SAJC 2008 H2-Bio-TYS-ANS

2008 ‘A’ Level H2 Biology Mark Scheme (updated 10 Jan 2015) 1 2008 ‘A’ Level H2 Biology Mark Scheme PAPER 1 (MCQ) 1 A 21 D 2 B 22 A 3 C 23 C 4 A 24 B 5 B 25 C 6 A 26 B 7 B 27 D 8 A 28 D 9 C 29 A 10 D 30 A 11 B 31 C 12 A 32 B 13 C 33 C 14 B 34 A 15 B 35 D 16 D 36 A 17 D 37 B 18 C 38 B 19 C 39 D 20 C 40 C

2008 ‘A’ Level H2 Biology Mark Scheme (updated 10 Jan 2015) 2 PAPER 2 (CORE) QUESTION 1 (a) (i) 1 Metaphase 2 Anaphase (ii) 1 centromeres divide after 15 min R centromeres split 2 chromatids separate to become chromosomes which are then pulled apart by spindle fibres to opposite poles of the cell REJECT vague references to moving of chromatids without reference to spindle fibres REJECT vague references to ends of the cells rather than poles of the cell (iii) 1 poles move closer together and then move further apart 2 ref supporting data 3 role of pole-to-pole fibres (polar microtubules/polar spindle fibres) in elongating the centre of the spindle so that poles are pushed further apart REJECT poor reference to curve C moving (b) 1 repetitive DNA 2 involved in chromatids adhesion followed by division during anaphase 3 ref to kinetochore formation on centromeres, being proteins which bind onto centromeres to anchor spindle fibres so that they could be pulled to the poles 4 involved in chromatid alignment and separation (c) 1 homologous chromosomes form bivalents in meiosis but not mitosis 2 crossing over occurs in meiosis but not in mitosis 3 homologous pairs line up at the equator in meiosis but in mitosis chromosomes line up singly 4 homologous chromosomes separate in anaphase 1 of meiosis but chromatids separate in anaphase of mitosis 5 meiosis results in the haploid number of chromosomes but mitosis maintains the diploid number. 6 there are 2 divisions in meiosis but only one in mitosis. Examiner’s comments: Note that each mark requires a difference to be stated between the behaviour of chromosomes in meiosis and mitosis.

2008 ‘A’ Level H2 Biology Mark Scheme (updated 10 Jan 2015) 3 QUESTION 2 (a) 1 (base) substitution REJECT missense mutation, point mutation (b) 1 tertiary structure of the molecule changes; 2 ref different R groups of hydrophobic valine (substituting hydrophilic glutamic acid) involved in protein folding; 3 effect on HbS occurs at low oxygen concentration, hydrophobic regions on different molecules stick together, solubility of deoxygenated HbS decreases 4 HbS will polymerize into fibres, ability to carry oxygen decreases. (c) 1 Affected red blood cells (REJECT HbS) will adopt a sickle shape; 2 ref tendency for red blood cells to stick/clump together; 3 sickle shaped red blood cells clog capillaries, preventing other cells from moving through capillaries; obstruction of blood flow to organs, organ damage occurs 4 shorter lifespan of red blood cells resulting in anaemia in patients. Examiner’s comments: A large number of candidates thought that once the HbS had polymerised it could not carry oxygen again. (d) (i) 1 Gene for HbF would be transcribed/activated/expressed; 2 Resulting in translation of mRNA to produce HbF. (ii) 1 Less HbS in red blood cells and so there would be less polymerization 2 Less sickling of red blood cells would in turn lead to longer lifespans of red blood cell 3 HbF has a higher affinity for oxygen 4 as more oxygen can be transported

2008 ‘A’ Level H2 Biology Mark Scheme (updated 10 Jan 2015) 4 QUESTION 3 (a) 1 Haemagglutinin (HA), a glycoprotein found on the viral envelope, recognises and bind to specific receptor molecules (sialic acid) on the cell surface membrane of epithelial cells (adsorption) 2 The virus then enters via a vesicle to form endocytic vesicles (endocytosis); viral envelope then fuses with the membrane of endocytic vesicle 3 capsid released into the cytosol enzymatically removed and genome enters nucleus (b) 1 Use of host cell’s ribosomes on rough endoplasmic reticulum (rER) to translate the viral (+) mRNA into viral proteins Examiner’s comments: Reject vague reference to the use of the host cell machinery (c) 1 encodes RNA polymerases which are not present in host cell; 2 Conversion of viral negative strand RNA to positive strand RNA for translation into viral proteins (Only positive strand RNA can be used) (d) 1 neuraminidase enables enzymatic removal of sialic acid (from the viral envelope to prevent agglutination of the enveloped virus; 2 if neuraminidase is inhibited, new viruses cannot emerge from the infected cells to infect other cells

2008 ‘A’ Level H2 Biology Mark Scheme (updated 10 Jan 2015) 5 QUESTION 4 (a) 1 ionizing radiation like UV light, gamma rays, X-rays / chemical carcinogens like tar in cigarette smoke; 2 may induce mutations in DNA; 3 gain-of-function mutation in proto-oncogene leads to activation of oncogene 4 resulting in increased cell division (REJECT cell growth) 5 loss-of-function mutation in tumour suppressor genes leads to inactivation of both tumour suppressor genes 6 resulting in inability to slow down cell cycle in presence of DNA damage (b) (i) 1 involved in normal cell division; 2 encodes proteins involved as transcription factors stimulating expression of other genes / encodes signal transduction molecules that stimulate cell division / encodes cell cycle regulators. Examiner’s comments: Candidates often failed to gain credit because their responses were not precise enough. Often candidates confused cell growth with cell division. (ii) 1 mutated form of proto-oncogene; 2 results in uncontrolled cell division, leading to cancer (c) 1 translocation of Myc proto-oncogene near to a gene regulatory sequence (ref highly active promoter) / increased gene amplification 2 increased transcription of Myc gene, 3 therefore increased translation of Myc mRNA (d) 1 gene amplification results in more copies of Mdr1, leading to more copies of the transporter; 2 with more copies of the transporter, more drugs would be removed from the cell, causing the resistance to drugs Examiner’s comments: Many incorrectly suggested that with more transporters more drugs would enter the cell to cause the resistance. (e) 1 use of a inhibitor that will bind to and block transporter 2 drugs cannot be removed from the cell

2008 ‘A’ Level H2 Biology Mark Scheme (updated 10 Jan 2015) 6 QUESTION 5 (a) Parental phenotypes: Purple x Purple Parental genotype: RrEe RrEe Parental gametes: RE Re rE re RE Re rE re F1 genotypes: Punette square: R E R e r E r e RE RREE RREe RrEE RrEe Re RREe RRee RrEe Rree rE RrEE RrEe rrEE rrEe re RrEe Rree rrEe rree purple grains red grains white grains F1 phenotypes: Purple grains Red grains White grains F1 phenotypic ratio: 9 3 4 1 Parental genotype – RrEe 2 Parental gametes – RE, Re, rE, re 3 F1 genotypes 4 corresponds genotypes to phenotypes (legend for Punett square) 5 F1 phenotypic ratio 6 correct presentation of genetic diagram Examiners’ comments: A quick way to come up with the ratio is to divide the total number of the three phenotypes by 16. Phenotype Divide by 199.125 Approximate ratio Purple grains 1836 9.22 9 Red grains 578 2.90 3 White grains 772 3.87 4 Total 3186 Since the Punnett square has 16 genotypes, 3186 divided by 16 = 199.125. Next, divide each of the observed numbers by 199.125 to obtain the approximate ratios!

2008 ‘A’ Level H2 Biology Mark Scheme (updated 10 Jan 2015) 7 (b) 1 to determine if the difference between observed and expected results is significant or due to chance 2 to draw conclusions when there are differences from expected ratios (c) 1 Epistasis (d) 1 Both R and E alleles code for enzymes in a metabolic pathway when the products of one stage bec