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OCR A Level Biology

Topic questions Human physiology Respiration

Respiration

10 questions · 88 marks · about 2 hours

Respiration questions

OCR H420 Respiration 10 questions

Unified Biology June 2024, Question 4 7 marks

Respiration allows cells to carry out processes such as DNA replication.

(a)
Figure for question 4
[2]
Reveal mark scheme Mark scheme for question 4(a)
(b)

Outline why cellular respiration is necessary in cells that are carrying out DNA replication.

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

Chemiosmosis is a mechanism that is thought to occur in both respiration and photosynthesis.

(c)(i)

Evidence for chemiosmosis during photosynthesis can be demonstrated by:

• isolating chloroplasts from plant cells

• placing the chloroplasts in an acidic solution, in the dark

• shining light onto the chloroplasts in the solution

• measuring the final pH of the solution.

Explain what is likely to happen to the pH of the solution when light is shone on the chloroplasts.

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

André Jagendorf provided evidence for chemiosmosis in an experiment carried out in the 1960s.

In his experiment, Jagendorf:

• broke open chloroplasts to expose thylakoids, in the dark

• placed the thylakoids into a solution buffered at pH 4 (solution 1)

• after a short time period, transferred some of these thylakoids to a new solution buffered at pH 4 that contained ADP and Pi (solution 2)

• transferred other thylakoids from solution 1 to a new solution buffered at pH 8 that contained

Question 4, shown as an image

ATP production was measured in solution 2 and solution 3.

Explain what you would expect to observe in solution 2 and solution 3 after the transfer of the thylakoids.

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

Unified Biology June 2023, Question 5 9 marks

Also involves Aerobic respiration · Anaerobic respiration

Yeast can respire aerobically or anaerobically.

(a)

A student investigated how the type of respiratory substrate affects the rate of aerobic respiration in yeast.

Four different populations of yeast were provided with fructose, glucose, maltose or sucrose as a respiratory substrate. The student measured the CO2 produced by each population over 150 minutes.

The student’s results are shown in Fig. 5.1.

Figure for question 5
Question 5, shown as an image
Question 5, shown as an image
Question 5, shown as an image
Question 5, shown as an image
(a)(i)

Compare CO2 production when the substrate is glucose and CO2 production when the substrate is maltose.

[2]
Reveal mark scheme Mark scheme for question 5(a)(i)
(a)(ii)

The student wrote a note:

‘The CO2 production with fructose is approximately the same as the CO2 production with sucrose’.

Explain whether you agree with the student’s note.

[2]
Reveal mark scheme Mark scheme for question 5(a)(ii)
(b)

A student plans to compare the rate of anaerobic respiration in two species of yeast.

(b)(i)

The student uses glucose as the respiratory substrate in their experiment.

The student produces a 0.01 mol dm–3 glucose solution for each yeast population. This is the method that they use to produce each 0.01 mol dm–3 glucose solution:

• Make two 10‑fold dilutions from a 1.0 mol dm–3 stock solution.

• In each dilution, use a measuring cylinder to measure the volume of water and a dropping pipette to transfer the glucose solutions.

Suggest two ways in which the student can reduce the percentage error in their measurements when producing the glucose solutions.

[2]
Reveal mark scheme Mark scheme for question 5(b)(i)
(b)(ii)

Complete the sentences using appropriate words or phrases.

This is the method that the student uses in their experiment:

• To avoid contamination, place the yeast in glucose solutions that are produced

• Culture each species of yeast in different flasks.

• Ensure anaerobic conditions by using a........ flask.

• Use the rate of CO2 production as a measure of respiration rate.

• Standardise all other variables.

• Repeat the measurements with 10 populations from each species and calculate

two means and two........

• Analyse the data using a Student’s t ‑test.

[3]
Reveal mark scheme Mark scheme for question 5(b)(ii)

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 14 1 mark

A respirometer, shown below, can be used to investigate respiration.

Figure for question 14

A student placed 5 cm3 of potassium hydroxide solution in the left-hand tube of the respirometer. The student suspended a basket above the liquid and placed 10 g of respiring seeds in the basket.

Which of the following statements about this investigation is not correct?

A The level of the liquid in the left-hand arm of the capillary tube would decrease (move downwards) with time.

B The potassium hydroxide solution absorbs carbon dioxide.

C The rate of movement of the liquid in the capillary tube is a measure of the rate of respiration.

D The tap can be opened to equalise the pressure in each tube.

Question 14, shown as an image
[1]
Reveal mark scheme Mark scheme for question 14

Biological Processes November 2020, Question 11 1 mark

Figure for question 11

The diagram shows a respirometer used to compare respiration in two types of germinating seeds.

Figure for question 11

• The position of the meniscus was noted at the beginning of the experiment.

• The tube was left for 20 minutes.

• The new position of the meniscus was noted.

• The experiment was repeated with the other type of seed.

Which of the following would be necessary to ensure valid results?

1 Keeping the respirometer in the dark during the experiment. 2 Keeping the respirometer at the same temperature during the experiment. 3 Using the same dry mass of seeds each time.

A 1, 2 and 3

B only 1 and 2

C only 2 and 3

D only 1

Your answer

[1]
Reveal mark scheme Mark scheme for question 11

Biological Processes November 2020, Question 12 1 mark

Figure for question 12

The diagram shows a respirometer used to compare respiration in two types of germinating seeds.

Figure for question 12

• The position of the meniscus was noted at the beginning of the experiment.

• The tube was left for 20 minutes.

• The new position of the meniscus was noted.

• The experiment was repeated with the other type of seed.

The student investigated two types of seed, pea and sunflower:

• pea seeds store mainly starch

• sunflower seeds store mainly lipid.

Which of the following, A to D, describes the results you would expect with each type of seed?

A The meniscus would move to the left with pea seeds and further to the left with sunflower seeds.

B The meniscus would move to the left with sunflower seeds and to the right with pea seeds.

C The meniscus would move to the right with pea seeds and further to the right with sunflower seeds.

D The meniscus would not move.

Your answer

[1]
Reveal mark scheme Mark scheme for question 12

Biological Processes November 2020, Question 20 22 marks

Also involves Culturing microorganisms

A student investigated the effect of different sugars on the respiration rate in yeast using the following procedure.

• Prepare a stock solution of yeast containing 10 g of dried yeast in 250 cm3 of pH5 buffer.

• Prepare solutions of each sugar containing 5 g of sugar in 250 cm3 of distilled water.

• Keep the yeast and sugar solutions in a water bath at 35 ºC until required.

• Set up the apparatus with a 250 cm3 conical flask connected by a rubber tube to a 100 cm3 gas syringe.

• Add 25 cm3 of yeast solution and 25 cm3 of sugar solution to the flask, immediately connect the flask to the gas syringe and start the clock.

• Record the volume of gas produced after 30 s and then every 30 s for 7 min.

• Repeat the experiment 5 times for each different sugar.

• Prepare a fresh yeast solution for each set of sugars.

(a)(i)

Describe two precautions the student should take between each experiment to ensure repeatable results.

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

The student’s procedure did not include a negative control.

Describe one negative control that the student could have used.

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

The student calculated mean values for each sugar and plotted the results on a graph.

The graph is shown below.

Figure for question 20
Question 20, shown as an image
(b)(i)

Describe how the student could use the graph to calculate the rate of respiration for each type of sugar.

[3]
Reveal mark scheme Mark scheme for question 20(b)(i)
(b)(ii)

* The student concluded the following from the graph:

• The rates of respiration with glucose, maltose and sucrose were very similar.

• The yeast could not hydrolyse disaccharides.

Evaluate the student’s conclusions.

Additional answer space if required.

[6]
Reveal mark scheme Mark scheme for question 20(b)(ii)
(c)

The student wanted to study the effect of the different sugars on the rate of growth of a yeast population.

They used a colorimeter to measure the absorbance of a culture of yeast cells.

The absorbance of the yeast culture is proportional to the concentration of yeast cells.

As the yeast multiplied, it was necessary to dilute the sample to obtain a reading on the colorimeter.

(c)(i)

Describe how the student could use 1 cm3 pipettes and 10 cm3 measuring cylinders to (i) dilute the sample so that it was 10 000 times less concentrated.

[3]
Reveal mark scheme Mark scheme for question 20(c)(i)
(c)(ii)

A light microscope can be used to observe yeast cells.

State the equipment that would be needed, in addition to a microscope, to measure the average diameter of yeast cells.

[2]
Reveal mark scheme Mark scheme for question 20(c)(ii)
(c)(iii)

The student prepared a starter culture using 2.5 × 10−3 g yeast cells in 1 dm3 nutrient (iii) broth.

The average mass of a yeast cell is 2.0 × 10−11 g.

Calculate the number of cells in the starter culture.

Give your answer in standard form.

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

The population of this yeast doubles every 90 minutes when growing under ideal conditions.

A different starter culture was prepared containing 1 × 107 cells.

Use the axes below to sketch the growth curve you would expect for the yeast culture over the first 15 h.

Figure for question 20
[3]
Reveal mark scheme Mark scheme for question 20(c)(iv)

Biological Processes June 2018, Question 16 11 marks

(a)

Gestational diabetes is a medical condition that affects pregnant women. It results in high levels of glucose in the blood, even though the woman produces normal levels of insulin.

(a)(i)

Gestational diabetes is most similar to which other type of diabetes?

Explain your answer.

[2]
Reveal mark scheme Mark scheme for question 16(a)(i)
(a)(ii)

Suggest two ways a woman with gestational diabetes can manage her condition.

[2]
Reveal mark scheme Mark scheme for question 16(a)(ii)
(b)(i)

Skeletal muscle is one of the main tissues where glucose is removed from the blood in response to insulin.

Name the other tissue.

[1]
Reveal mark scheme Mark scheme for question 16(b)(i)
(b)(ii)

Explain why glucose is required for the contraction of skeletal muscle.

[3]
Reveal mark scheme Mark scheme for question 16(b)(ii)
(c)

During late pregnancy, women find ventilation more difficult, as the developing foetus reduces the volume of the thorax. This can lead to tiredness and difficulty breathing.

A student used a spirometer to measure ventilation in a woman who was 36 weeks pregnant.

Fig. 16.1 shows the trace produced.

Figure for question 16

Mean oxygen uptake rate at rest in women is around 0.020 dm3 s–1.

Using these data, the student made the following conclusion:

My data show that being pregnant reduces rate of oxygen uptake by up to 20%.

Evaluate this claim, using the data in Fig. 16.1.

[3]
Reveal mark scheme Mark scheme for question 16(c)

Biological Processes June 2018, Question 22 8 marks

(a)

A scientist used a respirometer to investigate the rate of respiration and photosynthesis of maize in different light intensities.

• The scientist placed ten maize seedlings in a respirometer and kept it in the dark for three hours.

• The respirometer contained soda-lime to remove any CO2 produced by the seedlings.

• The scientist placed ten maize seedlings in a separate respirometer without soda-lime and placed it in different light intensities for three hours at a time.

Figure for question 22
(a)(i)

The diameter of the capillary tubing was 0.1 mm.

The volume of a cylinder can be calculated using the following formula:

Question 22, shown as an image

Calculate the rate of oxygen uptake by the seedlings in the dark. Give your answer to two significant figures. Show your working.

Answer =........ mm3 h–1

[3]
Reveal mark scheme Mark scheme for question 22(a)(i)
(a)(ii)

1700 lux is a typical light intensity on a cloudy day in the UK. Calculate the percentage increase in gas production between 1700 and 2000 lux. Show your working.

[2]
Reveal mark scheme Mark scheme for question 22(a)(ii)
(a)(iii)

Suggest why soda-lime was not placed in the respirometer with the seedlings grown in the light.

[1]
Reveal mark scheme Mark scheme for question 22(a)(iii)
(b)

The scientist made the following claim:

These results suggest that, in maize seedlings, the rate of photosynthesis only exceeds the rate of respiration when the light intensity is above 1510 lux.

Use the data in Table 22.1 to explain why the scientist made this claim.

[2]
Reveal mark scheme Mark scheme for question 22(b)

Unified Biology June 2017, Question 2 15 marks

Bread contains a mixture of polypeptides known as gluten.

Gluten consists of two types of polypeptide: gliadins and glutenins.

(a)
(a)(i)

The table below contains statements about the structures of gluten polypeptides.

In the boxes next to each statement, write the level of protein structure (primary, secondary, tertiary, or quaternary) to which the statement refers.

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

Coeliac disease is caused by an immune reaction to gliadins in a person’s digestive system. The immune system produces antibodies that bind to part of the gliadin polypeptides, which causes inflammation.

Some people who stop eating foods that contain gluten still occasionally experience the symptoms of coeliac disease.

What can you conclude about:

• the structure of the antibody that causes coeliac disease; and

• what the antibody binds to when producing the symptoms of coeliac disease?

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

Gluten helps to trap carbon dioxide within bread dough. This enables bread to rise when it is baked.

The carbon dioxide is produced by baker’s yeast, Saccharomyces cerevisiae. This species of yeast is able to convert ethanol to acetyl CoA at low glucose concentrations.

Fig. 2 shows the oxygen consumption and carbon dioxide production of a population of S. cerevisiae grown in batch culture. The population was provided with glucose as their only initial source of carbon.

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

Fig. 2

(b)(i)

Suggest and explain what conclusions can be drawn from Fig. 2 about the factors that affected the rate and type of respiration carried out by S. cerevisiae in this batch culture.

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

Describe two practical considerations to ensure the S. cerevisiae population grows successfully when the initial culture is established.

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

A group of students were designing an experiment to investigate the effect of temperature on the respiration rate of S. cerevisiae.

Their planned method included the following:

• S. cerevisiae yeast suspension will be divided into six equal volumes to form the experimental groups.

• Six temperatures will be tested: 15 °C, 20 °C, 25 °C, 30 °C, 35 °C and 40 °C.

• Beakers of S. cerevisiae will be placed in water baths to control the temperature.

• Respiration rate will be measured by using a pH probe to monitor changes in the pH of the suspensions.

• The experiment will be repeated four times.

Evaluate whether the students’ method is likely to produce valid results.

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

The students used a Student’s t-test to compare the results at 30 °C and 35 °C.

They calculated a t value of 2.200.

The critical value for p = 0.05 is 2.306.

Assuming their final method was valid, what can the students conclude from the result of the t-test?

[1]
Reveal mark scheme Mark scheme for question 2(b)(v)

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