RI 2023 Prelim H2 Biology P1234 Answers

H2 Biology ● RI Prelim 2023 PAPER 1 1. B 7. B 13. D 19. C 25. D 2. B 8. A 14. C 20. A 26. C 3. B 9. D 15. D 21. B 27. A 4. A 10. A 16. D 22. A 28. C 5. C 11. C 17. D 23. C 29. C 6. D 12. D 18. D 24. C 30. C PAPER 2 1(a) Name the structures A, B and C. [2] A mitochondria; B centrioles; C lysosome/Golgi vesicle/ secretory vesicle; A: vesicle 2 marks if all correct, 1 mark if 2 correct (b) Name one organelle that would also be present in a prokaryotic cell. [1] Ribosome; R: 80S ribosome (c) Describe the structure of B and its role. [4] 1. A pair of hollow cylinders made up of 9 triplets of microtubules (hollow tubes made of protein tubulin) each; 2. The tw o rod-like cylinders are positioned at right angles to each other; 3. Found in a region called centrosome which is microtubule organising centre (MTOC); MAX 2m 4. During mitosis/ meiosis/ nuclear division, centrioles replicate and move to opposite poles of cells; 5. To determine polarity of the cell; 6. They help to organize microtubules to form spindle fibre; 7. To ensure proper separation of chromosomes during nuclear division; MAX 2m

(d) Sodium ions cross cell surface membranes using facilitated diffusion or active transport. Explain why sodium ions cross cell surface membranes by these mechanisms and not by simple diffusion. [3] 1. Sodium ions are charged and hydrophilic; 2. They cannot pass through the hydrophobic core* of the phospholipid bilayer of membrane; 3. So must pass through transport proteins/carrier proteins w ith hydrophilic pore/channel embedded on cell surface membrane involved in facilitated diffusion or active transport. [Total: 10] 2(a) Describe two differences in DNA behaviour during binary fission in a prokaryotic cell and mitosis in a eukaryotic cell. [2] Binary fission Mitosis 1. DNA replication of circular bacterial chromosome occurs during binary fission; DNA replication of chromosomes occurs during the S phase of interphase prior to mitosis; 2. Circular chromosome does not condense prior to separation; All chromosomes in the nucleus will condense during prophase*; 3. Chromosome does not attach to spindle fibres; Chromosomes attach to spindle fibres via kinetochore proteins; 4. Single chromosome does not line up across the equator of the cell; Chromosomes line up in a single row at the metaphase plate; 5. Chromosome does not separate into chromatids; Chromosomes separate into sister chromatids during anaphase*; 6. AVP (b) Suggest why the flasks were incubated at 35°C. [1] This is the optimum temperature for the bacterial enzymes resulting in its fastest rate of growth/ division; Fig. 2.2 is a graph which shows the results of the investigation. Fig. 2.2 number of living bacteria remaining after 3 hours (x 106) concentration of antibiotic / µg cm-3 This is not a flat line. The number of living bacteria decreases from 38 x 10 6 (at 50µg cm-3) to 36 x 106 (at 100µg

(c) A student concluded from these results that an antibiotic dose equivalent to 50 μg cm -3 will effectively treat infections caused by S. aureus. Evaluate his conclusion. [3] Quoting of data (max 1 mark): 1. As concentration of antibiotic increase from 0 to 50 µg cm-3, number of living bacteria decreases from 100 to 38 x 10 6; (other quoting of relevant data, where correct, will also be given credit) Agree: 2. At 50 μgcm-3, there was only 38 x 10 6 bacteria remaining and reduction of most of bacteria (62%) meant that 50 μgcm-3 was an effective dose; or When antibiotic concentration increased from 50 μgcm -3 to 100 μgcm -3, there was only a very marginal decrease in bacteria (from 38 x 106 to 36 x 106) which was not significant. Hence 50 μgcm -3 was considered an effective dose without negative effects of higher antibiotic concentrations (beyond 50 μgcm-3); Disagree: 3. When antibiotic concentration increased from 50 μgcm -3 to 100 μgcm-3, there was a decrease in bacteria (from 38 x 106 to 36 x 106), and this means that 50 μgcm-3 was not as effective as 100 μgcm-3; 4. While the results are valid based on the experimental data, the effectiveness of the antibiotic was not studied in humans; 5. Number of living bacteria may be lower at higher antibiotic concentrations if incubated for longer than 3 hours; 6. Other concentrations can also be considered to treat the infection effectively as it causes a decrease in bacterial growth; 7. At 50 μgcm-3, there is still 38 x 10 6 bacteria living, hence a large enough figure to cause disease, so ineffective; OR At 50 μgcm-3, there is still 38 x 10 6 bacteria living, not all bacteria removed, hence ineffective; (d) The antibiotic used in the experiment was penicillin. Penicillin acts as an inhibitor to an enzyme found in S. aureus. (i) Name the enzyme inhibited by penicillin. [1] Transpeptidase (ii) Describe how penicillin inhibits this enzyme. [2] 1. Penicillin acts as a competitive inhibitor* to transpeptidase; 2. It has a similar shape to substrate/complementary in shape to active site of the enzyme and will bind to active site* of the enzyme; 3. Since penicillin binds and blocks active site*, the substrate is unable to bind to active site;

(iii) Explain how penicillin reduces bacterial growth as a result of this inhibition. [3] 1. Penicillin inhibits bacterial cell wall synthesis/ disrupts peptidoglycan synthesis; 2. By competitively binding to and inhibiting transpeptidase, it will inhibit formation of cross-links between adjacent chains/ peptidoglycan chains; 3. As a result, weakening the bacterial cell wall of dividing bacterial cells; 4. Because of high osmotic pressure inside the bacterium / when bacteria take in water by osmosis; 5. increased turgor pressure against weakened cell wall causes bacterium to swell and lyse; [Total: 12] 3(a) (i) What are the unique features of stem cells that made them suitable to treat this disease? [2] 1. A stem cell is an undifferentiated / unspecialized* cell capable of undergoing proliferation* and self-renewal*; 2. Ability to differentiate* into specialized cells; 3. This is a result of differential switching on of genes which occurs when appropriate molecular signals*; (ii) Explain why the mice in group SC -X needed to have their immune systems suppressed. [1] 1. To avoid stimulation of their immune system / prevent rejection of introduced cells; 2. because donor stem cells are recognised as foreign; (b) (i) With reference to Tables 3.1 and 3.2, compare the effects of the different treatments given to the mice in groups SC-X and SC-S. [2] 1. Levels of expression of normal allele for IDUA is higher in group SC -X compared to group SC-S; 2. Ther e is improvement in health of 2 mice in group SC -X but none in group SC- S; 3. indicating that high levels of expression (+++ at least) of normal allele of IDUA was required for improvement to health; A: reference to low expression (++ and below did not lead to improvement to health; R: merely copying table’s data as +++ without interpreting this as “high” 4. Both showed gene expression;

(ii) Suggest why the expression of the normal allele of IDUA is different in the stem cells given to the mice in groups SC-X and SC-S. [2] 1. Stem cells from an unaffected donor in group SC -X likely to have 2 copies of normal functional allele IDUA in its genome, while transduced stem cells in group SC-S likely to have at most 1 copy of normal functional allele IDUA integrated into its genome resulting in a lower level of expression; 2. Normal functional allele IDUA in group SC-S may be linked to, and is regulated by a weaker promoter* as compared to that in group SC-X; 3. Normal functional allele in transduced stem cells in SC -S may not be stably integrated* into genome and is gradually lost with ea