2010 Prelim 2 Essay Compilation

SAJC 2010

8(a) Describe the eukaryotic processing of pre-mRNA in terms of intron splicing, polyadenylation and 5’capping. [6]

Capping
1 modified guanosine nucleotide / 7-methylguanosine / added to 5’ end of mRNA;
2 enzymes (e.g. guanyl / methyl transferase) catalyses reaction between 5’ end of the RNA transcript and GTP;
3 ref. association of cap-binding protein

Polyadenylation
4 addition of 30-200 adenine sequences to the 3’ end of the pre-mRNA;
5 after polyadenylation signal has been transcribed;
6 catalysed by poly(A) polymerase / polyadenylate polymerase;
7 ref. poly-A binding protein binds to poly(A) tail

Splicing
8 Splicing is catalyzed by snRNPs / small nuclear ribonucleoproteins
9 snRNPs are made up of proteins & snRNA (small nuclear RNA)
10 snRNPs + other proteins join together to form a complex called spliceosomes

11 spliceosomes is first responsible for folding the pre-mRNA into the correct orientation for splicing

12 splicing involves cleavage of the 5’ end of the intron
13 and its attachment to the branchpoint sequence
14 to form a tailed loop structure called a lariat
15 intron is then excised/released

16 by being cleaved at its 3’ end
17 and the exons are brought together and ligated
18 spliceosomes dissociates once splicing is completed

8(b) Compare structure and organisation of prokaryotic and eukaryotic chromosome. [6]

[½ mark per correct comparison]

Prokayotic genome Eukaryotic genome
1. Genome size Smaller genomes, 0.6 to 10Mb. Large genomes, being less than 10 Mb – 100,000 Mb
2. Gene length Shorter gene sequences
/ more compact genetic organisation Longer gene sequences
/ presence of more intergenic spaces
3. Chromosome number Single chromosome
/ Haploid Many chromosomes
/ Diploid or polyploid
4. Chromosome structure Circular DNA molecule.
Linear DNA, each contained in a different chromosome.
5. Location Chromosome found in the nucleoid region of the cytoplasm Chromosomes are enclosed within a double-membrane bound nucleus
6. Packaging of DNA Does not form chromatin.
(Prokaryotic DNA is organized into a DNA-protein complex called the nucleoid. Eukaryotic DNA is complexed with histones and other proteins to form chromatin.
7. Introns Coding sequence proceeds from start to finish without interruption by introns. Presence of introns within genes.
8. Repetitive sequences Few repetitive DNA sequences. Many highly repetitive DNA sequences
9. Coding and non-coding DNA Most of DNA are coding sequences (codes for protein, tRNA, or rRNA. Most of DNA are non-coding.

10. Regulatory sequences Small amount of non-coding
DNA consists mainly of regulatory sequences, such as promoters) More complex regulatory sequences (eg. enhancers and silencers)
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

12. Origins of replication One
Many
13. Presence of extrachromosomal DNA Independent small, circular, molecules called plasmids. Circular, double-stranded DNA in mitochondria / choloroplasts.
14. Telomeres Absent Present
15. Centromere Absent Present


8(c) Outline the differences between prokaryotic control of gene expression with the eukaryotic model. [8]

For every comparison:
½ mark for correct comparison
½ mark for correct information

Chromosomal Level (max 2)
Prokaryotes Eukaryotes
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
(Eukaryotic DNA is complexed with histones and other proteins to form chromatin / associated with histones)
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
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.

Transcriptional Level (max 2)
Prokaryotes Eukaryotes
4. One RNA polymerase consisting of five subunits
/ All RNAs synthesized by the same RNA polymerase;



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
/ Three different classes of RNA each synthesized by a different RNA polymerase
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
6. Related genes are transcribed together as operons
/ only 1 promoter
/ polycistronic mRNA
No operon
/ each gene has own promoter / monocistronic mRNA



Post-transcriptional Level (max 2)
Prokaryotes Eukaryotes
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).

8. Primary transcripts are the actual mRNAs
/no post-transcriptional modification
Primary transcripts undergo processing to produce mature mRNAs
/ post-transcriptional modification
9. Triphosphate start at the 5’ end
/ No tail at the 3’ end
/ No splicing Methylated guanosine cap at the 5’ end
/ Poly-A tail at the 3’ end
/ Splicing occurs (also alternative splicing)
10. Lower stability of transcript
/ degradation within seconds or minutes
/ mRNAs shorter half life to rapidly respond to environmental changes Higher stability of transcript
/ prevent transcript degradation
/ mRNAs longer half-life remaining much longer to orchestrate protein synthesis prior to their degradation by nucleases in the cell

Translational level (max 2)
Prokaryotes Eukaryotes
11. Control at this level is unlikely; due to simultaneous transcription and translation Control at translational level:
phosphorylation of ribosomal translation initiation factors
/ negative translational control through regulatory proteins
/ cytoplasmic elongation of poly (A) tails
/ mRNA degradation
/ RNA interference and microRNA

12. Unique initiator tRNA carries formylmethionine Initiator tRNA carries methionine
13. Smaller ribosomes
/ less complex rRNA and protein components Occurs on ribosomes that are larger
/ rRNA and protein components are more complex than those of prokaryotes
14. mRNAs have multiple ribosome binding sites
/ direct the synthesis of several different polypeptides mRNAs have only one start site
/ direct synthesis of only one kind of polypeptide
15. Small ribosomal subunit immediately binds to the mRNA’s ribosome binding site
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


Post-translational Level (max 2)
Prokaryotes Eukaryotes
16. no/minimal post-translational modifications occur Post-translational modifications determine the functional abilities of the protein
17. Proteolysis: Processing eukaryotic polypeptides to yield functional protein molecules e.g. cleavage of pro-insulin to form the active insulin hormone
18. Chemical modification of proteins to yield functional protein molecules
19. Phosphorylation of proteins to increase or decrease its function
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
21. Ubiquitination marks protein for degradation, ref to ubiquitin & proteasome

9(a) Describe the general structure of a named animal virus, and explain why viruses are obligate parasites. [6]
1 example of animal viruses: influzena, HIV, herpes virus etc
2 animal viruses composed of phospholipids/glycoprotein envelope
3 similar in nature to cell surface membrane of animal cell
4 with glycoprotein spikes (for attachment to host cell membrane)
5 enclosed in the viral envelope is the capsid
6 which is composed of capsomeres / protein subunits
7 which contains the viral genome
8 either DNA or RNA, but never both
9 either single or segmented, linear or circular, single-stranded or double-stranded, etc
10 ref. viral enzymes (eg. reverse transcriptase in HIV)

11 acellular / absence of cytoplasm and cellular organelles
12 lacks ribosomes / protein-synthesising apparatus
13 hijacks host cell’s host’s protein synthesis machinery (transcription and translation machineries) to produce own viral proteins [REJECT ‘metabolic machinery’]
14 ref. replication of viral genome
15 metabolically inert / do not grow or divide on their own
16 can only multiply inside living cells, and not on inanimate media

9(b) Outline how the influenza virus is able to bypass the human defense mechanism to cause disease. [4]
1 ref. antigenic drift
2 8 single-stranded RNA segments
3 spontaneous genetic mutation /mutation during replication
4 lack of proof-reading during RNA replication

5 ref. antigenic shift
6 genetic reassortment of the RNA segments between two strains of viruses
7 novel glycoproteins (spikes) produced on viral envelope
/modified neuraminidase and haemaglutinin
8 cannot be recognized by previous antibodies


9(c) Compare and contrast the reproductive cycles of the Lambda phage and the Human Immunodeficiency Virus (HIV). [10]

[1 mark per similarity] (max 4)
1 Life cycle of both viruses involves the stages attachment, penetration, replication, maturation, and release
2 Both attach to their host cell at receptor sites on the host cell’s plasma membrane;
3 Both introduce their viral nucleic acids into their host cell
4 Both are obligate intracellular parasites/make use of their host cell’s resources for synthesis of viral proteins and nucleic acids
5 Both integrate their viral DNA into host genome/viral DNA replicates as part of host’s DNA every time the cell divides

[1 mark per difference] (max 6)
Feature of comparison Lambda phage HIV
Host cell 6 Infects bacterial cells Infects human T-cells
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;
8 Phage injects only its ds DNA; HIV releases its nucleocapsid (ssRNA and reverse transcriptase) into cytoplasm;
Uncoating 9 No uncoating required as capsid does not enter host cell; Uncoating of nucleocapsid to release genome and enzymes;
Fate of viral nucleic acids 10 No reverse transcription
/ 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;




11 Can enter lytic phase or lysogenic phrase (prophage); Viral genome incorporated into host cell chromosomes (provirus);
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;
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;
Release / Exit 14 Host cell is lysed to release the new viruses; The new viruses bud off from host cell’s plasma membrane;







CJC 2010

8 (a) Describe the role of glucagon in regulating blood glucose. [6]

1. when blood glucose levels low, glucagon released from alpha cells in pancreas which acts on liver cells;
2. breakdown of glycogen to glucose;
3. use of fatty acids in respiration;
4. production of glucose from other compounds / fats / amino acids / via gluconeogenesis;
5. liver releases glucose into blood and glucose levels rise and return to normal at 90mg/100ml;
6. switching off glucagon secretion
7. effects of glucagon are antagonistic to insulin;

(b) Describe the reproductive cycles of bacteriophages that reproduce via a lytic cycle, using a named example. [14]

1. Bacteriophages do not randomly attach to the surface of a host cell, T4 phages attach to specific surface structures called receptor sites.
2. They use cell wall lipopolysaccharides or proteins as receptors.
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.
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.
5. Specific strains of bacteriophages can only adsorb to specific strain of host bacteria, this is known as viral specificity.

6. After the baseplate is firmly on the cell surface, conformation changes occur in the baseplate and sheath.
7. Sheath contracts and tail tube penetrates outer membrane.
8. A phage enzyme (lysozyme) “drills” a hole in the bacterial wall.
9. Pilot protein helps phage DNA to cross inner membrane.
10. Phage DNA enters bacteria cytosol. This marks the start of the eclipse period.
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.
12. Replication of phage DNA occurs using the bacterium’s metabolic machinery.
13. Transcription of phage DNA.
14. Synthesis of phage proteins, enzymes and structural componenets using bacterium’s metabolic machinery, e.g ribosomes.
15. Assembly of new bacteriophages around the genomes (spontaneous assembly).
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.
[20 marks]









9 (a) Describe the main ways in which a globular protein differs from DNA. [6]

Structural Differences

1a. a linear, single chain made up of amino acids
1b. double stranded, made up of nucleotides

2a. made up of monomers that consist of up to 20 amino acids
2b. only 4 bases in DNA

3a. ref to details of amino acid structure
3b. ref. to details of nucleotide structure

4a. amino acids held together by peptide linkages
4b. nucleotides held together by phosphodiester links

5a. conformation of a folded tertiary structure
5b. double helix of DNA; reference to association with histone proteins

6a. may contain contain sulphur
6b. DNA contain phosphorus

7 suitable reference to size, quoting appropriate units / location within the cell where they are synthesized or replicated / appropriate named function


(b) Explain how the structure of collagen and haemoglobin are related to their function. [8]
Collagen
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

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

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

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

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
Haemoglobin
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

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

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

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

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

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)

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.

(c) Describe, with examples, the roles of proteins in membranes. [6]

With appropriate reference to integral / peripheral / transmembrane proteins;

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)

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

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

As transport proteins that are either specific or non-specific, transporting ions / molecules / solutes via passive facilitated diffusion / active transport

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

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
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

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


AJC 2010

8 (a) Describe how the structures of RNA are adapted for translation. [8]

Messenger RNA (mRNA)
• Single-stranded;
• One codon codes for one amino acid;
• mRNA (5’UTR) has ribosome attachment site;
• Codons are non-overlapping
• Has spliceosome recognition sites to allow for excising of introns;
• Mature mRNA consists of exons/ no introns;
• Has START/ INITIATOR (AUG) codon and STOP (UAA, UGA, UAG) codons;
• 5’ 7-methyl guanosine cap and 3’ poly-A tail to ensure mRNA stability;
• Complementary base pairing between codon and anticodons; (award only once in either mRNA or tRNA discussion)
(4 marks max)

Transfer RNA (tRNA)
• Has (3’ CCA) end to attach to amino acid/ has amino acid attachment site;
• At least 20 different tRNAs - one for each amino acid
• Has anticodons;
• Has shape complementary to amino acyl tRNA synthetase for activation of amino acid;
• Internal hydrogen bonds to form clover leaf shape/ have specific 3D configuration/ stabilize the molecule;
(2 marks max)

Ribosomal RNA (rRNA)
• A structural component of ribosome;
• Due to hydrogen bonding/ complementary base pairing within;
• Gives rise to a variety of ribosomal types (e.g. 70S, 80S, 50S, 30S, etc);
• Involved in peptidyl transferase activity;
(2 marks max)






(b) Distinguish between chromosomal and gene mutations. [8]

Gene Mutation Chromosomal Mutation/ Aberration
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
Gene locus • Involves one gene locus • Involves a few gene loci
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.
No of chromosome involved • Within one chromosome • Involves one or a few chromosomes
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
Effects on allele • Gives rise to a new allele, resulting in new protein which may have a novel function
• Does not reshuffle alleles • No new allele arisen;
• Only results in reshuffling of alleles on a chromosome.
Ploidy • No change in ploidy • Could also result in polypoidy or aneuploidy
Frequency • More frequent because genes outnumber chromosomes by several thousand to one • Less frequent
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
Visibility under light microscope • Not visible • Usually visible
Examples • Sickle cell anemia (due to substitution of one nucleotide) • Down’s syndrome (due to an extra chromosome 21)
Any valid comparisons;

(c) Explain how the Meselsohn and Stahl experiment supports the semi-conservative model of DNA replication. [4]

Reference to experimental details
• DNA labeled with heavy nitrogen / 15N, then cells transferred to light nitrogen/ 14N for two generations;
• Reference to the three different bands after (CsCl) centrifuge; accept labelled diagram;
• Parental strands unzip/ hydrogen bonds broken;
(1-2 marks max)

How experiment supports
• Parental strands act as template;
• Complementary base pairing;
• New DNA molecule consists of one parental strand and one daughter strand;
• Daughter DNA are genetically identical to each other;
(2-3 marks max)

9(a) Discuss how signal amplification is illustrated by the effect of hormone on glycogenolysis.[8]

• Binding of one / a molecule of epinephrine / glucagon to GPCR causeds GPRC will undergo a conformation change ;;

• to allow the activation of several / few G-protein by displacing GDP for GTP ;;


• 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.
[R: increase cAMP, activation of cAMP]

• These cAMP in turn binds and activates large number of protein kinase A ;;

• Each activated protein kinase A will initiate a sequential phosphorylation and activation of kinases / phosphorylation cascade ;;

• 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.

• lead to the activation of large number of glycogen phosphorylase

• each will catalyse for the breakdown of large amount of glycogen into glucose.


• the number of activated product is always greater than those in the preceding step as one move down the cascade ;;
• binding of 1 glucagon to receptors will lead to the hydrolysis of large number of glycogen;;

* If less than 3 of the highlighted points are present, 1 mark will be deducted from the overall marks.

9 (b) Describe the similarities between the interaction of a substrate with an enzyme and the interaction of a ligand with a receptor. [6]

• Both bind to specific regions of protein (idea of specific region/portion is impt);
• Active site for enzyme;
• Binding site for receptor;
• Both are complementary in shape to sites that they bind to;
• Both will bind to protein via H bonds, ionic bonds, hydrophobic interactions;
• Both can induce a conformational change in the protein when they bind;
• Ref. to induced fit hypothesis for enzymes;
• Ref. to activation of receptor;
• Both interactions are not permanent;
• Ligand dissociates from binding site of receptor / Ligand-receptor complex are removed;
• ES complex formed will be converted to product, which is released;


9(c) Compare between nervous and hormonal control. [6]
Similarities:
Both serve as means of communication/ allow living organisms to respond to stimulus
Stimulus  transmission of message  effector / stimulus elicit a response

Differences:
Endocrine Nervous system
Nature of transmission Chemical transmission (hormones) through blood system Electrical and chemical transmission (nerve impulses and chemical across synapses)mil and electrical
Transmission speed Slow
Slower transmission and relatively slow-acting (adrenaline an exception) Fast
Rapid transmission and response
Pathway of transmission Specific (via nerve cells ) Not specific (blood around whole body)
Strength of message Strength dependent on amount of hormones Impulse size same regardless of stimulus
Length of response Often long-term changes (except adrenaline) Often short-term changes nothing action
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

IJC 2010
9 (a) Describe the structure of DNA and its organisation in a eukaryotic chromosome. [8]

1. 2 polymers of deoxyribonucleotides chains twisted into a double helix;;
2. held by H- bonds between complementary bases, A=T, C≡G;;
3. a. purine-pyrimidine: 2 nm diameter;
3. b. 0.34nm/10 bp per complete turn;;
4. distinct polarity 3’ end with free hydroxyl group at 3C and 5’ end with phosphate group at 5C;;
5. DNA molecule associated 1¾ turns with 8 histone proteins/octomer, formed nucleosome core;;
6. histones positively charged, formed ionic bonds with negatively charged DNA;;
7. associate with H1 histone to form nucleosome, linked by spacer/linker DNA forming 10nm ‘beads-on-string’ structure;;
8. further coiled into a solenoid forming a 30nm fibre;;
9. futher looping in presence of scaffold proteins to form metaphase chromosomes;;
AVP;;

(b) Describe the control of gene expression in eukaryotes. [12]
Gene Amplification
1. duplication of genes to increase the number of copies;;
Chromatin remodelling
2. acetylation removes +ve charges from histones, loosen –ve charged DNA, allow GTF to bind;;
3. demethylation removes methyl grps from proximal promoters, e.g. CpG islands;;
Transcriptional control
4. cis-acting elements, GTFs binding to promoters and proximal promoters;;
5. binding of STF e.g. activators to enhancer, repressors to silencers á and â transcription rate;;
Post-transcriptional control
6. ss pre-mRNA, with 5’ capping with modified guanine/methylated guanine;;
7. 3’ polyadenylation for stability/ nuclear export of mature mRNA;;
8. splicing of introns out and joining of exons forms mature mRNA by spliceosome;;
9. association with proteins for nuclear export;;
10. mRNA editing, the ∆ in codon results in different functional polypeptide;;
Translational control
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;;
12. gene silencing by miRNA, forming complementary dsRNA in DICER cplx prevent ribosomal attachment/ cleave mRNA;;
Post-translational control
13. proteins not required are tagged for ubiquination by protease;;
14. AVP;;
[Total: 20]

10 (a) Explain the principles of homeostasis. [4]

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;;
2. The regulated variable is usually at a set point which can be changed by a stimulus;;
3. change detect by a receptor which sends signals to control center/regulator;;
4. activates effector resulting in a response to restore set point via positive/negative feedback;;

(b) Describe the roles of insulin and glucagon in regulation of blood glucose. [10]

Insulin
1. [bld glu] á above norm, β cells in islets of Langerhans detect the á;;
2. insulin released to bind to rtk receptors of effectors e.g. liver cells;;
3. results in increased no. of glucose transporters in the liver cells to increase uptake of glucose;;
4. convert excess glucose to glycogen in glycogenesis by activating glycogen synthetase;;
5. Depresses rate of synthesis of glucose from non-carbohydrate source via gluconeogenesis;;
Glucagon
6. [bld glu] â below nom, α cells in islets of Langerhans detect the drop;;
7. glucagons released bind to GPCR on effector e.g. liver cells;;
8. results in activation of glycogen phosphorylase to convert glycogen into glucose;;
9. reached set point, neg feedback mechanism prevents further release of hormones by pancreas;;
10. AVP
(c) Describe the molecular events when glucagon binds to its target cell. [6]

1. Glucagon binds to the complementary ligand-binding site on GPCR receptor causing a conformational change to receptor;;
2. GTP from the cytosol then displaces GDP from the nucleotide-binding site on Ga;;
3. Ga simultaneously dissociates from its Gβγ, translocate along the plasma membrane to activate the enzyme adenylyl cyclase;;
4. Adenylyl cyclase catalyses the conversion of ATP to cyclic AMP (cAMP);;
5. cAMP activate relay protein, protein kinase A (PKA);;
6. Activated PKA then phosphorylates other relay proteins and enzymes, activates other enzymes in phosphorylation cascade;;
7. The effect of cAMP is removed by phosphodiesterase which converts cAMP back to AMP;;
8. AVP;;
[Total: 20]

JJC 2010
8 (a) Describe the process of ATP production in the chloroplast. [7]

1. Photosynthetic pigments such as chlorophyll a, chlorophyll b and carotenoids;
2. are embedded on the thylakoid membranes;
3. are arranged in photosystems;
4. Two photosystems PSI (P700) and PSII (P680) which absorbs wavelengths of 700nm and 680nm respectively;
5. Photon of light absorbed by both PS will excite electrons in special chlorophyll a in reaction centres to higher energy levels;
6. These excited electrons are accepted by primary electron carriers;
7. And pass on to electron carriers on the electron transport chain (ETC);
8. Electron carriers on the ETC are of progressively lowered energy level;
9. As electrons move down the ETC, energy is released;
10. The energy released is used to pump H+;
11. From the stroma into the thylakoid lumen;
12. Accumulation of H+ generate potential energy;
13. H+ return to the stroma by diffusing down their concentration gradient
14. through H+ channel;
15. Energy is released and used to couple the phosphorylation of ADP + Pi to produce ATP;
16. Catalyzed by ATPase;
Max 7

(b) Give an overview of the Krebs cycle and its significance in respiration. [7]
Overview
1. Occurs in matrix of the mitochondrion;;
2. Per cycle, one molecule of acetyl CoA (2C) undergo oxidative decarboxylation;;
OR
Citrate (6C) undergoes oxidative decarboxylation to form α–ketoglutarate (5C);
α–ketoglutarate (5C) undergoes oxidative decarboxylation to form succinate (4C);
3. Acetyl CoA (2C) and oxaloacetate (4C) are condensed to form citrate (6C);;
4. Description of the decarboxylation process i.e. 6C to 5C to 4C intermediates;;
i. Citrate (6C) undergoes oxidative decarboxylation to form α–ketoglutarate (5C)
ii. α–ketoglutarate (5C) undergoes oxidative decarboxylation to form succinate (4C)
iii. Succinate (4C) undergoes oxidation to form fumarate (4C)
iv. Fumarate (4C) is converted to malate (4C)
v. Malate (4C) undergoes oxidation to form oxaloacetate (4C)
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;;
6. At the end of the cycle, one molecule of oxaloacetate (4C) is regenerated to receive more acetyl groups and the cycle continues;;
Max 5 marks

Significance
7. Completes the oxidative breakdown of glucose to CO2 and H2O and release sufficient energy;;
8. Release of hydrogens (NADH produced) which can be used in oxidative phosphorylation to provide energy to produce ATP;;
9. Carbohydrate intermediates can be converted to amino acids;;
10. Permits amino acids to enter Krebs cycle when glucose is in short supply;;
Max 3 marks

(c) Compare and contrast oxidative phosphorylation and photophosphorylation. [6]
Similarities:
1. involve transfer of electrons between electron carriers;;
2. energy released from electron transport is used to generate proton gradient;;
3. potential energy of proton gradient is harnessed for ATP synthesis;;

Differences:
Features Photophosphorylation Oxidative Phosphorylation
Sources of energy for ATP synthesis energy for making ATP comes from light energy for making ATP comes from glucose oxidation processes;;
Location
Thylakoid membrane of chloroplast Inner membrane of mitochondria;;
Involvement of light energy Required to energize the electrons in special chl a Not required;;
Electron donors
For non-cyclic reaction: water
For cyclic reaction: PS I NADH, FADH2;;

Electron acceptors
For non-cyclic reaction: NADP+
For cyclic reaction: PS I Oxygen;;

Establishment of proton gradient
H+ pumped inwards, from stroma to thylakoid space
H+ pumped outwards, from mitochondrial matrix to intermembrane space;;
Max 6

9 (a) Describe the important events that occur during Meiosis I and explain the significance of each event. [10]
Important events during Meiosis I Significance
1. Prophase I;
2. Synapsis ;
3. pairing of homologous chromosomes;
4. to form tetrads / bivalents ; 5. Allows for alignment of genes on chromosomes ;
6. to prepare for crossing over between the homologous regions later ;
7. Crossing over ;
8. between non-sister chromatids of homologous chromosomes ; formation of chiasmata ; 9. Allows for exchange of genetic material ;
10. results in new combination of alleles ;
11. genetic variation in the gametes later ;
12. Metaphase I;
13. Homologous chromosomes are arranged along equator in pairs ;
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 ;
16. genetic variation in gametes later ;
17. Anaphase I;
18. Homologous chromosomes are pulled apart to opposite poles ;
19. movement of each homologous pair is independent of other homologous pairs / random segregation ; 20. Allows gametes to contain a haploid set of chromosomes ;
21. for restoration of diploid no. of chromosomes ; following fertilization ;
22. Allows gametes to contain a random combination of paternal and maternal chromosomes ;
23. genetic variation in gametes later ;
24. Telophase I & cytokinesis;
25. Division of cytoplasm (and organelles) ;
26. furrowing (animal cells) / cell-plate formation (plant cells) ; 27. Allows formation of 2 haploid daughter cells ;
28. Preparation of cells for Meiosis II ;

(b) Describe the significance of gene amplification. [6]

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;;
2. These amplified genes can be transcribed and translated OR leads to an overproduction of the corresponding mRNAs and proteins;;

3. Meet the needs of cells resulting in higher level of mRNA and polypeptide synthesis at different development stages of cells;;
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;;
5. Cause cancer due to over-expression of proteins (leading to development of malignant tumours);;
6. If an oncogene is amplified, then the resulting over-expression of that gene can lead to de-regulated cell growth;;
7. An example is the amplification of the ErbB-2 oncogene in breast and ovarian cancers;;

8. Genes related to drug resistance in these cancer cells are increased; conferring them with drug resistance;
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;
10. Confer selective advantage which allow organisms to survive in a particular environment;; Max 6
(c) Outline the end-replication problem in eukaryotic chromosomes. [4]

1. DNA polymerase;
2. can only add nucleotides to the 3′ end;
3. of a elongating/pre-existing polynucleotide/primer;
4. primer at 5’ end;
5. removed;
6. but cannot be replaced with DNA/ gap cannot be filled in;
7. because no 3’-OH available (to which a DNA nucleotide can be added);
8. 5’ end of the DNA becomes shorter relative to that of the previous generation;


YJC 2010
8
(a) Describe how the presence of lactose results in the transcription of the Lac operon in prokaryotes.
[8]
Correct diagram of the Lac operon (in particular, the position of the regulatory and structural genes, as well as promoter and operator);
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)
Trace amounts of lactose permease (minimally transcribed / expressed) on cell surface membrane of prokaryote permits the entry of lactose;
Trace amounts of β-galactosidase (minimally transcribed / expressed) converts lactose to allolactose as a side reaction;
The Lac repressor is a protein that is transcribed and translated from LacI gene (regulatory gene) located in front of the promoter site;
The repressor has two binding sites – DNA binding site and repressor binding site (referred to as allosteric site);
Allolactose binds to the Lac repressor bound to the operator of the Lac operon;
Causes a conformational change in the repressor, preventing it from being attached to the operator site (releasing it from the DNA binding site);
Operator site overlaps with promoter site;
RNA polymerase binds to the promoter site and transcribes the structural genes;
LacZ is transcribed and translated to produce β-galactosidase, LacY produces lactose permease, and LacA produces transacetylase;
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;
Accurate reference to glucose and CAP-binding site: 1 mark max;


(b) Compare and contrast the structure and organisation of prokaryotic and eukaryotic chromosomes.
[6]
Similarities (max 3)
They both have origins of replication that allow for them to be replicated;
They consist essentially of the same types of nucleotides – adenine, thymine, cytosine and guanine;
They are composed of DNA double helix structure;
Similar kinds of bonds can be found in the chromosomes – phosphodiester, hydrogen, glycosidic;
Both are compacted and associated with proteins that help to pack them;
Differences (max 5)
Prokaryotic chromosomes are circular while eukaryotic chromosomes are linear;
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;
Eukaryotic chromosomes have multiple origins of replication while prokaryotic chromosomes have only one;
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;
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;
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;
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;



(c) Explain how mutations related to a named tumour suppressor gene lead to cancer.
[6]

E.g. of named tumour suppressor gene: P53;
Normal functions of tumour suppressor gene:
Activate DNA repair;
Inducing mitotic arrest (stopping of cell cycle);
Initiate apoptosis (programmed cell death);
Types of mutations causing loss of function
Gene mutation (any named kind);
Mutation in regulatory sequence;
Both copies of alleles to be mutated in order for the cancer phenotype to be expressed (loss of function);
Description of cell activity resulting from loss of function
DNA damage not repaired for proteins regulating cell cycle leads to cell division;
Checkpoints of mitosis not regulated resulting in continuous cell division;
Cancer cells do not undergo apoptosis and carries on dividing;

[20]


9 (a) Explain the main ways in which a globular protein differs from a sequence of DNA.
[8]

Pt of comparison Globular proteins DNA
1 Shape Compact & round
Reject Globular Linear in eukaryotic cells

Note that prokaryotic cells have circular DNA.
2 Solubility in water Yes No/less soluble;
3 Monomers Amino acids deoxyribonucleotides;
4 Types of monomers 20 used in humans 4 types (A,T,G,C);
5 Bonding (1) Linked by peptide bonds Linked by phosphodiester bonds;
6 Number of subunits One or more Always 2 under physiological conditions;
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;
8 Function Multiple functions: Enzymatic, structural, transport, energy source (any 2) Carry hereditary information/code for synthesizing products;
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;
10 Levels of organization 4 levels 1 structure/a-helix;
11 Charge Can be of any charge (positive, negative or neutral) Negatively charged;
12 Location Everywhere/both inside and outside the cell Only inside the cell /in nucleus;

13 Process of synthesis transcription and translation DNA replication
14 Template for synthesis mRNA DNA /parent DNA




(b) Outline the main differences in structure and function between starch, cellulose and glycogen.
[6]
Similarities
All three are polymers made up of glucose molecules

Differences
Starch glycogen Cellulose
Types of bonds and significance to function
Amylose is a planar polymer of glucose linked mainly by α(1→4) glycosidic bonds

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.

The bonds can be broken down by hydrolysis easily- suitable as a energy store
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

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.








The bonds cannot be broken down by hydrolysis easily- suitable as structural carbohydrates.

Packing and relation to function Coil helically with multiple branching into a compact structure

thus can be stored in large amounts within a fixed volume. Form fibrils / fibers,


thus providing tensile strength for cell wall.




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.

Amylose forms a colloidal dispersion in hot water (which helps to thicken gravies) whereas amylopectin is completely insoluble.

Insoluble in water, thus does not affect the water potential of cells.
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) .


Large intermolecular spaces, thus allow water to move through the cell wall





PJC 2010
8 (a) Compare the structure and organization between prokaryotic and eukaryotic genome. [8]
(b) Describe the eukaryotic processing of pre-mRNA. [6]
(c) Describe various ways in which gene expression may be controlled at translational and post-translational level.
[6]
9 (a) Describe the structure of DNA and tRNA. [6]
(b) Describe the process of DNA replication in prokaryotes. [8]
(c) Describe the experimental evidence for semi-conservative replication. [6]

Compare the structure and organization between prokaryotic and eukaryotic genome.
Similarities:
1. Both consist of double stranded DNA;
2. Both shows different levels of packing to compact the genome such that they can fit into the cell;

Differences between prokaryotic and eukaryotic genome
Prokaryote Eukaryote
Genome description Circular/closed loop linear chromosome;
Size Shorter. (Contains a few million bp.) Typically longer chromosome. (Contains tens of millions to hundreds of millions of base pairs (bp) in length.)
Chromosome location Chromosome not in nucleus, found in nucleoid region, (in contact with cytoplasm); Chromosome found in nucleus;
Packing of DNA (2 marks) Packing also take place but not as extensive.
The chromosomal DNA must be compacted about 1000-fold in order to fit into the bacterial cell.
There are two main ways:
1. Formation of looped domain – a segment of DNA that is folded into a loop-like structure.
DNA-binding proteins are involved in holding the loops in place.
AND
2. Supercoiling of the looped DNA.

Multi-levels of packing.



Basic level  “Beads-on-a-string” model where DNA wind around histone proteins to form nucleosomes.


Nucleosomes undergo further packing to form 30 nm fibre/solenoid  300 nm fibre

Introns Introns not found within genes.
(mRNA formed during transcription need not be processed. They can be translated immediately.) Introns (non-coding, repetitive sequences) found interspersed within a gene.
(This arise the need for post-transcriptional modification  RNA splicing.)
No of ori 1 origin of replication on 1 chromosome; Each chromosome consists of several origin of replication;
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
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.;
plasmid An extrachromosomal plasmid is usually found in prokaryotes. Absence of plasmid
Differences 7 max

1. Post- transcription modification is required to process the pre-mRNA to form mature mRNA and to facilitate the export from nucleus to cytoplasm.
2. The 5’ end, the end made first during transcription, is immediately capped off with a modified form of a guanine nucleotide;
3. The enzymes required for capping is grouped to form the Capping Enzyme Complex (CEC) and;
4. is bound to the RNA polymerase II before transcription starts;
5. At the 3’ end, where there is polyadenylation sequence present
6. an enzyme, poly (A) polymerase (PAP);
7. makes a poly(A) tail consisting of 50 to 250 adenine nucleotides;
8. The signals for RNA splicing are short nucleotide sequences at the ends of introns;
9. Particles called small nuclear ribonucleoproteins or snRNPs, recognize these splice sites.
10. snRNPs are located in the cell nucleus and are composed of RNA and protein molecules;
11. The RNA in a snRNP particle is called a small nuclear RNA (snRNA), each molecule is about 150 nucleotides long;
12. Several different snRNPs join with additional proteins to form an even larger assembly called a spliceosome;
13. The spliceosome interacts with the splice sites at the ends of an intron, intron region was looped out
14. It cuts at specific points to release the intron,
15. then immediately joins together two exons that flanked the intron;
16. snRNA plays a role in the catalytic process as well as in spliceosome assembly and splice site recognition;
max 6

(c) Gene expression may be controlled at translational level via these processes:
Process involved Mechanism Regulation
Gene expression control of during translation initiation 1. Regulatory proteins binds to specific sequence within the 5’ UTR;
2. Prevent attachment of small ribosomal subunit to mRNA; 15. Block translation initiation;
Global control
translation initiation 3. Activation /inactivation of one or more of TIF(translation initiation factors) (by phosphorylation/dephosph);
4. Promote/ Inhibit the formation of translation initiation complex; 16. Translation will be initiated for ALL the mRNA in a cell;
mRNA degradation 5. Enzymes (3’5’ exonucleases) breaks down poly (A) tail first;
6. Followed by enzyme removal of 5’ end (decapping)  critical step;
7. Allows nuclease enzymes to rapidly digest the mRNA; 17. Control the lifespan of mRNA in the cytoplasm  the amount of proteins formed;

Gene expression may be controlled at post translational level via these processes:
Process involved Mechanism Regulation
Post translational modification
Protein processing (during translation),
Chemical modification 8. A polypeptide will coil/fold forming function 3D prot with secondary & tertiary structure (cisternae of RER);
9. Cleavage of initial pro-insulin (polypeptide) into active hormones;
10. Chemical modification- adding of sugar/lipids/phosphate group;
11. Cell surface proteins must be transported to target destination to function; 18. Protein processing control the type of proteins to be formed;
Protein degradation
12. To mark a protein for destruction, ubiquitin protein is tagged;
13. Leads to recognition by large proteasomes;
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;

Structure:
1. The DNA molecule consists of two deoxyribonucleotide chains
2. The two chains coil round each other to form a double helix.
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,
4. with the nitrogenous bases arranged as side groups off the chains (project into the centre of the double helix).
5. The two chains run in opposite directions (antiparallel) and
6. are held together by hydrogen bonds between their nitrogenous bases.
7. Hydrogen bonding can occur only between a purine and a pyrimidine.
8. Thus the sequence of bases in one chain determines that in the other chain. The two chains are said to be complementary.
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.
10. Along the axis of the molecule the distance between adjacent nucleotides is 0.34nm
11. One complete turn of the double helix is 3.4 nm and comprises 10 base pairs

tRNA
Structure
1. Single stranded RNA
2. Wound into clover-leaf shape with 3 loops
3. 3’ end of the tRNA always ends in the base sequence of CCA
4. and binds an amino acid
5. 5’ end of the tRNA always ends in a base, guanine.
6. Acts as an intermediate molecule between the triplet code of mRNA and amino acid sequence of polypeptide chain.
7. Carries the correct amino acid to the forming polypeptide chain on the ribosome.
8. One loop contains a triplet of bases called an anticodon which pairs with the complementary codon on mRNA.
9. There are more than 20 different tRNA molecules in a given cell, each carrying a specific amino acid.

(b)Unwinding DNA double helix
1. Mechanism is by semi-conservative replication.
2. Activated deoxyribonucleotides and transported into the nucleus via pores in the nuclear envelope
3. Replication of DNA begins at specific sites called the origins of replication (Ori).
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.
5. Topoisomerase helps to prevent over-straining of the replication fork.
6. The separating DNA results in a “replication bubble”, allowing replication to take place at both ends, known as a replication fork (bidirectional)
7. The unwinding DNA chains provide templates for the synthesis of new DNA chains.
8. After parental strand separation, single-stranded binding protein (SSB) binds to the unpaired DNA strands, stabilizing them.


Formation of RNA primer
9. The enzyme primase catalyses the synthesis of a short RNA primer, (about ten ribonucleotides long),
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.
11. The new DNA strand will start from the 3’end of the RNA primer.
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

Synthesis of new DNA strand
13. DNA polymerase III recognises the bases and selects free deoxyribonucleotides that are complementary to those on the template strand.
14. Adenine pairs with thymine and vice versa. Similarly, cytosine pairs with guanine and vice versa.
15. Both daughter strands are synthesized in the 5' to 3' direction.
16. One of the daughter strands, called the leading strand, is synthesized continuously/ towards replication fork
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.
18. The DNA strand is called the lagging strand, synthesised in the form of short fragments called Okazaki fragments
19. The enzyme, DNA ligase catalyses the formation of phosphodiester bonds between two Okazaki fragments.
20. At the end of the replication, both the parental and daughter strands rewind into a double helix.
21. The new DNA molecule consist of 1 parental and 1 newly synthesized daughter strand;

(c)
1. Semi-conservation replication  both DNA strands are template for DNA synthesis of new strands and
2. The new DNA molecule consist of 1 parental and 1 newly synthesized daughter strand;
3. Messelson and Stahl experiment
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.
5. The cells were then transferred to another medium containing normal nitrogen (14N) and
6. allowed to divide once and therefore the DNA replicates once.
7. Cells were then allowed to divide a second and third time.
8. DNA, extracted from cells of A, B, C and D, was then centrifuged
9. with CsCl2
10. DNA in the centrifuge tubes appeared as narrow bands when examined under ultraviolet light
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)
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
13. Tube B contains only hybrid DNA with one band occupying the intermediate density position
14. Tube C contains both light and hybrid DNA in the ratio 1:1 and
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.
(points 12 to 15 – need to explain or annotated diagram)


TPJC 2010
QUESTION 8
(a) Explain how mitosis produces genetically identical cells. [6]
1. Replication of DNA during S-phase of interphase
2. To produce genetically identical sister chromatids attached via centromeres
3. Kinetochore microtubules attach to kinetochores on the centromere
4. Exact alignment of sister chromatids on metaphase / equatorial plate
5. Centromeres divide resulting in separation of sister chromatids
6. Sister chromatids, now as chromosomes, move to opposite poles of the cell
7. Two genetically identical cells are produced when the cell membrane furrows in,
8. Definition of genetically identical cells – some number and type of chromosomes;
Max 6

(b) Describe the role of meiosis in natural selection. [10]
1. Meiosis produces genetically varied gametes
2. Leading to formation of genetically varied organisms due to random fusion of gametes
3. Prophase 1: crossing over between homologous chromosomes result in new combination of alleles
4. Metaphase 1/anaphase 1: independent assortment of chromosomes occur to result in gametes randomly distributed to either poles of the cells
5. Anaphase 1: segregation of homologous chromosomes to both poles will result in the formation of haploid gametes
6. For natural selection to occur, there must be a varied gene pool
7. In the presence of an environmental change, the environment will select the organisms which has selective advantage
8. This organisms will be able to survive, reproduce and pass on their alleles to their offspring
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
10. Evolution will not occur OR promotes speciation
11. Award marks for example given: variation amidst finches allow speciation etc


(c) Explain how one named factor increases the chances of cancerous growth. [4]
1. Named factor: ionizing radiation / UV light / tar in cigarette smoke etc
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.
3. Damage to DNA which cannot be repaired
4. Mutation to proto-oncogenes / tumour suppressor genes
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;]

Question 9
(a) Explain briefly what is meant by the terms gene mutation and chromosome mutation.[10]

Definition of mutation:
• Change in the amount, arrangement or structure of the DNA of an organism.
• Produces a change in the genotype and may result in the change in appearance of a characteristic in a population.

Definition of Chromosomal mutations
• Result of changes in the number and structure of chromosomes.

Chromosome mutation: Changes in chromosome number
• These changes may involve the loss or gain of single chromosomes, a condition called aneuploidy,
• Or the increase in entire haploid sets of chromosomes, a condition called euploidy (polypoidy).

Aneuploidy
• Loss of gain or single chromosomes eg. (n+l), (2n+l), (n-1), (2n-1)
• 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.

Any named example:
1. Down’s syndrome (Mongolism)
 Presence of an extra chromosome no. 21 in the cells.
 Symptoms – mental retardation, short stocky body, thick neck, characteristic folds of skin over the inner corner of the eye.
 Related to the age of the mother’s egg cells.

2. Klinefelter’s syndrome (XXY)
 Failure of X chromosome to separate during oogenesis in female.
 Symptoms – male with female characteristic, sterile, testes very small, little facial hair, breasts may develop, low intelligence.

3. Turner’s syndrome (XO)
 Sterile female.
 Lacking normal secondary sexual characteristics.

4. XXX
 Female, normal appearance
 Fertile but mentally retarded

5. XYY
 Male, variable intelligence, may possess psychopathic traits or tendency for petty criminal acts

Euploidy (Polyploidy)
• Cells contain multiples of the haploid number of chromosomes ie. polyploids 2n, 3n.....
• More common in plants than animals
• Associated with hybrid vigour
• 2 forms:
a. Autopolyploids
 An individual with more than 2 sets of chromosomes all derived from the same species.
 Can be induced by using colchicine.
 Autopolyploids can be fertile if they have an even number of chromosome sets.

b. Allopolyploids
 An individual with increased chromosome number derived from different haploid sets.
 Many inter-species hybrids are sterile because chromosomes do not pair up at meiosis - diploid hybrid ( sterile ).
 If chromosomes are doubled, each member has another to pair up with and meiosis proceeds normally -- fertile, allopolyploid.

Chromosome mutation: Changes in structure
1. Deletion: removes a chromosomal segment
2. Inversion: reverses a segment within a chromosome
3. Translocation: moves a segment from one chromosome to another, non-homologous one
4. Duplication: repeats a segment

Any named example:
1. Cri-du-chat syndrome:
• deletion in the short arm of chromosome 5
• child is physically and mentally retarded

2. Chronic myelogenous leukaemia (CML):
• Translocation
• Most of the chromosome 22 has been transolcated onto the long arm of chromosome 9.
• In addition, the small distal portion of chromosome 9 is translocated to chromosome 22.
• Caused increased cell proliferation, reduced apoptosis  cancer

3. Acute myelogenous leukaemia (AML)
• Chromosomal inversion on chromosome 16
• Results in cancer

4. Charcot-Marie-Tooth syndrome (CMT)
• In one form, chromosomal duplication on chromosome 17
• High gene dosage of myelin sheath protein resulting in abnormal structure and function of the myelin sheath
• Symptoms: weakness of lower foot, loss of balance, poor motor skills and muscle atrophy.


Gene Mutation
Definition of Gene mutation
• Sudden and spontaneous changes
• Change in the nucleotide sequence of DNA:
Duplication, insertion, deletion, inversion or substitution of bases

Effects:
• Alteration in a sequence of nucleotides  Change in amino acid sequence of polypeptide chain  affect phenotype of individual ( may be inheritable
Types:
(a) Substitution
-1 nucleotide is replaced by another

Effects:
(i) Silent Mutation
• Due to degeneracy of genetic code
• E.g. DNA: 3’-CCG-5’ ( 3’-CCA-5’
mRNA: 5’-GGC-3’ ( 5’-GGU-3’
• Glycine is still produced
• OR if new amino acid have properties similar to that of replaced amino acid

(ii) Missense Mutations
• If mutation occurs in crucial areas. E.g. active site of enzymes
• Result in useless/ less active protein that impairs cellular function. E.g. Sickle Cell Anaemia

(iii) Non-sense Mutations
• Cause a change in codon for amino acid to stop codon
• Result in premature termination of translation, non-functional protein is formed

(b) Inversion










(C) Frameshift Mutation
• Due to the insertion/ deletion of nucleotide pairs
• Have a more disastrous effect as it cause ribosome to read incorrect triplets from point of mutation
• Result in shorter chain ad/ or mostly non-functional proteins

E.g.
mRNA 5’ AUG AAG UUU GGC UAA 3’
Protein met– Lys– Phe – Gly

(i) Frameshift mutation causing immediate nonsense
mRNA 5’ AUG UAA GUU UGG CUA A 3’
Protein met– STOP

(ii) Frameshift mutation causing extensive nonsense
mRNA 5’ AUG AAG UU|G GCU AA 3’
Protein met– Lys– Leu- Ala-

(iii) +/ - 3 nucleotides: no Frameshift BUT additional/ missing amino acid
mRNA 5’ AUG UUU GGC UAA 3’
Protein met– Phe – Gly

Case Study: Sickle Cell Anaemia
Normal Adult: Haemoglobin (Hb A)
• Quaternary protein with tetramer consisting of 2 different types of polypeptide chain, 2α and 2β globin chains
• Chains are encoded by 2 different genes on 2 different chromosomes
Normal S.C.A
Gene 3’-CTT-5’ 3’-CAT-5’
Codon on mRNA 5’-GAA-3’ 5’-GUA-3’
Polypeptide -Glu- -Val-
Solubility Soluble Less soluble
At low [O2] Remain soluble Crystallize into rod-like fibres
Shape of RBC Round, biconcave Sickle shape


(b) Explain how mutations for antibiotic resistance spread so rapidly among bacteria. [4]
• largely because bacteria reproduce very quickly;
• mutation can be quickly passed on to large numbers of descendents (usually) via plasmids during binary fission;
• through conjugation, whereby transfer of antibiotic resistance genes occurs across a mating bridge;
• through transformation, whereby antibiotic resistance genes taken up by competent bacterial cells through transient pores in the bacterial cell wall;
• through transduction, whereby (defective) bacteriophages transfer bacterial genes / antibiotic resistance genes;
Max. [4]

(c) Outline the role of isolating mechanisms in the evolution of new species. [6]
• ref. to definition of species;
• ref. allopatric;
• geographical isolation;
• ref. to examples e.g. islands/ lakes/ mountain chains / idea of barrier;
• ref. to example organism;
• ref. to populations prevented from interbreeding;
• isolated populations subjected to differential selection pressure/ conditions;
• over time sufficient differences to prevent interbreeding;
• ref. sympatric;
• ref. to reproductive isolation;
• ref. behavioural barriers (within a population);
• e.g. day active/ night active;
• correct ref. to gene pool;
• change to allele frequencies;
[ 6 max ]

2010 YJC