BIO PAPER 1 2018

Question 1

0 1 Figure 1 shows all the chromosomes present in one human cell during mitosis. A
scientist stained and photographed the chromosomes. In Figure 2, the scientist has
arranged the images of these chromosomes in homologous pairs.
Figure 1 Figure 2
0 1 . 1 Give two pieces of evidence from Figure 1 that this cell was undergoing mitosis.
Explain your answers.
[2 marks]

Answer 1

1. The (individual) chromosomes are
visible because they have condensed;
2. (Each) chromosome is made up of two
chromatids because DNA has
replicated;
3. The chromosomes are not arranged in
homologous pairs, which they would be
if it was meiosis;

Question 2

0 1 box . 2 Tick ( ) one box that gives the name of the stage of mitosis shown in Figure 1.
[1 mark]
Anaphase
Interphase
Prophase
Telophase

Answer 2

 prophase

Question 3

1 . 3 When preparing the cells for observation the scientist placed them in a solution that
had a slightly higher (less negative) water potential than the cytoplasm. This did not
cause the cells to burst but moved the chromosomes further apart in order to reduce
the overlapping of the chromosomes when observed with an optical microscope.
Suggest how this procedure moved the chromosomes apart.
[2 marks]

Answer 3

1. Water moves into the cells/cytoplasm by
   osmosis;
2. Cell/cytoplasm gets bigger;

Question 4

0 1 box . 4 The dark stain used on the chromosomes binds more to some areas of the
chromosomes than others, giving the chromosomes a striped appearance.
Suggest one way the structure of the chromosome could differ along its length to
result in the stain binding more in some areas.
[1 mark]

Answer 4

Differences in base sequences
OR
Differences in histones/interaction with
histones
OR
Differences in condensation/(super)coiling

Question 01.4 tested Assessment Objective 2 (application of knowledge) and there were several
parts of the specification from which students could select material to support their answer. 42.2%
did this successfully; those who did not often did not describe sufficiently how a feature would be
different along the length of the chromosome to result in the striped appearance. For example,
mentioning “histones” or “bases” alone was not creditworthy.

Question 5

1 . 5 In Figure 2 the chromosomes are arranged in homologous pairs.
What is a homologous pair of chromosomes?
[1 mark]

Answer 5

(Two chromosomes that) carry the same
genes;

Reject ‘same alleles’
Accept ‘same loci’ (plural) or
‘genes for the same
characteristics

In question 01.5, only 36.9% of students could define the term ‘homologous chromosome’. It is
likely that more students could have written about independent segregation or crossing over of
homologous chromosomes, but this question revealed that they did not fully understand this
biological term. Many students only referred to the origin of the chromosomes as paternal and
maternal.

Question 6

0 1 . 6 Give two ways in which the arrangement of prokaryotic DNA is different from the
arrangement of the human DNA in Figure 1.
[2 marks]

Answer 6

(Prokaryotic DNA) is
1. Circular (as opposed to linear);
2. Not associated with proteins/histones ;
3. Only one molecule/piece of DNA
OR
present as plasmids;

Max 1 if prokaryotic DNA only
found as plasmids OR if
prokaryotic DNA is single
stranded.
Ignore references to nucleus,
exons, introns or length of
DNA.
Do not credit converse
statements.
Ignore descriptions of
eukaryotic DNA alone.

For question 01.6, most students (83.2%) scored at least one mark, but there were many good
answers limited to one out of two by demonstration of fundamental misunderstanding. The most
common of these was that only eukaryotic DNA is a double helix and that prokaryotic DNA is
single-stranded.

Question 7

0 2 box A student investigated the effect of surface area on osmosis in cubes of potato.
• He cut two cubes of potato tissue, each with sides of 35 mm in length.
• He put one cube into a concentrated sucrose solution.
• He cut the other cube into eight equal-sized smaller cubes and put them into a
sucrose solution of the same concentration as the solution used for the large cube.
• He recorded the masses of the cubes at intervals.
His results are shown in Figure 3

0 2 . 1 Describe the method the student would have used to obtain the results in Figure 3.
Start after all of the cubes of potato have been cut. Also consider variables he should
have controlled.
[3 marks]

Answer 7

1. Method to ensure all cut surfaces of the eight
   cubes are exposed to the sucrose solution;
2. Method of controlling temperature;
3. Method of drying cubes before measuring;
4. Measure mass of cubes at stated time intervals;

This question was based on a variant of required practical activity 3 – “Production of a dilution
series of a solute to produce a calibration curve with which to identify the water potential of plant
tissue”.
Although question 02.1 was based on a more straightforward investigation of osmosis, very similar
principles apply to those used when carrying out the required practical. When asked for a method,
students must write out what actions need to be carried out as if instructing a fellow student.
Generic statements, such as “keep the temperature the same” or “record the mass at set intervals”,
were not credited. Instead, students should give specific instructions about how the results would
be obtained, for example using a water bath set to 30°C, removing the potato chips every 10
minutes to record their mass, and blotting the potato chips dry with a paper towel.

Question 8

0 2 box . 2 The loss in mass shown in Figure 3 is due to osmosis. The rate of osmosis between
0 and 40 minutes is faster in B (the eight small cubes) than in A (single large cube).
Is the rate of osmosis per mm2 per minute different between A and B during this time?
Use appropriate calculations to support your answer.
[3 marks]

Answer 8

Allow 1 mark for calculation of surface area of two
(sets of) cubes 7350 (mm2
) and 14700 (mm2
)
Allow 1 mark for calculation of both rates of
osmosis shown in first 40 minutes – between 0.12
and 0.13 and between 0.22 and 0.23
If surface area and/or rate of osmosis is incorrect
then, allow 1 mark for (their) calculated rate divided
by (their) calculated surface area

Question 02.2 was the first question on the paper testing mathematical skills (specifically MS 3.5
and MS 4.1). Students needed to calculate rate of change from a graph showing a linear
relationship; this was the most successfully completed part of the calculation, although using the
wrong part of the graph or incorrectly reading from the graph was not uncommon. Students also
needed to calculate the surface area of the one large cube and the eight smaller cubes, but only a
small number of students could do this correctly. Only 5% could correctly combine these to come
up with two correct rates per unit surface area. There was a small but sizeable minority of students
who did not show the steps in their calculations, but just gave the final (incorrect) answer, meaning
it was not possible to award any marks. A small minority wrote descriptions of the change in mass
shown by the graph, or explained why the cubes with large surface area would lose mass more
quickly, without any calculations, thereby ignoring the question stem. Disappointingly, 56.4% of
students failed to gain even one mark for this question.

Question 9

0 3 Bees are flying insects that feed on nectar made in box flowers. There are many different
species of bee.
Scientists investigated how biodiversity of bees varied in three different habitats
during a year. They collected bees from eight sites of each habitat four times per year
for three years.
The scientists’ results are shown in Figure 4 in the form they presented them.
Figure 4
0 3 . 1 What is meant by ‘species richness’?
[1 mark]

Answer 9

(A measure of) the number of (different) species in
a community; For ‘community’ accept
‘habitat/ecosystem/one
area/environment’
Reject ‘in a population’

67.1% of students could answer question 03.1, a test of recall from section 3.4.6 of the
specification. Once again, many students confused the terms population and community, and the
terms species richness and an index of diversity.

Question 10

0 3 box . 2 From the data in Figure 4, a student made the following conclusions.
1. The natural habitat is most favourable for bees.
2. The town is the least favourable for bees.
Do the data in Figure 4 support these conclusions? Explain your answer.
[4 marks]
1. The natural habitat is most favourable for bees.

2. The town is the least favourable for bees.

(4)

Answer 10

Yes, natural best, because
1. Peak of (mean) bee numbers in natural habitat
is highest;
2. The (mean) number of bees was higher in the
natural habitat until day 200;
3. (Mean) species richness in natural habitat
higher at all times;
No, natural not best, because
4. Lowest (mean) number of bees after day 220;
Yes, town worst, because
5. Peak of species richness higher in both natural
and farmland
OR
Species richness lowest in town from day 125;
No, town not worst, because
6. (Mean) species richness is lower in farmland
until day 125;
7. Similar (mean) number of bees to farmland;
OR
(Mean) number of bees lower in farmland until
day 140;
General, no, because
8. Index of diversity of bees not measured
OR
The number of bees of each species is not
known;

Question 03.2 was the first opportunity on the paper for students to interpret scientific evidence
and demonstrate Assessment Objective 3 skills. The question, ‘Do the data in Figure 4 support these conclusions?’, should have demonstrated that careful use of these data would be required in
the answer. Both conclusions were comparative statements, natural habitat being the most
favourable and town habitat being the least favourable, and so the marking points were linked to
comparative statements between the habitats. A description of one habitat alone was, thus, not
sufficient to gain credit. Numbers were provided on the axes and so quoting days of the year was
expected in the answers. This question provides a good example to demonstrate the importance
of using and quoting the correct evidence from the data when trying to support the conclusions. It
is also a good example of how times from the x-axis of the graph should be used to illustrate where
observed trends begin and end. Some 76.6% of students were able to gain at least one mark; only
7.9% scored four marks.

Question 11

0 3 box . 3 The scientists collected bees using a method that was ethical and allowed them to
identify accurately the species to which each belonged.
In each case, suggest one consideration the scientists had taken into account to
make sure their method
[2 marks]

1. was ethical.
2. allowed them to identify accurately the species to which each belonged.

Answer 11

1. Must not harm the bees
   OR
   Must allow the bee to be released
   unchanged;
2. Must allow close examination
   OR
   Use a key (to identify the species);

In part 03.3, the majority of students appreciated that the bees should not be harmed and so
gained the first mark. Many students got confused with mark-release-recapture techniques but,
since no population estimate was being made in this investigation, this was not relevant.

Question 12

3 . 4 Suggest and explain two ways in which the scientists could have improved the
method used for data collection in this investigation.
[2 marks]

Answer 12

1. Collect at more times of the year so
   more points on graph/better line (of
   best fit) on graph;
2. Counted number of individuals in each
   species so that they could calculate
   index of diversity;
3. Collected from more sites/more years
   to increase accuracy of (mean) data;

Question 13

0 3 box . 5 Three of the bee species collected in the farmland areas were Peponapis pruinosa,
Andrena chlorogaster and Andrena piperi.
What do these names suggest about the evolutionary relationships between these
bee species? Explain your answer.
[2 marks]

Answer 13

1. A. chlorogaster and A. piperi are more
   closely related (to each other than to
   P. pruinosa);
2. Because they are in the same genus;

03.5 was answered well, with most students (62.1%) understanding the binomial system for
naming of species and linking it to their evolutionary relationship. Some students suggested that
the two Andrena species were unrelated to Peponapis pruinosa and that they did not share a
common ancestor – this was not creditworthy

Question 14

0 4 . 1 Formation of an enzyme-substrate complex increases the rate of reaction.
Explain how.
[2 marks]

Answer 14

1. Reduces activation energy;
2. Due to bending bonds
   OR
   Without enzyme, very few substrates have
   sufficient energy for reaction;

In question 04.1, although many students could describe that the enzyme lowers the activation
energy, only 18% could explain that this was due to the enzyme bending the bonds in the
substrate. Some students gave lengthy descriptions of how the E-S complex forms, losing focus
on the question of how this results in an increase in the rate of reaction.

Question 15

0 4 . 2 A scientist measured the rate of removal of amino acids from a polypeptide with and
without an enzyme present. With the enzyme present, 578 amino acids were
released per second. Without the enzyme, 3.0 × 10−9 amino acids were released per
second.
Calculate by how many times the rate of reaction is greater with the enzyme present.
Give your answer in standard form.
[2 marks]

Answer 15

1.93 x 1011;;
Allow 1 max for
578/3.0 × 109
1.93 x 10x when x ≠11
Correct answer with incorrect standard form
e.g. 19.3 x 1010

Questions 04.2 and 04.3 assessed mathematical skills MS 0.2, MS 0.3 and MS 3.1. It was
pleasing to see standard form being used successfully, with 71.9% of students achieving both
marks for 04.2.

Question 16

Another scientist investigated an enzyme that catalyses the box following reaction.
ATP → ADP + Pi
The scientists set up two experiments, C and L.
Experiment C used
• the enzyme
• different concentrations of ATP.
Experiment L used
• the enzyme
• different concentrations of ATP
• a sugar called lyxose.
The scientists measured the rate of reaction in each experiment. Their results are
shown in Figure 5.

0 4 box . 3 Calculate the rate of reaction of the enzyme activity with no lyxose at 2.5 mmol dm−3 of
ATP as a percentage of the maximum rate shown with lyxose.
[2 marks]

Answer 16

Students found part 04.3 more challenging. Many thought they needed to
calculate a rate, not appreciating that the y-axis already gave the rate of reaction. Many did not
read the correct parts of the graph – at 2.5 mmol dm-3 for curve C and at the highest point for curve L. Credit of a single mark was given if a correct calculation was completed with readings from
incorrect parts of the graph, but all readings had to be accurate and this was often not the case.

Question 17

4 box . 4 Lyxose binds to the enzyme.
Suggest a reason for the difference in the results shown in Figure 5 with and without
lyxose.
[3 marks]

Answer 17

1. (Binding) alters the tertiary structure
of the enzyme ;
2. (This causes) active site to change
(shape);
3. (So) More (successful) E-S
complexes form (per minute)
OR
E-S complexes form more quickly
OR
Further lowers activation energy;

The key to answering question 04.4 successfully was to use the information provided – that the
lyxose binds to the enzyme and that the graph shows that lyxose increases the rate of reaction.
Many students started their answer with the idea of lyxose being an inhibitor and reducing the rate
of reaction – this limited them to one mark out of three. A significant number of students inferred
that lyxose is a respiratory substrate (rather than using the information that it binds to the enzyme),
so there would be an increase in ATP levels, so more substrate and hence a faster rate. It was a
shame that some students who worked out what was going on did not achieve full marks due to
imprecise use of language, for example referring to changing the 3-D shape of the enzyme rather
than changing the tertiary structure of its active site. Fewer than half the students scored any
marks on this question.

Question 18

0 5 box . 1 Draw the general structure of an amino acid.

[1 mark]

Answer 18

H
H2N —–c—COOH
R

64.5% of students could answer question 05.1, a test of recall from section 3.1.4.1 of the
specification.

Question 19

5 . 2 The genetic code is described as degenerate.
What is meant by this? Use an example from Table 1 to illustrate your answer.
[2 marks]

Answer 19

1. More than one codon codes for a single amino
   acid;
2. Suitable example selected from Table 1;

Question 20

A scientist investigated changes in the amino acid sequence of a human enzyme box
resulting from mutations. All these amino acid changes result from single base
substitution mutations.
This enzyme is a polypeptide 465 amino acids long.
Table 2 shows the result of three of the base substitutions.
Table 2
Amino acid
number
Correct amino
acid
Amino acid inserted as a result of
mutation
203 Val Ala
279 Glu Lys
300 Glu Lys
0 5 . 3 What is the minimum number of bases in the gene coding for this polypeptide?
[1 mark]

Answer 20

1395

05.3 was an interesting question as it could be interpreted as a straightforward 465 amino acids
multiplied by three to give the 1395 bases of the gene. Students could also add three or six bases
to this, for a start and/or stop codon. Some students appreciated that a gene would be made up of
double-stranded DNA so would comprise double this number of bases – this was also awarded
credit. 67% of students could complete this calculation.

Question 21

0 5 box . 4 Use information from Table 1 to tick ( ) one box that shows a single base substitution
mutation in DNA that would result in a change from Val to Ala at amino acid number
203

CAA → CGA
GUU → GCA
GUU → GUC
CAC → CGG 
[1 mark]

Answer 21

CAA → CGA

Question 05.4 showed that multiple-choice questions are not always simplistic or based on recall.
This question required students to interpret Table 1, convert the mRNA codon to a DNA base
triplet, and also notice in the question that this was a single base substitution. Only 9.6%
successfully achieved all of this. The majority of students ticked box 2 that did show a Val to Ala
change, but with a mutation in the mRNA (not DNA) and of 2 bases

Question 22

0 5 . 5 A change from Glu to Lys at amino acid 300 had no effect on the rate of reaction
catalysed by the enzyme. The same change at amino acid 279 significantly reduced
the rate of reaction catalysed by the enzyme.
Use all the information and your knowledge of protein structure to suggest reasons for
the differences between the effects of these two changes.
[3 marks]

Answer 22

1. (Both) negatively charged to positively charged
   change in amino acid;
2. Change at amino acid 300 does not change the
   shape of the active site
   OR
   Change at amino acid 300 does not change the
   tertiary structure
   OR
   Change at amino acid 300 results in a similar
   tertiary structure;
3. Amino acid 279 may have been involved in a
   (ionic, disulfide or hydrogen) bond and so the
   shape of the active site changes
   OR
   Amino acid 279 may have been involved in a
   (ionic, disulfide or hydrogen) bond and so the
   tertiary structure changed;
   OR
   Amino acid 279 may be in the active site and
   be required for binding the substrate;

The answers to question 05.5 revealed much misunderstanding of the principles of the link
between DNA, polypeptides and proteins, as many students discussed the changes in amino acid
in terms of silent mutations or frameshift mutations. The instruction to use all the information
required the students to look back at Table 1 and use the key showing the properties of the R
group of the amino acid. Very few students could apply this information and link the reduction in
rate of reaction with the R group of amino acid 279 changing from being negatively-charged to
being positively-charged. Imprecise biological language was also often seen, for example omitting
to reference the ‘shape’ of the active site and discussing the 3-D structure of the protein rather than
its tertiary structure. Just over half of the students (54.2%) managed to score at least one mark
here; only 3.2% gained maximum credit.

Question 23

0 6 box Figure 6 shows a faulty form of meiosis that can occur in some plant
s.

0 6 . 1 Complete Figure 7 to show the chromosome content of the cells that would result
from a normal meiotic division of the diploid parent cell shown in Figure 6.
[2 marks]

Answer 23

1. 1 long and 1 short chromosome, each
   made up of 2 chromatids held (by
   centromere), in each cell of 1st division;
2. 1 long and 1 short (separate) chromosome
   in each cell of 2nd division;

2. Allow ECF for correct
chromosomes shown in
each cell from candidate’s
1
st division cells.
2. Ignore drawing of
centromere

Question 24

0 6 box . 2 If two diploid (2n) gametes fuse at fertilisation, it can result in the growth of a tetraploid
plant which has 4 copies of each chromosome.
Red clover is a plant grown to produce cattle feed. Tetraploid red clover plants
produce a higher yield than diploid red clover plants.
Whether a red clover plant produces 2n gametes is genetically controlled.
Scientists investigated the possibility of breeding red clover plants that only produced
2n gametes.
• In breeding cycle 0, they grew red clover plants and identified plants that produced
2n gametes.
• In breeding cycle 1, they used the plants producing 2n gametes to produce
offspring.
• In breeding cycles 2 and 3, they identified plants producing 2n gametes and used
these to produce offspring.
Their results are shown in Table 3.
Table 3

The scientists used the following null hypothesis.
‘The proportion of plants that produce 2n gametes will not change from one breeding
cycle to the next.’
Complete Table 3 to show the expected number of plants that did not produce 2n
gametes and the expected number of plants that did produce 2n gametes after 1
cycle.
Give each answer to the nearest whole number.
[2 marks]

Answer 24

52 4

Allow 1 mark for numbers totalling 56 except
14/42 - repetition of observed values.
If table is blank, award 1 mark for evidence of
56.

Question 06.2 tested students’ understanding of the chi-squared test. The null hypothesis stated
‘the proportion of plants will not change from one breeding cycle to the next’. Students, therefore,
needed to maintain the proportions from breeding cycle 0 into breeding cycle 1, taking into account
that the total number of plants had changed from 54 to 56. 22.3% of students could successfully
combine this understanding of chi-squared with their maths skills, using proportions to achieve
both marks.

Question 25

0 6 . 3 The scientists tested their null hypothesis using the chi-squared statistical test.
After 1 cycle their calculated chi-squared value was 350
The critical value at P=0.05 is 3.841
What does this result suggest about the difference between the observed and
expected results and what can the scientists therefore conclude?
[2 marks]

Answer 25

1. There is a less than 0.05/5% probability
   that the difference(s) (between observed
   and expected) occurred by chance;
2. Calculated value is greater than critical
   value so the null hypothesis can be
   rejected;
3. (The scientists can conclude that) the
   proportion of plants that produce 2n
   gametes does change from one breeding
   cycle to the next;

06.3 showed that questions relating to P values are still not being answered well. Too many
students still write that the ‘results’ are due to chance, omitting the essential aspect of it being the
‘difference’ in the results that is or isn’t due to chance. It was rare to see the concepts of
probability and chance being appropriately applied. It may be that students have only used chisquared in relation to genetic crosses and this novel context caused confusion, but students should
be introduced to the use of all three required statistical tests in a range of contexts.

Question 26

0 6 . 4 Use your knowledge of directional selection to explain the results shown in Table 3.
[3 marks]

Answer 26

1. The scientists selected/used for breeding plants
   that produced 2n gametes;
2. (So these plants) passed on their alleles (for
   production of 2n gametes to the next
   generation);
3. The frequency of alleles for production of 2n
   gametes increased (in the population).

Question 06.4 tested section 3.4.4 of the specification, specifically ‘students should be able to use
unfamiliar information to explain how selection produces changes within a population of a species’.
Although the instruction to ‘use your knowledge of directional selection’ was intended to help
students focus their answer, it led to many answers that simply described the outcome of
directional selection with no explanation of how it comes about. When attempts were made to give
explanations, they were often not linked to this example. Many students suggested that the
environment had changed, so that the plants producing 2n gametes had a selective advantage,
instead of using the information provided that the scientists used only these specific plants for
breeding. The term ‘allele’, used in the correct context, was seen infrequently.

Question 27

0 7 box . 1 When a person is bitten by a venomous snake, the snake injects a toxin into the
person. Antivenom is injected as treatment. Antivenom contains antibodies against
the snake toxin. This treatment is an example of passive immunity.
Explain how the treatment with antivenom works and why it is essential to use passive
immunity, rather than active immunity.
[2 marks]

Answer 27

1. (Antivenom/Passive immunity) antibodies
   bind to the toxin/venom/antigen and
   (causes) its destruction;
2. Active immunity would be too slow/slower;

There were two parts to question 07.1 – how does an antibody work and what is the difference
between passive and active immunity? 28.1% of students could answer both of these correctly.
Confusion was demonstrated between antigens on a toxin and antibody binding to a pathogen.
Errors resulting in the first marking point not being awarded included describing binding, but failing
to discuss destruction, or discussing destruction without reference to binding. Commonly, students
used the idea of ‘complementary’ in place of binding. Those who were not awarded marking point
two tended not to make a comparative statement about active and passive immunity, and some
made vague statements about active immunity ‘taking time’, rather than expressing length of time.
A mark of zero was commonly the result of confusing active and passive immunity or treating the
antivenom as a vaccine which would then trigger an immune response.

Question 28

0 7 . 2 A mixture of venoms from several snakes of the same species is used.
Suggest why.
[2 marks]

Answer 28

1. May be different form of antigen/toxin
(within one species)
OR
Snakes (within one species) may have
different mutations/alleles;
2. Different antibodies (needed in the
antivenom)
OR
(Several) antibodies complementary (to
several antigens);

In 07.2, many students gave answers relating to not needing to identify the species of snake that
had bitten a person, as the antivenom would work against the venom of several species – this was
not creditworthy as the question is clearly related to several snakes of the same species. A

pleasing number of students understood the production of antivenom and appreciated the need for
several antibodies to be produced by the animal to be used in the patient. Some suggested that
one antibody could be effective against several antigens; this was not given credit. Pleasingly, few
students confused antigens and antibodies. Common errors included referring to different forms of
venom, which was indicated in the question stem, or failing to include reference to antibodies and
instead making vague statements about “being effective against” or “fighting off” or
“neutralising/counteracting” the venom. Roughly even proportions of students scored two, one and
zero marks for this question.

Question 29

0 7 box . 3 Horses or rabbits can be used to produce antivenoms.
When taking blood to extract antibody, 13 cm3 of blood is collected per kg of the
animal’s body mass.
The mean mass of the horses used is 350 kg and the mean mass of the rabbits used
is 2 kg
Using only this information, suggest which animal would be better for the production of
antivenoms.
Use a calculation to support your answer.
[2 marks]

Answer 29

In question 07.3, most students could complete the required calculation. Even so, many students
suggested the rabbit would be better as it would likely be safer for the rabbit as less blood was
removed or suggested that the same number of antibodies would be produced in a smaller volume
of blood. Of those who did pick the horse, many only stated that more blood could be collected,
rather than linking this to more antibodies/antivenom being collected from each animal. 81% of
students scored at least one mark here.

Question 30

0 7 . 4 During the procedure shown in Figure 8 the animals are under ongoing observation
by a vet.
Suggest one reason why.
[1 mark]

Answer 30

1. (So) the animal does not suffer from the
   venom/vaccine/toxin;
2. (So) the animal does not suffer
   anaemia/does not suffer as a result of
   blood collection;
3. (So) the animal does not have pathogen
   that could be transferred to humans;

Too many generalised answers were given to question 07.4. At this level, a specific reason why it
would be ethical to have veterinary supervision in this particular procedure was required.
Confusion was sometimes demonstrated here over whether the animals were being administered
venom or antivenom
Accept ‘To fulfil
licence/legal requirements’

Question 31

0 7 box . 5 During vaccination, each animal is initially injected with a small volume of venom.
Two weeks later, it is injected with a larger volume of venom.
Use your knowledge of the humoral immune response to explain this vaccination
programme.
[3 marks]

Answer 31

1. B cells specific to the venom reproduce by
   mitosis;
2. (B cells produce) plasma cells and memory
   cells;
3. The second dose produces antibodies (in
   secondary immune response) in higher
   concentration and quickly
   OR
   The first dose must be small so the animal
   is not killed;

Question 32

8 Scientists investigated the effect of a heat treatment on mass transport in barley box
plants.
• They applied steam to one short section of a leaf of the heat-treated plants. This
area is shown by the arrows in Figure 9.
• They did not apply steam to the leaves of control plants.
• They then supplied carbon dioxide containing radioactively-labelled carbon to each
plant in the area shown by the rectangular boxes in Figure 9.
• After 4 hours, they:
• found the position of the radioactively-labelled carbon in each plant. These
results are shown in Figure 9.
• recorded the water content of the parts of the leaf that were supplied with
radioactively-labelled carbon dioxide. These results are shown in Table 4.

0 8 box . 1 The scientists concluded that this heat treatment damaged the phloem.
Explain how the results in Figure 9 support this conclusion.
[2 marks]

Answer 32

EITHER
1. The radioactively labelled carbon is converted
into sugar/organic substances during
photosynthesis;
2. Mass flow/translocation in the phloem
throughout the plant only in plants that were
untreated/B/control
OR
Movement of sugar/organic substances in the
phloem throughout the plant only in plants that
were untreated/B/control;
OR
3. Movement in phloem requires living
cells/respiration/active transport/ATP;
4. Heat treatment damages living cells so
transport in the phloem throughout the plant
only in plants that were untreated/B/control
OR
Heat treatment stops respiration/active
transport/ATP production so transport in the
phloem throughout the plant only in plants that
were untreated/B/control;

Do not mix and match –
award either mp1 and
mp2 or mp3 and mp4.
1. For ‘organic
substance’ accept
named organic
substance, eg glucose,
sucrose, amino acid.
2. Accept
‘translocation/mass
transport in the phloem
past the heat treatment
only in the untreated
plant/B/control’.
2. Accept converse for
heat-treated plant/A
ie Movement of
sugar/organic
substances/mass
flow/translocation in the
phloem stops (beyond
the heat treatment) in
treated plants/A.

The majority of students limited their answer to question 08.1 to a description of the data rather
than an explanation. They simply stated that radioactive carbon dioxide was not visible throughout
the plant in A and therefore had not been transported through the phloem. As this question
required the students to explain the results, all the marking points required use of the students’
knowledge. The first point was for the idea that the radioactive carbon dioxide would be used in
the leaf to produce sugar in photosynthesis. The second point was then that sugar is transported
in the phloem. If the first point was not appreciated, then students could score the second marking
point for appreciating that mass flow/translocation happens in the phloem and that is what has
stopped in this case. The alternative valid approach to the question for marking points 3 and 4 was
not seen. Nearly 60% of students failed to make any headway in this question.

Question 33

0 8 . 2 The scientists also concluded that this heat treatment did not affect the xylem.
Explain how the results in Table 4 support this conclusion.
[2 marks]

Answer 33

1. (The water content of the leaves was) not
   different because (means ± 2) standard
   deviations overlap;
2. Water is (therefore) still being transported in the
   xylem (to the leaf)
   OR
   Movement in xylem is passive so unaffected by
   heat treatment;
3. For ‘not different’
   accept ‘difference is not
   significant’ or ‘difference
   due to chance’.

Question 34

8 box . 3 The scientists then investigated the movement of iron ions (Fe3+) from the soil to old
and young leaves of heat-treated barley plants and to leaves of plants that were not
heat treated. Heat treatment was applied half way up the leaves. The scientists
determined the concentration of Fe3+ in the top and lower halves of the leaves of each
plant.
Their results are shown in Figure 10.

What can you conclude about the movement of Fe3+ box in barley plants?
Use all the information provided.
[4 marks]

Answer 34

1. Heat treatment has a greater effect on young
   leaves than old;
2. Heat treatment damages the phloem;
3. Fe3+ moves up the leaf/plant;
4. (Suggests) Fe3+ is transported in the xylem in
   older leaf;
5. In young leaf, some in xylem, as some still
   reaches top part of leaf;
6. (Suggests) Fe3+ is (mostly) transported in phloem
   in young leaf
   OR
   Xylem is damaged in young leaf
   OR
   Xylem is alive in young leaf;
7. Higher ratio of Fe3+ in (all/untreated) old leaves
   than (all/untreated) young;
8. All ratios show there is less Fe3+ in the top than
   the lower part of leaves;
9. (But) no statistical test to show if the difference(s)
   is significant;

Question 35

0 9 box . 1 Describe the role of two named enzymes in the process of semi-conservative
replication of DNA.
[3 marks]

Answer 35

1. (DNA) helicase causes breaking of hydrogen/H
   bonds (between DNA strands);
2. DNA polymerase joins the (DNA) nucleotides;
3. Forming phosphodiester bonds;

Question 36

0 9 box . 2 Scientists investigated the function of a eukaryotic cell protein called cyclin A. This
protein is thought to be involved with the binding of one of the enzymes required at
the start of DNA replication.
The scientists treated cultures of cells in the following ways.
C – Control cells, untreated
D – Added antibody that binds specifically to cyclin A
E – Added RNA that prevents translation of cyclin A
F – Added RNA that prevents translation of cyclin A and added cyclin A protein
They then determined the percentage of cells in each culture in which DNA was
replicating.
Their results are shown in Table 5.
Table 5

Suggest explanations for the results in Table 5 box .
[3 marks]

Answer 36

1. (Treatment D Antibody binds to cyclin A so) it
   cannot bind to DNA/enzyme/initiate DNA
   replication;
2. (Treatment E) RNA interferes with
   mRNA/tRNA/ribosome/polypeptide formation
   (so cyclin A not made);
3. In Treatment F added cyclin A can bind to
   DNA/enzyme (to initiate DNA replication)
   OR
   Treatment F shows that it is the cyclin A that is
   being affected in the other treatments
   OR
   Treatment F shows that cyclin A allows the
   enzyme to bind (to DNA)
   OR
   (Some cells in D or E) can continue with DNA
   replication because they have a different cyclin
   A allele
   OR
   (Some cells in D or E) can continue with DNA
   replication because the antibody/RNA has not
   bound to all the cyclin A protein/mRNA
   OR
   (Some cells in E) can continue with DNA
   replication because they contain previously
   translated cyclin A;

Question 37

1 0 box . 1 Describe the gross structure of the human gas exchange system and how we breathe
in and out.
[6 marks]

Answer 37

1. Named structures – trachea, bronchi, bronchioles,
   alveoli;
2. Above structures named in correct order
   OR
   Above structures labelled in correct positions on a
   diagram;
3. Breathing in – diaphragm contracts and external
   intercostal muscles contract;
4. (Causes) volume increase and pressure decrease
   in thoracic cavity (to below atmospheric, resulting
   in air moving in);
5. Breathing out - Diaphragm relaxes and internal
   intercostal muscles contract;
6. (Causes) volume decrease and pressure increase
   in thoracic cavity (to above atmospheric, resulting
   in air moving out);

1, 2. Reject mp1 if
structures from other
physiological systems
are named but award
mp2 if the correct
structures are in the
correct order.
4. and 6. For thoracic
cavity accept ‘lungs’
or ‘thorax’.
4. and 6. Reference to
‘thoracic cavity’ only
required once.
If idea of thoracic
cavity is missing or
incorrect, allow ECF
for mark point 6.
5. Accept diaphragm
relaxes and (external)
intercostal muscles
relax and lung tissue
elastic (so recoils). 

Question 10.1 demanded recall from section 3.3.2 of the specification; 25.6% of students gained
five or six marks. Many, however, omitted any reference to the structure at all and it was surprising
that very few students elected to draw a labelled sketch to show the gross structure. Many lengthy
answers were seen detailing exchange of gases and the features of the alveolar epithelium, neither
of which was required by the wording in the question. Many answers included tracheoles as a part
of the human gas exchange system. Inhalation tended to be described in the best detail. Marking
points 3 and 5 were often not awarded because of a lack of precision in describing the role of the
pair of antagonistic intercostal muscles. Many students conflated the two, and referred to them
generically. The relaxation and contraction of the diaphragm and its corresponding shape were
frequently confused. For example, students referred to the ‘flattened’ or ‘domed’ shape of the
diaphragm without stating how that occurred. This question had the highest discrimination index of
the paper.

Question 38

1 0 . 2 Mucus produced by epithelial cells in the human gas exchange system contains
triglycerides and phospholipids.
Compare and contrast the structure and properties of triglycerides and phospholipids.
[5 marks]

Answer 38

1. Both contain ester bonds (between glycerol and
   fatty acid);
2. Both contain glycerol;
3. Fatty acids on both may be saturated or
   unsaturated;
4. Both are insoluble in water;
5. Both contain C, H and O but phospholipids also
   contain P;
6. Triglyceride has three fatty acids and
   phospholipid has two fatty acids plus phosphate
   group;
7. Triglycerides are hydrophobic/non-polar and
   phospholipids have hydrophilic and
   hydrophobic region;
8. Phospholipids form monolayer (on
   surface)/micelle/bilayer (in water) but
   triglycerides don’t

In question 10.2, although many students could demonstrate knowledge of phospholipids and
triglycerides individually, they struggled to complete the required ‘compare and contrast’ command.
When this command is used, every marking point requires a comparative statement that must be
clearly made by the student: examiners will not infer links between separate statements – in this case, separate descriptions of phospholipids and of triglycerides. Many students did not include
glycerol in their structure of the phospholipid. Some students were distracted into discussions of
applications for the molecules, and their energetic values.

Question 39

1 0 box . 3 Mucus also contains glycoproteins. One of these glycoproteins is a polypeptide with
the sugar, lactose, attached.
Describe how lactose is formed and where in the cell it would be attached to a
polypeptide to form a glycoprotein.
[4 marks]

Answer 39

1. Glucose and galactose;
2. Joined by condensation (reaction);
3. Joined by glycosidic bond;
4. Added to polypeptide in Golgi (apparatus);