2018 paper 1

Question 1

Give two pieces of evidence that the cell in figure 1 was undergoing mitosis

Answer 1

• individual chromosomes are visible because they have condensed
• each chromosome is made up of two chromatids because DNA has replicated
• the chromosomes are not arranged in homologous pairs, which they would be if it was meiosis

Question 2

Tick one box that gives the name of the stage in mitosis shown

Answer 2

Prophase

Question 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.
0 1 . 3
Suggest how this procedure moved the chromosomes apart.

Answer 3

• water moves into the cells / cytoplasm by osmosis

- cell / cytoplasm gets bigger

Question 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.

Answer 4

difference in base sequences
Difference in histones
- difference in coiling

Question 5

In Figure 2 the chromosomes are arranged in homologous pairs. What is a homologous pair of chromosomes?

Answer 5

Two chromosomes that carry the same genes

Question 6

Give two ways in which the arrangement of prokaryotic DNA is different from the
arrangement of the human DNA in Figure 1.

Answer 6

Prokaryotic dna is circular as opposed to linear , not associated with proteins/histones ,

Question 7

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.

Answer 7

Method to ensure all cut surfaces of the eight cubes are exposed to the sucrose solution;

2. Method of controlling temperature (water bath)
3. Method of drying cubes before measuring (paper towels )
4. Measure mass of cubes at stated time intervals;

Question 8

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.

Answer 8

Yes or No (no mark)
Calculation of rate per mm2 for both sets of data, accept answers in the range
1.6 × 105 to 1.8 × 105 and
1.5 × 105 to 1.6 × 105;;;
Both correct = 3
One correct = 2
Neither correct – look below for max 2
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 9

What is meant by ‘species richness’?

Answer 9

A measure of the number of different species in a community

Question 10

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.

Answer 10

1 (y) peak of mean bee numbers in natural habitat is highest/ mean species richness un natural habitat is higher at all time
(N) has lowest mean number of bees after day 220
2 (y) peaks of species richness higher both in natural habitat and farmland
(N) species richness is lower in farmland until day 125 / similar mean number of bees to farmland

Question 11

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

Answer 11

1- must not harm the bees

2-must allow close examination

Question 12

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

Answer 12

collect more times of the year so more points on the graph / better lobf on graph

• collect from more sites / more years to increase accuracy of mean data
• counted number of individuals in each species so they could calculate index of diversity

Question 13

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.

Answer 13

AC and AP are more closely related to each other than to PP because they are in the same genus

Question 14

Formation of an enzyme-substrate complex increases the rate of reaction. Explain how.

Answer 14

Reduces activation energy due to bending bonds

Question 15

without an enzyme present. With the enzyme present, 578 amino acids were
−9
released per second. Without the enzyme, 3.0 × 10 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.

Answer 15

578/ 3x10-9 = 1.93x10-11

Question 16

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.

Answer 16

31.4

Question 17

Lyxose binds to the enzyme.
Suggest a reason for the difference in the results shown in Figure 5 with and without
lyxose.

Answer 17

binding alters the tertiary structure of the enzymes

• this causes the active site to change shape
• so more successful es complexes form per minute

Question 18

Draw the general structure of an amino acid.

Answer 18

R
             |
H2N -C-COOH
             |
            H

Question 19

The genetic code is described as degenerate.

What is meant by this? Use an example from Table 1 to illustrate your answer

Answer 19

More than one codon codes for a single amino acid

Eg UUA and CUA both code for leu

Question 20

What is the minimum number of bases in the gene coding for this polypeptide?

Answer 20

1395

Question 21

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

Answer 21

CAA-CGA

Question 22

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 reduces the rate of reaction catalysed by the enzyme .
use all the info

Answer 22

2. Change at amino acid 300 does not change the shape of the active site/tertiary structure

Question 23

Figure 7

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;

Question 24

Picture

Answer 24

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 25

6 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?

Answer 25

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 26

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

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 27

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.

Answer 27

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

Question 28

A mixture of venoms from several snakes of the same species is used. Suggest why.

Answer 28

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 29

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.

Answer 29

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

Question 30

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

Answer 30

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

Question 31

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.

Answer 31

1.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

Picture The scientists concluded that this heat treatment damaged the phloem. Explain how the results in Figure 9 support this conclusion.

Answer 32

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

Question 33

The scientists also concluded that this heat treatment did not affect the xylem.
Explain how the results in Table 4 support this conclusion.

Answer 33

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 34

Picture What can you conclude about the movement of Fe3+ 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

Question 35

Describe the role of two named enzymes in the process of semi-conservative
replication of DNA.

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

Picture and Suggest explanations for the results in Table 5.

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)

Question 37

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

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);

Question 38

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.

Answer 38

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;

Question 39

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.

Answer 39

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