A level june 2018

Question 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

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 2

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

Question 3

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 3

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

Question 4

What is a homologous pair of chromosomes?

Answer 4

(Two chromosomes that) carry the same

genes;

Question 5

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 5

(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;

Question 6

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 6

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;

Question 7

What is meant by ‘species richness’?

[1 mark]

Answer 7

(A measure of) the number of (different) species in

a community

Question 8

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]
3. The natural habitat is most favourable for bees.
4. The town is the least favourable for bees.

Answer 8

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 9

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 9

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);

Question 10

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

Answer 10

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 11

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 11

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;

Question 12

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

Answer 12

1) Enzyme provides an alternative pathway with a lower activation energy.
2) Due to the bending of bonds.
OR
Without enzyme, very few substrates have
sufficient energy for reaction;

Question 13

Calculate by how many times the rate of reaction is greater with the enzyme present.
Give your answer in standard form.
[2 marks]

Answer 13

578/3.0 × 10 -9 =1.93x10 11

Question 14

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 14

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;

Question 15

The genetic code is described as degenerate.

What is meant by this? [2 marks]

Answer 15

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

Question 16

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 16

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;

Question 17

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 17

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;

Question 18

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

Answer 18

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 19

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 19

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

Question 20

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

Answer 20

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);

Question 21

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 21

1. Horses because more
antivenom/antibodies could be collected
(as more blood collected);
2. 4550 (cm3
) v 26 (cm3
) (blood collected);

Question 22

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

Answer 22

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;

Question 23

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 23

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 24

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

Answer 24

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;

Question 25

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 25

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;

Question 26

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

Answer 26

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 27

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

Answer 27

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

Question 28

Suggest explanations for the results in Table 5 box .

[3 marks

Answer 28

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 29

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

Answer 29

The human gas exchange system includes the trachea, the bronchioles and the alveoli. During inhalation (breathing in) the diaphragm contracts and the external intercostal muscles contract. This increases the volume and decreases the pressure in the thoracic cavity to lower than that of the atmosphere. Therefore, air moves in to the thoracic cavity (inhalation).During exhalation, the diaphragm relaxes and the internal intercostal muscles contract. This causes the volume to decrease and the pressure to increase in the thoracic cavity to greater than atmospheric pressure. Therefore, air moves out of the thoracic cavity (exhalation).

Question 30

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 30

Both types of molecule contain glycerol, contain ester bonds and contain the elements carbon, hydrogen and oxygen (but phospholipids also contain phosphorus). The fatty acids contained in both types of molecule may be saturated or unsaturated. Both are insoluble in water because of these fatty acid chains.In contrast, phospholipids have two fatty acid chains and a phosphate group versus triglycerides’ three fatty acid chains. Phospholipids have a hydrophilic region as well as a hydrophobic region (while triglycerides are entirely hydrophobic), and because of this, phospholipids form micelles or bilayers when in solution, which is something that triglycerides do not do.

Question 31

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 31

lactose is formed from glucose and galactose

joined by condensation (reaction)

joined by glycosidic bond;

added to polypeptide in Golgi (apparatus)