2017 Bio H2 MCQ Explanations

1. N17Q1 A We are told that the cell is “metabolically active”. 1 is mitochondrion and the chemical inhibits oxidative oxidation in the organelle and thus it cannot carry out its role for ATP production. 2 is heterochromatin and 3 is RER. E nergy in the form of ATP is needed by chromatin remodelling complexes to induce changes in conformation at the level of the nucleosome in 2 to form euchromatin, so that there is more transcription and subsequently more translation ( protein synthesis). RER ribosomes would have their functions impaired due to no protein synthesis. 2. N17Q2 A T4 phage has all 3 characteristics. For phage lambda, it would be 2 and 3 only. 3. N17Q3 A Molecule A, the 1,4 glycosidic bond formed between 2 alpha-glucose are beside each other. The 1,6 glycosidic bond is formed between the second and rightmost molecule. The rightmost molecule is a beta glucose as shown by the OH on C1 being above the ring. Molecule B has both 1,4 and 1,6 glycosidic bonds formed from 3 alpha glucose. Molecule C has all 3 beta glucose. Molecule D has 2 alpha glucose and 1 beta glucose (extreme right), but has both 4,4 and 1,6 glycosidic bonds (no 1,4) 4. N17Q4 C Carbohydrate chains on the cell surface membrane are responsible for cell -cell recognition and cell to cell adhesion which allow for assembling . 2,5,6 are glycoproteins and 4 is a glycolipid. 5. N17Q5 A Z must be quaternary as it has 2 polypeptides coded by 2 genes , X and Y are coded by only 1 gene each, so tertiary structure. Y must be globular as it a great number of different aa than X , hence not many repetitive amino acids . X is fibrous as there is a high percentage of commonest amino acids, suggesting repetitive amino acids. (Remember collagen has repetitive amino acids). 6. N17Q6 B GPCR is embedded on membrane hence has hydrophobic R groups associated with the hydrophobic core of the phospholipid bilayer. GPCR is not quaternary (Option D is wrong) Option A is for fibrous protein. 7. N17Q7 C Options A, B and D apply to all enzymes whether lock-and-key hypothesis is used or induced fit hypothesis is used. Only option C applies to induced fit, where the active site of the enzyme changes conformation to fit the substrate better once the substrate approaches the active site, so the ES complex is formed. In lock-and-key hypothesis, the active site and substrate are an exact fit and there is no further conformational change after substrate binds. 8. N17Q8 A Resistant insects do not die from insecticides sprayed o n them, because they have enzymes that metabolise (meaning “break down”) the insecticides, and so rendering the insecticides ineffective. Synergists are non-competitive inhibitors that bind to one of the enzymes in the insect’s body that metabolise (break down) the insecticides and inactivates it. Evidence 1: Resistant insects have high concentration of this enzyme but susceptible insects have low concentrations of this enzyme. Presence of synergist causes resistant insects to become susceptible. Conclusion – fewer available enzymes in resistant insects as some have been inhibited by synergist. Hence, lower concentration of enzymes (like in susceptible insects). Evidence 2: Synergists themselves have no effect on insects. Rules out options B and C. GCE A Level H2 Biology 9648 Biology November 2017

Evidence 3: When mixed with a synergist, less insecticide can be used without reducing the number of insects killed . Rules out option D. Also implies enzyme inhibition because at lower substrate concentration, same number of insects killed → less insecticide destroyed → less available enzymes. 9. N17Q9 B The E, P and A sites on the ribosomes will need to be larger to accommodate 4 nucleotides (on anticodon) instead of the usual 3. The anti -codon on tRNA will have to consist of 4 nucleotides instead of 3. mRNA, RNA polymerase and nucleotides can remain the same because the RNA polymerase just read the DNA template nucleotide bases (no need to read in threes during transcription, unlike translation), and make complementary mRNA bases. 10. N17Q10 D Since RNA polymerase I is involved in rRNA synthesis and RNA polymerase III is involved in tRNA synthesis and synthesis of some rRNA, RNA polymerase II must be involved in normal mRNA productions. mRNA is first transcribed as pre- mRNA and the introns are later removed and exons joined together to produce mature mRNA during post-transcriptional modification. 11. N17Q11 D 1 genes widely separated – genome is usually small and has very little noncoding sequences, hence genes are not widely separated. 2 code almost entirely for proteins – yes, mostly coding sequences, very few noncoding sequences 3 include small circular sequences of DNA – yes, plasmids 4 simultaneous transcription of related genes – yes, related genes arranged in operons under the control of a single promoter to make polycistronic mRNA. 12. N17Q12 A 1 prevent degradation of chromosomes by nucleases – yes, by forming the loops at the ends of the linear chromosomes, it prevents nucleases from attaching and hydrolyzing the DNA 2 prevent the ends of chromosomes fusing to each other – yes, by forming the loops at the ends of the linear chromosomes, the 3’ single -stranded sticky ends are not exposed and cannot complementary base pair with another sticky end from another chromosome. 3 prevent the loss of genetic information during DNA replication – yes, the shortening of DNA with each round of replication will cause the telomeres (found at the ends of linear chromosomes) to be shortened instead of genes found further in. 13. N17Q13 B Lac repressor is a protein (1), the enzymatic protein RNA polymerase (5) binds to lac promoter (non-coding DNA); the repressor protein (8) binds to operator (non-coding DNA); Lac Z gene (on DNA) is where RNA polymerase (11) reads and codes for protein beta-galactosidase enzyme (12). 14. N17Q14 C Fact 2 and 3 affect the half-life (stability) of mature mRNA. 1 and 4 deal with protein (or enzyme) after protein is made by translation. For 5: repressors bind to silencers, and for 6: activators bind to enhancers. 15. N17Q15 A During independent assortment, the arrangement of 1 pair of homologues is independent of the other pairs. Independent assortment results in different combination (and hence proportions) of paternal and maternal chromosomes to each pole (and to each gamete eventually). Hence the proportion of nuclear DNA from the maternal parent being passed on to the daughters is variable. B is incorrect. It will not be exactly 50%. C is incorrect. Each daughter will get one X chromosome from their father and one X chromosome from their mother.

D is incorrect: Keyword here is “proportion of nuclear DNA”. Both daughters will get 50% of their nuclear DNA from the mother and 50% of their nuclear DNA from their father. Hence, it is incorrect to say they have different proportion inherited from the mother. All of the DNA in an egg cell is derived from the woman producing that egg cell, whether or not crossing over in prophase I has occurred. Crossing over only affects the proportion of DNA in the egg cell that is derived from the maternal and paternal DNA of the previous generation (i.e. the grandparents of the mother’s daughters). 16. N17Q16 B Fact A is incorrect: Centromeres would have already bound to spindle microtubules earlier during metaphase. Picture A appears to show early telophase. Some students might consider this to be late anaphase. However, between options A and B, B seems to be a stronger answer, as it clearly shows metaphase (alignment along equator, and the microtubules begin to pull). C is incorrect. Picture 3 does not show condensed chromatin. The chromosomes are not visible. It is actually interphase. 4 is prophase as the chromosomes are condensed. 17. N17Q17 B Excess cyclin D will bind t