Hi I'm looking for any bio teacher to give the suggested answers for a level 2012 bio mcq H2.
Paper just completed on the 3-dec
Thankyou!! Tough paper eh.
These are my answers. Had to guess quite a few questions!
DADDA BBCBD BDADB CCADC BADBC DCAAD AACBB BBACA
Since no one has the answers here is a compiled answers from top bio students in top JC for bio mcq 2012 Alevels H2.
1.D
2.A
3.D
4.D
5.A
6.B
7.B
8.C
9.B
10.D
11.D
12.D
13.A
14.D
15.B
16.B
17.C
18.C
19.D
20.C
21.B
22.A
23.A
24.B
25.C
26.D
27.C
28.A
29.A
30.D
31.A
32.B
33.C
34.B
35.B
36.B
37.D
38.A
39.C
40.C
Some disputes:
Q4 – It is unclear if the ‘volume of oxygen released’ refers to the total volume of oxygen collected at the designated time interval (e.g. between 1s and 2s) or else, simply the volume of oxygen collected (from 1s to 2s). So, the answer can be either (B) or (D).
Q5 – The answer could also be (D). Adenoviruses can be used as a vehicle to deliver a functional TP53 gene to the tumour cells, rendering them susceptible to p53-mediated apoptosis (see http://www.ncbi.nlm.nih.gov/pubmed/11498763 for details). It would be incorrect to suggest that the mutant virus can only replicate inside tumour cells [option (A)], since the adenovirus can replicate in both the wildtype and mutant cells (N.B.: Entry of the adenovirus into the cell does not cause cell damage, since the adenovirus is endocytosed into the cell, hence this should not stimulate the wildtype cell to express p53 thereby triggering apoptosis).
Q6. Crossing over can occur between non-sister chromatids in a homologous pair of chromosomes (which is widely known) as well as between non-homologous chromosomes [a lesser known process known as ectopic recombination (see http://www.genetics.org/content/129/4/1085 for one such study investigating this phenomenon in D. melanogaster)]. Transposons may participate in the latter process as well, as they ‘jump’ (‘cross’) from one chromosome to another. The correct options should thus be options 1, 2, 3 and 5. However, since there is no such option in this context, either option (A) or option (B) would be valid in this context.
Q11. The correct answer should be (B).
Q23. The answer could also be (C). This is because even if the heterozygote is only able to produce half of the enzyme concentration as the homozygous dominant individual, enzymes are generally very efficient catalysts (with a very high turnover rate), hence they are able to generate a substantial amount of product even at low enzyme concentrations. (N.B. Inactive/non-functional enzymes are generally either subjected to the cell’s refolding mechanisms, or else broken down to prevent their accumulation in the cell). This allows the heterozygote to manifest a phenotype which is virtually indiscernible from the homozygous dominant individual, hence contributing to the difficulty to establish a causative relationship/link between the genotype and phenotype of the individual based on the observed parental and offspring phenotypes (as noted in the question), thereby satisfying the requirements in this context.
Q27. The answer should be (B). This is because despite numerous textbooks which claim that 2 ATP molecules are directly produced per molecule of glucose oxidised in the Kreb’s cycle (or else, 1 ATP molecule is produced per turn of the Kreb’s cycle), this is actually an anomaly, since the Kreb’s cycle generates a GTP molecule per turn of the cycle (during the conversion of succinyl CoA to succinate) (see Molecular Cell Biology, 6th edition, p. 489 by Lodish et al for details). Ironically, Molecular Cell Biology, 6th edition includes the same anomaly of equating GTP to ATP in this instance. In fact, GTP is a very different molecule from ATP, as may be clearly noted by its (GTP, not ATP) involvement in GPCR activation (for signal transduction), microtubule polymerization, etc. Likewise, ATP is utilised in enzymes such as adenylate cyclase to convert ATP to cAMP, while GTP cannot be utilised in this case by the same enzyme (i.e. adenylate cyclase), but rather by a different enzyme (guanylate cyclase). Hence, it is vital to note the distinction between ATP and GTP (and not to equate the two, although numerous correlations may be drawn between both molecules as described above). Thus, the correct answer to this question should be (B).
Q33. This question is relatively unclear, as there are several factors which affect variation of DNA and proteins (besides its sequence). For instance, in eukaryotes, variation in proteins is also due to alternative splicing, differences in folding, etc. Likewise DNA has a three dimensional structure as well, though this may only be eminent in the regulation of transcription. Hence, (D) is also a probable answer in this context.
Q37. The correct answer is (C). Heterozygous RFLP loci allow us to distinguish between the individual arms of a homologous chromosome pair, thus allowing for effective mapping of genes between 2 unique heterozygous restriction sites (thereby causing the formation of different fragment sizes for mutant and wildtype organisms). This is known as chromosome mapping. RFLPs cannot however be used to determine the base sequence of the genome (as there are insufficient palindromic sequences to do this) – for sequencing, we’d need to use Sanger dideoxy, pyrophosphate or 4th gen sequencing methodologies (which are well beyond the scope of the H2 A-Level syllabus).
Of course UCLES would likely have different perceptions with regards to the answers provided here (in which instance, I’d like to see their variations).