2023 ‘A’ Level Examinations

H2/9744 Biology
Paper 2

General comments by Cambridge examiners

The candidates’ responses were generally of a high standard and displayed very good subject
knowledge.

Most candidates were able to accurately interpret graphical information and apply their knowledge to
novel situations.

Candidates did not always use correct scientific terminology and there was a tendency for responses
to include irrelevant details that did not address the requirements of the question. Explanations were
sometimes not fully developed.

1
QUESTION 1
Fig. 1.1 shows the effect of increasing the external concentration of glucose or fatty acids on the
rate of uptake of these molecules into a cell.

Fig. 1.1

(a) With reference to Fig. 1.1, explain the effect of increasing the external concentration of glucose
on the rate of uptake of glucose into a cell. [3]

1. [describe] Rate of uptake into the cell via facilitated diffusion increases as external
concentration increases.

2. [explain] Due to increase in glucose concentration gradient.

3. [describe] Rate no longer increases beyond a certain external concentration.

4. [explain] The glucose transporters are all saturated with glucose.

Examiners’ Comments
Many responses clearly explained the relationship between glucose concentration and glucose
uptake.

Weak responses often did not consider the specific mechanism by which glucose molecules cross
cell surface membranes or recognise the role of specific membrane components in this process.

Some described the relationship shown in the graph rather than trying to explain it.

(b) Explain why there is no uptake of glucose into the cell at X even though glucose is present outside
the cell. [1]

1. The concentration gradient is not steep enough for glucose to diffuse into the cell / the
concentration of glucose outside and inside the cell are at equilibrium.

2. The affinity of glucose transporter is not high enough to bind to glucose at such low
concentrations.

Examiners’ Comments
Most candidates were able to give a correct explanation. Some responses incorrectly referred to
a lag time or a lack of ATP.

2
(c) One suggested mechanism for the transport of fatty acids into cells is shown in Fig. 1.2. This
mechanism is referred to as a flip-flop mechanism.

Fig. 1.2

3
With reference to Fig. 1.2, explain the flip-flop mechanism for the transport of fatty acid molecules
across the cell surface membrane, including the role of hydrogen ions. [5]

1. Fatty acid contains a long hydrophobic hydrocarbon and a carboxylate group / COO−

2. Due to the charged COO− group, fatty acid cannot interact with the hydrophobic fatty acid
tail of the membrane to be directly transported.

3. The hydrocarbon of the fatty acid first inserts into the outer phospholipid layer of the
membrane, with the COO− interacting with the phosphate group and aqueous environment.

4. Protons (H+) then reacts with COO− group to neutralize the charge (─COOH)

5. This allows the fatty acid to traverse through the hydrophobic fatty acid core of the
membrane to be inserted into the inner phospholipid layer of the membrane.

6. The proton then dissociates from the fatty acid to restore its COO− group.

7. The fatty acid is then released into the cytosol.

Examiners’ Comments
Strong responses made effective use of scientific terminology to explain the alignment of fatty
acid molecules in the phospholipid bilayer and the involvement of hydrogen ions in the flip-flop
mechanism.
Some candidates were unclear about the distinction between phospholipids, triglycerides and fatty
acids. Weaker responses often referred to a phosphate head or glycerol on the fatty acid molecule.

[Total: 9]

4
QUESTION 2
Hydrogen peroxide is a harmful by-product of many metabolic processes in cells and is quickly broken
down into oxygen and water by the enzyme catalase.

Fig. 2.1 shows the energy changes during the breakdown of hydrogen peroxide into water and oxygen
with, and without, catalase.

Fig. 2.1

(a) Explain how the change in activation energy shown in Fig. 2.1 affects the rate of a catalase-
controlled reaction. [2]

1. Catalase lowers the activation energy to speed up the rate of reaction.

2. As lesser energy is needed for the substrate to overcome to reach transition state.

Note: the x-axis “progress of reaction” is NOT time. It refers to the state of the reaction: substrate to the
unstable intermediates and eventually to products.

Examiners’ Comments
Most candidates interpreted the graph correctly and recognised the effect of enzymes on
activation energy. Strong responses provided an explanation for the effect of changes in activation
energy on the rate of reaction.

Weak responses often simply quoted figures from the graph without addressing the question.

5
(b) The effect of storage temperature on the activity of catalase was investigated over a period of 1.6
hours.

The results are shown in Fig. 2.2.

Fig. 2.2

(i) With reference to Fig. 2.2, state what can be concluded from the results of the investigation.
[3]
1. The higher the storage temperature, the faster the relative activity of catalase falls.

2. As the storage time increases, the relative activity of catalase continues to decrease.

3. However, the relative activity of catalase does not fall by the same magnitude with every
5C increase / decrease in relative activity is the most drastic when the storage
temperature is increased from 40C to 45C, e.g. @0.6h, from 0.7 to 0.2.

4. At 50C, relative catalase activity reaches 0 by 1.2h, the fastest drop among the storage
temperatures. [Accept: 1.1h to 1.4h]

Examiners’ Comments
Many candidates found this a challenging question and simply listed data from the graph rather
than drawing conclusions. Several did not appreciate that the question was about storage
temperature of the enzyme and answered the question in terms of the effect of temperature on
enzyme activity.

6
(ii) Explain the effect of storage temperature on the activity of catalase, as shown in Fig. 2.2. [5]

1. The higher the temperature, the greater the kinetic energy of catalase

2. Higher rate of breakage of hydrogen bonds and ionic bonds between R-groups

3. Disrupts the 3D conformation / tertiary structure of catalase

4. Leads to denaturation of catalase

5. Change in shape of the active site

6. Unable to bind to substrate to form enzyme-substrate complex, (hence decrease in

relative activity as temperature increases).

Examiners’ Comments
This was generally well answered, with many candidates expressing themselves clearly and
showing a good understanding of the reasons why different storage temperatures affect catalase
activity.

Weaker responses often omitted to consider relevant details such as how storage temperature
affected the tertiary structure of the enzyme. Some provided descriptions of the graph rather than
developing an explanation.

[Total: 10]

7
QUESTION 3
Bacteria can take up DNA fragments from the environment and incorporate them into their genomes.
This process is known as transformation. As a result of transformation, bacteria may acquire new
genetic traits such as resistance to antibiotics.

The DNA fragments usually come from dead bacteria lysing (splitting open) and releasing their
contents into the surrounding area.

Fig. 3.1 shows the main features of environmental DNA uptake by bacteria.

Fig. 3.1

Fig. 3.2 represents an enlargement of the region labelled 'uptake of DNA' in Fig. 3.1. It shows in detail
how environmental DNA enters the recipient cell.

Fig. 3.2

8
(a) Name the parts of Fig. 3.2 labelled B, C and D. [3]

B peptidoglycan cell wall

C cytosol / cytoplasm

D transport protein / channel protein

Examiners’ Comments
Few candidates were able to name all of the labelled parts. Most responses correctly identified
part C, but parts B and D were recognised much less frequently.

(b) With reference to Fig. 3.2, suggest how environmental DNA fragments enter the bacterial cell.
[3]
1. Double stranded DNA binds to protein A (a DNA-binding protein).

2. Protein A digests / hydrolyses one of the strands of the dsDNA

3. The remaining DNA strand enters the cytosol via protein D.

Examiners’ Comments
Strong responses made effective use of the diagram to develop an explanation that referenced
the structures labelled in the diagram. Weaker responses only referred vaguely to the structures
shown in Fig. 3.2 and did not consider the interactions between these structures and the DNA
molecule.

(c) Two other ways in which DNA can enter bacterial cells are transduction and conjugation.

Describe how DNA enters bacterial cells by transduction and conjugation. [4]

transduction

1. DNA is packaged into the icosahedral capsid head of a daughter bacteriophage.

2. This bacteriophage attaches to and injects the DNA into another bacterial cell.

conjugation

3. F+ cell forms a sex pilus (a cytoplasmic bridge) with a F− cell.

4. One of the DNA strands of the F plasmid in the F+ cell is nicked and transferred, 5’ end first,
into the F− cell.

5. The strand then acts as a template for the synthesis of the complementary strand in the F−
cell, forming dsDNA.

Examiners’ Comments
This was generally well answered, although some responses included unnecessary details.

Many candidates correctly described the role of a bacteriophage in transduction, but further details
were often omitted.

The majority of candidates correctly described conjugation. Weaker responses were vague and
often omitted to use the expected scientific terminology. Some referred only to F plasmids with no
mention of the bacterial cells.

[Total: 10]

9
QUESTION 4
(a) The annealing temperature for the polymerase chain reaction must be close to, but below, the
temperature (melting temperature) at which the primer and template DNA separate.

One way to calculate the melting temperature of a DNA sequence is to use this equation:

Tm = 64.9 + [41  (nG + nC − 16.4) / (nA + nT + nG + nC)]

key

Tm = melting temperature in C
nA = number of nucleotides containing the base adenine
nT = number of nucleotides containing the base thymine
nC = number of nucleotides containing the base cytosine
nG = number of nucleotides containing the base guanine

(i) Use the equation to calculate the melting temperature at which this primer DNA sequence
separates from template DNA: [1]

TCGACTTCCTCGAACC

Tm = 64.9 + [41  (2 + 7 − 16.4) / (3 + 4 + 2 + 7)]

Tm = 64.9 + [41  (−7.4) / (16)]
Tm = 64.9 + [−18.9625]
Tm = 45.9C

Melting temperature = ………………………….C

(ii) The equation predicts that the melting temperature of DNA sequences of the same length will
increase as the proportion of nucleotides containing cytosine and guanine increases.

Use your knowledge of DNA structure to suggest why an increased proportion of bases
containing cytosine and guanine will increase the melting temperature. [2]

1. Three hydrogen bonds between C and G as compared to two between A and T, hence
stronger.

2. More heat energy (hence higher melting temperature) is required to break the hydrogen
bonds between C and G.

Examiners’ Comments
Many candidates correctly identified the difference in the number of hydrogen bonds between the
different base pairs; fewer went on to suggest how this could affect the melting temperature.

Some responses incorrectly referred to differences in the strength of hydrogen bonds without any
consideration of the number of bonds.

10
 (iii) DNA sequences with very low melting temperatures are unsuitable for use as primers for the
polymerase chain reaction. This is because they require the use of low annealing
temperatures and, at low temperatures, base pairing can occur between non-complementary
nucleotides.

Suggest the consequence for the results of a polymerase chain reaction of using low annealing
temperatures. [1]

• The primers will bind non-specifically to multiple regions of the initial template DNA,
resulting in the amplification of unintended regions of the DNA.

Examiners’ Comments
Many responses provided rational suggestions. Weak responses often repeated the question
stem without providing any further details.

(b) Describe the advantages of using the polymerase chain reaction. [3]

1. Sensitivity: PCR can detect minute amounts of DNA, even from a small sample, making it
highly sensitive in identifying pathogens or genetic material.

2. Specificity: primers allow PCR to target and amplify specific DNA sequences, allowing for
precise detection and analysis of particular gene or sequence.

3. Speed: PCR can amplify DNA exponentially within hours, enabling quick analysis compared
to traditional methods that might take days or weeks.

4. Versatility: It has diverse applications such as diagnosing diseases, genetic testing, forensics,
studying evolutionary relationships.

5. Quantification: PCR can also be used for quantitative analysis, determining the initial amount
of DNA present in a sample through methods like real-time PCR.

6. Automation: Modern PCR machines are automated, reducing human error and increasing
efficiency in handling multiple samples simultaneously.

(c) State the limitations of the polymerase chain reaction. [3]

1. PCR is sensitive, so contamination of sample with extraneous (foreign) DNA can result in
unintended sequences being amplified.

2. Prior knowledge of the sequence flanking the target DNA needs to be known for suitable DNA
primers to be synthesized.

3. Taq DNA polymerase has low replication fidelity since it lacks a 3’ to 5’ exonuclease
proofreading activity, hence may cause mutations in the PCR products.

4. The length of DNA that can be reliably amplified is limited by the polymerase's processivity
and the efficiency of the amplification process, hence long DNA fragments might not amplify
efficiently.

Examiners’ Comments
Candidate knowledge of the limitations of PCR was not as strong as their knowledge of its
advantages, with many omitting any mention of well-known problems. Good responses were
expressed clearly and included relevant details.

[Total: 10]

11
QUESTION 5
Fig. 5.1 shows the behaviour of one pair of homologous chromosomes during two stages of meiosis
I and two stages of meiosis II.

Fig. 5.1

(a) Name the stages of meiosis shown in Fig. 5.1. [2]

stage P metaphase I

stage Q anaphase I

stage R metaphase II

stage S anaphase II

Examiners’ Comments
The majority of responses identified stages Q, R and S correctly. Several incorrectly identified
stage P as prophase I.

12
(b) Describe similarities and differences between stage Q and stage S. [3]

similarities

1. Both involves the shortening of pole-to-centromere spindle fibers

2. Both involves the separation of chromosomes to opposite poles of the cell

differences

3. Stage Q separates homologous chromosomes but stage S separates sister chromatids.

4. The centromere does not divide in stage Q but divides in stage S.

5. Two spindle fibres attach to each (double-arm) chromosome in stage Q while one spindle fibre
attaches to each (single-arm) chromosome in stage S.

6. In stage Q, cells are still diploid but in stage S, cells are haploid.

Examiners’ Comments
The majority of candidates correctly described some similarities between stage Q and stage S
and many also described at least one difference. One common error was for candidates to refer
to separation of non-sister chromatids rather than separation of sister chromatids..

(c) Stage P and stage R are both important in the production of gametes that are genetically different.

In stage R, genetic variation could not be generated without an event that occurs during an earlier
stage of meiosis.

(i) Name this event and the stage of meiosis in which it occurs. [2]

event crossing over

stage prophase I

Examiners’ Comments
Most candidates were able to recall the name of the event but not all recognised the stage of
meiosis in which it occurs. Some responses incorrectly identified the event as independent
assortment.

(ii) Explain why stage R does not generate genetic variation in the absence of this event. [1]

• In the absence of crossing over, the sister chromatids of each chromosome will remain
genetically identical (since crossing over produces non-genetically identical sister
chromatids).

Examiners’ Comments
Many responses correctly explained why genetic variation did not occur. A minority explained how
stage R generates genetic variation rather than explaining why this would not happen in the
absence of the event named in (c)(i).

13
(d) Fig. 5.2 represents the production of an ovum by meiosis. A single pair of homologous
chromosomes with one gene having two alleles, A and a, is shown in the original cell.

As a result of meiosis, one ovum and two polar bodies are produced..

Complete Fig. 5.2 to show the chromosomes and alleles that are present in the daughter cell
from the first meiotic division, the first polar body and the second polar body.

You should assume that the event named in (c)(i) has not occurred. [2]

[Total: 10]

14
QUESTION 6
The inheritance of coat colour in labrador retriever dogs is controlled by two unlinked genes, each
with a dominant allele and a recessive allele. The recessive allele of each gene produces no functional
product.

Fig. 6.1 shows how the presence of the dominant alleles of these two genes, E and B, affects the coat
colour of labrador retriever dogs.

Fig. 6.1

(a) State the coat colour shown by labrador retrievers with the following genotypes. [1]

eeBb yellow

EeBb black

Examiners’ Comments
The majority of candidates answered this correctly. Weaker responses incorrectly gave the
genotype eeBb as black coat.

(b) Explain the coat colour of labrador retrievers with the genotype Eebb. [3]

1. Brown coat

2. One copy of allele E codes for a functional enzyme that converts yellow pigment to brown
pigment.

3. Two copies allele b results in no functional enzyme to further convert the brown pigment to
black pigment.

4. allele e of gene E/e is epistatic over allele B of gene B/b.

Examiners’ Comments
Strong responses provided concise and reasoned explanations. Weak explanations were
muddled or only discussed the coat colour rather than the production of a pigment.

15
(c) Two black labrador retrievers were mated and the resulting litter of puppies included yellow, brown
and black coat colours.

Draw a genetic diagram to explain the results of this cross. Indicate the expected phenotypic ratio
of the offspring.

Use the symbols B, b, E and e to represent the alleles. [5]

Parent phenotype: black x black

Parent genotype: EeBb x EeBb

Gametes: EB Eb eB eb EB Eb eB eb

F2 genotypes and phenotypes:

EB Eb eB eb

EEBB EEBb EeBB EeBb

EB black coat black coat black coat black coat

EEBb EEbb EeBb Eebb

Eb black coat brown coat black coat brown coat

EeBB EeBb eeBB eeBb

eB
black coat black coat yellow coat yellow coat

EeBb Eebb eeBb eebb

eb black coat brown coat yellow coat yellow coat

F2 phenotypic ratio: 9 black coat : 3 brown coat : 4 yellow coat

Examiners’ Comments
The majority of candidates answered this well. A few errors were seen in the completion of the
Punnett square offspring genotypes and some correct genotypes were assigned incorrect
phenotypes.

(d) Name the type of gene interaction shown in Fig. 6.1. [1]

• (Recessive) epistasis

[Total: 10]

16
QUESTION 7
Fig. 7.1 shows the arrangement of the large protein complexes involved in oxidative phosphorylation
in mitochondria.

Fig. 7.1

(a) With reference to Fig. 7.1, explain the movement of protons into the intermembrane space using
the large protein complexes. [4]

1. Reduced NAD is oxidised to release high-energy electrons (e−)

2. The e− is transported down a series of electron carriers of progressively lower energy level

3. This releases energy that is used by H+ pump to transport H+ from mitochondrial matrix to the
intermembane space

4. Against the H+ concentration gradient.

Examiners’ Comments
Many candidates were able to correctly explain the movement of protons into the intermembrane
space.

One common error was the improper use of terms as in ‘generation’ or ‘production’ of energy.

Other errors included the misconception that the movement of protons took place through channel
proteins via facilitated diffusion, or that proton movement required ATP.

17
(b) Explain the role of oxygen in oxidative phosphorylation. [3]

1. As the final electron acceptor

2. Combines with electron and protons to form water

3. Enables electron transport chain to continue transporting electron to maintain high proton
concentration in the intermembrane space.

4. Also ensures that NAD+ and FAD are regenerated for glyvolysis, link reaction and Kreb cycle.

Examiners’ Comments
Most responses correctly explained some aspects of the role of oxygen in oxidative
phosphorylation. Few provided full explanations that considered what would happen in the
absence of oxygen.

(c) NAD is an important molecule in aerobic respiration and NADP is an important molecule in
photosynthesis.

Compare the role of reduced NAD in aerobic respiration with the role of reduced NADP in
photosynthesis. [3]

1. Both are coenzymes.

2. Both are electron donors.

3. NADH donates electron to the electron transport chain during oxidative phosphorylation
while NADPH reduces 1,3-bisphosphoglycerate to glyceraldehyde-3-phosphate in Calvin
cycle.

4. Upon oxidation, the proton from NADH contributes to the proton concentration in the
mitochondrial matrix, while the proton in NADPH is used to reduce 1,3-bisphosphoglycerate
to glyceraldehyde-3-phosphate in Calvin cycle

Examiners’ Comments
Candidates found this question challenging; many responses lacked a clear structure and omitted
significant details. There were few clear statements about the roles of reduced NAD and reduced
NADP. Reduced NAD and reduced NADP were often referred to incorrectly as electron acceptors.

Strong responses included details of the use of reduced NADP in the Calvin cycle with reference
to the various compounds involved, and the use of reduced NAD in oxidative phosphorylation.

[Total: 10]

18
QUESTION 8
Glucagon receptors are examples of G-protein linked receptors and are found in the cell surface
membrane of hepatocytes (liver cells).

Fig. 8.1 represents the molecular structure of a glucagon receptor attached to its ligand, glucagon,
and shows how the receptor is inserted in the cell surface membrane of a hepatocyte.

Fig. 8.1

19
(a) With reference to Fig. 8.1, explain how the molecular structure of the G-protein linked receptor
relates to its function. [5]

1. The transmembrane domain interacts with the hydrophobic fatty acid tails of the membrane
via hydrophobic interaction, and the ligand-binding domain and cytoplasmic domain
interact with water via hydrogen bonding or ion-dipole interaction. This allow GPLR to span
the membrane.

2. The ligand-binding domain faces the external and has a complementary shape to bind to
glucagon

3. The cytoplasmic domain binds to a G protein.

4. Ligand binding induces a series of conformational changes in the GPLR structure.

5. These conformational changes allow the signal from glucagon to be transduced into the cell.

Examiners’ Comments
The majority of responses addressed some aspects of the molecular structure of the G-protein
linked receptor and how these related to its functions. Details were sometimes incomplete. For
example, when considering how the protein spans the cell surface membrane, explanations were
often limited only to hydrophobic interactions.

Strong responses considered several aspects of the G-protein linked receptor in relation to effects
of ligand binding.

(b) Binding of glucagon to the glucagon receptor (G-protein linked receptor) leads to the release of
cAMP as the second messenger.

Explain how the release of cAMP leads to a cellular response. [5]

1. cAMP binds to and activates protein kinase A (PKA)…

2. …by binding to its regulatory subunits, causing them to dissociate from the catalytic subunits.

3. The catalytic subunits of PKA phosphorylates and (de)activate various target proteins,
leading to a cellular response.

4. The phosphorylated target protein either changes conformation or binds to other proteins to
become (de)activated.

5. e.g. Phosphorylation can activate enzymes involved in glycogen breakdown.

6. e.g. Phosphorylation can activate transcription factors that regulate expression of genes
involved in gluconeogenesis / glycogenolysis.

Examiners’ Comments
Strong responses provided a concise, sequential account of the stages involved, using correct
scientific terminology.

A common error was for responses to include a description of the events that occur before cAMP
is made. This was not required by the question. Other commonly seen errors included confusion
over which enzymes were activated or inactivated and between the terms glucagon and glycogen.

Key terms were not always spelt correctly, sometimes obscuring which term was intended.

[Total: 10]

20
QUESTION 9
Lord Howe Island is a small isolated oceanic volcanic island in the Tasman Sea, 600 km from the
Australian mainland. Two species of palm tree, Howea forsteriana and H. belmoreana, have evolved
by sympatric speciation on the island since it was formed 6.9 million years ago.

Fig. 9.1 shows the flowering times of the two species.

Fig. 9.1

(a) With reference to Fig. 9.1, suggest how these two species of palm tree may have evolved by
sympatric speciation. [5]

1. Fig 9.1 shows that the two species today have different peak flowering times
(week 3 vs week 9)

2. This suggests that their common ancestor exhibited variation in flowering times, hence
started to be temporally isolated.

3. Reduced gene flow between the populations with different flowering times, as they were not
pollinated at the same time / could not cross-pollinate each other.

4. Mutations occurred independently from each other, so allele frequencies changed differently.

5. Over time, both populations accumulated sufficient genetic differences (genetic

divergence).

6. Eventually, the two populations no longer interbred, leading to the two different species that
are reproductively isolated today.

7. Since these two species are on the same isolated oceanic island, they are not
geographically isolated, hence evolved by sympatric speciation.

Examiners’ Comments
Many responses demonstrated a sound understanding of sympatric speciation; fewer related this
understanding to the context of these two palm tree species evolving on an isolated oceanic island.

Some responses referred to the two species as though they had always existed, rather than
referring to how the two species had evolved from a common ancestor.

21
(b) Fig. 9.2 shows a phylogenetic tree for the two species of Howea from Lord Howe Island and their
closest relatives from mainland Australia (Linospadix and Laccospadix).

Fig. 9.2

A student concluded that Fig. 9.2 shows that Lord Howe Island was originally colonised by a
species of plant that was a common ancestor for Howea, Laccospadix and Linospadix.

Explain whether or not this conclusion is correct. [2]

1. Incorrect, since the island was formed only 6.9 million years ago.

2. Fig. 9.2 shows that the common ancestor for Howea, Laccospadix and Linospadix diverged
11.5 million years ago.

Examiners’ Comments
Many responses incorrectly stated that the conclusion was correct. Few related the information in
the phylogenetic tree to the age of the island.

(c) State the types of evidence that can be used to establish the phylogenetic relationships between
species. [3]

1. Nucleotide sequence of homologous DNA/gene shared by all the species.

2. Amino acid sequence of homologous protein shared by all the species.

3. Biogeography – studying the geographical distributions of animals and plants, including

fossils.

4. Carbon dating of fossils records to estimate the time of existence of the extinct common
ancestor.

Examiners’ Comments
Most responses included consideration of at least two types of evidence. Some mistakenly
referred to fossil fuels as evidence.

[Total: 10]

22
QUESTION 10
Fig. 10.1 shows the specific T cell and antibody response in people with mild and severe symptoms
caused by infection by the same virus.

Fig. 10.1

(a) With reference to Fig. 10.1, describe the differences in the immune response of infected people
with mild symptoms and infected people with severe symptoms. [3]

feature people with mild symptoms people with severe symptoms

Appears earlier on day 2 from onset Appears later on day 12 from onset
of symptoms of symptoms
relative
numbers of Increases quickly and peaks at day
virus-specific Increases very slowly till day 30
7, then gradually decline till day 30
T cells
Higher at every time point compared Lower at every time point compared
to severe symptoms to mild symptoms
Appears earlier on day 11 from Appears later on day 14 from onset
Relative onset of symptoms of symptoms
numbers of
virus-specific
antibodies peaks at day 16 peaks later at day 21

Examiners’ Comments
Many candidates interpreted the data correctly. Weaker candidates confused the key to the
graphs and referred to mild symptoms instead of severe or referred to antibodies instead of T
cells.

23
(b) Suggest and explain what vaccination for this virus needs to achieve in order to give long-term
protection against severe symptoms. [2]

1. Active immunity - vaccination with attenuated/weakened form of the whole virus OR surface
glycoproteins of the virus.

2. This allows memory T cells to form.

3. Allows rapid clonal expansion of virus-specific T cells, which is found to be delayed in

people with severe symptoms (Fig. 10.1).

Examiners’ Comments
Some responses described the type of vaccine that would be used, or the need for herd immunity,
which did not address the requirements of the question.

Many referred to memory cells, but few made full use of the information in Fig. 10.1 to identify the
most important factors needed to achieve long-term protection against severe symptoms.

[Total: 5]

24
QUESTION 11
Alexander von Humboldt visited Mount Chimborazo in Ecuador in 1802. Humboldt took
measurements of the heights up the mountain at which glaciers and vegetation occurred.

Fig. 11.1 compares the heights at which glaciers and vegetation occurred on Mount Chimborazo in
1802 with the heights at which glaciers and vegetation occurred on Mount Chimborazo in 2012. Only
heights above 4400 m are shown. Below this height, no glaciers occurred and vegetation cover was
continuous.

Fig. 11.1

(a) Describe the changes shown in Fig. 11.1 from 1802 to 2012. [2]

1. The range in which glaciers occurred decreases from between 4820m and 5360m in 1802
to between 5270m and 5360m in 2012.

2. The range in which vegetation occurred increases from between 0m and 4600m in 1802
to between 0m and 5120m in 2012.

Examiners’ Comments
Many candidates correctly quoted comparative figures to describe the changes shown in Fig. 11.1.
A number of responses referred incorrectly to a decrease in the height of glaciers rather than a
decrease in the height range over which glaciers occur.

25
(b) Suggest how human activity may account for the changes shown in Fig. 11.1. [4]

1. Ref to any anthropogenic activities that increases greenhouse gases.

a. Deforestation for the purpose of other land uses
b. Burning of fossil fuels for electricity and heating
c. Livestock farming due to increased meat consumption demand

2. Enhanced greenhouse gases trap more heat in Earth’s atmosphere, leading to global
warming.

3. Higher altitudes that were once colder becomes warmer.

4. Hence, glaciers at lower altitude have melted, and thus limited to only the highest altitude.

5. Hence, vegetations have extended to higher altitudes where temperatures are now suitable
for vegetation growth.

Examiners’ Comments
The majority of responses demonstrated a good understanding of the human activities that lead
to increased global warming. Weak responses did not consider that it was the increase in various
relevant human activities that was important.

[Total: 6]

END OF PAPER 2

26