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Biology B · Edexcel 9BI0 · Paper 3: General and Practical Principles in Biology · November 2020

Edexcel A Level Biology B Paper 3: General and Practical Principles in Biology November 2020 question paper 9BI0/03

2h 30m 120 marks

Edexcel 9BI0 · Paper 3

10 questions
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Question 1

(a)

Describe how oxygen from the air is able to reach insect muscle cells.

(3)

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

Give the units for a valid comparison of the rate of respiration of these two species of insect.

(1)

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

Other factors may affect the results from this respirometer.

Describe how one named factor could be controlled when using this respirometer.

(2)

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

Explain why the soda lime in the tube must be replaced when the other species of insect is placed in the respirometer.

(2)

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

Question 2

The egg of a hen contains albumin, a globular protein.

(a)

Describe the tertiary structure of a globular protein.

(3)

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

Pepsin is an enzyme that digests albumin.

Albumin can be put into a 50 mm long capillary tube and then made solid by boiling in water for 10 minutes.

The diagram shows a 50 mm capillary tube containing solid albumin.

Figure for question 2

Devise an investigation into the effect of pepsin concentration on the digestion of solid albumin.

(5)

Answer
Mark scheme for question 2(b)

Question 3

Water fleas are small animals that live in pond water.

A student observed a water flea using a light microscope.

The diagram shows the student’s drawing of the water flea.

Figure for question 3

The diagram shows an eyepiece graticule and a stage micrometer used by the student to measure the diameter of the heart in this water flea.

Figure for question 3

(Source: http://biology4alevel.blogspot.com/2014/07/2-cell-structure-microscopy.html)

Each small division on the stage micrometer is 0.01 mm.

The diameter of the heart was found to be 0.2 mm.

(a)

Determine the number of eyepiece units that equal the diameter of the water flea heart.

(1)

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

The water flea has a spherical heart with a diameter of 0.2 mm.

Cardiac output is the volume of blood pumped out of a heart in one minute.

The resting heart rate of a water flea is 200 beats min−1.

The volume of a sphere is calculated using the formula

Question 3, shown as an image

where π = 3.142

(b)(i)
Figure for question 3 Figure for question 3

(3)

Answer
Mark scheme for question 3(b)(i)
(b)(ii)

The cardiac output of a human heart at rest is 5.0 dm3min−1.

Give one reason why the cardiac output of a human heart at rest needs to be greater than the cardiac output of a water flea at rest.

(1)

Answer
Mark scheme for question 3(b)(ii)
(c)
Figure for question 3
(c)(i)

Calculate the standard deviation for the mean heart rate in the pond water.

Use the formula

Figure for question 3

(2)

Answer
Mark scheme for question 3(c)(i)
(c)(ii)

Plot a graph to show the data for the mean heart rate and SD in each treatment.

(2)

Answer
Mark scheme for question 3(c)(ii)
(c)(iii)

Explain why abiotic variables should be controlled in this investigation.

(2)

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

Question 4

A scientist investigated the effect of temperature on the time it takes for blood to clot.

The graph shows the results of this investigation.

Figure for question 4
(a)

Explain how these results show that enzymes are involved in blood clotting.

(3)

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

Prothrombin is involved in the blood clotting process.

The F 2 gene codes for the synthesis of prothrombin.

This gene is located from base pair 46 719 191 to base pair 46 739 504 on chromosome 11.

Determine the number of codons in this gene.

(1)

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

A mutation of the F 2 gene causes thrombophilia, a condition that results in the production of excess prothrombin.

In this gene mutation, guanine is replaced with adenine.

(c)(i)

Name this type of mutation.

(1)

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

People without this mutation have a 1 in 1000 risk of producing a blood clot in an artery.

The mutation increases this risk by 20 times.

State the probability of producing a blood clot for people with this mutation.

(1)

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

Explain why thrombophilia increases the risk of producing a blood clot in an artery.

(3)

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

A genetic test can be used to find out if a person has thrombophilia.

The test involves using a restriction endonuclease to obtain genetic material from white blood cells.

This genetic material is then used in the polymerase chain reaction (PCR).

(d)(i)

State the role of a restriction endonuclease.

(1)

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

Describe the process of PCR.

(3)

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

Question 5

Gibberellin stimulates cells in the aleurone layer of cereal grains, such as barley, to produce the enzyme amylase.

The diagram shows the location of the aleurone layer in a barley grain.

Figure for question 5
(a)

The part labelled X contains triploid cells.

Name the part labelled X.

(1)

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

The diagram shows a cell from the aleurone layer with some of the structures involved in the production of amylase.

Figure for question 5

Gibberellin binds to a protein receptor in the cell surface membrane and this stimulates transcription in the nucleus.

(b)(i)

Describe the process of transcription in the nucleus of this cell.

(2)

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

Describe the processes occurring after transcription that result in the release of amylase from the cell shown in the diagram.

(5)

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

Global warming may cause sea levels to rise and salty seawater to flood land used for growing crops.

A student read that salt (NaC l) inhibits gibberellin synthesis.

This could affect amylase production in cereal grains.

Devise an investigation to show the effect of salt concentration on amylase production in cereal grains.

(5)

Answer
Mark scheme for question 5(c)

Question 6

Ultraviolet (UV) radiation damages DNA and is used in microbiology to produce aseptic conditions.

The UV radiation changes the bonding found in DNA.

This change in bonding creates thymine dimers in which covalent bonds form between adjacent thymine nucleotides.

Thymine dimers bend the DNA backbone. This prevents replication of DNA.

The diagram shows the change in bonding caused by UV radiation.

Figure for question 6
(a)

Name the bonds that are lost when thymine dimers are formed.

(1)

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

Explain why preventing the replication of DNA produces aseptic conditions.

(2)

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

Aseptic conditions are used when growing bacteria in laboratories.

Justify two methods, other than UV radiation, for ensuring aseptic conditions.

(2)

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

The effectiveness of UV radiation was investigated, using three species of pathogenic bacteria.

The table provides information about the bacteria used in this investigation.

Figure for question 6
(d)(i)

Compare and contrast the structure of the wall of Gram positive bacteria and Gram negative bacteria.

(3)

Answer
Mark scheme for question 6(d)(i)
(d)(ii)
Figure for question 6

(2)

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

A student concluded from this investigation that if food is exposed to UV radiation, it would remove all risk of food poisoning.

Criticise the validity of this conclusion.

(4)

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

Question 7

The photograph shows a wood mouse, Apodemus sylvaticus.

Figure for question 7

A scientist measured the population size of these mice living in a woodland habitat.

The scientist used the capture-mark-recapture method (CMR) to sample the population.

This is the CMR method used by the scientist:

● randomly place mammal traps in the woodland for one day

● put a coloured band on a leg of each captured mouse (S1)

● release these marked mice

● after one week place mammal traps in the woodland again for one day

● count the number of mice captured (S2)

● count the number of mice with leg bands in this second sample (L).

The table shows some of the results.

Number of mice captured
S1S2
200250
(a)

The scientist calculated that the number of mice in the woodland was 1000, using the formula

Question 7, shown as an image

Determine the number of mice that had leg bands (L) in the second sample.

(1)

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

The diagram shows a section of the trap used by the scientist.

Figure for question 7

Explain how the design of the trap is effective for capturing mice.

(3)

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

Wood mice are small mammals.

They cannot maintain their body temperature and will die if left in the trap for a long period of time.

Explain why wood mice cannot maintain their body temperature if left in the trap.

(4)

Answer
Mark scheme for question 7(c)
(d)

Comment on the claim that the CMR method produces an accurate measure of the population size of mice in the woodland.

(5)

Answer
Mark scheme for question 7(d)

Question 8

Figure for question 8
(a)

State the location of the electron transport chain in a mitochondrion.

(1)

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

Hydrogen cyanide gas is a poison that can kill an animal if it is inhaled.

Cyanide is a non-competitive inhibitor of the enzyme cytochrome oxidase.

Cytochrome oxidase is the last carrier in the electron transport chain.

(b)(i)

Explain why hydrogen cyanide gas can kill an animal.

(5)

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

A dose of 1.0 mg of hydrogen cyanide per kilogram of body mass is fatal.

An animal with a body mass of 10 kg inhaled air containing 0.24 mg dm−3 of hydrogen cyanide.

The animal breathed at a rate of 15 dm3 of air per minute.

Calculate how many minutes it would take for this animal to obtain a fatal dose.

Give your answer to three significant figures.

(3)

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

The effect of cyanide on oxygen uptake by mitochondria was investigated using the following method:

● suspend mitochondria in a flask containing a buffer solution

● measure the concentration of oxygen in the flask for the first five minutes

● then add a respiratory substrate and ADP to the flask

● measure the concentration of oxygen in the flask for the next 10 minutes

● then add cyanide solution to the flask

● measure the concentration of oxygen in the flask for the next 10 minutes.

The graph shows the results for the first five minutes.

Complete the graph to show the results for the next 20 minutes.

Figure for question 8

(2)

Answer
Mark scheme for question 8(c)

Question 9

Biodiversity is monitored by conservationists.

Biodiversity can be calculated using the formula

Question 9, shown as an image
(a)

Give the symbol in this formula that represents the total number of organisms of all species.

(1)

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

The photograph shows a hedge.

Figure for question 9

Many fields in the UK are separated by hedges.

Conservationists investigated biodiversity in hedges of different ages.

Hedges of different ages were sampled at random and the number of species of woody plants was counted.

The graph shows the results of this investigation.

Figure for question 9

Analyse the data to explain why conservationists object to the destruction of hedgerows to increase the size of fields.

(4)

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

State why random sampling was used in this investigation.

(1)

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

The conservationists observed that the plants in older hedges contained more wood than the plants in younger hedges.

Wood consists of xylem tissue.

(d)(i)

Give an advantage for older plants of containing more wood.

(1)

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

Describe a method to show that the stem of an older plant contains more wood.

(3)

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

A student thought that the age of a plant affects the water potential of its tissue.

The student used potatoes of different ages to test this hypothesis.

Justify each step of a method the student could use to compare the water potential of potato tissue of different ages.

(5)

Answer
Mark scheme for question 9(e)

Question 10

The diagram shows the flow of energy in an ecosystem.

The numbers represent the energy in kJ m−2yr−1 in each category.

Figure for question 10
(a)(i)

Calculate the percentage efficiency of energy transfer from sunlight to the producers.

(2)

Answer
Mark scheme for question 10(a)(i)
(a)(ii)

Give one reason why the energy transfer from sunlight to the producers is less than 100% efficient.

(1)

Answer
Mark scheme for question 10(a)(ii)
(a)(iii)

Explain the energy difference between the producers and the primary consumers.

(2)

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

Plant productivity is affected by the abiotic and biotic factors that exist at different latitudes.

The diagram shows latitudes of the Earth.

Figure for question 10

The table shows the productivity of plants growing on land at different latitudes in the Northern hemisphere.

Latitude / degreesProductivity / g m−2yr−1
0 to 10 (equatorial conditions)900 to 1500
30 to 40 (warm temperate conditions)400 to 800
40 to 60 (cool temperate conditions)200 to 600
60 to 70 (alpine to arctic conditions)0 to 200

A student concluded that primary productivity is determined by abiotic factors that affect the light-dependent and light-independent stages of photosynthesis.

Discuss the validity of this conclusion.

(9)

Answer
Mark scheme for question 10(b)

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