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Biology B · Edexcel 9BI0 · Paper 3: General and Practical Principles in Biology · June 2025

Edexcel A Level Biology B Paper 3: General and Practical Principles in Biology June 2025 question paper 9BI0/03

2h 30m 120 marks

Edexcel 9BI0 · Paper 3

11 questions
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Question 1

Saccharomyces cerevisiae (yeast) is a single-celled organism widely used in brewing and baking.

The diagram shows a drawing of a yeast cell, as seen using an electron microscope.

Figure for question 1
(a)

The actual size of the cell measured from point A to point B is 6 µ m.

Calculate the magnification of this yeast cell.

Give your answer to the nearest whole number.

(2)

Answer
Mark scheme for question 1(a)
(b)

Draw this yeast cell as it would be seen using a light microscope at high magnification.

Labels are not required.

(2)

Answer
Mark scheme for question 1(b)
(c)

The information in the diagram can be used to classify yeast.

(c)(i)

State why yeast is classified as eukaryotic not prokaryotic.

(1)

Answer
Mark scheme for question 1(c)(i)
(c)(ii)

State why yeast is not classified as a plant.

(1)

Answer
Mark scheme for question 1(c)(ii)

Question 2

Where two neurones meet they are linked by a synapse.

(a)

The diagram shows the structure of a synapse.

Figure for question 2
(a)(i)

Draw an arrow on the diagram to show the direction in which the nerve impulse travels.

(1)

Answer
Mark scheme for question 2(a)(i)
(a)(ii)

Describe the role of acetylcholine in synaptic transmission.

(3)

Answer
Mark scheme for question 2(a)(ii)
(a)(iii)

State the role of the mitochondria in synaptic transmission.

(1)

Answer
Mark scheme for question 2(a)(iii)
(b)

Some neurones have a myelin sheath around them.

Explain how the myelin sheath speeds up the transmission of nerve impulses.

(2)

Answer
Mark scheme for question 2(b)

Question 3

The wavelength of light affects the rate of photosynthesis.

(a)
Figure for question 3

(2)

Answer
Mark scheme for question 3(a)
(b)

The diagram shows apparatus that can be used to study the effect of light on photosynthesis in pondweed.

Figure for question 3

Describe how you could modify and use this apparatus to find the optimum wavelength of light for photosynthesis of the pondweed.

(5)

Answer
Mark scheme for question 3(b)
(c)

Explain why plants have a variety of photosynthetic pigments.

(2)

Answer
Mark scheme for question 3(c)

Question 4

Red-green colour blindness is a sex-linked condition.

(a)

In humans, the gene for colour vision is located on the X chromosome.

The allele for red-green vision (XB) is dominant to the allele for red-green colour blindness (Xb).

(a)(i)

State all the possible genotypes of females.

(1)

Answer
Mark scheme for question 4(a)(i)
(a)(ii)

State all the possible genotypes of males.

(1)

Answer
Mark scheme for question 4(a)(ii)
(a)(iii)

The diagram shows a family tree for a family where some individuals have this condition.

Figure for question 4 Figure for question 4

Give all the possible genotypes for each of the following individuals.

(4)

Answer
Mark scheme for question 4(a)(iii)
(b)

Squirrel monkeys are found in the forests of Central and South America.

The photograph shows a squirrel monkey.

Figure for question 4

(Source: Adrian Hepworth / Alamy Stock Photo)

Squirrel monkeys have a varied diet, including leaves, fruits and insects.

In squirrel monkeys, the production of the pigments needed for red and green vision is determined by two alleles of a single gene.

This gene is located on the X chromosome.

Female squirrel monkeys have two X chromosomes, male squirrel monkeys have one X chromosome and one Y chromosome.

The colour red can be seen only if the allele XR is present.

The colour green can be seen only if the allele XG is present.

Scientists have studied the inheritance of colour vision in squirrel monkeys.

(b)(i)

A female monkey able to see both red and green is mated with a male monkey able to see only green.

Determine the ratio of phenotypes you would expect in their offspring.

Use a genetic diagram to support your answer.

(2)

Answer
Mark scheme for question 4(b)(i)
(b)(ii)

In separate crosses two monkeys were mated.

Half of the offspring were females with the ability to see both red and green colours and half were males able to see red only.

Determine the genotypes of these parents.

Use a genetic diagram to support your answer.

(3)

Answer
Mark scheme for question 4(b)(ii)
(b)(iii)

State why it would be an advantage for monkeys living in forests in South America to be able to see both red and green colours.

(1)

Answer
Mark scheme for question 4(b)(iii)
(b)(iv)

In 2009, scientists successfully used gene technology on some squirrel monkeys unable to see the colour red.

The scientists identified the gene for the protein that is sensitive to red light in humans. This gene was transferred to cells in the retina of these monkeys.

Describe how the scientists may have carried out this gene technology.

(3)

Answer
Mark scheme for question 4(b)(iv)

Question 5

The photograph shows a germinating pollen grain, on a slide, as seen using a light microscope.

Figure for question 5

(Source: CAROLINA BIOLOGICAL SUPPLY COMPANY/SCIENCE PHOTO LIBRARY)

(a)

Describe how you would use a light microscope to measure the actual length of this pollen tube.

(2)

Answer
Mark scheme for question 5(a)
(b)
Figure for question 5
(b)(i)

Calculate the mean rate of growth of the pollen tube in the 500 mg dm–3 calcium nitrate solution.

Give your answer in µ m per minute, to two decimal places.

........ µm minute

(2)

Answer
Mark scheme for question 5(b)(i)
(b)(ii)

Give three conclusions that can be drawn from this investigation.

(3)

Answer
Mark scheme for question 5(b)(ii)
(c)

Devise a method that could have been used to collect valid data in this investigation.

(4)

Answer
Mark scheme for question 5(c)

Question 6

(a)

The photograph shows a dandelion plant.

Figure for question 6

A student wanted to carry out a survey to estimate the number of dandelion plants growing in a large field.

The student carried out a pilot study on an area of grassland 5 m by 10 m, to find the size of quadrat which would give the most accurate estimate.

The diagram shows the position of each dandelion plant in the pilot study.

Figure for question 6

Three sizes of quadrat were tested:

Question 6, shown as an image
Question 6, shown as an image
Question 6, shown as an image

The quadrats were placed randomly and the number of dandelion plants in each quadrat was recorded.

The total number for all quadrats was then calculated.

The table shows the results obtained.

Figure for question 6
(a)(i)

Describe how the quadrats would be placed randomly.

(2)

Answer
Mark scheme for question 6(a)(i)
(a)(ii)

The total number of dandelion plants in this area of grassland was 53.

Analyse the data to determine the size of quadrat that gives the most accurate estimate of the number of dandelion plants.

(4)

Answer
Mark scheme for question 6(a)(ii)
(b)

In a different survey, the student wanted to measure the biodiversity of the grassland, using a biodiversity index.

Describe the data that must be collected so that the biodiversity index can be calculated.

(2)

Answer
Mark scheme for question 6(b)
(c)
(c)(i)

State the meaning of the term species.

(1)

Answer
Mark scheme for question 6(c)(i)
(c)(ii)

Describe how scientists can use DNA from two similar looking plants to investigate whether they are closely related.

(2)

Answer
Mark scheme for question 6(c)(ii)

Question 7

The activity of enzymes is affected by a number of factors.

(a)
Figure for question 7
(a)(i)

Describe how you would ensure that the concentration of the enzyme extract was the same for all students carrying out the investigation.

(2)

Answer
Mark scheme for question 7(a)(i)
(a)(ii)

Describe how you would ensure that the colorimeter readings were comparable for each of the students.

(2)

Answer
Mark scheme for question 7(a)(ii)
(b)

The table shows the results obtained in this investigation.

Temperature / °C1020304050607080
Absorbance of light0.250.330.390.460.530.550.500.33
(b)(i)

Explain the effect of temperature on the activity of phosphatase.

(3)

Answer
Mark scheme for question 7(b)(i)
(b)(ii)

For enzyme-controlled reactions at temperatures between 0°C and 40°C, a 10°C rise in temperature will double the rate of reaction.

The data collected in this investigation do not show this pattern, suggesting that there may be errors in the method used.

The students repeated the investigation with these four changes:

• adding buffer solution to the phenolphthalein phosphate

• placing separate tubes containing phenolphthalein phosphate and enzyme extract into water baths for 10 minutes before mixing together

• using smaller temperature intervals than 10°C

• stopping the reaction after 5 minutes instead of 20 minutes.

Justify why the students repeated the investigation with these four changes.

(4)

Answer
Mark scheme for question 7(b)(ii)
(c)

The phosphatase enzyme catalyses the removal of phosphate groups from organic phosphate molecules, releasing phosphate ions.

Give two examples of biochemical processes in plant cells requiring phosphate ions.

(2)

Answer
Mark scheme for question 7(c)

Question 8

The photograph shows tar spots on a sycamore leaf.

Figure for question 8

(Source: Nigel Cattlin / Alamy Stock Photo)

Tar spots are caused by a fungus, Rhytisma acerinum, growing on the sycamore leaf.

Fungal spores spend the winter in dead leaves on the ground. When new leaves emerge in the spring, they are infected by these spores and tar spots develop.

The fungus is known to be sensitive to sulfur dioxide and oxides of nitrogen in the air.

The fungus grows more slowly when these gases are present.

(a)

A student investigated whether there were fewer tar spots on sycamore leaves close to busy roads where air pollution is higher.

This is the method the student used.

Step 1 Choose one sycamore tree close to a busy road and one sycamore tree in parkland away from busy roads.

Step 2 Choose 15 leaves from each tree.

Step 3 Record the number of tar spots on each leaf.

(a)(i)

Give a suitable null hypothesis for this investigation.

(1)

Answer
Mark scheme for question 8(a)(i)
(a)(ii)

The table shows the data the student recorded.

LocationNumber of tar spots per leafMeanSD
Near road8 12 1817 5 2221 3 244 7 1615 13 912.936.78
Parkland20 10 1417 21 2313 16 2114 13 1524 18 1717.074.06

The student analysed the data with a t-test.

Calculate the value of t.

Figure for question 8

maths symbol is the mean for each set of data

n is the number of samples in each set of data

S is the standard deviation for each set of data

(3)

Answer
Mark scheme for question 8(a)(ii)
(a)(iii)

The table shows the critical values of t for different degrees of freedom.

Degrees of freedomp = 0.10p = 0.05p = 0.01
141.762.142.98
151.752.132.95
161.752.122.92
171.742.112.90
181.732.102.88
191.732.092.86
201.722.092.85
211.722.082.83
221.722.072.82
231.712.072.81
241.712.062.80
251.712.062.79
261.712.062.78
271.702.052.77
281.702.052.76
291.702.052.76
301.702.042.75

Analyse the data to comment on the results of this investigation.

(3)

Answer
Mark scheme for question 8(a)(iii)
(b)

Justify two improvements to the method used by the student.

(2)

Answer
Mark scheme for question 8(b)

Question 9

Transpiration is the loss of water from the leaves of plants.

(a)

Water moves through xylem vessels in a plant during transpiration.

Explain one way in which xylem vessels are adapted for their function.

(2)

Answer
Mark scheme for question 9(a)
(b)

The graph shows the rate of transpiration in oat seedlings during one day. Windspeed and humidity were controlled.

Figure for question 9
(b)(i)

Calculate the mean rate of change in transpiration from five to thirteen hours.

Give your answer to one decimal place.

........ g m–2 hour

(1)

Answer
Mark scheme for question 9(b)(i)
(b)(ii)

Explain the shape of the graph over the 22-hour period.

(3)

Answer
Mark scheme for question 9(b)(ii)
(c)

The diagrams show two types of potometer, A and B.

Figure for question 9

Evaluate which type of potometer would be better to measure the rate of transpiration.

(5)

Answer
Mark scheme for question 9(c)
(d)

A scientist investigated the movement of sugar from the leaf to the roots of a plant.

One leaf was supplied with carbon dioxide containing the radioisotope14C.

The diagram shows part of the apparatus used by the scientist.

Figure for question 9

The sugar produced by photosynthesis in this leaf is radioactive.

After a few hours, a section was cut through the plant stem and placed on photographic film.

Structures containing radioactive material will cause darkening of the film.

The diagram shows the structure of the stem when a section is cut.

Figure for question 9
(d)(i)

Shade the areas you would expect to be darker because they are transporting radioactive sugar.

(1)

Answer
Mark scheme for question 9(d)(i)
(d)(ii)

Name the radioactive sugar causing this darkening on the film.

(1)

Answer
Mark scheme for question 9(d)(ii)

Question 10

Oceans cover 71% of the Earth’s surface. Marine and coastal habitats are thought to contain about one quarter of all species worldwide.

(a)

Some parts of the UK coastline have been designated Marine Conservation Zones (MCZ). These are part of a wider network of sites protected by law, called Marine Protected Areas (MPA).

Large parts of the UK coastline and the sea surrounding it are protected in this way. There are more than 370 protected coastal or marine sites around the UK.

The regulations in these areas protect coastal and marine habitats, and the communities and species which live in them, from damage by humans.

This is an example of in-situ conservation.

Comment on the benefits and difficulties of in-situ conservation in coastal and marine sites.

(4)

Answer
Mark scheme for question 10(a)
(b)

*(b) Seagrasses are flowering plants that live underwater in coastal habitats.

Seagrass meadows are one of the world’s most threatened habitats, but only 26% are within Marine Protected Areas.

In the past, seagrass meadows were common worldwide. Over the last 90 years the area covered by seagrass has declined dramatically, with around 7% being lost each year.

These losses are due to water pollution, damage from fishing, dredging and building ports.

Worldwide, seagrass covers 0.2% of the ocean, but is estimated to be responsible for 10% of its carbon storage. It can store at least 0.5 tonnes of carbon per hectare per year, which is more than a tropical rainforest.

Seagrass meadows protect the coastline from wave erosion. They also provide food and shelter for young fish, including commercially important species like cod.

Around the UK, 90% of seagrass meadows have been lost and only 8 500 hectares remain. A pilot project will plant two hectares of seagrass meadow off the coast of Wales.

Different planting methods have been trialled at different sites. The diagram shows the most successful method, where 4% of seeds germinate after eight months.

Figure for question 10

Analyse the information to evaluate the importance of seagrass as a habitat, and the importance of this pilot project.

(9)

Answer
Mark scheme for question 10(b)

Question 11

The growth of bacterial populations can be studied if they are grown in liquid culture.

(a)

One type of bacteria is grown in nutrient broth under aseptic conditions.

The number of bacteria in the culture at a particular time can be determined using serial dilution and plating.

The diagram shows the dilution series used.

Question 11, shown as an image
Figure for question 11

One sample of 0.5 cm3 is taken from each tube (A to F). Each sample is spread onto a separate agar plate using aseptic technique.

The plates are incubated at 25 °C for 48 hours and the number of colonies on each plate is counted.

Each colony represents a single bacterium in the original culture.

The table shows the results obtained.

TubeABCDEF
Number of colonies on agar plate**1261310
Serial dilution factor1 in 10

* = too many colonies to count

(a)(i)

Calculate the number of bacteria per cm3 in the original culture.

Give your answer in standard form.

........ per cm3

(2)

Answer
Mark scheme for question 11(a)(i)
(a)(ii)

The graph shows the population of bacteria in a culture, as determined using this method.

Figure for question 11

Explain the shape of this graph.

(4)

Answer
Mark scheme for question 11(a)(ii)
(a)(iii)
Figure for question 11

(2)

Answer
Mark scheme for question 11(a)(iii)
(a)(iv)

Calculate the mean generation time, g (time for the population of bacteria to double, in minutes). Use your value for k and the formula k = fraction

(1)

Answer
Mark scheme for question 11(a)(iv)
(b)

An alternative method to dilution plating is to use a colorimeter to measure the turbidity of a culture.

Turbidity is an indicator of the growth of the bacterial population.

Assess the advantages of each of these methods to measure the growth of a bacterial population.

(4)

Answer
Mark scheme for question 11(b)

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