GCE A Level H2 Biology
9744 Biology November 2020
1. N20Q1 C The three tenets of cell theory are:
1. Cells are the smallest unit of life.
2. All cells come from pre-existing cells.
3. Living organisms are composed of cells.
2. N20Q2 A A centriole is indeed a hollow cylinder formed by a ring of microtubules.
A lysosome is a spherical sac surrounded by a single membrane, not a double
membrane.
A mitochondrion is a double membrane organelle with the inner membrane
folded, rather than enclosing a folded membrane.
A ribosome is made up of two different sized subunits (one large and one small).
It can be attached to the rER membrane or it can be free.
3. N20Q3 C Statement 1 is wrong because amylopectin, just like glycogen, is branched.
Since we are told that due to the branching enzyme action, the amylopectin is
made more soluble than amylose (unbranched), this mean that, equally, the
glycogen is also made more soluble.
Soluble molecules will always have an osmotic effect in the cell, so statement 3
is wrong.
Statement 2 about the molecule being more compact and hence occupies less
space and statement 4 about the branching helping the molecule to have more
free ends for hydrolysis to break down into glucose faster, are both good
properties of glycogen being a suitable storage molecule
4. N20Q4 D Follow the sequence given in the question, asp-ala-gly-lys
Synthesis of polypeptide always occur in the N to C terminus, and NH3+ will form
peptide bond with COO- thus it starts with 9 bonded to 7, 6 bonded to 2, 1 bonded
to 5
5. N20Q5 B Haemoglobin
Statement 1 is wrong, because Haemoglobin has haem group that contain iron
and porphyrin ring
Statemnt 4 is wrong, because repeating pattern of 3 type of amino acids is
referring glycine X Y of collagen and not Haemoglobin
Collagen
Statement 3 is wrong, because it is insoluble. It does not have hydrophilic amino
acid on it is surface. The R of it amino acids project into the core of it is triple
helical structures.
6. N20Q6 C 2,3-BPG functions similar to inhibitor of an allosteric enzyme (note that
haemoglobin is not an enzyme). At low concentrations of oxygen, 2,3-BPG has
a higher affinity and will bind to haemoglobin, stabilizing it in inactive
conformation, oxygen is unable to compete with it for oxygen binding sites.
A and B incorrect – carbon dioxide and oxygen do not bind to the same binding
site in haemoglobin, in any case it should not cause pH in red blood cell to fall
(as stated in statement A).
D – statements provide no evidence of any interaction between carbon dioxide
and 2,3-BPG
7. N20Q7 D Enzyme Z is an allosteric enzyme and graph shape should be sigmoidal.
In the presence of an inhibitor, the graph should shift to the right, meaning a
higher concentration of substrate is required to reach the same rate of reaction
compared to when the inhibitor is absent.
8. N20Q8 A A person with Tay-Sachs cannot break down the gangliosides and therefore the
concentration of gangliosides should remain high. The concentration of
gangliosides in a normal person should decrease with time. Since the
breakdown of gangliosides is an enzyme catalysed reaction, we expect the
shape of the graph to have a steep concentration (indicating higher rate of
reaction) at first but become more gradual as the amount of substrate
(gangliosides) will decrease with time.
9. N20Q9 C guanine and adenine are purines
cytosine, thymine and uracil are pyrimidines
10. N20Q10 B In semi-conservative DNA replication, the RNA primers at the 5’ end of the
daughter strands are removed but cannot be replaced with deoxyribonucleotides
due to the lack of a 3’OH group for DNA polymerase to bind to. This end-
replication problem results in daughter DNA molecules having 3’ overhangs.
11. N20Q11 B Option 1 – Correct. These transcription factors are called general transcription
factors which recruit RNA polymerase to the promoter.
Option 2 – Correct. These are specific transcription factors known as activators.
Option 3 – Correct.
Option 4 – Incorrect. The processing of pre-mRNA in eukaryotes (post-
transcriptional modifications) occur in the nucleus, but it only occurs after
transcription is complete.
12. N20Q12 A Centromeres and telomeres consist of tandem repeat sequences. Some introns
contain short interspersed (non-tandem) repeats.
13. N20Q13 D cAMP binds to regulator protein (CAP: catabolite activator protein) and activates
it. The activated regulator protein (CAP) then binds to the CAP binding site on
the promoter. This increases the affinity of RNA polymerase to the promoter.
Hence the answer is D.
A and C are incorrect. cAMP does not prevent regulator protein (CAP) from
binding to promoter and so it does not prevent expression of the lac operon when
glucose concentration is high.
B is incorrect. cAMP does not activate the inducer (allolactose).
14. N20Q14 A
15. N20Q15 D This is anaphase of mitosis. There will be a pair of centrioles at each pole. So
eliminate A and C.
B is incorrect because the centriole to centriole microtubules (i.e. non-
kinetochore microtubules) will elongate as they slide pass each other. So they
become longer, not shorter.
D is correct as during anaphase the kinetochore microtubules will shorten
leading the daughter chromosomes to opposite pole of the cell, led by the
centromeres.
16. N20Q16 B All but statement 4 are supported by the findings.
Statement 4 is incorrect because it is people whose JAK2 gene mutates before
the TET2 gene are more likely to go on to develop leukemia. People with
mutations in the JAK2 gene alone are not more likely to develop leukemia than
people with mutations in the TET2 gene.
17. N20Q17 C CB DNA need not be replicated to be expressed which rules out option A. CB
DNA is not directly translated, only mRNA is translated so this rules out option
B. Only one CB allele needs to be expressed for its phenotype to be seen as it is
a dominant allele – this rules out option D.
Correct sequence is that the CB allele is first transcribed to form CB mRNA which
is then translated to form the enzyme needed to synthesise the pigment melanin.
18. N20Q18 A Heterozygous black dog (BbDd) crossed with a fawn dog (bbdd) will produce the
following offspring:
1 BbDb : 1 Bbdd : 1bbDd : 1 bbdd
1 black dog : 1 grey dog : 1 red dog : 1 fawn dog
Hence, probability of getting 2 black puppies is ¼ x ¼ = 0.0625
19. N20Q19 B Phenotypes must be influenced by genes. Since bushes with white flowers only
produce white flowers and not pink or blue, the white flower phenotype is only
determined only by genes.
However bushes that produce pink or blue flowers depend on the genotype
which determines if the flowers are pink or blue at first and they also depend on
changing soil pH levels (environment) which can cause the flower colour to
change.
20. N20Q20 A F2 generation results when two heterozygous F1 offspring are crossed with each
other.
In cross 1, the ratio produced is 3 dark red grains : 1 white grain. This arises
when Aa x Aa, where A codes for dark red grains and a codes for white grains.
This is an example of monohybrid inheritance involving one gene.
In cross 2, the ratio produced is 15 dark red grains : 1 white grain. This arises
when AaBb x AaBb and epistasis is involved where the dominant allele A masks
the effect of B/b locus and the dominant allele B masks the effect of the A/a
locus. Hence, the typical 9:3:3:1 ratio is modified to become 15:1.
In cross 3, the ratio produced is 63 dark red to pale pink grains : 1 white grain.
This arises when multiple genes are involved in coding for grain colour with each
gene having an additive effect on the colour. In addition, environment will also
play a part in influencing the grain colour.
21. N20Q21 A Options B and D were the most frequently selected incorrect answers. Option B
is incorrect as respiration is constantly taking place decreasing the amount of
carbon in the plant (at pt 3), even though rate of photosynthesis may be zero.
Option D is incorrect as rate of photosynthesis is limited by diffusion of carbon
dioxide at 2. Region 2 shows high concentrations of internal carbon dioxide, at
levels almost equivalent to higher than atmospheric concentrations
22. N20Q22 D D is the only one that explain small yield. ATP synthase is not required in
anaerobic respiration so A is wrong. B and C do not explain small yield
23. N20Q23 D Activated G protein with bound GTP has to bind to adenylyl cyclase in order for
it to be activated and catalyse the conversion from ATP to cAMP.
24. N20Q24 B Uptake of glucose is due to ligand insulin (not glucagon in option 2). It binds to
RTK (option 1) causing downstream cellular response such as increasing
GLUT4 glucose transporters (option 3) to cell surface in order to increase in
glucose intake via facilitated diffusion (option 4).
25. N20Q25 B The question is asking for explanation for natural selection. Of which 1 and 2
explains natural selection in terms of change in allele frequency and variation
due to mutation respectively. Point 3 does not fit the preamble that the tongue
length became shorter through the 40 years as the change in length was
gradual.
26. N20Q26 C The clue is that the question starts with, “a species of lizard”, which indicates
that they are still the same species among the populations on different islands.
Hence micro-evolution occurred. Between C and D, C is a better answer.
27. N20Q27 C Option 4 - Molecular homologies require suitable material, such as nucleic acids
or proteins, to be collected and analysed. This is not always present in fossils.
So you can only do that for some of the species. For those with fossil evidence,
anatomical homologies are used, hence option 1 is used. Option 3 – there is no
need to look at biogeography of living tetrapods as their molecular information
are easily available.
28. N20Q28 A Injection of P resulted in a rise in antibody concentration but not T-lymphocyte
concentration. This indicates passive immunity where antibodies are transferred
to the recipient without the participation of the recipient’s immune system. This
leaves options A and B as possible answers. Injection Q resulted in a rise in both
antibodies and T-lymphocytes, which suggests active immunity where the
individual’s own immune system is responding. As antibody production is part of
the adaptive (specific) immune response, and not the innate (non-specific)
immune response, option A is the correct answer.
29. N20Q29 D Option A - Class switching changes the constant region, not the variable region
of the heavy chain.
Option B – Class switching refers to DNA rearrangement at the constant gene
segment of Ig heavy chain gene locus. The changes at the DNA level would
apply to all immunoglobulin molecules produced by the particular B lymphocyte.
Option C – Antigen binding is influenced by the antigen-binding site which is not
a consequence of class switching.
Option D – Different classes of antibodies have different functions and will
therefore react with other components of the immune system in different ways.
30. N20Q30 C 1 – this statement can be justified as increase in population cannot account for
the huge increase in use of crude oil for energy between 1935 and 1975.
2 – cannot be justified as there are many other factors contributing to the
increase on atmospheric carbon dioxide concentration
3 – cannot be justified as this trend can only be observed for use of coal but not
crude oil.
