AS PAPER 1 2016

Question 1

0 1 . 1 Table 1 shows features of a bacterium and the human immunodeficiency virus
(HIV) particle.
Complete Table 1 by putting a tick () where a feature is present.
[2 marks]
Table 1
Feature Bacterium
Human
immunodeficiency
virus (HIV) particle
RNA
Cell wall
Enzyme molecules
Capsid

Answer 1

eature Bacterium Human
immunodeficiency
virus (HIV)
particle
RNA  
Cell wall 
Enzyme
molecules
 
Capsid 

The factual recall question, 1.1, proved far more challenging than intended. Only 5% of students
obtained both marks and 54% failed to score. There was no particular pattern to the wrong
answers

Question 2

0 1 . 2 When HIV infects a human cell, the following events occur.
• A single-stranded length of HIV DNA is made.
• The human cell then makes a complementary strand to the HIV DNA.
The complementary strand is made in the same way as a new complementary
strand is made during semi-conservative replication of human DNA.
Describe how the complementary strand of HIV DNA is made.
[3 marks]

Answer 2

1. (Complementary) nucleotides/bases pair
   OR
   A to T and C to G;
2. DNA polymerase;
3. Nucleotides join together (to form new
   strand)/phosphodiester bonds form;
4. & 3. Ignore ‘(DNA
   polymerase) forms
   base pairs/nucleotide
   pairs’
   If clearly writing rote
   answer about DNA
   replication 2 max e.g.
   helicase or separating
   strands

Question 1.2 discriminated very well, with 15% obtaining three marks and 21% scoring zero.
There were good, concise answers that scored three marks for including complementary base
pairing and the role of DNA polymerase in joining nucleotides together to form the new DNA
strand; often in two or three lines.
Many students failed to read the question carefully and did not answer the question as set. They
wrote at length about DNA replication, starting with DNA helicase. These answers were awarded a
maximum of two marks, because the question specifically asked how the complementary strand of
HIV DNA is made. Many students appeared to believe that DNA actively pulls free nucleotides into
place and makes them base pair; some even wrote about condensation reactions. There were
students who confused transcription with replication and gave accounts of mRNA production.
Some students appeared to focus on ‘HIV’ and ‘replication’ and gave an extended account of how
HIV infects cells, uses reverse transcriptase to make DNA, incorporates its DNA into host DNA,
takes over the cell, is replicated by the host cell, infects new cells and leads to AIDS. They often
went onto an additional page, or wrote their answer under 1.3 on the next page, in breach of
instructions given on the front of the exam paper. Many of these students may have found
themselves short of time for later questions.

Question 3

0 1 . 3 Contrast the structures of DNA and mRNA molecules to give three differences.
[3 marks]

Answer 3

1. DNA double stranded/double helix and
mRNA single-stranded;
2. DNA (very) long and RNA short;
3. Thymine/T in DNA and uracil/U in RNA;
4. Deoxyribose in DNA and ribose in RNA;
5. DNA has base pairing and mRNA doesn’t/
DNA has hydrogen bonding and mRNA
doesn’t;
6. DNA has introns/non-coding sequences
and mRNA doesn’t;

n 1.3, it was pleasing to find that many students did obey the command word to ‘contrast’ and
gave full statements about the differences between DNA and RNA. Many students knew enough
about the structures of DNA and mRNA to give correct contrasting features and 47% obtained all
three marks.

Question 4

0 2 . 1 Describe the difference between the structure of a triglyceride molecule and the
structure of a phospholipid molecule.
[1 mark]

Answer 4

1. In phospholipid, one fatty acid
   replaced by a phosphate;

About two thirds obtained the mark in 2.1 for a correct description of the difference between a
triglyceride and a phospholipid. Those who failed to score either did not know about the structure
of these molecules or just described the structure of a phospholipid (or triglyceride).

Question 5

0 2 . 2 Describe how you would test for the presence of a lipid in a sample of food.
[2 marks]

Answer 5

1. Add ethanol, then add water;
2. White (emulsion shows lipid);

1. Reject ethanal/ethonal
Accept ‘Alcohol/named
alcohol’
2. Accept milky – Ignore
‘cloudy’
Sequence must be correct
If heated then DQ point 1
Reject precipitate

In 2.1, about 40% of students could fully describe how to test for a lipid and obtained both marks. A
minority described tests for other biological molecules. Many made errors in their descriptions of
the emulsion test or of a positive result. These errors included: adding water before ethanol,
heating the mixture, the presence of a precipitate and failing to note that the colour of the emulsion
would be white.

Question 6

2 . 3 Animal fats contain triglycerides with a high proportion of saturated fatty acids.
If people have too much fat in their diet, absorption of the products of fat
digestion can increase the risk of obesity. To help people lose weight, fat
substitutes can be used to replace triglycerides in food.
Describe how a saturated fatty acid is different from an unsaturated fatty acid.
[1 mark]

Answer 6

Saturated single/no double bonds
(between carbons)
OR
Unsaturated has (at least one) double
bond (between carbons);

Accept hydrocarbon chain/R
group for ‘between carbons’
for either
Accept Sat = max number of
H atoms bound
‘It’ refers to saturated

Most students scored the mark in 2.3. Those who didn’t got saturated and unsaturated the wrong
way round in terms of carbon-carbon double bonds.

Question 7

Figure 1 shows the structure of a fat substitute.
Figure 1
0 2 . 4 This fat substitute cannot be digested in the gut by lipase.
Suggest why.
[2 marks]

Answer 7

1. (Fat substitute) is a different/wrong
shape/not complementary;
OR
Bond between glycerol/fatty acid
and propylene glycol different (to
that between glycerol and fatty
acid)/no ester bond;

2. Unable to fit/bind to (active site of)
   lipase/no ES complex formed;

In 2.4, 28% obtained both marks for stating that the fat substitute would not bind to the active site
of lipase because it has a different shape to a triglyceride. A similar percentage obtained one mark.
Those who failed to score often ignored the question’s reference to lipase and wrote about bile
salts, micelles and methods of absorption.

Question 8

0 2 . 5 This fat substitute is a lipid. Despite being a lipid, it cannot cross the cell-surface
membranes of cells lining the gut.
Suggest why it cannot cross cell-surface membranes.
[1 mark]

Answer 8

It is hydrophilic/is polar/is too large/is

too big;

Question 9

0 3 Cells constantly hydrolyse ATP to provide energy.
0 3 . 1 Describe how ATP is resynthesised in cells.
[2 marks]

Answer 9

1. From ADP and phosphate;
2. By ATP synthase;
3. During respiration/photosynthesis;

The specification (section 3.1.6) requires students to know that ATP is resynthesised by a
condensation reaction involving ADP and phosphate, catalysed by ATP synthase. The examiners
did not require reference to ‘a condensation reaction’ but only 26% of students obtained both
marks in 3.1. About 40% obtained one mark for reference to ADP and phosphate. Some failed to
score that mark because they wrote about phosphorus. Many wasted time writing about hydrolysis
of ATP but often went on to describe the reaction between ADP and phosphate

Question 10

0 3 . 2 Give two ways in which the hydrolysis of ATP is used in cells.
[2 marks]

Answer 10

1. To provide energy for other
   reactions/named process;
2. To add phosphate to other
   substances and make them more
   reactive/change their shape;

The same section of the specification (3.1.6) requires students to know that the hydrolysis of ATP
can be coupled to energy-requiring reactions, or the phoshorylation of other compounds (often
making them more reactive). In 3.2, 35% of students obtained one mark for reference to ATP
providing energy for a reaction (usually named). Only 3% did this and then made reference to
phosphorylation to obtain the second mark

Question 11

0 3 . 3 Figure 2 is a photograph (micrograph) of a mitochondrion taken using a
scanning electron microscope.
Figure 2
What is the evidence from Figure 2 that a scanning electron microscope was
used to take this photograph?
[1 mark]

Answer 11

(Can see) 3D image;

In 3.3, about 40% obtained the mark by noting that the image was 3-D. Many simply wrote that the
evidence was the black and white nature of the image.

Question 12

0 3 . 4 Name the part of the mitochondrion labelled X in Figure 2.
[1 mark]

Answer 12

03.4 Crista/cristae; 1 Ignore matrix

Question 13

0 4 . 1 The letters P, Q, R, S and T represent ways substances can move across
membranes.
• P – diffusion through the phospholipid bilayer
• Q – facilitated diffusion
• R – active transport
• S – co-transport
• T – osmosis
For each of the following examples of transport across membranes, select the
letter that represents the way in which the substance moves across the
membrane.
Write the appropriate letter in each box provided.
[3 marks]
Transport through a channel protein
Transport of small, non-polar molecules
Transport of glucose with sodium ions
Figure 3 shows how a plant cell produces its cell wall.
Figure 3

Answer 13

Q
P
S

Question 14

Figure 3 shows how a plant cell produces its cell wall.
Figure 3
9
*09*
Turn over ►
IB/M/Jun16/E3
Do not write
outside the
box
0 4 . 2 Y is a protein. One function of Y is to transport cellulose molecules across the
phospholipid bilayer.
Using information from Figure 3, describe the other function of Y.
[2 mark

Answer 14

1. (Y is) an enzyme/has active
site/forms ES complex;
2. That makes cellulose/attaches
substrate to cellulose/joins β
glucose;
OR
3. Makes cellulose/forms glycosidic
bonds;
4. From β glucose;

In both 4.2 and 4.3, students usually appeared to ignore the diagram in Figure 3 and the
information given, despite being instructed to use both in their answers. Students were told that
Figure 3 shows how a plant cell produces its cell wall. In the stem of 4.2, they were told that one
function of protein Y is to transport cellulose molecules across the bilayer. The figure shows
substrate molecules approaching the end of the cellulose molecules. Only 12% of students put all
this together and wrote that Y was an enzyme (one mark) that makes cellulose (one mark), or that Y makes cellulose (one mark) from β glucose (one mark). Some identified Y as an enzyme for one
mark but then had it breaking down cellulose into substrate. Most students (68%) wrote about Y
transporting various ions or other substances not mentioned in the information or figure, or about
membrane stability

Question 15

0 4 . 3 What is the evidence in Figure 3 that the phospholipid bilayer shown is part of
the cell-surface membrane?
[1 mark]

Answer 15

Cell wall forms outside cell-surface
membrane/has cellulose on it (on the
outside);

in 4.3, many students did not seem to appreciate that evidence is something to
be seen.

Question 16

0 4 . 4 In the cell wall, bonds hold the cellulose molecules together side by side.
Tick () one box that describes the type of bond that holds the cellulose
molecules together side by side.
[1 mark]

Answer 16

(Tick in box next to) Hydrogen;

Question 17

0 5 Scientists investigated the hydrolysis of sucrose in growing plant cells by an
enzyme called SPS.
0 5 . 1 Name the products of the hydrolysis of sucrose.
[2 marks]

Answer 17

1. Glucose;

2. Fructose

Question 18

0 5 . 2 The scientists grew plant cells in a culture for 12 days. At the start, there were
only a few cells in the culture. Each day, they determined the mass of sucrose
hydrolysed by SPS in the plant cells in 1 hour.
Table 2 shows their results.
Table 2
Day
Mass of sucrose
hydrolysed by SPS in
1 hour / µg
Rate of hydrolysis of sucrose
by SPS
0 0.07
2 0.09
4 0.11
6 0.15
8 0.20
10 0.24
12 0.24
For each day, calculate the rate per minute of the reaction catalysed by SPS.
Record the rates in standard form and plot a suitable graph of your processed
data.
[3 marks]

Answer 18

1. Line graph with rate on y axis and
days/time in days on x axis and
linear scales;
2. Correct units of µg min–1
/per
minute/minute-1
× 10–3
;
3. Rates correctly calculated and
plotted, with line connecting
points/line of best fit and no
extrapolation;

Correct answers × 10–3

1.17, 1.50, 1.83, 2.50, 3.33,
4.00, 4.00 (accept to 1DP)
2. Reject m-1
2. Reject if put 10-3
on axis
for each point
2. ‘/’ means separating units
from what goes before i.e.
accept sucrose hydrolysis
per min / gx10-3
3. Do not accept a ruled
straight line of best fit
Accept y axis starting at 1

In 5.2, it was heartening to see how many students could correctly calculate the rates and express
them in standard form. Graphing the data proved much more challenging. Only 6% of students
obtained all three marks. About 2% did not attempt to draw a graph. About 29% failed to score
after attempting the graph for reasons including: axes the wrong way round, non-linear scales and
failure to identify the axes. About 27% obtained 1 mark for putting rate on the y axis and time on
the x axis and using linear scales. About 37% obtained two marks for also correctly plotting correct
numerical rates and joining the points (with a line of best fit or point-to-point). Only about 6% could
correctly label the y axis, with rate as µg minute–1 times 10–3
. A very large number of students
appeared to think that rate = minute–1 (or ‘per minute). A rate is something per unit time, in this
case µg. The rate is proportional to one over time.

Question 19

0 5 . 3 What can you conclude about the growth of the plant cells from these data?
Explain how you reached your conclusions.
[3 marks]

Answer 19

1. Sucrose hydrolysis linked to some
aspect of growth;
2. Greater the rate of/faster
hydrolysis/more SPS activity as
plant grows/cells divide (up to 8/10
days);
3. Growth/division remains the
same/slows after 8/10 days
(because SPS activity is levelling
off);

1&2. Accept ‘breakdown’
2. Accept converse of greater
rate of growth, greater rate of
hydrolysis
2. Reject ‘sucrose broken
down’
3. Accept after 8 days/at 10
days growth rate
maximum/growth stops

In 5.3, students were asked what they could conclude about growth of plant cells from these data.
They could answer using either the raw data in the table, or their graph. Many simply described the
data and failed to score (38%). A similar number obtained one mark, either for noting that the rate
of hydrolysis of sucrose increased as growth occurred, or that growth levelled off after 8/10 days.
About 18% obtained two marks for mentioning both of these points. Only about 2% also linked
hydrolysis of sucrose to growth in some way and obtained a third mark.

Question 20

0 6 . 1 Describe the induced-fit model of enzyme action.

[2 marks]

Answer 20

1. (before reaction) active site not
complementary to/does not fit
substrate;
2. Shape of active site changes as
the substrate binds/as enzyme-substrate complex forms;
3. Stressing/distorting/bending bonds
(in substrate leading to reaction);

In 6.1, 29% obtained both marks. The idea of induced fit was generally known but often poorly
expressed. Some students wrote about the substrate having the same shape as the active site,
rather than a complementary shape. A minority had the active site on the substrate

Question 21

0 6 . 2 A scientist investigated the hydrolysis of starch.
He added amylase to a suspension of starch and measured the concentration of
maltose in the reaction mixture at regular intervals.
His results are shown in Figure 4.
Figure 4

0 6 . 3 Explain the results shown in Figure 4.
[2 marks]

Answer 21

1. (Rate of) increase in concentration
of maltose slows as
substrate/starch is used up
OR
High initial rate as plenty of
starch/substrate/more E-S
complexes;
2. No increase after 25 minutes/at
end/levels off because no
substrate/starch left;

In 6.3, many students simply described the curve, rather than explaining. Some attempting an
explanation incorrectly wrote about the enzyme being used up, or all its active sites being
occupied. 13% obtained both marks for writing about a rapid rate at the start because there is lots
of starch and the rate levelling off/falling to zero after 25/30 minutes as all the starch is used up.
Some 43% obtained one mark for making one or other of these points

Question 22

0 6 . 4 A quantitative Benedict’s test produces a colour whose intensity depends on the
concentration of reducing sugar in a solution. A colorimeter can be used to
measure the intensity of this colour.
The scientist used quantitative Benedict’s tests to produce a calibration curve of
colorimeter reading against concentration of maltose.
Describe how the scientist would have produced the calibration curve and used
it to obtain the results in Figure 4.
Do not include details of how to perform a Benedict’s test in your answer.
[3 marks]

Answer 22

1. Make/use maltose solutions of
known/different concentrations
(and carry out quantitative
Benedict’s test on each);
2. (Use colorimeter to) measure
colour/colorimeter value of each
solution and plot calibration
curve/graph described;
2.Axes must be correct if
axes mentioned,
concentration on x-axis and
colorimeter reading on y-axis
3. Find concentration of sample from
calibration curve;

Question 6.4 was based around skills that students were expected to have developed in required
practical 3; that is to say the production and use of a calibration curve. The purpose of the required
practicals is, generally, to allow students to practise and develop skills that can then be applied in
many different settings. 21% of students obtained one mark, either for noting that the scientist
would need solutions of maltose of known concentrations, or that a calibration curve would have
concentration on the x axis and colorimeter reading on the y axis. About 12% obtained both of these marks. Only 5% then went on to (briefly) describe how the calibration curve would be used to
determine the concentrations in the experiment.

Question 23

0 7 . 1 Human papilloma virus (HPV) is the main cause of cervical cancer. A vaccine
has been developed to protect girls and women from HPV.
Describe how giving this vaccine leads to production of antibody against HPV.
[4 marks]

Answer 23

1. Vaccine/it contains antigen (from
HPV);
2. Displayed on antigen-presenting cells;
3. Specific helper T cell (detects antigen
and) stimulates specific B cell;
4. B cell divides/goes through
mitosis/forms clone to give plasma
cells;
5. B cell/plasma cell produces antibody;
4 max 1. Term ‘antigen’ may be
first mentioned with point 2
2. Accept named example,
e.g.
macrophage/phagocyte/B
cells
3. Accept ‘helper T cell
with receptor on surface’ for
‘specific’ and B cells with
receptor/antibody on
surface that bind to antigen
for ‘specific’

7.1 was a very good discriminator. 19% obtained all four marks, 14% failed to score and equal
percentages obtained one, two or three marks. Many students had the idea that a vaccine contains
antigen and knowledge of antigen-presenting cells was common. There were also many correct
statements about plasma/B cells releasing antibodies. Fewer students had the idea of a B cell
dividing to form plasma cells. Not many students were able to express clearly the idea of a specific
helper T cell or B cell detecting, or responding to, a specific antigen. Quite a few students got
confused between the roles of T cells and B cells. Some students wrote at length about memory
cells and secondary responses, neither of which was required to answer the question. As in some
other questions, this inclusion of irrelevant material often generated additional pages and wasted
time that could have been spent on other questions

Question 24

7 . 2 Doctors investigated whether it was better to give two or three doses of the HPV
vaccine. They determined the mean concentration of antibody against HPV in
blood samples from girls who were given either two or three doses of the
vaccine.
• Girls given two doses received an initial vaccination, followed by a second at
6 months.
• Girls given three doses received an initial vaccination, followed by a second
at 1 month and a third at 6 months.
The doctors measured the concentration of antibody each month.
The results are shown in Figure 5.
Figure 5
What do these results suggest about whether it is better to give two or three
doses of the vaccine? Give reasons for your answer.
[2 marks]

Answer 24

1. Two (doses) because got more
antibody;
2. With three doses, second dose/dose
at 1 month doesn’t lead to production
of any more antibody (than the twodose group)/get same/similar
response;
3. Three doses would be more
expensive/less popular with
parents/girls (and serves no purpose);

Most students obtained one mark in 7.2 for suggesting greater antibody production with two
vaccinations. Almost none went on to make any other suggestion.

Question 25

0 7 . 3 The doctors carried out a statistical test to determine whether the antibody
concentrations were significantly different in girls given two doses of the vaccine,
compared with those given three doses. They determined the mean
concentrations of antibody 9 months after the first dose of vaccine.
What statistical test should the doctors have used? Give the reason for your
choice.
[1 mark]

Answer 25

t-test, because comparing two means;

In the mathematical requirements section of the specification (pages 62-66), selection and use of a
statistical test is not emboldened and so is required content for both AS level and full A-level. This
is different from the legacy specification and answers to 7.3 indicated that most students had not
learnt this. Only 6% of students could name the t-test and give the reason as testing the difference
between means. Further guidance about teaching statistics is provided in the support materials on
the AQA website.

Question 26

0 7 . 4 There is genetic diversity within HPV.
Give two ways doctors could use base sequences to compare different types of
HPV.
[2 marks]

Answer 26

1. Compare (base sequences of) DNA;
2. Look for mutations/named mutations
   (that change the base sequence);
3. Compare (base sequences of)
   (m)RNA;

Rather like 3.2 (two uses of ATP), the examiners were expecting statements from the specification
in answers to 7.4. In ‘Investigating diversity’ (section 3.4.7), genetic diversity is compared (amongst
other ways) by looking at base sequences of DNA and base sequences of mRNA. Very few
students (3%) came up with both of these but some (27%) managed to express one of them.

Question 27

0 8 . 1 The letters A, B, C, D and E represent stages in mitosis.
• A – anaphase
• B – interphase
• C – metaphase
• D – prophase
• E – telophase
Write one of the letters, A to E, in the box to complete the following statement.
[1 mark]

Chromosomes line up on the equator of the mitotic spindle in

Answer 27

C

Question 28

8 . 2 Scientists looking for treatments for cancer are investigating the use of
substances called kinesin inhibitors (KI). These inhibitors prevent successful
mitosis. Some kinesin inhibitors cause the development of a monopolar spindle
in mitosis.
Figure 6 shows chromosomes attached to a normal mitotic spindle and to a
monopolar mitotic spindle.
Figure 6
Suggest why the development of a monopolar mitotic spindle would prevent
successful mitosis.
[2 marks]

Answer 28

1. No separation of
chromatids/chromosomes/centromeres;
1. Accept anaphase
prevented
1. Accept nondisjunction
1.Reject homologous pairs

2. Chromatids/chromosomes all go to one
   pole/end/sides of cell/not pulled to
   opposite poles;

3. Doubles chromosome number in
cell/one daughter cell gets no
chromosomes or chromatids;
3. Accept DNA for
chromosomes
3. Accept ploidy
3. Ignore references to
‘genetic information’
Ignore simple descriptions
of what normally happens
in mitosis

Most students managed to score marks on 8.2. Most (about 45%) noted that the chromatids would
not separate and obtained one mark. Many (39%) went on to deduce that this would mean all of
the chromatids/chromosomes would go to one pole of the dividing cell and obtained both marks.
Some students got confused between sister chromatids and homologous chromosomes.

Question 29

0 8 . 3 Scientists investigated the effect of different concentrations of a kinesin inhibitor
(KI) on mitosis of human bone-cancer cells grown in a culture.
Table 3 shows the scientists’ results.
Table 3
Concentration of kinesin inhibitor /
nmol dm–3
Percentage of dividing human
bone-cancer cells showing a
monopolar mitotic spindle
0 0
1 0
10 8
100 93
1000 100
10 000 100
A student who saw these results concluded that in any future trials of this kinesin
inhibitor with people, a concentration of 100 nmol dm–3 would be most
appropriate to use.
Do these data support the student’s conclusion? Give reasons for your answer.
[4 marks]

Answer 29

1. (No, because) at 100 there are still
some (7%) cancer cells
dividing/undergoing mitosis;
2. So, cancer not destroyed/may continue
to grow/spread/form tumours;
3. Best concentration may be between
100 and 1000/need trials between 100
and 1000;
4. This research in culture, don’t know
effect of KI on people;
5. (Yes, because) above 100 produces
little increase in % of cells not
dividing/undergoing mitosis/at 100,
most (93%) cancer cells unable to
divide/dead;
6. Above 100 may be harmful (to body);
7. Higher concentrations more expensive;
8. (above 100) will have more effect on
(rapidly dividing) cancer cells;
4 max 1. Accept idea that all
division stops only at 1000
2. Must refer to cancer
spreading not cells dividing
4. Reject ‘not tested on
humans’
4.Reject ‘done in animals’
5. Must clearly link lack of
monopolar mitotic spindles
with cell division
6. Accept ‘above
100/high
concentrations
produce harmful side
effects/named effects’
8.Must relate to 100

8.3 was about the same context as 8.2, where the kinesin inhibitor was linked to prevention of
successful mitosis. It was hoped that students would appreciate that slowing or stopping mitosis
would be a good thing when treating cancer (section 3.2.2). Most students made no reference to
slowing or stopping mitosis when considering the results in 8.3. As a result, few went on to discuss
the prevention of spread of cancer with KI at 100 nmol dm–3
. Some even suggested that a low KI
dose would be better because it would allow more cell division. Many students described the
results at length. Students can usually be relied upon to comment on the method used in an
experiment, but here almost none of them noted that the doses of the drug were being tested on
cells grown in culture, not on people.

Question 30

0 8 . 4 At the start of their investigation, the scientists made a solution of kinesin
inhibitor (KI) with a concentration of 10 000 nmol dm–3. They used this to make
the other concentrations by a series of dilutions with water.
Describe how they made 100 cm3 of 1000 nmol dm–3 solution of kinesin inhibitor.
[2 marks]

Answer 30

1. 10 cm3
of 10 000 nmol dm–3
/ (original)
solution;
2. 90 cm3
of water;

8.4 tested a basic experimental skill. Broadly, 41% could describe how to make the required
volume of the required solution and 49% could not. A minority had the correct proportions of
solutions but the wrong volumes.

Question 31

0 9 Read the following passage.
Alzheimer’s disease leads to dementia. This involves small β-amyloid
proteins binding together to form structures called plaques in the brain.
Nerve cells in the brain produce a large protein called amyloid-precursor
protein that has a complex shape. This protein is the substrate of two
different enzymes, α-secretase and β-secretase. These enzymes are
normally produced in the brain. One product of the reaction catalysed by
β-secretase is a smaller protein that can lead to β-amyloid protein formation.
Many people with Alzheimer’s disease have mutations that decrease
α-secretase production, or increase β-secretase production.
One possible type of drug for treating Alzheimer’s disease is a competitive
inhibitor of β-secretase. When some of these types of drugs were trialled on
patients, the trials had to be stopped because some patients developed
serious side effects.
Use information from the passage and your own knowledge to answer the
following questions.
5
10
0 9 . 1 Suggest how amyloid-precursor protein can be the substrate of two different
enzymes, α-secretase and β-secretase (lines 3–5).
[2 marks]

Answer 31

1. Different parts/areas/amino acid
sequences (of amyloid-precursor)
protein;
2. Each enzyme is specific /fits/binds/
complementary to a different part of the
APP

In 9.1, the passage stated that amyloid-precursor protein is large, has a complex shape and is the
substrate of two different enzymes. About 14% of students correctly suggested that each enzyme
would have an active site that binds (specifically) to a different site on the amyloid-precursor
protein. (This would be rather similar to endo and exopeptidases acting on a protein – see
specification section 3.3.3.) Many students wrote about induced fit allowing an enzyme to adapt its
active site to fit a/any substrate, so both enzymes would bind to the same place on the substrate.
Others had active sites on the substrate that could adapt to the enzymes.

Question 32

0 9 . 2 One product of the reaction catalysed by β-secretase is a smaller protein
(lines 6–7).
Describe what happens in the hydrolysis reaction that produces the smaller
protein from amyloid-precursor protein.
[2 marks]

Answer 32

1. Peptide bond broken;
2. Using water;

Question 9.2 produced more discrimination than expected. About 27% obtained both marks, for
noting that a peptide bond is broken and water is used in the process. About 43% obtained one
mark for one or other of these points, usually the use of water. The others either didn’t know, or
suggested water was given off in the reaction

Question 33

0 9 . 3 Many people with Alzheimer’s disease have mutations that decrease
α-secretase production, or increase β-secretase production (lines 8–9).
Use the information provided to explain how these mutations can lead to
Alzheimer’s disease.
[3 marks]

Answer 33

1. Mutations prevent production of
enzyme(s)/functional enzyme;
2. (Increase in β-secretase) leads to
faster/more β-amyloid production OR
(Decrease in α-secretase) leads to
more substrate for β-secretase;
3. (Leads to) more/greater plaque
formation;

In 9.3, few students could string together all three points on the mark scheme. Many appeared to
be unable to use information from the passage to come up with any sensible suggestion of how the
mutations could lead to Alzheimer’s disease, such as a faster/greater production of plaque.

Question 34

0 9 . 4 One possible type of drug for treating Alzheimer’s disease is a competitive
inhibitor of β-secretase (lines 10–11).
Explain how this type of drug could prevent Alzheimer’s disease becoming
worse.
[2 marks]

Answer 34

1. (Inhibitor) binds to/blocks active site of
   β-secretase/enzyme;
2. Stops/reduces production of βamyloid/plaque;

In 9.4, about 46% obtained one mark for describing what a competitive inhibitor is and how it acts
by binding to the enzyme’s active site. About 30% were then able to go on and apply this
knowledge to the context in the passage and suggest that the inhibitor would reduce or stop βamyloid or plaque formation. Some students attempted to hedge their bets by saying that the
inhibitor changes the enzyme’s active site but not saying how. This approach was not given credit.
Some students thought that active sites are on substrates and failed to obtain the first mark point.
Students who could not remember what competitive inhibition of an enzyme entails were unable to
access 9.4. 23% of students failed to score on 9.4.

Question 35

0 9 . 5 When some of these types of drugs were trialled on patients, the trials were
stopped because some patients developed serious side effects (lines 11–13).
Using the information provided, suggest why some patients developed serious
side effects.
[1 mark]

Answer 35

1. Some β-amyloid required/needed (to
prevent side effects)
OR
(Some) β-secretase needed;
2. Leads to build-up of amyloid-precursor
protein (that causes harm)
OR
Too much product of α-secretase
(causes harm)

9.5 could only be answered using information from the passage. About 20% of students obtained
this mark by noting that the brain normally has the enzymes, and produces the proteins, mentioned
in the passage and they must, therefore, have some necessary function (that the drugs blocked).