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Edexcel IGCSE Biology

Topic questions Cell biology Osmosis

Osmosis

14 questions · 169 marks · about 3 hours

Osmosis questions

Edexcel 4BI1 Osmosis 14 questions

Biology Paper 1 June 2025 (R), Question 6 14 marks

A student uses this apparatus to investigate osmosis in a non-living system.

Figure for question 6

This is the student’s method.

• tie partially permeable tubing at one end with string to make a bag

• fill the bag with sugar solution

• tie the bag at the other end with string

• measure the mass of the bag, string, and sugar solution

• place the bag into a test tube filled with pure water

• after 5 minutes remove the bag and measure the mass

• return the bag to the test tube

• repeat the measurement of mass every 5 minutes for 30 minutes in total

The table shows the student’s results.

Figure for question 6
(a)
(a)(i)

Draw a line graph to show the student’s results. Join your points with straight lines.

[5]
Reveal mark scheme Mark scheme for question 6(a)(i)
(a)(ii)

Describe the change in mass of the bag during the 30 minutes.

[2]
Reveal mark scheme Mark scheme for question 6(a)(ii)
(a)(iii)

Explain the change in mass of the bag during the 30 minutes.

[2]
Reveal mark scheme Mark scheme for question 6(a)(iii)
(b)

The student decides to investigate how increasing the temperature would affect the results.

(b)(i)

Describe how the student could repeat the experiment safely at a higher temperature.

[2]
Reveal mark scheme Mark scheme for question 6(b)(i)
(b)(ii)

Explain how repeating the experiment at a higher temperature would affect the results.

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

Biology Paper 1 June 2024 (R), Question 1 7 marks

The image shows some cells from onion epidermis.

Figure for question 1

(Source: © TED KINSMAN / SCIENCE PHOTO LIBRARY)

(a)
(a)(i)

Some structures in these onion cells are also found in human red blood cells. Which row in the table is correct for human red blood cells?

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

The onion cells in the image are all epidermis cells. What is the term for a collection of cells of the same type?

Question 1, shown as an image
[1]
Reveal mark scheme Mark scheme for question 1(a)(ii)
(b)

The images show samples of red onion epidermis cells before and after being placed into a 5% sucrose solution for one hour.

Figure for question 1

(Source: © J.C. REVY, ISM / SCIENCE PHOTO LIBRARY) Before

Figure for question 1

(Source: © J.C. REVY, ISM / SCIENCE PHOTO LIBRARY) After

(b)(i)

Describe how to make 10 cm3 of a 5% sucrose solution from a 10% sucrose solution and pure water.

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

Explain the change in appearance of the onion cells shown in the images after being in the 5% sucrose solution for one hour.

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

Biology Paper 2 June 2024, Question 1 18 marks

Also involves Natural selection · Photosynthesis · Extinction

Figure for question 1
(a)

Megalodon was an animal.

Which of these features does megalodon share with the fungus yeast?

1 nuclei present in cells

2 nervous coordination

3 stores carbohydrate as glycogen

Question 1, shown as an image
[1]
Reveal mark scheme Mark scheme for question 1(a)
(b)

Scientists think that the giant, predatory toothed whales that replaced the extinct megalodon consumed the same sources of food.

Explain how the evolution of giant, predatory toothed whales may have caused the extinction of megalodon. (lines 5 to 7)

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

Volcanoes release sulfur dioxide into the atmosphere.

Describe the biological consequences of sulfur dioxide release. (lines 10 to 12)

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

The release of large amounts of dust into the atmosphere reduced light intensity.

Explain why this would cause a loss of food chains. (lines 9 to 11)

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

Explain why the cells in cryozoos are put in salt and sugar solution rather than pure water. (lines 16 to 18)

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

Describe how scientists could clone a banteng. (lines 18 to 20)

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

Explain why releasing cloned mammoths could have negative and positive effects on Arctic biodiversity. (lines 22 to 28)

[3]
Reveal mark scheme Mark scheme for question 1(e)

Biology Paper 1 November 2023, Question 2 10 marks

A student investigates the effect of different concentrations of sucrose on potato tissue. The student first makes a series of sucrose solutions using this dilution table.

Test tubeVolume of 1 molar sucrose solution in cm3Volume of distilled water in cm3Concentration of sucrose solution in mole per litre
12001.0
2160.8
31280.6
48120.4
54160.2
60200.0
(a)

Complete the table by writing the volume of distilled water for test tube 2.

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

This is the student’s method. Step 1 cut six equal-size cylinders of potato each 5 cm in length Step 2 dry the cut surfaces using filter paper Step 3 measure the mass of each cylinder Step 4 place one cylinder of potato in each of the six test tubes containing

• 20 cm3 of 1 molar (1.0 mole per litre) sucrose solution

• 20 cm3 of 0.8 molar sucrose solution

• 20 cm3 of 0.6 molar sucrose solution

• 20 cm3 of 0.4 molar sucrose solution

• 20 cm3 of 0.2 molar sucrose solution

Question 2, shown as an image

Step 5 put a bung in each test tube and leave the test tubes for one hour Step 6 remove the cylinders from each test tube Step 7 dry the cylinders with filter paper Step 8 measure the mass of each cylinder again

Explain the reason for step 7 in the student’s method.

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

The table shows the student’s results.

Concentration of sucrose solution in mole per litreOriginal mass in gFinal mass in gChange in mass in gPercentage change in mass (%)
1.02.11.8–0.3–14
0.82.11.9–0.2–9.5
0.62.22.20.00.0
0.42.22.3
0.22.12.20.14.8
0.02.12.30.29.5
(c)(i)

Calculate the percentage change in mass for the 0.4 mole per litre sucrose solution.

percentage change =........

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

Comment on the effect of the different concentrations of sucrose on the potato tissue.

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

Biology Paper 1 June 2023 (R), Question 1 8 marks

Also involves Eukaryotic and prokaryotic cells

Lactobacillus is a type of bacterium used to make yoghurt. The diagram shows a single Lactobacillus bacterium.

Figure for question 1
(a)
(a)(i)

What is the function of ribosomes?

Question 1, shown as an image
[1]
Reveal mark scheme Mark scheme for question 1(a)(i)
(a)(ii)

Which row of the table gives the components that are found in all bacteria and in all eukaryotic cells?

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

Lactobacillus cells do not burst when placed in distilled water. Red blood cells burst when placed in distilled water. Explain why the Lactobacillus cells do not burst but the red blood cells do burst.

[3]
Reveal mark scheme Mark scheme for question 1(a)(iii)
(b)

Describe how Lactobacillus bacteria produce yoghurt from milk.

[3]
Reveal mark scheme Mark scheme for question 1(b)

Biology Paper 1 June 2023, Question 3 12 marks

Also involves Units and scale in cells

The diagram shows a plant cell with some structures labelled.

Figure for question 3
(a)
(a)(i)

Which structure is the cell vacuole?

Question 3, shown as an image
[1]
Reveal mark scheme Mark scheme for question 3(a)(i)
(a)(ii)

Which structure is the site of photosynthesis?

Question 3, shown as an image
[1]
Reveal mark scheme Mark scheme for question 3(a)(ii)
(a)(iii)

Which structure is the cell wall?

Question 3, shown as an image
[1]
Reveal mark scheme Mark scheme for question 3(a)(iii)
(a)(iv)

Structure U is the site of protein synthesis in the cell. What is the name of structure U?

Question 3, shown as an image
[1]
Reveal mark scheme Mark scheme for question 3(a)(iv)
(b)

A cell is shaped like a cube. Each side has a length of 0.053 mm. Calculate the surface area to volume ratio of this cell. Give your answer in the form n: 1

surface area to volume ratio =........

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

Animal cells, unlike plant cells, do not have a cell wall.

(c)(i)

Explain how this difference affects red blood cells when placed in distilled water.

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

Explain how this difference affects red blood cells when placed in a concentrated salt solution.

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

Biology Paper 1 January 2023, Question 6 14 marks

Also involves Surface area to volume ratio

A student uses this method to investigate osmosis in potato tissue.

• cut three 5.0 cm long cylinders from a raw potato

• dry the cut surfaces using filter paper

• measure the mass of each cylinder using a balance

• place one cylinder in a test tube containing 10 cm3 of concentrated sucrose solution, one cylinder in a test tube containing 10 cm3 of distilled water and one cylinder in an empty test tube

• put a bung in each test tube and leave them for one hour

• remove the cylinders and dry them with filter paper

• measure the mass of each cylinder again

• measure their lengths using a ruler

Figure for question 6
(a)(i)

Describe what is meant by the term osmosis.

[2]
Reveal mark scheme Mark scheme for question 6(a)(i)
(a)(ii)

State the independent variable in the student’s investigation.

[1]
Reveal mark scheme Mark scheme for question 6(a)(ii)
(b)

Osmosis is affected by the surface area and volume of the cylinders, so the student keeps the original surface area and volume of each cylinder the same.

At the start of the experiment each cylinder had a radius of 0.25 cm and a length of 5.0 cm.

(b)(i)

Calculate the surface area of each cylinder using this formula.

Question 6, shown as an image

[ l = length of cylinder, r = radius of cylinder and π = 3.14]

[2]
Reveal mark scheme Mark scheme for question 6(b)(i)
(b)(ii)

Explain how surface area would affect the rate of osmosis.

[2]
Reveal mark scheme Mark scheme for question 6(b)(ii)
(b)(iii)

State another variable the student should control in the investigation.

[1]
Reveal mark scheme Mark scheme for question 6(b)(iii)
(c)

The table shows the student’s results for the mass of the cylinders.

Tube contentsOriginal mass in gFinal mass in gChange in mass in g
concentrated sucrose solution2.11.8–0.3
distilled water2.12.30.2
air2.22.1–0.1
(c)(i)

Explain the changes in the mass of the potato cylinder in each test tube.

[3]
Reveal mark scheme Mark scheme for question 6(c)(i)
(c)(ii)

The student also measured the change in length of each potato cylinder.

Assuming the length has the same percentage change as the mass, calculate the final length of the cylinder in the concentrated sucrose solution.

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

Biology Paper 1 June 2022, Question 6 11 marks

Also involves Turgor and plant support

The diagram shows a plant cell in distilled water and a plant cell in a concentrated solution of sodium chloride.

Figure for question 6

Cell in distilled water
Cell in concentrated solution of sodium chloride

(a)(i)

Which structure is labelled M?

Question 6, shown as an image
[1]
Reveal mark scheme Mark scheme for question 6(a)(i)
(a)(ii)

Which structure is labelled N?

Question 6, shown as an image
[1]
Reveal mark scheme Mark scheme for question 6(a)(ii)
(b)(i)

Give the name of the liquid found in the gap labelled O in the cell in the concentrated solution of sodium chloride.

[1]
Reveal mark scheme Mark scheme for question 6(b)(i)
(b)(ii)

Explain the differences in the appearance of the cell in distilled water and the cell in the concentrated solution of sodium chloride.

[4]
Reveal mark scheme Mark scheme for question 6(b)(ii)
(c)

Describe an experiment you could do to show how different concentrations of sodium chloride solution affect the appearance of plant cells.

[4]
Reveal mark scheme Mark scheme for question 6(c)

Biology Paper 2 June 2022, Question 1 16 marks

Also involves Units and scale in cells

Read the passage below. Use the information in the passage and your own knowledge to answer the questions that follow.

Supercharging plants to reduce global warming

The proportion of carbon dioxide in the atmosphere has increased in the last 100 years. In 2020, a mass of 727 gigatonnes of carbon dioxide was released into the atmosphere from natural processes, along with a mass of 37 gigatonnes from human activities. Scientists have estimated that plants naturally remove a mass of 746 gigatonnes of carbon dioxide from the atmosphere every year. The difference between what is removed and what is released causes atmospheric carbon dioxide to rise every year. Carbon dioxide is a greenhouse gas, and a significant rise will cause global warming.

To help solve the problem of rising concentrations of atmospheric carbon dioxide, scientists are planning to produce transgenic, supercharged plants that can remove atmospheric carbon dioxide and store it in their roots. The scientists estimate that if these plants can be developed, the plants could remove a mass of carbon dioxide equivalent to 50% of the emissions from human activities.

Coastal plants that have their roots in seawater contain a substance called suberin in the cell walls of the outer layer of the roots. Suberin is a waterproof substance that contains a high proportion of carbon atoms. Suberin is decomposed very slowly so remains in the soil for a long time.

The photograph shows a coastal plant called a mangrove.

Figure for question 1

To produce the supercharged plants, scientists intend to take the gene that codes for high suberin production from a coastal plant and insert it into crop plants. The crop plants used are perennial plants. Perennial plants live for many years rather than dying each winter. The transgenic crops would take in large amounts of carbon dioxide and use the carbon atoms to make suberin. The carbon would then be locked up and stored as suberin in the roots. After successfully producing one plant, they will use micropropagation rather than pollination to produce others.

These supercharged crop plants may have other uses. Suberin in roots helps to make them tolerant to soil with a high salt concentration, helping to produce higher crop yields in areas that have difficult growing conditions.

(a)(i)

Carbon dioxide is a greenhouse gas.

State the name of another greenhouse gas (line 7).

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

Calculate the increase in mass, in kg, of atmospheric carbon dioxide in 2020 (lines 2 to 7).

Give your answer in standard form.

[1 gigatonne = 1 000 000 000 000 kg]

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

State two consequences of global warming for the environment.

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

Explain why producing genetically engineered plants with additional suberin in their roots could reduce atmospheric carbon dioxide (lines 14 to 17 and lines 19 to 24).

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

Which enzyme is used to remove a gene from a section of DNA?

Question 1, shown as an image
[1]
Reveal mark scheme Mark scheme for question 1(c)
(d)

Explain why the additional suberin in the cell walls of the roots will make the transgenic plants tolerant to soil with a high salt concentration (lines 26 to 28).

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

Give three reasons why the scientists use micropropagation to reproduce the transgenic crop plants (lines 24 to 25).

[3]
Reveal mark scheme Mark scheme for question 1(e)

Biology Paper 2 January 2022, Question 1 17 marks

Also involves Units and scale in cells

Read the passage below. Use the information in the passage and your own knowledge to answer the questions that follow. Making Enough Blood for the World A blood transfusion is the transfer of blood from a donor into a patient. Blood transfusions are routine medical procedures that save the lives of millions of people every year.

Figure for question 1

The first successful blood transfusions were performed in the early twentieth century after scientists discovered that there are four main blood groups, A, B, AB and O. The blood groups are due to the presence of proteins, called antigens, on the surface of red blood cells. There are two main protein antigens called A and B. If a patient is given blood with antigens different to their own cells, their immune system will make antibodies against that antigen. The antigens present on the surface of red blood cells for each blood group are shown in the table.

Blood groupAntigens present
AA
BB
ABA and B
Oneither A nor B

Currently, blood transfusions are carried out with blood that has been donated by healthy people. The World Health Organisation calculates that there are 118.5 million blood donations collected globally every year. Of these donations, 40% are collected from a small group of high‑income countries. This means that there is a shortage of blood in many countries so the hunt is on to find an alternative.

Scientists have found a way to make artificial red blood cells. They made spheres of cell membranes filled with haemoglobin. These artificial cells are then suspended in sodium chloride solution. These artificial red blood cells have no proteins on their surface. Another way of making red blood cells is being developed in the United Kingdom. A research team has used stem cells to produce red blood cells with blood group O. The red blood cells produced are then suspended in sodium chloride solution. Both methods produce large quantities of safe red blood cells. There may be other advantages as well, artificial blood would always have the same concentration of solutes and will not clot when stored. Critics have pointed out that the artificial blood will only transport oxygen and that blood has many more functions.

(a)

Name the type of cell that produces antibodies. (Lines 8 and 9)

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

Human blood groups are controlled by three alleles, IA, IB and IO. The IA and IB alleles are codominant and the IO allele is recessive.

(b)(i)

State what is meant by the term codominant.

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

Two parents have genotypes of IAIO and IBIO. Which of these are all the possible blood groups of their children?

Question 1, shown as an image
[1]
Reveal mark scheme Mark scheme for question 1(b)(ii)
(c)

Calculate the number of blood donations collected per year from the high‑income countries. (Lines 14 and 15) Give your answer in standard form.

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

Some scientists have suggested that spherical artificial red blood cells transport oxygen less efficiently than normal human red blood cells. Explain why the shape of the artificial red blood cells reduces the efficiency of oxygen transport compared to normal human red blood cells. (Lines 18 and 19)

[3]
Reveal mark scheme Mark scheme for question 1(d)
(e)

Suggest why artificial blood does not clot when stored. (Lines 26 and 27)

[1]
Reveal mark scheme Mark scheme for question 1(e)
(f)

Explain why the artificial red blood cells are suspended in sodium chloride solution instead of in water. (Line 20)

[2]
Reveal mark scheme Mark scheme for question 1(f)
(g)(i)

Explain why stem cells can be used to make large quantities of red blood cells. (Lines 22 and 23)

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

Suggest why the scientists made red blood cells with blood group O. (Lines 22 and 23)

[2]
Reveal mark scheme Mark scheme for question 1(g)(ii)
(h)

Give two substances found in blood plasma that are not present in the artificial blood. (Lines 28 and 29)

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

Biology Paper 2 January 2022 (R), Question 3 12 marks

Also involves Units and scale in cells

The diagram shows a magnified image of a root hair cell from a young plant.

Figure for question 3
(a)

Give the names of structures labelled W, X, Y and Z.

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

The actual length of the cell, along the line between A and B, is 1000 µ m. Calculate the magnification of this drawing.

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

Explain the role of the root hair cell in absorption of water from the soil.

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

Sometimes gardeners give their plants too much water. The water fills up the air spaces in the soil around the plant roots. Explain how this can lead to plants failing to grow properly.

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

Biology Paper 2 January 2022 (R), Question 6 8 marks

A teacher carries out a demonstration to show the effect of different concentrations of salt solution on red blood cells. This is the teacher’s method.

• dilute a sample of blood using a salt solution that has the same concentration as blood plasma

• place 1 cm3 of the diluted blood into each of three test tubes labelled A, B and C

• add 10 cm3 of water to tube A

• add 10 cm3 of 1% sodium chloride solution to tube B

• add 10 cm3 of 5% sodium chloride solution to tube C

• leave each tube for 5 minutes

• compare the cloudiness of the solutions in the three test tubes

• take a drop of liquid from each tube and put on separate microscope slides

• observe each slide under a microscope

(a)

State the independent variable in this investigation.

[1]
Reveal mark scheme Mark scheme for question 6(a)
(b)

Give one variable that the teacher controls in this investigation.

[1]
Reveal mark scheme Mark scheme for question 6(b)
(c)

After 5 minutes, these are the teacher’s observations.

• tube A – a clear red solution

• tube B – a cloudy red suspension

• tube C – a cloudy red suspension

(c)(i)

Explain the differences in the teacher’s observations.

[2]
Reveal mark scheme Mark scheme for question 6(c)(i)
(c)(ii)

When the teacher looks down a microscope for cells on each slide, these are the teacher’s observations.

• slide from tube A – no cells are seen

• slide from tube B – normal biconcave red cells are seen

• slide from tube C – red cells are seen but the cells have shrunken edges The photographs show the teacher’s observations.

Figure for question 6

(Source: © Dennis Kunkel Microscopy / Science Photo Library) Explain the differences between the teacher’s observations of the slides from each tube.

[2]
Reveal mark scheme Mark scheme for question 6(c)(ii)
(d)

Blood samples can be separated into different layers using a centrifuge. This is a machine that spins blood at a high speed. A new sample of blood is shown after it has been spun in a centrifuge.

Figure for question 6

Describe how the blood in tubes A, B and C from the teacher’s demonstration would look after they had been spun in a centrifuge.

[2]
Reveal mark scheme Mark scheme for question 6(d)

Biology Paper 1 January 2021, Question 7 16 marks

Scientists investigate the effect of salt (NaC l) concentration on the germination of maize seeds.

Batches of seeds are watered with different concentrations of salt solution.

They count the number of seeds that germinate in each batch 12 days after sowing. They then determined the percentage germination.

(a)(i)

Describe how the scientists could tell if a maize seed had germinated.

[1]
Reveal mark scheme Mark scheme for question 7(a)(i)
(a)(ii)

The scientists measured the germination after 12 days.

State two other abiotic variables that the scientists need to control in their investigation.

[2]
Reveal mark scheme Mark scheme for question 7(a)(ii)
(b)

The table shows the scientists’ results.

Concentration of salt solution in mmolPercentage germination (%)
089
2083
4081
8073
12068
16055
20052
24049
28043
32038
(b)(i)

Plot a line graph to show the effect of concentration of salt solution on germination.

Use a ruler to join your points with straight lines.

[5]
Reveal mark scheme Mark scheme for question 7(b)(i)
(b)(ii)

Explain the effect of increasing the concentration of salt solution on germination.

[4]
Reveal mark scheme Mark scheme for question 7(b)(ii)
(c)

As plants grow, they produce roots and stems.

(c)(i)

Compare the responses of roots and stems to gravity.

[2]
Reveal mark scheme Mark scheme for question 7(c)(i)
(c)(ii)

Compare the responses of roots and stems to light.

[2]
Reveal mark scheme Mark scheme for question 7(c)(ii)

Biology Paper 1 June 2019, Question 4 6 marks

This apparatus can be used to show osmosis.

Figure for question 4
(a)

Explain what happens to the level of the sucrose solution in the glass tube.

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

Describe how this apparatus could be modified to measure the rate of osmosis at different temperatures.

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

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