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Biology B · Edexcel 9BI0 · Paper 1: Advanced Biochemistry, Microbiology and Genetics · June 2022

Edexcel A Level Biology B Paper 1: Advanced Biochemistry, Microbiology and Genetics June 2022 question paper 9BI0/01

1h 45m 90 marks

Edexcel 9BI0 · Paper 1

9 questions
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Question 1

The pathogenic effects of bacteria can be due to the toxins they release.

(a)

Endotoxins are released by Gram negative bacteria.

(a)(i)

Name one type of Gram negative bacteria that releases endotoxins.

(1)

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

Give one difference between the structure of Gram negative bacteria and Gram positive bacteria.

(1)

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

Endotoxins are usually less toxic than exotoxins.

(b)(i)

The LD50 value is the mass of the chemical per kg of body mass that would kill half the number of rodent animals.

The LD50 value can be used to indicate how toxic a chemical is.

One endotoxin has an LD50 value of 11 ng kg–1.

The mean body mass of a group of rodents is 28 g.

Calculate the mass of endotoxin given to each rodent that would kill half of the rodents in this group.

(1)

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

State two differences, other than toxicity, between endotoxins and exotoxins.

(2)

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

Question 2

A zygote is formed when gametes fuse at fertilisation.

(a)

Explain how meiosis results in genetic variation in the gametes.

(2)

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

Describe how the process of fertilisation results in the formation of a zygote from the gametes in humans.

(3)

Answer
Mark scheme for question 2(b)

Question 3

Malaria is a serious and sometimes fatal disease.

Scientists are constantly looking for new ways of controlling this disease.

(a)

Which row of the table shows the name of the pathogen that causes malaria, and its classification group?

Figure for question 3

(1)

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

One group of scientists has genetically modified a fungus to produce a spider toxin that kills mosquitoes.

Describe how a fungus could be genetically modified to produce spider toxin.

(3)

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

Another group of scientists has discovered a type of fungus that completely protects mosquitoes from infection by the pathogen that causes malaria.

This fungus does not kill the mosquitoes.

Explain why this approach is less controversial than the approach used by the scientists who are developing the genetically‑modified fungus.

(3)

Answer
Mark scheme for question 3(c)

Question 4

Bacteria are the host cells for λ (lambda) phage viruses.

(a)

Which is a description of a λ phage?

Question 4, shown as an image

(1)

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

The image shows phage viruses attacking a bacterium.

Figure for question 4

(Source: © nobeastsofierce Science/Alamy Stock Photo)

The length of this bacterium is 1.7 µm.

Calculate the length of the labelled phage.

Give your answer in nanometres (nm).

(2)

Answer
Mark scheme for question 4(b)
(c)
Figure for question 4
(c)(i)

Explain why there was a delay before the number of lysed cells started to increase.

(3)

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

Calculate the mean rate of increase in the actual number of lysed cells between 50 minutes and 90 minutes.

Question 4, shown as an image

(2)

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

A sketch has been made of this growth curve.

Figure for question 4

Complete this sketch to predict the shape of the growth curve after 120 minutes, assuming there is an excess of host cells.

(2)

Answer
Mark scheme for question 4(c)(iii)

Question 5

The light‑dependent stage and the light‑independent stage of photosynthesis both take place in the chloroplast.

The rate of photosynthesis is affected by a number of different factors, including carbon dioxide concentration and temperature.

(a)

The diagram shows a chloroplast.

Figure for question 5
(a)(i)

Where does the light‑dependent stage take place?

Question 5, shown as an image

(1)

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

The light‑dependent stage produces hydrogen ions.

Where do these hydrogen ions accumulate?

Question 5, shown as an image

(1)

Answer
Mark scheme for question 5(a)(ii)
(a)(iii)

Where does translation take place?

Question 5, shown as an image

(1)

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

Scientists measured the effect of two different concentrations of carbon dioxide on the rate of photosynthesis at different leaf temperatures, in one species of plant.

The results are shown in the graph.

Figure for question 5
(b)(i)

Which units are suitable for measuring the rate of photosynthesis in leaves?

Question 5, shown as an image

(1)

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

Analyse the data to identify three conclusions that can be made from this graph.

(3)

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

Explain the effects of carbon dioxide concentration and temperature on the rate of formation of GALP.

(4)

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

Question 6

The lipid content of the cell membranes of prokaryotic organisms changes in response to changes in the environmental temperature.

(a)

Phospholipids form a bilayer in the cell membranes of bacteria.

Describe the structure of a phospholipid.

(2)

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

The cell membranes of most organisms belonging to the domain Archaea are lipid monolayers.

The lipid that forms this monolayer is a bipolar lipid.

Explain why a bipolar lipid is a suitable molecule to form the cell membrane.

(2)

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

The percentage of branched‑chain fatty acids in membranes of bacteria that have different optimum growth temperatures was investigated.

The graph shows the results of this investigation.

Figure for question 6

Percentage of branched‑chain fatty acids (%)

(c)(i)

Calculate the mean percentage of branched‑chain fatty acids in bacteria whose optimum growth temperature is maths symbol20 °C.

(1)

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

Analyse the data to describe two conclusions that can be drawn from this investigation.

(2)

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

A change in temperature can affect the permeability and fluidity of the membrane.

Explain why it is important that the lipid composition of the membrane of prokaryotic organisms changes if the temperature changes.

(3)

Answer
Mark scheme for question 6(d)

Question 7

Macrophages are involved in response to infection.

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

Which diagram shows one antibody binding to the surface of a macrophage (Step 1)?

Figure for question 7

(1)

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

Actin is a fibrous protein.

Which row of the table describes the breakdown of actin (Step 2)?

Figure for question 7

(1)

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

Explain why water flows into the part of the cell where the soluble components are located (Step 3).

(2)

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

Describe the events that take place resulting in T helper cell activation, following the formation of pseudopodia by the macrophages (Step 4).

(3)

Answer
Mark scheme for question 7(a)(iv)
(b)
Figure for question 7 Figure for question 7

(6)

Answer
Mark scheme for question 7(b)

Question 8

Hypoxia is an inadequate supply of oxygen to tissues and cells that restricts their function.

(a)

The normal partial pressures of oxygen vary from tissue to tissue.

The table shows the normal partial pressure of oxygen in two tissues.

TissuePartial pressure of oxygen / kPa
Pulmonary arterial blood5.3
Other arterial blood13.3
(a)(i)

Calculate the ratio of the partial pressures of oxygen in these two tissues. Give your answer to one decimal place.

(1)

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

Explain why the partial pressures of oxygen in these two tissues are different.

(2)

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

The body responds to hypoxia by releasing hypoxia‑inducible transcription factors (HIF).

Investigations have shown that one effect of HIF is an increase in the rate of glycolysis in the affected cells.

(b)(i)

Explain how HIF could result in an increase in the rate of glycolysis.

(2)

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

Explain why cells need to respond to hypoxia with an increase in the rate of glycolysis.

(4)

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

The graph shows the changes in levels of two HIFs, HIF‑1 and HIF‑2, before and during hypoxia.

Question 8, shown as an image
Figure for question 8

Compare and contrast the changes in the levels of HIF‑1 and HIF‑2 during hypoxia.

(2)

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

Explain the changes in levels of HIF‑1 and HIF‑2 during hypoxia.

(2)

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

Question 9

Plants and animals are adapted for gas exchange.

(a)(i)

Which row of the table describes gas exchange in the root of a plant?

Figure for question 9

(1)

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

Which is the gas exchange surface in the stem of a woody plant?

Question 9, shown as an image

(1)

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

Spiracles are small openings in the exoskeletons of insects that allow air to enter the respiratory system.

Water can evaporate out of the spiracles when they are open. The insect can close the spiracles to reduce water loss.

In an investigation, the water loss from insects in air with different humidities was measured.

The insects were kept in air with 80% humidity and then moved into air with a lower humidity. Water loss was then measured.

The investigation was repeated in air high in carbon dioxide to keep the spiracles open.

The table shows the results of this investigation.

PercentageWater loss from insects / mg hr–1
humidity (%)Insects in airInsects in air high in carbon dioxide
00.100.90
200.130.68
400.150.50
600.130.35
800.070.07
(b)(i)

State how the water loss could have been measured in this investigation.

(1)

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

Calculate the percentage increase in water loss from the insects kept in air at 0% humidity compared with those kept at 80% humidity.

Give your answer to two decimal places.

(1)

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

Explain why the insects were kept in air with 80% humidity at the start of this investigation.

(2)

Answer
Mark scheme for question 9(b)(iii)
(b)(iv)

Explain the results of this investigation.

(3)

Answer
Mark scheme for question 9(b)(iv)
(c)
Figure for question 9
(c)(i)

Lamellae are present on both sides of the gill filaments.

The mean length of a gill filament is 25 mm.

Calculate the total number of lamellae on the gill filaments of Thunnus.

Express your answer in standard form.

(1)

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

*(ii) Analyse the data to explain the relationships between the activity of these fish and the structure of their gills.

(6)

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

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