2018 Bio H2 MCQ Explanations

1. N18Q1 B 2. N18Q2 A Envelope made of phospholipid bilayer which may contain embedded protein which contains nitrogen. Capsid is made of protein (which has nitrogen) and nucleic acids have nitrogen containing nitrogenous bases. 3. N18Q3 C A: wrong, as amylose being linear does not promote rapid digestion and amylopectin being branched does allow hydrolysis and NOT CONDENSATION. B: wrong, the part about gaps in amylose is wrong D: wrong, the close association does not promote rapid digestion. 4. N18Q4 A Glutamine has side-chains that are hydrophilic, hence hydrophobic interactions cannot occur between 2 glutamine amino acids. Since the bonds were formed between amino acids are in the same polypeptide chain, this implies that the polypeptide had folded back on itself and as a result, these bonds are likely to be within the inner core. 5. N18Q5 C A – incorrect. Not all ions are moved against the concentration gradient and the reason why they cannot pass through the phospholipid bilayer is not because they are too large but because they are charged. B – Incorrect. Glucose moves into cells via facilitated diffusion not active transport C – Correct. Oxygen is small enough to diffuse between the phospholipid molecules D – Incorrect. Water can also pass through the membrane by simple diffusion (in between the phospholipid molecules). Even though they are polar, they are small enough. GCE A Level H2 Biology 9648 Biology November 2018

6. N18Q6 B enzyme added to a solution of substrate J product Pt where enzyme acts E2 K J → K E2 and E3 M E2: J → K E3: K → L E1 and E2 K E2: J → K E1: M → N E1, E2 and E3 N E2: J→ K L → M E3: K → L E1: M → N Why not option A? If option A, when only E2 is added, the product obtained would be L not K. 7. N18Q7 C At low substrate concentration, rate of reaction is lower in the presence of the inhibitor. However, at high substrate concentration (200 mg cm -3), rate of reaction is almost the same in the presence or absence of inhibitor. This means that the effect of inhibition is overcome at higher substrate concentration and is characteristic of competitive inhibition. Also, it has to be reversible as the inhibitor must detach to allow the substrate to bind at higher concentrations. 8. N18Q8 D Statement 2 is true but does not explain why long non-coding RNAs are more likely to cause cancer in a bone marrow stem cell than mutations of p53. Inactivating tumour suppressor genes is a similarity in the effects of a p53 mutation and the presence of long non -coding RNAs. Inactivating tumour suppressor genes cannot therefore be used to account for differences in the consequences of p53 mutations and the presence of long non- coding RNAs. Statement 3 is true. Since it is mentioned in the question: the long non -coding RNAs i nactivate tumour suppressor genes and activate proto -oncogenes compared to p53 mutation which is only the tumour suppressor gene and requires two mutations. Statement 4 is true because a p53 mutation is random, spontaneous and may only occur in one cell. Since the long non -coding RNAs is present in all stem cells, chances of them interfering with cancer-causing genes is higher.

9. N18Q9 C Only 1 and 2 involve base pairing . 1 refers to codon base pairing with anti codon and ribonucleotide are added to mRNA via complementary base pairing to bases in DNA. In 2 transcription occurs and RNA polymerase is involved. In 3 activation of amino acid by aminoacyl tRNA synthetase, 4 refers to peptide bonds formed by peptidyl transferase which catalyse formation of peptide bond. 10. N18Q10 D Not conjugation cos involve 2 living bacteria, not transduction as no phage involved. So the DNA of heat-killed strain S must have been released and taken up by strain R by transformation. 11. N18Q11 A Options B and C may cause differentiation which will not retain potency. Option B activates differentiation; Option C is very vague ie determining cell specialisation (differentiation) may accelerate differentiation. Option D refers to pre-mRNA which means the gene is already transcribed which will not help with retain potency 12. N18Q12 A The lac operon is found in prokaryotic cells and does not have enhancer sequences. 13. N18Q13 C Trp (UGG)➔ Lys (AAG): 2 changes; Tyr (UAC)➔His(CAC): 1 change; His(CAU/CAC) ➔Cys(UGU/UGC): 2 changes 14. N18Q14 D One of the roles of the tumour suppressor genes, p53. The p53 protein will help in DNA repair. So ATM will increase the quantity of p53 protein so that it will be able to repair the DNA. The p53 protein functions at the checkpoints of the cell cycle, stopping it until the DNA repair is completed before proceeding. 15. N18Q15 D 1 All three events must happen for breast cancer to develop. No. Not necessary as breast cancer due to tumour suppressor genes not proto - oncogenes. 2 Breast cancer will only develop if mutated alleles of both BRCA1 and BRCA2 are inherited. No. The information given is that mutation of either BRCA1 or BRCA2 can increase the risk of cancer. 3 Inheritance of the mutated BRCA1 and BRCA2 alleles will only result in breast cancer when both alleles of a proto -oncogene are mutated. No. BRCA 1 and 2 are tumour suppressor genes not proto-oncogenes. 4 Mutations due to faulty DNA repair accumulate randomly with age. Yes.

16. N18Q16 B 17. N18Q17 B Since question says mother is heterozygous, her genotype for blood group A will be IAIO. And the genotype of the father with blood group O will be IOIO. Probability of baby having blood group O = 0.5 Mother is also a carrier for grey platelet syndrome (i.e. heterozygous e.g. Gg) and father has no family history for the syndrome (i.e GG). Probability of baby being a carrier for the grey platelet syndrome = 0.5 So, probability of baby having blood group O and being a carrier for the grey platelet syndrome = 0.5 x 0.5 = 0.25 = 25% 18. N18Q18 D Ratio of 12 black:3 purple:1 red indicates that genotype of red is aabb. So answer is D 19. N18Q19 C If two F1 are crossed, it is heterozygous crossed with heterozygous and if following normal Mendelian laws, the expected ratio should be 9:3:3:1. offspring phenotype observed numbers (O) expected numbers (E) (O – E) (O – E)2 (O – E)2/ E curly wings red eyes 40 45 -5 25 0.55 normal wings red eyes 20 15 5 25 1.67 curly wings purple eyes 16 15 1 1 0.07

normal wings purple eyes 4 5 -1 1 0.20 total 80 80 0 ꭕ2 = 2.49 =2.5 (to 1 dp) Fill up the table to 1 dp and add up the last column → answer is 2.5 20. N18Q20 A 1 true; electron cannot flow down to NADP, hence reduced NADP could not be formed hence reduction stage of Calvin cycle cannot proceed. 2 true; electron cannot flow down to NADP as flow of e- down ETC is blocked 3 true; photoactivation still occurs. 21. N18Q21 C 1 is false; the Y-axis is per unit surface area 4 is false; as a higher stomatal density would not be expected to lead to a decrease in the rate of photosynthesis since it would increase the rate of carbon dioxide diffusion into leaves. 22. N18Q22 D Since V has ethan al as final product thus there is an inhibitor for alcohol dehydrogenase preventing it from converting to alcohol. While W has inhibitor for (at least) pyruvate decarboxylase as pyruvate remains. 23. N18Q23 B Only 4 is wrong as electrons and protons are transferred from ETC to NADP 24. N18Q24 B A – false: the snakes are also able to sequester toxins and does not prevent the amphibians from being eaten by snakes. C – false: The toxins taken in via consumption and are not made by the amphibians. D – false: In this case predation still occurs as the snake still eats the amphibians. 25. N18Q25 A 1– True: there is still a change in frequency of alleles between the 7 types of salamanders. 2 – True: They are still considered the same species. Only v