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Biology B · Edexcel 9BI0 · Paper 1: Advanced Biochemistry, Microbiology and Genetics · June 2025

Edexcel A Level Biology B Paper 1: Advanced Biochemistry, Microbiology and Genetics June 2025 question paper 9BI0/01

1h 45m 90 marks

Edexcel 9BI0 · Paper 1

9 questions
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Question 1

Active transport is one type of cell transport mechanism.

(a)

What happens during active transport?

Question 1, shown as an image

(1)

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

Which substance is transported against a concentration gradient by active transport?

Question 1, shown as an image

(1)

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

The graph shows the change in concentration of molecule A and molecule B, as they are transported into the cytoplasm of a cell.

Figure for question 1

Explain the differences in the changes in concentration of molecule A and molecule B between 0 and 10 minutes and between 20 and 30 minutes.

(2)

Answer
Mark scheme for question 1(c)

Question 2

The diagram shows part of cellular respiration.

Figure for question 2
(a)

Which is molecule X ?

Question 2, shown as an image

(1)

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

Which row of the table shows the locations of glycolysis, the link reaction and the Krebs cycle in eukaryotic cells?

Figure for question 2

(1)

Answer
Mark scheme for question 2(b)
(c)

Cellular respiration yields 36 molecules of ATP per molecule of glucose.

(c)(i)

The oxidation of one glucose molecule yields 2 870 224 Joules.

The free energy stored in one ATP molecule is 30 566 Joules.

Calculate the efficiency of energy transfer in the formation of ATP.

Give your answer to two decimal places.

(2)

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

State why this process is not 100 % efficient.

(1)

Answer
Mark scheme for question 2(c)(ii)

Question 3

Figure for question 3
(a)

State why specifically-selected cell types have to be grown in Step 1.

(1)

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

Explain why the flask has to be incubated at an appropriate temperature for a few days in Step 3.

(2)

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

Explain why the viruses and not the cells will be present in the filtrate in Step 5.

(3)

Answer
Mark scheme for question 3(c)

Question 4

The permeability of the cell surface membrane to a particular molecule depends on a number of factors, including the properties of the membrane and the properties of the molecule moving through it.

(a)

The diagram shows some of the components of the cell surface membrane.

Figure for question 4
(a)(i)

Which row in the table describes parts R and S of the phospholipid?

Figure for question 4

(1)

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

Draw part of a cell surface membrane that has twelve phospholipids, one channel protein and one cholesterol molecule.

Use the components shown in the diagram.

(2)

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

Describe the role of channel proteins in cell transport.

(2)

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

Analyse the data to give two conclusions that can be made from this graph.

(2)

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

Explain the effect of temperature on membrane permeability at point P.

(2)

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

Question 5

Turner’s syndrome (TS) results from a chromosome mutation.

People with TS are at a higher risk of developing autoimmune disorders than people without TS.

An autoimmune disorder results when a person’s immune system attacks their own body.

(a)

Which of the following describes the chromosome mutation that causes TS?

Question 5, shown as an image

(1)

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

There were 613 936 babies born in the UK in 2020.

The ratio of baby boys to baby girls is 1.05: 1.

The incidence of TS is between 1 in 2 000 and 1 in 2 500 in baby girls.

Calculate the minimum number of baby girls born with TS in the UK in 2020.

(2)

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

Scientists studied a group of girls with TS and a group of girls without TS.

They recorded physical features and immunological markers such as antibodies and lymphocytes.

(c)(i)

Girls who had recently been vaccinated were excluded from this study.

Explain why these girls were excluded.

(3)

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

The table shows some of the physical features of the two groups of girls.

Physical featuresWith TSMean value ± SD for each group of girls Without TS
Age / years11.9 ± 4.112.5 ± 4.0
Mass / kg36.7 ± 14.845.2 ± 19.3
Height / cm134.8 ± 20.0146.3 ± 21.3
Standardised height score–2.16 ± 0.95–0.41 ± 0.50
Body mass index (BMI)19.2 ± 3.319.7 ± 4.6

Only one of these features was significantly different between the two groups of girls.

Explain which was the significantly different feature.

(2)

Answer
Mark scheme for question 5(c)(ii)
(c)(iii)
Figure for question 5

(2)

Answer
Mark scheme for question 5(c)(iii)
(c)(iv)

The scientists also determined the number of a type of cell called T regulatory cells in the blood of these two groups of girls.

They found that the numbers of these cells were very similar.

People with autoimmune diseases have been shown to have a decreased number of T regulatory cells.

Explain how this study could be modified to show that a decreased number of T regulatory cells is linked to the risk of autoimmune disease in girls with TS.

(2)

Answer
Mark scheme for question 5(c)(iv)

Question 6

Chlorella are single-celled photosynthetic algae.

(a)

The diagram shows part of a membrane in a chloroplast from Chlorella.

Figure for question 6
(a)(i)

Which membrane is shown in this diagram?

Question 6, shown as an image

(1)

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

Name reaction A.

(1)

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

What occurs in reaction B?

Question 6, shown as an image

(1)

Answer
Mark scheme for question 6(a)(iii)
(a)(iv)

Write the equation for reaction C.

(1)

Answer
Mark scheme for question 6(a)(iv)
(b)
Figure for question 6
(b)(i)

The mean growth rate for Chlorella grown in natural light during the fourteen- day period was 1.04 × 106 cells per day.

Calculate how many times faster Chlorella grew in alternating periods of light and dark.

Give your answer to two significant figures.

(2)

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

The cultures were observed for a further four days.

The table shows the observations.

Time inObservation
culture / daysChlorella grown in alternating periods of light and darkChlorella grown in natural light
17Dark green cells The number of cells had not increased since the previous dayLight green cells The number of cells had increased since the previous day
18Cells had entered the death phaseCells had not entered the death phase

Explain these observations.

(5)

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

The scientists chose to use Chlorella as a source of photosynthetic pigments for solar cells.

The graph shows the absorption spectra of four photosynthetic pigments found in Chlorella.

Figure for question 6

Explain why Chlorella cells are a suitable source of photosynthetic pigments for use in a solar cell.

(3)

Answer
Mark scheme for question 6(b)(iii)

Question 7

Living organisms have a range of gas exchange systems.

(a)

The photograph shows human lung tissue, as seen using a light microscope.

Figure for question 7
(a)(i)

The typical diameter of an adult alveolus is 200 µm.

There are 300 million alveoli in each lung.

Calculate the total surface area of a pair of adult lungs, using the formula:

surface area of one alveolus = 4πr2

Give your answer in m2.

(3)

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

Explain the importance of a large surface area in a gas exchange system.

(2)

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

Fluid passes out of the capillaries into the spaces around the alveoli.

Some of this fluid passes into the alveoli.

Explain the importance of having fluid inside the alveoli.

(2)

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

The photograph shows a marine sponge.

Figure for question 7

Sponges are aquatic animals that are covered in pores called ostia.

The diagram shows a section through a sponge.

Figure for question 7

Analyse the information in the diagram to explain how this sponge is adapted for gas exchange.

(3)

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

(4)

Answer
Mark scheme for question 7(c)

Question 8

The rate of photosynthesis in some plants can be affected by the level of nitrogen gas in the air.

The effects of low, moderate and high nitrogen levels on leaf structure and molecules present in the leaves were investigated.

(a)

Which pair of molecules contains nitrogen atoms?

Question 8, shown as an image

(1)

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

The diagram shows a section through a leaf.

Figure for question 8
(b)(i)

Which is the main gas exchange surface in a leaf?

Question 8, shown as an image

(1)

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

* (ii) The table shows some effects on the characteristics of leaves when plants are grown in low, moderate and high levels of nitrogen gas.

CharacteristicPlants grown in low levels of nitrogen gasEffect compared with plants grown in moderate levels of nitrogen gas Plants grown in high levels of nitrogen gas
Thickness of layerssmallerlarger
Size of chloroplastssmallerlarger
Surface area of chloroplasts exposed to intercellular air spacelowerhigher
RUBISCO activitylowerlower

Discuss how growing plants in low levels and high levels of nitrogen gas affects the rate of photosynthesis.

(6)

Answer
Mark scheme for question 8(b)(ii)
(c)
Figure for question 8
(c)(i)

Determine the effect of growing plants in low levels of nitrogen gas and in high levels of nitrogen gas on the changes in gene expression.

(3)

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

Describe how gene expression can be changed.

(2)

Answer
Mark scheme for question 8(c)(ii)

Question 9

Aptamers are single-stranded DNA or RNA sequences that fold into specific three-dimensional structures.

Aptamers bind tightly to targets such as proteins and viruses.

(a)

Compare and contrast the structure of single-stranded DNA with that of single-stranded RNA.

(3)

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

Aptamers can be amplified using PCR.

To amplify an RNA aptamer, reverse transcription-PCR (RT-PCR) must be used.

Describe how RT-PCR can be used to amplify RNA aptamers.

(4)

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

* (c) The diagram shows the structure of the influenza virus.

Haemagglutinin is a glycoprotein that binds to molecules on the cell membranes of the host cells of this virus.

Strains of influenza virus have evolved that have haemagglutinins on their surface, with different antigenic properties.

Figure for question 9

Aptamers (AT1, AT2 and AT3) have been developed that bind to haemagglutinins (H1, H3, H5, H7 and H9).

The strength of binding of these aptamers to the haemagglutinins and to four strains of influenza virus (W, X, Y and Z) was investigated.

The graphs show the results of this investigation.

Figure for question 9 Figure for question 9

Discuss the advantages and disadvantages of using aptamers to prevent infection with the influenza virus.

(6)

Answer
Mark scheme for question 9(c)

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