2019 paper 1

Question 1

Describe how a non-competitive inhibitor can reduce the rate of an enzyme-controlled reaction.

Answer 1

1. Attaches to the enzyme at a site other than the active site;
2. Changes (shape of) the active site
   OR
   Changes tertiary structure (of enzyme);
3. (So active site and substrate) no longer complementary so less/no substrate can fit/bind;

Question 2

The scientist concluded that pectin is a non-competitive inhibitor of the lipase enzyme.
Use Figure 1 to explain why the scientist concluded that pectin is a non-competitive
inhibitor

Answer 2

(With inhibitor) increase substrate/lipid (concentration) does not increase/affect/change rate of reaction
OR
(With inhibitor) increase substrate/lipid (concentration) does not increase/affect/change lipase activity
OR
High substrate (concentration) does not overcome inhibition
OR
High substrate (concentration) does not meet maximum rate of reaction/lipase activity;

Question 3

Calculate the maximum length of the large lipid droplet marked X in Figure 2.
Using a ruler with millimetre intervals always includes an uncertainty in the measurement. Use the uncertainty in your measurement to determine the uncertainty of your calculated maximum length.
The scientist also found that pectin stops the action of bile salts. He prepared two suspensions:
• suspension A – lipid and bile salts
• suspension B – lipid, bile salts and pectin.
He did not add lipase to either suspension.
He observed samples from the suspensions using an optical microscope.
Figure 2 shows what he saw in a typical sample from each suspension. Figure 2
You can assume there is no uncertainty in the magnification.

Answer 3

(Maximum length) 9 (μm); (Uncertainty) (±) 2 (μm);

Question 4

No large lipid droplets are visible with the optical microscope in the samples from suspension A.
Explain why.

Answer 4

1. Emulsification;

2. (Cannot be seen) due to resolution (of optical microscope);

Question 5

able 1 shows cell wall components in plants, algae, fungi and prokaryotes. Complete Table 1 by putting a tick () where a cell wall component is present

Answer 5

ell wall Plants Algae component
Cellulose   Murein
               Chitin
Fungi Prokaryotes



Question 6

student concluded from Figure 3 that eating an extra 10 g of fibre per day would significantly lower his risk of cardiovascular disease.
Evaluate his conclusion.

Answer 6

1. Negative correlation (between fibre eaten per day and risk of cardiovascular disease);
2. Original/current fibre intake (of student) not known;
3. (Idea of) significance linked to (2x) standard deviation overlap (at 10 g day-1 change);
4. If current intake between 5 and 30 (g day-1) then (eating 10g more results in a significant) decrease in risk
   OR
   If current intake between 30 and 50 (g day-1) then (eating 10g more results in) no significant decrease in risk;
5. Correlation does not mean causation
   OR
   Another named factor may be involved;
6. Little evidence/data for higher mass of fibre per day;
7. Large (2x) standard deviation at high/low mass of fibre makes (mean) less precise
   OR
   Large (2x) standard deviation at high/low amounts of fibre means there is a greater uncertainty;
8. No statistical test (to show if differences are significant);

Question 7

uggest one advantage of using the FFQ method and one disadvantage of using the
FFQ method compared with the alternative method.

Answer 7

1. Over longer period so more representative
   OR
   Diet over 24 hr may not be representative
   OR
   Diet may vary during the year/from day to day
   OR
   Person more likely to be honest on questionnaire (rather than speaking to nurse)
   OR
   More cost effective because fewer
   people/nurses required;
   (Disadvantage)
2. Relies on (long term) memory so may not be accurate
   OR
   Recall of 24 hr diet likely to be more accurate
   OR
   Estimation (from FFQ) may be less accurate (than details of last 24hrs)
   OR
   Person may be more honest when being interviewed;

Question 8

What data would the students need to collect to calculate their index of diversity in each habitat?
0 3 . 1
Do not include apparatus used for species sampling in your answer.

Answer 8

(Number of species and) number of individuals in each species (in each habitat)
OR
(Number of species and) population of each species (in each habitat);

Question 9

Give two ways the students would have ensured their index of diversity was representative of each habitat.

Answer 9

. Random samples;
2. Large number (of samples)
OR
(Continue sampling) until stable running mean;

Question 10

Modern farming techniques have led to larger fields and the removal of hedges between fields.
Use Figure 4 to suggest why biodiversity decreases when farmers use l

Answer 10

(Larger fields have relatively) More centre
OR
Less edge
OR
Less hedge
OR
Fewer species;

Question 11

rmers are now being encouraged to replant hedges on their land.
Suggest and explain one advantage and one disadvantage to a farmer of replanting
hedges on her farmland.

Answer 11

Advantage -
1. Greater (bio)diversity so increase in predators of pests
OR
Increase in predators of pests so more yield/income/less pesticides/less damage to crops
OR
Increase in pollinators so more yield/income
OR
May attract more tourists/subsidies to their farm so more income (from diversification);
Disadvantage -
2. Reduced land area for crop growth/income
OR
Greater (bio)diversity so increase pest
population
OR
Increase pest population so less yield/less income/(more) need for pesticides/(more) damage to crops
OR
Increased (interspecific) competition so less
yield/income
OR
More difficult to farm so less income;

Question 12

Use Figure 5 to explain how human mass at birth is affected by stabilising selection.

Answer 12

. (Most likely to be) transferred to a special care unit are those under 2800 g
OR
(Most likely to be) transferred to a special care unit are those over 4200 g;
2. Extreme mass babies least likely to survive (to reproduce) and so less likely to pass on their alleles (for extreme mass at birth);
3. Extreme mass at birth decreases in frequency (in the population)
OR
Alleles (for extreme mass at birth) decrease in frequency (in the population);
If neither 1 or 2 awarded allow correct stated mass less/more likely to survive for 1 mark

Question 13

KIR2DS1 is an (1) of the KIR gene, found at a
chromosome 19. KIR2DS1 is 14 021 bases long and is
that is 1101 bases long. This mRNA is then (4)
amino acids long. The polypeptide is then modified in the organelle, (5) , before forming its functional (6) protein structure.
Write the correct biological term beside each number below, that matches the space in
(2) on
(3)
into a polypeptide 304
into mRNA
the passage.
(1) (2) (3) (4) (5) (6)

Answer 13

1. Allele
2. Locus/loci
3. Transcribed
4. Translated
5. Golgi (apparatus)/Rough endoplasmic reticulum 6. Tertiary;;;

Question 14

Tick () one box that gives the name of the statistical test that the scientists should
use with the data in Table 2 to test this null hypothesis.

Answer 14

Automarked q –Chi-squared

Question 15

he scientists calculated a P value of 0.03 when testing their null hypothesis.
What can you conclude from this result? Explain your answer.

Answer 15

1. Probability that difference (in frequency of births above 4500 g) is due to chance is less than 0.05
   OR
   Probability that difference (in frequency of births above 4500 g) is due to chance is 0.03;
2. Reject null hypothesis;
3. Presence of KIR2DS1/allele does (significantly) affect the frequency of high birth mass

Question 16

escribe the structure of the human immunodeficiency virus (HIV).

Answer 16

1. RNA (as genetic material);
2. Reverse transcriptase;
3. (Protein) capsomeres/capsid; 4. (Phospho)lipid (viral) envelope
   OR
   Envelope made of membrane; 5. Attachment proteins;

Question 17

A test sample of 500 mm3 of blood is taken from an HIV controller to determine the viral load.
Tick () one box that shows the number of virus particles that would be present in a test sample of blood taken from an HIV controller with the median viral loa

Answer 17

106

Question 18

Use the data in Table 3 and your knowledge of the immune response to suggest why
HIV controllers do not develop symptoms of AIDS.

Answer 18

. (All) have more T helper/CD4 cells;
2. Lower viral load to infect/destroy helper T/CD4 cells;
3. (So more/continued) activation of B cells/cytotoxic T cells/phagocytes;
4. (With B cells more/continued) production of plasma cells/antibodies
OR
(With cytotoxic T cells more/continued) ability to kill virus infected cells;
5. (More able to) destroy other microbes/pathogens
OR
(More able to) destroy mutated/cancer cells;

Question 19

Describe and explain the data in Table 4

Answer 19

. (Trend of) slowing growth from before birth to 21 days
OR
(Trend of) decreasing percentage undergoing mitosis from before birth to 21 days
OR
(Trend of) decreasing percentage undergoing DNA replication from before birth to 21 days;
2. DNA replication happens before mitosis
OR
Heart growth slowing until (fully) developed OR
These cells lost the ability to divide;

Question 20

Describe how BrdU would be incorporated into new DNA during semi-conservative
replication.

Answer 20

1. DNA helicase;
2. Breaks hydrogen bonds (between 2 DNA strands);
3. BrdU complementary to adenine (on template strand)
   OR
   BrdU forms hydrogen bonds with adenine (on template strand);
4. DNA polymerase joins (adjacent) nucleotides (to incorporate BrdU into the new DNA strand);
5. Phosphodiester bonds form (between nucleotides);

Question 21

Cells with BrdU in their DNA are detected using an anti-BrdU antibody with an enzyme attached.
Use your knowledge of the ELISA test to suggest and explain how the scientists
identified the cells that have BrdU in their DNA.

Answer 21

1. Add antibody (anti-BrdU with enzyme attached) to cells/DNA
   OR
   Add cells/DNA to antibody (anti-BrdU with enzyme attached);
2. Wash (cells/DNA) to remove excess/unattached antibody
   OR
   Wash (immobilised antibody) to remove
   excess/unattached cells/DNA;
3. Add substrate to cause colour change;

Question 22

nlike plants, Ulva lactuca does not have xylem tissue.
Suggest how Ulva lactuca is able to survive without xylem tissue.
Figure 6 shows a diagram of one Ulva lactuca alga.

Answer 22

short diffusion pathway (to cells)
OR
It has a surface permeable (to water/ions into cells);

Question 23

On Figure 7 complete each box with an appropriate letter to show the type of cell division happening between each stage in the life cycle. Use ‘T’ to represent
mitosis and ‘E’ to represent meiosis.

Answer 23

E in top right box; (1 mark)

2. 3 x T in top and bottom left and bottom right boxes; (1 mark)

Question 24

Ulva prolifera also produces haploid, mobile single cells that can fuse to form a zygote.
Suggest and explain one reason why successful reproduction between Ulva prolifera
and Ulva lactuca does not happen.

Answer 24

1. They are different species;
2. (So) if fused together they would not produce fertile offspring
   OR
   (So) they have named characteristics that means they are reproductively isolated;

Question 25

The scientists needed solutions of known water potential to generate their calibration curve.
Table 5 shows how to make a sodium chloride solution with a water potential of −1.95 MPa
Complete Table 5 by giving all headings, units and volumes required to make 20 cm3
of this sodium chloride solution.

Answer 25

cm3, volume of water cm3, 0.8, 19.2

Question 26

There is a linear relationship between the water potential and the concentration of sodium chloride solution.
a water potential of −3.41 MPa
Answer = mol dm–3
Use the data in Table 6 to calculate the concentration of sodium chloride solution with
[2 marks]

Answer 26

Correct answer of 0.07 (mol dm–3) = 2 marks;; Incorrect answer 1 mark for any evidence of 48.6 to 48.8
OR
0.02
OR
0.7
OR
A final answer between 0.04 and 0.10
OR
A final answer of minus 0.07/-0.07;

Question 27

In addition to determining the water potential in the leaf cells, the scientists measured the growth of the leaves.
They recorded leaf growth as a percentage increase of the original leaf area. Their results are shown in Figure 9.
Figure 9
One leaf with an original area of 60 cm2 gave a voltage reading of −7 μV
Use Figure 8 (on page 28) and Figure 9 to calculate by how much this leaf increased in area.
Give your answer in cm2

Answer 27

Correct answer of 9 (cm2) = 2 marks;; Incorrect answer 1 mark for evidence of water
potential of between -1.85 and -1.95 (MPa)
OR
growth of 15%
OR
69 (cm2)
OR
A final answer between 8.7 and <9;

Question 28

Sunflowers are not xerophytic plants. The scientists repeated the experiment with xerophytic plants.
Suggest and explain one way the leaf growth of xerophytic plants would be different
from the leaf growth of sunflowers in Figure 9.

Answer 28

EITHER
1. Low/slow growth;
2. Due to smaller number/area of stomata (for gas exchange);
OR
3. Growth may continue at lower water potentials;
4. (Due to) adaptations in enzymes involved in photosynthesis/metabolic reactions;

Question 29

Use your knowledge of gas exchange in leaves to explain why plants grown in soil
with very little water grow only slowly.

Answer 29

1. Stomata close;

2. Less carbon dioxide (uptake) for less photosynthesis/glucose production;

Question 30

Plot the haemoglobin saturation data from Table 7 and use these points to sketch the
full oxyhaemoglobin dissociation curves for a horse and a mouse

Answer 30

1. y axis 0 – 100 in linear scale and x axis minimum 1 to 8 in linear scale and both axes use at least half size of grid;
2. Correct plots for 50% and 25% for both animals;
3. Both curves levelling off (at higher partial pressures and at percentage saturations ≤100%);

Question 31

The following equation can be used to estimate the metabolic rate of an animal. Metabolic rate = 63 × BM–0.27
BM = body mass in grams
Use this equation to calculate how many times faster the metabolic rate of a mouse is
0 9 . 3
Answer = times faster
The data in Table 7 show differences between the oxyhaemoglobin dissociation curve for a mouse and the oxyhaemoglobin dissociation curve for a horse.

Answer 31

correct answer of 15 (times faster) = 2marks ;;
If ≥3sf given, accept answers in the range 15.0 to 15.4 (times faster) = 2marks;;
Incorrect answer 1 mark for evidence of: 23–0.27 divided by 550 000–0.27
OR
0.42888777
OR
0.02819045
OR
Between 27 and 27.1
OR
Between 1.77599861 and 1.8
OR
0.06

Question 32

Suggest how these differences allow the mouse to have a higher metabolic rate than
than the metabolic rate of a horse.

Answer 32

1. Mouse haemoglobin/Hb has a lower affinity for oxygen
   OR
   For the same pO2 the mouse haemoglobin/Hb is less saturated
   OR
   At oxygen concentrations found in tissue mouse haemoglobin/Hb is less saturated;
2. More oxygen can be dissociated/released/unloaded (for metabolic reactions/respiration);

Question 33

Mammals such as a mouse and a horse are able to maintain a constant body temperature.
Use your knowledge of surface area to volume ratio to explain the higher metabolic
rate of a mouse compared to a horse

Answer 33

Mouse

1. (Smaller so) larger surface area to volume ratio;
2. More/faster heat loss (per gram/in relation to body size);
3. (Faster rate of) respiration/metabolism releases heat;

Question 34

Explain five properties that make water important for organisms.

Answer 34

1. A metabolite in condensation/hydrolysis/ photosynthesis/respiration;
2. A solvent so (metabolic) reactions can occur OR
   A solvent so allowing transport of substances;
3. High heat capacity so buffers changes in
   temperature;
4. Large latent heat of vaporisation so provides a cooling effect (through evaporation);
5. Cohesion (between water molecules) so supports columns of water (in plants);
6. Cohesion (between water molecules) so produces surface tension supporting (small) organisms;

Question 35

Describe the biochemical tests you would use to confirm the presence of lipid,
non-reducing sugar and amylase in a sample.

Answer 35

Lipid
1. Add ethanol/alcohol then add water and shake/mix
OR
Add ethanol/alcohol and shake/mix then pour into/add water;
2. White/milky emulsion
OR
emulsion test turns white/milky;
Non-reducing sugar
3. Do Benedict’s test and stays blue/negative;
4. Boil with acid then neutralise with alkali;
5. Heat with Benedict’s and becomes red/orange (precipitate);
Amylase
6. Add biuret (reagent) and becomes purple/violet/mauve/lilac;
7. Add starch, (leave for a time), test for reducing sugar/absence of starch;

Question 36

Describe the chemical reactions involved in the conversion of polymers to monomers and monomers to polymers.
10.3
Give two named examples of polymers and their associated monomers to illustrate
your answer.

Answer 36

1. A condensation reaction joins monomers together and forms a (chemical) bond and releases water;
2. A hydrolysis reaction breaks a (chemical) bond between monomers and uses water;
3. A suitable example of polymers and the monomers from which they are made;
4. A second suitable example of polymers and the monomers from which they are made;
5. Reference to a correct bond within a named polymer;