2019 Bio H2 MCQ Explanations

1. N19Q1 D Not only should any correct option be biologically correct, but it must also answer the question. Statement 4 does not explain why cells are considered to be the smallest unit of all life since it only relates to multicellular organisms. 2. N19Q2 B Options A and D were the most frequently selected incorrect answers. Information indicates that the stain is specific for phospholipids rather than proteins. An explanation for the presence of the red -coloured structures must account for the removal of the red colour from most of the cell. 3. N19Q3 A Options B and D are incorrect because nuclear envelope and 80S ribosomes are found in eukaryotic cells not in E. coli (prokaryotic cell). Option C is incorrect because RNA polymerase is used to transcribe DNA to mRNA not translate mRNA to protein. 4. N19Q4 B Statement 1 is true. At optimum temp of 86°C for thermophilic bacteria, the percentage unfolding of the enzyme is more than half at 58%. Statement 2 is true. The psychrophilic bacteria begins unfolding at 38°C and is 100% unfolded at 48°C which is within the 15°C range. Statement 3 is false. At 62°C, for mesophilic bacteria, the enzyme is 100% unfolded but the activity of the enzyme remains high at about 97%. 5. N19Q5 C Graph X reaches a higher concentration of product within the same time limit. Hence, it has a higher rate of enzyme reaction which could be due to higher enzyme concentration or higher substrate concentration. (Hence, options narrowed to B and C). Graph Y has a steeper initial gradient but reaches the same concentration of product as the original reaction. Hence, it has the same concentration of substrate but higher rate of reaction due to higher temperature (greater number of effective collisions between E and S). This excludes option B and leaves answer as option C. To confirm, graph Z has a lower initial rate of reaction and reaches a lower maximum product concentration in the same time frame. Hence, graph Z has a lower substrate concentration. 6. N19Q6 C From the figure, it can be seen that the structure of arabinose and glucose are similar except for slight differences. The graph shows that as concentration of glucose (substrate) increases, the effects of arabinose inhibition can be overcome and the graph will reach the same V max as the uninhibited enzyme graph. Hence, this is an example of competitive inhibition. Note: option B is not correct because arabinose does not have the same structure as glucose but a similar structure. 7. N19Q7 D Only totipotent stem cells can give rise to genetically modified individuals 8. N19Q8 B H: refers to the entire DNA strand including the nitrogenous bases, hence it is the polynucleotide strands. J: enzyme adding free dNTPs to the 3’ end of the leading strand → DNA polymerase K: enzyme joining the ends of the Okazaki fragments together → DNA ligase 9. N19Q9 A H7N10 does not exist, hence options B & C cannot be the answer. For option D, H2N2 is no longer found at the present so it cannot be the answer. Also, influenza does not remain in the bodies of those who have recovered from it. That leaves only A as the correct answer. 10. N19Q10 A Statement 1: True. Both cells must come in contact via a mating bridge before conjugation can occur. Statement 2: True. In Hfr cells, part of the bacterial chromosome may be transferred over during conjugation. GCE A Level H2 Biology 9744 Biology November 2019

Statement 3: True. Upon completion of conjugation, the recipient cell now has an F plasmid and is known as a F+ cell. Statement 4: True. The genes involved in pilus formation are found on the F plasmid which will be transferred over during conjugation. 11. N19Q11 C inducer = allolactose (carbohydrate) promoter = DNA repressor = protein 12. N19Q12 D Statement 1 is correct but cannot be concluded from the findings above. Statement 2 is correct because the initial maternal mRNA has a short poly(A) tail which can then be lengthened when needed for embryonic development, hence not encoded in the maternal DNA. Statement 3 is correct because when maternal mRNA is needed to be translated during embryonic development, the poly(A) tail is extended which implies it is needed for translation to begin. Statements that were true but could not be concluded from the information provided should have been eliminated. 13. N19Q13 D These are the steps listed in sequential order for a Southern blot and nucleic acid hybridization experiment to take place. 14. N19Q14 A Option 1: p16 is said to be a tumour suppressor gene. It was found methylated in cancer cells. Since loss-of-function of tumour suppressor genes result in cancer, it can be concluded that p16 gene methylation switches off the gene. Option 2: The gene was not mutated, only methylated. Tumour formation resulted with methylation of the p16 gene, so this statement is true. Option 3: The methyl groups were said to be derived from the diet (environmental factor). Since methylation of p16 gene was often found in cancer cells, we can infer that the environment contributes to the development of tumours. 15. N19Q15 C Statement 1 – ability to migrate and form new colonies → metastasis Statement 2 – divides indefinitely (immortalisation) but haven’t formed tumour yet Statement 3 – loss of contact inhibition – forming primary tumour (transformation) 16. N19Q16 D As it is an extra Y chromosome (for XYY genotype), non-disjunction must occur during sperm formation. In Meiosis I, the XY chromosomes separate normally into daughter cells. In Meiosis II, the Y chromosomes do not separate and both Y chromosomes enter one sperm cell. When this YY sperm fuse with an X egg, it will result in an XYY zygote. 17. N19Q17 A Since it involves 2 genes for two enzymes P and Q respectively = dihybrid inheritance Both genes are found on chromosome 7 = linkage (linked genes) Enzyme P and enzyme Q belong to the same metabolic pathway and enzyme Q catalyses the later step in the pathway. So if enzyme P is non -functional or absent, whether enzyme Q is present or not, it cannot carry out its function. Hence, pp is epstatic of the Q/q locus = epistasis

18. N19Q18 C Since both have mothers who have lilac phenotype (bbdd), they will each have b and d alleles in their genotype. Seal male = BbDd Blue female = Bbdd Parental phenotype: Seal cat x Blue cat Parental genotype: BbDd x Bbdd Gametes: BD Bd Bd bd bD bd Fertilisation BD Bd bD bd Bd BBDd seal cat BBdd blue cat BbDd seal cat Bbdd blue cat bd BbDd seal cat Bbdd blue cat bbDd chocolate cat bbdd lilac cat Offspring genotype: 3 B_D_ : 3 B_dd : 1 bbD_ : 1 bbdd Offspring phenotype: 3 seal cats : 3 blue cats : 1 chocolate cat: 1 lilac cat 19. N19Q19 B The observed numbers will help to calculate the total no. of offspring present. The expected numbers can be calculated from the expected ratio and total no. of offspring. Since O and E are known, it is possible to calculate ꭕ2 value. To determine if the results are significantly different or not from expected values, we need to use the ꭕ2 value to find the probability in a ꭕ2 distribution table. 20. N19Q20 B 3 false, as chlorophyll appears green as it reflects green light 21. N19Q21 C A, Wrong as reduced NADP not synthesized in light-independent stage. B, wrong, PS I is active even in low light intensity D, wrong, in low light intensity, there should be low rate of e- flow, hence accumulation of proton in thylakoid space should be low. Hence rate of ATP and NADPH syn is low. That will result in decrease rate of reducti