BIO PAPER 1

Question 1

0 1 . 1 Give the two types of molecule from which a ribosome is made.
[1 mark]

Answer 1

One of RNA/ribonucleic acid(s)/nucleotide(s)/nucleic
acid(s)/rRNA/ribosomal RNA/ribosomal ribonucleic
acid
and
one of protein(s)/polypeptide(s)/amino acid(s)/
peptide(s)/ribosomal protein;

01.1 Only 22% of students could correctly name the types of molecule that make up a ribosome.

Question 2

0 1 . 2 Describe the role of a ribosome in the production of a polypeptide. Do not include
transcription in your answer.
[3 marks]

Answer 2

1. mRNA binds to ribosome;
2. Idea of two codons/binding sites;
3. (Allows) tRNA with anticodons to bind/associate;

4. (Catalyses) formation of peptide bond between
amino acids (held by tRNA molecules);

5. Moves along (mRNA to the next
   codon) /translocation described;

01.2 This question was answered much more successfully than 01.1, with 48% of students
gaining all three marks.

Question 3

1 . 3 Table 1 shows the base sequence of part of a pre-mRNA molecule from a
eukaryotic cell.
Complete the table with the base sequence of the DNA strand from which this
pre-mRNA was transcribed.
[1 mark]

Answer 3

TGCGTAATA;
Any errors = 0 marks

01.3 This question proved to be very accessible, with 96.3% of students gaining the mark.

Question 4

0 1 . 4 In a eukaryotic cell, the base sequence of the mRNA might be different from the
sequence of the pre-mRNA.
Explain why.
[2 marks]

Answer 4

1. Introns (in pre-mRNA);
2. Removal of sections of (pre-mRNA)/splicing;

01.4 Although 83% of students scored 2 marks for this question, there were some incorrect
answers, e.g., indicating that the introns removed from the pre-mRNA were made of DNA.

Question 5

0 2 In mammals, in the early stages of pregnancy, a developing embryo exchanges
substances with its mother via cells in the lining of the uterus. At this stage, there
is a high concentration of glycogen in cells lining the uterus.
0 2 . 1 Describe the structure of glycogen.
[2 marks]

Answer 5

1. Polysaccharide of α-glucose;
OR
polymer of α-glucose;
2. (Joined by) glycosidic bonds
 OR
Branched structure;

02.1 The majority of students scored mark point 2 for the idea that glycogen is a branched
molecule or that it contains glycosidic bonds. Although many students knew that glycogen
is a ‘chain’ of alpha glucose, fewer could use the correct terminology to state that it is a
polysaccharide or a polymer. Students must take care to specify alpha glucose or αglucose rather than a-glucose. It was not uncommon for students to describe glycogen as
if it were identical to starch and to include references to amylose and amylopectin in their
answers

Question 6

2 . 2 During early pregnancy, the glycogen in the cells lining the uterus is an important
energy source for the embryo.
Suggest how glycogen acts as a source of energy.
Do not include transport across membranes in your answer.
[2 marks]

Answer 6

1. Hydrolysed (to glucose);
2. Glucose used in respiration;

02.2 This question required a little application of biological principles, with students’ knowledge
of glycogen. Many students started their answers with reference to branching molecules
providing many ‘ends’ for enzymes – this was not relevant and is indicative of students
rushing to write all they know, rather than considering what is relevant to answer the
specific question asked. As with question 02.1, students often knew the basic principle of
breaking down glycogen to release the glucose, but did not use the appropriate term of
‘hydrolysis’ to gain the first mark point. Many students suggested that this hydrolysis of
glycosidic bonds releases energy, rather than thinking about glucose as the substrate for
respiration in all human cells.

Question 7

0 2 . 3 Suggest and explain two ways the cell-surface membranes of the cells lining the
uterus may be adapted to allow rapid transport of nutrients.
[2 marks]

Answer 7

1. Membrane folded so increased/large surface
   area;
   OR
   Membrane has increased/large surface area
   for (fast) diffusion/facilitated diffusion/active
   transport/co-transport;
2. Large number of protein channels/carriers (in
   membrane) for facilitated diffusion;
3. Large number of protein carriers (in membrane)
   for active transport;
4. Large number of protein (channels/carriers in
   membrane) for co-transport;

02.3 Although this question gave the context of the adaptation of the cell-surface membranes of
cells lining the uterus, making it an AO2 question, this should have been fairly
straightforward application of knowledge from 3.2.3. Many students did not gain full marks
because they seemed not to read the question sufficiently carefully. It required reference to
the cell-surface membranes only, so answers relating to “one cell thick”, “many
mitochondria”, or “a good blood supply” were not relevant. The question also required a
suggestion and an explanation to gain each mark. It was not uncommon to see reference
to “thin membranes”, indicating a clear confusion between the phospholipid bilayer
structure of a cell-surface membrane and the epithelial layer of an exchange surface.
Reference to ‘villi’ also demonstrated similar confusion and prevented a student from
obtaining mark point 1. It is important to note that, when referring to protein
carriers/channels, only carriers can be used for active transport; either can be involved in
facilitated diffusion. Also, for a membrane to be adapted for rapid transport, it must have
more of these proteins, rather than them just being present.

Question 8

2 . 4 In humans, after the gametes join at fertilisation, every cell of the developing
embryo undergoes mitotic divisions before the embryo attaches to the uterus lining.
• The first cell division takes 24 hours.
• The subsequent divisions each take 8 hours.
After 3 days, the embryo has a total volume of 4.2 × 10–3 mm3
.
What is the mean volume of each cell after 3 days? Express your answer in
standard form.
Show your working.
[2 marks]

Answer 8

3.3 x 10–5 OR 3.28 x 10–5 OR 3.281 x 10–5
;;
1 mark for
Evidence of 128 (cells)
Correct numerical calculation but not in standard
form gains 1 mark (0.00003281 OR 0.0000328 OR
0.000033);

02.4 As with all calculations that meet the national requirement to be at the standard of higher
tier GCSE Mathematics, this required a multi-step process. Firstly, students needed to
work out that there would be 128 cells after 3 days: 2 cells after the first day, then another 6
divisions. This was where many students struggled. It was pleasing to see how many
students could correctly use standard form to express their final answer. It should be noted
that the convention for standard form to be correct is to have a number greater than 1 and
less than 10, multiplied by 10 to a power, i.e. the decimal point should be after the first
digit.

Question 9

0 3 . 1 Sodium ions from salt (sodium chloride) are absorbed by cells lining the gut. Some
of these cells have membranes with a carrier protein called NHE3.
NHE3 actively transports one sodium ion into the cell in exchange for one proton
(hydrogen ion) out of the cell.
Use your knowledge of transport across cell membranes to suggest how NHE3
does this.
[3 marks]

Answer 9

1. Co-transport;
2. Uses (hydrolysis of) ATP;
3. Sodium ion and proton bind to the protein;
4. Protein changes shape (to move sodium ion
   and/or proton across the membrane);

03.1 There were some excellent answers here, but many students did not gain marks because
their answers failed to focus on the NHE3 carrier protein. Many tried to explain why the
ions would move, often describing how movement of one ion would generate an
electrochemical gradient allowing the other to move (by facilitated diffusion), rather than
sticking with the question of how the active transport (as stated in the stem of the question)
of these two ions could work.

Question 10

0 3 . 2 Scientists investigated the use of a drug called Tenapanor to reduce salt absorption
in the gut. Tenapanor inhibits the carrier protein, NHE3.
The scientists fed a diet containing a high concentration of salt to two groups of
rats, A and B.
• The rats in Group A were not given Tenapanor (0 mg kg–1).
• The rats in Group B were given 3 mg kg–1 Tenapanor.
One hour after treatment, the scientists removed the gut contents of the rats and
immediately weighed them.
Their results are shown in Table 2.
Table 2
Concentration of Tenapanor
/ mg kg–1
Mean mass of contents of the gut
/ g
0 2.0
3 4.1
The scientists carried out a statistical test to see whether the difference in the
means was significant. They calculated a P value of less than 0.05.
They concluded that Tenapanor did reduce salt absorption in the gut.
Use all the information provided and your knowledge of water potential to explain
how they reached this conclusion.
[4 marks]

Answer 10

1. Tenapanor/(Group)B/drug causes a significant
increase;
OR
There is a significant difference with
Tenapanor/drug/between A and B;

2. There is a less than 0.05 probability that the
   difference is due to chance;
3. (More salt in gut) reduces water potential in gut
   (contents) ;
4. (so) less water absorbed out of gut (contents)
   by osmosis
   OR
   Less water absorbed into cells by osmosis
   OR
   Water moves into the gut (contents) by
   osmosis.
   OR
   (so) water moves out of cells by osmosis;

03.2 This question was the first test of AO3 skills in this paper and gave students their first
opportunity to interpret and evaluate scientific information. There were two parts to the
mark scheme. The first required students to explain why an increase in mean mass of
contents in the gut would allow scientists to reach the conclusion that salt absorption had
been reduced. Many students successfully achieved these two marks, although some
showed confusion between the contents in the lumen of the gut and in the cells lining the
gut wall. The second required students to interpret the statistical test; this was often not
attempted by students and those who did were rarely completely successful. Even though
the stem contained a statement to help students (The scientists carried out a statistical test
to see whether the difference in the means was significant), many referred to significant
‘results’, and very few could correctly describe what the P value showed with appropriate
use of the words ‘probability’ and ‘chance’.

Question 11

0 3 . 3 High absorption of salt from the diet can result in a higher than normal
concentration of salt in the blood plasma entering capillaries. This can lead to a
build-up of tissue fluid.
Explain how.
[2 marks]

Answer 11

1. (Higher salt) results in lower water potential of
   tissue fluid;
2. (So) less water returns to capillary by osmosis
   (at venule end);
   OR
3. (Higher salt) results in higher blood
   pressure/volume;
4. (So) more fluid pushed/forced out (at arteriole
   end) of capillary;

03.3 This question required students to apply their knowledge of topic 3.3.4.1 and only 15%
could do this with sufficient detail to score both marks. The question was carefully worded
to lead students to the high salt concentration in the blood entering the capillaries, with the
hope that an explanation linked to decrease in water potential of the tissue fluid would be
given. Those who went down this route often scored this mark, but then referred to less
‘tissue fluid’ being reabsorbed at the venule end, rather than less water reabsorbed by
osmosis. Students are not expected to have knowledge of how salt affects the blood, but
may have some knowledge of the fact that a high salt diet can lead to an increase in blood
pressure. Any explanation of how a high salt concentration increased blood pressure was
ignored, and only the part of the answer focusing on what happens at the capillary bed to
cause a build-up of tissue fluid was marked. Consequently, many students scored one
mark for salt increasing the blood pressure (although it had to be clearly higher than
normal; statements referring to ‘high’ pressure at the arteriole end were insufficient to gain
credit). Only a few could go on to suggest that, as a result, more fluid would move out at
the arteriole end of the capillary.

Question 12

4 . 1 Bacteria are often used in industry as a source of enzymes. One reason is
because bacteria divide rapidly, producing a large number of them in a short time.
Describe how bacteria divide.
[2 marks]

Answer 12

1. Binary fission;
2. Replication of (circular) DNA;
3. Division of cytoplasm to produce 2 daughter
   cells;
4. Each with single copy of (circular) DNA;

4.1 There were many excellent answers to this question, with over 50% of students achieving
both marks. The majority of students knew that bacteria divide by binary fission, but often
did not include sufficient, or specific enough, detail in their description of what happens
during this process to score the second mark. There were several statements about
mitosis and some descriptions of viral replication

Question 13

0 4 . 2 Washing powders often contain enzymes from bacteria. These enzymes include
proteases that hydrolyse proteins in clothing stains.
Figure 1 shows the effect of temperature on a protease that could be used in
washing powder.
Figure 1
11
11
Turn over ►
IB/M/Jun17/E5
Do not write
outside the
box
Explain the shape of the curves at 50 °C and 60 °C.
[4 marks]

Answer 13

1. Both denatured (by high temperature);
2. Denaturation faster at 60 °C due to more
   (kinetic) energy;
3. Breaks hydrogen/ionic bonds (between amino
   acids/R groups);
4. Change in shape of the active site/active site no
   longer complementary so fewer enzymesubstrate complexes formed/substrate does not
   fit

04.2 This question required an explanation and so there were no marks for a description of the
graph. Many students gave lengthy descriptions and, even if they went on to write
creditworthy statements after this, they had wasted much time and energy. The first
requirement was for students to realise that, since the 30 ºC line stays at 100%, both lines
at 50 ºC and 60 ºC show denaturation of the enzymes. Many students got this far, but their explanations of what happens to the protein during denaturation were often incomplete or
imprecise. For example, some stated that ‘bonds’ would be broken or that ‘hydrogen, ionic,
disulfide and peptide bonds’ would be broken, so not achieving mark point 3. The mark
least often awarded was for the explanation that a temperature of 60 ºC would denature the
enzyme more quickly because of the increased (kinetic) energy at this temperature.

Question 14

4 . 3 Some proteases are secreted as extracellular enzymes by bacteria.
Suggest one advantage to a bacterium of secreting an extracellular protease in its
natural environment.
Explain your answer.
[2 marks]

Answer 14

1. To digest protein;
2. (So) they can absorb amino acids for
   growth/reproduction/protein synthesis/synthesis
   of named cell component;
   OR
   (So) they can destroy a toxic substance/protein;

V04.3 / 04.4 Both of these questions required careful reading of the question stems to access the
mark points; many students got half-way through their answers, but gave insufficient detail
to score full marks. For question 04.3, many students appreciated that the extracellular
proteases would digest protein and this might protect the bacteria in some way, but their
answers were not specific enough to gain credit for this idea. Similarly, many appreciated
that extracellular protein digestion would provide useful products for the bacteria but did not
go on to state that these products would be amino acids that could be used for
growth/protein synthesis within the bacteria.

Question 15

4 . 4 Mammals have some cells that produce extracellular proteases. They also have
cells with membrane-bound dipeptidases.
Describe the action of these membrane-bound dipeptidases and explain their
importance.
[2 marks]

Answer 15

1. Hydrolyse (peptide bonds) to release amino
   acids;
2. Amino acids can cross (cell) membrane;
   OR
   Dipeptides cannot cross (cell) membrane;
   OR
   Maintain concentration gradient of amino acids
   for absorption;
   OR
   Ensure (nearly) maximum yield from protein
   breakdown;

For question 04.4, it was pleasing to see how
many students could state that the dipeptidases would hydrolyse bonds in dipeptides, but
many were not confident what these bonds were, or what the products of this hydrolysis
would be. Many did not continue their answer to state the importance of this hydrolysis with
reference to the passage of amino acids across the cell-surface membrane into the cells for
absorption.

Question 16

0 5 Scientists investigated treatment of a human bladder infection caused by a species
of bacterium. This species of bacterium is often resistant to the antibiotics currently
used for treatment.
They investigated the use of a new antibiotic to treat the bladder infection. The new
antibiotic inhibits the bacterial ATP synthase enzyme.
0 5 . 1 Place a tick () in the appropriate box next to the equation which represents the
reaction catalysed by ATP synthase.
[1 mark]

Answer 16

ADP + Pi —- ATP + H2O

Question 17

0 5 . 2 The new antibiotic is safe to use in humans because it does not inhibit the ATP
synthase found in human cells.
Suggest why human ATP synthase is not inhibited and bacterial synthase is
inhibited.
[1 mark]

Answer 17

Human ATP synthase has a different tertiary
structure to bacterial ATP synthase
OR
Human ATP synthase has a different shape
active site to bacterial ATP synthase
OR
Antibiotic cannot enter human cells/mitochondria
OR
Antibiotic not complementary (to human ATP
synthase);

05.2 Some good understanding was demonstrated of how this universal enzyme could be
different in different species. Some students demonstrated understanding but their
responses were insufficiently precise to fulfil the marking criteria; they failed to reference
the ‘tertiary’ structure or the ‘shape’ of the active site. Some students successfully
suggested that the human enzyme would be found in the mitochondria and so be
inaccessible to the antibiotic.

Question 18

0 5 . 3 The scientists tested the new antibiotic on mice with the same bladder infection.
They divided these mice into three groups, C, R and A.
• Group C was the control (untreated).
• Group R was treated with an antibiotic currently used against this bladder
infection.
• Group A was treated with the new antibiotic.
They removed samples from the bladder of these mice after treatment and
estimated the total number of bacteria in the bladder.
Their results are shown in Figure 2.
Figure 2

The antibiotics were given to the mice at a dose of 25 mg kg–1 per day.
Calculate how much antibiotic would be given to a 30 g mouse each day.
Show your working.
[2 marks]

Answer 18

0.75;;
One mark for showing 30 g = 0.03 kg;
One mark for showing 0.025 mg g05.3 67% of students scored 2 marks. Those who did not, usually could not convert 30g to
0.03kg.-1

Question 19

0 5 . 4 Calculate the percentage difference in actual numbers of bacteria in group A
compared with group R. The actual number of bacteria can be calculated from the
log10 value by using the 10× function on a calculator.
Show your working.
[2 marks]

Answer 19

Answer in range 97.0 – 97.8%;;
OR
Answer in range 3288 – 4368%;;

4 There were two ranges of correct answer here, crediting a percentage difference compared
with either group A or group R. Ranges were calculated to allow for a tolerance of half a
2x2 mm square when reading from the graph, and for rounding at different stages. It was
pleasing to note that 46% of students could correctly complete this calculation. Some
clearly were unfamiliar with using the 10x function on their scientific calculators. 14% of
students could correctly read from the graph and convert these values into actual numbers
of bacteria, but then could not calculate the percentage difference.

Question 20

0 5 . 5 The scientists suggested that people newly diagnosed with this bladder infection
should be treated with both the current antibiotic and the new antibiotic.
Explain why the scientists made this suggestion.
Use information from Figure 2 and your knowledge of evolution of antibiotic
resistance in bacteria in your answer.
[3 marks]

Answer 20

1. (From Fig 2) New/old antibiotic does not kill
   all bacteria;
   OR
   (From Fig 2) Some bacteria are resistant to
   the new/old antibiotic;
2. Resistant bacteria will reproduce to produce
   (more) resistant bacteria;
3. (Use of both) one antibiotic will kill bacteria
   resistant to the other antibiotic;
   OR
   Unlikely that bacteria are resistant to both the
   new and the old antibiotic;
   OR
   Use of both antibiotics (likely to) kill all/most
   bacteria;

Question 21

6 2,4-D is a selective herbicide that kills some species of plants but not others. 2,4-D
disrupts cell-surface membranes but the extent of disruption differs in different
species.
Scientists investigated the effect of 2,4-D on wheat plants (a crop) and on wild oat
plants (a weed).
They grew plants of both species in glasshouses. They put plants of each species
into one of two groups, W and H, which were treated as follows:
• Group W – leaves sprayed with water
• Group H – leaves sprayed with a solution of 2,4-D.
After spraying, they cut 40 discs from the leaves of plants in each group and
placed them in flasks containing 10 cm3 de-ionised water. After 5 minutes, they
calculated the disruption to cell-surface membranes by measuring the
concentration of ions released into the water from the leaf discs.
Their results are shown in Table 3.
The lowest significant difference (LSD), is the smallest difference between two
means that would be significant at P≤0.05
Table 3

Table 3
Group Treatment
Mean concentration of ions in water
/ arbitrary units
Wheat Wild oats
W Water 26 45
H 2,4-D 27 70
Lowest significant
difference (LSD) 7 10
0 6 . 1 Give three environmental variables that should be controlled when growing the
plants before treatment with the different sprays.
[2 marks]

Answer 21

1. Concentration of mineral ion/named mineral
ion in soil;
2. Soil pH;
3. Temperature;
4. Light intensity/wavelength/duration;
5. Distance between seeds/plants;
6. Volume of water given;
7. CO2 concentration;
8. Humidity;

1 and 2. Allow
‘growing solution’ for
‘soil’.
2. pH alone is
insufficient.
3. Allow ‘colour of
light’
Reject ‘amount’ for
mps 1, 4, 6 and 7.
Ignore O2
concentration
Three correct = 2
marks
Two correct = 1
mark
One or none correct
= 0 marks

06.1 Some good understanding was demonstrated here, but students often failed to gain credit
due to poor use of language. “Amount” and “level” are not accepted units; pH and light
must be qualified, e.g. ‘soil pH’ and ‘light intensity’, and ‘nutrients’ is not an acceptable
alternative to mineral ion concentration at this level.

Question 22

0 6 . 2 Evaluate the use of 2,4-D as a herbicide on a wheat crop that contains wild oats as
a weed. Use all the information provided.
[4 marks]

Answer 22

1. 2,4-D causes an increase in release of ions
   from wild oat cells and 2,4-D does not
   affect/has little effect on the release of ions
   from wheat cells;
2. (For wheat) Difference is less than LSD/7 so
   difference is not significant;
   OR
   (For wild oats) Difference is more than LSD/10
   so difference is significant;
3. Loss of ions from cells (likely to) lead to
   cell/plant death/damage;
   OR
   Disruption of cell membrane (likely to) lead to
   cell/plant death/damage;
4. No evidence here about death of plants as a
   result of this ion loss;
5. No evidence here of other
   ecological/environmental impact;

06.2 A very small number of students achieved 4 marks here. The most commonly awarded
mark was for identifying that 2,4-D increased the release of ions from wild oats but had very
little effect on wheat (mark point 1). There was quite a lot to read and understand in the
stem of the question, and some students did not appreciate that disruption of the cellsurface membranes was linked to loss of ions, and so damage to the plant itself. Some
students confused these ions from leaves with ions in the soil and consequently suggested
greater ion concentration in the water would allow more ion uptake, to be used for growth.
The lowest significant difference (LSD) was a novel context, and many students did not
read/understand the explanation, and assumed the LSD was a form of standard deviation

Question 23

0 6 . 3 The scientists incubated the flasks containing the leaf discs at 26 °C and gently
shook the flasks.
Suggest one reason why the scientists ensured the temperature remained constant
and one reason why the leaf discs were shaken.
[2 marks]
Temperature
Shaken

Answer 23

1. (Maintain temperature) so that the rate of
   diffusion (of ions out of cells) remains constant
   OR
   (Maintain temperature) so no change in fluidity
   of phospholipids/kinetic energy of
   phospholipids;
   OR
   (Maintain temperature) so no change in
   shape/structure/denaturation of membrane
   proteins;
2. (Shaking) So all surfaces of the leaf discs are
   exposed (to water)/so all submerged;
   OR
   To maintain diffusion/concentration gradient
   (for ions out of leaf discs);

06.3 Unfortunately, many students did not fully link the context of this question to the practical
they had carried out, and so were not confident in their suggestions here. Many mistakenly
thought that the 2,4-D was in the water at this point. Those who did appreciate that the
temperature must remain constant, to avoid further changes in membrane permeability
(rather than enzyme activity), were often too vague in their responses to gain credit. Better
students often appreciated that temperature would affect the rate of diffusion of ions out of
the leaf discs. Why the leaf discs were shaken was generally better answered, although
those who showed some idea of maintaining the diffusion gradient for ions often gave
vague, imprecise answers that could not be given credit

Question 24

0 7 . 1 Describe how phagocytosis of a virus leads to presentation of its antigens.
[3 marks]

Answer 24

1. Phagosome/vesicle fuses with lysosome;
2. (Virus) destroyed by lysozymes/hydrolytic
   enzymes;
3. Peptides/antigen (from virus) are displayed on
   the cell membrane;

07.1 Many students did not refer to the antigen being presented on the surface membrane of the
phagocyte, so could not be awarded mark point 3.

Question 25

0 7 . 2 Describe how presentation of a virus antigen leads to the secretion of an antibody
against this virus antigen.
[3 marks]

Answer 25

1. Helper T cell/TH cell binds to the antigen (on
   the antigen-presenting cell/phagocyte);
2. This helper T/TH cell stimulates a specific B
   cell;
3. B cell clones
   OR
   B cell divides by mitosis;
4. (Forms) plasma cells that release antibodies;

07.2 Some students described, once again, how the virus would be presented. Mark points 3
and 4 were more often awarded than mark points 1 and 2, suggesting better understanding
of the actions of B cells in the immune response than T cell involvement. Mark point 1 could be awarded if the student stated that a T cell binds to the antigen and then
differentiates into a T helper cell.

Question 26

0 7 . 3 Collagen is a protein produced by cells in joints, such as the knee.
Rheumatoid arthritis (RA) is an auto-immune disease. In an auto-immune disease,
a person’s immune system attacks their own cells. RA causes pain, swelling and
stiffness in the joints.
Scientists have found a virus that produces a protein very similar to human
collagen.
Suggest how the immune response to this viral protein can result in the
development of RA.
[2 marks]

Answer 26

1. The antibody against virus (antigen) will bind
   to collagen;
2. This results in the destruction of the (human)
   cells/collagen;

07.3 The responses to this question revealed much misunderstanding of the immune response
as a whole, with many references to ‘thinking’ immune systems. Many students simply
repeated phrases from the question stem; for example, “since the virus protein and the
human collagen have a similar shape, the immune system will attack the human collagen”.
Students needed to identify that the part of the immune response which would ‘attack’ the
collagen would be the binding of specific antibodies, and then to use their knowledge of
how an antigen-antibody complex leads to the destruction of the antigen (section 3.2.4 of
the specification), i.e., human collagen in this case. Credit could be gained for reference to
agglutination or phagocytosis as methods of ‘attacking’ the human collagen, since these
are the methods of antigen destruction named in the specification

Question 27

0 8 Figure 3 shows two different ways of classifying the same three species of snake.
• Classification X is based on the frequency of observable characteristics
• Classification Y is based on other comparisons of genetic characteristics.
All three species of snake belong to the Python family.
Figure 3
0 8 . 1 What do these classifications suggest about the evolutionary relationships between
these species of snake?
[2 marks]
Classification X
Classification Y

Answer 27

1. (Without genetic analysis/X) mackloti and
   olivaceus have a more recent common
   ancestor with each other (than with papuana);
2. (Genetic analysis indicates/Y) papuana and
   mackloti have a more recent common
   ancestor with one another (than with
   olivaceus);
   2 Accept ‘more closely
   related to’ for ‘more
   recent common
   ancestor’

This question is based on sections 3.4.5 and 3.4.7 of the specification. For question 08.4, only
14% of students could name the three comparisons that are stated in the specification for
investigating genetic diversity (other than the observable characteristics that had been used to
generate Classification X). Although this method is no longer in the specification, DNA
hybridisation was accepted as a method of comparing DNA base sequences.

Question 28

0 8 . 2 Complete Table 4 below to show the missing names of the taxa when classifying
these snakes.
[1 mark]

Answer 28

Domain Eukaryote
Kingdom Animal
Phylum Chordata
Class Reptilia
Order Squamata
Family Python

Question 29

0 8 . 3 There is a debate about the name of one of these species of snake. Some
scientists name it Liasis papuana and other scientists name it Apodora papuana.
Give the name of the taxon about which the scientists disagree.
[1 mark]

Answer 29

Genus/genera;

Question 30

0 8 . 4 State three comparisons of genetic diversity that the scientists used in order to
generate Classification Y.
[3 marks]

Answer 30

1. The (base) sequence of DNA;
2. The (base) sequence of mRNA;
3. The amino acid sequence (of proteins);

14% of students could name the three comparisons that are stated in the specification for
investigating genetic diversity (other than the observable characteristics that had been used to
generate Classification X). Although this method is no longer in the specification, DNA
hybridisation was accepted as a method of comparing DNA base sequences.

Question 31

0 9 Figure 4 shows the stages of development of an insect called a damselfly.
Figure 4
0 9 . 1 The adult damselfly uses a tracheal system for gas exchange.
Explain three ways in which an insect’s tracheal system is adapted for efficient gas
exchange.
[3 marks]

Answer 31

1. Tracheoles have thin walls so short
   diffusion distance to cells;
2. Highly branched/large number of
   tracheoles so short diffusion distance to
   cells;
3. Highly branched/large number of
   tracheoles so large surface area (for gas
   exchange);
4. Tracheae provide tubes full of air so fast
   diffusion (into insect tissues);
5. Fluid in the end of the tracheoles that
   moves out (into tissues) during exercise
   so faster diffusion through the air to the
   gas exchange surface;
   OR
   Fluid in the end of the tracheoles that
   moves out (into tissues) during exercise
   so larger surface area (for gas exchange);
6. Body can be moved (by muscles) to move
   air so maintains diffusion/concentration
   gradient for oxygen/carbon dioxide;

9.1 Even though the majority of students had some knowledge of the structure of the insect
tracheal system, it was surprising how few could explain how it is adapted for efficient gas
exchange. It should be noted that this is the context of section 3.3.2 of the specification –
adaptations of gas exchange surfaces…for efficient gas exchange and the limitation of
water loss. In order to explain successfully here, students needed to identify the feature of
the tracheal system, and go on to explain how this chosen feature allows for efficient gas
exchange. Many students stated that a short diffusion pathway is required, but could not
give the feature of the tracheal system that allowed this. Many students referred to “thin
tracheoles”, which was insufficient, or referred to “thin membranes of tracheoles”
(demonstrating a similar lack of clarity of understanding of the difference between a cellsurface membrane and epithelial cell layer of an exchange surface to that seen in question
02.3). Similarly, many students appreciated that a large surface area is required, but could
not sufficiently explain how this is achieved in an insect with highly branched/numerous
tracheoles. There were some very good descriptions of lactate formation, causing water to
be drawn out of the ends of the tracheoles, but few students could relate this to faster gas
exchange as a result of diffusion through air, or a larger surface area. It was common to
see those students, who identified ventilation as being important, then being unable to
explain how this maintained a diffusion gradient. Explanations such as “moving carbon
dioxide out and oxygen in” were not uncommon, and were not creditworthy. References to
spiracles were not given credit as these are essential for limiting water loss but are not
essential for efficient gas exchange.

Question 32

0 9 . 2 The damselfly larva is a carnivore that actively hunts prey. It has gills to obtain
oxygen from water.
Some other species of insect have larvae that are a similar size and shape to
damselfly larvae and also live in water. These larvae do not actively hunt prey and
do not have gills.
Explain how the presence of gills adapts the damselfly to its way of life.
[2 marks]

Answer 32

1. Damselfly larvae has high(er)
   metabolic/respiratory (rate);
2. (So) uses more oxygen (per unit time/per
   unit m09.2 This was designed to test understanding of the statement in section 3.3.1: ‘Students should
   be able to appreciate the relationship between surface area and volume ratio, and
   metabolic rate’. Most students saw this question as an opportunity to write about their
   knowledge of gills as a gas exchange system, rather than focusing on the way of life of the ass); damselfly larvae. Although this led some students to describe how more oxygen could be
   absorbed, the key to the question was about the increased activity of the damselfly and,
   therefore, its increased use of oxygen.

Question 33

0 9 . 3 A scientist measured the size of each gill lamella of the gills of 40 damselfly larvae.
His results are shown in Table 5.
Table 5
Mean width / mm
(± uncertainty / mm)
1.61
(± 0.19)
Mean length / mm
(± uncertainty / mm)
6.12
(± 0.41)
Calculate the mean surface area of one side of one gill lamella. Assume that a gill
lamella is rectangular and give your answer to an appropriate number of
significant figures.
Include the percentage error (uncertainty) of surface area in your answer.
Show your working.
[3 marks]

Answer 33

Mean SA = 9.85 mm2
/9.9 mm2
;;
Percentage uncertainty of SA = 18.5/18.7/19;;
If both answers incorrect 1 mark for
Percentage uncertainty of dimensions 11.8/12
and 6.70/6.7
Surface area correctly calculated with correct
units but not rounded to appropriate sf
(9.8532 mm2
)
Surface area correct (with appropriate sf) but no/incorrect unit given

09.3 Only 7% of students gained all three marks here. Dealing with uncertainties is new to this
specification, as is using an appropriate number of significant figures (specification section
6.2, MS 1.1 and MS 1.11). There are clear explanations of using uncertainties and
choosing an appropriate number of significant figures in the online Practical Handbook,
Sections K and L

Question 34

0 9 . 4 A student used an optical microscope to observe part of a damselfly larva gill.
Figure 5 shows the drawing the student produced.
Figure 5
29
*29*
Turn over ►
IB/M/Jun17/E5
Do not write
outside the
box
Suggest two ways the student could improve the quality of her scientific drawing of
this gill.
[2 marks]

Answer 34

1. Don’t use shading;
2. Only use single lines/don’t use sketching
   (lines)/ensure lines are
   continuous/connected;
3. Add further labels/annotations;
4. Don’t cross label lines;
5. Add magnification/scale (bar);

09.4 This question tests PS 4.1 from section 8.3 of the specification, linked to required practical
activity 5 and apparatus and techniques skill AT e. The most common answers that did not
gain credit were suggestions that the drawing should contain more detail, or an electron
microscope should be used to enable more detail to be drawn. References to adding more
labels were given credit even if the extra parts to be labelled were named incorrectly.

Question 35

1 0 . 1 Contrast how an optical microscope and a transmission electron microscope work
and contrast the limitations of their use when studying cells.
[6 marks]

Answer 35

1. TEM use electrons and optical use light;
2. TEM allows a greater resolution;
3. (So with TEM) smaller organelles/named cell
   structure can be observed
   OR
   greater detail in organelles/named cell structure
   can be observed;
4. TEM view only dead/dehydrated specimens and
   optical (can) view live specimens;
5. TEM does not show colour and optical (can);
6. TEM requires thinner specimens;
7. TEM requires a more complex/time consuming
   preparation;
8. TEM focuses using magnets and optical uses
   (glass) lenses;
   6 max 3. ‘clearer’ is not
   equivalent to ‘detail’
9. Accept ‘Only optical
   can view live
   specimens’
10. Accept ‘Only optical
    can show colour’
11. Accept ‘TEM
    requires a more
    difficult preparation’
    Ignore references to
    artefacts

10.1 The command word here was contrast and so statements showing clear differences
between the use of the two microscopes were required to gain credit. Most students
demonstrated sound knowledge of the optical and electron microscopes, but few managed
to gain all six marks for relevant contrasting statements. Many suggested that no
organelles could be seen with an optical microscope, rather than only larger organelles
being visible. Some referred to SEMs and 3D images; neither was relevant here.

Question 36

1 0 . 2 Figure 6 shows an image from an optical microscope of meiosis occurring in a
flower bud of a flowering plant. W and Z are undergoing meiosis.
Figure 6
Explain the appearance of W and Z.
[4 marks]

Answer 36

1. W has 4 cells/nuclei since it is at the (end of) 2
   nd
   division (of meiosis);
2. Z has 2 cells/nuclei since it is at the (end of) 1
   st
   division (of meiosis);
3. W shows haploid cells/cells containing n
   chromosomes;
4. (Cells in) W contain half the (mass of) DNA of
   (Cells in) Z;
   OR
   (between Z and W) chromatids have separated;
   OR
   In Z homologous chromosomes have separated;

10.2 There were many good answers here, with over 70% of students scoring 2 marks and over
40% gaining 3 marks. Mark point 4 was the least often awarded. Students often failed to
gain mark point 3 because they stated that W contained half the genetic material of Z,
rather than specifying half the DNA. Several students discussed these cells as if they were
human cells rather than plant cells, mentioning 23 pairs of chromosomes originally, or 23
chromosomes in cells in W. Some students named stages of meiosis in their answer.
These were ignored as this is not expected knowledge; we were only looking for the
outcomes of 1st and 2nd meiotic divisions

Question 37

0 . 3 An environmental scientist investigated a possible relationship between air
pollution and the size of seeds produced by one species of tree.
He was provided with a very large number of seeds collected from a population of
trees in the centre of a city and also a very large number of seeds collected from a
population of trees in the countryside.
Describe how he should collect and process data from these seeds to investigate
whether there is a difference in seed size between these two populations of trees.
[5 marks]

Answer 37

1. Use random sample of seeds (from each
   population);
2. Use (large enough) sample to be representative
   of whole population;
3. Indication of what size was measured eg mass;
4. Calculate a mean and standard deviation (for
   each population);
5. Use the (Student’s) t-test;
6. Analyse whether there is a significant difference
   between (the means of) the two populations;
   5 max 1. Accept described,
   suitable method of
   random sampling.
7. Reject description
   of inappropriate
   method of random
   sampling (eg random
   coordinates in the
   field/use of quadrats)
8. Accept ‘running
   mean does not
   change’
9. For representative
   accept ‘reliable,
   reproducible,
   repeatable’ OR a
   mean close to the true
   value.
10. Accept ‘Use 95%
    confidence limits’
11. Reject unqualified
    references to results
    being significant

0.3 The full range of 0 – 5 marks was seen in similar proportions here (although slightly fewer
with full marks). The main reason for the discrimination was how fully each student used
the information from the question stem. The two groups of seeds to be investigated had
been collected already, and the investigation that was needed was simply to find out if there
was a difference in size of these seeds. Many students gave extensive answers relating to
how to collect the seeds from the environment, and how to measure the pollution – neither
of which was relevant to the question. Many students selected an inappropriate statistical
test or listed several options, which could not be given credit, even if t-test was amongst the
list of possibilities (see note about the ‘list rule’ in the final paragraph of “General
Comments” above)