AQA A-Level Biology: Paper 1 2021

Question 1

1.1 Describe the induced fit model of enzyme action and how an enzyme acts as a catalyst.

Answer 1

• The substrate binds to the active site and forms an ES complex.
• The active site changes shape slightly so it’s complimentary to the substrate.
• This reduces the activation energy.

Question 2

1.2 Tick one box to show which are the substrate the scientists must add to the reaction mixture is to produce ATP.

Answer 2

• Adenosine diphosphate.

Question 3

1.3 Suggest and explain a procedure the scientist could have used to stop each reaction.

Answer 3

• The nature of the enzyme by putting it in ice.

Question 4

1.4 Explain the change in ATP concentration with increasing inorganic phosphate concentration

Answer 4

• With increasing phosphate concentration more ES complex is form.
• And at 40 mol dm3, all active sites are occupied.

Question 5

2.1 Explain the advantage for large animals of having a specialised system that facilitates oxygen uptake.

Answer 5

• Large organisms have a smaller surface area to volume ratio.
• This means they overcome a long diffusion pathway.

Question 6

2.2 Suggest how the environmental conditions have resulted in adaptations of systems using model A rather than model B.

Answer 6

Water is more dense than air.

- So it supports the gills.

Question 7

2.3 A student studies figure 3 and concluded that the fish gas exchange system is more efficient than the human gas exchange system. Use figure 3 to justify this conclusion.

Answer 7

• The difference of oxygen concentration between the artery and vein is larger in the fish than the human.
• So the fish remove a greater amount of oxygen that they intake.

Question 8

2.4 Explain how the countercurrents principle allows sufficient oxygen uptake in the fish gas exchange system.

Answer 8

• Blood and water continuously flow in opposite directions.

- So the concentration gradient is maintained along the length of the lamellae

Question 9

3.1 Describe how one amino acid is added to a polypeptide that is being formed at a ribosome during translation.

Answer 9

• TRNA brings the specific amino acid to the ribosome.
• The anticodon on the tRNA binds to the codon on the mRNA.
• The amino acids join to form a peptide bond using ATP.

Question 10

3.2 Use information in table to to suggest why this amino acid replacement changes the properties of crystallin.

Answer 10

• The hydrogen bonds form instead of the ionic bonds.

- And it changes the tertiary structure of the crystallin.

Question 11

3.3 MRNA codon for the non-mutant triplets.
mutated mRNA codon.
Mutated DNA triplet

Answer 11

• AGG
• GGG
• CCC

Question 12

4.1 Suggest two ways the student could improve the quality of his scientific drawing of the blood vessels in this dissection

Answer 12

• Add labels.

- Don’t use shading

Question 13

4.2 Describe one feature that allows you to identify the blood vessels.

Blood vessel X
Blood vessel Y
Feature

Answer 13

• Artery
• Vein
• They have different wall thicknesses.

Question 14

4.3 Describe two precautions the student should take when clearing away after the dissection.

Answer 14

• Disinfect all instruments.

- Disinfect hands

Question 15

5.1 Describe how a sample of chloroplasts could be isolated from leaves

Answer 15

• Break open the cells and filter it.
• Keep it in the cold to ensure it has the same water potential.
• Keep it in a controlled pH.
• Centrifuge it to remove cell debris.
• And centrifuge at increase in speeds so the chloroplast can settle out.

Question 16

5.2 Give one feature of the chloroplast that allows proteins to be synthesised inside the chloroplast and describe one difference between this feature in the chloroplast and similar features in the rest of the cell.

Answer 16

• Feature: Ribosomes.

- Difference: In chloroplasts there are 70s ribosomes. In the cytoplasm there are 80s ribosomes.

Question 17

5.4 Use figure 6 to suggest why iron deficient plants have reduced growth rates.

Answer 17

• There are less thylakoids membranes so there is a smaller surface area of chloroplasts.
• As there is less chlorophyll less light is absorbed so that are slower rates of photosynthesis.

Question 18

6.2 Describe the role of the spindle fibres and the behaviour of the chromosome during each of these phases.

Answer 18

• C: Prophase AND
• D: Metaphase AND
• E: Anaphase.
• In prophase the chromosomes condense.
• The centromeres attach to the spindle fibres.
• In metaphase the chromosomes lineup at the equator of the cell.
• In anaphase the centromere divides.

Question 19

7.1 Use your knowledge of phagocytosis to describe how an ADC enters and kills the tumour cell.

Answer 19

• The cell engulfs the antibody.
• The lysome fuses with the vesicle containing ADC.
• The lysosomes breakdown the antibody to release the drug.

Question 20

7.2 Some of the antigens found on the surface of tin cells are also found on the surface of healthy human cells. Use this information to explain why treatments with ADC often causes side-effects.

Answer 20

• ADC will bind to healthy cells.

- Cause a decrease in them.

Question 21

7.4 Suggest one reason why there are no data for group G and group H after day eight.

Answer 21

Because the mice died.

Question 22

7.5 Suggest and explain to file investigations that should be done before this ADC is tested on human breast cancer patients.

Answer 22

• Use other animals to check for side-effects for example a rat.
• Use healthy humans check for side-effects.

Question 23

8.1 describe how a triglyceride molecule is formed

Answer 23

• Condensation reaction that removes 3 water molecules.
• And joins 3 fatty acids and one glycerol.
• Joined by ester bonds

Question 24

8.2 put a tick in one box that contains the correct information about one of these fatty acids.

Answer 24

• Palmitoleic acid is an unsaturated fatty acid represented by the diagram K.

Question 25

8.3 suggest one reason why they use the data logger and explain why this was important in this investigation.

Answer 25

• Increases accuracy.

- Reduces chance of human error.

Question 26

8.4 Use this information on table 6 to show how each population is best adapted for its natural environment when compared with other population.

Answer 26

ROOT LENGTH

• POP 1 grew longer roots in warm temp.
• POP 2 grew longer roots in cool temp.

STANDARD DEVIATION
- Standard deviation doesn’t overlap so the difference in mean isn’t likely to be due to chance.

FATTY ACIDS

• POP 1 is better adapted to warm weather because it has more SATURATED fatty acids so more every is available = more growth.
• POP 2 is better adapted to cool weather because it has more UNSATURATED fatty acids so there’s more lipase activity = more growth.

Question 27

8.5 Explain why they are both given this name - Helianthus annuus.

Answer 27

• Because they are from the same species.

Question 28

9.3 Give the letter and name of the molecule supported and explain why the results do not support other molecules.

Answer 28

• Model: Q
• Name: Semi-conservative DNA replication.
• Explanation for first UNSUPPORTED model: P. Only has one peak is shown in generation 1.
• Explanation for second UNSUPPORTED model: R. During generation 3, there should be an overlapping peak.

Question 29

10.1 Describe the structure of DNA.

Answer 29

• Polymer of nucleotides.
• Formed from deoxyribose sugar, a phosphate group and nitrogenous base.
• The nucleotides are joined by phophodiester bonds.
• Double helix strands are held together by hydrogen bonds.
• Between Adenine + Thymine and Cytosine + Guanine.

Question 30

10.2 Name and describe five ways substances can move across the cell surface membrane into a cell.

Answer 30

• DIFFUSION of small molecules down a conc gradient.
• FACILITATED DIFFUSION of molecules down a conc gradient using protein carriers.
• OSMOSIS of water molecules down a water potential gradient.
• ACTIVE TRANSPORT of molecules AGAINST a conc gradient via protein carriers using ATP.
• CO-TRANSPORT of two different substances using carrier proteins.

Question 31

10.3 Contrast the structure of the two cells visible in the electron micrographs shown in figure 14.

Answer 31

• Cell A is bigger than cell B.
• Cell A has LARGER ribosomes.
• Cell A has a nuclei whereas cell B had free floating DNA.
• Cell A had mitochondria, cell B doesn’t.
• Cell B has a capsule, cell A doesn’t.