2020 paper 1

Question 1

The action of the carrier protein X in Figure 1 is linked to a membrane-bound ATP hydrolase enzyme.
0 1 . 1
Explain the function of this ATP hydrolase

Answer 1

1. (ATP to ADP + Pi ) Releases energy;
2. (energy) allows ions to be moved against a concentration gradient
   OR
   (energy) allows active transport of ions;

Question 2

The movement of Na+ out of the cell allows the absorption of glucose into the cell lining the ileum.
Explain how.

Answer 2

1. (Maintains/generates) a concentration/diffusion gradient for Na+ (from ileum into cell);
2. Na+ moving (in) by facilitated diffusion, brings glucose with it
   OR
   Na+ moving (in) by co-transport, brings glucose with it;

Question 3

Describe and explain two features you would expect to find in a cell specialised for
absorption.1.

Answer 3

1. Folded membrane/microvilli so large surface area (for absorption);
2. Large number of co-transport/carrier/channel proteins so fast rate (of absorption)
   OR
   Large number of co-transport/carrier proteins
   for active transport
   OR
   Large number of co-transport/carrier/channel proteins for facilitated diffusion;
3. Large number of mitochondria so make (more) ATP (by respiration)
   OR
   Large number of mitochondria for aerobic respiration
   OR
   Large number of mitochondria to release energy for active transport;
4. Membrane-bound (digestive) enzymes so maintains concentration gradient (for fast absorption);

Question 4

Draw phospholipids on Figure 2 to show how the carrier protein, SGLT1, would fit into the cell-surface membrane.
Do not draw more than eight phospholipids.

Answer 4

1. Phospholipids drawn with head and two tails;

2. Correctly positioned as a bilayer on either side of SGLT1;

Question 5

igure 2 shows the SGLT1 polypeptide with NH2 at one end and COOH at the other end.
Describe how amino acids join to form a polypeptide so there is always NH2 at one end and COOH at the other end.
You may use a diagram in your answer.

Answer 5

1. One amine/NH2 group joins to a carboxyl/COOH group to form a peptide bond;
2. (So in chain) there is a free amine/NH2 group at one end and a free carboxyl/COOH group at the other
   OR
   Each amino acid is orientated in the same direction in the chain;

Question 6

Use your knowledge of lipid digestion to explain the differences in the results for samples A and B shown in Table 1.
You should assume that no absorption had occurred.

Answer 6

1. Triglycerides decrease because of the action of lipase
   OR
   Fatty acids increase because of the action of lipase;
2. Triglycerides decrease because of hydrolysis (of triglycerides)
   OR
   Fatty acids increase because of hydrolysis (of triglycerides);
3. Triglycerides decrease because of digestion of ester bonds (between fatty acid and glycerol)
   OR
   Fatty acids increase because of digestion of ester bonds (between fatty acid and glycerol);

Question 7

After collecting the samples, the scientist immediately heated them to 70 °C for 10 minutes.
Explain why.

Answer 7

1. To denature the enzymes/lipase;

2. So no further digestion/hydrolysis/catalysis occurred;

Question 8

Describe the role of micelles in the absorption of fats into the cells lining the ileum.

Answer 8

1. Micelles include bile salts and fatty acids;
2. Make the fatty acids (more) soluble in water;
3. Bring/release/carry fatty acids to cell/lining (of the ileum);
4. Maintain high(er) concentration of fatty acids to cell/lining (of the ileum);
5. Fatty acids (absorbed) by diffusion;

Question 9

At P on Figure 3, the pressure in the left ventricle is increasing. At this time, the rate of blood flow has not yet started to increase in the aorta.
0 3 . 1
Use evidence from Figure 3 to explain why.

Answer 9

1. Aortic/semi-lunar valves is closed;

2. Because pressure in aorta higher than in ventricle;

Question 10

At Q on Figure 3 there is a small increase in pressure and in rate of blood flow in the aorta.
Explain how this happens and its importance.

Answer 10

1. Elastic recoil (of the aorta wall/tissue);
2. Smooths the blood flow
   OR
   Maintains rate of blood flow OR
   Maintains blood pressure;

Question 11

A student correctly plotted the right ventricle pressure on the same grid as the left ventricle pressure in Figure 3.
Describe one way in which the student’s curve would be similar to and one way it
would be different from the curve shown in Figure 3

Answer 11

1. Peaks/contractions at the same/similar time OR

Same/similar pattern; 2. Lower pressure;

Question 12

Use information from Figure 3 to calculate the heart rate of this dog

Answer 12

167 (beats minute–1) OR
164 (beats minute–1) OR
171 (beats minute–1);

Question 13

Tick () one box that shows the most appropriate volumes she would use to make up
0 4 . 1
100 cm3 of extraction solvent E

Answer 13

69.3 cm3 solvent, 29.7 cm3 water, 1.0 cm3 acid (box 1 2);

Question 14

Name two other variables the student should have kept constant during this investigation.

Answer 14

1. Temperature;
2. Agitation/mixing/stirring;
3. Source/age/type of blueberries;
4. Crushing of the blueberries;
5. Rinsing of the blueberries prior to mixing; 6. Concentration of ethanol/acid;

Question 15

Use your knowledge of membrane structure to explain the results in Figure 4.

Answer 15

1. Higher absorbance indicates more anthocyanin OR
   Higher absorbance indicates more membrane damage/permeability
   OR
   (G not zero because) some anthocyanin released when blueberries are crushed
   OR
   (G not zero because) some membrane damage when blueberries are crushed;
2. More membrane damage/permeability results in more anthocyanin release
3. (E and F greater than water because) phospholipids dissolve in ethanol;
4. (E greater than F because) acid denatures membrane proteins;

Question 16

A different student did this investigation. He did not have a colorimeter.
Describe a method this student could use to prepare colour standards and use them
to give data for the total anthocyanin extracted.

Answer 16

1. Use known concentration of blueberry juice/extract
   OR
   Use known concentration of anthocyanin/pigment (solution)
   OR
   Use known concentration of (extraction) solvent to be added to blueberries;
2. Prepare dilution series;
3. Compare (results) with colour standards to give score/value/concentration;

Question 17

Describe the role of DNA polymerase in the semi-conservative replication of DN

Answer 17

1. Joins (adjacent DNA) nucleotides;
2. (Catalyses) condensation (reactions);
3. (Catalyses formation of) phosphodiester bonds (between adjacent nucleotides);

Question 18

It took less time for 25% of cells with cyclin D to be undergoing DNA replication than for 25% of cells without cyclin D.
Use Figure 5 to calculate this time difference as a percentage decrease. Show your working.
[2 marks]

Answer 18

Final answer with 2sf or 3sf in range 31.8 to 34.7%;;
1 mark for
5.5 to 6.1 hours
OR
Final answer with 2sf or 3sf in range 46.6 to 53.0% OR
Correct final answers rounded to more than 3sf OR
Final answer with 2sf or 3sf in range 30.8 to 31.7 or 34.8 to 35.6%.

Question 19

Cyclin D stimulates the phosphorylation of DNA polymerase, which activates the DNA polymerase.
0 5 . 3
Describe how an enzyme can be phosphorylated.

Answer 19

1. Attachment/association of (inorganic) phosphate (to the enzyme);
2. (Released from) hydrolysis of ATP OR
   (Released from) ATP to ADP + Pi;

Question 20

0 5 . 4
Some tumour cells contain higher than normal concentrations of cyclin D.
Use Figure 5 to suggest why higher than normal concentrations of cyclin D could
result in a tumour.

Answer 20

1. Shortens interphase OR
   Cells begin DNA replication earlier OR
   DNA replication (starts) faster;
2. Fast(er) cell cycle/division/multiplication/mitosis OR
   Uncontrolled cell division/mitosis;
3. (Resulting in) a mass/group of abnormal/excessive cells;

Question 21

Explain why death of alveolar epithelium cells reduces gas exchange in human lungs.

Answer 21

1. Reduced surface area;

2. Increased distance for diffusion; 3. Reduced rate of gas exchange;

Question 22

Do the data in Figure 6 show a linear relationship between concentration of particulate matter and percentage of dead cells?
Use suitable calculations to justify your answer.

Answer 22

1. 9 (percent per 5 μg cm–3);

2. 1.42/1.8 (percent per 5 μg cm–3

Question 23

Alpha-gal is a disaccharide found in red meat.
Alpha-gal is made of two galactose molecules. Galactose has the chemical formula C6H12O6
Give the chemical formula for the disaccharide, alpha-gal, and describe how it is
formed from two galactose molecules.

Answer 23

1. C12H22O11;
2. Condensation reaction
   OR
   With a glycosidic bond;

Question 24

Draw a labelled diagram of an antibody and identify the specific alpha-gal binding site.

Answer 24

1. Y shape showing two long and two short (polypeptide) chains correctly positioned;
2. (Alpha-gal) binding site labelled on the end of the branches of the Y of the antibody;
3. Variable region labelled OR
   Constant region labelled
   OR
   Disulfide bridge/bond labelled;

Question 25

Suggest how one antibody can be specific to tick protein and to alpha-gal.

Answer 25

1. (Part of tick protein and alpha-gal) have a similar shape/structure; 2. Antibody is complementary to both (tick protein and alpha-gal) OR Antigen-binding site is complementary to both (tick protein and alpha-gal) OR Antibody can form antigen-antibody complex with both (tick protein and alpha-gal);

Question 26

The scientists’ hypothesis was that an earlier immune response to tick protein causes the allergic reaction. Consider whether Figure 7 supports this hypothesis.

Answer 26

1. Exposure to tick (protein) is followed by increase in antibody (specific to alpha-gal); 2. (Later) greater/faster increase in antibody suggests there are memory cells; 3. Antibody (specific to alpha-gal) increases during/after allergic reaction; 4. During/after allergic reaction, total antibody increases more than alpha-gal antibody; 5. (So) may be other antibodies (that are causing allergic reaction);

Question 27

Complete Table 2 to show three differences between DNA in the nucleus of a plant cell and DNA in a prokaryotic cell.

Answer 27

Plant v prokaryote 1. (Associated with) histones/proteins v no histones/proteins; 2. Linear v circular; 3. No plasmids v plasmids; 4. Introns v no introns; 5. Long(er) v short(er);

Question 28

Scientists investigated the genetic diversity between several species of sweet potato. They studied non-coding multiple repeats of base sequences. Define ‘non-coding base sequences’ and describe where the non-coding multiple repeats are positioned in the genome.

Answer 28

1. DNA that does not code for protein/polypeptides OR DNA that does not code for (sequences of) amino acids OR DNA that does not code for tRNA/rRNA; 2. (Positioned) between genes;

Question 29

Use the information in Table 3 to complete the phylogenetic tree shown in Figure 8. Write the letter that represents the correct species into each box.

Answer 29

Top to bottom C T L R;

Question 30

Use Table 3 to evaluate how this information affects the validity of the phylogenetic tree.

Answer 30

1. (Supported) more similar than with any other species; 2. (Not supported) high (intraspecific) variation in species T (compared with variation between T and C); 3. Small sample OR Only five (individuals);

Question 31

Give your answer as number of stomata per mm2 | Show your working.

Answer 31

171 (per mm2);;

Question 32

Give a null hypothesis for this investigation and name a statistical test that would be appropriate to test your null hypothesis.

Answer 32

1. There is no association/correlation/relationship between the concentration of carbon dioxide and the stomatal density OR The concentration of carbon dioxide does not affect the stomatal density; 2. Correlation coefficient;

Question 33

From 1910 to 2000, the carbon dioxide concentration in the atmosphere increased from 300 parts per million to 365 parts per million. Use Figure 10 to calculate the mean rate of change in stomatal density from 1910 to 2000. Give your answer as number of stomata per mm2 per 10-year period. Show your working

Answer 33

inal answer in range 2. 6 to 2. 7 ;; 1 mark for Stomatal density decrease of 24 to 25

Question 34

A journalist saw Figure 10 and suggested that future increases in atmospheric carbon dioxide concentration could result in less transpiration. Evaluate his suggestion.

Answer 34

1. Increasing carbon dioxide (concentration) shows decreased stomatal density; 2. Fewer stomata means less transpiration OR Fewer stomata means less evaporation (of water from leaves) OR Fewer stomata means less diffusion of water vapour (from leaves); 3. Same (volume of) carbon dioxide can be absorbed for photosynthesis with smaller number of stomata; 4. Don’t know the size of the stomata; 5. Don’t know whether leaf size has changed; 6. Don’t know if this is true for all species (of plant); 7. Don’t know how long the stomata are open for; 8. Don’t know if this trend will continue (beyond the concentrations of carbon dioxide shown in Figure 10); 9. Other factors affect transpiration (rate);

Question 35

Describe how mRNA is formed by transcription in eukaryotes.

Answer 35

1. Hydrogen bonds (between DNA bases) break; 2. (Only) one DNA strand acts as a template; 3. (Free) RNA nucleotides align by complementary base pairing; 4. (In RNA) Uracil base pairs with adenine (on DNA) OR (In RNA) Uracil is used in place of thymine; 5. RNA polymerase joins (adjacent RNA) nucleotides; 6. (By) phosphodiester bonds (between adjacent nucleotides); 7. Pre-mRNA is spliced (to form mRNA) OR Introns are removed (to form mRNA);

Question 36

Describe how a polypeptide is formed by translation of mRNA.

Answer 36

1. (mRNA attaches) to ribosomes OR (mRNA attaches) to rough endoplasmic reticulum; 2. (tRNA) anticodons (bind to) complementary (mRNA) codons; 3. tRNA brings a specific amino acid; 4. Amino acids join by peptide bonds; 5. (Amino acids join together) with the use of ATP; 6. tRNA released (after amino acid joined to polypeptide); 7. The ribosome moves along the mRNA to form the polypeptide;

Question 37

Define ‘gene mutation’ and explain how a gene mutation can have: • no effect on an individual • a positive effect on an individual.

Answer 37

(Definition of gene mutation) 1. Change in the base/nucleotide (sequence of chromosomes/DNA); 2. Results in the formation of new allele; (Has no effect because) 3. Genetic code is degenerate (so amino acid sequence may not change); OR Mutation is in an intron (so amino acid sequence may not change); 4. Does change amino acid but no effect on tertiary structure; 5. (New allele) is recessive so does not influence phenotype; (Has positive effect because) 6. Results in change in polypeptide that positively changes the properties (of the protein) OR Results in change in polypeptide that positively changes a named protein; 7. May result in increased reproductive success OR May result in increased survival (chances);