2008 a Level H2 Biology P2 Ans

July 24, 2017 | Author: joannetzy | Category: Virus, Meiosis, Adenosine Triphosphate, Redox, Mitosis
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Paper 2 Section A 1. (a) (i) 0-15min: metaphase;; 15-30min: anaphase;; (a) (ii) sister chromatids are separated to become daughter chromosomes after separation of centromeres; (i.e. in graph 2, from 0 m to 50 m);; daughter chromosomes are pulled towards the poles by microtubules / spindle fibres (graph 1, from 20  m);; (a) (iii) 0-10 mins : poles initially move closer (45 - 40 m); as the spindle forms (to move chromosome to metaphase plate); 10-20 mins : poles remain at 40 m; alignment of chromosomes at metaphase plate; 20-30 mins : poles move further apart (40 m - 50 m); sliding action of polar microtubules past one another pushes the two poles apart/ interpolar microtubules lengthen to push the two spindle poles apart;

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2008 GCE A LEVEL EXAMINATION BIOLOGY PAPER 2 SOLUTIONS (b)  Reference to centromere as repetitive DNA sequence or satellite DNA;;  Role in chromatid adhesion;;  References to kinetochore, being proteins which bind onto centromeres to anchor spindle fibres;;  Centromere separates so that sister chromatids could be pulled to the poles;;  The role of centromeres in chromatid alignment/separation;; (c) Each mark required a difference to be stated between the behaviour of chromosomes in meiosis and mitosis, which the vast majority did. The major differences:  homologous chromosomes form bivalents in meiosis but not mitosis  crossing over occurs between non-sister chromatids of homologous chromosomes in meiosis but not in mitosis  homologous pairs line up at the equator in meiosis but in mitosis chromosomes line up singly  homologous chromosomes separate in anaphase 1 of meiosis but chromatids separate in anaphase of mitosis  meiosis results in the haploid number of chromosomes but mitosis maintains the diploid number.  there are 2 divisions in meiosis but only one in mitosis.

2. (a) base substitution;;

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2008 GCE A LEVEL EXAMINATION BIOLOGY PAPER 2 SOLUTIONS (b) Note: what the question required i.e. the consequences of replacing hydrophilic glutamic acid with hydrophobic valine.

the hydrophilic glutamic acid is replaced with hydrophobic valine;; Tertiary structure of the molecule would change;; at low oxygen concentration, hydrophobic areas on different molecules would stick together;; Consequently, HbS molecules would polymerise into fibres;; (c) Note: the point of the question is the change from HbA to HbS would result in changes to the red blood cells.

the change from HbA to HbS would result in changes to the red blood cells i.e. to become sickled shape;; the sickle shaped cells blocking capillaries;; resulting in organ damage as oxygen is not supplied to the organ;; causing the red blood cells to have a shorter life;; (d) the gene for HbF could be activated / transcribed;; resulting in translation to produce HbF;; (e) there would be less HbS in red blood cells; and so there would be less polymerization; The consequence would be less sickling of red blood cells; which would in turn have longer lives HbF has a greater affinity for oxygen;; 3. (a) haemagglutinin binds to receptors on the host cell membrane;; then the virus enters via a vesicle by the process of endocytosis;; the capsid is released into the cytosol;; OR haemagglutinin binds to receptors on the host cell membrane;; the viral envelope fuses with the cell surface membrane of the host cell;; the capsid passes into the cytosol;; (b) influenza virus made use of the host cell’s ribosomes to synthesise proteins;; Reject: use of host machinery (c) RNA dependent-RNA polymerases required by the virus are not be present in the host cell;; but these polymerases were required to replicate viral RNA as well as to transcribe viral RNA to mRNA for subsequent translation to make viral proteins;; (d) if neuraminidase is inhibited then new viruses cannot emerge from the infected cells;; thus unable to infect other cells;; 4. (a) named factor: ionising radiation/ UV light/ tar in cigarette smoke;; Gain of function mutation of proto-oncogenes, which stimulate cell division, to oncogenes;; Loss of function mutation of tumour suppressor genes, which prevent cell division;; The uncontrolled division of cells leads to cancerous growth;; (b) (i) the proto-oncogene codes for proteins that stimulate normal cell division;; (ii) the oncogene was the mutated form of the proto-oncogene;; an oncogene leads to an increase in either the amount of the proto-oncogene’s protein or intrinsic activity of each protein molecule;; uncontrolled cell division results in cancer;;

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2008 GCE A LEVEL EXAMINATION BIOLOGY PAPER 2 SOLUTIONS (c) chromosomal translocation;; near to a powerful gene regulatory sequence, e.g. strong promoter or enhancer, which caused increased transcription;; and therefore translation leading to increase in amount of Myc proteins;; OR increased gene amplification which increase the number of copies of protooncogenes in the cell;; hence more Myc proteins formed;; (d) gene amplification means more copies of Mdr1; gene amplification would give more copies of the transporter; with more copies of the transporter, more drugs would removed from the cells causing the resistance;; (e) an inhibitor that would block the transporter;; thus drug will not be able to bind to the transporter to be removed from the cell;; 5. (a) Parental phenotype Parental genotype Gametes

Purple plant


RrEe ;

Purple plant RrEe ;





RE ;





Gametes RE





RREE Purple

RREe Purple

RrEE Purple

RrEe Purple


RREe Purple

RRee Red

RrEe Purple

Rree Red


RrEE Purple

RrEe Purple

rrEE white

rrEe white

rrEe white

rree white


RrEe Rree Purple Red Correct genotypes;; Correct phenotypes;; Phenotypic ratio: 9 purple: 3 red: 4 white;;

(b) This will help to determine if the difference between the observed and expected is significant;; if the chi-squared value has a probability of greater than 0.05, the difference is not significant and does not deviate from expected ratios, the difference can be attributed to chance and vice versa;; OR if the chi-squared value has a probability of less than 0.05, the difference is significant and deviate significantly from expected ratios and vice versa;; (c) recessive epistasis;;

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2008 GCE A LEVEL EXAMINATION BIOLOGY PAPER 2 SOLUTIONS (d) allele R encodes an enzyme R which converts precursor (colourless compound) to red pigments;; red pigments are converted to purple pigments by enzyme E determined by allele E;; rr was epistatic over the E/e locus; a white phenotype was produced when the genotype is rr--; allele R allele E enzyme R precursor (colourless)

enzyme E red pigments

purple pigments

(e) The plant grown from a white grain had to be crossed with a red plant i.e. RRee or Rree in order to determine that it was homozygous recessive at both loci;; This cross would give either all red or red and white grains;; If any purple grains were seen, then the original plant grown from a white grain must have at least one E at the E/e locus;; 6. (a) stage A - binding of interferon with the receptors (reception); causes two separate receptor polypeptide chains to dimerize; associated Jaks are brought together so that they can cross-phosphorylate each other on tyrosines; They recruit different STATs (STAT1 and STAT2) and phosphorylate them; stage B - the STATs dissociate from the receptors; and dimerize via their SH2 domain;

Source: Molecular Biology of the Cell (b) interferon was too large or could not pass through because it was polar;; whereas cell surface membrane is made up of phospholipid bilayer, the hydrocarbon tails of the phospholipids is non-polar and hydrophobic;; (c) signal amplification;; and the ability to regulate or control the response;; specificity;; the ability of a molecule reaching a cell membrane to activate genes in the nucleus;; the activation of many cells simultaneously;; for a single signal molecule to trigger numerous cellular reactions at once;; [any three]

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2008 GCE A LEVEL EXAMINATION BIOLOGY PAPER 2 SOLUTIONS 7. (a) Environment factor: (1) the type of food;; i.e fruit/insects/different size and hardness of seeds; determine the beak size & shape that is adapted to the food type, e.g. short & wide beak is adapted to crushing hard/large seeds; (2) the availability of food or competition for food - select for finches with the suitable beak shape;; (b) increase in BMP4 expression increases beak height/depth and width or vice versa;; increase in CaM expression increase in beak length;; the extent of expression such as low/moderate etc. depends on the balance of expression of the 2 genes;; (c) Molecular evidence: 1. reference there are different levels of gene expression which in turn leads to variation;; 2. reference to BMP4 and CaM being present in the ancestor;; Explanation: reference to the selection pressure (i.e. the type /availability of food ) which favoured different phenotypes (e.g. different beak shape and size);; those finches that can feed will survive and reproduce;; allele being passed to offspring;; (d) island leads to a geographical barrier or isolation;; results in a lack of gene flow between population of finches;; the fact that there would be different conditions (e.g. environment) on the different islands;; Section B 8. (a) Site

Nature of process


Electron/hydrogen carrier Carbon dioxide


Type of reaction

Krebs cycle Matrix of the mitochondrion Occurs in all aerobically respiring cells Catabolic reaction (breakdown of pyruvate)

Calvin cycle Stroma of the chloroplast Occurs in plant cells/ aglae/ blue-green bacteria Anabolic reaction (formation of triose phosphate or starch)

Production of ATP by substrate Expenditure of ATP – in level phosphorylation reduction of GP to TP and in regeneration of RuBP NAD+ & FAD NADP+ Released by oxidative decarboxylation, 4 CO2 lost per glucose molecule Produced, 2 ATP by SLP, 22 ATP through NADH and FADH2 by OP per glucose molecule

Fixed by RuBP, 6 CO2 used to combine with 6 RuBP

Used in reduction of GP to GALP, and in regeneration of RuBP 18 ATP to form 1 glucose molecule Reactions are mainly oxidative Reactions are mainly as many dehydrogenation reductive where reduced reactions occur / many reduced NADP / NADPH is consumed. NAD formed.

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Oxidation of substrate (acetyl Reduction of substrate (CO2 ) CoA)

Starting material



[1m for each comparison; no mark if comparisons are not made, max 7 marks] (b) 1. 34 ATP from oxidative phosphorylation in the presence of oxygen;; 2. without oxygen as final electron acceptor of the electron transport chain and subsequently link reaction and Krebs cycle stopped;; 3. ETC unable to operate/ no oxidative phosphorylation;; 4. No ATP produced by oxidative phosphorylation;; 5. NAD+ will not be regenerated from NADH;;

6. Link reaction and Krebs cycle unable to operate;; 7. NADH used to reduce pyruvate to ethanol in yeast and lactate in mammalian cells;; 8. Only two ATP from glycolysis;; 9. Glucose not completely broken down;; (c) Similarities 1. involved an electron transport chain; electron carrier molecules transfer electrons down the electron transport chain; 2. the chain uses the exergonic flow of electrons to pump protons across the membrane; 3. the generation of a proton gradient across the membrane - inner membrane of mitochondria/thylakoid membrane of chloroplast; 4. presence of ATP synthetase; through which the protons flow down its concentration gradient; 5. proton-motive force is harnessed; to phosphorylate ADP, forming ATP; Reason for the similarities 6. they were once prokaryotes and were engulfed by another prokaryotes, living within the host cell – endosymbiontic origin;; 9 (a) Classification 1. classification as placing organisms into groups;; 2. the classification is hierarchical with each successive group containing more diverse kinds of organism; 3. Each group possesses unique features;; 4. The lowest or more exclusive taxon is the species; the highest or most inclusive is the kingdom;; 5. The taxa used in order of decreasing size are Kingdom, Phylum, Class, Order, Family, Genus and Species ;; 6. A binomial naming system was used to name the species;; to avoid confusion by common name 7. The first part of a binomial is the genus, the second part is the genus the species belong to;; Phylogeny 8. phylogeny was based on evolution; and involved passing genes from ancestors to descendants; 9. a phylogenetic tree can be constructed to reflect the evolutionary history;;

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2008 GCE A LEVEL EXAMINATION BIOLOGY PAPER 2 SOLUTIONS 10. use of DNA base sequences to assess relationships between species; or by comparing their anatomy; (b) 1. molecular methods were not dependent on subjective judgements or observations and that they involved quantitative differences;; 2. use of a molecular clock to date the time of divergence;; 3. these methods avoided the pitfalls of convergent evolution;; 4. able to measure the degree of relatedness;; 5. being able to compute probable relatedness;; 6. being able to analyse nucleotide sequences;; 7. major phenotypic differences may be due to small genetic differences, that some molecular differences are not visible e.g. supergenes - a group of neighbouring genes on a chromosome;; and can be used on living or dead material;; (c) 1. as the viruses depend on cells for their propagation therefore there is no common ancestor;; 2. viruses originated from fragments of cellular nucleic acids that could move from one cell to another;; 3. they replicate very quickly, providing a great deal of material for the engine of natural selection;; 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;; 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;; 6. plasmid and transposon are found in viral genome which help to transfer genome between cells;; 7. there are different types of virus, that viruses evolve with their host, to escaped genes, or to degenerate cells;;

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