div align="center" class="MsoNormal" style="text-align: center;"><b style="mso-bidi-font-weight: normal;"><span style="font-family: "Arial","sans-serif";">Wrong<o:p></o:p></span></b></div></td> <td style="border-left: none; border: solid windowtext 1.0pt; mso-border-alt: solid windowtext .5pt; mso-border-left-alt: solid windowtext .5pt; padding: 0cm 5.4pt 0cm 5.4pt; width: 231.7pt;" valign="top" width="309"> <div align="center" class="MsoNormal" style="text-align: center;"><b style="mso-bidi-font-weight: normal;"><span style="font-family: "Arial","sans-serif";">Correct<o:p></o:p></span></b></div></td> </tr>
<tr style="mso-yfti-irow: 1;"> <td style="border-top: none; border: solid windowtext 1.0pt; mso-border-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0cm 5.4pt 0cm 5.4pt; width: 194.4pt;" valign="top" width="259"> <div class="MsoNormal"><span style="font-family: "Arial","sans-serif";">Chromosomes are pulled to opposite ends of the cells by spindle fibres <o:p></o:p></span></div><div class="MsoNormal"><br />
</div></td> <td style="border-bottom: solid windowtext 1.0pt; border-left: none; border-right: solid windowtext 1.0pt; border-top: none; mso-border-alt: solid windowtext .5pt; mso-border-left-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0cm 5.4pt 0cm 5.4pt; width: 231.7pt;" valign="top" width="309"> <div class="MsoNormal"><span style="font-family: "Arial","sans-serif";">Chromosomes are pulled to opposite <b style="mso-bidi-font-weight: normal;">poles </b>of the cells by spindle fibres <o:p></o:p></span></div><div class="MsoNormal"><br />
</div></td> </tr>
<tr style="mso-yfti-irow: 2;"> <td style="border-top: none; border: solid windowtext 1.0pt; mso-border-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0cm 5.4pt 0cm 5.4pt; width: 194.4pt;" valign="top" width="259"> <div class="MsoNormal"><span style="font-family: "Arial","sans-serif";">Centromere split<o:p></o:p></span></div></td> <td style="border-bottom: solid windowtext 1.0pt; border-left: none; border-right: solid windowtext 1.0pt; border-top: none; mso-border-alt: solid windowtext .5pt; mso-border-left-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0cm 5.4pt 0cm 5.4pt; width: 231.7pt;" valign="top" width="309"> <div class="MsoNormal"><span style="font-family: "Arial","sans-serif";">Centromere <b style="mso-bidi-font-weight: normal;">divide </b><o:p></o:p></span></div><div class="MsoNormal"><br />
</div></td> </tr>
<tr style="mso-yfti-irow: 3;"> <td style="border-top: none; border: solid windowtext 1.0pt; mso-border-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0cm 5.4pt 0cm 5.4pt; width: 194.4pt;" valign="top" width="259"> <div class="MsoNormal"><span style="font-family: "Arial","sans-serif";">Chromosome replicate<o:p></o:p></span></div></td> <td style="border-bottom: solid windowtext 1.0pt; border-left: none; border-right: solid windowtext 1.0pt; border-top: none; mso-border-alt: solid windowtext .5pt; mso-border-left-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0cm 5.4pt 0cm 5.4pt; width: 231.7pt;" valign="top" width="309"> <div class="MsoNormal"><span style="font-family: "Arial","sans-serif";">DNA replicate (so each chromosome is now seen comprising of two sister chromatids)<o:p></o:p></span></div></td> </tr>
<tr style="mso-yfti-irow: 4;"> <td style="border-top: none; border: solid windowtext 1.0pt; mso-border-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0cm 5.4pt 0cm 5.4pt; width: 194.4pt;" valign="top" width="259"> <div class="MsoNormal"><span style="font-family: "Arial","sans-serif";">Mitosis has no homologous chromosomes<o:p></o:p></span></div></td> <td style="border-bottom: solid windowtext 1.0pt; border-left: none; border-right: solid windowtext 1.0pt; border-top: none; mso-border-alt: solid windowtext .5pt; mso-border-left-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0cm 5.4pt 0cm 5.4pt; width: 231.7pt;" valign="top" width="309"> <div class="MsoNormal"><span style="font-family: "Arial","sans-serif";">Homologous chromosomes are already in cells irregardless of whether they are undergoing cell and nuclear division. All humans have 23 pairs of homologous chromosomes.<o:p></o:p></span></div></td> </tr>
<tr style="mso-yfti-irow: 5;"> <td style="border-top: none; border: solid windowtext 1.0pt; mso-border-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0cm 5.4pt 0cm 5.4pt; width: 194.4pt;" valign="top" width="259"> <div class="MsoNormal"><span style="font-family: "Arial","sans-serif";">Diploid number = amount of DNA<o:p></o:p></span></div></td> <td style="border-bottom: solid windowtext 1.0pt; border-left: none; border-right: solid windowtext 1.0pt; border-top: none; mso-border-alt: solid windowtext .5pt; mso-border-left-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0cm 5.4pt 0cm 5.4pt; width: 231.7pt;" valign="top" width="309"> <div class="MsoNormal"><span style="font-family: "Arial","sans-serif";">Diploid number is the total number of chromosomes. DNA amount is twice in a chromosome with 2 sister chromatids compared to a chromosome that does not have sister chromatids.<o:p></o:p></span></div></td> </tr>
<tr style="mso-yfti-irow: 6;"> <td style="border-top: none; border: solid windowtext 1.0pt; mso-border-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0cm 5.4pt 0cm 5.4pt; width: 194.4pt;" valign="top" width="259"> <div class="MsoNormal"><span style="font-family: "Arial","sans-serif";">DNA replication occurs in the interphase between meiosis I and II<o:p></o:p></span></div></td> <td style="border-bottom: solid windowtext 1.0pt; border-left: none; border-right: solid windowtext 1.0pt; border-top: none; mso-border-alt: solid windowtext .5pt; mso-border-left-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0cm 5.4pt 0cm 5.4pt; width: 231.7pt;" valign="top" width="309"> <div class="MsoNormal"><span style="font-family: "Arial","sans-serif";">DNA replication occurs once only in the interphase before meiosis I<o:p></o:p></span></div></td> </tr>
<tr style="mso-yfti-irow: 7; mso-yfti-lastrow: yes;"> <td style="border-top: none; border: solid windowtext 1.0pt; mso-border-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0cm 5.4pt 0cm 5.4pt; width: 194.4pt;" valign="top" width="259"> <div class="MsoNormal"><span style="font-family: "Arial","sans-serif";">Organism is sterile because gametes have odd number of chromosomes<o:p></o:p></span></div></td> <td style="border-bottom: solid windowtext 1.0pt; border-left: none; border-right: solid windowtext 1.0pt; border-top: none; mso-border-alt: solid windowtext .5pt; mso-border-left-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0cm 5.4pt 0cm 5.4pt; width: 231.7pt;" valign="top" width="309"> <div class="MsoNormal"><span style="font-family: "Arial","sans-serif";">Organism is sterile because the organism itself have odd number of chromosomes and/or have even number of chromosomes that do not occur in homologous pairs <o:p></o:p></span></div></td> </tr>
</tbody></table><div class="MsoNormal"><br />
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</div><div class="MsoNormal"><b style="mso-bidi-font-weight: normal;"><u><span style="font-family: "Arial","sans-serif";">DNA and Genomics<o:p></o:p></span></u></b></div><div class="MsoNormal" style="margin-left: 21.0pt; mso-list: l0 level1 lfo1; tab-stops: list 21.0pt; text-indent: -21.0pt;"><!--[if !supportLists]--><span style="font-family: Wingdings; font-size: 8.0pt; mso-bidi-font-family: Wingdings; mso-fareast-font-family: Wingdings;"><span style="mso-list: Ignore;">l<span style="font: 7.0pt "Times New Roman";"> </span></span></span><!--[endif]--><span style="font-family: "Arial","sans-serif";">Unzip </span><span lang="ZH-CN" style="font-family: 宋体; mso-bidi-font-family: Arial;">≠</span><span style="font-family: "Arial","sans-serif";"> unwind<u><o:p></o:p></u></span></div><div class="MsoNormal" style="margin-left: 21.0pt;"><span style="font-family: "Arial","sans-serif";">Unzip = breaking of hydrogen bonds between complementary base pairs to separate the two DNA strands.<o:p></o:p></span></div><div class="MsoNormal" style="margin-left: 21.0pt;"><span style="font-family: "Arial","sans-serif";">Unwind = untwisting of the DNA double helix <o:p></o:p></span></div><div class="MsoNormal" style="margin-left: 21.0pt; mso-list: l0 level1 lfo1; tab-stops: list 21.0pt; text-indent: -21.0pt;"><!--[if !supportLists]--><span style="font-family: Wingdings; font-size: 8.0pt; mso-bidi-font-family: Wingdings; mso-fareast-font-family: Wingdings;"><span style="mso-list: Ignore;">l<span style="font: 7.0pt "Times New Roman";"> </span></span></span><!--[endif]--><!--[if gte vml 1]><v:shapetype id="_x0000_t202"
coordsize="21600,21600" o:spt="202" path="m,l,21600r21600,l21600,xe"> <v:stroke joinstyle="miter"/> <v:path gradientshapeok="t" o:connecttype="rect"/> </v:shapetype><v:shape id="_x0000_s1027" type="#_x0000_t202" style='position:absolute;
left:0;text-align:left;margin-left:271.7pt;margin-top:6.75pt;width:191.25pt;
height:21.75pt;text-indent:0;z-index:251658240' filled="f" stroked="f"/><![endif]--><!--[if !vml]--><span style="height: 33px; left: 0px; margin-left: 362px; margin-top: 9px; mso-ignore: vglayout; position: absolute; width: 259px; z-index: 251658240;"> </span></div><table cellpadding="0" cellspacing="0"><tbody>
<tr> <td height="33" style="vertical-align: top;" width="259"><!--[endif]--><!--[if !mso]--><span style="left: 0pt; mso-ignore: vglayout; position: absolute; z-index: 251658240;"> <table cellpadding="0" cellspacing="0"><tbody>
<tr> <td><!--[endif]--> <div class="shape" style="padding: 3.6pt 7.2pt 3.6pt 7.2pt;" v:shape="_x0000_s1027"> <div class="MsoNormal"><i style="mso-bidi-font-style: normal;"><span style="font-family: "Arial","sans-serif";">Be careful how you use the 2 terms!<o:p></o:p></span></i></div></div><!--[if !mso]--></td> </tr>
</tbody></table></span><!--[endif]--><!--[if !mso & !vml]--> <!--[endif]--><!--[if !vml]--></td> </tr>
</tbody></table><!--[endif]--><!--[if gte vml 1]><v:shapetype id="_x0000_t88" coordsize="21600,21600"
o:spt="88" adj="1800,10800" path="m,qx10800@0l10800@2qy21600@11,10800@3l10800@1qy,21600e"
filled="f"> <v:formulas> <v:f eqn="val #0"/> <v:f eqn="sum 21600 0 #0"/> <v:f eqn="sum #1 0 #0"/> <v:f eqn="sum #1 #0 0"/> <v:f eqn="prod #0 9598 32768"/> <v:f eqn="sum 21600 0 @4"/> <v:f eqn="sum 21600 0 #1"/> <v:f eqn="min #1 @6"/> <v:f eqn="prod @7 1 2"/> <v:f eqn="prod #0 2 1"/> <v:f eqn="sum 21600 0 @9"/> <v:f eqn="val #1"/> </v:formulas> <v:path arrowok="t" o:connecttype="custom" o:connectlocs="0,0;21600,@11;0,21600"
textboxrect="0,@4,7637,@5"/> <v:handles> <v:h position="center,#0" yrange="0,@8"/> <v:h position="bottomRight,#1" yrange="@9,@10"/> </v:handles> </v:shapetype><v:shape id="_x0000_s1026" type="#_x0000_t88" style='position:absolute;
left:0;text-align:left;margin-left:254.45pt;margin-top:4.5pt;width:9pt;
height:23.25pt;z-index:251657216'/><![endif]--><!--[if !vml]--><span style="height: 33px; left: 0px; margin-left: 338px; margin-top: 5px; mso-ignore: vglayout; position: absolute; width: 14px; z-index: 251657216;"><img height="33" src="file:///C:/Users/M300/AppData/Local/Temp/msohtmlclip1/01/clip_image001.gif" v:shapes="_x0000_s1026" width="14" /></span><!--[endif]--><span style="font-family: "Arial","sans-serif";">1 DNA <b style="mso-bidi-font-weight: normal;">strand</b> = 1 polynucleotide chain <o:p></o:p></span><br />
<div class="MsoNormal" style="margin-left: 21.0pt; mso-list: l0 level1 lfo1; tab-stops: list 21.0pt; text-indent: -21.0pt;"><!--[if !supportLists]--><span style="font-family: Wingdings; font-size: 8.0pt; mso-bidi-font-family: Wingdings; mso-fareast-font-family: Wingdings;"><span style="mso-list: Ignore;">l<span style="font: 7.0pt "Times New Roman";"> </span></span></span><!--[endif]--><span style="font-family: "Arial","sans-serif";">1 DNA <b style="mso-bidi-font-weight: normal;">molecule</b> is made up of 2 DNA strands <o:p></o:p></span></div><div class="MsoNormal" style="margin-left: 21.0pt; mso-list: l0 level1 lfo1; tab-stops: list 21.0pt; text-indent: -21.0pt;"><!--[if !supportLists]--><span style="font-family: Wingdings; font-size: 8.0pt; mso-bidi-font-family: Wingdings; mso-fareast-font-family: Wingdings;"><span style="mso-list: Ignore;">l<span style="font: 7.0pt "Times New Roman";"> </span></span></span><!--[endif]--><span style="font-family: "Arial","sans-serif";">The triplet of 3 bases that forms the genetic code is known as <o:p></o:p></span></div><div class="MsoNormal" style="margin-left: 21.0pt;"><span style="font-family: "Arial","sans-serif";">Base triplet in DNA <o:p></o:p></span></div><div class="MsoNormal" style="margin-left: 21.0pt;"><span style="font-family: "Arial","sans-serif";">Codon in mRNA (<i style="mso-bidi-font-style: normal;">some textbooks </i>refer codon to DNA too, but we’ll keep to mRNA for now)<o:p></o:p></span></div><div class="MsoNormal" style="margin-left: 21.0pt;"><span style="font-family: "Arial","sans-serif";">Anticodon in tRNA <o:p></o:p></span></div><div class="MsoNormal"><br />
</div><table border="1" cellpadding="0" cellspacing="0" class="MsoTableGrid" style="border-collapse: collapse; border: none; mso-border-alt: solid windowtext .5pt; mso-padding-alt: 0cm 5.4pt 0cm 5.4pt; mso-yfti-tbllook: 480;"><tbody>
<tr style="mso-yfti-firstrow: yes; mso-yfti-irow: 0;"> <td style="border: solid windowtext 1.0pt; mso-border-alt: solid windowtext .5pt; padding: 0cm 5.4pt 0cm 5.4pt; width: 213.05pt;" valign="top" width="284"> <div align="center" class="MsoNormal" style="text-align: center;"><b style="mso-bidi-font-weight: normal;"><span style="font-family: "Arial","sans-serif";">Wrong<o:p></o:p></span></b></div></td> <td style="border-left: none; border: solid windowtext 1.0pt; mso-border-alt: solid windowtext .5pt; mso-border-left-alt: solid windowtext .5pt; padding: 0cm 5.4pt 0cm 5.4pt; width: 213.05pt;" valign="top" width="284"> <div align="center" class="MsoNormal" style="text-align: center;"><b style="mso-bidi-font-weight: normal;"><span style="font-family: "Arial","sans-serif";">Correct<o:p></o:p></span></b></div></td> </tr>
<tr style="mso-yfti-irow: 1;"> <td style="border-top: none; border: solid windowtext 1.0pt; mso-border-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0cm 5.4pt 0cm 5.4pt; width: 213.05pt;" valign="top" width="284"> <div class="MsoNormal"><span style="font-family: "Arial","sans-serif";">Non-coding genes <o:p></o:p></span></div><div class="MsoNormal"><br />
</div></td> <td style="border-bottom: solid windowtext 1.0pt; border-left: none; border-right: solid windowtext 1.0pt; border-top: none; mso-border-alt: solid windowtext .5pt; mso-border-left-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0cm 5.4pt 0cm 5.4pt; width: 213.05pt;" valign="top" width="284"> <div class="MsoNormal"><span style="font-family: "Arial","sans-serif";">Non-coding sequences/ regions <o:p></o:p></span></div><div class="MsoNormal"><span style="font-family: "Arial","sans-serif";">Genes are already defined as coding DNA<o:p></o:p></span></div></td> </tr>
<tr style="mso-yfti-irow: 2; mso-yfti-lastrow: yes;"> <td style="border-top: none; border: solid windowtext 1.0pt; mso-border-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0cm 5.4pt 0cm 5.4pt; width: 213.05pt;" valign="top" width="284"> <div class="MsoNormal"><span style="font-family: "Arial","sans-serif";">In leading strand, DNA is synthesized in the 5’ to 3’ direction, while in lagging strand, DNA is synthesized in the 3’ to 5’ direction<o:p></o:p></span></div></td> <td style="border-bottom: solid windowtext 1.0pt; border-left: none; border-right: solid windowtext 1.0pt; border-top: none; mso-border-alt: solid windowtext .5pt; mso-border-left-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0cm 5.4pt 0cm 5.4pt; width: 213.05pt;" valign="top" width="284"> <div class="MsoNormal"><span style="font-family: "Arial","sans-serif";">DNA polymerase III synthesizes DNA daughter strands in <b style="mso-bidi-font-weight: normal;">5’ to 3’ direction only</b>. Leading strand is formed due to replication towards replication fork, while the lagging strand is formed due to replication away from replication fork.<o:p></o:p></span></div></td> </tr>
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</div><div class="MsoNormal"><b style="mso-bidi-font-weight: normal;"><u><span style="font-family: "Arial","sans-serif";">Control of Eukaryotic and Prokaryotic Genome<o:p></o:p></span></u></b></div><div class="MsoNormal"><br />
</div><table border="1" cellpadding="0" cellspacing="0" class="MsoTableGrid" style="border-collapse: collapse; border: none; mso-border-alt: solid windowtext .5pt; mso-padding-alt: 0cm 5.4pt 0cm 5.4pt; mso-yfti-tbllook: 480;"><tbody>
<tr style="mso-yfti-firstrow: yes; mso-yfti-irow: 0;"> <td style="border: solid windowtext 1.0pt; mso-border-alt: solid windowtext .5pt; padding: 0cm 5.4pt 0cm 5.4pt; width: 213.05pt;" valign="top" width="284"> <div align="center" class="MsoNormal" style="text-align: center;"><b style="mso-bidi-font-weight: normal;"><span style="font-family: "Arial","sans-serif";">Wrong<o:p></o:p></span></b></div></td> <td style="border-left: none; border: solid windowtext 1.0pt; mso-border-alt: solid windowtext .5pt; mso-border-left-alt: solid windowtext .5pt; padding: 0cm 5.4pt 0cm 5.4pt; width: 213.05pt;" valign="top" width="284"> <div align="center" class="MsoNormal" style="text-align: center;"><b style="mso-bidi-font-weight: normal;"><span style="font-family: "Arial","sans-serif";">Correct<o:p></o:p></span></b></div></td> </tr>
<tr style="mso-yfti-irow: 1;"> <td style="border-top: none; border: solid windowtext 1.0pt; mso-border-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0cm 5.4pt 0cm 5.4pt; width: 213.05pt;" valign="top" width="284"> <div class="MsoNormal"><span style="font-family: "Arial","sans-serif";">Control elements are proteins/ transcription factors are DNA<o:p></o:p></span></div><div class="MsoNormal"><br />
</div></td> <td style="border-bottom: solid windowtext 1.0pt; border-left: none; border-right: solid windowtext 1.0pt; border-top: none; mso-border-alt: solid windowtext .5pt; mso-border-left-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0cm 5.4pt 0cm 5.4pt; width: 213.05pt;" valign="top" width="284"> <div class="MsoNormal"><span style="font-family: "Arial","sans-serif";">Control elements (promoter, silence and enhancer) are non-coding DNA sequences. Transcription factors (which are proteins) bind to control elements to regulate transcription.<o:p></o:p></span></div></td> </tr>
<tr style="mso-yfti-irow: 2;"> <td style="border-top: none; border: solid windowtext 1.0pt; mso-border-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0cm 5.4pt 0cm 5.4pt; width: 213.05pt;" valign="top" width="284"> <div class="MsoNormal"><span style="font-family: "Arial","sans-serif";">Prokaryotes have control elements<o:p></o:p></span></div></td> <td style="border-bottom: solid windowtext 1.0pt; border-left: none; border-right: solid windowtext 1.0pt; border-top: none; mso-border-alt: solid windowtext .5pt; mso-border-left-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0cm 5.4pt 0cm 5.4pt; width: 213.05pt;" valign="top" width="284"> <div class="MsoNormal"><span style="font-family: "Arial","sans-serif";">Eukaryotes have control elements, while prokaryotes have regulatory sequences (like operator) and regulatory genes (like <i style="mso-bidi-font-style: normal;">lacI</i> gene).<o:p></o:p></span></div></td> </tr>
<tr style="mso-yfti-irow: 3; mso-yfti-lastrow: yes;"> <td style="border-top: none; border: solid windowtext 1.0pt; mso-border-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0cm 5.4pt 0cm 5.4pt; width: 213.05pt;" valign="top" width="284"> <div class="MsoNormal"><span style="font-family: "Arial","sans-serif";">Bacteria divides by mitosis<o:p></o:p></span></div></td> <td style="border-bottom: solid windowtext 1.0pt; border-left: none; border-right: solid windowtext 1.0pt; border-top: none; mso-border-alt: solid windowtext .5pt; mso-border-left-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0cm 5.4pt 0cm 5.4pt; width: 213.05pt;" valign="top" width="284"> <div class="MsoNormal"><span style="font-family: "Arial","sans-serif";">Bacteria divides by binary fission.<o:p></o:p></span></div></td> </tr>
</tbody></table><div class="MsoNormal"><br />
</div><div class="MsoNormal"><b style="mso-bidi-font-weight: normal;"><u><span style="font-family: "Arial","sans-serif";">Energetics<o:p></o:p></span></u></b></div><div class="MsoNormal"><br />
</div><table border="1" cellpadding="0" cellspacing="0" class="MsoTableGrid" style="border-collapse: collapse; border: none; mso-border-alt: solid windowtext .5pt; mso-padding-alt: 0cm 5.4pt 0cm 5.4pt; mso-yfti-tbllook: 480;"><tbody>
<tr style="mso-yfti-firstrow: yes; mso-yfti-irow: 0;"> <td style="border: solid windowtext 1.0pt; mso-border-alt: solid windowtext .5pt; padding: 0cm 5.4pt 0cm 5.4pt; width: 213.05pt;" valign="top" width="284"> <div align="center" class="MsoNormal" style="text-align: center;"><b style="mso-bidi-font-weight: normal;"><span style="font-family: "Arial","sans-serif";">Wrong<o:p></o:p></span></b></div></td> <td style="border-left: none; border: solid windowtext 1.0pt; mso-border-alt: solid windowtext .5pt; mso-border-left-alt: solid windowtext .5pt; padding: 0cm 5.4pt 0cm 5.4pt; width: 213.05pt;" valign="top" width="284"> <div align="center" class="MsoNormal" style="text-align: center;"><b style="mso-bidi-font-weight: normal;"><span style="font-family: "Arial","sans-serif";">Correct<o:p></o:p></span></b></div></td> </tr>
<tr style="mso-yfti-irow: 1;"> <td style="border-top: none; border: solid windowtext 1.0pt; mso-border-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0cm 5.4pt 0cm 5.4pt; width: 213.05pt;" valign="top" width="284"> <div class="MsoNormal"><span style="font-family: "Arial","sans-serif";">One of the electron carriers in the electron transport chain is a proton pump.<o:p></o:p></span></div></td> <td style="border-bottom: solid windowtext 1.0pt; border-left: none; border-right: solid windowtext 1.0pt; border-top: none; mso-border-alt: solid windowtext .5pt; mso-border-left-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0cm 5.4pt 0cm 5.4pt; width: 213.05pt;" valign="top" width="284"> <div class="MsoNormal"><span style="font-family: "Arial","sans-serif";">The electron carriers in the electron transport chain are <b style="mso-bidi-font-weight: normal;">coupled to</b> proton pumps.<o:p></o:p></span></div></td> </tr>
<tr style="mso-yfti-irow: 2;"> <td style="border-top: none; border: solid windowtext 1.0pt; mso-border-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0cm 5.4pt 0cm 5.4pt; width: 213.05pt;" valign="top" width="284"> <div class="MsoNormal"><span style="font-family: "Arial","sans-serif";">Electrochemical gradient is the same as concentration gradient<o:p></o:p></span></div></td> <td style="border-bottom: solid windowtext 1.0pt; border-left: none; border-right: solid windowtext 1.0pt; border-top: none; mso-border-alt: solid windowtext .5pt; mso-border-left-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0cm 5.4pt 0cm 5.4pt; width: 213.05pt;" valign="top" width="284"> <div class="MsoNormal"><span style="font-family: "Arial","sans-serif";">Electrochemical gradient includes<span style="mso-spacerun: yes;"> </span>concentration gradient AND electric potential gradient (so be careful how you use the term!)<o:p></o:p></span></div></td> </tr>
<tr style="mso-yfti-irow: 3;"> <td style="border-top: none; border: solid windowtext 1.0pt; mso-border-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0cm 5.4pt 0cm 5.4pt; width: 213.05pt;" valign="top" width="284"> <div class="MsoNormal"><span style="font-family: "Arial","sans-serif";">When Cl<sup>-</sup> ions enter the neurone, it leads to repolarization.<o:p></o:p></span></div></td> <td style="border-bottom: solid windowtext 1.0pt; border-left: none; border-right: solid windowtext 1.0pt; border-top: none; mso-border-alt: solid windowtext .5pt; mso-border-left-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0cm 5.4pt 0cm 5.4pt; width: 213.05pt;" valign="top" width="284"> <div class="MsoNormal"><span style="font-family: "Arial","sans-serif";">When Cl<sup>-</sup> ions enter the neurone, it leads to hyperpoloarization.<o:p></o:p></span></div></td> </tr>
<tr style="mso-yfti-irow: 4; mso-yfti-lastrow: yes;"> <td style="border-top: none; border: solid windowtext 1.0pt; mso-border-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0cm 5.4pt 0cm 5.4pt; width: 213.05pt;" valign="top" width="284"> <div class="MsoNormal"><span style="font-family: "Arial","sans-serif";">Light dependent reactions occur in the presence of light while light independent reaction occurs in the dark<o:p></o:p></span></div></td> <td style="border-bottom: solid windowtext 1.0pt; border-left: none; border-right: solid windowtext 1.0pt; border-top: none; mso-border-alt: solid windowtext .5pt; mso-border-left-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0cm 5.4pt 0cm 5.4pt; width: 213.05pt;" valign="top" width="284"> <div class="MsoNormal"><span style="font-family: "Arial","sans-serif";">Light dependent reactions occur in the presence of light while light independent reaction can occur in the presence and absence of light. <o:p></o:p></span></div></td> </tr>
</tbody></table><div class="MsoNormal"><br />
</div><div class="MsoNormal"><b style="mso-bidi-font-weight: normal;"><u><span style="font-family: "Arial","sans-serif";">Genetic Basis of Variation<o:p></o:p></span></u></b></div><div class="MsoNormal"><br />
</div><table border="1" cellpadding="0" cellspacing="0" class="MsoTableGrid" style="border-collapse: collapse; border: none; mso-border-alt: solid windowtext .5pt; mso-padding-alt: 0cm 5.4pt 0cm 5.4pt; mso-yfti-tbllook: 480;"><tbody>
<tr style="mso-yfti-firstrow: yes; mso-yfti-irow: 0;"> <td style="border: solid windowtext 1.0pt; mso-border-alt: solid windowtext .5pt; padding: 0cm 5.4pt 0cm 5.4pt; width: 213.05pt;" valign="top" width="284"> <div align="center" class="MsoNormal" style="text-align: center;"><b style="mso-bidi-font-weight: normal;"><span style="font-family: "Arial","sans-serif";">Wrong<o:p></o:p></span></b></div></td> <td style="border-left: none; border: solid windowtext 1.0pt; mso-border-alt: solid windowtext .5pt; mso-border-left-alt: solid windowtext .5pt; padding: 0cm 5.4pt 0cm 5.4pt; width: 213.05pt;" valign="top" width="284"> <div align="center" class="MsoNormal" style="text-align: center;"><b style="mso-bidi-font-weight: normal;"><span style="font-family: "Arial","sans-serif";">Correct<o:p></o:p></span></b></div></td> </tr>
<tr style="mso-yfti-irow: 1; mso-yfti-lastrow: yes;"> <td style="border-top: none; border: solid windowtext 1.0pt; mso-border-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0cm 5.4pt 0cm 5.4pt; width: 213.05pt;" valign="top" width="284"> <div class="MsoNormal"><span style="font-family: "Arial","sans-serif";">The two genes are linked so closely together that crossing over does not happen.<o:p></o:p></span></div></td> <td style="border-bottom: solid windowtext 1.0pt; border-left: none; border-right: solid windowtext 1.0pt; border-top: none; mso-border-alt: solid windowtext .5pt; mso-border-left-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0cm 5.4pt 0cm 5.4pt; width: 213.05pt;" valign="top" width="284"> <div class="MsoNormal"><span style="font-family: "Arial","sans-serif";">The two genes are linked so closely together that crossing over <b style="mso-bidi-font-weight: normal;">to separate the two genes</b> does not happen.<o:p></o:p></span></div></td> </tr>
</tbody></table><div class="MsoNormal"><br />
</div><div class="MsoNormal"><br />
</div><div class="MsoNormal"><b style="mso-bidi-font-weight: normal;"><u><span style="font-family: "Arial","sans-serif";">Diversity and Evolution<o:p></o:p></span></u></b></div><div class="MsoNormal" style="tab-stops: 87.75pt;"><span style="font-family: "Arial","sans-serif";"><span style="mso-tab-count: 1;"> </span><o:p></o:p></span></div><table border="1" cellpadding="0" cellspacing="0" class="MsoTableGrid" style="border-collapse: collapse; border: none; mso-border-alt: solid windowtext .5pt; mso-padding-alt: 0cm 5.4pt 0cm 5.4pt; mso-yfti-tbllook: 480;"><tbody>
<tr style="mso-yfti-firstrow: yes; mso-yfti-irow: 0;"> <td style="border: solid windowtext 1.0pt; mso-border-alt: solid windowtext .5pt; padding: 0cm 5.4pt 0cm 5.4pt; width: 194.4pt;" valign="top" width="259"> <div align="center" class="MsoNormal" style="text-align: center;"><b style="mso-bidi-font-weight: normal;"><span style="font-family: "Arial","sans-serif";">Wrong<o:p></o:p></span></b></div></td> <td style="border-left: none; border: solid windowtext 1.0pt; mso-border-alt: solid windowtext .5pt; mso-border-left-alt: solid windowtext .5pt; padding: 0cm 5.4pt 0cm 5.4pt; width: 231.7pt;" valign="top" width="309"> <div align="center" class="MsoNormal" style="text-align: center;"><b style="mso-bidi-font-weight: normal;"><span style="font-family: "Arial","sans-serif";">Correct<o:p></o:p></span></b></div></td> </tr>
<tr style="mso-yfti-irow: 1;"> <td style="border-top: none; border: solid windowtext 1.0pt; mso-border-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0cm 5.4pt 0cm 5.4pt; width: 194.4pt;" valign="top" width="259"> <div class="MsoNormal"><span style="font-family: "Arial","sans-serif";">Traits are passed on to the next generation <o:p></o:p></span></div><div class="MsoNormal"><br />
</div></td> <td style="border-bottom: solid windowtext 1.0pt; border-left: none; border-right: solid windowtext 1.0pt; border-top: none; mso-border-alt: solid windowtext .5pt; mso-border-left-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0cm 5.4pt 0cm 5.4pt; width: 231.7pt;" valign="top" width="309"> <div class="MsoNormal"><span style="font-family: "Arial","sans-serif";">Genes/alleles are passed on to the next generation<o:p></o:p></span></div></td> </tr>
<tr style="mso-yfti-irow: 2;"> <td style="border-top: none; border: solid windowtext 1.0pt; mso-border-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0cm 5.4pt 0cm 5.4pt; width: 194.4pt;" valign="top" width="259"> <div class="MsoNormal"><span style="font-family: "Arial","sans-serif";">Organisms mutate to adapt to the environment<o:p></o:p></span></div></td> <td style="border-bottom: solid windowtext 1.0pt; border-left: none; border-right: solid windowtext 1.0pt; border-top: none; mso-border-alt: solid windowtext .5pt; mso-border-left-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0cm 5.4pt 0cm 5.4pt; width: 231.7pt;" valign="top" width="309"> <div class="MsoNormal"><span style="font-family: "Arial","sans-serif";">Mutations are random, they cause variation within populations<o:p></o:p></span></div></td> </tr>
<tr style="mso-yfti-irow: 3;"> <td style="border-top: none; border: solid windowtext 1.0pt; mso-border-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0cm 5.4pt 0cm 5.4pt; width: 194.4pt;" valign="top" width="259"> <div class="MsoNormal"><span style="font-family: "Arial","sans-serif";">Mutations occur at a constant rate in neutral theory<o:p></o:p></span></div></td> <td style="border-bottom: solid windowtext 1.0pt; border-left: none; border-right: solid windowtext 1.0pt; border-top: none; mso-border-alt: solid windowtext .5pt; mso-border-left-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0cm 5.4pt 0cm 5.4pt; width: 231.7pt;" valign="top" width="309"> <div class="MsoNormal"><span style="font-family: "Arial","sans-serif";">Neutral mutations have been observed/ found to occur at a constant rate in certain DNA sequences or genes<o:p></o:p></span></div></td> </tr>
<tr style="mso-yfti-irow: 4; mso-yfti-lastrow: yes;"> <td style="border-top: none; border: solid windowtext 1.0pt; mso-border-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0cm 5.4pt 0cm 5.4pt; width: 194.4pt;" valign="top" width="259"> <div class="MsoNormal"><span style="font-family: "Arial","sans-serif";">Only silent mutation is involved in neutral theory<o:p></o:p></span></div></td> <td style="border-bottom: solid windowtext 1.0pt; border-left: none; border-right: solid windowtext 1.0pt; border-top: none; mso-border-alt: solid windowtext .5pt; mso-border-left-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0cm 5.4pt 0cm 5.4pt; width: 231.7pt;" valign="top" width="309"> <div class="MsoNormal"><span style="font-family: "Arial","sans-serif";">Neutral theory involves silent mutation and any mutation that results in a phenotype which is selectively neutral.<o:p></o:p></span></div></td> </tr>
</tbody></table><div class="MsoNormal"><br />
</div><div class="MsoNormal"><br />
</div><div class="MsoNormal"><b style="mso-bidi-font-weight: normal;"><u><span style="font-family: "Arial","sans-serif";">Applications <o:p></o:p></span></u></b></div><div class="MsoNormal"><br />
</div><table border="1" cellpadding="0" cellspacing="0" class="MsoTableGrid" style="border-collapse: collapse; border: none; mso-border-alt: solid windowtext .5pt; mso-padding-alt: 0cm 5.4pt 0cm 5.4pt; mso-yfti-tbllook: 480;"><tbody>
<tr style="mso-yfti-firstrow: yes; mso-yfti-irow: 0;"> <td style="border: solid windowtext 1.0pt; mso-border-alt: solid windowtext .5pt; padding: 0cm 5.4pt 0cm 5.4pt; width: 213.05pt;" valign="top" width="284"> <div align="center" class="MsoNormal" style="text-align: center;"><b style="mso-bidi-font-weight: normal;"><span style="font-family: "Arial","sans-serif";">Wrong<o:p></o:p></span></b></div></td> <td style="border-left: none; border: solid windowtext 1.0pt; mso-border-alt: solid windowtext .5pt; mso-border-left-alt: solid windowtext .5pt; padding: 0cm 5.4pt 0cm 5.4pt; width: 213.05pt;" valign="top" width="284"> <div align="center" class="MsoNormal" style="text-align: center;"><b style="mso-bidi-font-weight: normal;"><span style="font-family: "Arial","sans-serif";">Correct<o:p></o:p></span></b></div></td> </tr>
<tr style="mso-yfti-irow: 1;"> <td style="border-top: none; border: solid windowtext 1.0pt; mso-border-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0cm 5.4pt 0cm 5.4pt; width: 213.05pt;" valign="top" width="284"> <div class="MsoNormal"><span style="font-family: "Arial","sans-serif"; mso-bidi-font-size: 10.5pt;">Gene therapy is a technique for introducing healthy/ therapeutic gene <span style="mso-spacerun: yes;"> </span><o:p></o:p></span></div></td> <td style="border-bottom: solid windowtext 1.0pt; border-left: none; border-right: solid windowtext 1.0pt; border-top: none; mso-border-alt: solid windowtext .5pt; mso-border-left-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0cm 5.4pt 0cm 5.4pt; width: 213.05pt;" valign="top" width="284"> <div class="MsoNormal"><span style="font-family: "Arial","sans-serif"; mso-bidi-font-size: 10.5pt;">…..introducing <b style="mso-bidi-font-weight: normal;">normal </b>allele/gene <o:p></o:p></span></div><div class="MsoNormal"><br />
</div></td> </tr>
<tr style="mso-yfti-irow: 2;"> <td style="border-top: none; border: solid windowtext 1.0pt; mso-border-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0cm 5.4pt 0cm 5.4pt; width: 213.05pt;" valign="top" width="284"> <div class="MsoNormal"><span style="font-family: "Arial","sans-serif"; mso-bidi-font-size: 10.5pt;">Package the normal allele into liposomes/ virus<o:p></o:p></span></div></td> <td style="border-bottom: solid windowtext 1.0pt; border-left: none; border-right: solid windowtext 1.0pt; border-top: none; mso-border-alt: solid windowtext .5pt; mso-border-left-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0cm 5.4pt 0cm 5.4pt; width: 213.05pt;" valign="top" width="284"> <div class="MsoNormal"><span style="font-family: "Arial","sans-serif"; mso-bidi-font-size: 10.5pt;">Package the <b style="mso-bidi-font-weight: normal;">DNA containing the normal allele/gene</b> into liposomes/ virus<o:p></o:p></span></div></td> </tr>
<tr style="mso-yfti-irow: 3;"> <td style="border-top: none; border: solid windowtext 1.0pt; mso-border-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0cm 5.4pt 0cm 5.4pt; width: 213.05pt;" valign="top" width="284"> <div class="MsoNormal"><span style="font-family: "Arial","sans-serif"; mso-bidi-font-size: 10.5pt;">Deletion of a codon results in phenylalanine not to be produced/ gene lacking the amino acid<o:p></o:p></span></div></td> <td style="border-bottom: solid windowtext 1.0pt; border-left: none; border-right: solid windowtext 1.0pt; border-top: none; mso-border-alt: solid windowtext .5pt; mso-border-left-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0cm 5.4pt 0cm 5.4pt; width: 213.05pt;" valign="top" width="284"> <div class="MsoNormal"><span style="font-family: "Arial","sans-serif"; mso-bidi-font-size: 10.5pt;">Deletion of a codon results in phenylalanine not to be <b style="mso-bidi-font-weight: normal;">coded</b>/ <b style="mso-bidi-font-weight: normal;">protein</b> lacking the amino acid<o:p></o:p></span></div></td> </tr>
<tr style="mso-yfti-irow: 4;"> <td style="border-top: none; border: solid windowtext 1.0pt; mso-border-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0cm 5.4pt 0cm 5.4pt; width: 213.05pt;" valign="top" width="284"> <div class="MsoNormal"><span style="font-family: "Arial","sans-serif"; mso-bidi-font-size: 10.5pt;">Insertional inactivation can be used to distinguish recombinant plasmids from reannealed plasmids<o:p></o:p></span></div></td> <td style="border-bottom: solid windowtext 1.0pt; border-left: none; border-right: solid windowtext 1.0pt; border-top: none; mso-border-alt: solid windowtext .5pt; mso-border-left-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0cm 5.4pt 0cm 5.4pt; width: 213.05pt;" valign="top" width="284"> <div class="MsoNormal"><span style="font-family: "Arial","sans-serif"; mso-bidi-font-size: 10.5pt;">Insertional inactivation can be used to distinguish <b style="mso-bidi-font-weight: normal;">bacteria containing</b> recombinant plasmids from <b style="mso-bidi-font-weight: normal;">bacteria containing</b> reannealed plasmids.<o:p></o:p></span></div></td> </tr>
<tr style="mso-yfti-irow: 5;"> <td style="border-top: none; border: solid windowtext 1.0pt; mso-border-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0cm 5.4pt 0cm 5.4pt; width: 213.05pt;" valign="top" width="284"> <div class="MsoNormal"><span style="font-family: "Arial","sans-serif";">In plant cloning, clones have same genes/ genetically similar to the same parent<o:p></o:p></span></div></td> <td style="border-bottom: solid windowtext 1.0pt; border-left: none; border-right: solid windowtext 1.0pt; border-top: none; mso-border-alt: solid windowtext .5pt; mso-border-left-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0cm 5.4pt 0cm 5.4pt; width: 213.05pt;" valign="top" width="284"> <div class="MsoNormal"><span style="font-family: "Arial","sans-serif";">In plant cloning, clones are <b style="mso-bidi-font-weight: normal;">genetically identical</b> to the same parent<o:p></o:p></span></div></td> </tr>
<tr style="mso-yfti-irow: 6;"> <td style="border-top: none; border: solid windowtext 1.0pt; mso-border-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0cm 5.4pt 0cm 5.4pt; width: 213.05pt;" valign="top" width="284"> <div class="MsoNormal"><span style="font-family: "Arial","sans-serif";">Stem cells repair damaged cells<o:p></o:p></span></div></td> <td style="border-bottom: solid windowtext 1.0pt; border-left: none; border-right: solid windowtext 1.0pt; border-top: none; mso-border-alt: solid windowtext .5pt; mso-border-left-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0cm 5.4pt 0cm 5.4pt; width: 213.05pt;" valign="top" width="284"> <div class="MsoNormal"><span style="font-family: "Arial","sans-serif";">Stem cells <b style="mso-bidi-font-weight: normal;">replace</b> damaged cells<o:p></o:p></span></div><div class="MsoNormal"><span style="font-family: "Arial","sans-serif";">Stem cells <b style="mso-bidi-font-weight: normal;">repair damaged organs</b> <o:p></o:p></span></div></td> </tr>
<tr style="mso-yfti-irow: 7;"> <td style="border-top: none; border: solid windowtext 1.0pt; mso-border-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0cm 5.4pt 0cm 5.4pt; width: 213.05pt;" valign="top" width="284"> <div class="MsoNormal"><span style="font-family: "Arial","sans-serif";">In autoradiography, expose the nitrocellulose membrane to X-ray<o:p></o:p></span></div></td> <td style="border-bottom: solid windowtext 1.0pt; border-left: none; border-right: solid windowtext 1.0pt; border-top: none; mso-border-alt: solid windowtext .5pt; mso-border-left-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0cm 5.4pt 0cm 5.4pt; width: 213.05pt;" valign="top" width="284"> <div class="MsoNormal"><span style="font-family: "Arial","sans-serif";">In autoradiography, place a piece of X-ray <b style="mso-bidi-font-weight: normal;">film</b> over the nitrocellulose membrane, which will reveal the bands <o:p></o:p></span></div></td> </tr>
<tr style="mso-yfti-irow: 8; mso-yfti-lastrow: yes;"> <td style="border-top: none; border: solid windowtext 1.0pt; mso-border-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0cm 5.4pt 0cm 5.4pt; width: 213.05pt;" valign="top" width="284"> <div class="MsoNormal"><span style="font-family: "Arial","sans-serif";">Southern Blotting includes gel electrophoresis<o:p></o:p></span></div></td> <td style="border-bottom: solid windowtext 1.0pt; border-left: none; border-right: solid windowtext 1.0pt; border-top: none; mso-border-alt: solid windowtext .5pt; mso-border-left-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0cm 5.4pt 0cm 5.4pt; width: 213.05pt;" valign="top" width="284"> <div class="MsoNormal"><span style="font-family: "Arial","sans-serif";">Southern Blotting is carried out AFTER gel electrophoresis.<o:p></o:p></span></div></td> </tr>
</tbody></table><div class="MsoNormal"><br />
</div><div class="MsoNormal" style="margin-left: 21.0pt; mso-list: l0 level1 lfo1; tab-stops: list 21.0pt; text-indent: -21.0pt;"><!--[if !supportLists]--><span style="font-family: Wingdings; font-size: 8.0pt; mso-bidi-font-family: Wingdings; mso-fareast-font-family: Wingdings;"><span style="mso-list: Ignore;">l<span style="font: 7.0pt "Times New Roman";"> </span></span></span><!--[endif]--><span style="font-family: "Arial","sans-serif";">When describing gel electrophoresis, write “ <b style="mso-bidi-font-weight: normal;"><u>direct </u></b>current was applied” <o:p></o:p></span></div><div class="MsoNormal" style="margin-left: 21.0pt; mso-list: l0 level1 lfo1; tab-stops: list 21.0pt; text-indent: -21.0pt;"><!--[if !supportLists]--><span style="font-family: Wingdings; font-size: 8.0pt; mso-bidi-font-family: Wingdings; mso-fareast-font-family: Wingdings;"><span style="mso-list: Ignore;">l<span style="font: 7.0pt "Times New Roman";"> </span></span></span><!--[endif]--><span style="font-family: "Arial","sans-serif";">In nucleic aid hybridization: if <b style="mso-bidi-font-weight: normal;">radioactive probe </b>is used, <b style="mso-bidi-font-weight: normal;">autoradiography </b>is used to visualise the DNA bands on X-ray film<o:p></o:p></span></div><div class="MsoNormal" style="margin-left: 21.0pt; mso-list: l0 level1 lfo1; tab-stops: list 21.0pt; text-indent: -21.0pt;"><!--[if !supportLists]--><span style="font-family: Wingdings; font-size: 8.0pt; mso-bidi-font-family: Wingdings; mso-fareast-font-family: Wingdings;"><span style="mso-list: Ignore;">l<span style="font: 7.0pt "Times New Roman";"> </span></span></span><!--[endif]--><span style="font-family: "Arial","sans-serif";">In gel electrophoresis: If <b style="mso-bidi-font-weight: normal;">ethidium bromide</b> is use, <b style="mso-bidi-font-weight: normal;">UV</b> light is used to visualise the bands on the gel <o:p></o:p></span></div><div class="MsoNormal" style="margin-left: 21.0pt; mso-list: l0 level1 lfo1; tab-stops: list 21.0pt; text-indent: -21.0pt;"><!--[if !supportLists]--><span style="font-family: Wingdings; font-size: 8.0pt; mso-bidi-font-family: Wingdings; mso-fareast-font-family: Wingdings;"><span style="mso-list: Ignore;">l<span style="font: 7.0pt "Times New Roman";"> </span></span></span><!--[endif]--><span style="font-family: "Arial","sans-serif";">DNA ligase forms phosphodiester bonds <u><o:p></o:p></u></span></div><div class="MsoNormal" style="margin-left: 21.0pt; mso-list: l0 level1 lfo1; tab-stops: list 21.0pt; text-indent: -21.0pt;"><!--[if !supportLists]--><span style="font-family: Wingdings; font-size: 8.0pt; mso-bidi-font-family: Wingdings; mso-fareast-font-family: Wingdings;"><span style="mso-list: Ignore;">l<span style="font: 7.0pt "Times New Roman";"> </span></span></span><!--[endif]--><span style="font-family: "Arial","sans-serif";">Restriction enzymes breaks/cleave phosphodiester bonds. <i style="mso-bidi-font-style: normal;">Always write, restriction enzyme <b style="mso-bidi-font-weight: normal;"><u>cuts</u></b> (not cleave, degrade, break down) DNA at restriction sites</i><u><o:p></o:p></u></span></div><div class="MsoNormal" style="margin-left: 21.0pt; mso-list: l0 level1 lfo1; tab-stops: list 21.0pt; text-indent: -21.0pt;"><!--[if !supportLists]--><span style="font-family: Wingdings; font-size: 8.0pt; mso-bidi-font-family: Wingdings; mso-fareast-font-family: Wingdings;"><span style="mso-list: Ignore;">l<span style="font: 7.0pt "Times New Roman";"> </span></span></span><!--[endif]--><span style="font-family: "Arial","sans-serif";">Ligation = formation of phosphodiester bonds between nucleotides <u><o:p></o:p></u></span></div><div class="MsoNormal" style="margin-left: 21.0pt; mso-list: l0 level1 lfo1; tab-stops: list 21.0pt; text-indent: -21.0pt;"><!--[if !supportLists]--><span style="font-family: Wingdings; font-size: 8.0pt; mso-bidi-font-family: Wingdings; mso-fareast-font-family: Wingdings;"><span style="mso-list: Ignore;">l<span style="font: 7.0pt "Times New Roman";"> </span></span></span><!--[endif]--><span style="font-family: "Arial","sans-serif";">Annealing = formation of hydrogen bonds between complementary base pairs <u><o:p></o:p></u></span></div><div class="MsoNormal" style="margin-left: 21.0pt; mso-list: l0 level1 lfo1; tab-stops: list 21.0pt; text-indent: -21.0pt;"><!--[if !supportLists]--><span style="font-family: Wingdings; font-size: 8.0pt; mso-bidi-font-family: Wingdings; mso-fareast-font-family: Wingdings;"><span style="mso-list: Ignore;">l<span style="font: 7.0pt "Times New Roman";"> </span></span></span><!--[endif]--><b style="mso-bidi-font-weight: normal;"><span style="font-family: "Arial","sans-serif";">Southern Blotting</span></b><span style="font-family: "Arial","sans-serif";"> simply means putting the gel into an alkali so the alkali can denature the DNA into single-strands. Place a nitrocellulose/ nylon membrane over the gel under heavy weight so that the single-stranded DNA can bind to the membrane after travelling by capillary action.<u><o:p></o:p></u></span></div><div class="MsoNormal" style="margin-left: 21.0pt; mso-list: l0 level1 lfo1; tab-stops: list 21.0pt; text-indent: -21.0pt;"><!--[if !supportLists]--><span style="font-family: Wingdings; font-size: 8.0pt; mso-bidi-font-family: Wingdings; mso-fareast-font-family: Wingdings;"><span style="mso-list: Ignore;">l<span style="font: 7.0pt "Times New Roman";"> </span></span></span><!--[endif]--><b style="mso-bidi-font-weight: normal;"><span style="font-family: "Arial","sans-serif";">Nucleic acid hybridization</span></b><span style="font-family: "Arial","sans-serif";"> simply means complementatary base pairing through hydrogen bonds between single-stranded, radioactive and specific probe and single-stranded DNA.<u><o:p></o:p></u></span></div><div class="MsoNormal" style="margin-left: 21.0pt; mso-list: l0 level1 lfo1; tab-stops: list 21.0pt; text-indent: -21.0pt;"><!--[if !supportLists]--><span style="font-family: Wingdings; font-size: 8.0pt; mso-bidi-font-family: Wingdings; mso-fareast-font-family: Wingdings;"><span style="mso-list: Ignore;">l<span style="font: 7.0pt "Times New Roman";"> </span></span></span><!--[endif]--><b style="mso-bidi-font-weight: normal;"><span style="font-family: "Arial","sans-serif";">Autoradiography </span></b><span style="font-family: "Arial","sans-serif";">simply means placing a piece of photographic or X-ray <b style="mso-bidi-font-weight: normal;">film</b> over the nitrocellulose membrane, which will reveal the bands.<u><o:p></o:p></u></span></div><div class="MsoNormal" style="margin-left: 21.0pt; mso-list: l0 level1 lfo1; tab-stops: list 21.0pt; text-indent: -21.0pt;"><!--[if !supportLists]--><span style="font-family: Wingdings; font-size: 8.0pt; mso-bidi-font-family: Wingdings; mso-fareast-font-family: Wingdings;"><span style="mso-list: Ignore;">l<span style="font: 7.0pt "Times New Roman";"> </span></span></span><!--[endif]--><span style="font-family: "Arial","sans-serif";">How to use<b style="mso-bidi-font-weight: normal;"> “RFLP” </b>term? This can be used in e.g. ….perform <b style="mso-bidi-font-weight: normal;">RFLP analysis</b>…/… if <b style="mso-bidi-font-weight: normal;">RFLP band fragment </b>is seen…/… <b style="mso-bidi-font-weight: normal;">RFLP locus</b> is located so close to disease allele…<u><o:p></o:p></u></span></div><div class="MsoNormal"><br />
</div>United JC Unionhttp://www.blogger.com/profile/06131630985986252085noreply@blogger.com2tag:blogger.com,1999:blog-4274038432688951963.post-25030776746010201692011-02-03T01:42:00.001-08:002011-02-03T01:42:14.986-08:00Independent Assortment of ChromosomesSynapsis<br />
Crossing-over<br />
Chiasmata<br />
Bivalents<br />
THREE Sources of Variation<br />
Crossing Over at Prophase I<br />
Crossing Over at Prophase I<br />
Independent Assortment at Metaphase I<br />
Independent Assortment at Metaphase I<br />
Possible combinations = 2n, n = haploid number.<br />
Humans: n = 23.<br />
Possible combinations = 223 = 8.4 million! <br />
Independent Assortment is also known as Mendel’s Second Law<br />
Random Fertilization<br />
Each Gamete <br />
One out of about 8.4 million (223) possible chromosome combinations due to independent assortment. <br />
2 gametes fuse<br />
Zygote with any of about 70 trillion (223 X 223) diploid combinations!!<br />
Mendel’s First Law<br />
Law of Segregation<br />
<br />
A somatic cell carries two alleles at any one locus.<br />
Alleles of a gene pair segregate during anaphase I of meiosis<br />
Half of the gametes carry one allele of a gene pair; the other half carry the other allele.<br />
Summary of meiosis and mitosis<br />
Assignment<br />
Explain the need for the production of genetically identical cells & fine control of replication<br />
<br />
Since the functions of mitosis are to produce new cells for growth, repair of body tissues, and asexual reproduction, <br />
essential that the cells are identical so that the daughter cells would continue to code for essential proteins that can function properly<br />
<br />
How to Get Seedless Watermelon?<br />
<br />
Double the diploid number (2n) in a normal watermelon plant by the use of the chemical colchicine to tetraploid (4n). <br />
Colchicine’s function?<br />
Tetraploid’s gamete = 2n<br />
Normal watermelon’s gamete = n<br />
Fuse the tetraploid’s gamete (2n) with normal water melon’s gamete (n)<br />
<br />
Why is triploid watermelon sterile?<br />
During meiosis the normal pairing of chromosomes cannot properly take place since one set will have no homologous set to pair withUnited JC Unionhttp://www.blogger.com/profile/06131630985986252085noreply@blogger.com1tag:blogger.com,1999:blog-4274038432688951963.post-15667051899451731942011-02-03T01:41:00.000-08:002011-02-03T01:41:07.335-08:00Meiosis II SummaryPROPHASE II<br />
A spindle apparatus forms.<br />
Chromosomes composed of two chromatids associated at the centromere<br />
METAPHASE II<br />
Chromosomes are positioned on the metaphase plate as in mitosis.<br />
Crossing over in meiosis <br />
Two sister chromatids of each chromosome are not genetically identical <br />
ANAPHASE II<br />
Centromeres divide or replicate to allow separation of sister chromatids;<br />
The chromatids move toward opposite poles as individual chromosomes.<br />
ANAPHASE II<br />
Add:<br />
Direction of chromosome movement in Anaphase I at right angles to that in Anaphase II. <br />
TELOPHASE II AND CYTOKINESIS<br />
Nuclei form<br />
Chromosomes begin decondensing<br />
Cytokinesis occurs<br />
4 daughter haploid cells<br />
Each genetically distinct from the other <br />
It’s Video Time!United JC Unionhttp://www.blogger.com/profile/06131630985986252085noreply@blogger.com1tag:blogger.com,1999:blog-4274038432688951963.post-61779149811866473772011-02-03T01:29:00.000-08:002011-02-03T01:29:17.323-08:00Plant Cloning (Micropropagation)Plant Cloning (Micropropagation)<br />
And Revision<br />
Tissue Culture<br />
Vegetative propagation:<br />
Stem, leaf or root cuttings <br />
Micropropagation (tissue culture)<br />
What are needed in micropropagation?<br />
Tissue culture nutrient medium<br />
(1) essential elements, or mineral ions, supplied as a complex mixture of salts;<br />
(2) an organic supplement supplying vitamins and/or amino acids; and<br />
(3) a source of fixed carbon: sucrose.<br />
2. Plant growth regulators (Plant hormones)<br />
Auxin<br />
Cytokinin<br />
3. Carefully controlled environment<br />
Aseptic environment<br />
No bacteria, fungus, virus, etc.<br />
<br />
Step 1: Surface sterilization of explants<br />
Small pieces of tissue (explants) removed from any part of the plant<br />
May be meristematic (such as the shoot tip, the root tip), or non-meristematic (leaf parts). <br />
First surface sterilized <br />
Transferred in sterile surroundings to a sterile medium <br />
<br />
Step 2: Callus formation (induction)<br />
A callus is a mass of undifferentiated cell mass formed by mitosis <br />
<br />
During callus formation, there is dedifferentiation <br />
Consequence lose the ability to photosynthesise. <br />
Plant tissue culture cells have a small vacuole, lack chloroplasts and photosynthetic pathways and the structural or chemical features <br />
Callus can later be induced to re-differentiate into whole plants by alterations to the growth media <br />
Step 3: Subculturing<br />
Aim:<br />
callus can be subdivided to increase the number of plants eventually produced<br />
Step 4: Shoots and roots formation<br />
Auxin to cytokinin ratio<br />
Step 5: Acclimatization followed by transplanting<br />
Acclimatization allows environmental and physiological adaptation of tissue cultured plants to a greenhouse or a field environment <br />
Before transplanting the plantlets<br />
<br />
Advantages<br />
extremely high multiplication rates. <br />
identical copies of plants with desirable traits <br />
Both seeds and pollinators are not required. <br />
Plant cells can be genetically modified. <br />
Limitations<br />
aseptic condition requirements <br />
A virus infection would quickly destroy a batch of plants <br />
Suitable techniques of micropropagation are not available for many valuable species <br />
Somaclonal variation may arise during in vitro culture<br />
Revision/ Clarification<br />
Tutorial: Enzymes<br />
The initial rate of a reaction catalysed by an enzyme was measured at various substrate concentrations. Which graph shows the effect of a low concentration of non-competitive inhibitor on the reaction?United JC Unionhttp://www.blogger.com/profile/06131630985986252085noreply@blogger.com1tag:blogger.com,1999:blog-4274038432688951963.post-39580343162248370362011-02-03T01:16:00.000-08:002011-02-03T01:16:29.612-08:00SPA Isolating, Cloning and Sequencing DNA<div class="WordSection1"> <div align="center" class="MsoNormal" style="text-align: center;"><b><span style="font-family: "Arial","sans-serif"; font-size: 11.0pt;">Topic: Isolating, Cloning and Sequencing DNA<o:p></o:p></span></b></div><div align="center" class="MsoNormal" style="text-align: center;"><b><span style="font-family: "Arial","sans-serif"; font-size: 11.0pt;">Practical: </span></b><b style="mso-bidi-font-weight: normal;"><span style="font-family: "Arial","sans-serif"; font-size: 11.0pt;">DNA fingerprinting<span style="mso-bidi-font-weight: bold;"><o:p></o:p></span></span></b></div><div align="center" class="MsoHeader" style="layout-grid-mode: char; text-align: center;"><br />
</div><div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;"><b><span style="font-family: "Arial","sans-serif"; font-size: 11.0pt; mso-text-raise: 3.0pt; position: relative; top: -3.0pt;">Activity:</span></b><span style="font-family: "Arial","sans-serif"; font-size: 11.0pt; mso-text-raise: 3.0pt; position: relative; top: -3.0pt;"><span style="mso-tab-count: 1;"> </span></span><span style="font-family: "Arial","sans-serif"; font-size: 11.0pt;">To <u><span style="color: red;">establish identity</span></u> of confiscated alligator meat.<span style="mso-text-raise: 3.0pt; position: relative; top: -3.0pt;"><span style="mso-tab-count: 1;"> </span><span style="mso-spacerun: yes;"> </span><span style="mso-tab-count: 1;"> </span><o:p></o:p></span></span></div><div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;"><!--[if gte vml 1]><v:shapetype
id="_x0000_t202" coordsize="21600,21600" o:spt="202" path="m,l,21600r21600,l21600,xe"> <v:stroke joinstyle="miter"/> <v:path gradientshapeok="t" o:connecttype="rect"/> </v:shapetype><v:shape id="_x0000_s1026" type="#_x0000_t202" style='position:absolute;
left:0;text-align:left;margin-left:95pt;margin-top:4.35pt;width:385pt;
height:45pt;z-index:251657728' filled="f" stroked="f" strokecolor="#030"/><![endif]--><!--[if !vml]--><span style="mso-ignore: vglayout;"> </span></div><table align="left" cellpadding="0" cellspacing="0"><tbody>
<tr> <td height="6" width="127"></td> </tr>
<tr> <td></td> <td height="64" style="vertical-align: top;" width="517"><!--[endif]--><!--[if !mso]--><span style="left: 0pt; mso-ignore: vglayout; position: absolute; z-index: 251657728;"> <table cellpadding="0" cellspacing="0"><tbody>
<tr> <td><!--[endif]--> <div class="shape" style="padding: 3.6pt 7.2pt 3.6pt 7.2pt;" v:shape="_x0000_s1026"> <div class="MsoNormal"><b style="mso-bidi-font-weight: normal;"><span style="color: blue; font-family: "Arial","sans-serif"; font-size: 10.0pt;">Key procedure: <span style="mso-tab-count: 1;"> </span>DNA fingerprinting <o:p></o:p></span></b></div><div class="MsoNormal"><b style="mso-bidi-font-weight: normal;"><span style="color: blue; font-family: "Arial","sans-serif"; font-size: 10.0pt;">Key words:<span style="mso-tab-count: 2;"> </span>VNTRs, restriction digestion<o:p></o:p></span></b></div></div><!--[if !mso]--></td> </tr>
</tbody></table></span><!--[endif]--><!--[if !mso & !vml]--> <!--[endif]--><!--[if !vml]--></td> </tr>
</tbody></table><!--[endif]--><span style="font-family: "Arial","sans-serif"; mso-text-raise: 3.0pt; position: relative; top: -3.0pt;"><o:p> </o:p></span><br />
<br clear="ALL" style="mso-ignore: vglayout;" /> <div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;"><b style="mso-bidi-font-weight: normal;"><span style="font-family: "Arial","sans-serif"; font-size: 11.0pt;">Skill A – Planning<o:p></o:p></span></b></div><div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;"><br />
</div><div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;"><span lang="EN" style="font-family: "Arial","sans-serif"; font-size: 11.0pt; mso-ansi-language: EN;">An <span style="mso-bidi-font-weight: bold;">alligator</span> is a reptile in the genus <i>Alligator</i> of the family Alligatoridae</span><span style="font-family: "Arial","sans-serif"; font-size: 11.0pt;">. <span style="mso-spacerun: yes;"> </span></span><span lang="EN" style="font-family: "Arial","sans-serif"; font-size: 11.0pt; mso-ansi-language: EN;">There are two extant alligator species: (i) the American alligator (<i>Alligator mississippiensis</i>); and (ii) the Chinese alligator (<i>Alligator sinensis</i>).<span style="mso-spacerun: yes;"> </span>The Chinese alligator currently is found only in the <st1:place w:st="on">Yangtze River</st1:place> valley and is extremely endangered, with only a few dozen believed to be left in the wild. Indeed, far more Chinese alligators live in zoos around the world than can be found in the wild! <span style="mso-spacerun: yes;"> </span>Hence, the Chinese alligator is listed as a CITES Appendix I species, which puts extreme restrictions on its trade throughout the world.<span style="mso-spacerun: yes;"> </span><o:p></o:p></span></div><div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;"><br />
</div><div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;"><span style="color: black; font-family: "Arial","sans-serif"; font-size: 11.0pt;">A consignment of suspected Chinese alligator meat, which is prized for its medicinal properties, has been confiscated by the Agri-Food and Veterinary Authority (AVA).<span style="mso-spacerun: yes;"> </span>As its Chief Scientist, you have been tasked to </span><span style="font-family: "Arial","sans-serif"; font-size: 11.0pt;">plan, but not carry out, an investigation to <span style="color: black;">verify if the meat originated from a poached Chinese alligator.<o:p></o:p></span></span></div><div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;"><br />
</div><div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;"><span style="font-family: "Arial","sans-serif"; font-size: 11.0pt;">Your planning must be based on the assumption that you have been provided with the following equipment and materials which you must use:<o:p></o:p></span></div><div class="MsoNormal" style="margin-left: 17.4pt; mso-list: l8 level1 lfo1; tab-stops: list 17.4pt; text-align: justify; text-indent: -14.4pt; text-justify: inter-ideograph;"><!--[if !supportLists]--><span style="font-family: Symbol; font-size: 11.0pt; mso-bidi-font-family: Symbol; mso-fareast-font-family: Symbol;"><span style="mso-list: Ignore;">·<span style="font: 7.0pt "Times New Roman";"> </span></span></span><!--[endif]--><span style="font-family: "Arial","sans-serif"; font-size: 11.0pt;">muscle tissues from (i) confiscated consignment; (ii) an American alligator; (iii) a Chinese alligator<o:p></o:p></span></div><div class="MsoNormal" style="margin-left: 17.4pt; mso-list: l8 level1 lfo1; tab-stops: list 17.4pt; text-align: justify; text-indent: -14.4pt; text-justify: inter-ideograph;"><!--[if !supportLists]--><span style="font-family: Symbol; font-size: 11.0pt; mso-bidi-font-family: Symbol; mso-fareast-font-family: Symbol;"><span style="mso-list: Ignore;">·<span style="font: 7.0pt "Times New Roman";"> </span></span></span><!--[endif]--><span style="font-family: "Arial","sans-serif"; font-size: 11.0pt;">laboratory blender<o:p></o:p></span></div><div class="MsoNormal" style="margin-left: 17.4pt; mso-list: l8 level1 lfo1; tab-stops: list 17.4pt; text-align: justify; text-indent: -14.4pt; text-justify: inter-ideograph;"><!--[if !supportLists]--><span style="font-family: Symbol; font-size: 11.0pt; mso-bidi-font-family: Symbol; mso-fareast-font-family: Symbol;"><span style="mso-list: Ignore;">·<span style="font: 7.0pt "Times New Roman";"> </span></span></span><!--[endif]--><span style="font-family: "Arial","sans-serif"; font-size: 11.0pt;">DNA extraction buffer solution <o:p></o:p></span></div><div class="MsoNormal" style="margin-left: 17.4pt; mso-list: l8 level1 lfo1; tab-stops: list 17.4pt; text-align: justify; text-indent: -14.4pt; text-justify: inter-ideograph;"><!--[if !supportLists]--><span style="font-family: Symbol; font-size: 11.0pt; mso-bidi-font-family: Symbol; mso-fareast-font-family: Symbol;"><span style="mso-list: Ignore;">·<span style="font: 7.0pt "Times New Roman";"> </span></span></span><!--[endif]--><span style="font-family: "Arial","sans-serif"; font-size: 11.0pt;">micropipettors<o:p></o:p></span></div><div class="MsoNormal" style="margin-left: 17.4pt; mso-list: l8 level1 lfo1; tab-stops: list 17.4pt; text-align: justify; text-indent: -14.4pt; text-justify: inter-ideograph;"><!--[if !supportLists]--><span style="font-family: Symbol; font-size: 11.0pt; mso-bidi-font-family: Symbol; mso-fareast-font-family: Symbol;"><span style="mso-list: Ignore;">·<span style="font: 7.0pt "Times New Roman";"> </span></span></span><!--[endif]--><span style="font-family: "Arial","sans-serif"; font-size: 11.0pt;">microcentrifuge tubes<o:p></o:p></span></div><div class="MsoNormal" style="margin-left: 17.4pt; mso-list: l8 level1 lfo1; tab-stops: list 17.4pt; text-align: justify; text-indent: -14.4pt; text-justify: inter-ideograph;"><!--[if !supportLists]--><span style="font-family: Symbol; font-size: 11.0pt; mso-bidi-font-family: Symbol; mso-fareast-font-family: Symbol;"><span style="mso-list: Ignore;">·<span style="font: 7.0pt "Times New Roman";"> </span></span></span><!--[endif]--><span style="font-family: "Arial","sans-serif"; font-size: 11.0pt;">centrifuge<o:p></o:p></span></div><div class="MsoNormal" style="margin-left: 17.4pt; mso-list: l8 level1 lfo1; tab-stops: list 17.4pt; text-align: justify; text-indent: -14.4pt; text-justify: inter-ideograph;"><!--[if !supportLists]--><span style="font-family: Symbol; font-size: 11.0pt; mso-bidi-font-family: Symbol; mso-fareast-font-family: Symbol;"><span style="mso-list: Ignore;">·<span style="font: 7.0pt "Times New Roman";"> </span></span></span><!--[endif]--><span style="font-family: "Arial","sans-serif"; font-size: 11.0pt;">restriction enzyme<o:p></o:p></span></div><div class="MsoNormal" style="margin-left: 17.4pt; mso-list: l8 level1 lfo1; tab-stops: list 17.4pt; text-align: justify; text-indent: -14.4pt; text-justify: inter-ideograph;"><!--[if !supportLists]--><span style="font-family: Symbol; font-size: 11.0pt; mso-bidi-font-family: Symbol; mso-fareast-font-family: Symbol;"><span style="mso-list: Ignore;">·<span style="font: 7.0pt "Times New Roman";"> </span></span></span><!--[endif]--><span style="font-family: "Arial","sans-serif"; font-size: 11.0pt;">agarose gel<o:p></o:p></span></div><div class="MsoNormal" style="margin-left: 17.4pt; mso-list: l8 level1 lfo1; tab-stops: list 17.4pt; text-align: justify; text-indent: -14.4pt; text-justify: inter-ideograph;"><!--[if !supportLists]--><span style="font-family: Symbol; font-size: 11.0pt; mso-bidi-font-family: Symbol; mso-fareast-font-family: Symbol;"><span style="mso-list: Ignore;">·<span style="font: 7.0pt "Times New Roman";"> </span></span></span><!--[endif]--><span style="font-family: "Arial","sans-serif"; font-size: 11.0pt;">power pack, i.e. suitable source of current<o:p></o:p></span></div><div class="MsoNormal" style="margin-left: 17.4pt; mso-list: l8 level1 lfo1; tab-stops: list 17.4pt; text-align: justify; text-indent: -14.4pt; text-justify: inter-ideograph;"><!--[if !supportLists]--><span style="font-family: Symbol; font-size: 11.0pt; mso-bidi-font-family: Symbol; mso-fareast-font-family: Symbol;"><span style="mso-list: Ignore;">·<span style="font: 7.0pt "Times New Roman";"> </span></span></span><!--[endif]--><span style="font-family: "Arial","sans-serif"; font-size: 11.0pt;">nitrocellulose membrane<o:p></o:p></span></div><div class="MsoNormal" style="margin-left: 17.4pt; mso-list: l8 level1 lfo1; tab-stops: list 17.4pt; text-align: justify; text-indent: -14.4pt; text-justify: inter-ideograph;"><!--[if !supportLists]--><span style="font-family: Symbol; font-size: 11.0pt; mso-bidi-font-family: Symbol; mso-fareast-font-family: Symbol;"><span style="mso-list: Ignore;">·<span style="font: 7.0pt "Times New Roman";"> </span></span></span><!--[endif]--><span style="font-family: "Arial","sans-serif"; font-size: 11.0pt;">radioactive probe<o:p></o:p></span></div><div class="MsoNormal" style="margin-left: 17.4pt; mso-list: l8 level1 lfo1; tab-stops: list 17.4pt; text-align: justify; text-indent: -14.4pt; text-justify: inter-ideograph;"><!--[if !supportLists]--><span style="font-family: Symbol; font-size: 11.0pt; mso-bidi-font-family: Symbol; mso-fareast-font-family: Symbol;"><span style="mso-list: Ignore;">·<span style="font: 7.0pt "Times New Roman";"> </span></span></span><!--[endif]--><span style="font-family: "Arial","sans-serif"; font-size: 11.0pt;">autoradiography equipment<o:p></o:p></span></div><div class="MsoNormal" style="margin-left: 3.0pt; text-align: justify; text-justify: inter-ideograph;"><br />
</div><div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;"><br />
</div><div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;"><span style="font-family: "Arial","sans-serif"; font-size: 11.0pt;">Your plan should: have a clear and helpful structure to include<o:p></o:p></span></div><div class="MsoNormal" style="margin-left: 14.4pt; mso-list: l4 level1 lfo2; tab-stops: list 14.4pt; text-align: justify; text-indent: -14.4pt; text-justify: inter-ideograph;"><!--[if !supportLists]--><span style="font-family: Symbol; font-size: 11.0pt; mso-bidi-font-family: Symbol; mso-fareast-font-family: Symbol;"><span style="mso-list: Ignore;">·<span style="font: 7.0pt "Times New Roman";"> </span></span></span><!--[endif]--><span style="font-family: "Arial","sans-serif"; font-size: 11.0pt;">an explanation of theory to support your practical procedure<o:p></o:p></span></div><div class="MsoNormal" style="margin-left: 14.4pt; mso-list: l4 level1 lfo2; tab-stops: list 14.4pt; text-align: justify; text-indent: -14.4pt; text-justify: inter-ideograph;"><!--[if !supportLists]--><span style="font-family: Symbol; font-size: 11.0pt; mso-bidi-font-family: Symbol; mso-fareast-font-family: Symbol;"><span style="mso-list: Ignore;">·<span style="font: 7.0pt "Times New Roman";"> </span></span></span><!--[endif]--><span style="font-family: "Arial","sans-serif"; font-size: 11.0pt;">a description of the method used, including the scientific reasoning behind the method<o:p></o:p></span></div><div class="MsoNormal" style="margin-left: 14.4pt; mso-list: l4 level1 lfo2; tab-stops: list 14.4pt; text-align: justify; text-indent: -14.4pt; text-justify: inter-ideograph;"><!--[if !supportLists]--><span style="font-family: Symbol; font-size: 11.0pt; mso-bidi-font-family: Symbol; mso-fareast-font-family: Symbol;"><span style="mso-list: Ignore;">·<span style="font: 7.0pt "Times New Roman";"> </span></span></span><!--[endif]--><span style="font-family: "Arial","sans-serif"; font-size: 11.0pt;">the type of data generated by the experiment<o:p></o:p></span></div><div class="MsoNormal" style="margin-left: 14.4pt; mso-list: l4 level1 lfo2; tab-stops: list 14.4pt; text-align: justify; text-indent: -14.4pt; text-justify: inter-ideograph;"><!--[if !supportLists]--><span style="font-family: Symbol; font-size: 11.0pt; mso-bidi-font-family: Symbol; mso-fareast-font-family: Symbol;"><span style="mso-list: Ignore;">·<span style="font: 7.0pt "Times New Roman";"> </span></span></span><!--[endif]--><span style="font-family: "Arial","sans-serif"; font-size: 11.0pt;">how the results will be analysed including how the origin of the organism can be determined<o:p></o:p></span></div><div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;"><br />
</div><div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;"><br />
</div><div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;"><br />
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</div><div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;"><b><span style="color: black; font-family: "Arial","sans-serif"; font-size: 11.0pt;">Pre-Task Survey<o:p></o:p></span></b></div><div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;"><span style="color: black; font-family: "Arial","sans-serif"; font-size: 11.0pt; mso-bidi-font-weight: bold;">SPA Planning tasks that are designed around Application Syllabus topics might not follow the usual style as those designed around Core Syllabus topics.<span style="mso-spacerun: yes;"> </span>Read the task given and complete the following (i) <u>self-assessment</u>; and (ii) <u>flow chart</u>.<span style="mso-spacerun: yes;"> </span>Complete and submit them to your tutor’s pigeonhole by <u>Tuesday, 17 August</u>.<o:p></o:p></span></div><div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;"><br />
</div><div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;"><br />
</div><div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;"><b><span style="color: black; font-family: "Arial","sans-serif"; font-size: 11.0pt;">Self Assessment</span></b><span style="color: black; font-family: "Arial","sans-serif"; font-size: 11.0pt; mso-bidi-font-weight: bold;"> <o:p></o:p></span></div><div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;"><span style="color: black; font-family: "Arial","sans-serif"; font-size: 11.0pt; mso-bidi-font-weight: bold;">Attempt this immediately, i.e. <u>before</u> you proceed to the next section.<span style="mso-spacerun: yes;"> </span>Do take time to complete this so that your tutors can </span><span lang="EN-GB" style="color: black; font-family: "Arial","sans-serif"; font-size: 11.0pt; mso-ansi-language: EN-GB; mso-bidi-font-weight: bold;">customise</span><span style="color: black; font-family: "Arial","sans-serif"; font-size: 11.0pt; mso-bidi-font-weight: bold;"> their lessons to aid your understanding. <i style="mso-bidi-font-style: normal;"><span style="mso-spacerun: yes;"> </span><o:p></o:p></i></span></div><div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;"><br />
</div><table border="1" cellpadding="0" cellspacing="0" class="MsoTableGrid" style="border-collapse: collapse; border: none; mso-border-alt: solid windowtext .5pt; mso-padding-alt: 0cm 5.4pt 0cm 5.4pt; mso-table-layout-alt: fixed; mso-yfti-tbllook: 480;"><tbody>
<tr style="mso-yfti-firstrow: yes; mso-yfti-irow: 0;"> <td rowspan="2" style="border: solid windowtext 1.0pt; mso-border-alt: solid windowtext .5pt; padding: 0cm 5.4pt 0cm 5.4pt; width: 158.85pt;" width="212"> <div align="center" class="MsoNormal" style="text-align: center;"><span style="color: black; font-family: "Arial","sans-serif"; font-size: 10.0pt; mso-bidi-font-weight: bold;">Criteria<o:p></o:p></span></div></td> <td colspan="4" style="border-left: none; border: solid windowtext 1.0pt; mso-border-alt: solid windowtext .5pt; mso-border-left-alt: solid windowtext .5pt; padding: 0cm 5.4pt 0cm 5.4pt; width: 317.8pt;" valign="top" width="424"> <div align="center" class="MsoNormal" style="text-align: center;"><span style="color: black; font-family: "Arial","sans-serif"; font-size: 10.0pt; mso-bidi-font-weight: bold;">My Comfort Level with Application Topics Planning Task*<o:p></o:p></span></div></td> </tr>
<tr style="mso-yfti-irow: 1;"> <td style="border-bottom: solid windowtext 1.0pt; border-left: none; border-right: solid windowtext 1.0pt; border-top: none; mso-border-alt: solid windowtext .5pt; mso-border-left-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0cm 5.4pt 0cm 5.4pt; width: 69.1pt;" valign="top" width="92"> <div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;"><span style="color: black; font-family: "Arial","sans-serif"; font-size: 10.0pt; mso-bidi-font-weight: bold;">Comfortable<o:p></o:p></span></div></td> <td style="border-bottom: solid windowtext 1.0pt; border-left: none; border-right: solid windowtext 1.0pt; border-top: none; mso-border-alt: solid windowtext .5pt; mso-border-left-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0cm 5.4pt 0cm 5.4pt; width: 69.15pt;" valign="top" width="92"> <div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;"><span style="color: black; font-family: "Arial","sans-serif"; font-size: 10.0pt; mso-bidi-font-weight: bold;">Somewhat comfortable<o:p></o:p></span></div></td> <td style="border-bottom: solid windowtext 1.0pt; border-left: none; border-right: solid windowtext 1.0pt; border-top: none; mso-border-alt: solid windowtext .5pt; mso-border-left-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0cm 5.4pt 0cm 5.4pt; width: 78.3pt;" valign="top" width="104"> <div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;"><span style="color: black; font-family: "Arial","sans-serif"; font-size: 10.0pt; mso-bidi-font-weight: bold;">Uncomfortable<o:p></o:p></span></div></td> <td style="border-bottom: solid windowtext 1.0pt; border-left: none; border-right: solid windowtext 1.0pt; border-top: none; mso-border-alt: solid windowtext .5pt; mso-border-left-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0cm 5.4pt 0cm 5.4pt; width: 101.25pt;" valign="top" width="135"> <div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;"><span style="color: black; font-family: "Arial","sans-serif"; font-size: 10.0pt; mso-bidi-font-weight: bold;">Remarks, i.e. <o:p></o:p></span></div><div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;"><span style="color: black; font-family: "Arial","sans-serif"; font-size: 10.0pt; mso-bidi-font-weight: bold;">What are some difficulties that I face? <o:p></o:p></span></div><div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;"><span style="color: black; font-family: "Arial","sans-serif"; font-size: 10.0pt; mso-bidi-font-weight: bold;">Why so?<o:p></o:p></span></div></td> </tr>
<tr style="height: 108.0pt; mso-yfti-irow: 2;"> <td style="border-top: none; border: solid windowtext 1.0pt; height: 108.0pt; mso-border-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0cm 5.4pt 0cm 5.4pt; width: 158.85pt;" valign="top" width="212"> <div class="MsoNormal" style="margin-left: 18.0pt; mso-list: l1 level1 lfo5; tab-stops: list 18.0pt; text-indent: -18.0pt;"><!--[if !supportLists]--><span style="color: black; font-family: "Arial","sans-serif"; font-size: 10.0pt; mso-bidi-font-weight: bold; mso-fareast-font-family: Arial;"><span style="mso-list: Ignore;">1.<span style="font: 7.0pt "Times New Roman";"> </span></span></span><!--[endif]--><span style="color: black; font-family: "Arial","sans-serif"; font-size: 10.0pt; mso-bidi-font-weight: bold;">Providing theoretical basis for my suggested procedures <o:p></o:p></span></div></td> <td style="border-bottom: solid windowtext 1.0pt; border-left: none; border-right: solid windowtext 1.0pt; border-top: none; height: 108.0pt; mso-border-alt: solid windowtext .5pt; mso-border-left-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0cm 5.4pt 0cm 5.4pt; width: 69.1pt;" valign="top" width="92"> <div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;"><br />
</div></td> <td style="border-bottom: solid windowtext 1.0pt; border-left: none; border-right: solid windowtext 1.0pt; border-top: none; height: 108.0pt; mso-border-alt: solid windowtext .5pt; mso-border-left-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0cm 5.4pt 0cm 5.4pt; width: 69.15pt;" valign="top" width="92"> <div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;"><br />
</div></td> <td style="border-bottom: solid windowtext 1.0pt; border-left: none; border-right: solid windowtext 1.0pt; border-top: none; height: 108.0pt; mso-border-alt: solid windowtext .5pt; mso-border-left-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0cm 5.4pt 0cm 5.4pt; width: 78.3pt;" valign="top" width="104"> <div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;"><br />
</div></td> <td style="border-bottom: solid windowtext 1.0pt; border-left: none; border-right: solid windowtext 1.0pt; border-top: none; height: 108.0pt; mso-border-alt: solid windowtext .5pt; mso-border-left-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0cm 5.4pt 0cm 5.4pt; width: 101.25pt;" valign="top" width="135"> <div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;"><br />
</div></td> </tr>
<tr style="height: 108.0pt; mso-yfti-irow: 3;"> <td style="border-top: none; border: solid windowtext 1.0pt; height: 108.0pt; mso-border-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0cm 5.4pt 0cm 5.4pt; width: 158.85pt;" valign="top" width="212"> <div class="MsoNormal" style="margin-left: 18.0pt; mso-list: l1 level1 lfo5; tab-stops: list 18.0pt left 41.45pt; text-align: justify; text-indent: -18.0pt; text-justify: inter-ideograph;"><!--[if !supportLists]--><span style="color: black; font-family: "Arial","sans-serif"; font-size: 10.0pt; mso-bidi-font-weight: bold; mso-fareast-font-family: Arial;"><span style="mso-list: Ignore;">2.<span style="font: 7.0pt "Times New Roman";"> </span></span></span><!--[endif]--><span style="color: black; font-family: "Arial","sans-serif"; font-size: 10.0pt; mso-bidi-font-weight: bold;">Identifying what I want to observe / measure<o:p></o:p></span></div></td> <td style="border-bottom: solid windowtext 1.0pt; border-left: none; border-right: solid windowtext 1.0pt; border-top: none; height: 108.0pt; mso-border-alt: solid windowtext .5pt; mso-border-left-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0cm 5.4pt 0cm 5.4pt; width: 69.1pt;" valign="top" width="92"> <div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;"><br />
</div></td> <td style="border-bottom: solid windowtext 1.0pt; border-left: none; border-right: solid windowtext 1.0pt; border-top: none; height: 108.0pt; mso-border-alt: solid windowtext .5pt; mso-border-left-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0cm 5.4pt 0cm 5.4pt; width: 69.15pt;" valign="top" width="92"> <div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;"><br />
</div></td> <td style="border-bottom: solid windowtext 1.0pt; border-left: none; border-right: solid windowtext 1.0pt; border-top: none; height: 108.0pt; mso-border-alt: solid windowtext .5pt; mso-border-left-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0cm 5.4pt 0cm 5.4pt; width: 78.3pt;" valign="top" width="104"> <div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;"><br />
</div></td> <td style="border-bottom: solid windowtext 1.0pt; border-left: none; border-right: solid windowtext 1.0pt; border-top: none; height: 108.0pt; mso-border-alt: solid windowtext .5pt; mso-border-left-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0cm 5.4pt 0cm 5.4pt; width: 101.25pt;" valign="top" width="135"> <div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;"><br />
</div></td> </tr>
<tr style="height: 108.0pt; mso-yfti-irow: 4;"> <td style="border-top: none; border: solid windowtext 1.0pt; height: 108.0pt; mso-border-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0cm 5.4pt 0cm 5.4pt; width: 158.85pt;" valign="top" width="212"> <div class="MsoNormal" style="margin-left: 18.0pt; mso-list: l1 level1 lfo5; tab-stops: list 18.0pt; text-align: justify; text-indent: -18.0pt; text-justify: inter-ideograph;"><!--[if !supportLists]--><span style="color: black; font-family: "Arial","sans-serif"; font-size: 10.0pt; mso-bidi-font-weight: bold; mso-fareast-font-family: Arial;"><span style="mso-list: Ignore;">3.<span style="font: 7.0pt "Times New Roman";"> </span></span></span><!--[endif]--><span style="color: black; font-family: "Arial","sans-serif"; font-size: 10.0pt; mso-bidi-font-weight: bold;">Identifying what I want to conclude<o:p></o:p></span></div></td> <td style="border-bottom: solid windowtext 1.0pt; border-left: none; border-right: solid windowtext 1.0pt; border-top: none; height: 108.0pt; mso-border-alt: solid windowtext .5pt; mso-border-left-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0cm 5.4pt 0cm 5.4pt; width: 69.1pt;" valign="top" width="92"> <div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;"><br />
</div></td> <td style="border-bottom: solid windowtext 1.0pt; border-left: none; border-right: solid windowtext 1.0pt; border-top: none; height: 108.0pt; mso-border-alt: solid windowtext .5pt; mso-border-left-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0cm 5.4pt 0cm 5.4pt; width: 69.15pt;" valign="top" width="92"> <div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;"><br />
</div></td> <td style="border-bottom: solid windowtext 1.0pt; border-left: none; border-right: solid windowtext 1.0pt; border-top: none; height: 108.0pt; mso-border-alt: solid windowtext .5pt; mso-border-left-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0cm 5.4pt 0cm 5.4pt; width: 78.3pt;" valign="top" width="104"> <div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;"><br />
</div></td> <td style="border-bottom: solid windowtext 1.0pt; border-left: none; border-right: solid windowtext 1.0pt; border-top: none; height: 108.0pt; mso-border-alt: solid windowtext .5pt; mso-border-left-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0cm 5.4pt 0cm 5.4pt; width: 101.25pt;" valign="top" width="135"> <div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;"><br />
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<tr style="height: 108.0pt; mso-yfti-irow: 5; mso-yfti-lastrow: yes;"> <td style="border-top: none; border: solid windowtext 1.0pt; height: 108.0pt; mso-border-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0cm 5.4pt 0cm 5.4pt; width: 158.85pt;" valign="top" width="212"> <div class="MsoNormal" style="margin-left: 18.0pt; mso-list: l1 level1 lfo5; tab-stops: list 18.0pt; text-align: justify; text-indent: -18.0pt; text-justify: inter-ideograph;"><!--[if !supportLists]--><span style="color: black; font-family: "Arial","sans-serif"; font-size: 10.0pt; mso-bidi-font-weight: bold; mso-fareast-font-family: Arial;"><span style="mso-list: Ignore;">4.<span style="font: 7.0pt "Times New Roman";"> </span></span></span><!--[endif]--><span style="color: black; font-family: "Arial","sans-serif"; font-size: 10.0pt; mso-bidi-font-weight: bold;">Putting together the methods / procedure so that I can make my observations<o:p></o:p></span></div></td> <td style="border-bottom: solid windowtext 1.0pt; border-left: none; border-right: solid windowtext 1.0pt; border-top: none; height: 108.0pt; mso-border-alt: solid windowtext .5pt; mso-border-left-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0cm 5.4pt 0cm 5.4pt; width: 69.1pt;" valign="top" width="92"> <div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;"><br />
</div></td> <td style="border-bottom: solid windowtext 1.0pt; border-left: none; border-right: solid windowtext 1.0pt; border-top: none; height: 108.0pt; mso-border-alt: solid windowtext .5pt; mso-border-left-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0cm 5.4pt 0cm 5.4pt; width: 69.15pt;" valign="top" width="92"> <div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;"><br />
</div></td> <td style="border-bottom: solid windowtext 1.0pt; border-left: none; border-right: solid windowtext 1.0pt; border-top: none; height: 108.0pt; mso-border-alt: solid windowtext .5pt; mso-border-left-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0cm 5.4pt 0cm 5.4pt; width: 78.3pt;" valign="top" width="104"> <div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;"><br />
</div></td> <td style="border-bottom: solid windowtext 1.0pt; border-left: none; border-right: solid windowtext 1.0pt; border-top: none; height: 108.0pt; mso-border-alt: solid windowtext .5pt; mso-border-left-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt; padding: 0cm 5.4pt 0cm 5.4pt; width: 101.25pt;" valign="top" width="135"> <div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;"><br />
</div></td> </tr>
</tbody></table><div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;"><span style="color: black; font-family: "Arial","sans-serif"; font-size: 10.0pt; mso-bidi-font-weight: bold;">* Put a tick in the box correlating to your comfort level<o:p></o:p></span></div><div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;"><br />
</div></div><b><span style="color: black; font-family: "Arial","sans-serif"; font-size: 11.0pt; mso-ansi-language: EN-US; mso-bidi-language: AR-SA; mso-fareast-font-family: "Times New Roman"; mso-fareast-language: EN-US;"><br clear="all" style="mso-break-type: section-break; page-break-before: always;" /> </span></b> <div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;"><b><span style="color: black; font-family: "Arial","sans-serif"; font-size: 11.0pt;">Draft I <o:p></o:p></span></b></div><div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;"><br />
</div><div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;"><span style="color: black; font-family: "Arial","sans-serif"; font-size: 11.0pt; mso-bidi-font-weight: bold;">Follow the guidelines given to come up with the first draft – this should be the mental picture that you should form within 1 min of analysing the exam question </span><b><span style="color: blue; font-family: "Arial","sans-serif"; font-size: 11.0pt;"><o:p></o:p></span></b></div><div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;"><br />
</div><div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;"><br />
</div><div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;"><b><span style="color: blue; font-family: "Arial","sans-serif"; font-size: 11.0pt;">Guidelines:<o:p></o:p></span></b></div><div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;"><br />
</div><div class="MsoNormal" style="margin-left: 14.4pt; mso-list: l5 level1 lfo3; tab-stops: list 0cm; text-align: justify; text-indent: -14.4pt; text-justify: inter-ideograph;"><!--[if !supportLists]--><span style="color: blue; font-family: "Arial","sans-serif"; font-size: 11.0pt; mso-fareast-font-family: Arial;"><span style="mso-list: Ignore;">1.<span style="font: 7.0pt "Times New Roman";"> </span></span></span><!--[endif]--><span style="color: blue; font-family: "Arial","sans-serif"; font-size: 11.0pt;">An <u>explanation of theory</u> to support your practical procedure - theoretical consideration or rationale of the plan to justify the practical procedure<o:p></o:p></span></div><div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;"><br />
</div><div class="MsoNormal" style="margin-left: 14.4pt; mso-list: l5 level1 lfo3; tab-stops: list 0cm; text-align: justify; text-indent: -14.4pt; text-justify: inter-ideograph;"><!--[if !supportLists]--><span style="color: blue; font-family: "Arial","sans-serif"; font-size: 11.0pt; mso-fareast-font-family: Arial;"><span style="mso-list: Ignore;">2.<span style="font: 7.0pt "Times New Roman";"> </span></span></span><!--[endif]--><span style="color: blue; font-family: "Arial","sans-serif"; font-size: 11.0pt;">A <u>description of the method</u> used including the <u>scientific reasoning behind the method</u><o:p></o:p></span></div><div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;"><br />
</div><div class="MsoNormal" style="margin-left: 14.4pt; text-align: justify; text-justify: inter-ideograph;"><span style="color: black; font-family: Wingdings; font-size: 20.0pt; mso-ascii-font-family: Arial; mso-bidi-font-family: Arial; mso-char-type: symbol; mso-hansi-font-family: Arial; mso-symbol-font-family: Wingdings;"><span style="mso-char-type: symbol; mso-symbol-font-family: Wingdings;">I</span></span><span style="color: black; font-family: "Arial","sans-serif"; font-size: 20.0pt;"> PAUSE</span><span style="color: black; font-family: "Arial","sans-serif"; font-size: 11.0pt;"><o:p></o:p></span></div><div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;"><br />
</div><ol start="1" style="margin-top: 0cm;" type="a"><li class="MsoNormal" style="color: black; mso-list: l0 level1 lfo4; tab-stops: list 36.0pt; text-align: justify; text-justify: inter-ideograph;"><span style="font-family: "Arial","sans-serif"; font-size: 11.0pt;">Identify the key procedure.<span style="mso-spacerun: yes;"> </span>You might have many different sub-procedures / steps, some of which merely supports the key procedure.<span style="mso-spacerun: yes;"> </span>The trick lies in the identification of the key.<o:p></o:p></span></li>
</ol><div class="MsoNormal" style="margin-left: 36.0pt; text-align: justify; text-justify: inter-ideograph;"><u><span style="color: black; font-family: "Arial","sans-serif"; font-size: 11.0pt;">Hint</span></u><span style="color: black; font-family: "Arial","sans-serif"; font-size: 11.0pt;">: <span style="mso-tab-count: 1;"> </span></span><span lang="EN-GB" style="color: black; font-family: "Arial","sans-serif"; font-size: 11.0pt; mso-ansi-language: EN-GB;">Scrutinise</span><span style="color: black; font-family: "Arial","sans-serif"; font-size: 11.0pt;"> the aim of experiment.<span style="mso-spacerun: yes;"> </span><o:p></o:p></span></div><div class="MsoNormal" style="margin-left: 72.0pt; text-align: justify; text-justify: inter-ideograph;"><span style="color: black; font-family: "Arial","sans-serif"; font-size: 11.0pt;">The command word in the aim is </span><u><span style="color: red; font-family: "Comic Sans MS"; font-size: 11.0pt; mso-bidi-font-family: Arial;">to establish identity</span></u><b style="mso-bidi-font-weight: normal;"><span style="color: black; font-family: "Arial","sans-serif"; font-size: 11.0pt;"><o:p></o:p></span></b></div><div class="MsoNormal" style="margin-left: 18.0pt; text-align: justify; text-indent: 18.0pt; text-justify: inter-ideograph;"><span style="color: black; font-family: "Arial","sans-serif"; font-size: 11.0pt;"><span style="mso-tab-count: 1;"> </span>What is one key procedure that relates to the command word?<span style="mso-spacerun: yes;"> </span>Jot this down!<o:p></o:p></span></div><div class="MsoNormal" style="margin-left: 18.0pt; text-align: justify; text-indent: 18.0pt; text-justify: inter-ideograph;"><b style="mso-bidi-font-weight: normal;"><span style="color: red; font-family: "Comic Sans MS"; font-size: 11.0pt; mso-bidi-font-family: Arial;"><span style="mso-tab-count: 1;"> </span></span></b><u><span style="color: red; font-family: "Comic Sans MS"; font-size: 11.0pt; mso-bidi-font-family: Arial;">DNA fingerprinting through RFLP analysis of VNTRs<o:p></o:p></span></u></div><div class="MsoNormal" style="margin-left: 18.0pt; text-align: justify; text-indent: 18.0pt; text-justify: inter-ideograph;"><br />
</div><ol start="2" style="margin-top: 0cm;" type="a"><li class="MsoNormal" style="color: black; mso-list: l0 level1 lfo4; tab-stops: list 36.0pt; text-align: justify; text-justify: inter-ideograph;"><span style="font-family: "Arial","sans-serif"; font-size: 11.0pt;">Justify the procedure / method, i.e. what’s the scientific basis? <o:p></o:p></span></li>
</ol><div class="MsoNormal" style="margin-left: 36.0pt; text-align: justify; text-justify: inter-ideograph;"><u><span style="color: black; font-family: "Arial","sans-serif"; font-size: 11.0pt;">Hint</span></u><span style="color: black; font-family: "Arial","sans-serif"; font-size: 11.0pt;">:<span style="mso-tab-count: 1;"> </span>Think about <u>three main theoretical concepts</u> surround the procedure that you are proposing.<span style="mso-spacerun: yes;"> </span>Jot down the key words.<o:p></o:p></span></div><div class="MsoNormal" style="margin-bottom: .0001pt; margin-bottom: 0cm; margin-left: 72.0pt; margin-right: 0cm; margin-top: 6.0pt; mso-list: l2 level1 lfo7; tab-stops: list 72.0pt; text-align: justify; text-indent: -36.0pt; text-justify: inter-ideograph;"><!--[if !supportLists]--><span style="color: red; font-family: "Comic Sans MS"; font-size: 11.0pt; mso-bidi-font-family: "Comic Sans MS"; mso-fareast-font-family: "Comic Sans MS";"><span style="mso-list: Ignore;">1.<span style="font: 7.0pt "Times New Roman";"> </span></span></span><!--[endif]--><span style="color: red; font-family: "Comic Sans MS"; font-size: 11.0pt; mso-bidi-font-family: Arial;">VNTRs are <u>DNA sequences</u>, which are <u>repeated in tandem</u> <u>a variable number of times</u> at certain loci<o:p></o:p></span></div><div class="MsoNormal" style="margin-bottom: .0001pt; margin-bottom: 0cm; margin-left: 72.0pt; margin-right: 0cm; margin-top: 6.0pt; mso-list: l2 level1 lfo7; tab-stops: list 72.0pt; text-align: justify; text-indent: -36.0pt; text-justify: inter-ideograph;"><!--[if !supportLists]--><span style="color: red; font-family: "Comic Sans MS"; font-size: 11.0pt; mso-bidi-font-family: "Comic Sans MS"; mso-fareast-font-family: "Comic Sans MS";"><span style="mso-list: Ignore;">2.<span style="font: 7.0pt "Times New Roman";"> </span></span></span><!--[endif]--><span style="color: red; font-family: "Comic Sans MS"; font-size: 11.0pt; mso-bidi-font-family: Arial;">Restriction enzymes recognise and cut <u>sites flanking the VNTR regions</u>, producing <u>restriction fragments of different lengths<o:p></o:p></u></span></div><div class="MsoNormal" style="margin-bottom: .0001pt; margin-bottom: 0cm; margin-left: 72.0pt; margin-right: 0cm; margin-top: 6.0pt; mso-list: l2 level1 lfo7; tab-stops: list 72.0pt; text-align: justify; text-indent: -36.0pt; text-justify: inter-ideograph;"><!--[if !supportLists]--><span style="color: red; font-family: "Comic Sans MS"; font-size: 11.0pt; mso-bidi-font-family: "Comic Sans MS"; mso-fareast-font-family: "Comic Sans MS";"><span style="mso-list: Ignore;">3.<span style="font: 7.0pt "Times New Roman";"> </span></span></span><!--[endif]--><u><span style="color: red; font-family: "Comic Sans MS"; font-size: 11.0pt; mso-bidi-font-family: Arial;">Alligator species</span></u><span style="color: red; font-family: "Comic Sans MS"; font-size: 11.0pt; mso-bidi-font-family: Arial;"> exhibit <u>tandem repeat polymorphism</u>, hence each alligator species can be identified through the unique pattern of restriction fragments<o:p></o:p></span></div><div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;"><br />
</div><div class="MsoNormal" style="margin-left: 14.4pt; mso-list: l5 level1 lfo3; tab-stops: list 0cm; text-align: justify; text-indent: -14.4pt; text-justify: inter-ideograph;"><!--[if !supportLists]--><span style="color: blue; font-family: "Arial","sans-serif"; font-size: 11.0pt; mso-fareast-font-family: Arial;"><span style="mso-list: Ignore;">3.<span style="font: 7.0pt "Times New Roman";"> </span></span></span><!--[endif]--><span style="color: blue; font-family: "Arial","sans-serif"; font-size: 11.0pt;">The type of data generated by the experiment<o:p></o:p></span></div><div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;"><br />
</div><div class="MsoNormal" style="margin-left: 14.4pt; text-align: justify; text-justify: inter-ideograph;"><span style="color: black; font-family: Wingdings; font-size: 20.0pt; mso-ascii-font-family: Arial; mso-bidi-font-family: Arial; mso-char-type: symbol; mso-hansi-font-family: Arial; mso-symbol-font-family: Wingdings;"><span style="mso-char-type: symbol; mso-symbol-font-family: Wingdings;">I</span></span><span style="color: black; font-family: "Arial","sans-serif"; font-size: 20.0pt;"> PAUSE</span><span style="color: black; font-family: "Arial","sans-serif"; font-size: 11.0pt;"><o:p></o:p></span></div><div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;"><br />
</div><ol start="3" style="margin-top: 0cm;" type="a"><li class="MsoNormal" style="color: black; mso-list: l0 level1 lfo4; tab-stops: list 36.0pt; text-align: justify; text-justify: inter-ideograph;"><span style="font-family: "Arial","sans-serif"; font-size: 11.0pt;">Based procedure / method suggested, identify the observation / measurement that you can make.<span style="mso-spacerun: yes;"> </span>How will these help you achieve the aim? <o:p></o:p></span></li>
</ol><div class="MsoNormal" style="margin-bottom: .0001pt; margin-bottom: 0cm; margin-left: 72.0pt; margin-right: 0cm; margin-top: 6.0pt; mso-list: l6 level1 lfo8; tab-stops: list 72.0pt; text-align: justify; text-indent: -36.0pt; text-justify: inter-ideograph;"><!--[if !supportLists]--><b style="mso-bidi-font-weight: normal;"><span style="color: red; font-family: "Comic Sans MS"; font-size: 10.0pt; mso-bidi-font-family: "Comic Sans MS"; mso-bidi-font-size: 11.0pt; mso-fareast-font-family: "Comic Sans MS";"><span style="mso-list: Ignore;">1.<span style="font: 7.0pt "Times New Roman";"> </span></span></span></b><!--[endif]--><u><span style="color: red; font-family: "Comic Sans MS"; font-size: 11.0pt; mso-bidi-font-family: Arial;">Number</span></u><span style="color: red; font-family: "Comic Sans MS"; font-size: 11.0pt; mso-bidi-font-family: Arial;"> and <u>length</u> of restriction fragments from each muscle tissue<o:p></o:p></span></div><div class="MsoNormal" style="margin-bottom: .0001pt; margin-bottom: 0cm; margin-left: 72.0pt; margin-right: 0cm; margin-top: 6.0pt; mso-list: l6 level1 lfo8; tab-stops: list 72.0pt; text-align: justify; text-indent: -36.0pt; text-justify: inter-ideograph;"><!--[if !supportLists]--><b style="mso-bidi-font-weight: normal;"><span style="color: red; font-family: "Comic Sans MS"; font-size: 10.0pt; mso-bidi-font-family: "Comic Sans MS"; mso-bidi-font-size: 11.0pt; mso-fareast-font-family: "Comic Sans MS";"><span style="mso-list: Ignore;">2.<span style="font: 7.0pt "Times New Roman";"> </span></span></span></b><!--[endif]--><u><span style="color: red; font-family: "Comic Sans MS"; font-size: 11.0pt; mso-bidi-font-family: Arial;">Assumption</span></u><span style="color: red; font-family: "Comic Sans MS"; font-size: 11.0pt; mso-bidi-font-family: Arial;">: <o:p></o:p></span></div><div class="MsoNormal" style="margin-left: 72.0pt; text-align: justify; text-justify: inter-ideograph;"><span style="color: red; font-family: "Comic Sans MS"; font-size: 11.0pt; mso-bidi-font-family: Arial;">Confiscated consignment should have the same pattern of restriction fragments as that of the Chinese alligator, i.e. same number of restriction fragments, same length for each type of fragment <o:p></o:p></span></div><div class="MsoNormal" style="margin-left: 36.0pt; text-align: justify; text-justify: inter-ideograph;"><br />
</div><div class="MsoNormal" style="margin-left: 36.0pt; text-align: justify; text-justify: inter-ideograph;"><br />
</div><div class="MsoNormal" style="margin-left: 36.0pt; text-align: justify; text-justify: inter-ideograph;"><br />
</div><ol start="4" style="margin-top: 0cm;" type="a"><li class="MsoNormal" style="color: black; mso-list: l0 level1 lfo4; tab-stops: list 36.0pt; text-align: justify; text-justify: inter-ideograph;"><span style="font-family: "Arial","sans-serif"; font-size: 11.0pt;">Identify <u>two</u> further sub-procedures / steps that you need to take to make the data meaningful. <span style="mso-spacerun: yes;"> </span><o:p></o:p></span></li>
</ol><div class="MsoNormal" style="margin-left: 54.0pt; mso-list: l0 level2 lfo4; tab-stops: list 54.0pt; text-align: justify; text-indent: -18.0pt; text-justify: inter-ideograph;"><!--[if !supportLists]--><b style="mso-bidi-font-weight: normal;"><span style="color: red; font-family: "Comic Sans MS"; font-size: 11.0pt; mso-bidi-font-family: "Comic Sans MS"; mso-fareast-font-family: "Comic Sans MS";"><span style="mso-list: Ignore;">0.<span style="font: 7.0pt "Times New Roman";"> </span></span></span></b><!--[endif]--><span style="color: red; font-family: "Comic Sans MS"; font-size: 11.0pt; mso-bidi-font-family: Arial;">Homogenise muscle tissues using ice-cold DNA extraction buffer in a blender<o:p></o:p></span></div><div class="MsoNormal" style="margin-left: 54.0pt; mso-list: l0 level2 lfo4; tab-stops: list 54.0pt; text-align: justify; text-indent: -18.0pt; text-justify: inter-ideograph;"><!--[if !supportLists]--><b style="mso-bidi-font-weight: normal;"><span style="color: red; font-family: "Comic Sans MS"; font-size: 11.0pt; mso-bidi-font-family: "Comic Sans MS"; mso-fareast-font-family: "Comic Sans MS";"><span style="mso-list: Ignore;">1.<span style="font: 7.0pt "Times New Roman";"> </span></span></span></b><!--[endif]--><span style="color: red; font-family: "Comic Sans MS"; font-size: 11.0pt; mso-bidi-font-family: Arial;">Agarose gel electrophoresis </span><span style="color: white; font-family: "Comic Sans MS"; font-size: 11.0pt; mso-bidi-font-family: Arial;">to separate restriction fragments based on length</span><span style="color: red; font-family: "Comic Sans MS"; font-size: 11.0pt; mso-bidi-font-family: Arial;"> <o:p></o:p></span></div><div class="MsoNormal" style="margin-left: 54.0pt; mso-list: l0 level2 lfo4; tab-stops: list 54.0pt; text-align: justify; text-indent: -18.0pt; text-justify: inter-ideograph;"><!--[if !supportLists]--><b style="mso-bidi-font-weight: normal;"><span style="color: red; font-family: "Comic Sans MS"; font-size: 11.0pt; mso-bidi-font-family: "Comic Sans MS"; mso-fareast-font-family: "Comic Sans MS";"><span style="mso-list: Ignore;">2.<span style="font: 7.0pt "Times New Roman";"> </span></span></span></b><!--[endif]--><span style="color: red; font-family: "Comic Sans MS"; font-size: 11.0pt; mso-bidi-font-family: Arial;">Probe with complementary sequence </span><span style="color: white; font-family: "Comic Sans MS"; font-size: 11.0pt; mso-bidi-font-family: Arial;">to identify fragments bearing the VNTRs</span><span style="color: red; font-family: "Comic Sans MS"; font-size: 11.0pt; mso-bidi-font-family: Arial;"> <o:p></o:p></span></div><div class="MsoNormal" style="margin-left: 36.0pt; text-align: justify; text-justify: inter-ideograph;"><br />
</div><div class="MsoNormal" style="tab-stops: 299.55pt; text-align: justify; text-justify: inter-ideograph;"><span style="color: blue; font-family: "Arial","sans-serif"; font-size: 11.0pt;"><span style="mso-tab-count: 1;"> </span><o:p></o:p></span></div><div class="MsoNormal" style="margin-left: 14.4pt; mso-list: l5 level1 lfo3; tab-stops: list 0cm; text-align: justify; text-indent: -14.4pt; text-justify: inter-ideograph;"><!--[if !supportLists]--><span style="color: blue; font-family: "Arial","sans-serif"; font-size: 11.0pt; mso-fareast-font-family: Arial;"><span style="mso-list: Ignore;">4.<span style="font: 7.0pt "Times New Roman";"> </span></span></span><!--[endif]--><span style="color: blue; font-family: "Arial","sans-serif"; font-size: 11.0pt;">How the results will be analysed including how the identity of the organism can be determined<o:p></o:p></span></div><div class="MsoNormal" style="margin-left: 57.6pt; mso-list: l7 level1 lfo9; tab-stops: list 57.6pt; text-align: justify; text-indent: -21.6pt; text-justify: inter-ideograph;"><!--[if !supportLists]--><span style="color: blue; font-family: Symbol; font-size: 11.0pt; mso-bidi-font-family: Symbol; mso-bidi-font-weight: bold; mso-fareast-font-family: Symbol;"><span style="mso-list: Ignore;">·<span style="font: 7.0pt "Times New Roman";"> </span></span></span><!--[endif]--><u style="text-underline: thick;"><span style="color: red; font-family: "Comic Sans MS"; font-size: 11.0pt; mso-bidi-font-family: Arial; mso-bidi-font-weight: bold;">Compare</span></u><u><span style="color: red; font-family: "Comic Sans MS"; font-size: 11.0pt; mso-bidi-font-family: Arial; mso-bidi-font-weight: bold;"> restriction fragments</span></u><span style="color: red; font-family: "Comic Sans MS"; font-size: 11.0pt; mso-bidi-font-family: Arial; mso-bidi-font-weight: bold;"> obtained from </span><span style="color: red; font-family: "Comic Sans MS"; font-size: 11.0pt; mso-bidi-font-family: Arial;">(i) confiscated consignment; (ii) American alligator; (iii) Chinese alligator;<span style="mso-spacerun: yes;"> </span><span style="mso-bidi-font-weight: bold;"><o:p></o:p></span></span></div><div class="MsoNormal" style="margin-left: 57.6pt; mso-list: l7 level1 lfo9; tab-stops: list 57.6pt; text-align: justify; text-indent: -21.6pt; text-justify: inter-ideograph;"><!--[if !supportLists]--><span style="color: blue; font-family: Symbol; font-size: 11.0pt; mso-bidi-font-family: Symbol; mso-bidi-font-weight: bold; mso-fareast-font-family: Symbol;"><span style="mso-list: Ignore;">·<span style="font: 7.0pt "Times New Roman";"> </span></span></span><!--[endif]--><u><span style="color: red; font-family: "Comic Sans MS"; font-size: 11.0pt; mso-bidi-font-family: Arial; mso-bidi-font-weight: bold;">Identify the restriction fragments</span></u><span style="color: red; font-family: "Comic Sans MS"; font-size: 11.0pt; mso-bidi-font-family: Arial; mso-bidi-font-weight: bold;"> that </span><span style="color: red; font-family: "Comic Sans MS"; font-size: 11.0pt; mso-bidi-font-family: Arial;">confiscated consignment<span style="mso-bidi-font-weight: bold;"> <u>has in common with each of the two species</u>; <o:p></o:p></span></span></div><div class="MsoNormal" style="margin-left: 57.6pt; mso-list: l7 level1 lfo9; tab-stops: list 57.6pt; text-align: justify; text-indent: -21.6pt; text-justify: inter-ideograph;"><!--[if !supportLists]--><span style="color: blue; font-family: Symbol; font-size: 11.0pt; mso-bidi-font-family: Symbol; mso-bidi-font-weight: bold; mso-fareast-font-family: Symbol;"><span style="mso-list: Ignore;">·<span style="font: 7.0pt "Times New Roman";"> </span></span></span><!--[endif]--><u><span style="color: red; font-family: "Comic Sans MS"; font-size: 11.0pt; mso-bidi-font-family: Arial; mso-bidi-font-weight: bold;">Determine the number of the restriction fragments</span></u><span style="color: red; font-family: "Comic Sans MS"; font-size: 11.0pt; mso-bidi-font-family: Arial; mso-bidi-font-weight: bold;"> that </span><span style="color: red; font-family: "Comic Sans MS"; font-size: 11.0pt; mso-bidi-font-family: Arial;">confiscated consignment<span style="mso-bidi-font-weight: bold;"> <u>has in common with each of the two species</u>; <o:p></o:p></span></span></div><div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;"><br />
</div><div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;"><br />
</div><div class="MsoNormal" style="margin-left: 14.4pt; mso-list: l5 level1 lfo3; tab-stops: list 0cm; text-align: justify; text-indent: -14.4pt; text-justify: inter-ideograph;"><!--[if !supportLists]--><span style="color: blue; font-family: "Arial","sans-serif"; font-size: 11.0pt; mso-fareast-font-family: Arial;"><span style="mso-list: Ignore;">5.<span style="font: 7.0pt "Times New Roman";"> </span></span></span><!--[endif]--><span style="color: blue; font-family: "Arial","sans-serif"; font-size: 11.0pt;">The correct use of technical and scientific terms<o:p></o:p></span></div><div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;"><br />
</div><div class="MsoNormal" style="margin-left: 14.4pt; text-align: justify; text-justify: inter-ideograph;"><span style="color: black; font-family: Wingdings; font-size: 20.0pt; mso-ascii-font-family: Arial; mso-bidi-font-family: Arial; mso-char-type: symbol; mso-hansi-font-family: Arial; mso-symbol-font-family: Wingdings;"><span style="mso-char-type: symbol; mso-symbol-font-family: Wingdings;">I</span></span><span style="color: black; font-family: "Arial","sans-serif"; font-size: 20.0pt;"> PAUSE</span><span style="color: black; font-family: "Arial","sans-serif"; font-size: 11.0pt;"><o:p></o:p></span></div><div class="MsoNormal" style="margin-left: 18.0pt; text-align: justify; text-justify: inter-ideograph;"><br />
</div><ol start="5" style="margin-top: 0cm;" type="a"><li class="MsoNormal" style="color: black; mso-list: l0 level1 lfo4; tab-stops: list 36.0pt; text-align: justify; text-justify: inter-ideograph;"><span style="font-family: "Arial","sans-serif"; font-size: 11.0pt;">In a nutshell, identify the supporting sub-procedures / steps as instructed in <b style="mso-bidi-font-weight: normal;">d</b> </span><span style="font-family: Wingdings; font-size: 11.0pt; mso-ascii-font-family: Arial; mso-bidi-font-family: Arial; mso-char-type: symbol; mso-hansi-font-family: Arial; mso-symbol-font-family: Wingdings;"><span style="mso-char-type: symbol; mso-symbol-font-family: Wingdings;">à</span></span><span style="font-family: "Arial","sans-serif"; font-size: 11.0pt;"> draw a flow chart of critical steps.<span style="mso-spacerun: yes;"> </span>You may want to consider the following details later for the actual write up:<o:p></o:p></span></li>
</ol><div class="MsoNormal" style="margin-bottom: .0001pt; margin-bottom: 0cm; margin-left: 54.0pt; margin-right: 0cm; margin-top: 6.0pt; mso-list: l3 level2 lfo6; tab-stops: list 54.0pt; text-align: justify; text-indent: -18.0pt; text-justify: inter-ideograph;"><!--[if !supportLists]--><span style="color: black; font-family: Symbol; font-size: 10.0pt; mso-bidi-font-family: Symbol; mso-bidi-font-size: 11.0pt; mso-fareast-font-family: Symbol;"><span style="mso-list: Ignore;">·<span style="font: 7.0pt "Times New Roman";"> </span></span></span><!--[endif]--><span style="color: black; font-family: "Arial","sans-serif"; font-size: 11.0pt;">how would you <u>prepare</u> the materials based on what you have been given? why do you select these conditions?<o:p></o:p></span></div><div class="MsoNormal" style="margin-bottom: .0001pt; margin-bottom: 0cm; margin-left: 54.0pt; margin-right: 0cm; margin-top: 6.0pt; mso-list: l3 level2 lfo6; tab-stops: list 54.0pt; text-align: justify; text-indent: -18.0pt; text-justify: inter-ideograph;"><!--[if !supportLists]--><span style="color: black; font-family: Symbol; font-size: 10.0pt; mso-bidi-font-family: Symbol; mso-bidi-font-size: 11.0pt; mso-fareast-font-family: Symbol;"><span style="mso-list: Ignore;">·<span style="font: 7.0pt "Times New Roman";"> </span></span></span><!--[endif]--><span style="color: black; font-family: "Arial","sans-serif"; font-size: 11.0pt;">what are the steps involved in each procedure?<span style="mso-spacerun: yes;"> </span>what is the scientific basis behind each step?<o:p></o:p></span></div><div class="MsoNormal" style="margin-bottom: .0001pt; margin-bottom: 0cm; margin-left: 54.0pt; margin-right: 0cm; margin-top: 6.0pt; mso-list: l3 level2 lfo6; tab-stops: list 54.0pt; text-align: justify; text-indent: -18.0pt; text-justify: inter-ideograph;"><!--[if !supportLists]--><span style="color: black; font-family: Symbol; font-size: 10.0pt; mso-bidi-font-family: Symbol; mso-bidi-font-size: 11.0pt; mso-fareast-font-family: Symbol;"><span style="mso-list: Ignore;">·<span style="font: 7.0pt "Times New Roman";"> </span></span></span><!--[endif]--><span style="color: black; font-family: "Arial","sans-serif"; font-size: 11.0pt;">what observations / measurements would you want to take?<span style="mso-spacerun: yes;"> </span><o:p></o:p></span></div><div class="MsoNormal" style="margin-bottom: .0001pt; margin-bottom: 0cm; margin-left: 54.0pt; margin-right: 0cm; margin-top: 6.0pt; mso-list: l3 level2 lfo6; tab-stops: list 54.0pt; text-align: justify; text-indent: -18.0pt; text-justify: inter-ideograph;"><!--[if !supportLists]--><span style="color: black; font-family: Symbol; font-size: 10.0pt; mso-bidi-font-family: Symbol; mso-bidi-font-size: 11.0pt; mso-fareast-font-family: Symbol;"><span style="mso-list: Ignore;">·<span style="font: 7.0pt "Times New Roman";"> </span></span></span><!--[endif]--><span style="color: black; font-family: "Arial","sans-serif"; font-size: 11.0pt;">how will these observations / measurements help you in identification?<o:p></o:p></span></div><div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;"><br />
</div><div class="MsoNormal" style="text-align: justify; text-justify: inter-ideograph;"><u><span style="color: black; font-family: "Arial","sans-serif"; font-size: 11.0pt;">Basic Plan of the Experiment</span></u><span style="color: black; font-family: "Arial","sans-serif"; font-size: 11.0pt;"> (Answer parts <b style="mso-bidi-font-weight: normal;">a</b> to <b style="mso-bidi-font-weight: normal;">e</b> in a <b style="mso-bidi-font-weight: normal;">flow chart)</b><o:p></o:p></span></div><div class="MsoNormal" style="margin-left: 15.0pt; text-align: justify; text-justify: inter-ideograph;"><br />
</div><div class="MsoNormal" style="margin-left: 15.0pt; text-align: justify; text-justify: inter-ideograph;"><span style="color: red; font-family: "Comic Sans MS"; font-size: 11.0pt; mso-bidi-font-family: Arial;">Homogenisation:<span style="mso-spacerun: yes;"> </span>using ice-cold DNA extraction buffer and a blender<o:p></o:p></span></div><div class="MsoNormal" style="margin-left: 15.0pt; text-align: justify; text-justify: inter-ideograph;"><span style="color: blue; font-family: "Comic Sans MS"; font-size: 11.0pt; mso-bidi-font-family: Arial;">What does the ice-cold DNA extraction buffer contain?<span style="mso-spacerun: yes;"> </span>Why must you use this buffer?<o:p></o:p></span></div><div class="MsoNormal" style="margin-left: 15.0pt; text-align: justify; text-justify: inter-ideograph;"><span style="color: red; font-family: "Comic Sans MS"; font-size: 11.0pt; mso-bidi-font-family: Arial;"><span style="mso-spacerun: yes;"> </span><span style="mso-tab-count: 1;"> </span></span><span style="color: red; font-family: Wingdings; font-size: 11.0pt; mso-ascii-font-family: "Comic Sans MS"; mso-bidi-font-family: Arial; mso-char-type: symbol; mso-hansi-font-family: "Comic Sans MS"; mso-symbol-font-family: Wingdings;"><span style="mso-char-type: symbol; mso-symbol-font-family: Wingdings;">ò</span></span><span style="color: red; font-family: "Comic Sans MS"; font-size: 11.0pt; mso-bidi-font-family: Arial;"><o:p></o:p></span></div><div class="MsoNormal" style="margin-left: 15.0pt; text-align: justify; text-justify: inter-ideograph;"><span style="color: red; font-family: "Comic Sans MS"; font-size: 11.0pt; mso-bidi-font-family: Arial;">Centrifuge and transfer supernatant to a clean Eppendorf tube.<span style="mso-spacerun: yes;"> </span>Add ice-cold ethanol. Centrifuge again and keep the pellet.<o:p></o:p></span></div><div class="MsoNormal" style="margin-left: 15.0pt; text-align: justify; text-justify: inter-ideograph;"><span style="color: blue; font-family: "Comic Sans MS"; font-size: 11.0pt; mso-bidi-font-family: Arial;">Why do you need to centrifuge to obtain supernatant?<span style="mso-spacerun: yes;"> </span>Why do you need to ice-cold ethanol?<o:p></o:p></span></div><div class="MsoNormal" style="margin-left: 15.0pt; text-align: justify; text-indent: 21.0pt; text-justify: inter-ideograph;"><span style="color: red; font-family: Wingdings; font-size: 11.0pt; mso-ascii-font-family: "Comic Sans MS"; mso-bidi-font-family: Arial; mso-char-type: symbol; mso-hansi-font-family: "Comic Sans MS"; mso-symbol-font-family: Wingdings;"><span style="mso-char-type: symbol; mso-symbol-font-family: Wingdings;">ò</span></span><span style="color: red; font-family: "Comic Sans MS"; font-size: 11.0pt; mso-bidi-font-family: Arial;"><o:p></o:p></span></div><div class="MsoNormal" style="margin-left: 18.0pt; text-align: justify; text-justify: inter-ideograph;"><span style="color: red; font-family: "Comic Sans MS"; font-size: 11.0pt; mso-bidi-font-family: Arial;">Resuspend pellet in buffer.<span style="mso-spacerun: yes;"> </span>Add restriction enzyme and incubate<o:p></o:p></span></div><div class="MsoNormal" style="margin-left: 15.0pt; text-align: justify; text-indent: 21.0pt; text-justify: inter-ideograph;"><span style="color: red; font-family: Wingdings; font-size: 11.0pt; mso-ascii-font-family: "Comic Sans MS"; mso-bidi-font-family: Arial; mso-char-type: symbol; mso-hansi-font-family: "Comic Sans MS"; mso-symbol-font-family: Wingdings;"><span style="mso-char-type: symbol; mso-symbol-font-family: Wingdings;">ò</span></span><span style="color: red; font-family: "Comic Sans MS"; font-size: 11.0pt; mso-bidi-font-family: Arial;"><o:p></o:p></span></div><div class="MsoNormal" style="margin-left: 18.0pt; text-align: justify; text-justify: inter-ideograph;"><span style="color: red; font-family: "Comic Sans MS"; font-size: 11.0pt; mso-bidi-font-family: Arial;">Agarose gel electrophoresis <o:p></o:p></span></div><div class="MsoNormal" style="margin-left: 18.0pt; text-align: justify; text-justify: inter-ideograph;"><span style="color: blue; font-family: "Comic Sans MS"; font-size: 11.0pt; mso-bidi-font-family: Arial;">What’s the basis of using AGE in relation to restriction digestion?<span style="mso-spacerun: yes;"> </span><o:p></o:p></span></div><div class="MsoNormal" style="margin-left: 15.0pt; text-align: justify; text-indent: 21.0pt; text-justify: inter-ideograph;"><span style="color: red; font-family: Wingdings; font-size: 11.0pt; mso-ascii-font-family: "Comic Sans MS"; mso-bidi-font-family: Arial; mso-char-type: symbol; mso-hansi-font-family: "Comic Sans MS"; mso-symbol-font-family: Wingdings;"><span style="mso-char-type: symbol; mso-symbol-font-family: Wingdings;">ò</span></span><span style="color: red; font-family: "Comic Sans MS"; font-size: 11.0pt; mso-bidi-font-family: Arial;"><o:p></o:p></span></div><div class="MsoNormal" style="margin-left: 18.0pt; text-align: justify; text-justify: inter-ideograph;"><span style="color: red; font-family: "Comic Sans MS"; font-size: 11.0pt; mso-bidi-font-family: Arial;">Southern blot, probing & autoradiography<o:p></o:p></span></div><div class="MsoNormal" style="margin-left: 18.0pt; text-align: justify; text-justify: inter-ideograph;"><span style="color: blue; font-family: "Comic Sans MS"; font-size: 11.0pt; mso-bidi-font-family: Arial;">What’<span class="MsoPageNumber"><span style="font-family: "Comic Sans MS"; mso-bidi-font-family: Arial;">s the basis behind these steps?</span></span></span><span style="color: purple; font-family: "Arial","sans-serif"; font-size: 11.0pt;"> <o:p></o:p></span></div>United JC Unionhttp://www.blogger.com/profile/06131630985986252085noreply@blogger.com0tag:blogger.com,1999:blog-4274038432688951963.post-47002997157844488322011-02-02T18:46:00.001-08:002011-02-02T18:46:59.196-08:00FAQ on PhotosynthesisWhat is a light harvesting complex (LHC)?<br />
A1. LHC is a protein complex containing photosynthetic pigments (chlorophyll and<br />
carotenoids) which surrounds a reaction centre (see Fig 24, pg 23 of lecture notes).<br />
The photosynthetic pigments are important in light-harvesting complexes as they<br />
capture the light energy and transfer it via resonance energy transfer to the<br />
reaction centre of the photosystem. More details will be covered in the lecture<br />
scheduled on 22 Jan 2010 (Fri). <br />
<br />
Q2. What is chlorosis?<br />
A2. Chlorosis is a condition in leaves whereby the leaves produce no or little<br />
chlorophyll. In the context of the lecture, chlorosis results when there is a deficiency<br />
of magnesium, which is integral in the structure of chlorophyll (Fig 10, pg 14 of<br />
lecture notes). In fact, chlorosis is derived from the word Chloris or from the Greek<br />
Khloros meaning "greenish-yellow," "pale green," "pale," "pallid" or "fresh". Examples<br />
of how leaves look like when they are suffering from chlorosis are shown as follows:United JC Unionhttp://www.blogger.com/profile/06131630985986252085noreply@blogger.com0tag:blogger.com,1999:blog-4274038432688951963.post-34800812436245046252011-02-02T18:44:00.001-08:002011-02-02T18:44:40.216-08:00Photosynthesis SPA Alga BeadsUnited JC Unionhttp://www.blogger.com/profile/06131630985986252085noreply@blogger.com0tag:blogger.com,1999:blog-4274038432688951963.post-57067293524709765092011-02-02T18:43:00.001-08:002011-02-02T18:43:19.626-08:00Photosynthesis SPATopic: Photosynthesis<br />
Practical 3E: Chromatography<br />
<br />
Activity: To conduct a chromatographic investigation of the pigments present in a leaf extract. <br />
<br />
Skill A – Planning<br />
<br />
You are required to plan, but not carry out, a chromatographic investigation of the pigments present in a leaf extract.<br />
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A leaf extract can be made by mixing finely ground leaf with extraction solution. <br />
<br />
A thin line of leaf extract is painted across the silica gel strip. The strip is a sheet of plastic coated with a thin layer of absorbent material, silica gel. The silica gel strip should be placed in the chromatography solvent to conduct the experiment.<br />
<br />
Your planning must be based on the assumption that you have been provided with the following equipment and materials which you must use:<br />
• fresh spinach leaves<br />
• sharp knife<br />
• mortar and pestle<br />
• extraction solution <br />
• chromatography solvent <br />
• pure photosynthetic pigments e.g. chlorophyll, carotenoid and xanthophylls<br />
• silica gel strip<br />
• boiling tube with a rubber bung<br />
• ruler<br />
• brush<br />
• a variety of different sized beakers, measuring cylinders, syringes and pipettes for measuring volumes<br />
<br />
Your plan should: have a clear and helpful structure to include<br />
• an explanation of theory to support your practical procedure<br />
• a description of the method used including the scientific reasoning behind the method<br />
• an explanation of the dependent and independent variables involved<br />
• relevant, clearly labeled diagrams<br />
• how you will record your results and ensure they are as accurate and reliable as possible<br />
• proposed layout of results tables and graphs with clear headings and labels<br />
• the correct use of technical and scientific terms<br />
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1. An explanation of theory to support your practical procedure <br />
a. Reaction: a variety of pigments are involved in photosynthesis, i.e. chlorophyll a, chlorophyll b, carotenoids.<br />
<br />
Chromatography is a means of separating one type of molecule from another. A strip of absorptive silica gel strip carrying a concentrated line of the leaf extract is dipped into a suitable solvent. The solvent moves up the silica gel strip sweeping the pigments with it. The pigments have different solubilities in the solvent and will therefore travel at different speeds and become separated from each other. <br />
<br />
b. Relate to method of measurement: The identification of the molecule is made on the basis of the distance travelled by the substance in relation to the distance moved by the solvent. Each molecule can then be refferred to by its Rf value expressed as:<br />
Rf = Distance travelled by a compound<br />
Distance travelled by solvent front<br />
<br />
2. A description of the method used including the scientific reasoning behind the method<br />
a. Preparation of material: <br />
• chloroplast extraction, use of knife/pestle and mortar/sharp sands <br />
• use of extraction solution<br />
• filter to remove the debris<br />
b. Experiment set up: method of spotting leaf extract on silica gel strip, e.g. paint a thin line across the strip and allow this to dry; place the strip in a beaker filled with 10 cm3 of solvent; the spot should be above the solvent; cover the beaker; <br />
c. Other variables to be controlled for replicates: mass of leaves used; <br />
<br />
d. Standard: carry chromatography using pure photosynthetic pigments, one strip for each pigment; <br />
<br />
e. Taking measurement: Measure the distance traveled by the pigments and solvent; calculate Rf value (formula); compare Rf with standard to identify the pigments; <br />
<br />
3. an explanation of the dependent and independent variables involved relevant, clearly labeled diagrams<br />
N.A.<br />
<br />
4. Relevant, clearly labeled diagrams <br />
<br />
5. How you will record your results and ensure they are as accurate and reliable as possible<br />
a. At least 3 replicates <br />
<br />
6. Proposed layout of results tables and graphs with clear headings and labels <br />
Rf value of pure photosynthetic pigments<br />
Pigment Rf<br />
<br />
A <br />
B <br />
C <br />
D <br />
E <br />
<br />
<br />
<br />
<br />
<br />
<br />
<br />
<br />
Table showing the Rf value of the unknown pigments<br />
<br />
Pigment Distance traveled, d/cm Rf Identity of pigment<br />
D1 D 2 D 3 <br />
<br />
1 <br />
2 <br />
3 <br />
4 <br />
5 <br />
7. The correct use of technical and scientific termsUnited JC Unionhttp://www.blogger.com/profile/06131630985986252085noreply@blogger.com0tag:blogger.com,1999:blog-4274038432688951963.post-62548281845545993632011-02-02T18:42:00.001-08:002011-02-02T18:42:06.187-08:002010 Prelim 3 Essay Compilation2010 CJC<br />
4 (a) Describe how gel electrophoresis can be used to analysis nucleic acids. [8]<br />
1. A loading dye is added to the DNA samples, which allows the process of electrophoresis to be tracked / helps DNA to sink into the wells. <br />
2. DNA fragments are placed in wells at one end of an agarose gel with a micropipette. <br />
3. Gel is submerged in a buffer solution that will conduct electricity. <br />
4. An electric current is applied though electrodes at opposite ends of the gel. <br />
5. DNA is negatively charged and moves towards the anode/positive electrode.<br />
6. Rate of movement / speed / distance traveled in a given time depends on size of DNA fragment.<br />
7. Gel electrophoresis separates DNA fragments by size / molecular weights, with the smallest fragments moving the fastest and furthest as the smaller fragments have lower resistance in moving through the pores of the gel, whereas the larger fragments move slower due to higher resistance.<br />
8. A DNA ladder using known lengths of DNA can be used to calibrate the gel. <br />
9. Ethidium bromide or fluorescent dyes can be used to bind to the invisible DNA fragments and UV light is used to visualize the DNA bands.<br />
10. Complementary radioactive or fluorescent probes can be used to identify specific bands or sequences. <br />
11. This enables identification of different alleles or mutations of genes and allows sequencing of DNA. <br />
<br />
<br />
(b) With reference to a named example, explain how RFLP can be used in disease detection. [7]<br />
1. Human DNA is cut with a restriction enzyme (DdeI) was subjected to agarose gel electrophoresis followed by Southern blotting to transfer the DNA to a nylon membrane.<br />
2. This is followed by hybridisation with a radioactive probe specific for the β-globin gene. <br />
3. The membrane is laid under photographic film, allowing any radioactive areas to expose the film (autoradiography). <br />
4. Black bands on the film correspond to the locations on the membrane where DNA has hybridized to the probe. <br />
5. Differences in DNA fragment sizes (polymorphisms) are detected by autoradiography depending on where the probe binds to, i.e. it binds to complementary sequences in the β- globin gene. <br />
6. A healthy person with two copies of the normal β-globin allele (βAβA) shows two bands at 175 and 201 bp. <br />
7. A person with sickle cell anaemia (βSβS) would have two copies of the mutant β-globin allele and show one band at 376 bp. <br />
8. A person considered a “carrier” would have one normal and one defective β-globin allele (AS) but would not have sickle cell anaemia because there is one functioning copy of the β-globin allele. For the carrier, bands show at 376, 201 and 175 bp.<br />
<br />
<br />
(c) Discuss the ethical concerns that have arisen about the human genome project. [5] <br />
<br />
Statements made must be accompanied with some discussion. <br />
1) Privacy and confidentiality of genetic information<br />
• difficult to determine who owns and controls genetic information;<br />
<br />
2) Fairness in the use of genetic information <br />
• by insurance companies, employers, courts, schools, adoption agencies, and the military;<br />
• difficult to determine who should have access to personal genetic information and how it will be used;<br />
3) Commercialization of products <br />
• including property rights (patents, copyrights, and trade secrets) and accessibility of data and materials;<br />
• difficult to determine who owns genes and other pieces of DNA;<br />
• difficult to ascertain if patenting DNA sequences will limit their accessibility and development into useful products;<br />
<br />
4) Fairness in access to advanced genomic technologies. <br />
• difficult to ascertain who will benefit or if it will result in major worldwide inequities;<br />
<br />
5) Clinical issues<br />
• including the education of doctors and other health-service providers, people identified with genetic conditions, and the general public about capabilities, limitations, and social risks; and implementation of standards and quality control measures;<br />
• problems associated with the evaluation and regulation for accuracy, reliability and utility of genetic tests;<br />
• the need to prepare healthcare professionals for the new genetics as well as the public to make informed choices;<br />
<br />
6) Uncertainties associated with gene tests for susceptibilities and complex conditions <br />
• e.g., heart disease, diabetes, and Alzheimer’s disease – questions related to whether:<br />
tests should be performed when no treatment is available;<br />
parents have the rights to have their minor children tested for adult-onset diseases;<br />
genetic tests are reliable and interpretable by the medical community;<br />
<br />
7) Reproductive issues <br />
• including adequate and informed consent and use of genetic information in reproductive decision making; <br />
• possibility of improper counselling of parents by healthcare personnel about the risks and limitations of genetic technology;<br />
• difficult to determine the reliability and usefulness of fetal genetic tests;<br />
<br />
8) Psychological impact, stigmatization, and discrimination <br />
• due to an individual’s genetic differences;<br />
• possibility that personal genetic information as perceived by society, may affect an individual in various ways;<br />
<br />
9) Conceptual and philosophical implications<br />
• regarding human responsibility, free will vs genetic determinism, and concepts of health and disease;<br />
• some of the related perceptions and uncertainties include:<br />
whether behaviour is determined by genes;<br />
what is the acceptable level of diversity;<br />
whether there is a clear distinction between medical treatment and enhancement;<br />
[Total: 20] <br />
<br />
2010 YJC<br />
5 (a) Discuss the uses of plant tissue culture. [4]<br />
The term plant tissue culture refers to the aseptic growing of excised plant parts in vitro.<br />
<br />
1. Rapid production; of large numbers; of transgenic plants (transgenic/normal) from just one or a few stock plants can be achieved; through asexual means; <br />
<br />
2. Plant diseases can be avoided. Production of virus-free plants; can be obtained by selecting only meristematic tissue of the transgenic plant for propagation; <br />
<br />
3. If a new gene is introduced into a plant cell by genetic engineerin;, the modified cell can be grown into a whole plant; and then cloned to produce many new plants/mass production;, all containing that gene (transgenic plants) and are genetically identical;.<br />
<br />
Recall that for transgenic plants to be made: <br />
4. These transgenic plants can be produced at any time of the year;<br />
a. can be put in cold storage, taking up relatively little space;<br />
b. combined with rapid production, this gives great flexibility in supplying consumer demand;<br />
c. plants can be produced out of season and people can buy plants at lower prices than usual; <br />
<br />
5. Land are needed to grow these transgenic plants through tissue culture is much less; save space;. <br />
<br />
6. Reliability; and quality control can be achieved; since growing conditions are standardized; and batch after batch of standard plants are produced. <br />
<br />
<br />
As compared to normal reproduction: _____________________<br />
7. Prevents “loss” of transgene due to sexual reproduction / through gamete formation; maintenance of transgene within entire crop;<br />
<br />
8. Prevents transgenic plants from being completely wiped out by changing environmental factors; as the plants are always available as stored in the cold storage;<br />
<br />
The transgenic plantlets produced are light and small in size; so they can be air-freighted and transported easily and cheap;, thus increasinginternational trade<br />
<br />
(b) Discuss briefly, with specific examples, desirable genes that have been genetically engineered into crop plants and animals used as food. [7]<br />
<br />
Genetic engineering serves to bring in desirable characteristics from other species (unrelated) to add characteristics to enhance the quality of the organism. It can be to <br />
<br />
Choose any three <br />
development of resistance to viral, bacterial, and fungal diseases; E.g. Development of plants resistant to pest / BTcorn resistant to corn borers (or Bt tomato). Plants will express the Bt toxin gene from the bacterium B. thuringiensis <br />
<br />
Transgenic plants with important traits such as herbicide resistance can be introduced. 1. E.g., glyphosate-resistant soybean (Round-Up Ready ™) <br />
<br />
The plants have been engineered to readily degrade the herbicide. Spraying of herbicides will kill weeds but not affect crop thus reducing competition for nutrients/light. Herbicide resistant plant can spray herbicides to kill weeds without affecting crops <br />
<br />
<br />
Enhancement of nutritional values. E.g., Golden rice with genes for enzymes converting a natural compound in rice to beta-carotene (e.g., daffodil) rich in beta-carotene. Golden rice has been genetically engineered to produce beta-carotene a pre-cursor of vitamin A is a such a product. Vitamin A deficiency leads to blindness and susceptibility to disease <br />
<br />
<br />
(c) Discuss the social and ethical concerns pertaining to the use of genetically modified animals. [4]<br />
<br />
1. Environmental protection from GMOs; safeguards of products with GM components may not be adequate; to prevent transfer of genes/pollen from GM organisms to wild-type organisms; <br />
<br />
2. Unknown effects of GM products on human health and disease; effects may only be known after long period of exposure; allergic reactions may occur if people unknowingly consume products containing introduced genes; <br />
<br />
3. Tampering with nature / “playing God”; by mixing genes among species; genetic modifications of animals especially resulting in enhancement; draws parallels that human genetic modification for enhancement one day may occur; <br />
<br />
4. Improper human use of animals - violations of animal rights; cruelty to animals; results in animal deformity / disease; (OWTTE) GM animals may suffer unnecessarily.E.g., oncomouse which may increased susceptibility to cancer. <br />
<br />
5. Introduction of animal genes to plants; / porcine genes to other organisms; may result in objections from vegetarian; / specific religious groups; respectively especially when GM foods are unlabelled<br />
<br />
6. Accountability of biotechnological and agricultural firms wrt to GMOs; labeling issues; Ownerships issues – unfair for large multinational companies to patent GM animals.<br />
<br />
7. Release (accidental or otherwise) of GM animals into the wild may result in GM animals outcompeting wild types such that ecological balance is disrupted.E.g., larger transgenic salmon may be preferably selected as mates over smaller wild types. <br />
<br />
8. Introduction of foreign gene(s) may result in production of secondary metabolites that may be toxic to animals themselves and/or livestock/humans that consume them.New proteins in GM animals may be potentially allergenic to humans that consume them<br />
<br />
All other logical points accepted if appropriate elaboration given <br />
2010 DHS<br />
4 (a) Discuss the advantages and evolutionary consequences of using plant tissue culture to propagate transgenic plants. [10]<br />
(b) Explain, with reference to Bt corn, the significance of genetic engineering in improving the world food situation. [6]<br />
(c) Discuss the ethical issues involved with the production of genetically modified organisms. [4]<br />
<br />
Advantages <br />
1. Tthe modified cell can be grown into a whole plant and then cloned to produce many new plants/mass production, all containing that gene (transgenic plants) and are genetically identical. <br />
2. Rapid production of large numbers of transgenic plants from just one or a few stock plants can be achieved through asexual means. <br />
3. Plant diseases can be avoided / production of virus-free plants. <br />
4. These transgenic plants can be produced at any time of the year and put in cold storage, taking up relatively little space. Combined with rapid production, this gives great flexibility in supplying consumer demand as plants can be produced out of season and people can buy plants at lower prices than usual. <br />
5. Reliability and quality control can be achieved since growing conditions are standardised and batch after batch of standard plants are produced. <br />
6. Land area needed to grow these transgenic plants through tissue culture is much less /saves space. <br />
7. Prevents “loss” of transgene due to sexual reproduction / through gamete formation /maintenance of transgene within entire crop. <br />
8. Prevents transgenic plants from being completely wiped out by changing environmental factors as the plants are always available as stored in the cold storage. <br />
9. The transgenic plantlets produced are light and small in size so they can be air-freighted and transported easily and cheaply. <br />
10. Produces rooted plantlets that are ready for growth. <br />
11. Plantlets of a particular desired sex can be grown selectively. <br />
Max 6 <br />
<br />
Evolutionary consequences <br />
12. Desirable / beneficial genes are quickly passed on to subsequent generations / ensures integrity and maintenance of genetic composition as the transgenic plants can propagate rapidly under favourable environment. <br />
13. Lack of genetic variation posed a problem to the survival of a species / unable to respond to unfavourable changes in selection pressure as a result of changing environment. Low evolutionary potential / No variation unless mutation occurs, thus no speciation / natural selection cannot occur without variation. <br />
14. Thus there is a high risk of extinction of that species as all individuals are genetically identical / equally susceptible to pathogens / succumb to outbreak of epidemics / disease. <br />
15. Decrease in gene pool due to loss of beneficial alleles. <br />
16. Such transgenic plants produced, when reintroduced into the wild, might cross-pollinate and these might cause production of pesticide/herbicide resistance / generating “superweeds” that will disrupt the ecological balance / ecosystem.<br />
Max 5 <br />
Overall max 10 <br />
<br />
(b)<br />
1. Genetic engineering allows these organisms to acquire by genes from the same or different species that improves crop yield or nutritional value. <br />
2. Bt delta endotoxin genes from Bacillus thuringiensis is inserted into corn crops via use of Ti plasmid of Agrobacterium tumefaciens. <br />
3. Protein produced kills the larvae of the Lepidoptera species, including the European corn borer due to damage caused to the gut wall; hence Bt endotoxin specifically acts on species that ingest corn. <br />
4. Hence they are unable to cause further damage to crops, this prevents yield loss due to corn borers. <br />
5. Use of Bt corn reduces need for pesticide applications, hence reducing cost, labour, environmental damage and non-specific harming of other insect species. <br />
6. This allows farmers to concentrate resources on aspects like fertilizers, improved farming practices etc. that result in further improved crop yield. <br />
7. Implementation in developing countries, however, may not alleviate problems of world food situation due to unequal food distribution / prohibitive cost of GM seeds / public distrust of GMOs. <br />
Max 6<br />
<br />
(c)<br />
1. Environmental protection / safeguards of products with GM components may not be adequate to prevent transfer of genes/pollen from GM organisms to wild-type organisms. <br />
2. Unknown effects of GM products on human health and disease – effects may only be known after long period of exposure / allergic reactions may occur if people unknowingly consume products containing introduced genes; <br />
3. Tampering with nature by mixing genes among species / genetic modifications of animals especially resulting in enhancement draws parallels that human genetic modification for enhancement one day may occur. <br />
4. Improper human use of animals - violations of animal rights / cruelty to animals / results in animal deformity / disease. <br />
5. Introduction of animal genes to plants / porcine genes to other organisms may result in objections from vegetarian / specific religious groups respectively. <br />
6. Accountability of biotechnological and agricultural firms with relation to GMOs. <br />
7. Labeling issues. <br />
Max 4<br />
<br />
2010 IJC <br />
5 (a) Explain how gel electrophoresis is used to analyse DNA. [6]<br />
1. agarose gel acting as molecular sieve to separate fragments based on molecular size;;<br />
2. distance travelled by band is inversely proportional to molecular size;;<br />
3. presence of TAE/TBE buffer provide the ions to support conductivity;;<br />
4. application of direct current enables the neg charged DNA to travel to the positive electrode;;<br />
5. gel stained with methylene blue to reveal positions of DNA bands;;<br />
6. relative thickness of band is an indication of the amount of DNA with that molecular size;;<br />
(b) Outline the process of nucleic acid hybridisation and explain how it can be used to detect and analyse restriction fragment length polymorphism (RFLP). [5]<br />
1. Lay nitrocellulose paper on the electrophoresed gel submerged in NaOH with a stack of paper towels;;<br />
2. dsDNA denatured into ssDNA, adhere to nitrocellulose paper;;<br />
3. placed into bag with radioactive ssDNA probe with complementary sequences to target sequences;;<br />
4. Excess probe washed off and placed on X-ray film, hybridized areas will show up as dark bands;;<br />
5. banding patterns is an indication of number of fragments and fragment length that can be compared to other DNA samples;;<br />
<br />
(c) Explain how RFLP analysis facilitated the process of linkage mapping, diseases detection and DNA fingerprinting. [9]<br />
1. RFLP is based on variation in no. & length of restriction fragments btw individuals visible through gel electrophoresis / nucleic acid hybridization / Southern blotting;;<br />
Linkage Map<br />
2. shows the relationship between 2 or more genes located on the same chromosome<br />
3. tightly linked RFLP with genes are used as DNA markers for respective genes where linkage are calculated based on the recombinant phenotype freq;;<br />
Disease detection<br />
4. disease causing allele differs from normal allele due to mutation, may result in the lost/creation of a restriction site in the mutated allele;;<br />
5. RFLP marker is closely linked to or within disease causing allele, resultant fragments from RE digestion can indicate presence / inheritance of specific allele;;<br />
DNA fingerprinting<br />
6. DNA profiling process identifying individuals based on diff in repetitive DNA sequence e.g. VNTR<br />
7. Different number of repeats within a restriction site results variable length of fragments formed after RE digestion;;<br />
8. Analysis is based on matching DNA fingerprint between samples;; <br />
9. perfect match in parentage identification/ criminal analysis or degree of similarity in evolutionary studies;;<br />
<br />
[Total: 20]<br />
<br />
2010 JJC<br />
(a) With reference to the analysis of DNA molecules, describe the process and explain the use of Gel Electrophoresis and Nucleic Acid Hybridisation. [10]<br />
<br />
Gel Electrophoresis<br />
1. DNA fragments are loaded in a well;<br />
2. at one end of the agarose gel;<br />
3. Electric field or voltage applied across the gel;<br />
4. DNA being negatively charged; <br />
5. will move towards the positive electrode/anode;<br />
6. DNA ladder is added to calibrate the size of DNA;<br />
7. Distance traveled in a given time depends on the molecular weight of the DNA with larger DNA fragments travel slower and smaller fragments traveling faster; <br />
8. More resistance for the larger fragment to move through the pores of the gel;<br />
9. Ethidium Bromide and UV light/ fluorescent dyes are use to see the DNA bands;<br />
10. This allows for separation of DNA fragments and identification of different alleles/ two different DNA molecules;;<br />
[max 4m for description and 1 m for analysis of DNA]<br />
<br />
Nuclei Acid Hybridisation<br />
11. DNA fragments in the gel is denatured by alkaline solution; <br />
12. resulting in single stranded DNA;<br />
13. then transferred to the nitrocellulose membrane via capillary action; <br />
14. a single stranded probe;<br />
15. which consists of a radioisotope and short nucleotide sequence/oligonucleotide radioactively labeled probe;<br />
16. which is complementary to the DNA sequence of the target DNA fragment is added; <br />
17. the probe hybridise to the target fragment; <br />
18. excess probe is wash off; <br />
19. the membrane is then exposed to a X-ray film; <br />
20. the region where probe bind to will give a dark band;<br />
21. This allows for identification of different alleles/ two different DNA molecules/ detect and analyse Restriction Fragment Length Polymorphisms;;<br />
[Any 4 points -4m for description and 1 m for analysis of DNA]<br />
<br />
(b) Discuss the uses of Restriction Fragment Length Polymorphisms (RFLP) analysis.[10]<br />
<br />
1. RFLP – refers to the genetic variation that is observed in the size of the DNA fragments that is obtained after the DNA (of different individuals) is digested by a specific restriction enzyme;<br />
2. Different sized fragments can be separated by gel electrophoresis;<br />
3. Followed by nucleic acid hybridization to make the bands visible;<br />
Max 1<br />
<br />
Disease detection;<br />
4. Indirect screening for disease-causing alleles;<br />
5. Due to loss or gain of restriction sites/ changes in DNA sequences that affects a restriction site;<br />
6. Generating different size restriction fragments;<br />
7. Use of RFLP as proxy for genetic diseases because the marker is closely linked to disease-causing allele;<br />
8. Inherited together as crossing over is unlikely;<br />
9. inheritance of the a particular RFLP can indicate the inheritance of the disease causing allele; <br />
Max 3<br />
<br />
Genomic mapping;<br />
10. Use of RFLP as genetic markers;<br />
11. The frequency with which two RFLP markers or an RFLP marker and a certain allele for a gene is inherited together;<br />
12. is a measure of the closeness of the 2 gene loci on a chromosome /relative distance between the 2 gene loci;<br />
13. Based on recombination frequencies/ frequency of crossing over between 2 RFLP markers/ cross-over values; <br />
14. Can be used to construct linkage map; <br />
Max 3<br />
<br />
DNA fingerprinting/ genetic fingerprinting/ DNA profiling; <br />
15. Ref to VNTR (variable number of tandem repeats or STR (short tandem repeats);<br />
16. Different VNTR have different lengths;<br />
17. VNTR highly polymorphic;<br />
18. Each individual’s combination is unique, having inherited one VNTR locus from each parent; <br />
19. Use to identify criminals/ determine paternity/ identify endangered animals etc;<br />
Reject answers that are not clear eg. genetic fingerprinting use for criminal cases <br />
Max 3<br />
[Total: 20]<br />
<br />
2010 PJC<br />
<br />
<br />
(a) Question 4<br />
<br />
Explain how RFLP analysis is used to support the process of linkage mapping.<br />
<br />
1. A linkage map shows the relative location;<br />
2. or the order of genes along a chromosome;<br />
3. constructed on the assumption that the probability of a crossover between two genetic loci is proportional to the distance separating the loci;<br />
4. Linkage maps are usually constructed with several thousand known genetic markers spaced evenly throughout the genome. <br />
5. The RFLP markers can be any genes or any other identifiable DNA sequences, such as variation number tandem repeats (VNTR). <br />
6. Linkage map must be done based on experimental crosses; <br />
7. RFLP from parental and offspring organisms are analysed;<br />
8. using Southern blot;<br />
9. RFLP pattern obtained are used to calculate the total percentage of recombinants;<br />
10. Give an indication of the distance between the RFLP sequences based on the recombination frequencies obtained;<br />
11. the farther apart the two RFLP sequences are, the higher the probability that a crossover will occur between them and therefore<br />
12. the higher the recombination frequency;<br />
13. For example, if 70% of the progeny produced are parental and 30% were recombinant, the RFLP loci are 30 centi-Morgans (cM) apart from each other (1 mark for example)<br />
<br />
(b) Describe how the micropropagation technique is used in cloning of plants.<br />
<br />
1. In micropropagation, plant tissues or explants eg: meristematic cells are removed from plants ® explants only;<br />
2. To produce thousands of plantlets which are clones of the original plant;<br />
3. Surface of explants sterilised with dilute sodium hypochlorite (Clorox) to kill bacterial and fungal pathogens or organisms;<br />
4. And grown in a containing sterile media containing nutrients and hormones needed for plant growth; <br />
5. Eg: inorganic ions such as nitrogen,magnesium, iron, potassium), vitamins, carbohydrate source and <br />
6. plant growth substances such as auxin and cytokinins<br />
7. Containers holding the explants are then sealed and incubated for 1-9 weeks.<br />
8. Conditions like temperature, light intensity and humidity are carefully controlled;<br />
9. During this period, the cultured cells divide by mitosis<br />
10. to form a mass of undifferentiated tissue called a callus.<br />
11. As callus increases in size, pieces of callus is sliced off and grown on new medium composition (subculturing); <br />
12. By adjusting concentration of plant hormones in growth medium, <br />
13. cells in callus can be induced to differentiate into roots and shoots; <br />
14. Further growth being encouraged by the use of plant growth substances until plantlets are large enough<br />
15. To be weaned and planted in sterile soil in green house<br />
16. After acclimatization in green house, the plant are transferred to soil for field planting<br />
<br />
(c) Explain how gel electrophoresis can be used to provide evidence of molecular homology.<br />
<br />
Must make the link to explain how you can compare samples from different organisms in order to establish molecular homology.<br />
<br />
1. Define molecular homology: the similarities in DNA nucleotides and/or amino acid sequence of organisms.<br />
2. the more closely related organisms are, the more similarities they share;<br />
3. Gel electrophoresis is used as a tool to analyse the similarities in DNA sequences between organisms;<br />
4. Isolate genomic DNA from different organisms; <br />
5. Obtain cytochrome c or haemoglobin DNA as starting material using PCR using specific primers; <br />
6. Cut amplified sample DNA using the same restriction enzyme<br />
7. A loading dye is added to the DNA samples, which allows the process of electrophoresis to be tracked/ helps DNA to sink into the wells;<br />
8. DNA standard / ladder is loaded to enable estimation of the sizes of the DNA bands obtained<br />
9. DNA samples loaded into wells at negative end of an agarose gel;<br />
10. Gel is submerged in a buffer solution that will conduct electricity;<br />
11. An electric current is applied through electrodes at opposite ends of the gel;<br />
12. Negatively-charged DNA fragments migrate to the positively charged electrode/anode;<br />
13. Gel electrophoresis separates DNA fragments by size/ molecular weight;<br />
14. with the smallest fragments moving the fastest and furthest;<br />
15. as the smaller fragments have lower resistance in moving through the pores of the gel;<br />
16. whereas the larger fragments move slower due to higher resistance;<br />
17. invisible DNA bands can be viewed under UV light by staining the gel with a dye such as ethidium bromide;<br />
18. Two samples of DNA with similar maps for the locations of restriction sites will produce similar banding patterns;<br />
19. In contrast, two genomes that have diverged extensively since their last common ancestor will have a very different distribution of restriction sites, and the DNA will not match closely in the sites of restriction fragments. <br />
<br />
<br />
<br />
<br />
2010 SAJC<br />
(a) Discuss with specific examples, how desirable genes have been genetically engineered into crop plants and animals that are used as food. [7]<br />
<br />
1 ref. gene transfer techniques for plants, eg. electroporation, liposomes on protoplast, Ti plasmid (not in syllabus), gene gun (not in syllabus)<br />
2 ref. gene transfer techniques for animals, eg. microinjection, electroporation, liposomes<br />
3 ref. plant tissue culture techniques<br />
4 ref. insertion of gene construct into animal embryos<br />
<br />
Animals with faster growth rate<br />
5 Atlantic salmon with more active salmon growth hormone gene <br />
6 grow to its full length in a shorter period of time<br />
OR<br />
7 Cattle with bigger muscles<br />
8 more meat per animal<br />
<br />
Pest-resistant plants<br />
9 Eg. Corn / potato / broccoli / tomato plants which can produce the Bt toxin;<br />
10 express the Bt toxin gene from the bacteria Bacillus thuringiensis;<br />
11 specifically kills insect pests<br />
12 crop losses can be reduced, leading increased profits in agriculture<br />
<br />
Herbicide resistant plants<br />
13 E.g. Glyphosate-resistant soybean / tomato<br />
14 glyphosate works by inhibiting an enzyme EPSP synthetase, which plants require to make essential aromatic amino acids<br />
15 can spray herbicides to kill weeds without affecting crops<br />
16 crop losses can be reduced, leading increased profits in agriculture<br />
<br />
Plants with improved nutritional qualities <br />
17 E.g. Golden rice enriched with beta-carotene<br />
18 produced by transplanting genes from daffodil and bacteria<br />
19 help prevent Vitamin A deficiency which leads to blindness and susceptibility to disease<br />
<br />
Plants with delayed ripening<br />
20 Eg. Flavr-Savr tomato<br />
21 polygalacturonase normally responsible for the ripening process<br />
22 has antisense gene of enzyme polygalacturonase binds to the polygalacturonase mRNA, preventing it from being expressed<br />
23 delay in fruit ripening / spoiling during transport / improved shelf life<br />
24 larger and has greater flavour<br />
<br />
OR<br />
25 blocking the biosynthetic pathway for ethene<br />
26 ethene is hormone responsible for fruit ripening<br />
27 fruits can be allowed to grow on the vine for a longer period of time, <br />
28 delay in fruit ripening / spoiling during transport / improved shelf life<br />
29 larger and has greater flavour<br />
<br />
<br />
(b) Discuss the social and ethical concerns pertaining to the use of genetically modified animals. [8]<br />
1 release (accidental or otherwise) of GM animals into the wild;<br />
2 may result in GM animals out competing wild types;<br />
3 ecological balance is disrupted / severe impacts on the food-chain;<br />
4 destabilises & hence threatens biodiversity;<br />
5 E.g. larger transgenic salmon may be preferably selected as mates over smaller wild types;<br />
<br />
6 introduction of foreign gene(s);<br />
7 may result in production of secondary metabolites;<br />
8 maybe toxic to animals themselves and/or livestock/humans that consume them;<br />
9 new proteins in GM animals;<br />
10 may be potentially allergenic to humans that consume them;<br />
<br />
11 animal rights – GM animals may suffer unnecessarily;<br />
12 E.g. oncomouse;<br />
13 which may increased susceptibility to cancer;<br />
14 rearing animals for transplant purposes;<br />
<br />
15 religious implications in food choice;<br />
16 especially when GM foods are unlabelled;<br />
17 E.g. incorporation of pig genes into cows;<br />
18 ownership issues – unfair for large multinational companies to patent GM animals / increased dependence of undeveloped countries on rich developed countries<br />
<br />
2010 TPJC<br />
(a) Explain the use of Restriction Fragment Length Polymorphism in disease detection and genomic mapping. [8] <br />
Disease detection<br />
1 RFLP is based on variation in the length of DNA sequence (variation in nucleotide sequences or VNTR) in an allele or non-coding region; <br />
2 The different patterns of restriction fragments/genetic fingerprint of individuals is detected by the use of restriction enzymes, gel electrophoresis / Southern Blotting with some proper description ;<br />
3 Can be used for comparison for similarity between affected and unknown individuals;<br />
Indirect screening: <br />
4 RFLP marker that is tightly-linked (located close to disease causing allele) can serve as a indicator for the disease-causing allele;<br />
5 Crossing over between the marker and allele is unlikely to occur during meiosis, hence the marker and allele would always be inherited together;<br />
Direct screening for disease allele:<br />
6 A disease-causing allele differs from the normal allele due to mutation which alters the ability of the restriction enzyme to recognise and cut that site;<br />
7 Mutation within in a restriction site within an allele may cause the restriction site to be destroyed;<br />
8 resulting in the appearance of a larger fragment and the loss of the two smaller fragments;<br />
OR<br />
9 Mutation occurring in the gene may create a restriction site;<br />
10 resulting in the loss of a larger fragment and the appearance of two smaller fragments;<br />
[max 5]<br />
Genomic mapping<br />
11 RFLP markers are used to construct linkage / genetic maps of a several thousands markers evenly spaced throughout the chromosomes<br />
12 RFLP markers serve as a genetic markers to order the sequence of genes;<br />
13 RFLP markers can be used to determine the relative distances of genes or between genetic markers on chromosomes<br />
14 based on the recombination frequency of two RFLP markers / an RFLP marker and a certain allele of a gene;<br />
15 Recombination frequency is a measure of closeness of the two loci on the chromosome e.g. distance in cM; or give example: the higher the frequency of the two loci inherited together, the closer are the two loci / vice versa;<br />
[max 3]<br />
<br />
<br />
(b) Describe the genetic basis for X-linked Severe Combined Immunodeficiecy disease (SCID) <br />
and explain how gene therapy can be used to treat the disease. [8] <br />
<br />
Genetic basis for X-linked Severe Combined Immunodeficiecy disease<br />
1. Due to mutations in the gene interleukin-2 receptor gamma chain (IL2RG) encoding the common gamma chain, a protein that is shared by the receptors for some interleukins. <br />
2. These interleukins and their receptors are responsible for the differentiation/production of T and B cells.<br />
3. Mutations in gene that result in non functional common gamma chain will result in defects in interleukin signaling deficient immune system.<br />
4. Gene that code for the common gamma chain is situated on the X chromosome.<br />
5. X-linked SCID affects only males;<br />
[ max 4 marks]<br />
<br />
Gene therapy to treat the disease<br />
<br />
6. Gene therapy is a technique for introducing normal / functional copy of the IL2RG allele / gene into cells of patients;<br />
7. restores target cell to normal state / corrects function of cell / correct phenotype by the expression of functional IR2RG protein.<br />
8. Preparation of genetically engineered retrovirus vector to contain a normal copy of the IL2RG gene<br />
9. Retrovirus infects T lymphocytes/blood stem cells (can be ex vivo or in vitro) and transfers IL2RG gene into nucleus of cells. <br />
10. May be integrated into the chromosomal DNA of the lymphocytes by integrase (depend on the type of virus used) .<br />
11. Expression of functional interleukin gamma chain functional interleukin receptor functional T cells and B cells production competent immune system.<br />
[ max 4 marks]<br />
<br />
<br />
(c) One of the treatments for SCID is bone marrow transplant.<br />
Explain how the properties of hematopoietic stem cells allow them to be effectively used to fight a disease such as SCID. [4]<br />
<br />
1. Haematopoietic stem cells from donor’s bone marrow contains a normal/functional copy of the allele ADA / Interleukin 2 receptor gamma chain which can be expressed to produce a functional protein ADA / IR2RG.<br />
2. HSCs are multipotent which has the ability to differentiate into a variety of specialised blood cells which include functional T lymphocytes and B lymphocytes. <br />
3. To replace diseased cells lacking ADA/IR2RG or give rise to competent immune system to fight SCID<br />
4. HSCs are capable of self-renewal and can divide by mitosis;<br />
5. to continue supplying functional cells for long-term treatment;<br />
6. reference to asymmetric division; Any 4 marksUnited JC Unionhttp://www.blogger.com/profile/06131630985986252085noreply@blogger.com1tag:blogger.com,1999:blog-4274038432688951963.post-70944275986930304112011-02-02T18:39:00.001-08:002011-02-02T18:39:50.634-08:002010 Prelim 2 Essay CompilationSAJC 2010<br />
<br />
8(a) Describe the eukaryotic processing of pre-mRNA in terms of intron splicing, polyadenylation and 5’capping. [6]<br />
<br />
Capping<br />
1 modified guanosine nucleotide / 7-methylguanosine / added to 5’ end of mRNA;<br />
2 enzymes (e.g. guanyl / methyl transferase) catalyses reaction between 5’ end of the RNA transcript and GTP;<br />
3 ref. association of cap-binding protein<br />
<br />
Polyadenylation<br />
4 addition of 30-200 adenine sequences to the 3’ end of the pre-mRNA;<br />
5 after polyadenylation signal has been transcribed;<br />
6 catalysed by poly(A) polymerase / polyadenylate polymerase;<br />
7 ref. poly-A binding protein binds to poly(A) tail<br />
<br />
Splicing<br />
8 Splicing is catalyzed by snRNPs / small nuclear ribonucleoproteins<br />
9 snRNPs are made up of proteins & snRNA (small nuclear RNA) <br />
10 snRNPs + other proteins join together to form a complex called spliceosomes <br />
<br />
11 spliceosomes is first responsible for folding the pre-mRNA into the correct orientation for splicing<br />
<br />
12 splicing involves cleavage of the 5’ end of the intron<br />
13 and its attachment to the branchpoint sequence <br />
14 to form a tailed loop structure called a lariat<br />
15 intron is then excised/released<br />
<br />
16 by being cleaved at its 3’ end<br />
17 and the exons are brought together and ligated<br />
18 spliceosomes dissociates once splicing is completed<br />
<br />
8(b) Compare structure and organisation of prokaryotic and eukaryotic chromosome. [6]<br />
<br />
[½ mark per correct comparison]<br />
<br />
Prokayotic genome Eukaryotic genome<br />
1. Genome size Smaller genomes, 0.6 to 10Mb. Large genomes, being less than 10 Mb – 100,000 Mb <br />
2. Gene length Shorter gene sequences<br />
/ more compact genetic organisation Longer gene sequences<br />
/ presence of more intergenic spaces<br />
3. Chromosome number Single chromosome<br />
/ Haploid Many chromosomes<br />
/ Diploid or polyploid<br />
4. Chromosome structure Circular DNA molecule. <br />
Linear DNA, each contained in a different chromosome.<br />
5. Location Chromosome found in the nucleoid region of the cytoplasm Chromosomes are enclosed within a double-membrane bound nucleus<br />
6. Packaging of DNA Does not form chromatin.<br />
(Prokaryotic DNA is organized into a DNA-protein complex called the nucleoid. Eukaryotic DNA is complexed with histones and other proteins to form chromatin.<br />
7. Introns Coding sequence proceeds from start to finish without interruption by introns. Presence of introns within genes. <br />
8. Repetitive sequences Few repetitive DNA sequences. Many highly repetitive DNA sequences <br />
9. Coding and non-coding DNA Most of DNA are coding sequences (codes for protein, tRNA, or rRNA. Most of DNA are non-coding.<br />
<br />
10. Regulatory sequences Small amount of non-coding<br />
DNA consists mainly of regulatory sequences, such as promoters) More complex regulatory sequences (eg. enhancers and silencers)<br />
11. Presence and absence of operons Two or more genes may be expressed and regulated as a unit (genomes arranged in operons / polycistronic genes) Absence of operons / monocistronic genes<br />
<br />
12. Origins of replication One<br />
Many <br />
13. Presence of extrachromosomal DNA Independent small, circular, molecules called plasmids. Circular, double-stranded DNA in mitochondria / choloroplasts.<br />
14. Telomeres Absent Present<br />
15. Centromere Absent Present<br />
<br />
<br />
8(c) Outline the differences between prokaryotic control of gene expression with the eukaryotic model. [8]<br />
<br />
For every comparison:<br />
½ mark for correct comparison<br />
½ mark for correct information<br />
<br />
Chromosomal Level (max 2)<br />
Prokaryotes Eukaryotes<br />
1. DNA and histone modification cannot occur (Prokaryotic DNA does not form chromatin / not associated with histones) DNA and histone modification can occur, resulting in conversion between euchromatin and heterochromatin, and hence the ease of transcription<br />
(Eukaryotic DNA is complexed with histones and other proteins to form chromatin / associated with histones)<br />
2. DNA sequences, including operators and activators, serve as the on/off switch Structure of chromatin – euchromatin, ready to be transcribed, or heterochromatin and not available – is the major on/off switch for gene regulation<br />
3. Prokaryotic DNA not organised into chromatin. Initiation of transcription of eukaryotic genes requires that the compact chromatin fibre, characterised by nucleosome coiling, to be uncoiled and the DNA made accessible to RNA polymerase and other regulatory proteins.<br />
<br />
Transcriptional Level (max 2)<br />
Prokaryotes Eukaryotes<br />
4. One RNA polymerase consisting of five subunits <br />
/ All RNAs synthesized by the same RNA polymerase; <br />
<br />
<br />
<br />
Occurs within the nucleus under the direction of three separate forms of RNA polymerases, each containing 10 or more subunits; different polymerases transcribe different genes <br />
/ Three different classes of RNA each synthesized by a different RNA polymerase<br />
5. Simple regulatory sequence: Transcriptional regulatory protein / Regulator protein binds to DNA-binding sites upstream of the cluster of structural genes to regulate initiation of transcription. Complex regulatory sequence: More extensive interaction between upstream DNA sequences and protein factors involved to stimulate and initiate transcription. In addition to promoters, enhancers and silencers control rate of transcription initiation<br />
6. Related genes are transcribed together as operons <br />
/ only 1 promoter<br />
/ polycistronic mRNA<br />
No operon <br />
/ each gene has own promoter / monocistronic mRNA<br />
<br />
<br />
<br />
Post-transcriptional Level (max 2)<br />
Prokaryotes Eukaryotes<br />
7. Translation is often coupled to transcription (Transcription and translation take place in the same cellular compartment simultaneously) No direct coupling of transcription and translation. (mRNA must pass across nuclear envelope before translation in the cytoplasm. RNA transcript is not free to associate with ribosomes prior to the completion of transcription).<br />
<br />
8. Primary transcripts are the actual mRNAs<br />
/no post-transcriptional modification<br />
Primary transcripts undergo processing to produce mature mRNAs<br />
/ post-transcriptional modification<br />
9. Triphosphate start at the 5’ end<br />
/ No tail at the 3’ end<br />
/ No splicing Methylated guanosine cap at the 5’ end<br />
/ Poly-A tail at the 3’ end<br />
/ Splicing occurs (also alternative splicing)<br />
10. Lower stability of transcript <br />
/ degradation within seconds or minutes <br />
/ mRNAs shorter half life to rapidly respond to environmental changes Higher stability of transcript<br />
/ prevent transcript degradation <br />
/ mRNAs longer half-life remaining much longer to orchestrate protein synthesis prior to their degradation by nucleases in the cell<br />
<br />
Translational level (max 2)<br />
Prokaryotes Eukaryotes<br />
11. Control at this level is unlikely; due to simultaneous transcription and translation Control at translational level:<br />
phosphorylation of ribosomal translation initiation factors<br />
/ negative translational control through regulatory proteins<br />
/ cytoplasmic elongation of poly (A) tails<br />
/ mRNA degradation<br />
/ RNA interference and microRNA<br />
<br />
12. Unique initiator tRNA carries formylmethionine Initiator tRNA carries methionine<br />
13. Smaller ribosomes<br />
/ less complex rRNA and protein components Occurs on ribosomes that are larger<br />
/ rRNA and protein components are more complex than those of prokaryotes<br />
14. mRNAs have multiple ribosome binding sites <br />
/ direct the synthesis of several different polypeptides mRNAs have only one start site<br />
/ direct synthesis of only one kind of polypeptide<br />
15. Small ribosomal subunit immediately binds to the mRNA’s ribosome binding site<br />
Small ribosomal subunits bind first to the methylated cap at the 5’ end of the mature mRNA and then scans the mRNA to find the ribosome binding site, the AUG start codon<br />
<br />
<br />
Post-translational Level (max 2)<br />
Prokaryotes Eukaryotes<br />
16. no/minimal post-translational modifications occur Post-translational modifications determine the functional abilities of the protein<br />
17. Proteolysis: Processing eukaryotic polypeptides to yield functional protein molecules e.g. cleavage of pro-insulin to form the active insulin hormone<br />
18. Chemical modification of proteins to yield functional protein molecules<br />
19. Phosphorylation of proteins to increase or decrease its function<br />
20. Transportation of proteins to target destinations in the cell where it functions is mediated by signal sequences at N-terminus of some proteins. Once transported to destination, signal sequence is enzymatically removed from the proteins<br />
21. Ubiquitination marks protein for degradation, ref to ubiquitin & proteasome<br />
<br />
9(a) Describe the general structure of a named animal virus, and explain why viruses are obligate parasites. [6]<br />
1 example of animal viruses: influzena, HIV, herpes virus etc <br />
2 animal viruses composed of phospholipids/glycoprotein envelope <br />
3 similar in nature to cell surface membrane of animal cell <br />
4 with glycoprotein spikes (for attachment to host cell membrane) <br />
5 enclosed in the viral envelope is the capsid <br />
6 which is composed of capsomeres / protein subunits<br />
7 which contains the viral genome <br />
8 either DNA or RNA, but never both<br />
9 either single or segmented, linear or circular, single-stranded or double-stranded, etc <br />
10 ref. viral enzymes (eg. reverse transcriptase in HIV)<br />
<br />
11 acellular / absence of cytoplasm and cellular organelles<br />
12 lacks ribosomes / protein-synthesising apparatus <br />
13 hijacks host cell’s host’s protein synthesis machinery (transcription and translation machineries) to produce own viral proteins [REJECT ‘metabolic machinery’]<br />
14 ref. replication of viral genome<br />
15 metabolically inert / do not grow or divide on their own <br />
16 can only multiply inside living cells, and not on inanimate media<br />
<br />
9(b) Outline how the influenza virus is able to bypass the human defense mechanism to cause disease. [4]<br />
1 ref. antigenic drift <br />
2 8 single-stranded RNA segments<br />
3 spontaneous genetic mutation /mutation during replication<br />
4 lack of proof-reading during RNA replication<br />
<br />
5 ref. antigenic shift<br />
6 genetic reassortment of the RNA segments between two strains of viruses<br />
7 novel glycoproteins (spikes) produced on viral envelope <br />
/modified neuraminidase and haemaglutinin<br />
8 cannot be recognized by previous antibodies <br />
<br />
<br />
9(c) Compare and contrast the reproductive cycles of the Lambda phage and the Human Immunodeficiency Virus (HIV). [10]<br />
<br />
[1 mark per similarity] (max 4)<br />
1 Life cycle of both viruses involves the stages attachment, penetration, replication, maturation, and release<br />
2 Both attach to their host cell at receptor sites on the host cell’s plasma membrane;<br />
3 Both introduce their viral nucleic acids into their host cell<br />
4 Both are obligate intracellular parasites/make use of their host cell’s resources for synthesis of viral proteins and nucleic acids<br />
5 Both integrate their viral DNA into host genome/viral DNA replicates as part of host’s DNA every time the cell divides <br />
<br />
[1 mark per difference] (max 6)<br />
Feature of comparison Lambda phage HIV<br />
Host cell 6 Infects bacterial cells Infects human T-cells<br />
Penetration / Entry 7 Phage does not undergo fusion with host cell’s plasma membrane / contracts its tail sheath; HIV envelope fuses with the host cell’s plasma membrane;<br />
8 Phage injects only its ds DNA; HIV releases its nucleocapsid (ssRNA and reverse transcriptase) into cytoplasm;<br />
Uncoating 9 No uncoating required as capsid does not enter host cell; Uncoating of nucleocapsid to release genome and enzymes; <br />
Fate of viral nucleic acids 10 No reverse transcription <br />
/ ds DNA is either immediately used as template for synthesis of viral proteins and nucleic acids (lytic cycle) or integrated into bacterial chromosome; Its ssRNA is converted to dsDNA, using reverse transcriptase;<br />
<br />
<br />
<br />
<br />
11 Can enter lytic phase or lysogenic phrase (prophage); Viral genome incorporated into host cell chromosomes (provirus); <br />
12 Prophage is excised from host cell’s chromosome upon spontaneous induction; Provirus is not excised as viral mRNA is transcribed from viral DNA together with host cell’s genes;<br />
Integration of viral glycoproteins into cell membrane 13 No integration of viral glycoproteins into host’s cell membrane; Integration of viral glycoproteins (gp120 and gp41) into host’s cell membrane;<br />
Release / Exit 14 Host cell is lysed to release the new viruses; The new viruses bud off from host cell’s plasma membrane;<br />
<br />
<br />
<br />
<br />
<br />
<br />
<br />
CJC 2010<br />
<br />
8 (a) Describe the role of glucagon in regulating blood glucose. [6]<br />
<br />
1. when blood glucose levels low, glucagon released from alpha cells in pancreas which acts on liver cells;<br />
2. breakdown of glycogen to glucose;<br />
3. use of fatty acids in respiration; <br />
4. production of glucose from other compounds / fats / amino acids / via gluconeogenesis;<br />
5. liver releases glucose into blood and glucose levels rise and return to normal at 90mg/100ml;<br />
6. switching off glucagon secretion<br />
7. effects of glucagon are antagonistic to insulin;<br />
<br />
(b) Describe the reproductive cycles of bacteriophages that reproduce via a lytic cycle, using a named example. [14]<br />
<br />
1. Bacteriophages do not randomly attach to the surface of a host cell, T4 phages attach to specific surface structures called receptor sites.<br />
2. They use cell wall lipopolysaccharides or proteins as receptors. <br />
3. Attachment sites on the tail fibres of T4 phage recognize and adsorb to receptor sites on the host bacterium, via weak interaction between tail fibres and lipopolysaccharide (receptor site) of host bacterium.<br />
4. As more tail fibres make contact, the phage tail pins attach via strong interaction to receptor (outer membrane proteins) on surface of outer membrane of host cell.<br />
5. Specific strains of bacteriophages can only adsorb to specific strain of host bacteria, this is known as viral specificity.<br />
<br />
6. After the baseplate is firmly on the cell surface, conformation changes occur in the baseplate and sheath.<br />
7. Sheath contracts and tail tube penetrates outer membrane.<br />
8. A phage enzyme (lysozyme) “drills” a hole in the bacterial wall.<br />
9. Pilot protein helps phage DNA to cross inner membrane.<br />
10. Phage DNA enters bacteria cytosol. This marks the start of the eclipse period.<br />
11. Phage genome is expressed and enzymes are produced. These enzymes coded by the phage genome degrade host DNA, shutting down the bacterium’s gene expression and macromolecular synthesis (protein, RNA and DNA). This provides raw material for virus DNA synthesis. <br />
12. Replication of phage DNA occurs using the bacterium’s metabolic machinery.<br />
13. Transcription of phage DNA.<br />
14. Synthesis of phage proteins, enzymes and structural componenets using bacterium’s metabolic machinery, e.g ribosomes.<br />
15. Assembly of new bacteriophages around the genomes (spontaneous assembly).<br />
16. Usually, a phage-encoded lysozyme breaks down the bacterial peptidoglycan causing osmotic lysis and release of the intact new bacteriophages leading to destruction of host cell. <br />
[20 marks]<br />
<br />
<br />
<br />
<br />
<br />
<br />
<br />
<br />
<br />
9 (a) Describe the main ways in which a globular protein differs from DNA. [6]<br />
<br />
Structural Differences<br />
<br />
1a. a linear, single chain made up of amino acids <br />
1b. double stranded, made up of nucleotides<br />
<br />
2a. made up of monomers that consist of up to 20 amino acids <br />
2b. only 4 bases in DNA<br />
<br />
3a. ref to details of amino acid structure<br />
3b. ref. to details of nucleotide structure<br />
<br />
4a. amino acids held together by peptide linkages<br />
4b. nucleotides held together by phosphodiester links <br />
<br />
5a. conformation of a folded tertiary structure<br />
5b. double helix of DNA; reference to association with histone proteins<br />
<br />
6a. may contain contain sulphur<br />
6b. DNA contain phosphorus<br />
<br />
7 suitable reference to size, quoting appropriate units / location within the cell where they are synthesized or replicated / appropriate named function <br />
<br />
<br />
(b) Explain how the structure of collagen and haemoglobin are related to their function. [8]<br />
Collagen<br />
1. reference to collagen as a fibrous protein / quaternary structure: made up of a staggered array of tropocollagen molecules, each consisting of 3 polypeptide chains tightly twisted and wound together into a triple helix, held by H-bonds hydrogen bonds between glycine and proline or hydroxyproline residues in different polypeptides<br />
<br />
2. Each polypeptide has a regular, repetitive sequence of amino acids, often follows the pattern Gly-Pro-X or Gly-X-Hyp, where almost every third amino acid sequence is a glycine with high proportion of proline / hydroxyproline residues and where X may be any of various other amino acid residues<br />
<br />
3. glycine is the smallest amino acid and this together with proline allow the polypeptide chain to be wound into a tightly coiled, straight and unbranched (left-handed) helix that has a rigid kinked conformation, allowing the 3 helices in the tropocollgen to be tightly twisted into a triple helix <br />
<br />
4. the triple-stranded tropocollagen molecules run parallel one another and disulfide cross-linkages between the R-groups of the amino acid lysine holds the molecules together, forming long parallel fibres<br />
<br />
5. due to H-bonds between the 3 polypeptides, the tight winding of 3 polypeptides together to form a triple helix and presence of strong covalent bonds between tropocollagen molecules - make collage a very stable structure with high tensile strength, hence its important structural roles in connective tissues, bones and as support in tendons, ligaments for attachment of muscle to bone<br />
Haemoglobin<br />
1. reference to haemoglobin as a globular protein / quaternary structure: made up of 2 and 2 globin subunits, which are interlocked by hydrophobic and ionic interactions<br />
<br />
2. There is a highly irregular sequences of amino acids in the polypeptide chains; each polypeptide chain undergo extensive folding to form a compact, spherical shaped protein due to hydrophobic interactions between the (hydrophobic) R groups; all 4 polypeptide chains are linked to form a roughly globular molecule <br />
<br />
3. Being globular, haemoglobin is soluble in water to form a colloidal suspension; hydrophilic amino acid residues directed outwards on the outer surface, where they form hydrogen bonds with water<br />
<br />
4. Each subunit / polypeptide is attached to a Fe2+-containing haem group, hence is a conjugated protein; each haem groups binds to 1 moelcule of O2 so each haemoglobin molecules can bind up to a maximum of 4 O2 molecules<br />
<br />
5. the haem group binds reversibly to oxygen to enable oxygen to be taken up and released readily - at high O2 tension O2 binds as molecular oxygen to the Fe of the haem group; at low O2 tension, O2 is released; important in the metabolic role of Hb in red blood cells as an O2 carrier – able to bind with and pick up oxygen in O2-rich environment e.g. alveoli in the lungs and releasing O2 in O2 -low environment e.g. to actively respiring cells<br />
<br />
6. reference to co-operative binding allosteric effect; where the binding of 1st O2 molecule causes a change in the conformational state of the haemoglobin molecule / a change in the position of the haem groups, increases the affinity of the haemoglobin for oxygen and hence facilitates the binding of the 2nd O2 molelcule and so on. (affinity of haemoglobin for the 4th O2 molecule is approximately 300 times that for the 1st)<br />
<br />
Generalised / motherhood statement such as protein made up of long chains of amino acids linked together by peptide bonds – to be credited only once. No credits for non-specific reference to other bonds such as hydrogen bonds, disulphide linkages, hydrophobic and hydrophilic interactions, ionic bonds being involved.<br />
<br />
(c) Describe, with examples, the roles of proteins in membranes. [6]<br />
<br />
With appropriate reference to integral / peripheral / transmembrane proteins; <br />
<br />
1. As membrane-bound enzymes – involved in metabolic functions e.g. ATPase phosphorylates ADP to ATP using energy released as H+ ions follow down a concentration gradient / cytochrome oxidase in ETC (or any appropriate example) <br />
<br />
2. As electrons acceptors / carriers of the ETC on inner mitochondrial membrane / thylakoid membrane in chloroplast, using the energy released during the electron transfer as electrons travel downhill in energy terms from one carrier to another, to generate a proton gradient across membrane<br />
<br />
3. As surface receptors for binding of ligands / neurotransmitters / hormones e.g. G-protein linked receptors for glucagon / tyrosine kinase receptors on muscle cells for insulin for cell-to-cell signaling and signal transduction / ionotropic receptors that are linked to (ion) channel proteins e.g. acetylcholine receptors associated with Na+ ion channels on postsynaptic membrane for acetylcholine for nervous transmission across synapse <br />
<br />
As transport proteins that are either specific or non-specific, transporting ions / molecules / solutes via passive facilitated diffusion / active transport <br />
<br />
4. As channel proteins, with specific hydrophilic domains / channels to allow passage of ions and hydrophilic / charged, polar molecules that cannot transverse the hydrophobic core of the phospholipids bilayer, moving down a concentration gradient / as non-gated channel – providing an open passage way for movement e.g. auaporins / as gated channel can open / close to allow cells to regulate the movement, e.g. voltage-gated Na+ channels, voltage-gate Ca2+ ion channels <br />
<br />
5. As carrier proteins, allowing only specific molecules to enter/exit; bind the solute on one side of the membrane which then causing the protein to undergo a conformational change that brings the solute to the opposite side of the membrane<br />
e.g. Na+-K+ pump – a type of protein carrier that usually use ATP to move solutes against a concentration gradient (from a low solute concentration to a high solute concentration); e.g. glucose transporter GLUT1, found across membranes of red blood cells for facilitated diffusion of glucose<br />
<br />
6. Glycoproteins as surface markers for cell-to-cell recognition to identify cell type – significant in immune response/transplant / for attachment purposes: as in cell-to-cell adhesion to allow aggregation and association of ‘like’ cells into tissues; as anchors on the cytoplasmic surface for attachment to cytoskeleton<br />
<br />
<br />
AJC 2010<br />
<br />
8 (a) Describe how the structures of RNA are adapted for translation. [8]<br />
<br />
Messenger RNA (mRNA)<br />
• Single-stranded;<br />
• One codon codes for one amino acid;<br />
• mRNA (5’UTR) has ribosome attachment site;<br />
• Codons are non-overlapping<br />
• Has spliceosome recognition sites to allow for excising of introns;<br />
• Mature mRNA consists of exons/ no introns;<br />
• Has START/ INITIATOR (AUG) codon and STOP (UAA, UGA, UAG) codons;<br />
• 5’ 7-methyl guanosine cap and 3’ poly-A tail to ensure mRNA stability;<br />
• Complementary base pairing between codon and anticodons; (award only once in either mRNA or tRNA discussion)<br />
(4 marks max)<br />
<br />
Transfer RNA (tRNA)<br />
• Has (3’ CCA) end to attach to amino acid/ has amino acid attachment site;<br />
• At least 20 different tRNAs - one for each amino acid<br />
• Has anticodons;<br />
• Has shape complementary to amino acyl tRNA synthetase for activation of amino acid;<br />
• Internal hydrogen bonds to form clover leaf shape/ have specific 3D configuration/ stabilize the molecule;<br />
(2 marks max)<br />
<br />
Ribosomal RNA (rRNA)<br />
• A structural component of ribosome;<br />
• Due to hydrogen bonding/ complementary base pairing within;<br />
• Gives rise to a variety of ribosomal types (e.g. 70S, 80S, 50S, 30S, etc);<br />
• Involved in peptidyl transferase activity;<br />
(2 marks max)<br />
<br />
<br />
<br />
<br />
<br />
<br />
(b) Distinguish between chromosomal and gene mutations. [8]<br />
<br />
Gene Mutation Chromosomal Mutation/ Aberration<br />
Definition • Change in structure of DNA/ change in (nucleotide) base sequence/ different types of gene mutations including point, missense, nonsense, frameshift, silent • Change in structure or number of chromosomes<br />
Gene locus • Involves one gene locus • Involves a few gene loci<br />
Protein products • May or may not affect protein products (depending on whether silent mutation or in non-coding DNA regions). • Usually affect many gene products since so many gene loci involved.<br />
No of chromosome involved • Within one chromosome • Involves one or a few chromosomes<br />
Mechanisms • Brought about by deletion, insertion, inversion or substitution of one or more nucleotides • Brought about by deletion, duplication, inversion or translocation of several gene loci on a chromosome<br />
Effects on allele • Gives rise to a new allele, resulting in new protein which may have a novel function<br />
• Does not reshuffle alleles • No new allele arisen;<br />
• Only results in reshuffling of alleles on a chromosome.<br />
Ploidy • No change in ploidy • Could also result in polypoidy or aneuploidy<br />
Frequency • More frequent because genes outnumber chromosomes by several thousand to one • Less frequent<br />
Importance • Of evolutionary importance because acquisition of new alleles increase gene pool for natural selection • Of lower evolutionary importance because it only reshuffles alleles already existing in the gene pool<br />
Visibility under light microscope • Not visible • Usually visible<br />
Examples • Sickle cell anemia (due to substitution of one nucleotide) • Down’s syndrome (due to an extra chromosome 21)<br />
Any valid comparisons;<br />
<br />
(c) Explain how the Meselsohn and Stahl experiment supports the semi-conservative model of DNA replication. [4]<br />
<br />
Reference to experimental details <br />
• DNA labeled with heavy nitrogen / 15N, then cells transferred to light nitrogen/ 14N for two generations;<br />
• Reference to the three different bands after (CsCl) centrifuge; accept labelled diagram;<br />
• Parental strands unzip/ hydrogen bonds broken;<br />
(1-2 marks max)<br />
<br />
How experiment supports <br />
• Parental strands act as template;<br />
• Complementary base pairing;<br />
• New DNA molecule consists of one parental strand and one daughter strand;<br />
• Daughter DNA are genetically identical to each other;<br />
(2-3 marks max)<br />
<br />
9(a) Discuss how signal amplification is illustrated by the effect of hormone on glycogenolysis.[8] <br />
<br />
• Binding of one / a molecule of epinephrine / glucagon to GPCR causeds GPRC will undergo a conformation change ;; <br />
<br />
• to allow the activation of several / few G-protein by displacing GDP for GTP ;; <br />
<br />
<br />
• Each activated G-protein activate enzyme adenylyl cyclase, each of which is able to catalyse the conversion of large number of ATP to cAMP ;; hence amplifying the signal. <br />
[R: increase cAMP, activation of cAMP] <br />
<br />
• These cAMP in turn binds and activates large number of protein kinase A ;;<br />
<br />
• Each activated protein kinase A will initiate a sequential phosphorylation and activation of kinases / phosphorylation cascade ;;<br />
<br />
• At each phosphorylation step/cascade, each activated kinase is able to activate a large number of the next relay molecule/kinase ;; hence the signal is further amplified. <br />
<br />
• lead to the activation of large number of glycogen phosphorylase<br />
<br />
• each will catalyse for the breakdown of large amount of glycogen into glucose. <br />
<br />
<br />
• the number of activated product is always greater than those in the preceding step as one move down the cascade ;; <br />
• binding of 1 glucagon to receptors will lead to the hydrolysis of large number of glycogen;; <br />
<br />
* If less than 3 of the highlighted points are present, 1 mark will be deducted from the overall marks. <br />
<br />
9 (b) Describe the similarities between the interaction of a substrate with an enzyme and the interaction of a ligand with a receptor. [6] <br />
<br />
• Both bind to specific regions of protein (idea of specific region/portion is impt); <br />
• Active site for enzyme; <br />
• Binding site for receptor; <br />
• Both are complementary in shape to sites that they bind to; <br />
• Both will bind to protein via H bonds, ionic bonds, hydrophobic interactions; <br />
• Both can induce a conformational change in the protein when they bind; <br />
• Ref. to induced fit hypothesis for enzymes; <br />
• Ref. to activation of receptor; <br />
• Both interactions are not permanent; <br />
• Ligand dissociates from binding site of receptor / Ligand-receptor complex are removed; <br />
• ES complex formed will be converted to product, which is released; <br />
<br />
<br />
9(c) Compare between nervous and hormonal control. [6] <br />
Similarities: <br />
Both serve as means of communication/ allow living organisms to respond to stimulus <br />
Stimulus transmission of message effector / stimulus elicit a response <br />
<br />
Differences: <br />
Endocrine Nervous system<br />
Nature of transmission Chemical transmission (hormones) through blood system Electrical and chemical transmission (nerve impulses and chemical across synapses)mil and electrical<br />
Transmission speed Slow <br />
Slower transmission and relatively slow-acting (adrenaline an exception) Fast <br />
Rapid transmission and response<br />
Pathway of transmission Specific (via nerve cells ) Not specific (blood around whole body) <br />
Strength of message Strength dependent on amount of hormones Impulse size same regardless of stimulus <br />
Length of response Often long-term changes (except adrenaline) Often short-term changes nothing action <br />
Target Response may be very widespread, e.g. growth et organs at different parts of body Response often very localized, e.g. one muscle target organ <br />
<br />
IJC 2010<br />
9 (a) Describe the structure of DNA and its organisation in a eukaryotic chromosome. [8]<br />
<br />
1. 2 polymers of deoxyribonucleotides chains twisted into a double helix;;<br />
2. held by H- bonds between complementary bases, A=T, C≡G;;<br />
3. a. purine-pyrimidine: 2 nm diameter;<br />
3. b. 0.34nm/10 bp per complete turn;;<br />
4. distinct polarity 3’ end with free hydroxyl group at 3C and 5’ end with phosphate group at 5C;;<br />
5. DNA molecule associated 1¾ turns with 8 histone proteins/octomer, formed nucleosome core;;<br />
6. histones positively charged, formed ionic bonds with negatively charged DNA;;<br />
7. associate with H1 histone to form nucleosome, linked by spacer/linker DNA forming 10nm ‘beads-on-string’ structure;;<br />
8. further coiled into a solenoid forming a 30nm fibre;;<br />
9. futher looping in presence of scaffold proteins to form metaphase chromosomes;;<br />
AVP;;<br />
<br />
(b) Describe the control of gene expression in eukaryotes. [12]<br />
Gene Amplification<br />
1. duplication of genes to increase the number of copies;;<br />
Chromatin remodelling<br />
2. acetylation removes +ve charges from histones, loosen –ve charged DNA, allow GTF to bind;;<br />
3. demethylation removes methyl grps from proximal promoters, e.g. CpG islands;;<br />
Transcriptional control<br />
4. cis-acting elements, GTFs binding to promoters and proximal promoters;;<br />
5. binding of STF e.g. activators to enhancer, repressors to silencers á and â transcription rate;;<br />
Post-transcriptional control<br />
6. ss pre-mRNA, with 5’ capping with modified guanine/methylated guanine;;<br />
7. 3’ polyadenylation for stability/ nuclear export of mature mRNA;;<br />
8. splicing of introns out and joining of exons forms mature mRNA by spliceosome;;<br />
9. association with proteins for nuclear export;;<br />
10. mRNA editing, the ∆ in codon results in different functional polypeptide;;<br />
Translational control<br />
11. stability of mRNA in the cytoplasm controlled by degradation in response to presence of RNA sequence signalling degradation, concentration of the translated product, extracellular signals like hormones;;<br />
12. gene silencing by miRNA, forming complementary dsRNA in DICER cplx prevent ribosomal attachment/ cleave mRNA;;<br />
Post-translational control<br />
13. proteins not required are tagged for ubiquination by protease;;<br />
14. AVP;;<br />
[Total: 20]<br />
<br />
10 (a) Explain the principles of homeostasis. [4]<br />
<br />
1. Homeostasis is the maintenance of a constant, stable internal environment within an organism regardless of changes in external environment via a self-regulating mechanism;;<br />
2. The regulated variable is usually at a set point which can be changed by a stimulus;;<br />
3. change detect by a receptor which sends signals to control center/regulator;;<br />
4. activates effector resulting in a response to restore set point via positive/negative feedback;;<br />
<br />
(b) Describe the roles of insulin and glucagon in regulation of blood glucose. [10]<br />
<br />
Insulin<br />
1. [bld glu] á above norm, β cells in islets of Langerhans detect the á;;<br />
2. insulin released to bind to rtk receptors of effectors e.g. liver cells;;<br />
3. results in increased no. of glucose transporters in the liver cells to increase uptake of glucose;;<br />
4. convert excess glucose to glycogen in glycogenesis by activating glycogen synthetase;;<br />
5. Depresses rate of synthesis of glucose from non-carbohydrate source via gluconeogenesis;;<br />
Glucagon<br />
6. [bld glu] â below nom, α cells in islets of Langerhans detect the drop;;<br />
7. glucagons released bind to GPCR on effector e.g. liver cells;;<br />
8. results in activation of glycogen phosphorylase to convert glycogen into glucose;;<br />
9. reached set point, neg feedback mechanism prevents further release of hormones by pancreas;;<br />
10. AVP<br />
(c) Describe the molecular events when glucagon binds to its target cell. [6]<br />
<br />
1. Glucagon binds to the complementary ligand-binding site on GPCR receptor causing a conformational change to receptor;;<br />
2. GTP from the cytosol then displaces GDP from the nucleotide-binding site on Ga;;<br />
3. Ga simultaneously dissociates from its Gβγ, translocate along the plasma membrane to activate the enzyme adenylyl cyclase;;<br />
4. Adenylyl cyclase catalyses the conversion of ATP to cyclic AMP (cAMP);;<br />
5. cAMP activate relay protein, protein kinase A (PKA);;<br />
6. Activated PKA then phosphorylates other relay proteins and enzymes, activates other enzymes in phosphorylation cascade;;<br />
7. The effect of cAMP is removed by phosphodiesterase which converts cAMP back to AMP;;<br />
8. AVP;;<br />
[Total: 20]<br />
<br />
JJC 2010<br />
8 (a) Describe the process of ATP production in the chloroplast. [7]<br />
<br />
1. Photosynthetic pigments such as chlorophyll a, chlorophyll b and carotenoids; <br />
2. are embedded on the thylakoid membranes;<br />
3. are arranged in photosystems;<br />
4. Two photosystems PSI (P700) and PSII (P680) which absorbs wavelengths of 700nm and 680nm respectively;<br />
5. Photon of light absorbed by both PS will excite electrons in special chlorophyll a in reaction centres to higher energy levels;<br />
6. These excited electrons are accepted by primary electron carriers; <br />
7. And pass on to electron carriers on the electron transport chain (ETC);<br />
8. Electron carriers on the ETC are of progressively lowered energy level;<br />
9. As electrons move down the ETC, energy is released;<br />
10. The energy released is used to pump H+;<br />
11. From the stroma into the thylakoid lumen;<br />
12. Accumulation of H+ generate potential energy;<br />
13. H+ return to the stroma by diffusing down their concentration gradient <br />
14. through H+ channel;<br />
15. Energy is released and used to couple the phosphorylation of ADP + Pi to produce ATP;<br />
16. Catalyzed by ATPase;<br />
Max 7 <br />
<br />
(b) Give an overview of the Krebs cycle and its significance in respiration. [7]<br />
Overview<br />
1. Occurs in matrix of the mitochondrion;;<br />
2. Per cycle, one molecule of acetyl CoA (2C) undergo oxidative decarboxylation;;<br />
OR<br />
Citrate (6C) undergoes oxidative decarboxylation to form α–ketoglutarate (5C);<br />
α–ketoglutarate (5C) undergoes oxidative decarboxylation to form succinate (4C);<br />
3. Acetyl CoA (2C) and oxaloacetate (4C) are condensed to form citrate (6C);;<br />
4. Description of the decarboxylation process i.e. 6C to 5C to 4C intermediates;; <br />
i. Citrate (6C) undergoes oxidative decarboxylation to form α–ketoglutarate (5C)<br />
ii. α–ketoglutarate (5C) undergoes oxidative decarboxylation to form succinate (4C)<br />
iii. Succinate (4C) undergoes oxidation to form fumarate (4C)<br />
iv. Fumarate (4C) is converted to malate (4C)<br />
v. Malate (4C) undergoes oxidation to form oxaloacetate (4C)<br />
5. Two molecules of CO2, three molecules of NADH, one molecule of FADH2 and one molecule of ATP through substrate level phosphorylation are released per cycle;;<br />
6. At the end of the cycle, one molecule of oxaloacetate (4C) is regenerated to receive more acetyl groups and the cycle continues;;<br />
Max 5 marks<br />
<br />
Significance<br />
7. Completes the oxidative breakdown of glucose to CO2 and H2O and release sufficient energy;;<br />
8. Release of hydrogens (NADH produced) which can be used in oxidative phosphorylation to provide energy to produce ATP;;<br />
9. Carbohydrate intermediates can be converted to amino acids;;<br />
10. Permits amino acids to enter Krebs cycle when glucose is in short supply;;<br />
Max 3 marks<br />
<br />
(c) Compare and contrast oxidative phosphorylation and photophosphorylation. [6] <br />
Similarities:<br />
1. involve transfer of electrons between electron carriers;;<br />
2. energy released from electron transport is used to generate proton gradient;;<br />
3. potential energy of proton gradient is harnessed for ATP synthesis;;<br />
<br />
Differences:<br />
Features Photophosphorylation Oxidative Phosphorylation<br />
Sources of energy for ATP synthesis energy for making ATP comes from light energy for making ATP comes from glucose oxidation processes;;<br />
Location<br />
Thylakoid membrane of chloroplast Inner membrane of mitochondria;;<br />
Involvement of light energy Required to energize the electrons in special chl a Not required;;<br />
Electron donors<br />
For non-cyclic reaction: water<br />
For cyclic reaction: PS I NADH, FADH2;;<br />
<br />
Electron acceptors<br />
For non-cyclic reaction: NADP+<br />
For cyclic reaction: PS I Oxygen;;<br />
<br />
Establishment of proton gradient<br />
H+ pumped inwards, from stroma to thylakoid space<br />
H+ pumped outwards, from mitochondrial matrix to intermembrane space;;<br />
Max 6<br />
<br />
9 (a) Describe the important events that occur during Meiosis I and explain the significance of each event. [10] <br />
Important events during Meiosis I Significance<br />
1. Prophase I; <br />
2. Synapsis ; <br />
3. pairing of homologous chromosomes; <br />
4. to form tetrads / bivalents ; 5. Allows for alignment of genes on chromosomes ; <br />
6. to prepare for crossing over between the homologous regions later ;<br />
7. Crossing over ; <br />
8. between non-sister chromatids of homologous chromosomes ; formation of chiasmata ; 9. Allows for exchange of genetic material ; <br />
10. results in new combination of alleles ; <br />
11. genetic variation in the gametes later ; <br />
12. Metaphase I; <br />
13. Homologous chromosomes are arranged along equator in pairs ; <br />
14. alignment of each homologous pair is independent of other homologous pairs / independent assortment ; 15. Allows gametes to contain a random combination of paternal and maternal chromosomes ; <br />
16. genetic variation in gametes later ;<br />
17. Anaphase I; <br />
18. Homologous chromosomes are pulled apart to opposite poles ; <br />
19. movement of each homologous pair is independent of other homologous pairs / random segregation ; 20. Allows gametes to contain a haploid set of chromosomes ; <br />
21. for restoration of diploid no. of chromosomes ; following fertilization ;<br />
22. Allows gametes to contain a random combination of paternal and maternal chromosomes ; <br />
23. genetic variation in gametes later ;<br />
24. Telophase I & cytokinesis; <br />
25. Division of cytoplasm (and organelles) ; <br />
26. furrowing (animal cells) / cell-plate formation (plant cells) ; 27. Allows formation of 2 haploid daughter cells ;<br />
28. Preparation of cells for Meiosis II ;<br />
<br />
(b) Describe the significance of gene amplification. [6]<br />
<br />
1. Gene amplification refers to the process of (selectively) increasing the number of copies of a particular gene (located at a particular region of the chromosome), without a proportional increase in other genes;;<br />
2. These amplified genes can be transcribed and translated OR leads to an overproduction of the corresponding mRNAs and proteins;;<br />
<br />
3. Meet the needs of cells resulting in higher level of mRNA and polypeptide synthesis at different development stages of cells;; <br />
4. E.g. In the developing ova of eukaryotes, million or more additional copies of rRNA genes are synthesized. This allows the ova to make enormous numbers of ribosomes resulting in a burst of protein synthesis once the ova are fertilized;;<br />
5. Cause cancer due to over-expression of proteins (leading to development of malignant tumours);;<br />
6. If an oncogene is amplified, then the resulting over-expression of that gene can lead to de-regulated cell growth;;<br />
7. An example is the amplification of the ErbB-2 oncogene in breast and ovarian cancers;;<br />
<br />
8. Genes related to drug resistance in these cancer cells are increased; conferring them with drug resistance;<br />
9. and the ability to prevent absorption of therapeutic drugs; thus drug-resistant tumors can continue to grow and spread even in the presence of chemotherapy drugs;<br />
10. Confer selective advantage which allow organisms to survive in a particular environment;; Max 6<br />
(c) Outline the end-replication problem in eukaryotic chromosomes. [4]<br />
<br />
1. DNA polymerase;<br />
2. can only add nucleotides to the 3′ end;<br />
3. of a elongating/pre-existing polynucleotide/primer;<br />
4. primer at 5’ end; <br />
5. removed;<br />
6. but cannot be replaced with DNA/ gap cannot be filled in;<br />
7. because no 3’-OH available (to which a DNA nucleotide can be added);<br />
8. 5’ end of the DNA becomes shorter relative to that of the previous generation;<br />
<br />
<br />
YJC 2010<br />
8<br />
(a) Describe how the presence of lactose results in the transcription of the Lac operon in prokaryotes. <br />
[8]<br />
Correct diagram of the Lac operon (in particular, the position of the regulatory and structural genes, as well as promoter and operator);<br />
Description of negative inducible mechanism, whereby the operon is regulated via a repressor (negative), and presence of lactose induces the transcription of the operon (normally not transcribed - inducible)<br />
Trace amounts of lactose permease (minimally transcribed / expressed) on cell surface membrane of prokaryote permits the entry of lactose; <br />
Trace amounts of β-galactosidase (minimally transcribed / expressed) converts lactose to allolactose as a side reaction;<br />
The Lac repressor is a protein that is transcribed and translated from LacI gene (regulatory gene) located in front of the promoter site;<br />
The repressor has two binding sites – DNA binding site and repressor binding site (referred to as allosteric site);<br />
Allolactose binds to the Lac repressor bound to the operator of the Lac operon;<br />
Causes a conformational change in the repressor, preventing it from being attached to the operator site (releasing it from the DNA binding site);<br />
Operator site overlaps with promoter site;<br />
RNA polymerase binds to the promoter site and transcribes the structural genes;<br />
LacZ is transcribed and translated to produce β-galactosidase, LacY produces lactose permease, and LacA produces transacetylase;<br />
Transcription (and translation) continues until all/most of the lactose are digested to produce glucose and galactose, and repressor binds to the operator site again;<br />
Accurate reference to glucose and CAP-binding site: 1 mark max;<br />
<br />
<br />
(b) Compare and contrast the structure and organisation of prokaryotic and eukaryotic chromosomes. <br />
[6]<br />
Similarities (max 3)<br />
They both have origins of replication that allow for them to be replicated;<br />
They consist essentially of the same types of nucleotides – adenine, thymine, cytosine and guanine;<br />
They are composed of DNA double helix structure;<br />
Similar kinds of bonds can be found in the chromosomes – phosphodiester, hydrogen, glycosidic;<br />
Both are compacted and associated with proteins that help to pack them;<br />
Differences (max 5)<br />
Prokaryotic chromosomes are circular while eukaryotic chromosomes are linear;<br />
Prokaryotic chromosomes are usually singular and carry the entire genome while usually several eukaryotic chromosomes make up the entire genome, with each chromosome carrying a small proportion of the eukaryotic genome;<br />
Eukaryotic chromosomes have multiple origins of replication while prokaryotic chromosomes have only one;<br />
The structural genes of prokaryotic chromosomes are organised in operons, where several genes are under the control of a single promoter, while the structural genes of eukaryotic chromosomes are each under the control of a single promoter;<br />
Eukaryotic chromosomes contain a high proportion of non-coding DNA such as introns, centromeres, telomeres, and other intergenic regions such as satellite DNA, which are almost absent in prokaryotic chromosomes;<br />
Eukaryotic chromosomes are made highly compact with the involvement of many histones while comparatively, prokaryotic chromosomes are loosely packed with the involvement of some histone-like proteins;<br />
Eukaryotic chromosomes are bounded by nuclear envelope whilst prokaryotic chromosomes are not enclosed by the envelope (found in the nucleoid region) and are usually attached / anchored to cell wall;<br />
<br />
<br />
<br />
(c) Explain how mutations related to a named tumour suppressor gene lead to cancer. <br />
[6]<br />
<br />
E.g. of named tumour suppressor gene: P53;<br />
Normal functions of tumour suppressor gene:<br />
Activate DNA repair;<br />
Inducing mitotic arrest (stopping of cell cycle);<br />
Initiate apoptosis (programmed cell death);<br />
Types of mutations causing loss of function<br />
Gene mutation (any named kind);<br />
Mutation in regulatory sequence;<br />
Both copies of alleles to be mutated in order for the cancer phenotype to be expressed (loss of function);<br />
Description of cell activity resulting from loss of function<br />
DNA damage not repaired for proteins regulating cell cycle leads to cell division;<br />
Checkpoints of mitosis not regulated resulting in continuous cell division;<br />
Cancer cells do not undergo apoptosis and carries on dividing;<br />
<br />
[20]<br />
<br />
<br />
9 (a) Explain the main ways in which a globular protein differs from a sequence of DNA. <br />
[8]<br />
<br />
Pt of comparison Globular proteins DNA<br />
1 Shape Compact & round <br />
Reject Globular Linear in eukaryotic cells<br />
<br />
Note that prokaryotic cells have circular DNA. <br />
2 Solubility in water Yes No/less soluble;<br />
3 Monomers Amino acids deoxyribonucleotides;<br />
4 Types of monomers 20 used in humans 4 types (A,T,G,C);<br />
5 Bonding (1) Linked by peptide bonds Linked by phosphodiester bonds;<br />
6 Number of subunits One or more Always 2 under physiological conditions;<br />
7 Bonding (2) Polypeptide chains/subunits linked by ionic bonds, covalent disulphide bonds, hydrophobic interactions and hydrogen bonding Only hydrogen bonding links the two DNA chains;<br />
8 Function Multiple functions: Enzymatic, structural, transport, energy source (any 2) Carry hereditary information/code for synthesizing products;<br />
9 What confers functionality Its function is usually linked to its 3D shape/conformation and configuration Its function is dependent only on the order in which the monomers are arranged in DNA chain;<br />
10 Levels of organization 4 levels 1 structure/a-helix;<br />
11 Charge Can be of any charge (positive, negative or neutral) Negatively charged;<br />
12 Location Everywhere/both inside and outside the cell Only inside the cell /in nucleus; <br />
<br />
13 Process of synthesis transcription and translation DNA replication<br />
14 Template for synthesis mRNA DNA /parent DNA<br />
<br />
<br />
<br />
<br />
(b) Outline the main differences in structure and function between starch, cellulose and glycogen. <br />
[6]<br />
Similarities <br />
All three are polymers made up of glucose molecules<br />
<br />
Differences<br />
Starch glycogen Cellulose <br />
Types of bonds and significance to function <br />
Amylose is a planar polymer of glucose linked mainly by α(1→4) glycosidic bonds<br />
<br />
Amylopectin is a highly branched polymer of glucose. Glucose units are linked in a linear way with α(1→4) glycosidic bonds. Branching takes place with α(1→6) glycosidic bonds occurring every 24 to 30 glucose units.<br />
<br />
The bonds can be broken down by hydrolysis easily- suitable as a energy store <br />
glycogen is highly similair to amylopectin, it is a highly branched polymer of glucose. Glucose units are linked in a linear way with α(1→4) glycosidic bonds. Branching takes place with α(1→6) glycosidic bonds occurring every 24 to 30 glucose units<br />
<br />
The bonds can be broken down by hydrolysis easily- suitable as a energy store Cellulose is a polysaccharide consisting of a linear chain of several hundred to over ten thousand β(1→4) glycosidic links.<br />
<br />
<br />
<br />
<br />
<br />
<br />
<br />
<br />
The bonds cannot be broken down by hydrolysis easily- suitable as structural carbohydrates.<br />
<br />
Packing and relation to function Coil helically with multiple branching into a compact structure<br />
<br />
thus can be stored in large amounts within a fixed volume. Form fibrils / fibers, <br />
<br />
<br />
thus providing tensile strength for cell wall.<br />
<br />
<br />
<br />
<br />
Association with water and relation to function The hydroxyl groups (-OH) on the glucose residues on amylopectin are used to form hydrogen bonds with adjacent chains.<br />
<br />
Amylose forms a colloidal dispersion in hot water (which helps to thicken gravies) whereas amylopectin is completely insoluble. <br />
<br />
Insoluble in water, thus does not affect the water potential of cells. <br />
Highly branched structure (more so than starch) increases its solubility in water and allows for the rapid synthesis and degradation of glycogen. Due to position of the hydroxyl groups (-OH) on the glucose residues, each successive glucose residue in the chain is rotated at 180° to allow the formation of β(1→4) .<br />
<br />
<br />
Large intermolecular spaces, thus allow water to move through the cell wall<br />
<br />
<br />
<br />
<br />
<br />
PJC 2010<br />
8 (a) Compare the structure and organization between prokaryotic and eukaryotic genome. [8]<br />
(b) Describe the eukaryotic processing of pre-mRNA. [6]<br />
(c) Describe various ways in which gene expression may be controlled at translational and post-translational level. <br />
[6]<br />
9 (a) Describe the structure of DNA and tRNA. [6]<br />
(b) Describe the process of DNA replication in prokaryotes. [8]<br />
(c) Describe the experimental evidence for semi-conservative replication. [6]<br />
<br />
Compare the structure and organization between prokaryotic and eukaryotic genome.<br />
Similarities:<br />
1. Both consist of double stranded DNA;<br />
2. Both shows different levels of packing to compact the genome such that they can fit into the cell; <br />
<br />
Differences between prokaryotic and eukaryotic genome<br />
Prokaryote Eukaryote<br />
Genome description Circular/closed loop linear chromosome; <br />
Size Shorter. (Contains a few million bp.) Typically longer chromosome. (Contains tens of millions to hundreds of millions of base pairs (bp) in length.)<br />
Chromosome location Chromosome not in nucleus, found in nucleoid region, (in contact with cytoplasm); Chromosome found in nucleus; <br />
Packing of DNA (2 marks) Packing also take place but not as extensive.<br />
The chromosomal DNA must be compacted about 1000-fold in order to fit into the bacterial cell.<br />
There are two main ways:<br />
1. Formation of looped domain – a segment of DNA that is folded into a loop-like structure.<br />
DNA-binding proteins are involved in holding the loops in place.<br />
AND<br />
2. Supercoiling of the looped DNA.<br />
<br />
Multi-levels of packing. <br />
<br />
<br />
<br />
Basic level “Beads-on-a-string” model where DNA wind around histone proteins to form nucleosomes.<br />
<br />
<br />
Nucleosomes undergo further packing to form 30 nm fibre/solenoid 300 nm fibre <br />
<br />
Introns Introns not found within genes.<br />
(mRNA formed during transcription need not be processed. They can be translated immediately.) Introns (non-coding, repetitive sequences) found interspersed within a gene.<br />
(This arise the need for post-transcriptional modification RNA splicing.)<br />
No of ori 1 origin of replication on 1 chromosome; Each chromosome consists of several origin of replication;<br />
Location of repetitive sequences Repetitive sequences interspersed throughout the chromosome; Repetitive sequences are commonly found near; centromeric and telomeric regions but may also be throughout the chromosome<br />
Genes organization Coordinately controlled genes are often clustered into an operon. ; (A transcriptional unit consisting of coordinately regulated clusters of genes with related functions.) Coordinately controlled genes are organized not in clusters but interspersed throughout the genome, each with own promoter.;<br />
plasmid An extrachromosomal plasmid is usually found in prokaryotes. Absence of plasmid<br />
Differences 7 max<br />
<br />
1. Post- transcription modification is required to process the pre-mRNA to form mature mRNA and to facilitate the export from nucleus to cytoplasm.<br />
2. The 5’ end, the end made first during transcription, is immediately capped off with a modified form of a guanine nucleotide; <br />
3. The enzymes required for capping is grouped to form the Capping Enzyme Complex (CEC) and;<br />
4. is bound to the RNA polymerase II before transcription starts; <br />
5. At the 3’ end, where there is polyadenylation sequence present <br />
6. an enzyme, poly (A) polymerase (PAP);<br />
7. makes a poly(A) tail consisting of 50 to 250 adenine nucleotides;<br />
8. The signals for RNA splicing are short nucleotide sequences at the ends of introns;<br />
9. Particles called small nuclear ribonucleoproteins or snRNPs, recognize these splice sites.<br />
10. snRNPs are located in the cell nucleus and are composed of RNA and protein molecules;<br />
11. The RNA in a snRNP particle is called a small nuclear RNA (snRNA), each molecule is about 150 nucleotides long;<br />
12. Several different snRNPs join with additional proteins to form an even larger assembly called a spliceosome;<br />
13. The spliceosome interacts with the splice sites at the ends of an intron, intron region was looped out <br />
14. It cuts at specific points to release the intron, <br />
15. then immediately joins together two exons that flanked the intron;<br />
16. snRNA plays a role in the catalytic process as well as in spliceosome assembly and splice site recognition;<br />
max 6<br />
<br />
(c) Gene expression may be controlled at translational level via these processes: <br />
Process involved Mechanism Regulation<br />
Gene expression control of during translation initiation 1. Regulatory proteins binds to specific sequence within the 5’ UTR; <br />
2. Prevent attachment of small ribosomal subunit to mRNA; 15. Block translation initiation; <br />
Global control <br />
translation initiation 3. Activation /inactivation of one or more of TIF(translation initiation factors) (by phosphorylation/dephosph);<br />
4. Promote/ Inhibit the formation of translation initiation complex; 16. Translation will be initiated for ALL the mRNA in a cell; <br />
mRNA degradation 5. Enzymes (3’5’ exonucleases) breaks down poly (A) tail first; <br />
6. Followed by enzyme removal of 5’ end (decapping) critical step; <br />
7. Allows nuclease enzymes to rapidly digest the mRNA; 17. Control the lifespan of mRNA in the cytoplasm the amount of proteins formed; <br />
<br />
Gene expression may be controlled at post translational level via these processes: <br />
Process involved Mechanism Regulation<br />
Post translational modification <br />
Protein processing (during translation),<br />
Chemical modification 8. A polypeptide will coil/fold forming function 3D prot with secondary & tertiary structure (cisternae of RER);<br />
9. Cleavage of initial pro-insulin (polypeptide) into active hormones;<br />
10. Chemical modification- adding of sugar/lipids/phosphate group;<br />
11. Cell surface proteins must be transported to target destination to function; 18. Protein processing control the type of proteins to be formed;<br />
Protein degradation <br />
12. To mark a protein for destruction, ubiquitin protein is tagged;<br />
13. Leads to recognition by large proteasomes;<br />
14. Enzymes in the proteasome cut the proteins into small peptides degrade by other enzymes in cytosol; 19. Protein degradation affects the amount of protein present in cell; <br />
<br />
Structure:<br />
1. The DNA molecule consists of two deoxyribonucleotide chains<br />
2. The two chains coil round each other to form a double helix.<br />
3. Each chain consists of a sugar-phosphate backbone, held together by phosphodiester bonds, between the sugar of one nucleotide and the phosphate group of the next nucleotide, <br />
4. with the nitrogenous bases arranged as side groups off the chains (project into the centre of the double helix). <br />
5. The two chains run in opposite directions (antiparallel) and<br />
6. are held together by hydrogen bonds between their nitrogenous bases. <br />
7. Hydrogen bonding can occur only between a purine and a pyrimidine. <br />
8. Thus the sequence of bases in one chain determines that in the other chain. The two chains are said to be complementary. <br />
9. The width between the two chains is constant (2nm) and is equal to the width of one base pair, ie. the width of a purine plus a pyrimidine. <br />
10. Along the axis of the molecule the distance between adjacent nucleotides is 0.34nm <br />
11. One complete turn of the double helix is 3.4 nm and comprises 10 base pairs<br />
<br />
tRNA<br />
Structure<br />
1. Single stranded RNA<br />
2. Wound into clover-leaf shape with 3 loops<br />
3. 3’ end of the tRNA always ends in the base sequence of CCA <br />
4. and binds an amino acid<br />
5. 5’ end of the tRNA always ends in a base, guanine. <br />
6. Acts as an intermediate molecule between the triplet code of mRNA and amino acid sequence of polypeptide chain. <br />
7. Carries the correct amino acid to the forming polypeptide chain on the ribosome.<br />
8. One loop contains a triplet of bases called an anticodon which pairs with the complementary codon on mRNA.<br />
9. There are more than 20 different tRNA molecules in a given cell, each carrying a specific amino acid.<br />
<br />
(b)Unwinding DNA double helix<br />
1. Mechanism is by semi-conservative replication.<br />
2. Activated deoxyribonucleotides and transported into the nucleus via pores in the nuclear envelope<br />
3. Replication of DNA begins at specific sites called the origins of replication (Ori). <br />
4. The enzyme helicase causes the DNA molecule to unwind and unzip/the hydrogen bonds between the bases to break, causing the two DNA chains to separate.<br />
5. Topoisomerase helps to prevent over-straining of the replication fork.<br />
6. The separating DNA results in a “replication bubble”, allowing replication to take place at both ends, known as a replication fork (bidirectional)<br />
7. The unwinding DNA chains provide templates for the synthesis of new DNA chains.<br />
8. After parental strand separation, single-stranded binding protein (SSB) binds to the unpaired DNA strands, stabilizing them.<br />
<br />
<br />
Formation of RNA primer<br />
9. The enzyme primase catalyses the synthesis of a short RNA primer, (about ten ribonucleotides long), <br />
10. Such primers are necessary because the enzyme, DNA polymerase III (DNA pol III), that is responsible for the synthesis of DNA, cannot initiate synthesis of a polynucleotide; they can only add nucleotides to the end of an already existing chain that is base-paired with the template strand.<br />
11. The new DNA strand will start from the 3’end of the RNA primer.<br />
12. Once synthesis of the new DNA chain starts, the RNA primer is hydrolysed (removed) by enzyme DNA polymerase I (DNA pol I). Removal of RNA primer leaves substantial gaps between the DNA fragments. The filling of these gaps by deoxyribonucleotides is also catalysed by DNA pol I<br />
<br />
Synthesis of new DNA strand<br />
13. DNA polymerase III recognises the bases and selects free deoxyribonucleotides that are complementary to those on the template strand.<br />
14. Adenine pairs with thymine and vice versa. Similarly, cytosine pairs with guanine and vice versa.<br />
15. Both daughter strands are synthesized in the 5' to 3' direction. <br />
16. One of the daughter strands, called the leading strand, is synthesized continuously/ towards replication fork<br />
17. To elongate the other strand in a 5’ to 3’ direction, DNA pol III must work along the other template strand in the direction away from the replication fork.<br />
18. The DNA strand is called the lagging strand, synthesised in the form of short fragments called Okazaki fragments<br />
19. The enzyme, DNA ligase catalyses the formation of phosphodiester bonds between two Okazaki fragments. <br />
20. At the end of the replication, both the parental and daughter strands rewind into a double helix.<br />
21. The new DNA molecule consist of 1 parental and 1 newly synthesized daughter strand;<br />
<br />
(c)<br />
1. Semi-conservation replication both DNA strands are template for DNA synthesis of new strands and <br />
2. The new DNA molecule consist of 1 parental and 1 newly synthesized daughter strand;<br />
3. Messelson and Stahl experiment<br />
4. When cultures of bacteria are grown for many generations in a medium containing heavy isotope of nitrogen (15N), all the DNA molecules became labeled with 15N.<br />
5. The cells were then transferred to another medium containing normal nitrogen (14N) and<br />
6. allowed to divide once and therefore the DNA replicates once. <br />
7. Cells were then allowed to divide a second and third time.<br />
8. DNA, extracted from cells of A, B, C and D, was then centrifuged <br />
9. with CsCl2<br />
10. DNA in the centrifuge tubes appeared as narrow bands when examined under ultraviolet light<br />
11. The width of the DNA bands and their positions in the centrifuge tubes reflect the amount of the various types of DNA molecules (or diagram)<br />
12. Tube A contains only heavy DNA because the bacteria are grown for many generations in a medium (A) containing 15N, which is used to make the nitrogenous bases so all the DNA became labeled with 15N<br />
13. Tube B contains only hybrid DNA with one band occupying the intermediate density position<br />
14. Tube C contains both light and hybrid DNA in the ratio 1:1 and <br />
15. in tube D the ratio is 3:1. Both tubes have 2 bands, 1 band on top at the same density position as the light DNA, and the other at the intermediate density position.<br />
(points 12 to 15 – need to explain or annotated diagram)<br />
<br />
<br />
TPJC 2010<br />
QUESTION 8<br />
(a) Explain how mitosis produces genetically identical cells. [6] <br />
1. Replication of DNA during S-phase of interphase<br />
2. To produce genetically identical sister chromatids attached via centromeres<br />
3. Kinetochore microtubules attach to kinetochores on the centromere<br />
4. Exact alignment of sister chromatids on metaphase / equatorial plate<br />
5. Centromeres divide resulting in separation of sister chromatids<br />
6. Sister chromatids, now as chromosomes, move to opposite poles of the cell<br />
7. Two genetically identical cells are produced when the cell membrane furrows in, <br />
8. Definition of genetically identical cells – some number and type of chromosomes; <br />
Max 6<br />
<br />
(b) Describe the role of meiosis in natural selection. [10]<br />
1. Meiosis produces genetically varied gametes<br />
2. Leading to formation of genetically varied organisms due to random fusion of gametes<br />
3. Prophase 1: crossing over between homologous chromosomes result in new combination of alleles<br />
4. Metaphase 1/anaphase 1: independent assortment of chromosomes occur to result in gametes randomly distributed to either poles of the cells<br />
5. Anaphase 1: segregation of homologous chromosomes to both poles will result in the formation of haploid gametes<br />
6. For natural selection to occur, there must be a varied gene pool<br />
7. In the presence of an environmental change, the environment will select the organisms which has selective advantage<br />
8. This organisms will be able to survive, reproduce and pass on their alleles to their offspring <br />
9. Lack of genetic variation may lead to death of all members in the population due to the selection pressure present OR variation allows continuation of the species in the presence of chances in the environment / increase frequency of alleles in the population<br />
10. Evolution will not occur OR promotes speciation <br />
11. Award marks for example given: variation amidst finches allow speciation etc<br />
<br />
<br />
(c) Explain how one named factor increases the chances of cancerous growth. [4]<br />
1. Named factor: ionizing radiation / UV light / tar in cigarette smoke etc<br />
2. Gives any form of example of how DNA can be mutated e.g. point mutation (e.g.), translocation of DNA to active promoters etc. <br />
3. Damage to DNA which cannot be repaired<br />
4. Mutation to proto-oncogenes / tumour suppressor genes <br />
5. Results in the synthesis of mutated gene products which leads to uncontrolled cell division leading to cancer/increases chances of cancerous growth [R: uncontrolled cell division;]<br />
<br />
Question 9 <br />
(a) Explain briefly what is meant by the terms gene mutation and chromosome mutation.[10]<br />
<br />
Definition of mutation:<br />
• Change in the amount, arrangement or structure of the DNA of an organism.<br />
• Produces a change in the genotype and may result in the change in appearance of a characteristic in a population.<br />
<br />
Definition of Chromosomal mutations<br />
• Result of changes in the number and structure of chromosomes.<br />
<br />
Chromosome mutation: Changes in chromosome number<br />
• These changes may involve the loss or gain of single chromosomes, a condition called aneuploidy,<br />
• Or the increase in entire haploid sets of chromosomes, a condition called euploidy (polypoidy).<br />
<br />
Aneuploidy<br />
• Loss of gain or single chromosomes eg. (n+l), (2n+l), (n-1), (2n-1)<br />
• Result from the failure of a pair, or pairs of homologous chromosomes to separate during anaphase I of meiosis. This is known as non-disjunction.<br />
<br />
Any named example:<br />
1. Down’s syndrome (Mongolism)<br />
Presence of an extra chromosome no. 21 in the cells.<br />
Symptoms – mental retardation, short stocky body, thick neck, characteristic folds of skin over the inner corner of the eye.<br />
Related to the age of the mother’s egg cells.<br />
<br />
2. Klinefelter’s syndrome (XXY)<br />
Failure of X chromosome to separate during oogenesis in female.<br />
Symptoms – male with female characteristic, sterile, testes very small, little facial hair, breasts may develop, low intelligence.<br />
<br />
3. Turner’s syndrome (XO)<br />
Sterile female.<br />
Lacking normal secondary sexual characteristics.<br />
<br />
4. XXX<br />
Female, normal appearance<br />
Fertile but mentally retarded<br />
<br />
5. XYY<br />
Male, variable intelligence, may possess psychopathic traits or tendency for petty criminal acts<br />
<br />
Euploidy (Polyploidy)<br />
• Cells contain multiples of the haploid number of chromosomes ie. polyploids 2n, 3n.....<br />
• More common in plants than animals<br />
• Associated with hybrid vigour<br />
• 2 forms:<br />
a. Autopolyploids<br />
An individual with more than 2 sets of chromosomes all derived from the same species. <br />
Can be induced by using colchicine. <br />
Autopolyploids can be fertile if they have an even number of chromosome sets.<br />
<br />
b. Allopolyploids<br />
An individual with increased chromosome number derived from different haploid sets. <br />
Many inter-species hybrids are sterile because chromosomes do not pair up at meiosis - diploid hybrid ( sterile ).<br />
If chromosomes are doubled, each member has another to pair up with and meiosis proceeds normally -- fertile, allopolyploid.<br />
<br />
Chromosome mutation: Changes in structure <br />
1. Deletion: removes a chromosomal segment<br />
2. Inversion: reverses a segment within a chromosome<br />
3. Translocation: moves a segment from one chromosome to another, non-homologous one<br />
4. Duplication: repeats a segment<br />
<br />
Any named example:<br />
1. Cri-du-chat syndrome: <br />
• deletion in the short arm of chromosome 5<br />
• child is physically and mentally retarded<br />
<br />
2. Chronic myelogenous leukaemia (CML): <br />
• Translocation<br />
• Most of the chromosome 22 has been transolcated onto the long arm of chromosome 9. <br />
• In addition, the small distal portion of chromosome 9 is translocated to chromosome 22.<br />
• Caused increased cell proliferation, reduced apoptosis cancer<br />
<br />
3. Acute myelogenous leukaemia (AML)<br />
• Chromosomal inversion on chromosome 16<br />
• Results in cancer<br />
<br />
4. Charcot-Marie-Tooth syndrome (CMT)<br />
• In one form, chromosomal duplication on chromosome 17<br />
• High gene dosage of myelin sheath protein resulting in abnormal structure and function of the myelin sheath<br />
• Symptoms: weakness of lower foot, loss of balance, poor motor skills and muscle atrophy.<br />
<br />
<br />
Gene Mutation<br />
Definition of Gene mutation<br />
• Sudden and spontaneous changes<br />
• Change in the nucleotide sequence of DNA:<br />
Duplication, insertion, deletion, inversion or substitution of bases<br />
<br />
Effects:<br />
• Alteration in a sequence of nucleotides Change in amino acid sequence of polypeptide chain affect phenotype of individual ( may be inheritable<br />
Types:<br />
(a) Substitution<br />
-1 nucleotide is replaced by another<br />
<br />
Effects:<br />
(i) Silent Mutation<br />
• Due to degeneracy of genetic code<br />
• E.g. DNA: 3’-CCG-5’ ( 3’-CCA-5’<br />
mRNA: 5’-GGC-3’ ( 5’-GGU-3’<br />
• Glycine is still produced<br />
• OR if new amino acid have properties similar to that of replaced amino acid<br />
<br />
(ii) Missense Mutations<br />
• If mutation occurs in crucial areas. E.g. active site of enzymes<br />
• Result in useless/ less active protein that impairs cellular function. E.g. Sickle Cell Anaemia<br />
<br />
(iii) Non-sense Mutations<br />
• Cause a change in codon for amino acid to stop codon<br />
• Result in premature termination of translation, non-functional protein is formed<br />
<br />
(b) Inversion<br />
<br />
<br />
<br />
<br />
<br />
<br />
<br />
<br />
<br />
<br />
(C) Frameshift Mutation<br />
• Due to the insertion/ deletion of nucleotide pairs<br />
• Have a more disastrous effect as it cause ribosome to read incorrect triplets from point of mutation<br />
• Result in shorter chain ad/ or mostly non-functional proteins<br />
<br />
E.g. <br />
mRNA 5’ AUG AAG UUU GGC UAA 3’<br />
Protein met– Lys– Phe – Gly<br />
<br />
(i) Frameshift mutation causing immediate nonsense<br />
mRNA 5’ AUG UAA GUU UGG CUA A 3’<br />
Protein met– STOP<br />
<br />
(ii) Frameshift mutation causing extensive nonsense<br />
mRNA 5’ AUG AAG UU|G GCU AA 3’<br />
Protein met– Lys– Leu- Ala- <br />
<br />
(iii) +/ - 3 nucleotides: no Frameshift BUT additional/ missing amino acid<br />
mRNA 5’ AUG UUU GGC UAA 3’<br />
Protein met– Phe – Gly<br />
<br />
Case Study: Sickle Cell Anaemia<br />
Normal Adult: Haemoglobin (Hb A)<br />
• Quaternary protein with tetramer consisting of 2 different types of polypeptide chain, 2α and 2β globin chains<br />
• Chains are encoded by 2 different genes on 2 different chromosomes<br />
Normal S.C.A<br />
Gene 3’-CTT-5’ 3’-CAT-5’<br />
Codon on mRNA 5’-GAA-3’ 5’-GUA-3’<br />
Polypeptide -Glu- -Val-<br />
Solubility Soluble Less soluble<br />
At low [O2] Remain soluble Crystallize into rod-like fibres<br />
Shape of RBC Round, biconcave Sickle shape<br />
<br />
<br />
(b) Explain how mutations for antibiotic resistance spread so rapidly among bacteria. [4]<br />
• largely because bacteria reproduce very quickly;<br />
• mutation can be quickly passed on to large numbers of descendents (usually) via plasmids during binary fission;<br />
• through conjugation, whereby transfer of antibiotic resistance genes occurs across a mating bridge;<br />
• through transformation, whereby antibiotic resistance genes taken up by competent bacterial cells through transient pores in the bacterial cell wall;<br />
• through transduction, whereby (defective) bacteriophages transfer bacterial genes / antibiotic resistance genes;<br />
Max. [4]<br />
<br />
(c) Outline the role of isolating mechanisms in the evolution of new species. [6]<br />
• ref. to definition of species;<br />
• ref. allopatric;<br />
• geographical isolation;<br />
• ref. to examples e.g. islands/ lakes/ mountain chains / idea of barrier;<br />
• ref. to example organism;<br />
• ref. to populations prevented from interbreeding;<br />
• isolated populations subjected to differential selection pressure/ conditions;<br />
• over time sufficient differences to prevent interbreeding;<br />
• ref. sympatric;<br />
• ref. to reproductive isolation;<br />
• ref. behavioural barriers (within a population);<br />
• e.g. day active/ night active;<br />
• correct ref. to gene pool;<br />
• change to allele frequencies;<br />
[ 6 max ]<br />
<br />
2010 YJCUnited JC Unionhttp://www.blogger.com/profile/06131630985986252085noreply@blogger.com0tag:blogger.com,1999:blog-4274038432688951963.post-78007942888442374492011-02-02T18:38:00.003-08:002011-02-02T18:38:52.109-08:002010 Prelim 2 Essay CompilationSAJC 2010<br />
<br />
8(a) Describe the eukaryotic processing of pre-mRNA in terms of intron splicing, polyadenylation and 5’capping. [6]<br />
<br />
Capping<br />
1 modified guanosine nucleotide / 7-methylguanosine / added to 5’ end of mRNA;<br />
2 enzymes (e.g. guanyl / methyl transferase) catalyses reaction between 5’ end of the RNA transcript and GTP;<br />
3 ref. association of cap-binding protein<br />
<br />
Polyadenylation<br />
4 addition of 30-200 adenine sequences to the 3’ end of the pre-mRNA;<br />
5 after polyadenylation signal has been transcribed;<br />
6 catalysed by poly(A) polymerase / polyadenylate polymerase;<br />
7 ref. poly-A binding protein binds to poly(A) tail<br />
<br />
Splicing<br />
8 Splicing is catalyzed by snRNPs / small nuclear ribonucleoproteins<br />
9 snRNPs are made up of proteins & snRNA (small nuclear RNA) <br />
10 snRNPs + other proteins join together to form a complex called spliceosomes <br />
<br />
11 spliceosomes is first responsible for folding the pre-mRNA into the correct orientation for splicing<br />
<br />
12 splicing involves cleavage of the 5’ end of the intron<br />
13 and its attachment to the branchpoint sequence <br />
14 to form a tailed loop structure called a lariat<br />
15 intron is then excised/released<br />
<br />
16 by being cleaved at its 3’ end<br />
17 and the exons are brought together and ligated<br />
18 spliceosomes dissociates once splicing is completed<br />
<br />
8(b) Compare structure and organisation of prokaryotic and eukaryotic chromosome. [6]<br />
<br />
[½ mark per correct comparison]<br />
<br />
Prokayotic genome Eukaryotic genome<br />
1. Genome size Smaller genomes, 0.6 to 10Mb. Large genomes, being less than 10 Mb – 100,000 Mb <br />
2. Gene length Shorter gene sequences<br />
/ more compact genetic organisation Longer gene sequences<br />
/ presence of more intergenic spaces<br />
3. Chromosome number Single chromosome<br />
/ Haploid Many chromosomes<br />
/ Diploid or polyploid<br />
4. Chromosome structure Circular DNA molecule. <br />
Linear DNA, each contained in a different chromosome.<br />
5. Location Chromosome found in the nucleoid region of the cytoplasm Chromosomes are enclosed within a double-membrane bound nucleus<br />
6. Packaging of DNA Does not form chromatin.<br />
(Prokaryotic DNA is organized into a DNA-protein complex called the nucleoid. Eukaryotic DNA is complexed with histones and other proteins to form chromatin.<br />
7. Introns Coding sequence proceeds from start to finish without interruption by introns. Presence of introns within genes. <br />
8. Repetitive sequences Few repetitive DNA sequences. Many highly repetitive DNA sequences <br />
9. Coding and non-coding DNA Most of DNA are coding sequences (codes for protein, tRNA, or rRNA. Most of DNA are non-coding.<br />
<br />
10. Regulatory sequences Small amount of non-coding<br />
DNA consists mainly of regulatory sequences, such as promoters) More complex regulatory sequences (eg. enhancers and silencers)<br />
11. Presence and absence of operons Two or more genes may be expressed and regulated as a unit (genomes arranged in operons / polycistronic genes) Absence of operons / monocistronic genes<br />
<br />
12. Origins of replication One<br />
Many <br />
13. Presence of extrachromosomal DNA Independent small, circular, molecules called plasmids. Circular, double-stranded DNA in mitochondria / choloroplasts.<br />
14. Telomeres Absent Present<br />
15. Centromere Absent Present<br />
<br />
<br />
8(c) Outline the differences between prokaryotic control of gene expression with the eukaryotic model. [8]<br />
<br />
For every comparison:<br />
½ mark for correct comparison<br />
½ mark for correct information<br />
<br />
Chromosomal Level (max 2)<br />
Prokaryotes Eukaryotes<br />
1. DNA and histone modification cannot occur (Prokaryotic DNA does not form chromatin / not associated with histones) DNA and histone modification can occur, resulting in conversion between euchromatin and heterochromatin, and hence the ease of transcription<br />
(Eukaryotic DNA is complexed with histones and other proteins to form chromatin / associated with histones)<br />
2. DNA sequences, including operators and activators, serve as the on/off switch Structure of chromatin – euchromatin, ready to be transcribed, or heterochromatin and not available – is the major on/off switch for gene regulation<br />
3. Prokaryotic DNA not organised into chromatin. Initiation of transcription of eukaryotic genes requires that the compact chromatin fibre, characterised by nucleosome coiling, to be uncoiled and the DNA made accessible to RNA polymerase and other regulatory proteins.<br />
<br />
Transcriptional Level (max 2)<br />
Prokaryotes Eukaryotes<br />
4. One RNA polymerase consisting of five subunits <br />
/ All RNAs synthesized by the same RNA polymerase; <br />
<br />
<br />
<br />
Occurs within the nucleus under the direction of three separate forms of RNA polymerases, each containing 10 or more subunits; different polymerases transcribe different genes <br />
/ Three different classes of RNA each synthesized by a different RNA polymerase<br />
5. Simple regulatory sequence: Transcriptional regulatory protein / Regulator protein binds to DNA-binding sites upstream of the cluster of structural genes to regulate initiation of transcription. Complex regulatory sequence: More extensive interaction between upstream DNA sequences and protein factors involved to stimulate and initiate transcription. In addition to promoters, enhancers and silencers control rate of transcription initiation<br />
6. Related genes are transcribed together as operons <br />
/ only 1 promoter<br />
/ polycistronic mRNA<br />
No operon <br />
/ each gene has own promoter / monocistronic mRNA<br />
<br />
<br />
<br />
Post-transcriptional Level (max 2)<br />
Prokaryotes Eukaryotes<br />
7. Translation is often coupled to transcription (Transcription and translation take place in the same cellular compartment simultaneously) No direct coupling of transcription and translation. (mRNA must pass across nuclear envelope before translation in the cytoplasm. RNA transcript is not free to associate with ribosomes prior to the completion of transcription).<br />
<br />
8. Primary transcripts are the actual mRNAs<br />
/no post-transcriptional modification<br />
Primary transcripts undergo processing to produce mature mRNAs<br />
/ post-transcriptional modification<br />
9. Triphosphate start at the 5’ end<br />
/ No tail at the 3’ end<br />
/ No splicing Methylated guanosine cap at the 5’ end<br />
/ Poly-A tail at the 3’ end<br />
/ Splicing occurs (also alternative splicing)<br />
10. Lower stability of transcript <br />
/ degradation within seconds or minutes <br />
/ mRNAs shorter half life to rapidly respond to environmental changes Higher stability of transcript<br />
/ prevent transcript degradation <br />
/ mRNAs longer half-life remaining much longer to orchestrate protein synthesis prior to their degradation by nucleases in the cell<br />
<br />
Translational level (max 2)<br />
Prokaryotes Eukaryotes<br />
11. Control at this level is unlikely; due to simultaneous transcription and translation Control at translational level:<br />
phosphorylation of ribosomal translation initiation factors<br />
/ negative translational control through regulatory proteins<br />
/ cytoplasmic elongation of poly (A) tails<br />
/ mRNA degradation<br />
/ RNA interference and microRNA<br />
<br />
12. Unique initiator tRNA carries formylmethionine Initiator tRNA carries methionine<br />
13. Smaller ribosomes<br />
/ less complex rRNA and protein components Occurs on ribosomes that are larger<br />
/ rRNA and protein components are more complex than those of prokaryotes<br />
14. mRNAs have multiple ribosome binding sites <br />
/ direct the synthesis of several different polypeptides mRNAs have only one start site<br />
/ direct synthesis of only one kind of polypeptide<br />
15. Small ribosomal subunit immediately binds to the mRNA’s ribosome binding site<br />
Small ribosomal subunits bind first to the methylated cap at the 5’ end of the mature mRNA and then scans the mRNA to find the ribosome binding site, the AUG start codon<br />
<br />
<br />
Post-translational Level (max 2)<br />
Prokaryotes Eukaryotes<br />
16. no/minimal post-translational modifications occur Post-translational modifications determine the functional abilities of the protein<br />
17. Proteolysis: Processing eukaryotic polypeptides to yield functional protein molecules e.g. cleavage of pro-insulin to form the active insulin hormone<br />
18. Chemical modification of proteins to yield functional protein molecules<br />
19. Phosphorylation of proteins to increase or decrease its function<br />
20. Transportation of proteins to target destinations in the cell where it functions is mediated by signal sequences at N-terminus of some proteins. Once transported to destination, signal sequence is enzymatically removed from the proteins<br />
21. Ubiquitination marks protein for degradation, ref to ubiquitin & proteasome<br />
<br />
9(a) Describe the general structure of a named animal virus, and explain why viruses are obligate parasites. [6]<br />
1 example of animal viruses: influzena, HIV, herpes virus etc <br />
2 animal viruses composed of phospholipids/glycoprotein envelope <br />
3 similar in nature to cell surface membrane of animal cell <br />
4 with glycoprotein spikes (for attachment to host cell membrane) <br />
5 enclosed in the viral envelope is the capsid <br />
6 which is composed of capsomeres / protein subunits<br />
7 which contains the viral genome <br />
8 either DNA or RNA, but never both<br />
9 either single or segmented, linear or circular, single-stranded or double-stranded, etc <br />
10 ref. viral enzymes (eg. reverse transcriptase in HIV)<br />
<br />
11 acellular / absence of cytoplasm and cellular organelles<br />
12 lacks ribosomes / protein-synthesising apparatus <br />
13 hijacks host cell’s host’s protein synthesis machinery (transcription and translation machineries) to produce own viral proteins [REJECT ‘metabolic machinery’]<br />
14 ref. replication of viral genome<br />
15 metabolically inert / do not grow or divide on their own <br />
16 can only multiply inside living cells, and not on inanimate media<br />
<br />
9(b) Outline how the influenza virus is able to bypass the human defense mechanism to cause disease. [4]<br />
1 ref. antigenic drift <br />
2 8 single-stranded RNA segments<br />
3 spontaneous genetic mutation /mutation during replication<br />
4 lack of proof-reading during RNA replication<br />
<br />
5 ref. antigenic shift<br />
6 genetic reassortment of the RNA segments between two strains of viruses<br />
7 novel glycoproteins (spikes) produced on viral envelope <br />
/modified neuraminidase and haemaglutinin<br />
8 cannot be recognized by previous antibodies <br />
<br />
<br />
9(c) Compare and contrast the reproductive cycles of the Lambda phage and the Human Immunodeficiency Virus (HIV). [10]<br />
<br />
[1 mark per similarity] (max 4)<br />
1 Life cycle of both viruses involves the stages attachment, penetration, replication, maturation, and release<br />
2 Both attach to their host cell at receptor sites on the host cell’s plasma membrane;<br />
3 Both introduce their viral nucleic acids into their host cell<br />
4 Both are obligate intracellular parasites/make use of their host cell’s resources for synthesis of viral proteins and nucleic acids<br />
5 Both integrate their viral DNA into host genome/viral DNA replicates as part of host’s DNA every time the cell divides <br />
<br />
[1 mark per difference] (max 6)<br />
Feature of comparison Lambda phage HIV<br />
Host cell 6 Infects bacterial cells Infects human T-cells<br />
Penetration / Entry 7 Phage does not undergo fusion with host cell’s plasma membrane / contracts its tail sheath; HIV envelope fuses with the host cell’s plasma membrane;<br />
8 Phage injects only its ds DNA; HIV releases its nucleocapsid (ssRNA and reverse transcriptase) into cytoplasm;<br />
Uncoating 9 No uncoating required as capsid does not enter host cell; Uncoating of nucleocapsid to release genome and enzymes; <br />
Fate of viral nucleic acids 10 No reverse transcription <br />
/ ds DNA is either immediately used as template for synthesis of viral proteins and nucleic acids (lytic cycle) or integrated into bacterial chromosome; Its ssRNA is converted to dsDNA, using reverse transcriptase;<br />
<br />
<br />
<br />
<br />
11 Can enter lytic phase or lysogenic phrase (prophage); Viral genome incorporated into host cell chromosomes (provirus); <br />
12 Prophage is excised from host cell’s chromosome upon spontaneous induction; Provirus is not excised as viral mRNA is transcribed from viral DNA together with host cell’s genes;<br />
Integration of viral glycoproteins into cell membrane 13 No integration of viral glycoproteins into host’s cell membrane; Integration of viral glycoproteins (gp120 and gp41) into host’s cell membrane;<br />
Release / Exit 14 Host cell is lysed to release the new viruses; The new viruses bud off from host cell’s plasma membrane;<br />
<br />
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<br />
<br />
<br />
CJC 2010<br />
<br />
8 (a) Describe the role of glucagon in regulating blood glucose. [6]<br />
<br />
1. when blood glucose levels low, glucagon released from alpha cells in pancreas which acts on liver cells;<br />
2. breakdown of glycogen to glucose;<br />
3. use of fatty acids in respiration; <br />
4. production of glucose from other compounds / fats / amino acids / via gluconeogenesis;<br />
5. liver releases glucose into blood and glucose levels rise and return to normal at 90mg/100ml;<br />
6. switching off glucagon secretion<br />
7. effects of glucagon are antagonistic to insulin;<br />
<br />
(b) Describe the reproductive cycles of bacteriophages that reproduce via a lytic cycle, using a named example. [14]<br />
<br />
1. Bacteriophages do not randomly attach to the surface of a host cell, T4 phages attach to specific surface structures called receptor sites.<br />
2. They use cell wall lipopolysaccharides or proteins as receptors. <br />
3. Attachment sites on the tail fibres of T4 phage recognize and adsorb to receptor sites on the host bacterium, via weak interaction between tail fibres and lipopolysaccharide (receptor site) of host bacterium.<br />
4. As more tail fibres make contact, the phage tail pins attach via strong interaction to receptor (outer membrane proteins) on surface of outer membrane of host cell.<br />
5. Specific strains of bacteriophages can only adsorb to specific strain of host bacteria, this is known as viral specificity.<br />
<br />
6. After the baseplate is firmly on the cell surface, conformation changes occur in the baseplate and sheath.<br />
7. Sheath contracts and tail tube penetrates outer membrane.<br />
8. A phage enzyme (lysozyme) “drills” a hole in the bacterial wall.<br />
9. Pilot protein helps phage DNA to cross inner membrane.<br />
10. Phage DNA enters bacteria cytosol. This marks the start of the eclipse period.<br />
11. Phage genome is expressed and enzymes are produced. These enzymes coded by the phage genome degrade host DNA, shutting down the bacterium’s gene expression and macromolecular synthesis (protein, RNA and DNA). This provides raw material for virus DNA synthesis. <br />
12. Replication of phage DNA occurs using the bacterium’s metabolic machinery.<br />
13. Transcription of phage DNA.<br />
14. Synthesis of phage proteins, enzymes and structural componenets using bacterium’s metabolic machinery, e.g ribosomes.<br />
15. Assembly of new bacteriophages around the genomes (spontaneous assembly).<br />
16. Usually, a phage-encoded lysozyme breaks down the bacterial peptidoglycan causing osmotic lysis and release of the intact new bacteriophages leading to destruction of host cell. <br />
[20 marks]<br />
<br />
<br />
<br />
<br />
<br />
<br />
<br />
<br />
<br />
9 (a) Describe the main ways in which a globular protein differs from DNA. [6]<br />
<br />
Structural Differences<br />
<br />
1a. a linear, single chain made up of amino acids <br />
1b. double stranded, made up of nucleotides<br />
<br />
2a. made up of monomers that consist of up to 20 amino acids <br />
2b. only 4 bases in DNA<br />
<br />
3a. ref to details of amino acid structure<br />
3b. ref. to details of nucleotide structure<br />
<br />
4a. amino acids held together by peptide linkages<br />
4b. nucleotides held together by phosphodiester links <br />
<br />
5a. conformation of a folded tertiary structure<br />
5b. double helix of DNA; reference to association with histone proteins<br />
<br />
6a. may contain contain sulphur<br />
6b. DNA contain phosphorus<br />
<br />
7 suitable reference to size, quoting appropriate units / location within the cell where they are synthesized or replicated / appropriate named function <br />
<br />
<br />
(b) Explain how the structure of collagen and haemoglobin are related to their function. [8]<br />
Collagen<br />
1. reference to collagen as a fibrous protein / quaternary structure: made up of a staggered array of tropocollagen molecules, each consisting of 3 polypeptide chains tightly twisted and wound together into a triple helix, held by H-bonds hydrogen bonds between glycine and proline or hydroxyproline residues in different polypeptides<br />
<br />
2. Each polypeptide has a regular, repetitive sequence of amino acids, often follows the pattern Gly-Pro-X or Gly-X-Hyp, where almost every third amino acid sequence is a glycine with high proportion of proline / hydroxyproline residues and where X may be any of various other amino acid residues<br />
<br />
3. glycine is the smallest amino acid and this together with proline allow the polypeptide chain to be wound into a tightly coiled, straight and unbranched (left-handed) helix that has a rigid kinked conformation, allowing the 3 helices in the tropocollgen to be tightly twisted into a triple helix <br />
<br />
4. the triple-stranded tropocollagen molecules run parallel one another and disulfide cross-linkages between the R-groups of the amino acid lysine holds the molecules together, forming long parallel fibres<br />
<br />
5. due to H-bonds between the 3 polypeptides, the tight winding of 3 polypeptides together to form a triple helix and presence of strong covalent bonds between tropocollagen molecules - make collage a very stable structure with high tensile strength, hence its important structural roles in connective tissues, bones and as support in tendons, ligaments for attachment of muscle to bone<br />
Haemoglobin<br />
1. reference to haemoglobin as a globular protein / quaternary structure: made up of 2 and 2 globin subunits, which are interlocked by hydrophobic and ionic interactions<br />
<br />
2. There is a highly irregular sequences of amino acids in the polypeptide chains; each polypeptide chain undergo extensive folding to form a compact, spherical shaped protein due to hydrophobic interactions between the (hydrophobic) R groups; all 4 polypeptide chains are linked to form a roughly globular molecule <br />
<br />
3. Being globular, haemoglobin is soluble in water to form a colloidal suspension; hydrophilic amino acid residues directed outwards on the outer surface, where they form hydrogen bonds with water<br />
<br />
4. Each subunit / polypeptide is attached to a Fe2+-containing haem group, hence is a conjugated protein; each haem groups binds to 1 moelcule of O2 so each haemoglobin molecules can bind up to a maximum of 4 O2 molecules<br />
<br />
5. the haem group binds reversibly to oxygen to enable oxygen to be taken up and released readily - at high O2 tension O2 binds as molecular oxygen to the Fe of the haem group; at low O2 tension, O2 is released; important in the metabolic role of Hb in red blood cells as an O2 carrier – able to bind with and pick up oxygen in O2-rich environment e.g. alveoli in the lungs and releasing O2 in O2 -low environment e.g. to actively respiring cells<br />
<br />
6. reference to co-operative binding allosteric effect; where the binding of 1st O2 molecule causes a change in the conformational state of the haemoglobin molecule / a change in the position of the haem groups, increases the affinity of the haemoglobin for oxygen and hence facilitates the binding of the 2nd O2 molelcule and so on. (affinity of haemoglobin for the 4th O2 molecule is approximately 300 times that for the 1st)<br />
<br />
Generalised / motherhood statement such as protein made up of long chains of amino acids linked together by peptide bonds – to be credited only once. No credits for non-specific reference to other bonds such as hydrogen bonds, disulphide linkages, hydrophobic and hydrophilic interactions, ionic bonds being involved.<br />
<br />
(c) Describe, with examples, the roles of proteins in membranes. [6]<br />
<br />
With appropriate reference to integral / peripheral / transmembrane proteins; <br />
<br />
1. As membrane-bound enzymes – involved in metabolic functions e.g. ATPase phosphorylates ADP to ATP using energy released as H+ ions follow down a concentration gradient / cytochrome oxidase in ETC (or any appropriate example) <br />
<br />
2. As electrons acceptors / carriers of the ETC on inner mitochondrial membrane / thylakoid membrane in chloroplast, using the energy released during the electron transfer as electrons travel downhill in energy terms from one carrier to another, to generate a proton gradient across membrane<br />
<br />
3. As surface receptors for binding of ligands / neurotransmitters / hormones e.g. G-protein linked receptors for glucagon / tyrosine kinase receptors on muscle cells for insulin for cell-to-cell signaling and signal transduction / ionotropic receptors that are linked to (ion) channel proteins e.g. acetylcholine receptors associated with Na+ ion channels on postsynaptic membrane for acetylcholine for nervous transmission across synapse <br />
<br />
As transport proteins that are either specific or non-specific, transporting ions / molecules / solutes via passive facilitated diffusion / active transport <br />
<br />
4. As channel proteins, with specific hydrophilic domains / channels to allow passage of ions and hydrophilic / charged, polar molecules that cannot transverse the hydrophobic core of the phospholipids bilayer, moving down a concentration gradient / as non-gated channel – providing an open passage way for movement e.g. auaporins / as gated channel can open / close to allow cells to regulate the movement, e.g. voltage-gated Na+ channels, voltage-gate Ca2+ ion channels <br />
<br />
5. As carrier proteins, allowing only specific molecules to enter/exit; bind the solute on one side of the membrane which then causing the protein to undergo a conformational change that brings the solute to the opposite side of the membrane<br />
e.g. Na+-K+ pump – a type of protein carrier that usually use ATP to move solutes against a concentration gradient (from a low solute concentration to a high solute concentration); e.g. glucose transporter GLUT1, found across membranes of red blood cells for facilitated diffusion of glucose<br />
<br />
6. Glycoproteins as surface markers for cell-to-cell recognition to identify cell type – significant in immune response/transplant / for attachment purposes: as in cell-to-cell adhesion to allow aggregation and association of ‘like’ cells into tissues; as anchors on the cytoplasmic surface for attachment to cytoskeleton<br />
<br />
<br />
AJC 2010<br />
<br />
8 (a) Describe how the structures of RNA are adapted for translation. [8]<br />
<br />
Messenger RNA (mRNA)<br />
• Single-stranded;<br />
• One codon codes for one amino acid;<br />
• mRNA (5’UTR) has ribosome attachment site;<br />
• Codons are non-overlapping<br />
• Has spliceosome recognition sites to allow for excising of introns;<br />
• Mature mRNA consists of exons/ no introns;<br />
• Has START/ INITIATOR (AUG) codon and STOP (UAA, UGA, UAG) codons;<br />
• 5’ 7-methyl guanosine cap and 3’ poly-A tail to ensure mRNA stability;<br />
• Complementary base pairing between codon and anticodons; (award only once in either mRNA or tRNA discussion)<br />
(4 marks max)<br />
<br />
Transfer RNA (tRNA)<br />
• Has (3’ CCA) end to attach to amino acid/ has amino acid attachment site;<br />
• At least 20 different tRNAs - one for each amino acid<br />
• Has anticodons;<br />
• Has shape complementary to amino acyl tRNA synthetase for activation of amino acid;<br />
• Internal hydrogen bonds to form clover leaf shape/ have specific 3D configuration/ stabilize the molecule;<br />
(2 marks max)<br />
<br />
Ribosomal RNA (rRNA)<br />
• A structural component of ribosome;<br />
• Due to hydrogen bonding/ complementary base pairing within;<br />
• Gives rise to a variety of ribosomal types (e.g. 70S, 80S, 50S, 30S, etc);<br />
• Involved in peptidyl transferase activity;<br />
(2 marks max)<br />
<br />
<br />
<br />
<br />
<br />
<br />
(b) Distinguish between chromosomal and gene mutations. [8]<br />
<br />
Gene Mutation Chromosomal Mutation/ Aberration<br />
Definition • Change in structure of DNA/ change in (nucleotide) base sequence/ different types of gene mutations including point, missense, nonsense, frameshift, silent • Change in structure or number of chromosomes<br />
Gene locus • Involves one gene locus • Involves a few gene loci<br />
Protein products • May or may not affect protein products (depending on whether silent mutation or in non-coding DNA regions). • Usually affect many gene products since so many gene loci involved.<br />
No of chromosome involved • Within one chromosome • Involves one or a few chromosomes<br />
Mechanisms • Brought about by deletion, insertion, inversion or substitution of one or more nucleotides • Brought about by deletion, duplication, inversion or translocation of several gene loci on a chromosome<br />
Effects on allele • Gives rise to a new allele, resulting in new protein which may have a novel function<br />
• Does not reshuffle alleles • No new allele arisen;<br />
• Only results in reshuffling of alleles on a chromosome.<br />
Ploidy • No change in ploidy • Could also result in polypoidy or aneuploidy<br />
Frequency • More frequent because genes outnumber chromosomes by several thousand to one • Less frequent<br />
Importance • Of evolutionary importance because acquisition of new alleles increase gene pool for natural selection • Of lower evolutionary importance because it only reshuffles alleles already existing in the gene pool<br />
Visibility under light microscope • Not visible • Usually visible<br />
Examples • Sickle cell anemia (due to substitution of one nucleotide) • Down’s syndrome (due to an extra chromosome 21)<br />
Any valid comparisons;<br />
<br />
(c) Explain how the Meselsohn and Stahl experiment supports the semi-conservative model of DNA replication. [4]<br />
<br />
Reference to experimental details <br />
• DNA labeled with heavy nitrogen / 15N, then cells transferred to light nitrogen/ 14N for two generations;<br />
• Reference to the three different bands after (CsCl) centrifuge; accept labelled diagram;<br />
• Parental strands unzip/ hydrogen bonds broken;<br />
(1-2 marks max)<br />
<br />
How experiment supports <br />
• Parental strands act as template;<br />
• Complementary base pairing;<br />
• New DNA molecule consists of one parental strand and one daughter strand;<br />
• Daughter DNA are genetically identical to each other;<br />
(2-3 marks max)<br />
<br />
9(a) Discuss how signal amplification is illustrated by the effect of hormone on glycogenolysis.[8] <br />
<br />
• Binding of one / a molecule of epinephrine / glucagon to GPCR causeds GPRC will undergo a conformation change ;; <br />
<br />
• to allow the activation of several / few G-protein by displacing GDP for GTP ;; <br />
<br />
<br />
• Each activated G-protein activate enzyme adenylyl cyclase, each of which is able to catalyse the conversion of large number of ATP to cAMP ;; hence amplifying the signal. <br />
[R: increase cAMP, activation of cAMP] <br />
<br />
• These cAMP in turn binds and activates large number of protein kinase A ;;<br />
<br />
• Each activated protein kinase A will initiate a sequential phosphorylation and activation of kinases / phosphorylation cascade ;;<br />
<br />
• At each phosphorylation step/cascade, each activated kinase is able to activate a large number of the next relay molecule/kinase ;; hence the signal is further amplified. <br />
<br />
• lead to the activation of large number of glycogen phosphorylase<br />
<br />
• each will catalyse for the breakdown of large amount of glycogen into glucose. <br />
<br />
<br />
• the number of activated product is always greater than those in the preceding step as one move down the cascade ;; <br />
• binding of 1 glucagon to receptors will lead to the hydrolysis of large number of glycogen;; <br />
<br />
* If less than 3 of the highlighted points are present, 1 mark will be deducted from the overall marks. <br />
<br />
9 (b) Describe the similarities between the interaction of a substrate with an enzyme and the interaction of a ligand with a receptor. [6] <br />
<br />
• Both bind to specific regions of protein (idea of specific region/portion is impt); <br />
• Active site for enzyme; <br />
• Binding site for receptor; <br />
• Both are complementary in shape to sites that they bind to; <br />
• Both will bind to protein via H bonds, ionic bonds, hydrophobic interactions; <br />
• Both can induce a conformational change in the protein when they bind; <br />
• Ref. to induced fit hypothesis for enzymes; <br />
• Ref. to activation of receptor; <br />
• Both interactions are not permanent; <br />
• Ligand dissociates from binding site of receptor / Ligand-receptor complex are removed; <br />
• ES complex formed will be converted to product, which is released; <br />
<br />
<br />
9(c) Compare between nervous and hormonal control. [6] <br />
Similarities: <br />
Both serve as means of communication/ allow living organisms to respond to stimulus <br />
Stimulus transmission of message effector / stimulus elicit a response <br />
<br />
Differences: <br />
Endocrine Nervous system<br />
Nature of transmission Chemical transmission (hormones) through blood system Electrical and chemical transmission (nerve impulses and chemical across synapses)mil and electrical<br />
Transmission speed Slow <br />
Slower transmission and relatively slow-acting (adrenaline an exception) Fast <br />
Rapid transmission and response<br />
Pathway of transmission Specific (via nerve cells ) Not specific (blood around whole body) <br />
Strength of message Strength dependent on amount of hormones Impulse size same regardless of stimulus <br />
Length of response Often long-term changes (except adrenaline) Often short-term changes nothing action <br />
Target Response may be very widespread, e.g. growth et organs at different parts of body Response often very localized, e.g. one muscle target organ <br />
<br />
IJC 2010<br />
9 (a) Describe the structure of DNA and its organisation in a eukaryotic chromosome. [8]<br />
<br />
1. 2 polymers of deoxyribonucleotides chains twisted into a double helix;;<br />
2. held by H- bonds between complementary bases, A=T, C≡G;;<br />
3. a. purine-pyrimidine: 2 nm diameter;<br />
3. b. 0.34nm/10 bp per complete turn;;<br />
4. distinct polarity 3’ end with free hydroxyl group at 3C and 5’ end with phosphate group at 5C;;<br />
5. DNA molecule associated 1¾ turns with 8 histone proteins/octomer, formed nucleosome core;;<br />
6. histones positively charged, formed ionic bonds with negatively charged DNA;;<br />
7. associate with H1 histone to form nucleosome, linked by spacer/linker DNA forming 10nm ‘beads-on-string’ structure;;<br />
8. further coiled into a solenoid forming a 30nm fibre;;<br />
9. futher looping in presence of scaffold proteins to form metaphase chromosomes;;<br />
AVP;;<br />
<br />
(b) Describe the control of gene expression in eukaryotes. [12]<br />
Gene Amplification<br />
1. duplication of genes to increase the number of copies;;<br />
Chromatin remodelling<br />
2. acetylation removes +ve charges from histones, loosen –ve charged DNA, allow GTF to bind;;<br />
3. demethylation removes methyl grps from proximal promoters, e.g. CpG islands;;<br />
Transcriptional control<br />
4. cis-acting elements, GTFs binding to promoters and proximal promoters;;<br />
5. binding of STF e.g. activators to enhancer, repressors to silencers á and â transcription rate;;<br />
Post-transcriptional control<br />
6. ss pre-mRNA, with 5’ capping with modified guanine/methylated guanine;;<br />
7. 3’ polyadenylation for stability/ nuclear export of mature mRNA;;<br />
8. splicing of introns out and joining of exons forms mature mRNA by spliceosome;;<br />
9. association with proteins for nuclear export;;<br />
10. mRNA editing, the ∆ in codon results in different functional polypeptide;;<br />
Translational control<br />
11. stability of mRNA in the cytoplasm controlled by degradation in response to presence of RNA sequence signalling degradation, concentration of the translated product, extracellular signals like hormones;;<br />
12. gene silencing by miRNA, forming complementary dsRNA in DICER cplx prevent ribosomal attachment/ cleave mRNA;;<br />
Post-translational control<br />
13. proteins not required are tagged for ubiquination by protease;;<br />
14. AVP;;<br />
[Total: 20]<br />
<br />
10 (a) Explain the principles of homeostasis. [4]<br />
<br />
1. Homeostasis is the maintenance of a constant, stable internal environment within an organism regardless of changes in external environment via a self-regulating mechanism;;<br />
2. The regulated variable is usually at a set point which can be changed by a stimulus;;<br />
3. change detect by a receptor which sends signals to control center/regulator;;<br />
4. activates effector resulting in a response to restore set point via positive/negative feedback;;<br />
<br />
(b) Describe the roles of insulin and glucagon in regulation of blood glucose. [10]<br />
<br />
Insulin<br />
1. [bld glu] á above norm, β cells in islets of Langerhans detect the á;;<br />
2. insulin released to bind to rtk receptors of effectors e.g. liver cells;;<br />
3. results in increased no. of glucose transporters in the liver cells to increase uptake of glucose;;<br />
4. convert excess glucose to glycogen in glycogenesis by activating glycogen synthetase;;<br />
5. Depresses rate of synthesis of glucose from non-carbohydrate source via gluconeogenesis;;<br />
Glucagon<br />
6. [bld glu] â below nom, α cells in islets of Langerhans detect the drop;;<br />
7. glucagons released bind to GPCR on effector e.g. liver cells;;<br />
8. results in activation of glycogen phosphorylase to convert glycogen into glucose;;<br />
9. reached set point, neg feedback mechanism prevents further release of hormones by pancreas;;<br />
10. AVP<br />
(c) Describe the molecular events when glucagon binds to its target cell. [6]<br />
<br />
1. Glucagon binds to the complementary ligand-binding site on GPCR receptor causing a conformational change to receptor;;<br />
2. GTP from the cytosol then displaces GDP from the nucleotide-binding site on Ga;;<br />
3. Ga simultaneously dissociates from its Gβγ, translocate along the plasma membrane to activate the enzyme adenylyl cyclase;;<br />
4. Adenylyl cyclase catalyses the conversion of ATP to cyclic AMP (cAMP);;<br />
5. cAMP activate relay protein, protein kinase A (PKA);;<br />
6. Activated PKA then phosphorylates other relay proteins and enzymes, activates other enzymes in phosphorylation cascade;;<br />
7. The effect of cAMP is removed by phosphodiesterase which converts cAMP back to AMP;;<br />
8. AVP;;<br />
[Total: 20]<br />
<br />
JJC 2010<br />
8 (a) Describe the process of ATP production in the chloroplast. [7]<br />
<br />
1. Photosynthetic pigments such as chlorophyll a, chlorophyll b and carotenoids; <br />
2. are embedded on the thylakoid membranes;<br />
3. are arranged in photosystems;<br />
4. Two photosystems PSI (P700) and PSII (P680) which absorbs wavelengths of 700nm and 680nm respectively;<br />
5. Photon of light absorbed by both PS will excite electrons in special chlorophyll a in reaction centres to higher energy levels;<br />
6. These excited electrons are accepted by primary electron carriers; <br />
7. And pass on to electron carriers on the electron transport chain (ETC);<br />
8. Electron carriers on the ETC are of progressively lowered energy level;<br />
9. As electrons move down the ETC, energy is released;<br />
10. The energy released is used to pump H+;<br />
11. From the stroma into the thylakoid lumen;<br />
12. Accumulation of H+ generate potential energy;<br />
13. H+ return to the stroma by diffusing down their concentration gradient <br />
14. through H+ channel;<br />
15. Energy is released and used to couple the phosphorylation of ADP + Pi to produce ATP;<br />
16. Catalyzed by ATPase;<br />
Max 7 <br />
<br />
(b) Give an overview of the Krebs cycle and its significance in respiration. [7]<br />
Overview<br />
1. Occurs in matrix of the mitochondrion;;<br />
2. Per cycle, one molecule of acetyl CoA (2C) undergo oxidative decarboxylation;;<br />
OR<br />
Citrate (6C) undergoes oxidative decarboxylation to form α–ketoglutarate (5C);<br />
α–ketoglutarate (5C) undergoes oxidative decarboxylation to form succinate (4C);<br />
3. Acetyl CoA (2C) and oxaloacetate (4C) are condensed to form citrate (6C);;<br />
4. Description of the decarboxylation process i.e. 6C to 5C to 4C intermediates;; <br />
i. Citrate (6C) undergoes oxidative decarboxylation to form α–ketoglutarate (5C)<br />
ii. α–ketoglutarate (5C) undergoes oxidative decarboxylation to form succinate (4C)<br />
iii. Succinate (4C) undergoes oxidation to form fumarate (4C)<br />
iv. Fumarate (4C) is converted to malate (4C)<br />
v. Malate (4C) undergoes oxidation to form oxaloacetate (4C)<br />
5. Two molecules of CO2, three molecules of NADH, one molecule of FADH2 and one molecule of ATP through substrate level phosphorylation are released per cycle;;<br />
6. At the end of the cycle, one molecule of oxaloacetate (4C) is regenerated to receive more acetyl groups and the cycle continues;;<br />
Max 5 marks<br />
<br />
Significance<br />
7. Completes the oxidative breakdown of glucose to CO2 and H2O and release sufficient energy;;<br />
8. Release of hydrogens (NADH produced) which can be used in oxidative phosphorylation to provide energy to produce ATP;;<br />
9. Carbohydrate intermediates can be converted to amino acids;;<br />
10. Permits amino acids to enter Krebs cycle when glucose is in short supply;;<br />
Max 3 marks<br />
<br />
(c) Compare and contrast oxidative phosphorylation and photophosphorylation. [6] <br />
Similarities:<br />
1. involve transfer of electrons between electron carriers;;<br />
2. energy released from electron transport is used to generate proton gradient;;<br />
3. potential energy of proton gradient is harnessed for ATP synthesis;;<br />
<br />
Differences:<br />
Features Photophosphorylation Oxidative Phosphorylation<br />
Sources of energy for ATP synthesis energy for making ATP comes from light energy for making ATP comes from glucose oxidation processes;;<br />
Location<br />
Thylakoid membrane of chloroplast Inner membrane of mitochondria;;<br />
Involvement of light energy Required to energize the electrons in special chl a Not required;;<br />
Electron donors<br />
For non-cyclic reaction: water<br />
For cyclic reaction: PS I NADH, FADH2;;<br />
<br />
Electron acceptors<br />
For non-cyclic reaction: NADP+<br />
For cyclic reaction: PS I Oxygen;;<br />
<br />
Establishment of proton gradient<br />
H+ pumped inwards, from stroma to thylakoid space<br />
H+ pumped outwards, from mitochondrial matrix to intermembrane space;;<br />
Max 6<br />
<br />
9 (a) Describe the important events that occur during Meiosis I and explain the significance of each event. [10] <br />
Important events during Meiosis I Significance<br />
1. Prophase I; <br />
2. Synapsis ; <br />
3. pairing of homologous chromosomes; <br />
4. to form tetrads / bivalents ; 5. Allows for alignment of genes on chromosomes ; <br />
6. to prepare for crossing over between the homologous regions later ;<br />
7. Crossing over ; <br />
8. between non-sister chromatids of homologous chromosomes ; formation of chiasmata ; 9. Allows for exchange of genetic material ; <br />
10. results in new combination of alleles ; <br />
11. genetic variation in the gametes later ; <br />
12. Metaphase I; <br />
13. Homologous chromosomes are arranged along equator in pairs ; <br />
14. alignment of each homologous pair is independent of other homologous pairs / independent assortment ; 15. Allows gametes to contain a random combination of paternal and maternal chromosomes ; <br />
16. genetic variation in gametes later ;<br />
17. Anaphase I; <br />
18. Homologous chromosomes are pulled apart to opposite poles ; <br />
19. movement of each homologous pair is independent of other homologous pairs / random segregation ; 20. Allows gametes to contain a haploid set of chromosomes ; <br />
21. for restoration of diploid no. of chromosomes ; following fertilization ;<br />
22. Allows gametes to contain a random combination of paternal and maternal chromosomes ; <br />
23. genetic variation in gametes later ;<br />
24. Telophase I & cytokinesis; <br />
25. Division of cytoplasm (and organelles) ; <br />
26. furrowing (animal cells) / cell-plate formation (plant cells) ; 27. Allows formation of 2 haploid daughter cells ;<br />
28. Preparation of cells for Meiosis II ;<br />
<br />
(b) Describe the significance of gene amplification. [6]<br />
<br />
1. Gene amplification refers to the process of (selectively) increasing the number of copies of a particular gene (located at a particular region of the chromosome), without a proportional increase in other genes;;<br />
2. These amplified genes can be transcribed and translated OR leads to an overproduction of the corresponding mRNAs and proteins;;<br />
<br />
3. Meet the needs of cells resulting in higher level of mRNA and polypeptide synthesis at different development stages of cells;; <br />
4. E.g. In the developing ova of eukaryotes, million or more additional copies of rRNA genes are synthesized. This allows the ova to make enormous numbers of ribosomes resulting in a burst of protein synthesis once the ova are fertilized;;<br />
5. Cause cancer due to over-expression of proteins (leading to development of malignant tumours);;<br />
6. If an oncogene is amplified, then the resulting over-expression of that gene can lead to de-regulated cell growth;;<br />
7. An example is the amplification of the ErbB-2 oncogene in breast and ovarian cancers;;<br />
<br />
8. Genes related to drug resistance in these cancer cells are increased; conferring them with drug resistance;<br />
9. and the ability to prevent absorption of therapeutic drugs; thus drug-resistant tumors can continue to grow and spread even in the presence of chemotherapy drugs;<br />
10. Confer selective advantage which allow organisms to survive in a particular environment;; Max 6<br />
(c) Outline the end-replication problem in eukaryotic chromosomes. [4]<br />
<br />
1. DNA polymerase;<br />
2. can only add nucleotides to the 3′ end;<br />
3. of a elongating/pre-existing polynucleotide/primer;<br />
4. primer at 5’ end; <br />
5. removed;<br />
6. but cannot be replaced with DNA/ gap cannot be filled in;<br />
7. because no 3’-OH available (to which a DNA nucleotide can be added);<br />
8. 5’ end of the DNA becomes shorter relative to that of the previous generation;<br />
<br />
<br />
YJC 2010<br />
8<br />
(a) Describe how the presence of lactose results in the transcription of the Lac operon in prokaryotes. <br />
[8]<br />
Correct diagram of the Lac operon (in particular, the position of the regulatory and structural genes, as well as promoter and operator);<br />
Description of negative inducible mechanism, whereby the operon is regulated via a repressor (negative), and presence of lactose induces the transcription of the operon (normally not transcribed - inducible)<br />
Trace amounts of lactose permease (minimally transcribed / expressed) on cell surface membrane of prokaryote permits the entry of lactose; <br />
Trace amounts of β-galactosidase (minimally transcribed / expressed) converts lactose to allolactose as a side reaction;<br />
The Lac repressor is a protein that is transcribed and translated from LacI gene (regulatory gene) located in front of the promoter site;<br />
The repressor has two binding sites – DNA binding site and repressor binding site (referred to as allosteric site);<br />
Allolactose binds to the Lac repressor bound to the operator of the Lac operon;<br />
Causes a conformational change in the repressor, preventing it from being attached to the operator site (releasing it from the DNA binding site);<br />
Operator site overlaps with promoter site;<br />
RNA polymerase binds to the promoter site and transcribes the structural genes;<br />
LacZ is transcribed and translated to produce β-galactosidase, LacY produces lactose permease, and LacA produces transacetylase;<br />
Transcription (and translation) continues until all/most of the lactose are digested to produce glucose and galactose, and repressor binds to the operator site again;<br />
Accurate reference to glucose and CAP-binding site: 1 mark max;<br />
<br />
<br />
(b) Compare and contrast the structure and organisation of prokaryotic and eukaryotic chromosomes. <br />
[6]<br />
Similarities (max 3)<br />
They both have origins of replication that allow for them to be replicated;<br />
They consist essentially of the same types of nucleotides – adenine, thymine, cytosine and guanine;<br />
They are composed of DNA double helix structure;<br />
Similar kinds of bonds can be found in the chromosomes – phosphodiester, hydrogen, glycosidic;<br />
Both are compacted and associated with proteins that help to pack them;<br />
Differences (max 5)<br />
Prokaryotic chromosomes are circular while eukaryotic chromosomes are linear;<br />
Prokaryotic chromosomes are usually singular and carry the entire genome while usually several eukaryotic chromosomes make up the entire genome, with each chromosome carrying a small proportion of the eukaryotic genome;<br />
Eukaryotic chromosomes have multiple origins of replication while prokaryotic chromosomes have only one;<br />
The structural genes of prokaryotic chromosomes are organised in operons, where several genes are under the control of a single promoter, while the structural genes of eukaryotic chromosomes are each under the control of a single promoter;<br />
Eukaryotic chromosomes contain a high proportion of non-coding DNA such as introns, centromeres, telomeres, and other intergenic regions such as satellite DNA, which are almost absent in prokaryotic chromosomes;<br />
Eukaryotic chromosomes are made highly compact with the involvement of many histones while comparatively, prokaryotic chromosomes are loosely packed with the involvement of some histone-like proteins;<br />
Eukaryotic chromosomes are bounded by nuclear envelope whilst prokaryotic chromosomes are not enclosed by the envelope (found in the nucleoid region) and are usually attached / anchored to cell wall;<br />
<br />
<br />
<br />
(c) Explain how mutations related to a named tumour suppressor gene lead to cancer. <br />
[6]<br />
<br />
E.g. of named tumour suppressor gene: P53;<br />
Normal functions of tumour suppressor gene:<br />
Activate DNA repair;<br />
Inducing mitotic arrest (stopping of cell cycle);<br />
Initiate apoptosis (programmed cell death);<br />
Types of mutations causing loss of function<br />
Gene mutation (any named kind);<br />
Mutation in regulatory sequence;<br />
Both copies of alleles to be mutated in order for the cancer phenotype to be expressed (loss of function);<br />
Description of cell activity resulting from loss of function<br />
DNA damage not repaired for proteins regulating cell cycle leads to cell division;<br />
Checkpoints of mitosis not regulated resulting in continuous cell division;<br />
Cancer cells do not undergo apoptosis and carries on dividing;<br />
<br />
[20]<br />
<br />
<br />
9 (a) Explain the main ways in which a globular protein differs from a sequence of DNA. <br />
[8]<br />
<br />
Pt of comparison Globular proteins DNA<br />
1 Shape Compact & round <br />
Reject Globular Linear in eukaryotic cells<br />
<br />
Note that prokaryotic cells have circular DNA. <br />
2 Solubility in water Yes No/less soluble;<br />
3 Monomers Amino acids deoxyribonucleotides;<br />
4 Types of monomers 20 used in humans 4 types (A,T,G,C);<br />
5 Bonding (1) Linked by peptide bonds Linked by phosphodiester bonds;<br />
6 Number of subunits One or more Always 2 under physiological conditions;<br />
7 Bonding (2) Polypeptide chains/subunits linked by ionic bonds, covalent disulphide bonds, hydrophobic interactions and hydrogen bonding Only hydrogen bonding links the two DNA chains;<br />
8 Function Multiple functions: Enzymatic, structural, transport, energy source (any 2) Carry hereditary information/code for synthesizing products;<br />
9 What confers functionality Its function is usually linked to its 3D shape/conformation and configuration Its function is dependent only on the order in which the monomers are arranged in DNA chain;<br />
10 Levels of organization 4 levels 1 structure/a-helix;<br />
11 Charge Can be of any charge (positive, negative or neutral) Negatively charged;<br />
12 Location Everywhere/both inside and outside the cell Only inside the cell /in nucleus; <br />
<br />
13 Process of synthesis transcription and translation DNA replication<br />
14 Template for synthesis mRNA DNA /parent DNA<br />
<br />
<br />
<br />
<br />
(b) Outline the main differences in structure and function between starch, cellulose and glycogen. <br />
[6]<br />
Similarities <br />
All three are polymers made up of glucose molecules<br />
<br />
Differences<br />
Starch glycogen Cellulose <br />
Types of bonds and significance to function <br />
Amylose is a planar polymer of glucose linked mainly by α(1→4) glycosidic bonds<br />
<br />
Amylopectin is a highly branched polymer of glucose. Glucose units are linked in a linear way with α(1→4) glycosidic bonds. Branching takes place with α(1→6) glycosidic bonds occurring every 24 to 30 glucose units.<br />
<br />
The bonds can be broken down by hydrolysis easily- suitable as a energy store <br />
glycogen is highly similair to amylopectin, it is a highly branched polymer of glucose. Glucose units are linked in a linear way with α(1→4) glycosidic bonds. Branching takes place with α(1→6) glycosidic bonds occurring every 24 to 30 glucose units<br />
<br />
The bonds can be broken down by hydrolysis easily- suitable as a energy store Cellulose is a polysaccharide consisting of a linear chain of several hundred to over ten thousand β(1→4) glycosidic links.<br />
<br />
<br />
<br />
<br />
<br />
<br />
<br />
<br />
The bonds cannot be broken down by hydrolysis easily- suitable as structural carbohydrates.<br />
<br />
Packing and relation to function Coil helically with multiple branching into a compact structure<br />
<br />
thus can be stored in large amounts within a fixed volume. Form fibrils / fibers, <br />
<br />
<br />
thus providing tensile strength for cell wall.<br />
<br />
<br />
<br />
<br />
Association with water and relation to function The hydroxyl groups (-OH) on the glucose residues on amylopectin are used to form hydrogen bonds with adjacent chains.<br />
<br />
Amylose forms a colloidal dispersion in hot water (which helps to thicken gravies) whereas amylopectin is completely insoluble. <br />
<br />
Insoluble in water, thus does not affect the water potential of cells. <br />
Highly branched structure (more so than starch) increases its solubility in water and allows for the rapid synthesis and degradation of glycogen. Due to position of the hydroxyl groups (-OH) on the glucose residues, each successive glucose residue in the chain is rotated at 180° to allow the formation of β(1→4) .<br />
<br />
<br />
Large intermolecular spaces, thus allow water to move through the cell wall<br />
<br />
<br />
<br />
<br />
<br />
PJC 2010<br />
8 (a) Compare the structure and organization between prokaryotic and eukaryotic genome. [8]<br />
(b) Describe the eukaryotic processing of pre-mRNA. [6]<br />
(c) Describe various ways in which gene expression may be controlled at translational and post-translational level. <br />
[6]<br />
9 (a) Describe the structure of DNA and tRNA. [6]<br />
(b) Describe the process of DNA replication in prokaryotes. [8]<br />
(c) Describe the experimental evidence for semi-conservative replication. [6]<br />
<br />
Compare the structure and organization between prokaryotic and eukaryotic genome.<br />
Similarities:<br />
1. Both consist of double stranded DNA;<br />
2. Both shows different levels of packing to compact the genome such that they can fit into the cell; <br />
<br />
Differences between prokaryotic and eukaryotic genome<br />
Prokaryote Eukaryote<br />
Genome description Circular/closed loop linear chromosome; <br />
Size Shorter. (Contains a few million bp.) Typically longer chromosome. (Contains tens of millions to hundreds of millions of base pairs (bp) in length.)<br />
Chromosome location Chromosome not in nucleus, found in nucleoid region, (in contact with cytoplasm); Chromosome found in nucleus; <br />
Packing of DNA (2 marks) Packing also take place but not as extensive.<br />
The chromosomal DNA must be compacted about 1000-fold in order to fit into the bacterial cell.<br />
There are two main ways:<br />
1. Formation of looped domain – a segment of DNA that is folded into a loop-like structure.<br />
DNA-binding proteins are involved in holding the loops in place.<br />
AND<br />
2. Supercoiling of the looped DNA.<br />
<br />
Multi-levels of packing. <br />
<br />
<br />
<br />
Basic level “Beads-on-a-string” model where DNA wind around histone proteins to form nucleosomes.<br />
<br />
<br />
Nucleosomes undergo further packing to form 30 nm fibre/solenoid 300 nm fibre <br />
<br />
Introns Introns not found within genes.<br />
(mRNA formed during transcription need not be processed. They can be translated immediately.) Introns (non-coding, repetitive sequences) found interspersed within a gene.<br />
(This arise the need for post-transcriptional modification RNA splicing.)<br />
No of ori 1 origin of replication on 1 chromosome; Each chromosome consists of several origin of replication;<br />
Location of repetitive sequences Repetitive sequences interspersed throughout the chromosome; Repetitive sequences are commonly found near; centromeric and telomeric regions but may also be throughout the chromosome<br />
Genes organization Coordinately controlled genes are often clustered into an operon. ; (A transcriptional unit consisting of coordinately regulated clusters of genes with related functions.) Coordinately controlled genes are organized not in clusters but interspersed throughout the genome, each with own promoter.;<br />
plasmid An extrachromosomal plasmid is usually found in prokaryotes. Absence of plasmid<br />
Differences 7 max<br />
<br />
1. Post- transcription modification is required to process the pre-mRNA to form mature mRNA and to facilitate the export from nucleus to cytoplasm.<br />
2. The 5’ end, the end made first during transcription, is immediately capped off with a modified form of a guanine nucleotide; <br />
3. The enzymes required for capping is grouped to form the Capping Enzyme Complex (CEC) and;<br />
4. is bound to the RNA polymerase II before transcription starts; <br />
5. At the 3’ end, where there is polyadenylation sequence present <br />
6. an enzyme, poly (A) polymerase (PAP);<br />
7. makes a poly(A) tail consisting of 50 to 250 adenine nucleotides;<br />
8. The signals for RNA splicing are short nucleotide sequences at the ends of introns;<br />
9. Particles called small nuclear ribonucleoproteins or snRNPs, recognize these splice sites.<br />
10. snRNPs are located in the cell nucleus and are composed of RNA and protein molecules;<br />
11. The RNA in a snRNP particle is called a small nuclear RNA (snRNA), each molecule is about 150 nucleotides long;<br />
12. Several different snRNPs join with additional proteins to form an even larger assembly called a spliceosome;<br />
13. The spliceosome interacts with the splice sites at the ends of an intron, intron region was looped out <br />
14. It cuts at specific points to release the intron, <br />
15. then immediately joins together two exons that flanked the intron;<br />
16. snRNA plays a role in the catalytic process as well as in spliceosome assembly and splice site recognition;<br />
max 6<br />
<br />
(c) Gene expression may be controlled at translational level via these processes: <br />
Process involved Mechanism Regulation<br />
Gene expression control of during translation initiation 1. Regulatory proteins binds to specific sequence within the 5’ UTR; <br />
2. Prevent attachment of small ribosomal subunit to mRNA; 15. Block translation initiation; <br />
Global control <br />
translation initiation 3. Activation /inactivation of one or more of TIF(translation initiation factors) (by phosphorylation/dephosph);<br />
4. Promote/ Inhibit the formation of translation initiation complex; 16. Translation will be initiated for ALL the mRNA in a cell; <br />
mRNA degradation 5. Enzymes (3’5’ exonucleases) breaks down poly (A) tail first; <br />
6. Followed by enzyme removal of 5’ end (decapping) critical step; <br />
7. Allows nuclease enzymes to rapidly digest the mRNA; 17. Control the lifespan of mRNA in the cytoplasm the amount of proteins formed; <br />
<br />
Gene expression may be controlled at post translational level via these processes: <br />
Process involved Mechanism Regulation<br />
Post translational modification <br />
Protein processing (during translation),<br />
Chemical modification 8. A polypeptide will coil/fold forming function 3D prot with secondary & tertiary structure (cisternae of RER);<br />
9. Cleavage of initial pro-insulin (polypeptide) into active hormones;<br />
10. Chemical modification- adding of sugar/lipids/phosphate group;<br />
11. Cell surface proteins must be transported to target destination to function; 18. Protein processing control the type of proteins to be formed;<br />
Protein degradation <br />
12. To mark a protein for destruction, ubiquitin protein is tagged;<br />
13. Leads to recognition by large proteasomes;<br />
14. Enzymes in the proteasome cut the proteins into small peptides degrade by other enzymes in cytosol; 19. Protein degradation affects the amount of protein present in cell; <br />
<br />
Structure:<br />
1. The DNA molecule consists of two deoxyribonucleotide chains<br />
2. The two chains coil round each other to form a double helix.<br />
3. Each chain consists of a sugar-phosphate backbone, held together by phosphodiester bonds, between the sugar of one nucleotide and the phosphate group of the next nucleotide, <br />
4. with the nitrogenous bases arranged as side groups off the chains (project into the centre of the double helix). <br />
5. The two chains run in opposite directions (antiparallel) and<br />
6. are held together by hydrogen bonds between their nitrogenous bases. <br />
7. Hydrogen bonding can occur only between a purine and a pyrimidine. <br />
8. Thus the sequence of bases in one chain determines that in the other chain. The two chains are said to be complementary. <br />
9. The width between the two chains is constant (2nm) and is equal to the width of one base pair, ie. the width of a purine plus a pyrimidine. <br />
10. Along the axis of the molecule the distance between adjacent nucleotides is 0.34nm <br />
11. One complete turn of the double helix is 3.4 nm and comprises 10 base pairs<br />
<br />
tRNA<br />
Structure<br />
1. Single stranded RNA<br />
2. Wound into clover-leaf shape with 3 loops<br />
3. 3’ end of the tRNA always ends in the base sequence of CCA <br />
4. and binds an amino acid<br />
5. 5’ end of the tRNA always ends in a base, guanine. <br />
6. Acts as an intermediate molecule between the triplet code of mRNA and amino acid sequence of polypeptide chain. <br />
7. Carries the correct amino acid to the forming polypeptide chain on the ribosome.<br />
8. One loop contains a triplet of bases called an anticodon which pairs with the complementary codon on mRNA.<br />
9. There are more than 20 different tRNA molecules in a given cell, each carrying a specific amino acid.<br />
<br />
(b)Unwinding DNA double helix<br />
1. Mechanism is by semi-conservative replication.<br />
2. Activated deoxyribonucleotides and transported into the nucleus via pores in the nuclear envelope<br />
3. Replication of DNA begins at specific sites called the origins of replication (Ori). <br />
4. The enzyme helicase causes the DNA molecule to unwind and unzip/the hydrogen bonds between the bases to break, causing the two DNA chains to separate.<br />
5. Topoisomerase helps to prevent over-straining of the replication fork.<br />
6. The separating DNA results in a “replication bubble”, allowing replication to take place at both ends, known as a replication fork (bidirectional)<br />
7. The unwinding DNA chains provide templates for the synthesis of new DNA chains.<br />
8. After parental strand separation, single-stranded binding protein (SSB) binds to the unpaired DNA strands, stabilizing them.<br />
<br />
<br />
Formation of RNA primer<br />
9. The enzyme primase catalyses the synthesis of a short RNA primer, (about ten ribonucleotides long), <br />
10. Such primers are necessary because the enzyme, DNA polymerase III (DNA pol III), that is responsible for the synthesis of DNA, cannot initiate synthesis of a polynucleotide; they can only add nucleotides to the end of an already existing chain that is base-paired with the template strand.<br />
11. The new DNA strand will start from the 3’end of the RNA primer.<br />
12. Once synthesis of the new DNA chain starts, the RNA primer is hydrolysed (removed) by enzyme DNA polymerase I (DNA pol I). Removal of RNA primer leaves substantial gaps between the DNA fragments. The filling of these gaps by deoxyribonucleotides is also catalysed by DNA pol I<br />
<br />
Synthesis of new DNA strand<br />
13. DNA polymerase III recognises the bases and selects free deoxyribonucleotides that are complementary to those on the template strand.<br />
14. Adenine pairs with thymine and vice versa. Similarly, cytosine pairs with guanine and vice versa.<br />
15. Both daughter strands are synthesized in the 5' to 3' direction. <br />
16. One of the daughter strands, called the leading strand, is synthesized continuously/ towards replication fork<br />
17. To elongate the other strand in a 5’ to 3’ direction, DNA pol III must work along the other template strand in the direction away from the replication fork.<br />
18. The DNA strand is called the lagging strand, synthesised in the form of short fragments called Okazaki fragments<br />
19. The enzyme, DNA ligase catalyses the formation of phosphodiester bonds between two Okazaki fragments. <br />
20. At the end of the replication, both the parental and daughter strands rewind into a double helix.<br />
21. The new DNA molecule consist of 1 parental and 1 newly synthesized daughter strand;<br />
<br />
(c)<br />
1. Semi-conservation replication both DNA strands are template for DNA synthesis of new strands and <br />
2. The new DNA molecule consist of 1 parental and 1 newly synthesized daughter strand;<br />
3. Messelson and Stahl experiment<br />
4. When cultures of bacteria are grown for many generations in a medium containing heavy isotope of nitrogen (15N), all the DNA molecules became labeled with 15N.<br />
5. The cells were then transferred to another medium containing normal nitrogen (14N) and<br />
6. allowed to divide once and therefore the DNA replicates once. <br />
7. Cells were then allowed to divide a second and third time.<br />
8. DNA, extracted from cells of A, B, C and D, was then centrifuged <br />
9. with CsCl2<br />
10. DNA in the centrifuge tubes appeared as narrow bands when examined under ultraviolet light<br />
11. The width of the DNA bands and their positions in the centrifuge tubes reflect the amount of the various types of DNA molecules (or diagram)<br />
12. Tube A contains only heavy DNA because the bacteria are grown for many generations in a medium (A) containing 15N, which is used to make the nitrogenous bases so all the DNA became labeled with 15N<br />
13. Tube B contains only hybrid DNA with one band occupying the intermediate density position<br />
14. Tube C contains both light and hybrid DNA in the ratio 1:1 and <br />
15. in tube D the ratio is 3:1. Both tubes have 2 bands, 1 band on top at the same density position as the light DNA, and the other at the intermediate density position.<br />
(points 12 to 15 – need to explain or annotated diagram)<br />
<br />
<br />
TPJC 2010<br />
QUESTION 8<br />
(a) Explain how mitosis produces genetically identical cells. [6] <br />
1. Replication of DNA during S-phase of interphase<br />
2. To produce genetically identical sister chromatids attached via centromeres<br />
3. Kinetochore microtubules attach to kinetochores on the centromere<br />
4. Exact alignment of sister chromatids on metaphase / equatorial plate<br />
5. Centromeres divide resulting in separation of sister chromatids<br />
6. Sister chromatids, now as chromosomes, move to opposite poles of the cell<br />
7. Two genetically identical cells are produced when the cell membrane furrows in, <br />
8. Definition of genetically identical cells – some number and type of chromosomes; <br />
Max 6<br />
<br />
(b) Describe the role of meiosis in natural selection. [10]<br />
1. Meiosis produces genetically varied gametes<br />
2. Leading to formation of genetically varied organisms due to random fusion of gametes<br />
3. Prophase 1: crossing over between homologous chromosomes result in new combination of alleles<br />
4. Metaphase 1/anaphase 1: independent assortment of chromosomes occur to result in gametes randomly distributed to either poles of the cells<br />
5. Anaphase 1: segregation of homologous chromosomes to both poles will result in the formation of haploid gametes<br />
6. For natural selection to occur, there must be a varied gene pool<br />
7. In the presence of an environmental change, the environment will select the organisms which has selective advantage<br />
8. This organisms will be able to survive, reproduce and pass on their alleles to their offspring <br />
9. Lack of genetic variation may lead to death of all members in the population due to the selection pressure present OR variation allows continuation of the species in the presence of chances in the environment / increase frequency of alleles in the population<br />
10. Evolution will not occur OR promotes speciation <br />
11. Award marks for example given: variation amidst finches allow speciation etc<br />
<br />
<br />
(c) Explain how one named factor increases the chances of cancerous growth. [4]<br />
1. Named factor: ionizing radiation / UV light / tar in cigarette smoke etc<br />
2. Gives any form of example of how DNA can be mutated e.g. point mutation (e.g.), translocation of DNA to active promoters etc. <br />
3. Damage to DNA which cannot be repaired<br />
4. Mutation to proto-oncogenes / tumour suppressor genes <br />
5. Results in the synthesis of mutated gene products which leads to uncontrolled cell division leading to cancer/increases chances of cancerous growth [R: uncontrolled cell division;]<br />
<br />
Question 9 <br />
(a) Explain briefly what is meant by the terms gene mutation and chromosome mutation.[10]<br />
<br />
Definition of mutation:<br />
• Change in the amount, arrangement or structure of the DNA of an organism.<br />
• Produces a change in the genotype and may result in the change in appearance of a characteristic in a population.<br />
<br />
Definition of Chromosomal mutations<br />
• Result of changes in the number and structure of chromosomes.<br />
<br />
Chromosome mutation: Changes in chromosome number<br />
• These changes may involve the loss or gain of single chromosomes, a condition called aneuploidy,<br />
• Or the increase in entire haploid sets of chromosomes, a condition called euploidy (polypoidy).<br />
<br />
Aneuploidy<br />
• Loss of gain or single chromosomes eg. (n+l), (2n+l), (n-1), (2n-1)<br />
• Result from the failure of a pair, or pairs of homologous chromosomes to separate during anaphase I of meiosis. This is known as non-disjunction.<br />
<br />
Any named example:<br />
1. Down’s syndrome (Mongolism)<br />
Presence of an extra chromosome no. 21 in the cells.<br />
Symptoms – mental retardation, short stocky body, thick neck, characteristic folds of skin over the inner corner of the eye.<br />
Related to the age of the mother’s egg cells.<br />
<br />
2. Klinefelter’s syndrome (XXY)<br />
Failure of X chromosome to separate during oogenesis in female.<br />
Symptoms – male with female characteristic, sterile, testes very small, little facial hair, breasts may develop, low intelligence.<br />
<br />
3. Turner’s syndrome (XO)<br />
Sterile female.<br />
Lacking normal secondary sexual characteristics.<br />
<br />
4. XXX<br />
Female, normal appearance<br />
Fertile but mentally retarded<br />
<br />
5. XYY<br />
Male, variable intelligence, may possess psychopathic traits or tendency for petty criminal acts<br />
<br />
Euploidy (Polyploidy)<br />
• Cells contain multiples of the haploid number of chromosomes ie. polyploids 2n, 3n.....<br />
• More common in plants than animals<br />
• Associated with hybrid vigour<br />
• 2 forms:<br />
a. Autopolyploids<br />
An individual with more than 2 sets of chromosomes all derived from the same species. <br />
Can be induced by using colchicine. <br />
Autopolyploids can be fertile if they have an even number of chromosome sets.<br />
<br />
b. Allopolyploids<br />
An individual with increased chromosome number derived from different haploid sets. <br />
Many inter-species hybrids are sterile because chromosomes do not pair up at meiosis - diploid hybrid ( sterile ).<br />
If chromosomes are doubled, each member has another to pair up with and meiosis proceeds normally -- fertile, allopolyploid.<br />
<br />
Chromosome mutation: Changes in structure <br />
1. Deletion: removes a chromosomal segment<br />
2. Inversion: reverses a segment within a chromosome<br />
3. Translocation: moves a segment from one chromosome to another, non-homologous one<br />
4. Duplication: repeats a segment<br />
<br />
Any named example:<br />
1. Cri-du-chat syndrome: <br />
• deletion in the short arm of chromosome 5<br />
• child is physically and mentally retarded<br />
<br />
2. Chronic myelogenous leukaemia (CML): <br />
• Translocation<br />
• Most of the chromosome 22 has been transolcated onto the long arm of chromosome 9. <br />
• In addition, the small distal portion of chromosome 9 is translocated to chromosome 22.<br />
• Caused increased cell proliferation, reduced apoptosis cancer<br />
<br />
3. Acute myelogenous leukaemia (AML)<br />
• Chromosomal inversion on chromosome 16<br />
• Results in cancer<br />
<br />
4. Charcot-Marie-Tooth syndrome (CMT)<br />
• In one form, chromosomal duplication on chromosome 17<br />
• High gene dosage of myelin sheath protein resulting in abnormal structure and function of the myelin sheath<br />
• Symptoms: weakness of lower foot, loss of balance, poor motor skills and muscle atrophy.<br />
<br />
<br />
Gene Mutation<br />
Definition of Gene mutation<br />
• Sudden and spontaneous changes<br />
• Change in the nucleotide sequence of DNA:<br />
Duplication, insertion, deletion, inversion or substitution of bases<br />
<br />
Effects:<br />
• Alteration in a sequence of nucleotides Change in amino acid sequence of polypeptide chain affect phenotype of individual ( may be inheritable<br />
Types:<br />
(a) Substitution<br />
-1 nucleotide is replaced by another<br />
<br />
Effects:<br />
(i) Silent Mutation<br />
• Due to degeneracy of genetic code<br />
• E.g. DNA: 3’-CCG-5’ ( 3’-CCA-5’<br />
mRNA: 5’-GGC-3’ ( 5’-GGU-3’<br />
• Glycine is still produced<br />
• OR if new amino acid have properties similar to that of replaced amino acid<br />
<br />
(ii) Missense Mutations<br />
• If mutation occurs in crucial areas. E.g. active site of enzymes<br />
• Result in useless/ less active protein that impairs cellular function. E.g. Sickle Cell Anaemia<br />
<br />
(iii) Non-sense Mutations<br />
• Cause a change in codon for amino acid to stop codon<br />
• Result in premature termination of translation, non-functional protein is formed<br />
<br />
(b) Inversion<br />
<br />
<br />
<br />
<br />
<br />
<br />
<br />
<br />
<br />
<br />
(C) Frameshift Mutation<br />
• Due to the insertion/ deletion of nucleotide pairs<br />
• Have a more disastrous effect as it cause ribosome to read incorrect triplets from point of mutation<br />
• Result in shorter chain ad/ or mostly non-functional proteins<br />
<br />
E.g. <br />
mRNA 5’ AUG AAG UUU GGC UAA 3’<br />
Protein met– Lys– Phe – Gly<br />
<br />
(i) Frameshift mutation causing immediate nonsense<br />
mRNA 5’ AUG UAA GUU UGG CUA A 3’<br />
Protein met– STOP<br />
<br />
(ii) Frameshift mutation causing extensive nonsense<br />
mRNA 5’ AUG AAG UU|G GCU AA 3’<br />
Protein met– Lys– Leu- Ala- <br />
<br />
(iii) +/ - 3 nucleotides: no Frameshift BUT additional/ missing amino acid<br />
mRNA 5’ AUG UUU GGC UAA 3’<br />
Protein met– Phe – Gly<br />
<br />
Case Study: Sickle Cell Anaemia<br />
Normal Adult: Haemoglobin (Hb A)<br />
• Quaternary protein with tetramer consisting of 2 different types of polypeptide chain, 2α and 2β globin chains<br />
• Chains are encoded by 2 different genes on 2 different chromosomes<br />
Normal S.C.A<br />
Gene 3’-CTT-5’ 3’-CAT-5’<br />
Codon on mRNA 5’-GAA-3’ 5’-GUA-3’<br />
Polypeptide -Glu- -Val-<br />
Solubility Soluble Less soluble<br />
At low [O2] Remain soluble Crystallize into rod-like fibres<br />
Shape of RBC Round, biconcave Sickle shape<br />
<br />
<br />
(b) Explain how mutations for antibiotic resistance spread so rapidly among bacteria. [4]<br />
• largely because bacteria reproduce very quickly;<br />
• mutation can be quickly passed on to large numbers of descendents (usually) via plasmids during binary fission;<br />
• through conjugation, whereby transfer of antibiotic resistance genes occurs across a mating bridge;<br />
• through transformation, whereby antibiotic resistance genes taken up by competent bacterial cells through transient pores in the bacterial cell wall;<br />
• through transduction, whereby (defective) bacteriophages transfer bacterial genes / antibiotic resistance genes;<br />
Max. [4]<br />
<br />
(c) Outline the role of isolating mechanisms in the evolution of new species. [6]<br />
• ref. to definition of species;<br />
• ref. allopatric;<br />
• geographical isolation;<br />
• ref. to examples e.g. islands/ lakes/ mountain chains / idea of barrier;<br />
• ref. to example organism;<br />
• ref. to populations prevented from interbreeding;<br />
• isolated populations subjected to differential selection pressure/ conditions;<br />
• over time sufficient differences to prevent interbreeding;<br />
• ref. sympatric;<br />
• ref. to reproductive isolation;<br />
• ref. behavioural barriers (within a population);<br />
• e.g. day active/ night active;<br />
• correct ref. to gene pool;<br />
• change to allele frequencies;<br />
[ 6 max ]<br />
<br />
2010 YJCUnited JC Unionhttp://www.blogger.com/profile/06131630985986252085noreply@blogger.com0tag:blogger.com,1999:blog-4274038432688951963.post-63116307562952572762011-02-02T18:37:00.001-08:002011-02-02T18:37:33.882-08:002010 SPA Planning Compilation2010 CJC<br />
5 You are required to plan, but not carry out, an investigation into the effect of light intensity on the rate of a light dependent reaction of photosynthesis in a leaf extract using the dye DCPIP. When DCPIP is reduced, it changes from a blue colour to colourless.<br />
DCPIP (blue) + electrons → reduced DCPIP (colourless)<br />
<br />
A leaf extract can be made by mixing finely ground leaf with cold buffer solution. This leaf extract should be kept cold and in the dark except when taking samples.<br />
When capillary tubes are dipped into this leaf extract, the solution rises up into the capillary tube and remains there.<br />
The leaf extract should be placed into the capillary tubes to conduct the experiment.<br />
Your planning must be based on the assumption that you have been provided with the following equipment and materials which you must use:<br />
• capillary tubes<br />
• fresh spinach leaves<br />
• sharp knife <br />
• mortar and pestle<br />
• cold buffer solution<br />
• ice<br />
• a light-proof container with a removable lid<br />
• light-proof aluminium foil<br />
• DCPIP solution (freshly made)<br />
• lamp (chosen so it does not generate heat)<br />
• ruler<br />
• fine sharp sand<br />
• a room which can be made dark<br />
• cloths<br />
• a variety of different sized beakers, measuring cylinders, syringes and pipettes for measuring volumes<br />
<br />
Your plan should have a clear and helpful structure to include:<br />
• an explanation of theory to support your practical procedure<br />
• a description of the method used including the scientific reasoning behind the method <br />
• an explanation of the dependent and independent variables involved<br />
• relevant, clearly labelled diagrams<br />
• how you will record your results and ensure they are as accurate and reliable as possible<br />
• proposed layout of results, tables and graphs with clear headings and labels<br />
• the correct use of technical and scientific terms<br />
[12 marks]<br />
<br />
Outline [1]<br />
1. Theoretical consideration or rationale of the plan to justify the practical procedure, including source of electrons for reduction of DCPIP; [1]<br />
<br />
Procedure [10]<br />
2. Chloroplast extraction - use of knife, cloths, pestle and mortar, and sharp sand;<br />
3. For storage of chloroplast – use cold (e.g. water bath with ice) and dark (e.g. light-proof container) conditions;<br />
4. To create leaf extract - use of buffer solution;<br />
5. Method of filling capillary tube - capillary tubes are dipped into this leaf extract, the solution rises up into the capillary tube;<br />
6. Description of control capillary tube - leaf extract and DCPIP wrapped in aluminium foil to keep out the light;<br />
7. Labelled diagram of equipment;<br />
8. Independent Variable: Five distances from the light source to vary intensity;<br />
9. Propose three replicates;<br />
10. Equilibration procedure – e.g. allow each capillary tube to rest on the tile for 5 mins prior to experiment so that it is at room temperature.<br />
11. Measurement of colour change - measure time taken to decolourise completely;<br />
12. Tabulation of data with correct headings;<br />
13. Processing of data – mean figures;<br />
<br />
Distance from lamp / cm Time taken to decolorise DCPIP completely Rate / s-1<br />
Replicate 1 Replicate 2 Replicate 3 Average <br />
10 <br />
20 <br />
30 <br />
40 <br />
50 <br />
14. Graphs with clear headings and labels; <br />
<br />
Language [1]<br />
15. The correct use of technical and scientific terms. [1]<br />
<br />
<br />
<br />
<br />
<br />
<br />
<br />
<br />
2010 AJC<br />
4 Write your answers to this question on the separate answer paper provided.<br />
<br />
You are required to plan, but not carry out, an investigation into the effect of temperature on the permeability of the cell membrane of beetroot tissue. Beetroot cells contain a water-soluble red pigment in their vacuoles. This pigment cannot pass through membranes unless the membrane is damaged. <br />
<br />
Your planning must be based on the assumption that you have been provided with the following equipment and material which you must use: <br />
<br />
• A large piece of beetroot tissue with the skin removed <br />
• Sharp knife <br />
• White tile <br />
• Ruler <br />
• Waterbath<br />
• Ice <br />
• Thermometer <br />
• Test-tubes <br />
• Stopwatch <br />
• Distilled water <br />
• Colorimeter <br />
• A variety of different-sized beakers , measuring cylinders and syringes for measuring volume <br />
<br />
Your plan should have a clear and helpful structure to include: <br />
• an explanation of the theory to support your practical procedure <br />
• a description of the method used, including the scientific reasoning behind the method <br />
• an explanation of the steps you have taken to ensure that the investigation is reliable and provides accurate quantitative data. <br />
• the type of data generated and how the results will be analysed <br />
[Total:12]<br />
½ mark for each point<br />
<br />
An explanation of the theory to support your practical procedure (4m) <br />
<br />
• Pigment molecule is large/hydrophilic<br />
• Cannot pass through hydrophobic core of phospholipid layer<br />
• Increasing temperature increases the kinetic energy of molecules in the membrane/ phospholipid bilayer more fluid<br />
• Above a certain temperature, there is denaturation / disruption of tertiary structure of <br />
• membrane proteins <br />
• disrupting the integrity/ regular arrangement of the membrane <br />
• allowing pigment to diffuse <br />
• through the vacuole membrane/tonoplast AND cell surface membrane <br />
• down a concentration/diffusion gradient <br />
[any 6]<br />
<br />
• Membrane permeability can be measured by measuring the extent of leakage of pigment from beetroot tissue under different temperature<br />
• Below room temperature, no leakage of color<br />
• As temperature increases (above room temperature) , more leakage, more intense red coloration detected <br />
[any 2 ]<br />
A description of the method used, including the scientific reasoning behind the method (4m) – apparatus must be mentioned <br />
• Beetroot tissue cut to same dimension into cubes/cylinders - white tile, sharp knife,ruler <br />
• Cubes/cylinders washed and rinsed in distilled water <br />
• to remove any pigments which may leak out during cutting <br />
• Place cube/cylinder into separate test tubes containing equal volume of distilled water <br />
• Each test-tube placed in a waterbath maintained at certain temperature – use of thermometer to check temperature <br />
• for a fixed time interval (accept 5-20mins) – use of stopwatch <br />
• At least 5 temperatures used <br />
• A range of temperature should at least cover 10oC – 80oC <br />
• Mention use of ice to achieve temperatures which are lower than room temperature <br />
• Solution measured for intensity of color using colorimeter<br />
[any 8]<br />
An explanation of the steps you have taken to ensure that the investigation is reliable and provides accurate quantitative data. (3m) <br />
<br />
Control experiment <br />
• Place cube/cylinder in a test tube containing distilled water at room temperature <br />
• Subject control to same experimental conditions <br />
• There should be no leakage of color from the beetroot tissue <br />
• Purpose: to ensure that any leakage of color from the beetroot tissue is due to the change in temperature and not any other external factors <br />
<br />
Replicates and Repeats <br />
• Replicate at least 3 times for each temperature;<br />
• Repeat entire experiment at least once; <br />
• to obtain the average absorbance; <br />
[any 6]<br />
The type of data generated and how the results will be analysed (1m) <br />
• Plot graph of absorbance/color intensity(dependent variable) against temperature (independent variable); <br />
• Determine range of temperature where leakage occurs <br />
<br />
2010 JJC<br />
You are given two extracts, A and B. One of these is an apple extract and the other one is a potato extract. <br />
<br />
You are required to plan, but not carry out, an investigation to identify the two extracts by using the reagents with which you have been provided to determine the presence of reducing sugars, starch, and proteins in these extracts.<br />
<br />
Your planning must be based on the assumption that you have been provided with the following equipment and materials which you must use:<br />
<br />
• 2 small beakers with solutions labeled A and B<br />
• Benedict’s solution<br />
• 1 % of KOH (potassium hydroxide)<br />
• 1 % of CuSO4 (copper(II) sulfate)<br />
• sodium hydrogen carbonate powder <br />
• 1 mol dm-3 HCl (hydrochloric acid)<br />
• Iodine solution<br />
• boiling-tubes <br />
• test-tubes<br />
• test-tube rack<br />
• test-tube holder<br />
• 5 cm3 syringe<br />
• 1 cm3 syringe<br />
• glassrod<br />
• waterbath<br />
• 500 cm3 beaker<br />
• White-tile<br />
• Dropper<br />
• Stopwatch<br />
• Filterfunnel<br />
• Filterpaper<br />
• Labels<br />
<br />
You plan should have a clear and helpful structure to include:<br />
• an explanation of theory to support your practical procedure<br />
• a description of the method used including the scientific reasoning behind the method<br />
• proposed layout of results tables with clear headings and labels<br />
• the correct use of technical and scientific terms<br />
<br />
<br />
Mark Scheme <br />
1 Theoretical consideration or rationale of the plan to justify the practical procedure<br />
2 Volume / number of drops of reagent to be kept the same<br />
3 Temperature of water bath for Benedict’s test is at boiling point <br />
4 Time of heating kept constant (not more than 2 minutes)<br />
5 Labeling of the test tubes<br />
6 Method of starch test<br />
7 Method of Benedict’s test<br />
8 Method of Biuret test<br />
9 Correct identification of biological molecules present in apple extract.<br />
10 Correct identification of biological molecules present in potato extract.<br />
11 Tabulation of results with correct headings<br />
12 The correct use of technical and scientific terms<br />
[Total: 12]<br />
<br />
The presence of starch indicates that extract is potato extract1. <br />
This is because potato is a plant storage1 and reproductive organ1 and its main role is to provide food for the growth of the plant. Hence, it will contain a large amount of the storage polysaccharide – starch hence positive result for starch test 1 and 10 (blue-black coloration).<br />
<br />
The higher amount of brick red ppt observed1 during the Benedict's test and the absence of starch indicates that extract is the apple extract1. <br />
This is because during the process of fruit ripening1, apple being a fruit will breakdown starch into glucose to be ready for animal seed dispersers. As ripening proceeds and starch is converted to sugar1 and 9, starch will be absence when iodine's test is performed and a large amount of brick red ppt will be expected during Benedict's test. <br />
<br />
1. Using only the reagents provided, carry out the appropriate tests to identify the contents of each of the extracts A and B.<br />
<br />
Test Method for test<br />
Reducing sugar7 Add 2 cm3 2 of extract A into a test tube labeled A5. Add equal volume2 of Benedict’s solution into the test tube. Shake and mix well. Place tube A into a boiling water bath3 for 2 minutes4. Repeat the procedure for extract B.<br />
Protein8 Add 2 cm3 2 of extract A into a test tube labeled A5. Add equal volume2 of 1% KOH into the test tube. Shake and mix well. Add 3 drops2 of 1% copper sulphate solution. Shake and mix well. No heating is required. Repeat the procedure for extract B.<br />
Starch6 Add 2 cm3 2 of extract A into a test tube labeled A5. Add 3 drops2 of iodine solution. Repeat the procedure for extract B.<br />
<br />
2. Carry out the tests on all the solutions.<br />
3. Record your observations of the tests on the solutions.<br />
11<br />
Test Observations<br />
Extract A Extract B<br />
Benedict’s test <br />
Biuret test <br />
Starch test <br />
2010 PJC <br />
You are required to plan, but not carry out, an investigation into the effect of light intensity on the rate of light dependent reaction of photosynthesis in a leaf extract using the dye DCPIP. When DCPIP is reduced, it changes from blue colour to colourless. <br />
<br />
DCPIP (blue) + electrons reduced DCPIP (colourless) <br />
<br />
A leaf extract can be made by mixing finely ground leaf with cold buffer solution. This lead extract should be kept cold and in the dark except when taking samples. <br />
<br />
When capillary tubes are dipped into this leaf extract, the solution rises up into the capillary tube and remains there. <br />
<br />
The leaf extract should be placed into the capillary tubes to conduct the experiment. <br />
<br />
Your planning must be based on the assumption that you have been provided with the following equipments and materials which you must use: <br />
<br />
• capillary tube<br />
• fresh spinach leaves<br />
• sharp knife<br />
• mortar and pestle<br />
• cold buffer solution<br />
• ice<br />
• a light-proof aluminium foil<br />
• DCPIP solution (freshly made)<br />
• lamp (chosen so it does not generate heat) <br />
• ruler<br />
• fine sharp sand<br />
• a room which can be made dark<br />
• cloths<br />
• a variety of different sized beakers, measuring cylinders, syringes and pipettes for measuring volumes <br />
<br />
Your plan should: have a clear and helpful structure to include<br />
<br />
• an explanation of theory to support your practical procedure<br />
• a description of the method used including the scientific reasoning behind the method<br />
• an explanation of the dependent and independent variables involved<br />
• relevant, clearly labeled diagrams<br />
• how you will record your results and ensure they are as accurate and reliable as possible<br />
• proposed layout of results tables and graphs with clear headings and labels<br />
• the correct use of technical and scientific terms<br />
<br />
Aim: <br />
To investigate the effect of light intensity on the rate of light dependent reaction of photosynthesis in a leaf extract using the dye DCPIP; <br />
<br />
Introduction: <br />
A1 (photoactivation): In light dependent rxn, when PSII in the chloroplast of leaf extract absorbs light, and light energy is transferred to a pair of special chlorophyll a of the reaction centre P680 where e- is excited to higher energy level. This e- will then be captured by primary e- acceptor (photoactivation).<br />
A2 (flow of electrons): Each photoexcited e- passes from primary e- acceptor of PSII to PSI via e- transport chain. When e- reaches the ‘bottom’ of e- transport chain, it fills the e- hole in P700. The primary e- acceptor of PSI passes the photoexcited electron to second electron transport chain (photophosphorylation)<br />
A3 (DCPIP accepts electrons): DCPIP accepts electrons from ETC and is being reduced. It turns from blue to colourless. DCPIP competes with NADP+ for electrons.<br />
A4 (measuring rate of PS): rate of PS is measured by the time needed for decolorisation of DCPIP.<br />
Hypothesis: <br />
A5: The rate of photosynthesis increase linearly with light intensity; <br />
Max 3<br />
<br />
<br />
Procedure: <br />
B1 (obtaining leaf extract): Fresh spinach green leaves are cut with sharp knife and homogenize with cold buffer solution with mortar and pestle and fine sand. <br />
B2 (cold and dark): The chloroplast extract is covered by a light-proof aluminium foil/place in dark room and placed in a petri dish. Extract has to be maintained at low temperature using ice. (2 underlined points)<br />
B3 (DCPIP and mix): Using pipette, add 3 drops of DCPIP to the leaf extract in the petri dish. Rock the petri dish to mix the liquids and replace the foil cover<br />
<br />
B4 (filling capillary tubes): Take five capillary tubes. Stand one end of each tube in the leaf extract/DCPIP mixture to collect some of the liquid and cover the capillary tube quickly in aluminium foil to prevent exposure to light. <br />
<br />
B5 (control): Prepare a control capillary tube wrapped with aluminium foil to keep out from the light<br />
<br />
B6 (experiment procedure): Place the first capillary tube 5cm away from the lamp (measure by ruler) on a white tile, remove the aluminium foil from capillary tube and immediately turn on the lamp. Record the time taken for DCPIP to decolorize. <br />
B7 (at least 5 distances): Steps 7 and 8 are repeated for the subsequent four capillary tubes with distance of 10cm, 15cm, 20cm and 25cm respectively.<br />
<br />
(diagram – 1 mark)<br />
B8 (replicates and repeats): Set up in replicates of 3<br />
Repeat the whole experiment twice<br />
<br />
5 max <br />
<br />
Variables: <br />
i) Independent variable: light intensity (distance between the lamp & capillary tube/cm)<br />
ii) Dependent variable: time taken to decolorize DCPIP/s<br />
iii) Other variables to be kept constant:<br />
• Carbon dioxide concentration (atmospheric CO2)<br />
• pH of solution (buffer is added to ensure constant pH throughout the experiment) <br />
• temperature (room temperature) <br />
<br />
1 max<br />
<br />
Table: tabulation of data with correct heading<br />
<br />
Distance from capillary tube to the lamp (cm) Time taken for decolorization (s) Rate of photosynthesis, 1/time (S-1)<br />
1st trial 2nd trial 3rd trial Average <br />
5 <br />
10 <br />
15 <br />
20 <br />
25 <br />
<br />
Table: Processing data to get mean results<br />
2 marks<br />
Graph: <br />
<br />
<br />
• Graph of 1/time (S-1) against distance from the capillary tube to the lamp (cm) is drawn. OR graph of time (s) against distance (cm)<br />
<br />
Evaluation: <br />
• a negative slope graph should be obtained <br />
• As the distance increases (light intensity decreases), the rate of light dependent reaction of photosynthesis decrease.<br />
<br />
<br />
2010 SAJC<br />
Respiration begins when a seed absorbs water. The enzymes inside seeds have to be in suspension to function. Activated enzymes start respiration and use organic matter stored in the seed during respiration to make ATP molecules needed for growth. Respiration continues in the seed after the sprout has emerged until the seed’s endospermic material is used up and the cotyledons have produced the sprout.<br />
[Source: http://www.ehow.com/about_6695705_respiration-germinating-seeds.html]<br />
<br />
The respiratory quotient (RQ) is the ratio of the amount of carbon dioxide produced, to the amount of oxygen used. RQ is used in the calculations of basal metabolic rate (BMR).<br />
<br />
Variety radiata and variety sublobata are two varieties of Vigna radiata (commonly known as green beans), an importance source of vegetable protein of the human diet in the tropics.<br />
<br />
You are required to compare respiration of this two varieties of germinating green bean seedings.<br />
<br />
Your planning must be based on the assumption that you have been provided with the following equipment and materials.<br />
<br />
• Syringe<br />
• Rubber connecting tube<br />
• Glass capillary tube with bore diameter of 0.4 mm<br />
• Soda lime granules<br />
• Germinating green bean seedlings (Variety radiata and Variety sublobata)<br />
• Glass beads<br />
• Coloured liquid (manometer fluid)<br />
• Ruler / Graph paper<br />
• Marker pen<br />
• Stop watch<br />
• Weighing balance<br />
• Paper towels<br />
<br />
• an explanation of the theory to support your practical procedure<br />
• a description of the method used, including the scientific reasoning behind the method<br />
• the type of data generated by the experiment<br />
• how the results will be analysed<br />
<br />
[Q4 Total: /12]<br />
<br />
<br />
Theoretical consideration and rationale of the plan <br />
Rate of respiration is estimated by measuring the rate of gas exchange. In aerobic respiration, oxygen is used as the final electron acceptor in the ETC. The protons and electrons recombine with oxygen to form water. Carbon dioxide is produced during oxidative decarboxylation during the link reaction and the Krebs cycle. <br />
<br />
In a closed vessel containing the respiring organism, when the organism takes in O2, it also gives off an equal volume of CO2. Hence, volume of gases remains constant. <br />
When a compound that absorbs CO2 is placed in the closed vessel, eg. soda lime, the pressure in the vessel decreases due to the uptake of O2 during respiration and air is sucked from the manometer to keep the pressure constant. The oxygen uptake is thus detected by the displacement of the manometric fluid. <br />
<br />
In the first experiment, volume of the O2 consumed by the germinating green beans is determined. In the second experiment, the soda lime in the respirometer is replaced by water.<br />
<br />
Change in volume = volume of CO2 produced + the volume of O2 consumed.<br />
Volume of CO2 produced = volume change in second experiment - volume change in first experiment.<br />
<br />
RQ can then be calculated. RQ = <br />
<br />
<br />
Procedure<br />
Remove the plunger from the syringe and place soda lime granules inside the syringe. Take 4 or 5 of the green bean seedlings and carefully remove and discard its testa (seed coat). Place the seedlings in the syringe barrel and replace the plunger by pushing it in until it is about 0.5 cm from the seedling.<br />
<br />
Connect the glass capillary tube securely to the syringe using a rubber connecting tubing. Dip the end of the glass capillary tube into the coloured liquid (also known as manometer fluid) so that a drop enter the capillary tube. Remove any excess liquid with paper towelling.<br />
<br />
Place the respirometer horizontally on a piece of graph paper. Wait for 3 minutes to ensure that the manometer fluid is moving smoothly towards the syringe. Without handling the apparatus, measure the distance travelled by the manometer fluid in 6 consecutive time intervals of 1 min. Do this by marking the position of the fluid on the graph paper and reading off the distances. (Alternatively, a piece of white paper can be used, and the markings measured using a ruler).<br />
<br />
Record these results in the following table.<br />
Experiment 1<br />
Minutes<br />
1 2 3 4 5 6 Average<br />
Distance travelled in each minute / mm <br />
<br />
Calculate the volume of oxygen (in mm3) consumed by the respiring green beans:<br />
Volume of O2 = x r2 x d<br />
r = internal radius of capillary tube<br />
d = distance moved by meniscus of manometer fluid <br />
<br />
Detach the syringe from the capillary tube by pulling it gently from the rubber connecting tube. Fit another empty syringe to the capillary tube and flush out the manometer fluid onto a piece of filter paper so that the bore of the capillary tube is empty.<br />
<br />
Compensation tube<br />
Return to the syringe containing the green bean seedlings and soda lime. Remove the syringe plunger, and keeping the syringe more or less horizontal, remove the seedlings and replace with an equal mass of glass beads. Replace the plunger to its original position in the syringe. Connect the glass capillary tube securely to the syringe and introduce the manometer fluid as before. Place the respirometer on a piece of graph paper. Wait for 3 minutes before measuring the distance travelled by the manometer fluid in 6 consecutive time intervals of 1 min.<br />
<br />
Record these results in the following table.<br />
Compensation tube experiment<br />
Minutes<br />
1 2 3 Average<br />
Distance travelled in each minute / mm <br />
<br />
For the second experiment, set up the same respirometer with the green bean seedlings, but replace the soda lime granules with water. Wait for 3 minutes before measuring the distance travelled by the manometer fluid in 6 consecutive time intervals of 1 min.<br />
<br />
Experiment 2<br />
Minutes<br />
1 2 3 4 5 6 Average<br />
Distance travelled in each minute / mm <br />
<br />
Change in volume = volume of CO2 produced + the volume of O2 consumed.<br />
Volume of CO2 produced = volume change in second experiment - volume change in first experiment.<br />
2010 YJC<br />
Homo sapiens (wise man) is the only extant (living) species of the genus Homo. There were other species that used to exist such as Homo erectus (upright man), Homo heidelbergensis (Heidelberg man) and Homo neanderthalis (Neanderthal man). Methods of classification have depended very much on morphological assessment of the skull and skeletal structures, as well as the analysis of bone and soil materials, so as to determine relationships between these groups. <br />
<br />
In 2003, a group of fossils were discovered in Indonesia, and were named Homo floresiensis (Flores man). The group was nicknamed “the hobbit” for the small size of the adult fossils. Controversy over the status of the group arose, with some scientists arguing that Flores man was actually a tribe of dwarf Homo sapiens or a population suffering from thyroid disorders. <br />
<br />
Plan an investigation to examine if the Flores man should be regarded as a different species, or to be regarded as a subspecies of modern man, based on molecular tools available, assuming that Homo sapiens and Homo neanderthalis are to be regarded as two different species.<br />
<br />
Your planning must be based on the assumption that you have been provided with the following equipment and materials.<br />
• tissue samples (skin) from museum specimens of Flores man and Homo neanderthalis, as well as that from human volunteers.<br />
• pestle and mortar<br />
• DNA extraction buffer solution<br />
• glass rods<br />
• microcentrifuge tubes and centrifuge<br />
• restriction enzyme<br />
• agarose or polyacrylamide gel plate and a suitable source of electric current<br />
• radioactive probe and autoradiography equipment<br />
• nitrocellulose membrane<br />
<br />
Your plan should have a clear and helpful structure to include:<br />
• an explanation of the theory to support your practical procedure<br />
• a description of the method used including the scientific reasoning behind the method<br />
• the type of data generated by the experiment<br />
• how the results will be analysed, including how the status of Flores man (as a species or subspecies) can be determined.<br />
[Total: 12]<br />
<br />
Mark Scheme for Specimen Planning Question<br />
1 Theoretical consideration or rationale of the plan to justify the practical procedure<br />
• What biological principles can we depend on?<br />
• Brief explanation of which overall technique (s) to use: RFLP analysis / VNTR analysis. Certain RFLP / VNTR are conserved within a species (DNA fingerprinting at the species level) but different between species. Differences in size / length of DNA sequence reflects distance in phylogeny / ancestry.<br />
• Good suggestion: mitochondrial DNA / essential genes and neutral theory; <br />
2 Method of DNA extraction including homogenisation and use of buffers<br />
• Mechanical breakage of tissues with pestle and mortar<br />
• Use of DNA extraction buffer solution <br />
o water <br />
o detergent (to dissolve membrane)<br />
o salt (prevent DNase activity)<br />
o protease (to digest histones and other enzymes)<br />
• Temperature treatment at 60 °C<br />
• Filtration with coarse paper<br />
• Ice-cold ethanol (to precipitate DNA – make insoluble, salt removal)<br />
3 Preparation of samples for electrophoresis, including use of centrifuge?<br />
• Re-dissolve DNA in water and centrifuge to remove pellet?<br />
• Mixing of DNA solution with loading dye<br />
• Preparation of agarose gel (…)<br />
4 Selection of restriction enzyme and reasons for the selection<br />
• Designed restriction enzyme must be able to cleave the RFLP / VNTR / DNA sequence at different sites / number of sites for all species, generating different sized fragments between species.<br />
5 Amplification of DNA fragments using PCR including detail of PCR <br />
• Primer design must flank gene to be amplified<br />
• Denaturation, annealing and elongation steps with temperature<br />
• 20 to 30 cycles<br />
6 Separation of fragments by gel electrophoresis and the principles behind the separation.<br />
• Loading of DNA into separate lanes for Dodo, unknown bird and several outgroups (e.g. pigeons, cuckoos)<br />
• 100V for 30min<br />
• Why DNA migrates toward the positive electrode (anode)<br />
7 Transfer of DNA onto nitrocellulose membrane<br />
• The gel is immersed in alkaline solution (sodium hydroxide) to denature double-stranded DNA<br />
• Pressing nitrocellulose membrane and several layers of paper towels using a heavy weight with a flat surface to transfer DNA onto membrane<br />
8 Hybridisation with radioactive labelled DNA probe<br />
• The nitrocellulose membrane is incubated with radioactive labelled DNA probe that hybridises to target sequence<br />
• Excess probes are washed off<br />
9 Autoradiography method<br />
• X-ray film laid over nitrocellulose blot<br />
• Expose the film to autoradiography to visualise the DNA position in the X-ray film relative to the position in the gel<br />
10 Method of visualisation<br />
• Positions of bands represent the sizes (lengths) of target sequences <br />
11 Significance of matching bands<br />
• Compare the fragment sizes of the Homo floresiensis to Homo sapiens and Homo neanderthalis – it could be closer to either one or different from both<br />
• Repeat steps 4-10 if necessary to analyse other genetic loci<br />
• Conclude to see if it is sufficiently different from Homo sapiens (at least as dissimilar as Homo neanderthalis) in terms of the DNA profile to conclude<br />
12 The correct use of technical and scientific terms<br />
• Cathode / anode, lanes, DNA fragments / sequence<br />
• Buffer solution<br />
• No confusion of terms (e.g. primers and probes)<br />
• Terms such as nucleic acid hybridisation, double-stranded / single-stranded DNA<br />
13 Bonus: Valid safety or reliability concerns;United JC Unionhttp://www.blogger.com/profile/06131630985986252085noreply@blogger.com1tag:blogger.com,1999:blog-4274038432688951963.post-47848084933135131262011-02-02T18:36:00.001-08:002011-02-02T18:36:16.671-08:00Diversity and Evolution1. Explain how homology (anatomical, embryological and molecular) supports Darwin's theory of natural selection. [8]<br />
<br />
Anatomical homology<br />
1. anatomical resemblances representing variations on a structural theme present in their common ancestor <br />
2. ref. examples: forelimbs of all mammals, including humans, cats, whales and bats show the same arrangement of bones from the shoulder to the tips of the digits/ vestigial legs in snakes<br />
3. even though these appendages can have very different functions (lifting, walking, swimming and flying) <br />
<br />
Embryology homology<br />
4. comparison of early stages of animal development reveals additional anatomical homologies not visible in adult organisms<br />
5. ref. examples: embryos of different types of vertebrates have a tail posterior to the anus/ pharyngeal (throat) pouches.<br />
6. these embryonic structures develop into homologous structures with very different functions, such as gill slits in fishes and parts of ears and throat in humans. <br />
<br />
Molecular homology<br />
7. all forms of life use the same genetic machinery of DNA and RNA, and the genetic code is essentially universal.<br />
8. DNA differences are accumulated as descendents evolve / closely related species share a greater portion of their DNA<br />
9. ref. examples: organisms as dissimilar as humans and bacteria share many genes inherited from a distant common ancestor <br />
*comparison of amino acid sequence of human haemoglobin with other vertebrates reveals the same pattern of evolutionary relationships as nonmolecular methods <br />
<br />
In relation to Darwin’s theory<br />
10. ref. similarities between individuals of the same species or among different species, that arose from common ancestry<br />
<br />
NOTES:<br />
Darwin’s theory = many modern species of organisms are descendents of ancestral species that were different from the species. As the descendents of that ancestral organism spilled into various habitats over millions of years, they accumulated diverse modifications, or adaptations, that fit them to specific ways of life.<br />
<br />
2(a) <br />
Classification<br />
1. Classification is placing organisms into groups<br />
2. The classification is hierarchical with each successive group containing more diverse kinds of organism<br />
3. Each group possesses unique features<br />
4. The lowest or more exclusive taxon is the species, the highest or most inclusive is the kingdom<br />
5. The taxa used in order of decreasing size are:<br />
Kingdom, Phylum, Class, Order, Family, Genus and Species <br />
6. A binomial naming system was used to name the species, to avoid confusion by common name<br />
7. The first part of a binomial is the genus, the second part is the genus the species belong to<br />
Phylogeny<br />
8. Phylogeny was based on evolution and involved passing genes from ancestors to descendants<br />
9. A phylogenetic tree can be constructed to reflect the evolutionary history<br />
10. Use of DNA base sequences to assess relationships between species or by comparing their anatomy<br />
<br />
(b) <br />
1. To assess phylogenetic relationships that cannot be measured by comparative anatomy and other nonmolecular methods.<br />
2. To compare species too closely related to display much divergence in morphology.<br />
3. To trace evolutionary relationships of species that are so different that there is little morphological homology<br />
4. Each nucleotide position along a stretch of DNA represents an inherited character in the form of one of the four DNA bases: A (adenine), G (guanine), C (cytosine), or T (thymine). Thus homologous regions of DNA from two species that are 1,000 nucleotides long provide 1,000 points of comparison.<br />
5. A systematist may compare several DNA regions to assess the relationship between two species.<br />
6. Molecular character states are unambiguous: A, C, G and T are easily recognizable and one cannot be confused with another.<br />
7. Possible to analyse the extensive quantity of genetic information and provides a quantitative tool for constructing cladograms. <br />
8. Molecular data are easily converted to numerical form and hence are amenable to mathematical and statistical analysis.<br />
9. Use of molecular clock to date the time of divergence.<br />
10. These methods avoided the pitfalls of convergent evolution.<br />
11. Can be used on living or dead material<br />
<br />
(c) <br />
1. as the viruses depend on cells for their propagation therefore there is no common ancestor polyphyletic evolution<br />
2. viruses originated from fragments of cellular nucleic acids that could move from one cell to another<br />
3. they replicate very quickly, providing a great deal of material for the engine of natural selection<br />
4. rapid genetic variation is acted on by powerful selective pressures provided by the host's adaptive immune system and by modern medicine, which destroy pathogens that fail to change<br />
5. error-prone replication mechanisms e.g. retroviruses acquire on average one point mutation every replication cycle, because the viral reverse transcriptase that produces DNA from the viral RNA genome cannot correct nucleotide misincorporation errors <br />
6. plasmid and transposon are found in viral genome which help to transfer genome between cells<br />
7. there are different types of virus, that viruses evolve with their host, to escaped genes, or to degenerate cellsUnited JC Unionhttp://www.blogger.com/profile/06131630985986252085noreply@blogger.com0tag:blogger.com,1999:blog-4274038432688951963.post-67832295650282322512011-02-02T18:35:00.001-08:002011-02-03T01:15:07.982-08:00Isolating, Cloning and Sequencing DNAApplications Topic: Isolating, Cloning and Sequencing DNA<br />
Section B: Free Response Questions<br />
1. Discuss<br />
a) the aims of the human genome project. [6]<br />
• to map all human genetic markers, i.e. by identification of their specific chromosomal location<br />
• to construct a detailed physical map of the entire human genome<br />
• to determine the base sequence of all 24 human chromosomes<br />
• to develop technology for the management of human genome information <br />
• to serve as an umbrella for similar mapping and sequencing projects on the genomes of other organisms, e.g. E.coli, yeast<br />
<br />
b) the benefits of the human genome project. [8]<br />
• improved diagnosis of disease and predisoposition to disease by genetic testing<br />
• better identification of disease carriers through genetic testing<br />
• better drugs can be designed using knowledge of protein structure<br />
• greater possibility of correcting genetic disorders using gene therapy<br />
• greater knowledge of family relationships through genetic testing, e.g. paternity testing in family courts<br />
• advances forensic science through analysis of DNA at crime scenes<br />
• improved knowledge of relationships between human and other organisms, which will help to develop better, more accurate classification systems.<br />
<br />
c) the ethical concerns that have arisen about the human genome project. [6]<br />
• unclear whether third parties should have rights to genetic test results – legislation is needed to ensure that there is no discrimination on the basis of genetic information<br />
• reproductive issues regarding use of genetic information in reproductive decision making and reproductive rights. Prenatal genetic testing could lead to genetic manipulation or a decision to abort based on undesirable traits disclosed by the tests.<br />
• Treatment vs enhancement of humans – no clear distinction between medical treatment and enhancement<br />
• Fairness in access to advanced genome technologies – difficult to ascertain who should benefit or it will results in major worldwide inequities<br />
• Commercialisation of products – difficult to determine who own genes and other pieces of DNA and to ascertain if patenting DNA sequences will limit the accessibility and development into useful products. [N08/P3/Q4] <br />
<br />
2. a) Distinguish between a genomic DNA library and a cDNA library [6]<br />
<br />
Feature Genomic DNA library cDNA library<br />
DNA fragment cloning of all DNA fragments representing the entire genome, including coding and non-coding regions contains DNA fragments representing only the coding region of a genome.<br />
Introns both the introns and the non - transcribed DNA are included in the clones intron sequences have been removed by RNA splicing during the formation of the mRNA, and a continuous coding sequence is therefore present in each clone.<br />
Method DNA is cut using RE <br />
DNA fragments are incorporated into vector mRNA is extracted from a particular cells then reverse transcribe to form complementary DNA before incorporated into vector<br />
Frequency All genes are represented equally in the library Genes that are transcribed abundantly will be represented more frequently in the library <br />
No. of clones Larger number of clones to screen Smaller no. of clones to screen (as it contains only the coding regions)<br />
Regulatory sequences Presence of regulatory sequences Absence of regulatory sequences<br />
<br />
<br />
b) Describe the properties of plasmids that allow them to be used as cloning vectors. [6]<br />
• Plasmids are circular, double-stranded DNA (dsDNA) molecules. <br />
a) small molecules, size: a few kb to more than 100 kb <br />
facilitates the vector’s entry into host cells and the biochemical manipulation of DNA.<br />
b) contain an origin of replication.<br />
The vector can replicate itself and the inserted gene of interest. <br />
During cell division, at least one copy of the plasmid DNA is segregated to each daughter cell, assuring continued propagation of the plasmid through successive generations of the host cell.<br />
c) possess several unique restriction sites. <br />
The vector can be cut to insert the gene of interest.<br />
d) contain genetic/ selectable markers.<br />
These are often resistance gene, which confer some well-defined phenotypes on the host organism which enables selection. <br />
i.e. allow us to identify the host cells that has taken up the recombinant DNA molecule.<br />
<br />
c) Outline the large-scale production of a named important protein by genetic engineering. <br />
[8] <br />
• DNA sequences encoding the A and B chains of human insulin were chemically synthesised<br />
• Each gene were placed under the control of the strong lac promoter and a part of the -galactosidase structural gene in the plasmid<br />
• Both recombinanat plasmids containing the two artificial genes were transformed separately into E.coli.<br />
• The 2 artificial genes were expressed independently as fusion proteins, consisting of the first few amino acid of -galactosidase for initiation of translation in bacteria<br />
• Lactose is added to induce transcription from the lac promoter <br />
• Insulin fragment and -galactosidase were separated by a methionine residue<br />
• Insulin polypeptides could be cleaved from the -galactosidase fragment by treatment with cyanogens bromide<br />
• Purified A and B chains were attached to each other by disulphide bond formation<br />
[N09/P3/Q4]United JC Unionhttp://www.blogger.com/profile/06131630985986252085noreply@blogger.com0tag:blogger.com,1999:blog-4274038432688951963.post-7218841928555384792011-02-02T18:33:00.001-08:002011-02-02T18:33:52.266-08:00GENETICS OF BACTERIAGENETICS OF BACTERIA<br />
Bacteria fossil<br />
<br />
Learning objectives<br />
Describe the structure of a bacterial chromosome including the arrangement of DNA within bacterial cells. <br />
<br />
Describe the process of binary fission, transformation, transduction and conjugation in bacteria and explain the role of F plasmids in bacterial conjugation. (Knowledge of Hfr is not required).<br />
<br />
Distinguish between structural and regulatory genes. A structural gene is a region of DNA that codes for a protein or RNA molecule that forms part of a structure or has an enzymatic function (e.g. lacY, lacZ, lacA, but excludes lacI). A regulatory gene codes for a specific protein product that regulates the expression of the structural genes (e.g. lacI).<br />
<br />
Distinguish between the concept of repressible and inducible systems of gene regulation using trp and lac operon as examples respectively (attenuation of trp operon is not required). <br />
<br />
Describe the concept of a simple operon (using lac operon as an example). <br />
GENERAL STRUCTURE OF PROKARYOTES<br />
GENERAL STRUCTURE OF PROKARYOTES<br />
GENERAL STRUCTURE OF PROKARYOTES<br />
GENERAL STRUCTURE OF PROKARYOTES<br />
GENERAL STRUCTURE OF PROKARYOTES<br />
GENERAL STRUCTURE OF PROKARYOTES<br />
GENERAL STRUCTURE OF PROKARYOTES<br />
GENERAL STRUCTURE OF PROKARYOTES<br />
GENERAL STRUCTURE OF PROKARYOTES<br />
GENERAL STRUCTURE OF PROKARYOTES<br />
Learning objective 1<br />
Describe the structure of a bacterial chromosome including the arrangement of DNA within bacterial cells. <br />
<br />
THE BACTERIAL GENOME<br />
The bacteria chromosome is part of the genome.<br />
<br />
What is a genome? <br />
the entire genetic information found in an organism. This includes both the genes and the non-coding sequences of the DNA <br />
<br />
THE BACTERIAL GENOME<br />
Bacterial chromosome<br />
Bacterial chromosome<br />
DNA binding proteins<br />
Bacterial chromosome<br />
THE BACTERIAL GENOME<br />
Plasmids<br />
Plasmids<br />
Plasmids can also be transferred between bacterial cells during conjugation. <br />
Plasmids<br />
Every plasmid contains origin of replication which enables plasmid to be duplicated independently from the chromosomal DNA. <br />
Roles of Plasmids<br />
F plasmids (Learning objective 2) facilitate genetic recombination, which may be advantageous in a changing environment that no longer favours existing strains in a bacterial population.<br />
<br />
Genetic recombination aids to increase genetic diversity <br />
<br />
Compare to meiosis in eukaryotic cells increase genetic diversity<br />
<br />
Genes of Plasmids<br />
Degrade complex macromolecules<br />
<br />
Provide resistance to antibiotics (R plasmids)<br />
Gene product coded by the gene in the R plasmid will protect the bacteria when it is exposed to antibiotics. <br />
<br />
Produce toxins (bacteriocins – to inhibit growth of similar or closely-related bacteria)<br />
<br />
Provide resistance to heavy metals<br />
<br />
For genetic engineering, <br />
Contain genetic markers, which confer well-defined phenotypes on the host cell. (see Appendix)<br />
Plasmids can also serve function as vectors in genetic engineering (see Appendix). <br />
Plasmids as vectors and markers<br />
Plasmids as vectors and markers<br />
Plasmids as vectors and markers<br />
Plasmids as vectors and markers<br />
CHECKPOINT pg 5<br />
What is a genome?<br />
the entire genetic information found in an organism. This includes both the genes and the non-coding sequences of the DNA <br />
<br />
What are the different genetic elements that constitute the bacteria genome?<br />
Bacterial chromosome, plasmid, transposon<br />
<br />
List the main differences between these genetic elements. <br />
only one bacterial chromosome vs many plasmids<br />
chromosome contain many more genes than plasmids<br />
plasmids can be transferred to other bacterial cells<br />
transposons contain IS sequences for insertion into chromosome/plasmid<br />
CHECKPOINT pg 5<br />
What is a genome?<br />
the entire genetic information found in an organism. This includes both the genes and the non-coding sequences of the DNA <br />
<br />
What are the different genetic elements that constitute the bacteria genome?<br />
Bacterial chromosome, plasmid, transposon<br />
<br />
List the main differences between these genetic elements. <br />
only one bacterial chromosome vs many plasmids<br />
chromosome contain many more genes than plasmids<br />
plasmids can be transferred to other bacterial cells<br />
transposons contain IS sequences for insertion into chromosome/plasmid<br />
CHECKPOINT<br />
What is a genome?<br />
the entire genetic information found in an organism. This includes both the genes and the non-coding sequences of the DNA <br />
<br />
What are the different genetic elements that constitute the bacteria genome?<br />
Bacterial chromosome, plasmid<br />
<br />
List the main differences between these genetic elements. <br />
only one bacterial chromosome vs many plasmids<br />
chromosome contain many more genes than plasmids<br />
plasmids can be transferred to other bacterial cells<br />
<br />
Learning objective 1 – pg 5<br />
Describe the structure of a bacterial chromosome including the arrangement of DNA within cells. <br />
<br />
Arrangement of DNA within cells<br />
Most of the genes found in chromosomal DNA are structural genes which codes for functional polypeptides. <br />
Structural gene functional polypeptide/protein carry out a specific role<br />
e.g. insulin, protein kinases, ATPase, collagen etc<br />
<br />
Only a very small amount of non-coding DNA is present.<br />
DNA is not transcribed into proteins<br />
<br />
<br />
Contains a single origin of replication – which allows bacteria to replicate by binary fission (page 9; Learning objective 2). <br />
Binary fission: asexual reproduction of bacteria cells to produce 2 genetically identical cells. <br />
<br />
<br />
<br />
Arrangement of DNA within cells<br />
Bacterial chromosome is commonly characterised with the presence of operons (pg30). These are regulatory regions with 2 or more structural genes and allow regulation of a group of genes that encode proteins with common functions.<br />
Arrangement of DNA within cells<br />
The typical E. coli cell has 4.6 x 106 bases.<br />
This would make a strand of DNA >1000 µm long, but <br />
E. coli is only 2-5 µm long. <br />
How does it fit? <br />
<br />
is achieved by <br />
(1) formation of loop domains as well<br />
(2)supercoiling of DNA. <br />
Arrangement of DNA within cells<br />
LOOP DOMAINS<br />
Arrangement of DNA within cells<br />
HU works with topoisomerase I to bind to DNA and introduce sharp bends in the chromosome to generate tension necessary for negative supercoiling.<br />
<br />
The folded DNA is organised into a variety of conformations that are supercoiled and wound around tetramers of the HU protein. <br />
<br />
The number of loops formed is dependent on the type of species and size of bacteria. <br />
<br />
<br />
<br />
<br />
Arrangement of DNA within cells<br />
SUPERCOILING<br />
Arrangement of DNA within cells<br />
SUPERCOILING<br />
<br />
Negative supercoiling – DNA topoiosomerase II (DNA gyrase)<br />
Type II topoisomerase (DNA gyrase) introduces negative supercoils by breaking TWO DNA strand and resealing the nick upon formation of a supercoil. <br />
Topoisomerase II – introduce supercoils<br />
Role of topoisomerases<br />
As bacterial chromosome is made up of approximately 50 supercoiled DNA domains, topoisomerase prevents the entire bacterial chromosomes from becoming relaxed every time a nick is made. <br />
I.e. a nick in the DNA in one of these domains does not relax the DNA in the others. <br />
Role of topoisomerases<br />
<br />
Action of topoisomerases allows DNA molecule alternate between supercoiled and relaxed state.<br />
Supercoiling for packing the DNA into the confines of a cell <br />
Relaxing for DNA to be replicated and transcribed. <br />
Role of topoisomerases<br />
<br />
Because topoisomerases play an important role in regulating the structure of DNA, it can be the target of antibiotics. <br />
<br />
- quinolones (e.g.nalidixic acid) <br />
- fluoroquinolones (e.g.ciprofloxacin) <br />
- novobiocin<br />
<br />
CHECKPOINT<br />
What are the two methods used by the bacteria to compact its genome?<br />
Loop Domain & Supercoiling<br />
<br />
Name the protein (for each method) that plays a crucial role to ensure that bacterial DNA is highly compacted. <br />
Loop Domain DNA binding proteins (HU and H-NS)<br />
Supercoiling DNA gyrase (Topoisomerase II)<br />
CHECKPOINT<br />
What are the two methods used by the bacteria to compact its genome?<br />
Loop Domain & Supercoiling<br />
<br />
Name the protein (for each method) that plays a crucial role to ensure that bacterial DNA is highly compacted. <br />
Loop Domain DNA binding proteins (HU and H-NS)<br />
Supercoiling DNA gyrase (Topoisomerase II)<br />
CHECKPOINT<br />
What are the two methods used by the bacteria to compact its genome?<br />
Loop Domain & Supercoiling<br />
<br />
Name the protein (for each method) that plays a crucial role to ensure that bacterial DNA is highly compacted. <br />
Loop Domain DNA binding proteins (HU and H-NS)<br />
Supercoiling DNA gyrase (Topoisomerase II)<br />
QUICK CHECK: Comparing eukaryotic and prokaryotic cells…<br />
Comparing eukaryotic and prokaryotic cells…<br />
Linear double-stranded DNA molecules that wound around proteins called histones to form structures called nucleosomes.<br />
<br />
As eukaryotic DNA is very long, supercoiling is important for DNA packaging. It reduces the space and allows for a lot more DNA to be packaged. <br />
<br />
Solenoidal supercoiling is achieved with histones to form a 10nm fiber. This fiber is further coiled into a 30nm fiber, and further coiled upon itself numerous times more. <br />
<br />
<br />
<br />
Learning objectives<br />
Describe the structure of a bacterial chromosome including the arrangement of DNA within bacterial cells. <br />
<br />
Describe the process of binary fission, transformation, transduction and conjugation in bacteria and explain the role of F plasmids in bacterial conjugation. (Knowledge of Hfr is not required).<br />
<br />
Distinguish between structural and regulatory genes. A structural gene is a region of DNA that codes for a protein or RNA molecule that forms part of a structure or has an enzymatic function (e.g. lacY, lacZ, lacA, but excludes lacI). A regulatory gene codes for a specific protein product that regulates the expression of the structural genes (e.g. lacI).<br />
<br />
Distinguish between the concept of repressible and inducible systems of gene regulation using trp and lac operon as examples respectively (attenuation of trp operon is not required). <br />
<br />
Describe the concept of a simple operon (using lac operon as an example). <br />
BINARY FISSION<br />
Bacteria cells divide by binary fission.<br />
<br />
Asexual reproduction (offspring arise from one parent)<br />
Leads to genetically identical cells.<br />
<br />
Bacterial chromosome must be replicated before actual division of the cell into daughter cell. <br />
BINARY FISSION<br />
Replication and partitioning (separation) of the chromosome occur as a concerted process. <br />
<br />
Compare to eukaryotic cell division: <br />
Replication of chromosomes during S phase THEN<br />
Mitosis takes place to separate the chromosome. <br />
<br />
BINARY FISSION<br />
<br />
As DNA is replicated, cell elongates as it grows larger in size. <br />
<br />
Plasma membrane invaginates when cell has grown to an appropriate size. Cell wall is then deposited on the membrane. <br />
Mechanism - BINARY FISSION<br />
Mechanism -BINARY FISSION<br />
Mechanism - BINARY FISSION<br />
<br />
<br />
Binary fission leads to formation of genetically identical bacteria. <br />
<br />
Is it advantageous for bacteria to stay genetically the same always?<br />
<br />
If the environment changes, what will happen to the population of bacteria which is genetically identical?<br />
<br />
Q: How to generate genetic diversity in prokaryotes?<br />
<br />
MUTATION<br />
GENETIC RECOMBINATION<br />
Mutation Genetic Diversity<br />
radiation, UV rays and various chemicals damages DNA increase the likelihood of mutation. <br />
This will result in some of the offspring possessing a slightly different genetic makeup.<br />
Because E.coli reproduces via binary bission very quickly, occasional mutations may lead to significant impact on genetic diversity. <br />
Genetic Recombination<br />
Genetic recombination can also generate diversity within bacterial populations. <br />
Bacteria can exchange DNA between different cells (with different genetic composition).<br />
Conjugation -requiring cell-to-cell contact<br />
Transduction - by means of viruses <br />
Transformation- bacteria can also pick up material from the environment<br />
Let’s compare to eukaryotes!<br />
What gives rise to genetic diversity in eukaryotes?<br />
Any genetic recombination events?<br />
Mutation<br />
Meiosis prophase I (crossing over at chiasma)<br />
<br />
Recap:<br />
Binary fission will lead to genetically identical cells. <br />
One cell giving rise to another cell; Bacteria chromosome replicate; Plasmids replicate; <br />
Asexual Reproduction<br />
<br />
What is the evidence that to prove that recombination is possible in bacteria?<br />
Detection of recombination<br />
E.coli requires both tryptophan and arginine (amino acids) to survive. <br />
It needs to synthesise these two enzymes, and hence requires the genes to code for the amino acid. <br />
Detection of recombination<br />
Detection of recombination<br />
<br />
Incubate the <br />
mutant strain arg+trp- and the <br />
mutant strain arg-trp+ together. <br />
<br />
Results in a recombinant strain<br />
arg+trp+<br />
Detection of recombination<br />
Bacterial colonies<br />
<br />
When bacteria from the two strains were incubated together, cells emerged that could grow on minimum medium (containing only glucose and salts), indicating that they made both tryptophan and arginine. <br />
<br />
These cells that could synthesise both amino acids must have acquired one or more genes from the other strain, by genetic recombination <br />
Learning Objective 2<br />
Describe the process of binary fission, transformation, transduction and conjugation in bacteria and explain the role of F plasmids in bacterial conjugation.<br />
Transformation<br />
Transformation<br />
Transformation is the alteration of a bacterial cell’s genotype and phenotype by the uptake of naked, foreign DNA from the surrounding environment. There is no requirement for cell to cell contact. <br />
<br />
This occurs naturally for some bacteria only e.g. Haemophilus, Streptococcus (not for E.coli) as a mean of gene transfer.<br />
Transformation<br />
In biotechnology, artificial transformation can be applied to introduce foreign genes into E.coli genome – genes coding for valuable proteins, such as human insulin or growth hormone. <br />
Evidence for transformation<br />
Griffith’s experiment – proves that naked DNA can indeed be taken up by other cells. <br />
<br />
Streptococcus pneumoniae has 2 forms<br />
R strain – is benign; does not kill host<br />
S strain – virulent; kills host<br />
Evidence for transformation<br />
Evidence for transformation<br />
<br />
Does R strain kill the mouse?<br />
Does not kill mouse<br />
Does S strain kill the mouse?<br />
Not treated with heat – kill mouse<br />
Heat treated S strain – does not kill mouse<br />
<br />
Does R strain kill the mouse?<br />
Does not kill mouse<br />
Does S strain kill the mouse?<br />
Not treated with heat – kill mouse<br />
Heat treated S strain – does not kill mouse<br />
Evidence for transformation<br />
Evidence for transformation<br />
Evidence for transformation<br />
Evidence for transformation<br />
<br />
How does Strain R (Streptococcus pneumoniae) pick take up DNA when it was released from heat-killed S-strain?<br />
<br />
Mechanism of transformation <br />
Learning objective 2: Describe the process of binary fission, transformation.<br />
Mechanism of transformation<br />
Mechanism of transformation<br />
Mechanism of transformation<br />
Mechanism of transformation<br />
Mechanism of transformation<br />
Crossing over in Prophase I<br />
Transformation<br />
http://del.icio.us/tpjcbiology <br />
<br />
http://www.sinauer.com/cooper/4e/animations0402.html<br />
<br />
http://highered.mcgraw-hill.com/sites/0072556781/student_view0/chapter13/animation_quiz_1.html <br />
Natural transformation<br />
For bacteria that carry out natural competence, they possess cell-surface proteins (known as competence-specific proteins) that first recognise and transport DNA from closely related species into the cell, which can then integrate the foreign DNA into the genome. <br />
Natural transformation<br />
Natural transformation<br />
1-3<br />
DNA binding proteins on surface binds to fragment of DNA<br />
<br />
Autolysins degrade cell wall.<br />
<br />
DNA penetrates through cell wall. <br />
Natural transformation<br />
4<br />
Nuclease cut bound DNA into fragments as DNA penetrates through the cell wall. <br />
<br />
Natural transformation<br />
5<br />
Other nucleases destroy one strand of DNA and allow one strand to enter the bacterium. <br />
<br />
Natural transformation<br />
5<br />
Other nucleases destroy one strand of DNA and allow one strand to enter the bacterium. <br />
<br />
Natural transformation<br />
6<br />
Single-stranded DNA is bound to a competence-specific protein, and reaches the bacteria chromosome where Rec A takes over. <br />
<br />
Natural transformation<br />
7<br />
Rec A protein promotes genetic exchange btw fragment of donor’s DNA and bacteria chromosome, allowing donor DNA to be integrated into it. <br />
Natural transformation<br />
7<br />
Rec A protein promotes genetic exchange btw fragment of donor’s DNA and bacteria chromosome, allowing donor DNA to be integrated into it. <br />
Natural transformation<br />
8. <br />
Uncombined DNA will be degraded. <br />
Natural transformation<br />
Natural transformation<br />
Natural transformation<br />
Competent cells<br />
Natural competence-<br />
A small percentage of bacteria are naturally capable of taking up DNA, either in laboratory conditions, or in their natural environments. Such species carry sets of genes specifying machinery for bringing DNA across the cell's membrane or membranes natural transformation<br />
<br />
The mechanism discussed on page 18 is natural transformation. <br />
Competent cells<br />
Artificial competence – <br />
induced by laboratory procedures in which cells are passively made permeable to DNA, using conditions that do not normally occur in nature, e.g. treating the cells with CaCl2 followed by heat, or via electroporation. <br />
The cell membrane is made temporarily permeable to DNA articificial transformation. <br />
Transformation - animation<br />
CHECKPOINT<br />
What is transformation?<br />
Uptake of naked DNA from the cell’s environment. <br />
<br />
What is the aim of transformation?<br />
Uptake of foreign DNA into the cell will promote genetic recombination between the foreign DNA and host cell’s DNA leads to genetic diversity<br />
<br />
Using the information from page 17 (figure 14) and 18 (description), identify the step that is critical in order to fulfill the aim of transformation. <br />
Step 7: integration of foreign DNA into host cell’s DNA – genetic exchange with the help of Rec A<br />
CHECKPOINT<br />
What is transformation?<br />
Uptake of naked DNA from the cell’s environment. <br />
<br />
What is the aim of transformation?<br />
Uptake of foreign DNA into the cell will promote genetic recombination between the foreign DNA and host cell’s DNA leads to genetic diversity<br />
<br />
Using the information from page 17 (figure 14) and 18 (description), identify the step that is critical in order to fulfill the aim of transformation. <br />
Step 7: integration of foreign DNA into host cell’s DNA – genetic exchange with the help of Rec A<br />
CHECKPOINT<br />
What is transformation?<br />
Uptake of naked DNA from the cell’s environment. <br />
<br />
What is the aim of transformation?<br />
Uptake of foreign DNA into the cell will promote genetic recombination between the foreign DNA and host cell’s DNA leads to genetic diversity<br />
<br />
Using the information from page 17 (figure 14) and 18 (description), identify the step that is critical in order to fulfill the aim of transformation. <br />
Step 7: integration of foreign DNA into host cell’s DNA – genetic exchange with the help of Rec A<br />
CHECKPOINT<br />
What is transformation?<br />
Uptake of naked DNA from the cell’s environment. <br />
<br />
What is the aim of transformation?<br />
Uptake of foreign DNA into the cell will promote genetic recombination between the foreign DNA and host cell’s DNA leads to genetic diversity<br />
<br />
Using the information from page 17 (figure 14) and 18 (description), identify the step that is critical in order to fulfill the aim of transformation. <br />
Step 7: integration of foreign DNA into host cell’s DNA – genetic exchange with the help of Rec AUnited JC Unionhttp://www.blogger.com/profile/06131630985986252085noreply@blogger.com0tag:blogger.com,1999:blog-4274038432688951963.post-40858149438733275322011-02-02T18:32:00.000-08:002011-02-02T18:32:10.011-08:00Genetic Basis of VariationExplain the terms locus, allele, dominant, recessive, codominant, homozygous, heterozygous, phenotype and genotype.<br />
Explain how genotype is linked to phenotype and how genes are inherited from one generation to the next via the germ cells or gametes.<br />
Explain, with examples, how the environment may affect phenotype.<br />
Use genetic diagrams to solve problems in dihybrid crosses, including those involving sex linkage, codominance and multiple alleles (involving autosomal linkage or epistasis).<br />
Use genetic diagrams to solve problems involving test crosses.<br />
Explain the meaning of the terms linkage and crossing-over and explain the effect of linkage and crossing over on the phenotypic ratios from dihybrid crosses.<br />
<br />
<br />
Explain, with examples, what is meant by the terms gene mutation and chromosome aberration.<br />
Describe the differences between continuous and discontinuous variation and explain the genetic basis of continuous (many, additive, genes control a characteristic) and discontinuous variation (one or few genes control a characteristic).<br />
Describe the causes of genetic variation in a population.<br />
Describe the interaction between loci (epistasis) and predict phenotypic ratios in problems involving epistasis.<br />
Use the chi-squared test to test the significance of differences between observed and expected results. (The formula for the chi-squared test will be provided.)<br />
<br />
Lecture Outline<br />
Introduction<br />
Monohybrid and dihybrid crosses<br />
Codominance<br />
Multiple alleles<br />
Epistasis<br />
Linkage and Crossing-over<br />
Sex determination and sex linkage<br />
Pedigree Study<br />
Chi-squared Test<br />
Polygenic Inheritance<br />
Variation<br />
Mutation<br />
<br />
Concept Map<br />
Introduction<br />
(Section 1)<br />
1. INTRODUCTION<br />
What is inheritance?<br />
http://learn.genetics.utah.edu/units/basics/tour/inheritance.swf<br />
<br />
Want to know more? Try this…<br />
http://www.dnaftb.org/dnaftb/<br />
<br />
Historical development of the gene concept<br />
<br />
Mendelian Genetics<br />
Mendel’s Laws<br />
Laws of Inheritance/Heredity<br />
Law of Segregation<br />
Law of Independent Assortment<br />
Law of Dominance<br />
<br />
Pea Plant (Pisum sativum)<br />
Significance of pea plants (Pisum sativum) used:<br />
No. of contrasting traits (e.g. tall vs dwarf)<br />
Self and cross pollinating possible<br />
Large no. of seeds <br />
Pea Characteristics<br />
Pea Characteristics<br />
Definition of terms<br />
<br />
<br />
<br />
<br />
<br />
<br />
<br />
<br />
<br />
Monohybrid and Dihybrid crosses<br />
(Section 2)<br />
MONOHYBRID CROSSES<br />
Refers to the inheritance of a single trait <br />
Mendel’s Work :Crosses were made with parents of pure line<br />
Results: <br />
All the F1 resemble the tall parent. <br />
F2 ratio of 3 tall: 1 dwarf. <br />
<br />
<br />
<br />
Mendel’s conclusions:<br />
Some factor present in the pea plants which determine inheritance of height. These factors occurred in pairs.<br />
One factor is dominant and may mask the expression of the other; one factor is recessive and is obscured by the dominant factor.<br />
When gametes are formed, factors separate from each other.<br />
<br />
Genetic Basis for Mendel’s Expts<br />
<br />
<br />
Another eg: Round & Wrinkled-seeded Plants<br />
<br />
Test Cross<br />
Simple breeding test, to verify the genotype of an individual. E.g. to determine if an individual is heterozygous.<br />
Applied when a homozygous dominant genotype has the same phenotype as the heterozygous genotype. E.g. Tall phenotype for both TT and Tt<br />
Test cross parent is always homozygous recessive for all genes under consideration.<br />
<br />
<br />
Example 2:<br />
Albinism: <br />
Recessive mutation<br />
Block in one of the chemical processes leading to production of the pigment melanin<br />
Features: white hair, light-coloured skin and pink eyes<br />
Hidden in heterozygotes ie. carriers<br />
<br />
<br />
<br />
<br />
<br />
(b) ♀ Carrier x albino ♂<br />
How about the offspring for ♀ Carrier x carrier ♂ ?<br />
<br />
(c) ♀ Carrier x carrier ♂<br />
<br />
<br />
Can you interpret the breeding pattern of the leopard using a genetic diagram ?<br />
<br />
Breeding pattern of the leopard:<br />
<br />
<br />
<br />
Breeding pattern of the leopard:<br />
<br />
<br />
DIHYBRID CROSSES<br />
Inheritance of 2 traits, each specified by a different pair of independently assorting autosomal genes.<br />
Mendel’s work: Crossed plants that differed in 2 pairs of alleles (seed shape and colour).<br />
<br />
<br />
Results:<br />
All individuals of the F1 generation produced round seeds with yellow cotyledons.<br />
<br />
The F2 generation showed the following phenotypes:<br />
9 produce round seeds with yellow cotyledons <br />
3 produce round seeds with green cotyledons <br />
3 produce wrinkled seeds with yellow cotyledons<br />
1 produces wrinkled seeds with green cotyledons <br />
<br />
Conclusions:<br />
Based on results Mendel was able to state that the two pairs of characteristics separate and behave independently from one another in subsequent generations. = Law of Segregation<br />
<br />
Forms basis of Mendel’s Law on Independent Assortment, which states that any one pair of a characteristic may combine with one of another pair.<br />
<br />
Chromosome theory of Heredity<br />
Each pair of factors is carried by a pair of homologous chromosomes, with each chromosome carrying one of the factors.<br />
Since the number of characteristics vastly outnumbers the chromosomes, as revealed by microscopy, each chromosome must carry many factors.<br />
Mendel’s Laws of Heredity<br />
The Law of Segregation: <br />
Each inherited trait is defined by a gene pair. <br />
Parental genes are randomly separated to the sex cells so that sex cells contain only one gene of the pair. <br />
Offspring therefore inherit one genetic allele from each parent when sex cells unite in fertilization.<br />
<br />
The Law of Independent Assortment: Genes for different traits are sorted separately from one another so that the inheritance of one trait is not dependent on the inheritance of another.<br />
<br />
<br />
The Law of Dominance: <br />
An organism with alternate forms of a gene will express the form that is dominant.<br />
<br />
Mendel’s Laws of Heredity<br />
The Law of Segregation: <br />
Each inherited trait is defined by a gene pair. <br />
Parental genes are randomly separated to the sex cells so that sex cells contain only one gene of the pair. <br />
Offspring therefore inherit one genetic allele from each parent when sex cells unite in fertilization.<br />
Law of segregation of factors<br />
Related to the separation (segregation) of homologous chromosomes which occurs during Anaphase I of meiosis. <br />
<br />
<br />
The Law of Independent Assortment: <br />
<br />
Genes for different traits are sorted separately from one another so that the inheritance of one trait is not dependent on the inheritance of another.<br />
Can be explained by the random distribution of alleles into gamete cells. <br />
As a result of independent arrangement and separation of homologous chromosomes, the genotype AaBb gives rise to all four possible combinations of alleles (AB, ab, Ab and aB), which occur equally frequently in the gametes.<br />
<br />
<br />
<br />
Recalling…<br />
Mendel’s work: Crossed plants that differed in 2 pairs of alleles (seed shape and colour).<br />
<br />
<br />
<br />
<br />
<br />
<br />
The 9:3:3:1 (dihybrid) ratio<br />
Depends on 2 conditions:<br />
The 2 different genes must not act on the same character. For instance, if the proteins encoded by the two genes are involved in the same biochemical pathway then the ratios of phenotypes resulting from the genotypes in the F2 generation will be altered. <br />
If 2 genes lie close together on the same chromosome the four classes of gamete are not produced at equal frequencies. <br />
Dihybrid testcross<br />
A test cross can be carried out on the F1 generation to find out which plant has the genotype of RrYy. <br />
RRYY RrYY<br />
RRYy RrYy<br />
<br />
Again, is involved crossing the F1 generation with a homozygous recessive pea plant with wrinkled seed and green cotyledon (rryy).<br />
<br />
<br />
CheckpointUnited JC Unionhttp://www.blogger.com/profile/06131630985986252085noreply@blogger.com0tag:blogger.com,1999:blog-4274038432688951963.post-88454843414423413692011-02-02T18:29:00.001-08:002011-02-02T18:29:42.246-08:00Enzyme Summary<b> ENZYMES</b><br />
Learning Objectives<br />
Explain the mode of action of enzymes in terms of an:<br />
<br />
active site <br />
enzyme-substrate complex<br />
lowering of activation energy, and <br />
enzyme specificity.<br />
<br />
<br />
Learning Objectives<br />
2. Follow the time course of an enzyme-catalysed reaction, by measuring :<br />
<br />
rate of formation of products (for example using catalase), or <br />
rate of disappearance of substrate (for example using amylase).<br />
<br />
Learning Objectives<br />
Investigate and explain the following effects on the rate of enzyme-catalysed reactions.<br />
temperature <br />
pH <br />
enzyme concentration, and <br />
substrate concentration <br />
Learning Objectives<br />
Explain the effects of <br />
- competitive and <br />
- non-competitive inhibitors <br />
(including allosteric inhibitors)<br />
on the rate of enzyme activity.<br />
What are Enzymes? – page 2<br />
What are Enzymes? – page 2<br />
MODE OF ACTION FOR ENZYMES<br />
2.1 Structure of Enzymes<br />
Enzymes are globular proteins.<br />
Mode of action of enzymes – page 3<br />
ENZYMES CAN CATALYSE THE FOLLOWING REACTIONS:<br />
<br />
Formation of 1 product by forming a new bond between 2 substrates. <br />
Anabolism – synthesis of molecules and usually require energy. Often involve condensation reactions. <br />
<br />
Formation of 2 products by breaking a bond in a particular substrate<br />
Catabolism – Breakdown of molecules and usually release energy. They often involve hydrolytic reactions. <br />
Why are enzymes so specific?<br />
Lock-and-Key Hypothesis<br />
<br />
Induced Fit Hypothesis<br />
<br />
Which of the above hypothesis suggests that the substrate and the active site are NOT complementary in shape?<br />
Enzymes carry out specific reactions<br />
Lock-and-Key Hypothesis<br />
Conformation of active site is complementary to that of substrate. <br />
<br />
Enzymes carry out specific reactions<br />
Induced Fit Hypothesis<br />
Initial conformation of the active site might not be complementary in shape to the substrate<br />
When substrate comes into contact with the active site, the substrate induces a conformational change in the shape of the enzyme<br />
Structure of Enzymes – pg 5<br />
Structure of Enzymes – pg 5<br />
Structure of Enzymes – Active Site<br />
Structure of Enzymes – Active Site<br />
Structure of Enzymes – Active Site (Contact Residues)<br />
Structure of Enzymes – Active Site (Contact Residues)<br />
Structure of Enzymes – Active Site (Catalytic Residues)<br />
Structure of Enzymes – globular shape maintained by other residues<br />
How does enzymes increase the rate of reactions? – page 6<br />
By lowering activation energy (Ea)<br />
Activation Energy is the energy that must be overcome in order for a chemical reaction to occur. <br />
Enzymes help to lower the Ea of the reaction energy barrier more easily overcome<br />
<br />
Enzymes increase the rate of reactions by lowering Ea – page 6<br />
Mechanisms used to lower Ea (pg 7) <br />
Enzymes help to maintain precise orientation of 2 (or more) substrates<br />
<br />
distort/stress the bonds <br />
<br />
Increase reactivity of substrates<br />
<br />
Active site provide microenvironment for reaction<br />
<br />
Active site directly participates in the chemical reaction. <br />
Mechanisms used to lower Ea (pg 7) <br />
Enzymes help to maintain precise orientation of 2 (or more) substrates at the active site so that reaction can occur<br />
Mechanisms used to lower Ea (pg 7) <br />
2. Inducing a stress in bonds of substrate. <br />
R groups of amino acids at active site are in close promixity to certain bonds of substrate<br />
chemical interaction between R groups and substrate distort/stress the bonds lower Ea to break bond<br />
Mechanisms used to lower Ea (pg 7) <br />
Increase reactivity of substrates<br />
Interaction between R groups of amino acids of the enzyme and substrate can cause changes within substrate that increases its reactivity<br />
E.g. charge on substrate is changed. <br />
Mechanisms used to lower Ea (pg 7) <br />
Active site provide microenvironemnt for reaction<br />
E.g. active site with acidic amino acids may transfer the H+ to the substrate<br />
<br />
5. Active site directly participates in the chemical reaction. <br />
May involve brief covalent bonding with the substrate<br />
Bond will be broken to restore active site in subsequent steps. <br />
Mechanisms used to lower Ea (pg 7) <br />
Enzymes help to maintain precise orientation of 2 (or more) substrates<br />
<br />
distort/stress the bonds <br />
<br />
Increase reactivity of substrates<br />
<br />
Active site provide microenvironment for reaction<br />
<br />
Active site directly participates in the chemical reaction. <br />
QUICK CHECK (page 8)<br />
QUICK CHECK (page 8)<br />
ENZYMES<br />
FACTORS AFFECTING RATE OF ENZYMATIC REACTIONS <br />
Learning Objectives<br />
Investigate and explain the following effects on the rate of enzyme-catalysed reactions.<br />
temperature <br />
pH <br />
enzyme concentration, and <br />
substrate concentration <br />
Rate of rxn is affected by temp – (Pg 10)<br />
Effects of Varying Temperature<br />
Effects of Varying Temperature<br />
Rate of rxn is affected by temp – (Pg 10)<br />
Rate of rxn is affected by temp – (Pg 10)<br />
Rate of rxn is affected by temp – (Pg 10)<br />
5.1 Temperature affecting Rate of reaction<br />
QUICK CHECK (page 15)<br />
<br />
Learning Objectives<br />
Investigate and explain the following effects on the rate of enzyme-catalysed reactions.<br />
temperature <br />
pH <br />
enzyme concentration, and <br />
substrate concentration <br />
pH affecting Rate of reaction- enz work at max rate at optimum pH<br />
pH affecting Rate of reaction- enz work at max rate at optimum pH<br />
At the optimum pH, do you think the bonds are broken?<br />
No, the intramolecular bonds, which maintain the secondary and tertiary structures of the enzyme are intact <br />
<br />
What would the conformation of the active site be like?<br />
the conformation of the active site is most ideal for binding of substrate.<br />
pH affecting Rate of reaction- enz work at max rate at optimum pH<br />
What is the frequency of successful collisions between enzyme and substrate molecules at the optimum pH? <br />
the frequency would be the highest. <br />
<br />
How fast would enzyme-substrate complexes will be formed?<br />
pH affecting Rate of reaction- when pH deviates from the optimum pH<br />
At pH lower or higher than the optimum, the concentration of hydrogen ions (H+) would have changed. <br />
<br />
pH affecting Rate of reaction- when pH deviates from the optimum pH<br />
5.2 pH affecting Rate of reaction- when pH deviates from the optimum pH<br />
What happens when the charges of the R groups change?<br />
Ionic bonds and Hydrogen bonds formed between the R groups will be disrupted.<br />
These bonds help to maintain the conformation of the enzyme molecule.<br />
<br />
What will happen to the active site when these bonds are broken?<br />
<br />
<br />
5.2 pH affecting Rate of reaction- when pH deviates from the optimum pH<br />
The conformation of the active site would be disrupted and the binding of substrate would be affected.<br />
<br />
Rate of enzyme-catalysed reactions will decrease, as pH deviates from the optimal pH. <br />
<br />
Therefore, enzymes work within a narrow range of pH<br />
<br />
pH affecting Rate of reaction<br />
pH affecting Rate of reaction<br />
If the pH is altered by a small extent from the optimum pH, are effects normally reversible?<br />
<br />
Yes. If the pH is restored to the optimum, the maximum activity of the enzyme will be restored. <br />
<br />
<br />
pH affecting Rate of reaction<br />
If the pH is altered by a small extent from the optimum pH, are effects normally reversible?<br />
<br />
Yes. If the pH is restored to the optimum, the maximum activity of the enzyme will be restored. <br />
<br />
<br />
Effect of pH change<br />
What happens to the conformation of an enzyme when pH is altered to a great extent? Will it be reversible?<br />
<br />
the conformation of the enzyme molecule would be severely affected. Denaturation of the enzyme might be irreversible. <br />
<br />
<br />
Learning Objectives<br />
Investigate and explain the following effects on the rate of enzyme-catalysed reactions.<br />
temperature <br />
pH <br />
substrate concentration <br />
enzyme concentration<br />
<br />
Substrate concentration affecting rate of reaction – page 13<br />
For a fixed enzyme concentration, the rate of reaction increases with increasing substrate concentration.<br />
Substrate concentration affecting rate of reaction<br />
An increase in the number of substrate molecules will result in an increase in the frequency of effective collisions between enzyme and substrate molecules. <br />
Substrate concentration affecting rate of reaction<br />
When the enzyme concentration is fixed, and the substrate concentration is increased,<br />
More enzyme-substrate complexes will be formed.<br />
The rate of reaction will increase (till the saturation point is reached)<br />
<br />
Substrate concentration affecting rate of reaction<br />
Rate of reaction will reach a maximum when the point of saturation is reached, whereby increasing concentration of substrates will not increase the rate of reaction any further.<br />
At the saturation point…<br />
fixed number of available active sites (as the enzyme concentration remains unchanged)<br />
<br />
the active sites of the enzyme molecules at any given moment are virtually saturated with substrate.<br />
<br />
Max no. of enzyme-substrate formed per unit time<br />
<br />
The enzyme/substrate complex has to dissociate before the active sites are free to accommodate more substrate.<br />
Learning Objectives<br />
Investigate and explain the following effects on the rate of enzyme-catalysed reactions.<br />
temperature <br />
pH <br />
substrate concentration <br />
enzyme concentration<br />
<br />
Enzyme concentration affecting rate of reaction<br />
In the presence of a large concentration of substrates, while pH and temperature are kept constant;<br />
<br />
When enzyme concentration increases <br />
Frequency of successful collisions between enzyme and substrate molecules increases<br />
no. of enzyme - substrate complexes forms increases<br />
Rate of reaction increases <br />
Enzyme concentration affecting rate of reaction<br />
CHECKPOINT<br />
Q: At low temperatures, what happens to enzymes?<br />
They are inactivated.<br />
<br />
Q: Why?<br />
Both enzymes and substrates do not have sufficient energy to have effective collisions with one another.<br />
CHECKPOINT<br />
CHECKPOINT<br />
Q: What kind of bonds are easily broken when enzymes are exposed to heat?<br />
Hydrogen bonds, ionic bonds and hydrophobic interactions<br />
Affects structure of active site affects formation of E-S complex<br />
<br />
Q: What changes occur in the enzyme when pH is changed?<br />
The charges of R groups of amino acid residues are altered.<br />
CHECKPOINT<br />
Q: What kind of bonds are affected?<br />
Ionic bonds and Hydrogen bonds<br />
<br />
Q: What will happen to the enzyme?<br />
The conformation at the active site will change and be unable to bind substrates.<br />
CHECKPOINT<br />
Q: In the presence of a fixed amount of enzyme, an increase in substrate amts would lead to an increase of ________?<br />
<br />
Frequency of effective substrate enzyme collisions. <br />
<br />
CHECKPOINT<br />
<br />
Q: Why does the rate of reaction reaches a max. in the presence of high [S]? <br />
<br />
Saturation Point is reached. All the active sites on the enzymes are fully occupied. Rate of reaction is limited by the number of active sites available. <br />
QUICK CHECK (page 15)<br />
ENZYMES<br />
Rate of rxn is affected by temp – (Pg 10)<br />
pH affecting Rate of reaction<br />
Substrate concentration affecting rate of reaction – page 13<br />
For a fixed enzyme concentration, the rate of reaction increases with increasing substrate concentration.<br />
Enzyme concentration affecting rate of reaction<br />
Inhibition of enzyme reaction<br />
Enzyme Inhibitors<br />
Inhibition of enzymatic reactions<br />
An inhibitor is a substance that prevents an enzyme from catalysing its reaction. <br />
<br />
It decreases the rate of an enzyme-catalysed reaction. <br />
<br />
The inhibitor combines with the enzyme to form an enzyme-inhibitor complex Enzyme cannot combine with the substrate molecule.<br />
<br />
Inhibitors can be divided into two main groups: Competitive inhibitors and non-competitive inhibitors.<br />
<br />
Competitive Inhibition <br />
Competitive Inhibition<br />
5.1 Competitive Inhibition<br />
5.1 Competitive Inhibition<br />
Competitive Inhibition<br />
Can the reaction still reach its maximum rate if a lot of substrate is added?<br />
At very high substrate concentrations, the rate of reaction can reach its maximum value.<br />
Inhibitors has no chance of binding to the active site<br />
<br />
Competitive Inhibition<br />
Competitive inhibition<br />
Non-Competitive Inhibition <br />
Non-Competitive Inhibition<br />
Non-Competitive Inhibition<br />
Non-Competitive Inhibition decreases the rate of reaction<br />
Overcoming non-competitive inhibition<br />
Non-competitive inhibition cannot be overcome even when substrate concentration increases. <br />
<br />
The rate of reaction is unable to achieve the theoretical Vmax, unlike with competitive inhibition. <br />
Comparing Competitive and Non-Competitive Inhibition <br />
<br />
<br />
<br />
<br />
<br />
<br />
<br />
Checkpoint<br />
What does a competitive inhibitor compete for?<br />
Active site of enzyme<br />
<br />
How can its inhibitory effect be decreased?<br />
Add more substrates.<br />
<br />
Can the maximum rate of reaction be reached if substrate concentration is high?<br />
Yes.<br />
Allosteric Enzymes<br />
7.1 What are Allosteric Enzymes?<br />
Allosteric inhibition is an example of reversible non-competitive inhibition. <br />
<br />
Allosteric inhibitors bind at a site away from the active site allosteric sites<br />
<br />
Leads to changes in the overall structure of the enzyme and of the active site<br />
<br />
Substrate is now unable to bind to active site of the enzyme. <br />
<br />
Decreases rate of reaction<br />
7.3 Mechanism of Allosteric Inhibition<br />
When the allosteric inhibitor is not bound to the allosteric site of an enzyme, the active site of the enzyme is able to bind the substrate and catalyse the formation of product(s)<br />
7.3 Mechanism of Allosteric Inhibition<br />
When the allosteric inhibitor binds to the enzyme at the allosteric site, the active site of the enzyme is altered and no substrate can bind to it.<br />
End Product Inhibition<br />
End Product Inhibition<br />
Enzymes whose activities are regulated by allosteric inhibitors tend to catalyse the first reaction in a biochemical pathway. <br />
<br />
The end-product of the pathway can act as an allosteric inhibitor of the first enzyme of the pathway, thereby stopping the synthesis of the end-product.<br />
Mechanism of Allosteric Inhibition: End Product Inhibition<br />
A metabolic pathway usually involves a series of reactions in which each is catalysed by an enzyme.<br />
<br />
When the end product of a metabolic pathway begins to accumulate, it may act as an inhibitor, usually on the enzyme catalysing the first reaction in the pathway <br />
<br />
<br />
7.3 Mechanism of Allosteric Inhibition: End Product Inhibition<br />
7.3 Mechanism of Allosteric Inhibition: End Product Inhibition<br />
Difference between Allosteric Inhibition and Non-Competitive Inhibition<br />
For both, the inhibitor bind to a site away from the active site<br />
<br />
<br />
For non-competitive inhibitor (generally) substrate may or may not still bind to active site no catalysis<br />
<br />
For allosteric inhibition substrate CANNOT bind to active site (conformation is changed) no catalysis<br />
Co-factors in enzymatic reactions<br />
What are enzyme Co-Factors?<br />
Non-protein components that are required by enzymes for their activities.<br />
May vary from simple inorganic ions to complex organic molecules.<br />
8. Cofactors<br />
8. Cofactors<br />
8. Cofactors<br />
http://www.clunet.edu/BioDev/omm/catalase/cat1.htm<br />
8. Cofactors<br />
8.3 Coenzymes<br />
Coenzymes are derived from vitamins. <br />
<br />
Eg. Nicotinamide Adenine Dinucleotide (NAD+) is an important coenzyme in respiration. It is derived from the vitamin nicotinic acid.<br />
<br />
8.3 NAD+ as coenzyme in respiration <br />
8.3 NAD+ as coenzyme in respirationUnited JC Unionhttp://www.blogger.com/profile/06131630985986252085noreply@blogger.com2tag:blogger.com,1999:blog-4274038432688951963.post-36563131381796628032011-02-02T18:25:00.000-08:002011-02-02T18:25:25.265-08:00DNA and RNA StructureCandidates should be able to:<br />
Describe two pieces of experimental evidence to support the conclusion that DNA is the genetic material.<br />
Outline the important events that led to the present knowledge of the nature of the DNA molecule.<br />
Describe the structure of RNA and DNA and explain the importance of base pairing and hydrogen bonding.<br />
<br />
<br />
1. Deoxyribonucleic acid (DNA)<br />
<br />
DNA carries the genetic information of a cell!<br />
<br />
Consist of thousands of genes!<br />
<br />
A. Pentose<br />
<br />
A sugar containing 5 carbon atoms<br />
<br />
In DNA, sugar deoxyribose is present<br />
In RNA, sugar ribose is present<br />
<br />
B. Phosphoric acid<br />
<br />
The formula for phosphoric acid is H3PO4<br />
<br />
Its structural formula is as follows: <br />
<br />
There are 2 categories of bases<br />
<br />
Purines<br />
<br />
2-ringed structure<br />
<br />
Pyrimidines<br />
<br />
1-ringed structure<br />
<br />
1-ringed structure<br />
<br />
Adenine, A<br />
Guanine, G<br />
<br />
Cytosine, C<br />
Thymine, T<br />
<br />
<div class="separator" style="clear: both; text-align: center;"><a href="http://www.learner.org/interactives/dna/images/project9.jpg" imageanchor="1" style="margin-left:1em; margin-right:1em"><img border="0" height="293" width="300" src="http://www.learner.org/interactives/dna/images/project9.jpg" /></a></div><br />
How is a polynucleotide formed?<br />
<br />
Sugar-phosphate backbone<br />
<br />
All units are identical<br />
along the chain<br />
<br />
Bases<br />
<br />
Base sequence changes <br />
along the length of chain<br />
<br />
<br />
Sequence of bases carry<br />
genetic information of <br />
the organism<br />
<br />
What makes up a DNA molecule?<br />
<br />
1. Consists of 2 <br />
Polynucleotide chains <br />
<br />
2. The 2 chains run in <br />
opposite directions<br />
<br />
3. Hydrogen bonds hold<br />
the chains together<br />
<br />
3. Complementary base<br />
pairs are formed by<br />
the hydrogen bonds<br />
<br />
More details about the DNA molecules<br />
<br />
1. A double helix of 2 polynucleotide chains <br />
<br />
2. The width between the backbones<br />
are constant and equal to the width <br />
of a purine and a pyrimidine<br />
<br />
3. The ratio of A to T is 1<br />
&<br />
the ratio of C to G is 1<br />
<br />
What makes up an RNA molecule?<br />
<br />
Its a single stranded molecule (except for some viruses).<br />
The pentose sugar is ribose.<br />
The nitrogenous base, instead of thymine, is uracil.<br />
There are 3 types of RNA<br />
Messenger RNA (mRNA)<br />
Ribosomal RNA (rRNA)<br />
Transfer RNA (tRNA)<br />
<br />
<div class="separator" style="clear: both; text-align: center;"><a href="http://click4biology.info/c4b/3/images/3.5/compare.gif" imageanchor="1" style="margin-left:1em; margin-right:1em"><img border="0" height="209" width="641" src="http://click4biology.info/c4b/3/images/3.5/compare.gif" /></a></div>United JC Unionhttp://www.blogger.com/profile/06131630985986252085noreply@blogger.com0tag:blogger.com,1999:blog-4274038432688951963.post-87110638886143225302011-02-02T18:07:00.000-08:002011-02-02T18:10:28.604-08:00Replication of DNA summaryReplication of DNA<br />
<br />
<br />
Section content<br />
<br />
Mechanism of DNA replication “3 hypotheses”<br />
Semi-conservative hypothesis “Evidence…”<br />
Semi-conservative DNA replication “Mechanism”<br />
Enzymes “What role do enzymes play?”<br />
<br />
<br />
1. Conservative<br />
hypothesis<br />
<br />
<br />
In this hypothesis…<br />
<br />
Both strands of DNA act as templates.<br />
The parent DNA is intact & goes into one daughter cell.<br />
The new DNA molecule goes into the other daughter cell<br />
<br />
<br />
. Dispersive<br />
hypothesis<br />
<br />
<br />
In this hypothesis…<br />
<br />
Parental DNA molecule breaks into short segments, which acts as templates for synthesis.<br />
The segments are then join together.<br />
Results in two DNA molecules with old & new segments in each strand.<br />
<br />
<br />
3. Semi-conservative<br />
hypothesis<br />
<br />
<br />
In this hypothesis…<br />
<br />
Both strands of parental DNA separate & act as templates.<br />
New bases join each old strand to become a new strand.<br />
Results in two DNA molecules that are hybrids of old & new strands.<br />
<br />
<br />
Evidence by<br />
Meselson & Stahl<br />
<br />
The result<br />
For the 1st generation cells, all DNA bands were between those of 14N & 15N. They were a hybrid of 14N & 15N strands<br />
<br />
<br />
For 2nd generation cells, half of the DNA molecules were of the hybrid type. The other half were pure 14N DNA.<br />
<br />
The results confirm that DNA replication is semi-conservative<br />
<br />
<div class="separator" style="clear: both; text-align: center;"><a href="http://mol-biol4masters.org/Prokaryotic_DNA_Replication1-Introduction_files/image007.jpg" imageanchor="1" style="clear:left; float:left;margin-right:1em; margin-bottom:1em"><img border="0" height="565" width="433" src="http://mol-biol4masters.org/Prokaryotic_DNA_Replication1-Introduction_files/image007.jpg" /></a></div><br />
2. Replication of DNA<br />
<br />
<br />
Free deoxyribonucleotides <br />
<br />
<br />
synthesised in the cytoplasm <br />
transported into the nucleoplasm via pores in the nuclear envelope<br />
<br />
<br />
Helicase (an Enzyme)<br />
<br />
<br />
Causes DNA molecule to unwind <br />
Begins at sites called origins of replication<br />
Hydrogen bonds between bases break<br />
<br />
<br />
Single-stranded DNA binding protein <br />
<br />
<br />
stabilises the unwound single strands of DNA. <br />
<br />
Therefore allows the unwound region to serve as a template.<br />
<br />
<div class="separator" style="clear: both; text-align: center;"><a href="http://www.prism.gatech.edu/~gh19/b1510/dnasynth.gif" imageanchor="1" style="clear:left; float:left;margin-right:1em; margin-bottom:1em"><img border="0" height="318" width="435" src="http://www.prism.gatech.edu/~gh19/b1510/dnasynth.gif" /></a></div><br />
Leading Strand<br />
Synthesized continuosly in 5’ to 3’ direction<br />
<br />
Primase (an enzyme)<br />
Catalyses the synthesis of a short RNA chain. <br />
RNA chain is called primer.<br />
Primer is complementary to DNA template.<br />
<br />
<br />
DNA polymerase III (an enzyme)<br />
Catalyses the bond formation between free deoxyribonucleotides & template.<br />
Ensures complementary base pairing.<br />
<br />
<br />
<br />
Lagging Strand<br />
Synthesized in short fragments<br />
known as Okazaki fragments<br />
<br />
<br />
DNA ligase (an enzyme)<br />
Catalyses the phosphodiester bonds between two adjacent Okazaki fragments.<br />
Requires a free hydroxyl group at 3’ end of one DNA chain.<br />
And a phosphate group at the 5’ end of the other DNA chain.<br />
New DNA<br />
New DNA results when the new parental-daughter pairs of DNA rewind’<br />
<br />
<div class="separator" style="clear: both; text-align: center;"><a href="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgcu7LPMRrJgistTymQKw42y8twaICxnWx0nCbVzRkMroKg98UubgkR-Xoqy6v28GrI_ui5-b0wUAyFIj1m5TOBfr5MzKV0ZihVJdmQ6J3mv5wnctoNR9qdBIuBjrj8J_izLpa-Xs6Pctem/s1600/DNA-20replication.gif" imageanchor="1" style="margin-left:1em; margin-right:1em"><img border="0" height="506" width="749" src="https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEgcu7LPMRrJgistTymQKw42y8twaICxnWx0nCbVzRkMroKg98UubgkR-Xoqy6v28GrI_ui5-b0wUAyFIj1m5TOBfr5MzKV0ZihVJdmQ6J3mv5wnctoNR9qdBIuBjrj8J_izLpa-Xs6Pctem/s1600/DNA-20replication.gif" /></a></div>United JC Unionhttp://www.blogger.com/profile/06131630985986252085noreply@blogger.com0tag:blogger.com,1999:blog-4274038432688951963.post-34873796940675698292011-02-02T00:37:00.000-08:002011-02-02T18:10:54.308-08:00Structure and Functions Of organelles in Plant & Animal cellsCell Structure:<br />
<br />
Candidates should be able to:<br />
Describe, interpret <br />
electron micrographs, <br />
rough and smooth endoplasmic reticulum, Golgi body, <br />
mitochondria, chloroplasts, <br />
ribosomes, lysosomes, <br />
cell surface membrane, <br />
nuclear envelope, centrioles, nucleus and nucleolus. <br />
<br />
Candidates should be able to:<br />
Outline <br />
functions of the membrane systems and organelles <br />
<br />
<br />
PROKARYOTES ( pro, before; karyon, nucleus ) lack true nuclei ie. their genetic material (DNA) is not enclosed by nuclear membranes, and lies free in the cytoplasm.<br />
eg. Bacteria.<br />
EUKARYOTES ( eu, true ) are more complex and are characterised by a true nucleus with their genetic material being enclosed by the nuclear envelope.<br />
eg. Protoctists, fungi, green plants, animals.<br />
<br />
<b>Are you able to tell the differences? </b><br />
<div class="separator" style="clear: both; text-align: center;"><a href="http://0.tqn.com/d/biology/1/0/Z/V/prokaryoticcell.jpg" imageanchor="1" style="clear: left; float: left; margin-bottom: 1em; margin-right: 1em;"><img border="0" height="326" src="http://0.tqn.com/d/biology/1/0/Z/V/prokaryoticcell.jpg" width="400" /></a></div><br />
<div class="separator" style="clear: both; text-align: center;"><a href="https://www.etap.org/demo/grade7_science/Image28.jpg" imageanchor="1" style="clear: left; float: left; margin-bottom: 1em; margin-right: 1em;"><img border="0" height="576" src="https://www.etap.org/demo/grade7_science/Image28.jpg" width="663" /></a></div><div class="separator" style="clear: both; text-align: center;"><a href="http://micro.magnet.fsu.edu/cells/plants/images/plantcell.jpg" imageanchor="1" style="clear: left; float: left; margin-bottom: 1em; margin-right: 1em;"><img border="0" height="356" src="http://micro.magnet.fsu.edu/cells/plants/images/plantcell.jpg" width="435" /></a></div><br />
Are you able to identify the different cell Organelles? <br />
<br />
<div class="separator" style="clear: both; text-align: center;"><a href="http://faculty.une.edu/com/abell/histo/golgilys.jpg" imageanchor="1" style="clear:left; float:left;margin-right:1em; margin-bottom:1em"><img border="0" height="336" width="432" src="http://faculty.une.edu/com/abell/histo/golgilys.jpg" /></a></div><br />
Cell Parts: <br />
<br />
NUCLEUS<br />
<br />
Found in eukaryotic cells. <br />
Largest cell organelle<br />
Controls cell activities<br />
<br />
NUCLEAR ENVELOPE<br />
NUCLEAR PORES<br />
PERINUCLEAR SPACE<br />
NUCLEOPLASM<br />
CHROMATIN – DNA, HISTONES<br />
CHROMOSOMES<br />
HETEROCHROMATIN<br />
EUCHROMATIN<br />
NUCLEOLUS<br />
<br />
CONTROLS CELL ACTIVITIES<br />
INVOLVED IN CELL DIVISION<br />
PROTEIN SYNTHESIS<br />
NUCLEOLUS MANUFACTURES RIBOSOMES<br />
<br />
WHICH ARE THE CELL ORGANELLES<br />
RELATED THROUGH THE<br />
ENDOMEMBRANE SYSTEM?<br />
<br />
Rough endoplasmic reticulum<br />
Smooth endoplasmic reticulum<br />
Golgi apparatus<br />
<br />
Consists of flattened, membrane-bound sacs called cisternae.<br />
Complex system of membranes running through the cytoplasm of all eukaryotic cells.<br />
Continuous with the nuclear membrane.<br />
<br />
2 types :<br />
ROUGH ENDOPLASMIC RETICULUM<br />
SMOOTH ENDOPLASMIC RETICULUM<br />
<br />
STRUCTURE OF ENDOPLASMIC RETICULUM<br />
<br />
ROUGH ER<br />
Covered with ribosomes<br />
Sheet-like<br />
<br />
SMOOTH ER<br />
No ribosomes<br />
More tubular<br />
<br />
RIBOSOMES<br />
<br />
<br />
RIBOSOMES<br />
Consists of a small subunit and a large subunit<br />
Ribosomes are found freely floating in the cytosol or attached to the endoplasmic reticulum<br />
Site of protein synthesis<br />
<br />
MICROSOMES<br />
Small membrane-bound sacs.<br />
Formed during the homogenisation procedure.<br />
Rough ER is broken into small pieces and they reseal into vesicles. <br />
ie. Microsomes<br />
Microsomes do not exist as such in intact cells<br />
<br />
FUNCTIONS OF ENDOPLASMIC RETICULUM<br />
<br />
ROUGH ER<br />
Transport of proteins -synthesised at the ribosomes on its surface<br />
Synthesise phospholipids<br />
<br />
SMOOTH ER<br />
Lipid synthesis<br />
Detoxification of drugs and poisons<br />
Carbohydrate synthesis<br />
Store calcium<br />
<br />
GOLGI APPARATUS <br />
STRUCTURE:<br />
Cisternae - stack of flattened, membrane-bound sacs <br />
Golgi vesicles.<br />
‘cis’ face – new cisternae formed<br />
‘trans’ face – cisternae break up into vesicles<div class="separator" style="clear: both; text-align: center;"><br />
<a href="http://creationrevolution.com/wp-content/uploads/2010/11/Golgi-Apparatus-and-ER.jpg" imageanchor="1" style="clear:left; float:left;margin-right:1em; margin-bottom:1em"><img border="0" height="920" width="1057" src="http://creationrevolution.com/wp-content/uploads/2010/11/Golgi-Apparatus-and-ER.jpg" /></a></div><br />
<div class="separator" style="clear: both; text-align: center;"><a href="http://micro.magnet.fsu.edu/cells/golgi/images/golgifigure1.jpg" imageanchor="1" style="clear:left; float:left;margin-right:1em; margin-bottom:1em"><img border="0" height="288" width="373" src="http://micro.magnet.fsu.edu/cells/golgi/images/golgifigure1.jpg" /></a></div><br />
FUNCTIONS OF GOLGI APPARATUS<br />
Transport and chemically modify the materials contained within it. <br />
ve.g. glycoproteins<br />
Sorts and targets completed materials to different parts of the cell/ for secretion out of the cell.<br />
Formation of lysosomes<br />
Vesicles budded off from the Golgi apparatus fuse with the plasma membrane<br />
replace membrane <br />
<br />
LYSOSOMES<br />
LYSIS : SPLITTING<br />
SOMA : BODY<br />
<br />
LYSOSOMES<br />
<br />
WHAT IS THE DIFFERENCE BETWEEN LYSOSOMES AND PEROXISOMES ?<br />
LYSOSOMES AND PEROXISOMES<br />
STRUCTURE OF LYSOSOMES<br />
USUALLY SPHERICAL SACS BOUNDED BY A SINGLE MEMBRANE<br />
<br />
0.2 – 0.5 um in diameter<br />
<br />
CONTAINS HYDROLYTIC ENZYMES<br />
<br />
CONTENTS ARE ACIDIC AND THE ENZYMES HAVE A LOW OPTIMUM PH<br />
STRUCTURE OF PEROXISOMES<br />
Spherical <br />
0.3 – 1.5 um in diameter<br />
Bounded by single membrane<br />
Derived from the ER<br />
Presence of the enzyme catalase which catalyses the decomposition of H2O2 to water and oxygen.<br />
Eg. Glyoxysomes, leaf peroxisome, non-specialised peroxisomes.<br />
<br />
LOW PH IN LYSOSOMES<br />
TYPES OF LYSOSOMES<br />
PRIMARY LYSOSOMES<br />
Enzymes are synthesised on rough ER and transported to the Golgi Apparatus. Golgi vesicles containing the processed enzymes later bud off and are called primary lysosomes.<br />
SECONDARY LYSOSOMES<br />
Primary lysosomes may fuse with vacuoles formed by endocytosis ( infolding of the plasma membrane ) to form secondary lysosomes.<br />
FUNCTIONS OF LYSOSOMES<br />
DIGESTION OF MATERIALS<br />
AUTOPHAGY<br />
RELEASE OF ENZYMES OUTSIDE THE CELL<br />
AUTOLYSIS<br />
<br />
ELECTRON MICROGRAPH OF MITOCHONDRIA<br />
Muscle Cell Mitochondrion (TEM x190,920). This image is copyright Dennis Kunkel at www.DennisKunkel.com, used with permission.<br />
STRUCTURE OF MITOCHONDRIA<br />
VARIABLE SHAPE AND SIZE<br />
BOUNDED BY TWO MEMBRANES<br />
INTERMEMBRANE SPACE<br />
SMOOTH OUTER UNIT MEMBRANE<br />
INNER MEMBRANE THROWN INTO FOLDS – CRISTAE<br />
CRISTAE – ELEMENTARY PARTICLES (with head piece, stalk and base)<br />
FLUID-FILLED INTRACRISTAL SPACE<br />
MATRIX – ribosomes, circular DNA, enzymes<br />
<br />
<br />
FUNCTION OF MITOCHONDRIA<br />
<br />
<br />
SITE OF AEROBIC RESPIRATION<br />
PRODUCTION OF ATP<br />
MATRIX – KREBS’ CYCLE<br />
CRISTAE – ELECTRON TRANSPORT<br />
What happens to old, worn-out mitochondria? <br />
CAN THE MITOCHONDRIA DIVIDE ?<br />
MITOCHONDRIAL DNA<br />
ORIGIN OF MITOCHONDRIA<br />
<br />
WHERE DO MITOCHONDRIA COME FROM ?<br />
<br />
MITOCHONDRIA AND ENDOSYMBIOSIS<br />
During the 1980s, Lynn Margulis proposed the theory of endosymbiosis to explain the origin of mitochondria and chloroplasts from permanent resident prokaryotes. According to this idea, a larger prokaryote (or perhaps early eukaryote) engulfed or surrounded a smaller prokaryote some 1.5 billion to 700 million years ago. <br />
<br />
The basic events in endosymbiosis. Image from Purves et al., Life: The Science of Biology, 4th Edition, by Sinauer Associates (www.sinauer.com) and WH Freeman (www.whfreeman.com), used with permission.<br />
<br />
<br />
<br />
Instead of digesting the smaller organisms the large one and the smaller one entered into a type of symbiosis known as mutualism, wherein both organisms benefit and neither is harmed. The larger organism gained excess ATP provided by the "protomitochondrion" and excess sugar provided by the "protochloroplast", while providing a stable environment and the raw materials the endosymbionts required. This is so strong that now eukaryotic cells cannot survive without mitochondria (likewise photosynthetic eukaryotes cannot survive without chloroplasts), and the endosymbionts can not survive outside their hosts. Nearly all eukaryotes have mitochondria. Mitochondrial division is remarkably similar to the prokaryotic methods that will be studied later in this course. <br />
MITOCHONDRIAL INHERITANCE <br />
<br />
WHAT HAVE YOU READ ABOUT MITOCHONDRIAL INHERITANCE ?<br />
PLANT CELLS UNDER LIGHT MICROSCOPE<br />
<br />
STRUCTURE OF CHLOROPLAST<br />
ELECTRON MICROGRAPH OF PLANT CELL <br />
ELECTRON MICROGRAPH OF CHLOROPLASTS <br />
ELECTRON MICROGRAPH OF CHLOROPLAST<br />
STRUCTURE OF CHLOROPLAST<br />
<br />
BICONVEX IN SECTION AND CIRCULAR IN SURFACE VIEW<br />
TWO MEMBRANES – CHLOROPLAST ENVELOPE<br />
INTERMEMBRANE SPACE<br />
THYLAKOIDS/LAMELLAE<br />
STROMA – circular DNA, ribosomes, enzymes, starch grains, lipid globules<br />
<br />
FUNCTION OF CHLOROPLAST<br />
<br />
Chloroplast provide sites on which the biochemical and photochemical reactions can occur without interference from those going in the rest of the cytoplasm.<br />
Membrane system : site of light reaction<br />
Stroma : site of dark reaction<br />
<br />
STRUCTURE OF CHLOROPLAST<br />
<br />
<br />
WHICH ARE THE ORGANELLES<br />
THAT FORM THE CELL’S INTERNAL <br />
SKELETON / OR ARE RELATED TO <br />
MOVEMENT?<br />
<br />
<br />
CENTRIOLES<br />
A pair of cylindrical, rod-like structures<br />
Long axes of centrioles are at 900 degrees to one another<br />
Each centriole contains nine triplets of microtubules arranged in a ring<br />
Found in region known as centrosome<br />
Located close to the nucleus<br />
During cell division, centrioles replicate and move to opposite ends of the cell.<br />
Cells of higher plants lack centrioles<br />
CENTRIOLES<br />
<br />
FUNCTION :<br />
Play a role in nuclear division in animal cells.<br />
Centrosome divides and a pair of centrioles move to opposite poles of the cell where they help to organise the formation of spindle fibres.<br />
<br />
IMPLICATION !<br />
CELLS OF HIGHER PLANTS LACK CENTRIOLES<br />
<br />
WHAT HAPPENS DURING NUCLEAR DIVISION ?<br />
Microtubule-organising centers<br />
Pericentriolar material surrounding the centrioles in the centrosome<br />
Contains ring-shaped structures composed of tubulin<br />
Can nucleate the assembly of microtubules in animal cells<br />
<br />
Centrosome of plants and fungi lack centrioles but still contain microtubule-organising centers.<br />
<br />
MOTILE CELLS<br />
Centrioles divide to produce basal bodies from which flagella and cilia develop.<br />
<br />
Cilia and flagella contain a characteristic “ 9 + 2 “ arrangement of microtubules.<br />
CENTRIOLES VS CILIA<br />
<br />
<br />
CYTOSKELETON<br />
Network of protein fibres<br />
Provides structural support<br />
Controls cell movement<br />
Provides anchorage for organelles and directs their movement within the cell.<br />
CYTOSKELETON<br />
Consists of:<br />
Microtubules<br />
Microfilaments<br />
Intermediate filaments<br />
<br />
<br />
<br />
WHAT IS THE OTHER <br />
NON-MEMBRANOUS<br />
CYTOPLASMIC INCLUSION ?<br />
<br />
<br />
RIBOSOMES<br />
STRUCTURE :<br />
Consists of 2 subunits<br />
Made of of ribosomal RNA and protein<br />
<br />
FUNCTION :<br />
Site of protein synthesis<br />
ER-bound ribosomes make proteins which are secreted at the cell surface<br />
Free ribosomes make proteins for use inside the cell<br />
<br />
RIBOSOMES<br />
<br />
TYPES:<br />
70S – found in prokaryotes<br />
80S – found in eukaryotes<br />
<br />
2 populations<br />
Free ribosomes<br />
ER-bound ribosomes <br />
<br />
<br />
<br />
ER-BOUND RIBOSOMES<br />
<br />
POLYSOMES<br />
<br />
<br />
Checkpoint:<br />
What is the site of enzyme synthesis in cells?<br />
Golgi apparatus<br />
Lysosomes<br />
Ribosome<br />
Smooth endoplasmic reticulum <br />
(J95 Q2)<br />
<br />
WHAT IS THE CYTOPLASMIC <br />
GROUND SUBSTANCE ?<br />
<br />
<br />
CYTOPLASMIC GROUND SUBSTANCE <br />
Aqueous ground substance containing <br />
cell organelles and other inclusions.<br />
<br />
CYTOSOL: <br />
soluble part of the cytoplasm<br />
90% water<br />
Forms a solution which contains all the fundamental biochemicals of life.<br />
Site of metabolic pathways<br />
‘CYTOPLASMIC STREAMING’ – active mass movement of cytoplasm<br />
<br />
WHAT ARE THE STRUCTURES CHARACTERISTIC OF <br />
PLANT CELLS ?<br />
<br />
<br />
FEATURE UNIQUE TO PLANT CELLS<br />
<br />
CHLOROPLASTS<br />
CELL WALL<br />
VACUOLE<br />
CELL WALL<br />
TYPES :<br />
<br />
PRIMARY WALL – cellulose microfibrils, matrix<br />
<br />
SECONDARY WALL – extra layers of cellulose, lignin<br />
<br />
SURFACE VIEW OF CELL WALL SHOWING CELLULOSE MICROFIBRILS<br />
<br />
CELL WALL<br />
FUNCTIONS :<br />
Mechanical strength and skeletal support<br />
Allows development of turgidity<br />
Limits and controls cell growth and shape<br />
Cuticle reduces water loss and risk of infection<br />
PLASMODESMATA<br />
<br />
PLASMODESMATA<br />
<br />
VACUOLES<br />
STRUCTURE :<br />
large central vacuole <br />
surrounded by tonoplast<br />
cell sap<br />
<br />
FUNCTIONS :<br />
osmotic uptake of water<br />
Contains pigments, hydrolytic enzymes, waste products, food reserves<br />
<br />
<br />
HOW ARE THE EUKARYOTIC ORGANELLES CLASSIFIED UNDER FOUR FUNCTIONAL CATEGORIES ? <br />
<br />
<br />
EUKARYOTIC ORGANELLES AND THEIR FUNCTIONSUnited JC Unionhttp://www.blogger.com/profile/06131630985986252085noreply@blogger.com1