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OCR A Level Biology

Topic questions Cell biology Factors affecting enzyme activity

Factors affecting enzyme activity

15 questions · 126 marks · about 3 hours

Factors affecting enzyme activity questions

OCR H420 Factors affecting enzyme activity 15 questions

Biological Processes June 2025, Question 17 15 marks

(a)

The diagram shows some of the steps in glycolysis.

Figure for question 17
Question 17, shown as an image
Figure for question 17

The enzyme phosphofructokinase (PFK) catalyses the conversion of fructose phosphate to fructose 1,6-bisphosphate (hexose bisphosphate).

(a)(i)

State the type of reaction that is catalysed by PFK.

[1]
Reveal mark scheme Mark scheme for question 17(a)(i)
(a)(ii)

The intermediate compound PEP is a reversible non-competitive inhibitor of PFK.

Explain how PEP inhibits PFK.

[3]
Reveal mark scheme Mark scheme for question 17(a)(ii)
(a)(iii)

ATP is also an inhibitor of PFK.

Suggest and explain why ATP cannot be a competitive inhibitor of PFK.

[2]
Reveal mark scheme Mark scheme for question 17(a)(iii)
(a)(iv)

Inhibition of PFK by ATP and PEP is an example of a negative feedback control mechanism.

State the name of this type of inhibition in metabolic pathways such as glycolysis.

[1]
Reveal mark scheme Mark scheme for question 17(a)(iv)
(a)(v)

ATP is converted to ADP and then AMP (adenosine monophosphate) when ATP provides energy for processes in the cell.

AMP reverses the inhibition of PFK by ATP.

Use this information to suggest how PFK acts as a key enzyme in the regulation of the rate of glycolysis.

[2]
Reveal mark scheme Mark scheme for question 17(a)(v)
(b)

Complete the following sentence.

Some inhibitors form strong covalent bonds with enzymes and are known as

[1]
Reveal mark scheme Mark scheme for question 17(b)
(c)

The chemical equation for aerobic respiration with glucose as substrate is:

Question 17, shown as an image
(c)(i)

State the name of the type of reaction in aerobic respiration that produces CO2.

[1]
Reveal mark scheme Mark scheme for question 17(c)(i)
(c)(ii)

State the names of the two stages in respiration where CO2 production occurs.

[2]
Reveal mark scheme Mark scheme for question 17(c)(ii)
(c)(iii)

Respiration of glucose involves a series of oxidation reactions.

Explain how oxidation occurs in glycolysis.

[2]
Reveal mark scheme Mark scheme for question 17(c)(iii)

Biological Processes June 2025, Question 19 14 marks

(a)

A student investigates the effect of light intensity on the rate of photosynthesis in algae that had been immobilised in calcium alginate to make algal beads.

The student places the algal beads in small screw-cap bottles containing hydrogencarbonate indicator.

The indicator turns from orange to red and then purple as CO2 is removed from the solution.

Figure for question 19

The student uses the time taken for the indicator to turn purple as a measure of the rate of photosynthesis.

(a)(i)

State two variables that the student should keep constant during this experiment.

[2]
Reveal mark scheme Mark scheme for question 19(a)(i)
(a)(ii)

The student places the bottles at different distances from a light source.

There are four replicates at each distance.

The student’s results are shown in the table.

DistanceRelativeTime taken for indicator to turn purple (min)
(D) from light source (cm)light intensity (1/D2)Replicate 1Replicate 2Replicate 3Replicate 4Mean
2501.60 × 10–56563666765.3
37577808278
5004.00 × 10–6103991019599.5
10001.00 × 10–6185190130195190.0

The student calculates relative light intensity using the following formula:

Question 19, shown as an image

Calculate the relative light intensity and mean for the 375 cm distance.

[2]
Reveal mark scheme Mark scheme for question 19(a)(ii)
(a)(iii)

Outline an improvement to the method that the student could use to get a more accurate measure of the rate of photosynthesis.

[2]
Reveal mark scheme Mark scheme for question 19(a)(iii)
(b)

The student wants to use the algal beads method to investigate the effect of CO2 concentration on the rate of photosynthesis.

(b)(i)

Explain why it would not be possible to use the hydrogencarbonate indicator method to investigate the effect of CO2 concentration.

[2]
Reveal mark scheme Mark scheme for question 19(b)(i)
(b)(ii)

Suggest how the student could modify the method so the effect of CO2 concentration on the rate of photosynthesis could be measured.

[2]
Reveal mark scheme Mark scheme for question 19(b)(ii)
(c)

The graph shows the effect of light intensity on the rate of photosynthesis at two temperatures

Figure for question 19
(c)(i)

Explain the evidence from the graph that shows that CO2 concentration is a limiting factor in these experiments.

[2]
Reveal mark scheme Mark scheme for question 19(c)(i)
(c)(ii)

A student looked at the graph and concluded that CO2 concentration has a greater effect on the rate of photosynthesis than temperature.

Use your knowledge of Q10 and a calculation to support the student’s conclusion.

[2]
Reveal mark scheme Mark scheme for question 19(c)(ii)

Biological Processes June 2023, Question 5 1 mark

Increasing the pH of an enzyme solution from 7.4 to 8.0 causes the rate of the reaction to fall by 90%. When the pH of the solution returns to 7.4 the rate of reaction returns to its original value.

Which of the statements about this enzyme-controlled reaction explains these observations?

A Increasing the pH from 7.4 to 8.0 breaks hydrogen bonds but returning the pH to 7.4 allows them to re-form.

B Increasing the pH from 7.4 to 8.0 changes the shape of the active site and so the substrate binds more tightly.

C Increasing the pH from 7.4 to 8.0 denatures the enzyme.

D Reducing the pH below 7.4 will reduce the rate of reaction.

Your answer

[1]
Reveal mark scheme Mark scheme for question 5

Biological Diversity June 2023, Question 17 17 marks

Figure for question 17
(a)(i)

State the independent variable in this investigation.

[1]
Reveal mark scheme Mark scheme for question 17(a)(i)
(a)(ii)

Describe how a student could produce an 8 a.u. solution from a 20 a.u. stock solution.

[2]
Reveal mark scheme Mark scheme for question 17(a)(ii)
(a)(iii)

Liver tissue contains a higher concentration of catalase than potato.

Suggest two reasons why the students chose potato rather than liver as a source of catalase.

[2]
Reveal mark scheme Mark scheme for question 17(a)(iii)
(a)(iv)

The potato cylinders were cut to equal lengths.

Suggest a further precaution the students should have taken when preparing them, to ensure the investigation was valid.

[1]
Reveal mark scheme Mark scheme for question 17(a)(iv)
(b)
Figure for question 17
(b)(i)

Calculate the standard deviation for the 60 s result at 20 a.u. hydrogen peroxide.

Question 17, shown as an image
[2]
Reveal mark scheme Mark scheme for question 17(b)(i)
(b)(ii)

Use the standard deviations to discuss the repeatability of the students’ results.

[2]
Reveal mark scheme Mark scheme for question 17(b)(ii)
(b)(iii)

The students calculated the initial rate of reaction over the first 30 s.

Calculate the rate of reaction over the first 30 s for the result at 2 a.u. hydrogen peroxide.

[2]
Reveal mark scheme Mark scheme for question 17(b)(iii)
(b)(iv)

When analysing the results, the students assumed that the volume of oxygen collected was the same as the volume of oxygen produced from breakdown of hydrogen peroxide.

Suggest two reasons why the value the students recorded for volume of oxygen collected might not accurately reflect the volume of oxygen produced.

[2]
Reveal mark scheme Mark scheme for question 17(b)(iv)
(c)

Fig. 17.2 shows a graph of the students’ results.

Fig. 17.2

Figure for question 17

Explain the results for 20 a.u. hydrogen peroxide.

[3]
Reveal mark scheme Mark scheme for question 17(c)

Unified Biology June 2023, Question 2 14 marks

Vitamins are molecules that are consumed in the diet of animals and have essential roles in the body.

(a)

Thin layer chromatography (TLC) was used to separate a mixture of vitamins from a vitamin supplement tablet.

(a)(i)

Most vitamins are colourless.

Suggest one method for visualising the vitamins in TLC.

[1]
Reveal mark scheme Mark scheme for question 2(a)(i)
(a)(ii)

The results of the TLC are shown in the chromatogram in Fig. 2.1.

Figure for question 2

Use Fig. 2.1 to calculate the Rf value of vitamin B2.

[2]
Reveal mark scheme Mark scheme for question 2(a)(ii)
(b)

Vitamin B9 is needed for the synthesis of molecule D, shown in Fig. 2.2.

Fig. 2.2

Figure for question 2

Molecule D is a component in the structure of DNA.

A deficiency of vitamin B9 in the diet can cause interphase to stop in some cells.

Use Fig. 2.2 to explain why a deficiency of vitamin B9 can cause interphase to stop.

[2]
Reveal mark scheme Mark scheme for question 2(b)
(c)

Vitamin B2 is used by the body to make FAD. Vitamin B3 is used by the body to make NAD.

(c)(i)

A mitochondrion is shown in Fig. 2.3.

Label Fig. 2.3 to show the locations where FAD gains H atoms and where FAD releases H atoms during respiration.

• Use the letter E to show where FAD gains H atoms.

• Use the letter F to show where FAD releases H atoms.

Figure for question 2
[2]
Reveal mark scheme Mark scheme for question 2(c)(i)
(c)(ii)

The table lists features that are correct for FAD, NAD, both or neither.

Complete the table by adding a tick (✓) to a box where the feature is correct and a cross (✗) where the feature is incorrect. You should add either a tick or a cross to every box in the table.

Figure for question 2
[2]
Reveal mark scheme Mark scheme for question 2(c)(ii)
(c)(iii)

ATP is produced in respiration. Some of these molecules of ATP are used to release energy for processes in respiration. For example, ATP is used in the early stages of glycolysis.

Suggest one other way in which ATP contributes to the process of respiration.

[1]
Reveal mark scheme Mark scheme for question 2(c)(iii)
(d)

Vitamin C affects the activity of enzymes in different ways.

(d)(i)

Vitamin C acts as a coenzyme for several enzymes in the synthesis of collagen.

Outline the role of coenzymes in biological reactions.

[2]
Reveal mark scheme Mark scheme for question 2(d)(i)
(d)(ii)

A scientist investigates the effect of vitamin C on the activity of the enzyme adenylyl cyclase.

Adenylyl cyclase catalyses the conversion of ATP to cAMP.

The scientist measures the activity of adenylyl cyclase without vitamin C and with 1 mmol dm–3 of vitamin C.

The scientist’s results are shown in Fig. 2.4.

Figure for question 2
Question 2, shown as an image
Question 2, shown as an image

Explain what can be concluded from the results in Fig. 2.4 about the effect of vitamin C on the activity of adenylyl cyclase.

[2]
Reveal mark scheme Mark scheme for question 2(d)(ii)

Biological Processes June 2022, Question 4 1 mark

The graph below shows the effect of changing substrate concentration on the rate of an enzyme-controlled reaction.

Figure for question 4
Question 4, shown as an image

Which graph shows how increasing the concentration of enzyme would affect the rate of this

Figure for question 4
Question 4, shown as an image
Figure for question 4
Question 4, shown as an image

Your answer

[1]
Reveal mark scheme Mark scheme for question 4

Biological Processes June 2022, Question 5 1 mark

Which statement about the effect of temperature on enzyme-controlled reactions is not correct?

A At temperatures above 40 °C the rate of all enzyme-controlled reactions decreases.

B Increasing the temperature above the optimum changes the tertiary structure of the active site.

C Increasing the temperature increases the kinetic energy of the enzyme and substrate.

D Increasing the temperature increases the probability of enzyme-substrate collisions.

Your answer

[1]
Reveal mark scheme Mark scheme for question 5

Biological Processes June 2022, Question 20 13 marks

(a)

Oplegnathus insignis is a fish species that lives in salt water. It may be suitable for use in fish farming.

Scientists investigated the growth of O. insignis at different temperatures to help decide on the optimum temperature for farming.

Fig. 20.1 shows a diagram of the respirometer the scientists used.

The diagram shows a single fish, but the respirometer is large enough to hold many.

Figure for question 20
(a)(i)

Suggest two modifications the researcher would need to make to this apparatus to investigate the effect of temperature on the rate of respiration.

[2]
Reveal mark scheme Mark scheme for question 20(a)(i)
(a)(ii)

Suggest one advantage of using a computer to collect data.

[1]
Reveal mark scheme Mark scheme for question 20(a)(ii)
(a)(iii)

A constant flow of water past the oxygen probe allows continuous monitoring of oxygen concentration.

Explain one other reason why it is necessary to maintain a constant flow of water in the respirometer.

[2]
Reveal mark scheme Mark scheme for question 20(a)(iii)
(a)(iv)

Suggest and explain one other precaution that the scientists should take to ensure that the results obtained using this apparatus are valid.

[2]
Reveal mark scheme Mark scheme for question 20(a)(iv)
(b)

The scientists studied the effect of living at different temperatures on respiration in young fish at different stages of growth.

High oxygen consumption in fish is associated with a fast growth rate.

Their results are shown in Fig. 20.2.

Figure for question 20
(b)(i)

Calculate the Q10 for respiration in the largest fish.

Give your answer to 3 significant figures.

[2]
Reveal mark scheme Mark scheme for question 20(b)(i)
(b)(ii)

The scientists concluded that the best temperature for farming of O. insignis was 18 °C.

Evaluate their conclusion.

[4]
Reveal mark scheme Mark scheme for question 20(b)(ii)

Biological Processes November 2021, Question 5 1 mark

The figure shows the metabolic pathways leading to synthesis of five amino acids: aspartate, lysine, methionine, threonine and isoleucine.

E1 to E8 represent different enzymes involved in these pathways.

Figure for question 5

aspartate semialdehyde fraction homoserine fraction threonine

Figure for question 5
Question 5, shown as an image
Figure for question 5

Which of the options is an example of end-product inhibition?

A aspartate inhibits E3 and E4

B isoleucine inhibits E7

C lysine inhibits E1 and E6

D methionine inhibits E8

Question 5, shown as an image
[1]
Reveal mark scheme Mark scheme for question 5

Biological Processes November 2021, Question 18 8 marks

The liver is an organ with a wide variety of functions.

(a)

Fig. 18.1, in the insert, shows a light micrograph of a section of normal human liver.

(a)(i)

Identify the structures labelled N and O in Fig. 18.1.

[2]
Reveal mark scheme Mark scheme for question 18(a)(i)
(a)(ii)

The walls of the structure labelled M in Fig. 18.1 are lined with cells.

Suggest and explain how these cells are adapted to carry out their function.

[2]
Reveal mark scheme Mark scheme for question 18(a)(ii)
(b)

Liver is a rich source of the enzyme catalase.

The table below describes examples of three terms from the list below.

coenzyme competitive inhibition enzyme-product complex

induced-fit lock and key non-competitive inhibition

prosthetic group secondary structure tertiary structure

Complete the table by choosing a term from the list that matches each example.

Figure for question 18
[3]
Reveal mark scheme Mark scheme for question 18(b)
(c)

Fig. 18.2 shows the effect of concentration of hydrogen peroxide on the rate of reaction of catalase.

Figure for question 18

Cyanide ions inhibit catalase, even at high concentrations of hydrogen peroxide.

Sketch a line on Fig. 18.2 to show the effect of repeating the experiment in the presence of cyanide ions.

[Answer on Fig. 18.2]

[1]
Reveal mark scheme Mark scheme for question 18(c)

Biological Diversity June 2019, Question 15 1 mark

The image below shows a diagram of an enzyme.

Figure for question 15

Which of the following could be a competitive inhibitor of this enzyme?

Figure for question 15

Your answer

[1]
Reveal mark scheme Mark scheme for question 15

Unified Biology June 2019, Question 3 12 marks

Icefish live in very cold water.

Icefish contain biological molecules that allow them to tolerate cold temperatures.

(a)

Adaptations can be grouped into three general categories.

Which category of adaptation is represented by cold-tolerant molecules?

[1]
Reveal mark scheme Mark scheme for question 3(a)
(b)
Figure for question 3
[2]
Reveal mark scheme Mark scheme for question 3(b)
(c)
Figure for question 3
(c)(i)

Suggest and explain two improvements that would increase the validity of the students’ investigation.

[4]
Reveal mark scheme Mark scheme for question 3(c)(i)
(c)(ii)

Suggest appropriate units to use to represent the rate of enzyme activity in this investigation.

[1]
Reveal mark scheme Mark scheme for question 3(c)(ii)
(c)(iii)

The students recorded the temperature that produced the fastest reaction rate in each of the five replicates. These results are shown in Table 3.

Figure for question 3

One of the students made the following statement:

I think the mean is a more accurate measure than the median or mode for these results.

Evaluate the student’s statement.

[2]
Reveal mark scheme Mark scheme for question 3(c)(iii)
(c)(iv)

The students wanted to know whether there was a difference between the reaction rates of the two forms of trypsin at 30 °C.

They performed a statistical test on the mean of the five replicates for human trypsin and the five replicates for icefish trypsin.

Suggest the most appropriate statistical test for the students to use and explain why this test is appropriate.

[2]
Reveal mark scheme Mark scheme for question 3(c)(iv)

Biological Processes June 2018, Question 5 1 mark

The following graph shows the rate of reaction of an enzyme in different substrate concentrations.

Figure for question 5

Which letter, A to D, shows the rate of reaction with a fixed quantity of competitive inhibitor?

Your answer

[1]
Reveal mark scheme Mark scheme for question 5

Unified Biology June 2018, Question 3 14 marks

Temperature and light intensity are two factors that affect the rate of photosynthesis.

A student investigated how temperature and light intensity affected the rate of photosynthesis in the aquatic plant Elodea canadensis. The rate of photosynthesis was measured by counting the number of bubbles produced by the plant per minute.

The student’s results are shown in Table 3.

Figure for question 3
(a)(i)

Identify the anomalous result in Table 3 and explain how this result could be confirmed as an anomaly.

[2]
Reveal mark scheme Mark scheme for question 3(a)(i)
(a)(ii)

* Describe how the student could improve their experimental method and the presentation of their data.

[6]
Reveal mark scheme Mark scheme for question 3(a)(ii)
(b)

Photosynthesis occurs in two stages: the light-dependent stage and the light-independent stage. The light-independent stage is affected by temperature more than the light-dependent stage.

Explain why temperature has a greater effect on the rate of the light-independent stage.

[2]
Reveal mark scheme Mark scheme for question 3(b)
(c)

Scientists are able to clone desirable plants that show a high rate of photosynthesis. The following passage describes how plants are cloned.

Complete the passage using the most appropriate words or phrases.

Cells are removed from the meristem tissue in axial buds or........

tips. The tissue sample that is removed is called the........ Ethanol can

be used to........ the plant tissue. Hormones are used to stimulate mitosis,

which produces a mass of cells called a........

[4]
Reveal mark scheme Mark scheme for question 3(c)

Unified Biology June 2017, Question 4 13 marks

Nitrogen cycling within ecosystems is controlled by various bacterial species. The table below lists four groups of bacterium that are involved in the nitrogen cycle.

(a)

Complete the table to show the locations of each type of bacterium in the cycle and the reactions they perform.

Type of bacteriaLocationReactant(s)ProductOxidation or reduction of nitrogen?
RhizobiumN2 and H+ ionsNH3reduction
Nitrosomonassoiloxidation
NitrobactersoilNO3–
Denitrifiying bacteriaNO3–
[4]
Reveal mark scheme Mark scheme for question 4(a)
(b)

Nitrogen fixation is an important part of the nitrogen cycle.

The rate of nitrogen fixation is reduced by the presence of oxygen.

Rhizobium uses the enzyme nitrogenase to fix atmospheric nitrogen.

Fig. 4 shows a simplified representation of the structure of nitrogenase and the reaction that it

Question 4, shown as an image
Figure for question 4

Overall reaction: N2 + 8H+ + 8e– + 16ATP fraction 2NH3 + H2 + 16ADP + 16Pi

Fig. 4

(b)(i)

What can you conclude from Fig. 4 about the molecules or ions that affect the functioning of the nitrogenase enzyme?

[4]
Reveal mark scheme Mark scheme for question 4(b)(i)
(b)(ii)

Leghaemoglobin is a molecule that improves the performance of nitrogenase. It has very similar properties to mammalian haemoglobin.

Suggest two ways in which leghaemoglobin improves the performance of the nitrogenase enzyme.

[2]
Reveal mark scheme Mark scheme for question 4(b)(ii)
(c)

Many species of bacteria act as decomposers within ecosystems by breaking down organic material.

Scientists analysed the energy flow within a grassland ecosystem.

They estimated that the energy in the decomposers’ trophic level was 950 000 J m–2 yr–1.

The energy within the producers’ trophic level was 800% greater than that of the decomposers.

(c)(i)

Calculate the energy in the producers’ trophic level in kJ m–2 yr–1.

Answer:........ kJ m–2 yr–1

[2]
Reveal mark scheme Mark scheme for question 4(c)(i)
(c)(ii)

Calculate the percentage efficiency of the energy transfer from producers to decomposers.

Give your answer to two significant figures.

[1]
Reveal mark scheme Mark scheme for question 4(c)(ii)

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