Author: philcolls76@outlook.com
Last saved by: Gareth Roberts

Carmel College

PAPER 3 practise questions                 NAME:

Q1.          A student was provided with two agar plates. She transferred a culture of bacterium A onto one plate and a culture of bacterium B to the second plate. She placed paper discs containing antibiotics on the surface of the agar. She then incubated the plates for 24 hours.
The diagram shows the agar plates before and after incubation.

(a)     The student used a pair of forceps to place the paper discs onto the surface of the agar.
Explain why she passed the forceps through a Bunsen flame before and after each time she used them.
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(b)     Explain the appearance of the agar plates after incubation.
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(Total 6 marks)
Q2.          S       The diagram shows apparatus used to measure the oxygen uptake of snails that live on the seashore. The apparatus was kept at a constant temperature.

(a)     (i)      Explain the purpose of the strip of filter paper in the potassium hydroxide solution.
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(ii)     The level of liquid in the right-hand side of the manometer went down during the experiment. Explain why.
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 (iii)     What measurements are needed to calculate the rate of oxygen uptake by the snails in mm^3 g^–1 h^–1?
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(b)     Two experiments were carried out using the apparatus shown in the diagram.
1     The oxygen uptake of batches of 10 seashore snails kept in moist air was measured at temperatures between 5 °C and 35 °C.
2     Experiment 1 was repeated but with batches of 10 seashore snails covered by aerated seawater.
The experiments were repeated several times and means and standard deviations calculated. The results are shown in the table. The values given are means plus or minus one standard deviation.

Temperature / °C

Oxygen uptake of
snails kept in moist
air / mm^3 g^–1 h^–1

Oxygen uptake of
snails kept in
seawater / mm^3 g^–1 h^–1

5 | 35 ± 2 | 28 ± 8
10 | 34 ± 6 | 32 ± 3
15 | 36 ± 3 | 35 ± 3
20 | 86 ± 8 | 52 ± 10
25 | 141 ± 13 | 96 ± 15
30 | 132 ± 14 | 108 ± 9
35 | 120 ± 16 | 79 ± 21

(i)      Describe one similarity and one difference between the pattern of mean oxygen uptake of the snails kept in moist air and those covered by seawater.
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(ii)     Explain why valid conclusions cannot be drawn about the trends in oxygen uptake at temperatures of 25 °C and above.
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(Total 10 marks)
 Q3.          In an investigation, the population density of plants in a regularly cut lawn was compared with that in a lawn which was only cut occasionally. The table shows the results.

Species

Mean population density /
number of plants per m^2

Result of
statistical test value of p

| Regularly cut lawn | Occasionally cut lawn |
Daisy | 36.0 | 18.6 | < 0.02
Dandelion | 10.8 | 3.4 | < 0.05
Field buttercup | 1.2 | 10.0 | < 0.01
Ribwort plantain | 4.3 | 2.8 | > 0.5
Greater plantain | 0.9 | 1.5 | > 0.5

(a)     Describe a practical technique which you could use to find the mean population density of daisies on a lawn.
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(b)     Give the null hypothesis for the statistical test on the population density of daisies.
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(c)     What conclusions can be drawn from the results of this investigation?
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(Total 7 marks)
Q4.          The percentage of light absorbed by an aquatic plant was measured when it was exposed to different wavelengths. The rate of photosynthesis was also measured at each wavelength of light. The results are shown in the graph.

(a)     Describe and explain the relationship between light absorption and the rate of photosynthesis for the wavelengths of light between 410 nm and 500 nm.
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(b)     Give one dependent variable you could measure in order to determine the rate of photosynthesis in an aquatic plant.
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(c)     Use the graph to identify the range of wavelengths of light that would be green in colour.
Give a reason for your answer.
Wavelengths ................... to ................... nm
Reason ........................................................................................................
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(d)     A suspension of chloroplasts was isolated from an aquatic plant and a reagent was added. The reagent is blue when oxidised and is colourless when reduced.
(i)      The suspension of chloroplasts in blue reagent was exposed to sunlight. The blue colour disappeared. Use your knowledge of the light-dependent reactions of photosynthesis to explain why.
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(ii)     Another suspension of chloroplasts was set up as before. Small quantities of ADP and phosphate ions were added and then the tube was exposed to light. The blue colour disappeared more quickly. Explain why.
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(Total 9 marks)
 
Q5.          (a)     The diagram shows the fluid-mosaic model of a cell surface membrane.

(i)      Name the molecules labelled A and B.
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(ii)     How does the bilayer formed by substance A affect entry and exit of substances into and out of a cell?
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(b)     A dialysis machine contains artificial membranes which enable urea to be removed from the blood of a person with kidney failure. The diagram shows a dialysis machine.

(i)      By what process does urea pass from the blood into the dialysis fluid?
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(ii)     Suggest two reasons for keeping the fluid in the dialysis machine at 40 °C rather than room temperature.
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(iii)     The blood and the dialysis fluid flow in opposite directions in the dialysis machine. Explain the advantage of this.
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(iv)    Blood flows through the dialysis machine at a rate of 200 cm^3 per minute.
Calculate the total volume which passes through the machine in 5 hours.
Give your answer in dm^3 and show your working.

 
 
Answer .................................... dm^3
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(Total 10 marks)

Q6.          (a)     Explain the meaning of these ecological terms.
Population ....................................................................................................
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Community ...................................................................................................
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(b)     Some students used the mark-release-recapture technique to estimate the size of a population of woodlice. They collected 77 woodlice and marked them before releasing them back into the same area. Later they collected 96 woodlice, 11 of which were marked.
(i)      Give two conditions necessary for results from mark-release-recapture investigations to be valid.
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(ii)     Calculate the number of woodlice in the area under investigation. Show your working.

 

 
 
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(c)     Explain how you would use a quadrat to estimate the number of dandelion plants in a field measuring 100 m by 150 m.
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(d)     Two similar species of birds (species A and species B) feed on slightly different sized insects and have slightly different temperature preferences. The diagram represents the response of each species to these factors.

(i)      Which of the numbered boxes describes conditions which represent
the niche of species A;                                                                              .............
the niche of species B;                                                                              .............
insects too small for species B and temperature too warm for species A; .............
insects too large for species A and temperature too cool for species B?  .............
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(ii)     These two species are thought to have evolved as a result of sympatric speciation. Suggest how this might have occurred.
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(Total 15 marks)

Q7.          A gene was broken into fragments using enzyme Z. The mixture of fragments produced was then separated by electrophoresis.
(a)     What type of enzyme is enzyme Z?
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The table shows the number of base pairs present in the fragments.
 
Fragment | Number of base pairs (× 10^3)
1 | 4.65
2 | 5.72
3 | 10.71
4 | 2.39
5 | 5.35
6 | 7.53

The diagram shows the electrophoresis gel used. The mixture of fragments was placed at the start point marked S and the process started. The boxes indicate the positions reached by the different fragments.

(b)     Explain why base pairs are a suitable way of measuring the length of a piece of DNA.
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(c)     (i)      Write 6 above the appropriate box on the diagram to show the position you would expect fragment 6 to have reached.
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(ii)     Explain how you arrived at your answer.
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(d)     Enzyme Z recognises a particular sequence of bases in the gene. How many times does this sequence appear in the DNA of this gene?
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(Total 6 marks)
 
Q8.          A student investigated the stages of mitosis in a garlic root. The root tip was placed on a microscope slide with a stain. A cover slip was placed on top and the root tip was firmly squashed.
(a)     Explain why
(i)      a root tip was used;
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(ii)     a stain was used;
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(iii)     the root tip was firmly squashed.
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(b)     The student examined the cells in the garlic root tip under the microscope, and obtained the following data.
 
| Stage | Number of cells
| Interphase | 872
| Prophase | 74
| Metaphase | 18
| Anaphase | 10
| Telophase | 8

(i)      Calculate the percentage of these cells in which the chromosomes are visible and would consist of a pair of chromatids joined together. Show your working.

 
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(ii)     A different set of results was obtained when the count was repeated on another occasion with a different garlic root tip. Give two reasons for the difference in results.
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(Total 7 marks)

Q9.          (a)     The biochemical pathway of aerobic respiration involves a number of different steps.
Name one step in which carbon dioxide is produced.
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In an investigation, scientists transferred slices of apple from air to anaerobic conditions in pure nitrogen gas. They measured the rate of carbon dioxide production.
(b)     The scientists kept the temperature constant throughout the investigation. Explain how a decrease in temperature would affect the rate of carbon dioxide production.
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(c)     When the apple slices were transferred to nitrogen, the following biochemical pathway took place.

Use this pathway to explain the part played by reduced NAD when the apple slices were transferred to nitrogen.
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(d)     The rate of carbon dioxide production was higher when the apple slices were in nitrogen than when they were in the air. Explain why.
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Q10.          Termites are insects. Some species live in colonies in the soil. Although most termites are wingless, winged termites are sometimes produced. The winged termites fly from the soil, mate and start new colonies.
A scientist studied the behaviour of winged termites. He divided these termites into three groups.
•        Group A had their eyes covered.
•        Group B had their antennae removed.
•        Group C was the control group.
He put individual winged termites on a sloping board that was illuminated from one side. The diagram shows the direction of movement of a typical termite from each of the three groups.

(a)     (i)      What type of behaviour was shown by the termite from group B?
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(ii)     Give the evidence for your answer.
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(b)     Explain what the results from group A suggest about the factors controlling the behaviour of winged termites.
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(c)     Suggest one advantage to the termites from group C of the behaviour shown in the investigation.
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Q11.A student investigated the rate of transpiration from a leafy shoot. She used a potometer to measure the rate of water uptake by the shoot. The diagram shows the potometer used by the student.
 
(a)     Give one environmental factor that the student should have kept constant during this investigation.
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(b)     The student cut the shoot and put it into the potometer under water. Explain why.
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(c)     The student wanted to calculate the rate of water uptake by the shoot in cm^3 per minute. What measurements did she need to make?
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(d)     The student assumed that water uptake was equivalent to the rate of transpiration.
Give two reasons why this might not be a valid assumption.
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(e)     The student measured the rate of water uptake three times.
(i)      Suggest how the reservoir allows repeat measurements to be made.
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(ii)     Suggest why she made repeat measurements.
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(Total 8 marks)

Q12.Scientists investigated the effect of drinking tea and coffee on reducing the risk of developing one type of brain cancer. The investigation involved 410 000 volunteers and was conducted in 10 European countries over a period of 8.5 years.
(a)     (i)      Apart from age, suggest two factors that the scientists should have considered when selecting volunteers for this trial.
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(ii)     Give two features of the design of this investigation that would ensure the reliability ofthe results obtained.
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(b)     The incidence for this type of brain cancer is 6 cases per 100 000 per year.
Use this information to calculate the expected number of volunteers developing this cancer during the 8.5 year period of this investigation. Show your working.

 

 
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(c)     In analysing the results of this investigation, the scientists took into account the age of the volunteers. Suggest why.
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(d)     During the investigation, the volunteers were asked to estimate the volume of tea and/or coffee that they drank each day. The types of tea and coffee consumed in different countries varied. When the data from all the countries were collected there was a correlation between drinking more than 100 cm^3of tea or coffee each day and a reduced risk of developing this type of brain cancer.
Tea and coffee contain caffeine. A newspaper reported the results of this investigation under the headline ‘Caffeine helps cut cancer risk’. Explain why scientists could not support this view solely on the basis of this investigation.
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(e)     Another group of scientists investigated the effect of caffeine on blood flow to certain parts of the brain. Volunteers were given different concentrations of caffeine solution to drink. A control group was also set up.
(i)      Describe how the control group should have been treated.
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(ii)     Volunteers who drank the same concentration of caffeine solution often had different concentrations of caffeine in their blood. Suggest one reason for the difference in concentration of caffeine in the blood of volunteers.
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(iii)    The investigation showed that caffeine reduces the blood flow to certain parts of the brain. Suggest one way in which this could lead to a reduced risk of brain cancers.
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(Total 15 marks)
Q13.Some mice have diabetes. The diabetes causes the blood glucose concentration to become very high after a meal. Scientists investigated the use of an inhibitor of amylase to treat diabetes.
The scientists took 30 mice with diabetes and divided them into two groups, A and B.
•        Group A was given yoghurt without the inhibitor of amylase each day.
•        Group B was given yoghurt with the inhibitor of amylase each day.
Apart from the yoghurt, all of the mice were given the same food each day.
The scientists measured the blood glucose concentration of each mouse, 1 hour after it had eaten. This was done on days 1, 10 and 20 after the investigation started.
The following figure shows the scientists’ results.
 
        Days after the investigation started
(a)     Group A acted as a control in this investigation.
Explain the purpose of this group.
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(b)     Apart from the yoghurt, it was important that all of the mice were given the same food each day. Give two reasons why it was important that all of the mice were given the same food each day.
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(c)     The scientists’ hypothesis was that adding the inhibitor of amylase to the food would lead to a lower blood glucose concentration.
Use your knowledge of digestion to suggest how the addition of the inhibitor could lead to a lower blood glucose concentration.
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(d)     Give one reason why these results may not support the use of the inhibitor of amylase to treat diabetes in mice.
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(Total 8 marks)
Q14.Plants have pigments that absorb light energy for photosynthesis. These pigments include two types of chlorophyll and a group of pigments known as carotenoids. Different species of plant contain different amounts of these pigments. The pigments that each plant species has are adaptations to where and how they live; their ecological niche.
Figure 1 shows the absorption of light of different wavelengths by chlorophyll a, chlorophyll b and carotenoids.
Figure 1

A scientist investigated the energy in light of different wavelengths reaching the ground in a forest. She measured the energy in
•        direct sunlight
•        sunlight that had passed through the leaves of trees.
Figure 2 shows her results.
Figure 2

Beech trees have two types of leaves called sun leaves and shade leaves. Sun leaves grow on branches exposed to direct sunlight, shade leaves grow on branches exposed to light that has passed through leaves. An ecologist collected sun leaves and shade leaves from beech trees and determined the mean mass of each photosynthetic pigment in both types of leaf. His results are shown in Figure 3.
Figure 3
 
| Photosynthetic pigment | Mean mass of each pigment per m^2 of leaf area / μg (± standard deviation)
| | Sun leaves | Shade leaves
| Chlorophyll a | 299.3 (± 2.1) | 288.9 (± 0.1)
| Chlorophyll b | 90.7 (± 2.1) | 111.1 (± 0.1)
| Carotenoids | 0.10 (± 0.01) | 0.07 (± 0.01)
 
(a)     It is an advantage to beech trees to produce more chlorophyll b in the shade leaves.
Suggest and explain why.
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(b)     There are two hypotheses about the advantage to plants of producing carotenoids.
Hypothesis 1
Carotenoids help shade leaves to absorb more light of wavelengths 480 nm to 520nm for photosynthesis.
Hypothesis 2
Carotenoids prevent damage to chlorophyll from very bright light.
(i)      Which hypothesis do the data provided on the resource sheet support?
Explain your answer.
Hypothesis ....................................
Explanation ............................................................................................
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(ii)     Suggest one other piece of experimental evidence you would need in order to be more confident about drawing your conclusion in (b)(i).
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(Total 5 marks)
Q15.A scientist examined the structure of mustard plant leaves. He viewed temporary mounts of leaf tissues with an optical microscope. The figure below shows a drawing of typical results.

(a)     Describe how temporary mounts are made.
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(b)     Calculate the distance in micrometres between G and H on the leaf.
Answer = .......................... µm(2)

(c)     Describe how the scientist could have used the temporary mounts of leaves to determine the mean number of chloroplasts in mesophyll cells of a leaf.
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Q16.          Doctors investigated babies who were bottle-fed with baby-formula milk and suffered from colic. Colic is a condition that affects the gut and makes babies cry.
Each mother was given two solutions to add to her baby’s milk. One solution contained the enzyme lactase, the other did not. The mother did not know which solution contained lactase. The mother added one of the solutions to her baby’s milk for a week and recorded how long it cried each day. The mother then used the other solution for the second week.
The table shows the results.
 
| Mean crying time / hours day^–1
Milk with lactase | 1.43
Milk without lactase | 2.57

(a)     Suggest an explanation for the results.
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(b)     The mothers were not told which solution contained lactase.
Suggest one reason why.
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(c)     Suggest one variable the doctors would have to control in this study to make it a fair test. Explain your answer.
Variable ........................................................................................................
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(d)     The doctors concluded that adding lactase to milk was, ‘A major breakthrough for babies with colic.’
Evaluate the evidence for this conclusion.
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(Total 8 marks)

Q17.          The diagram shows the human immunodeficiency virus (HIV).

(a)     Name
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B ..................................................................................................................
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The graph shows changes in the number of T-cells and HIV particles in the blood of a person following infection.

(b)     Explain why the number of HIV particles in the blood
(i)      rises during the first few months after infection
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(ii)     remains low between 1 and 7 years after infection.
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(c)     This person developed a large number of infections about 9 years after he first became infected with HIV. Using information from the graph, explain why.
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(Total 9 marks)
Q18.          Figure 1 shows sections through relaxed and contracted myofibrils of a skeletal muscle. The transverse sections are diagrams. The longitudinal sections are electron micrographs.
Figure 1

(a)     (i)      The electron micrographs are magnified 40 000 times.
Calculate the length of  band X in micrometres.
Show your working.

Length of band X =..................................... µm
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(ii)     Explain the difference in appearance between transverse sections A and C in Figure 1.
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(b)     Explain what leads to the differences in appearance between the relaxed myofibril and the contracted myofibril.
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(c)     Duchenne muscular dystrophy (DMD) is a condition caused by the recessive allele of a sex-linked gene. A couple have a son with DMD. They want to know the probability that they could produce another child with DMD. They consulted a genetic counsellor who produced a diagram showing the inheritance of DMD in this family.
This is shown in Figure 2.
Figure 2

The couple who sought genetic counselling are persons 6 and 7.

(i)      Give the evidence to show that DMD is caused by a recessive allele.
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(ii)     Give the numbers of two people in Figure 2 who are definitely carriers of muscular dystrophy.
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(iii)     Complete the genetic diagram to find the probability that the next child of couple 6 and 7 will be a son with muscular dystrophy. Use the following symbols:
X^D   = normal X chromosome
X^d   = X chromosome carrying the allele for muscular dystrophy
Y     = normal Y chromosome
 
| 6 | 7
Parental phenotypes | Unaffected | Unaffected
Parental genotypes | .............. | ..............
Gametes | .............. | ..............

 

Offspring genotypes          .....................................................................
Offspring phenotypes        .....................................................................
Probability of having a son with DMD ...................................................
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(d)     DMD is caused by a deletion mutation in the gene for a muscle protein called dystrophin. A deletion is where part of the DNA sequence of a gene is lost. People in different families may inherit mutations in different regions of this gene.
Scientists isolated the dystrophin gene from DNA samples taken from children 10, 11 and 12. They cut the gene into fragments using an enzyme. The scientists then used two DNA probes to identify the presence or absence of two of these fragments, called F and G. This allowed them to find the number of copies of each fragment in the DNA of a single cell from each child.
The table shows their results.
 
Child | Number of copies of gene fragment per cell
| F | G
10 (unaffected girl) | 2 | 1
11 (unaffected girl) | 2 | 2
12 (boy with DMD) | 1 | 0

(i)      The number of copies of gene fragments F and G shows that person 12 has DMD.
Explain how.
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(ii)     The number of copies of gene fragments F and G shows that person 12 is male.
Explain how.
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(iii)     The genetic counsellor examined the scientists' results. He concluded that person 10 is a carrier of DMD but her sister, 11, is not.
Describe and explain the evidence for this in the table.
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(e)     Person 12 took part in a trial of a new technique to help people with DMD.
Doctors took muscle cells from person 12’s father and grew them in tissue culture.
They suspended samples of the cultured cells in salt solution and injected them into a muscle in person 12’s left leg. They injected an equal volume of salt solution into the corresponding muscle in his right leg. Person 12 was given drugs to suppress his immune system throughout the trial.
Four weeks later, the doctors removed a muscle sample from near the injection site in each leg. They treated these samples with fluorescent antibodies. These antibodies were specific for the polypeptide coded for by gene fragment G of the dystrophin gene.
The results are shown in the table.

Location and
treatment

Percentage of muscle
fibres labelled with
antibody

Left leg - injected
with cultured cells
suspended in salt
solution

6.8

Right leg - injected
with salt solution

0.0

(i)      Why was it necessary to treat person 12 with drugs to suppress his immune system?
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(ii)     Explain why salt solution was injected into one leg and cultured cells suspended in salt solution into the other.
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(1)
(iii)     This technique is at an early stage in its development. The doctors suggested that further investigations need to be carried out to assess its usefulness for treating people with DMD.
Explain why they made this suggestion.
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(Total 25 marks)

Q19.          The graph shows changes in pressure in the aorta, left ventricle and left atrium during one heart beat.

(a)     The maximum pressure in the left atrium is lower than the maximum pressure in the left ventricle. What causes this difference in maximum pressure?
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......................................................................................................................(1)
(b)     A stethoscope can be used to listen to the sounds made by the heart.
(i)      What is the evidence from the graph that the first heart sound is caused by the atrioventricular valve closing?
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(1)
(ii)     What causes the second heart sound? Give the reason for your answer.
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(2)
(Total 4 marks)
Q20.          S       A medical officer investigated the effectiveness of five different types of influenza vaccine. A total of 1350 people agreed to be vaccinated. The medical officer divided these into five groups. The number who suffered from influenza in the following year was recorded. The results are shown in the table.
 
| Number of people vaccinated
Type of influenza vaccine | Suffered from influenza | Did not suffer from influenza | Total | Proportion suffering from influenza
I | 43 | 237 | 280 | 0.15
II | 52 | 198 | 250 | 0.21
III | 25 | 245 | 270 | 0.09
IV | | | 260 | 0.18
V | 57 | 233 | 290 | 0.20

(a)     Complete the spaces in the table for the people vaccinated with type IV vaccine.

(1)

(b)     The medical officer used a statistical test to assess the effectiveness of the five different vaccines.
(i)      What would be the null hypothesis?
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(1)
(ii)     The statistical test gave a probability of less than 0.05. What conclusion can be drawn from this?
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(1)
(c)     It was suggested that the raw data showed that the type III vaccine was the most effective. Give two reasons why this conclusion may not be reliable.
1 ....................................................................................................................
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2 ....................................................................................................................
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(2)
(Total 5 marks)
Q21.          Lacewings are insects that feed on aphids and mites, which are crop pests. The numbers of six species of lacewings, A to F, were counted on samples of apple and strawberry crops. The results are shown in the table.
 
Crop | Number of adults of each species of lacewing | Diversity index
| A | B | C | D | E | F |
Strawberry | 31 | 0 | 3 | 29 | 17 | 1 | 3.2
Apple | 10 | 1 | 1 | 7 | 0 | 1 |

The diversity index (d) is calculated from the formula

where N is the total number of organisms of all species
and n is the total number of organisms of each species.

(i)      Calculate the diversity index for lacewing species in the apple crop and write the figure in the table. Show your working.

 

(2)
(ii)      Suggest a reason why the diversity index for the lacewings is different between the two crops.
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(1)
(Total 3 marks)

Q22.          S       The diagram shows a single-celled organism called Chlamydomonas.

(a)     Chlamydomonas lives in fresh-water ponds. It uses its flagella to swim towards light of moderate intensity but away from very bright light. Using information in the diagram, explain the advantage of this behaviour.
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(2)

(b)     A Chlamydomonas cell has two flagella. These flagella contain a single sort of protein. A flagellum consists of a bundle of 242 filaments. Each filament consists of 7500 protein molecules. Each protein molecule contains 900 amino acid units.
(i)      What would be the minimum number of nucleotides in the coding region of the mRNA used to synthesise this protein?
.............................................................................................................
(1)
(ii)     In an investigation, a culture of Chlamydomonas was treated in a way that caused them to lose their flagella without any other damage to the cells. The flagella grew back to their original length in 60 minutes.
How many amino acid molecules would be incorporated into each growing flagellum per minute? Show your working.

Answer .........................................  (2)
(c)     The researchers investigated the rate at which the flagella grew in three different media.
1.      A medium containing actinomycin D, which prevents transcription by binding to the guanine in DNA
2.      A medium containing puromycin, which prevents translation by attaching to ribosomes
3.      A control medium
The results are shown in the graph.

(i)      Describe how the rate of growth was affected by puromycin.
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(2)
(ii)     The researchers concluded
1.       that the cells used mRNA that is already present in the cytoplasm for the regrowth of the flagella;
2.       that some of the regrowth uses protein molecules already present in the cell.
Explain the evidence for each of these conclusions.
1 ..........................................................................................................
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(4)
(Total 11 marks)

Q23.          Seals are aquatic mammals. They use lungs as organs of gas exchange so they do not breathe when they are under water during a dive.
The graph shows changes in oxygen and lactate concentration in the blood of a seal before, during and after a dive.

(a)     The concentration of oxygen in the blood fell during the dive. Explain why.
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......................................................................................................................
(1)

(b)     Use information in the graph to calculate how long it took from the end of the dive for the seal to recover fully.

 
 
Answer .......................................... minutes
(1)

(c)     Explain what causes the concentration of blood lactate to fall after a dive.
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(2)
(d)     Reducing the volume of blood pumped out by the heart reduces the rate of blood flow to the diaphragm muscles.
(i)      Give one other way in which blood flow into the diaphragm muscles may be reduced.
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(1)
(ii)     During a dive, blood flow to the diaphragm muscles of a seal is reduced. Suggest the advantage to the seal of maintaining some blood supply to the diaphragm muscles during a dive.
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(2)
(Total 7 marks)

Q24.          The graph shows the changes in pressure which take place in the aorta of a mouse during several heartbeats.

(a)     Which chamber of the heart produces the increase in pressure recorded in the aorta?
......................................................................................................................
(1)
(b)     The pressure of blood in the aorta decreases during each heartbeat but does not fall below 10 kPa. Explain what causes the pressure of blood to
(i)      decrease during each heartbeat;
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(1)
(ii)     stay above 10 kPa.
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(2)
(c)     The heart rate of a mouse is much higher than the heart rate of a human. Use the graph to calculate the heart rate of the mouse. Show your working.

 
 
Heart rate = .......................................... beats per minute
(2)
(d)     The cardiac output is the volume of blood pumped by a heart in one minute. The stroke volume is the volume of blood pumped by a heart in a single heartbeat.
cardiac output = stroke volume × heart rate
The cardiac output for a mouse with a heart rate of 550 beats per minute is 16.6 cm^3 per minute. Calculate the stroke volume for this mouse. Show your working.

 
 
Stroke volume = .......................................... cm^3
(2)(Total 8 marks)
Q25.          (a)     Describe how NADP is reduced in the light-dependent reaction of photosynthesis.
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(2)
(b)     In an investigation of the light-independent reaction, the amounts of glycerate
3-phosphate (GP) and ribulose bisphosphate (RuBP) in photosynthesising cells were measured under different environmental conditions.
Figure 1 shows the effect of reducing the carbon dioxide concentration on the amounts of glycerate 3-phosphate and ribulose bisphosphate in photosynthesising cells.
Figure 1

(i)      Explain why there is twice the amount of glycerate 3-phosphate as ribulose bisphosphate when the carbon dioxide concentration is high.
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(1)

(ii)     Explain the rise in the amount of ribulose bisphosphate after the carbon dioxide concentration is reduced.
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(1)

(c)     Figure 2 shows the results of an experiment in which photosynthesising cells were kept in the light and then in darkness.
Figure 2

(i)      In the experiment the cells were supplied with radioactively labelled ^14CO_2. Explain why the carbon dioxide used was radioactively labelled.
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(1)
(ii)     Explain how lack of light caused the amount of radioactively labelled glycerate 3-phosphate to rise.
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(2)

(iii)     Explain what caused the amount of radioactively labelled glucose to decrease after the light was switched off.
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(1)
(Total 8 marks)
Q26.          When coal is mined by open-cast mining, the top layer of soil is first scraped off and stored in a large heap. Once mining has finished, the area can be reclaimed. Soil from this store is then spread back over the surface.
Some of the bacteria living in the soil store respire aerobically and some respire anaerobically. Table 1 shows the numbers of aerobic and anaerobic bacteria found at different depths in a soil store.
 
Depth / cm | Mean number of bacteria per gram of soil (× 10^7)
| Aerobic bacteria | Anaerobic bacteria
| after 1 month | after 6 months | after 1 month | after 6 months
0 | 12.0 | 12.1 | 0.6 | 0.8
50 | 10.4 | 8.6 | 0.8 | 1.3
100 | 10.1 | 6.1 | 0.7 | 4.1
150 | 10.0 | 3.2 | 0.7 | 7.9
200 | 11.6 | 0.8 | 0.7 | 8.4
250 | 11.9 | 0.7 | 0.8 | 8.8
300 | 11.0 | 0.8 | 0.6 | 9.1

Table 1
(a)     Some of the soil used to determine bacterial numbers was collected from the surface of the soil store. Describe how you would ensure that this soil was collected at random.
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(2)

(b)     (i)      Describe how the numbers of aerobic bacteria after 6 months change with depth.
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(ii)     Explain the difference in the numbers of aerobic bacteria at a depth of 300 cm between 1 and 6 months.
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(2)
(c)     Explain how the changes in bacterial numbers which take place at 150 cm illustrate the process of succession.
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(3)

Dehydrogenase is an enzyme involved in aerobic respiration. Dehydrogenase activity in a soil sample can be used as a measure of the activity of aerobic bacteria. The graph shows the mean dehydrogenase activity of soil samples taken from the same depth in a soil store at different times. The bars on the graph represent two standard errors above and below the mean.

(d)     (i)      From what depth in the soil store would you expect these soil samples to have been taken? Use information from Table 1 to explain your answer.
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(2)
(ii)     How would you expect dehydrogenase activity to vary with depth after 6 months?
Use information from Table 1 to explain your answer.
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(3)

(e)     What do the error bars tell you about the difference between the mean dehydrogenase activity at 6 months and 3 years? Explain your answer in terms of probability and chance.
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(3)

(f)      Table 2 shows the dehydrogenase activity and the number of aerobic bacteria present in some soil samples.
 
Dehydrogenase activity / arbitrary units | Number of aerobic bacteria per gram of soil (× 10^7)
13.1 | 12.0
9.2 | 8.7
5.5 | 6.5
3.0 | 4.6
2.2 | 2.7
0.4 | 0.6
Table 2  A sample of soil was found to have dehydrogenase activity of 8.7 arbitrary units. Explain how you would use the data in Table 2 to predict the likely number of aerobic bacteria in 1 g of this soil sample.
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................................................................................................(3)(Total 20 marks)
Q27.          The graph shows dissociation curves for human oxyhaemoglobin at rest and during exercise.  Table 1 gives information about conditions in the body at rest and during exercise.

 
| Rest | Exercise
Plasma pH | 7.4 | 7.2
Blood temperature / °C | 37.0 | 39.0
Alveolar partial pressure of oxygen / kPa | 13.3 | 13.3
Tissue partial pressure of oxygen / kPa | 5.0 | 4.0

Table 1
(a)     What is meant by the term partial pressure?
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(1)

(b)     Use Table 1 and the graph to calculate the difference in the percentage saturation of haemoglobin in the tissues between rest and exercise.

Answer ............................ %
(1)
(c)     Explain the differences between the figures shown in Table 1 for rest and exercise.
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(4)

(d)     Explain the advantage of the difference in position of the dissociation curve during exercise.
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Table 2 shows how the oxygen concentration in the blood going to and from a muscle changes from rest to heavy exercise.

Oxygen concentration / cm^3 per 100 cm^3
blood

| Blood in arteries | Blood in veins
At rest | In solution | 0.3 | 0.2
| As oxyhaemoglobin | 19.5 | 15.0
| Total oxygen | 19.8 | 15.2
During heavy exercise | In solution | 0.3 | 0.1
| As oxyhaemoglobin | 20.9 | 5.3
| Total oxygen | 21.2 | 5.4

Table 2
(e)     By how many times is the volume of oxygen removed from the blood by the muscle in Table 2 during heavy exercise greater than the volume removed at rest?
Show your working.

 
 
Answer ..................................... times   (2)
(f)      Does enriching inspired air with oxygen have any effect on the amount of oxygen reaching the tissues? Support your answer with evidence from the graph and Table 2.
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 (3)

S       (g)     The change to the dissociation curve is one of a number of ways in which the total oxygen supplied to muscles is increased during exercise. Give two other ways in which the total oxygen supplied to muscles during exercise is increased.
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(2)
(Total 15 marks)

Q28.          (a)     The graph shows hourly blood pressure recordings from a group of 65 people.
 
(i)      Describe how the mean maximum arterial blood pressure changes over the period shown in the graph.
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.............................................................................................................(1)
(ii)     In each cardiac cycle, the arterial pressure has a maximum value. Explain the link between this maximum value and the events of the cardiac cycle.
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(iii)     The recordings shown in this graph were taken from an artery. Describe two ways in which you would expect blood pressure in a vein to differ from that in an artery.
1 ....................................................................…..................................
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2 ....................................................................…..................................
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(b)     Molecules of different substances differ in size. The relative molecular mass of a substance gives an indication of the size of its molecules. The table shows the relative permeability of the wall of a capillary to different molecules.
 
| Substance | Relative molecular mass | Relative permeability of capillary wall
| Water Urea Glucose Haemoglobin Albumin (plasma protein) Globulin (plasma protein) | 18 60 180 68 000 69 000 140 000 | 1.00 0.96 0.60 0.01 0 0

(i)      Describe the relationship between molecule size and the permeability of the capillary wall.
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(ii)     The water potential of the plasma at the venule end of the capillary is more negative than the water potential at the arteriole end. Use the table to explain why.
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(2)
(iii)     Although the capillary walls are slightly permeable to haemoglobin molecules, there is no haemoglobin in the tissue fluid. Explain what causes the absence of haemoglobin in tissue fluid.
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(1)
(Total 9 marks)

Q29.          People considered ‘at risk’ are offered a vaccination against influenza each year. The bar chart shows the number of people in the UK population aged 65 and over and the percentage of those who were vaccinated against influenza each winter.

(a)     Suggest one reason to explain the change in the percentage of people aged 65 and over being vaccinated.
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......................................................................................................................(1)
(b)     (i)      Calculate the change in the total number of people aged 65 and over being vaccinated between 1990/91 and 2000/01. Show your working.

 
Answer .................................(2)
(ii)     A student suggested that some people aged 65 and over were being vaccinated every year. Explain how the information in the bar chart supports this suggestion.
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(iii)     Suggest why it is advisable for people to be vaccinated against influenza every year.
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(2)
(c)     An influenza virus consists of a protein coat surrounding nucleic acid. The influenza vaccine consists only of the protein coat of the virus. Explain how the influenza vaccine produces immunity in the body.
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(Total 9 marks)
﻿
Q30.         Some students investigated the effect of light intensity in the environment on the size of leaves of nettles. They measured leaves on sixty plants in each of two sites. The results are summarised in the table.
 
Dimensions of leaves / mm | Site with high light intensity | Site with low light intensity
Length of longest leaf | 113 | 116
Length of shortest leaf | 41 | 42
Mean length | 86 | 92
Mean maximum width | 68 | 74

Standard deviation of lengths | 11 | 16
Standard deviation of maximum widths | 7 | 11
 
(a)     Each leaf to be measured was selected in the following way.
•        The top left hand corner of a quadrat frame was placed at coordinates given by a random number table; the nettle plant nearest the centre of the quadrat was selected,
•        The sixth leaf from the tip of the plant was selected.
Explain the importance of
(i)      the method of selecting the nettle plant;
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(2)
(ii)     measuring the sixth leaf.
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(2)
(b)      Use the data about the length of leaves in the two sites to explain why standard deviation is more useful than range as a measure of variation within a population.
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(4)
(c)     The area of a nettle leaf can be estimated using the formula
area   =   length × maximum width × 0.5
Calculate the ratio of the mean area of the leaves from the site with low intensity to the mean area of the leaves from the site with high light intensity. Show your working.

 
Answer .....................................(2)
(Total 10 marks)
Q31.          Relaxation of heart muscle is called diastole. Contraction is called systole. The diagram shows the periods of diastole and systole when the heart is beating.

(a)     At what time is the volume of blood in the ventricle at a maximum?
......................................................................................................................
(1)

(b)     Calculate the heart rate in beats per minute. Show your working.

 
 
Heart rate = ................................... beats per minute
(2)
(c)     The valves between the atria and the ventricles are closed between 0.1 s and 0.4 s.
(i)      Explain how pressure causes these valves to be shut.
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(1)
(ii)     Explain how closure of these valves is essential to the functioning of the heart.
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(1)

The graph shows the effect of the pressure of blood in the veins on the stroke volume of the heart.

(d)     Describe how venous blood pressure affects stroke volume.
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(2)
(e)     At a venous pressure of 2 kPa, the cardiac output is 5600 cm^3 per minute. Calculate the number of times the ventricle contracts in one minute. Show your working.

 

Answer ......................................
(2)
(f)      Explain how blood in a vein in the leg is returned to the heart.
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(6)
(Total 15 marks)
 
Q32.          (a)     Figure 1 shows the changes in membrane potential at one point on an axon when an action potential is generated.
Figure 1

The changes shown in Figure 1 are due to the movement of ions across the axon membrane. Complete the table by giving the letter (A to D) that shows where each process is occurring most rapidly.
 
Process | Letter
Active transport of sodium and potassium ions |
Diffusion of sodium ions |
Diffusion of potassium ions |

(2)
(b)     Figure 2 shows the relationship between axon diameter, myelination and the rate of conduction of the nerve impulse in a cat (a mammal) and a lizard (a reptile).
Figure 2

(i)      Explain the effect of myelination on the rate of nerve impulse conduction.
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(2)

S       (ii)     For the same diameter of axon, the graph shows that the rate of conduction of the nerve impulse in myelinated neurones in the cat is faster than that in the lizard. Suggest an explanation for this.
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.............................................................................................................(2)
Figure 3 shows how a stimulating electrode was used to change the potential difference across an axon membrane. Two other electrodes, P and Q, were used to record any potential difference produced after stimulation. The experiment was repeated six times, using a different stimulus potential each time. In experiments 1 to 4, the stimulating voltage made the inside of the axon less negative. In experiments 5 and 6, it made the inside of the axon more negative.
Figure 3

 
(c)     Explain the results of experiments 1 to 4.
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......................................................................................................................(5)
(d)     Figure 4 shows two neurones, X and Y, which each have a synapse with neurone Z.
Figure 4
 
Neurone X releases acetylcholine from its presynaptic vesicles. Neurone Y releases a different neurotransmitter substance which allows chloride ions (Cl^‑) to enter neurone Z. Use this information, and information from Figure 3, to explain how neurones X and Y have an antagonistic effect on neurone Z.
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(Total 15 marks)
Q33.          Taxol is a drug used to treat cancer. Research scientists investigated the effect of injecting taxol on the growth of tumours in mice. Some of the results are shown in Figure 1.
Figure 1
 
Number of days of treatment | Mean volume of tumour / mm^3
| Control group | Group injected with taxol in saline
1 | 1 | 1
10 | 7 | 2
20 | 21 | 11
30 | 43 | 20
40 | 114 | 48
50 | 372 | 87

(a)     Suggest how the scientists should have treated the control group.
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......................................................................................................................(2)
(b)     Suggest and explain two factors which should be considered when deciding the number of mice to be used in this investigation.
1 ...................................................................................................................
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(c)     The scientists measured the volume of the tumours. Explain the advantage of using volume rather than length to measure the growth of tumours.
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(1)

(d)     The scientists concluded that taxol was effective in reducing the growth rate of the tumours over the 50 days of treatment. Use suitable calculations to support this conclusion.

 
 
(2)
(e)     In cells, taxol disrupts spindle activity. Use this information to explain the results in the group that has been treated with taxol.
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(3)

(f)      The research scientists then investigated the effect of a drug called OGF on the growth of tumours in mice. OGF and taxol were injected into different mice as separate treatments or as a combined treatment. Figure 2 and Figure 3 show the results from this second investigation.
Figure 2

Figure 3

Treatment

Mean volume of tumour following 70 days treatment
/mm^3 (± standard deviation)

OGF | 322 (± 28.3)
Taxol | 207 (± 22.5)
OGF and taxol | 190 (± 25.7)
Control | 488 (± 32.4)

(i)      What information does standard deviation give about the volume of the tumours in this investigation?
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(1)

(ii)     Use Figure 2 and Figure 3 to evaluate the effectiveness of the two drugs when they are used separately and as a combined treatment.
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(4)
(Total 15 marks)

Q34.Scientists tested a claim that modified citrus pectin (MCP) increased the production of antibodies by the immune system.
•        They divided a large number of mice into five groups.
•        They gave the mice in each group a different amount of MCP in their food.
•        The scientists then stimulated antibody production in the mice. They did this by injecting them with a solution containing sheep red blood cells.
The results are shown in the graph.
 
(a)     The data obtained in this investigation have been plotted on a graph. How would you join the points? Give a reason for your answer.
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(b)     Use the graph to describe the effect of MCP on mean antibody production.
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(c)     Calculate the percentage increase in antibody production from when there was no MCP in the diet to when the dose is 1.0 g per kg.

 
Answer ....................................................%  (2)
(d)     The dose of MCP given to the mice was calculated in g per kg body mass. Explain why the dose was calculated per unit mass.
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(1)
(e)     Explain how antibodies were produced when the mice were injected with sheep red blood cells.
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(Extra space) .................................................................................................
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(3)
(f)      A newspaper suggested that these data show that taking MCP will give people increased resistance to disease. With reference to the data give two reasons why this conclusion may not be valid.
1 .....................................................................................................................
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2 .....................................................................................................................
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(2)
(Total 11 marks)

Q35.A scientist investigated the use of a new source of carbohydrate in the production of ethanol for biofuel. He wanted to find the optimum time to leave a mixture of yeast and this carbohydrate to produce ethanol. The scientist set up an airtight container containing yeast and this carbohydrate. He then measured the oxygen, carbon dioxide and ethanol concentrations over 8 hours.
The results of his investigation are shown in the graph below.

 
(a)     The scientist used a container that was airtight.
Give two explanations why the container had to be airtight.
1 ......................................................................................................................
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2 ......................................................................................................................
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(b)     Explain the relationship between the concentration of oxygen and the concentration of carbon dioxide between 0 and 3 hours.
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(2)
(c)     The scientist concluded that yeast starts to respire anaerobically when the oxygen concentration falls below a certain concentration. What is the oxygen concentration when the yeast starts to respire anaerobically? Explain your answer.
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(2)

(d)     (i)      The scientist worked for a biofuel company. Give two suggestions for further work he should do to make sure that the results he presented to the company were reliable.
Explain how each of your suggestions would make the results more reliable.
Suggestion ............................................................................................
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Explanation ...........................................................................................
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Suggestion ............................................................................................
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Explanation ...........................................................................................
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(4)

(ii)     The scientist recommended that when the ethanol is produced commercially as biofuel the reaction should be stopped at 6 hours. Use the graph to suggest why.
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(2)

(iii)    The scientist’s work was funded by a biofuel company. Explain why the source of funding can cause problems with scientific work.
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(2)
(Total 16 marks)

M1.          (a)     To sterilise/kill bacteria;
So that only one kind of bacteria present on agar plate/to prevent contamination
(by bacteria);
2

(b)     Clear zone / inhibition zone is where bacteria have not grown/been inhibited/killed;Antibiotic diffuses out of paper disc/into agar;Bacterium A inhibited/killed by tetracycline/tetracycline has little effect on
bacterium B;Bacterium B inhibited/killed by penicillin/bacterium A resistant to penicillin;Both kinds of bacteria resistant to streptomycin;
Q Ignore references to ‘immune’
4 max
[6]

M2.          (a)     (i)      to increase surface area (for carbon dioxide absorption);
1
(ii)     oxygen is used / carbon dioxide emitted is absorbed; 
so decrease in volume / pressure;
2
(iii)     change of level of (manometer) liquid over time;
bore of tube;
mass of snails;
time interval;
3 max

(b)     (i)      valid similarity taking into account SD e.g. between 5 and 15 °C
both show little effect of temperature / intakes similar
between 5 and 15 °C;
1
valid difference taking into account SD above 15 °C e.g.
rise at 20 °C and above is less when snails kept in sea water;
1
(ii)     standard deviations high;
means unreliable;
(accept 25 °C being out of normal range for snail / 
not enough temperature readings for 1 mark)
2
[10]

M3.          (a)     random sampling method; 
use of large numbers / many / 10 or more quadrats in each area;
counting daisies and dividing by area;
3
(b)     the cutting has no effect;
1

(c)     daisy, dandelion, buttercup show (statistically) significant differences;
no significant effect on plantains;
comment on relative significance of daisy / dandelion / buttercup;
regular cutting linked to significant increase in density of daisy / dandelion;
linked to significant decrease in density of buttercup;
(no marks if significance idea omitted)
3 max
[7]

M4.          (a)     the more light absorbed, the greater the rate of photosynthesis;
light provides the energy for light dependent reactions / photolysis / 
light independent reactions / production of reduced NADP / 
exciting electrons in chlorophyll;
(do not give credit if energy is used in photosynthesis)
2
(b)     count the number of bubbles / measure the volume of gas / measure the 
change in pH / carbon dioxide / hydrogen carbonate ions;
(credit oxygen produced)
1

(c)     530 – 630 nm;
(any values within this range)
limited absorption of light / (green) plants reflect green light / 
limited photosynthesis at these wavelengths of light;
(allow references to no light absorbed or no photosynthesis)
2
(d)     (i)      chlorophyll excited / reduced NADP formed;
electrons from chlorophyll / reduced NADP changes the dye colour;
2
(ii)     ADP and phosphate needed to produce ATP / ATP is a product of 
the light dependent reactions;
ADP levels are a limiting factor;
(must explain the idea of limiting factors – do not credit answers like more ADP causes more photosynthesis) 
2
[9]

M5.          (a)     (i)      A = phospholipid
B = protein;
(both correct)
1

(ii)     allows movement of lipid soluble / non-polar molecules / named
e.g. water / gases;
prevents movement of water soluble / polar molecules / named
e.g. ions / amino acids;
idea of selection / membrane partially / differentially permeable / 
large molecules do not move through, small molecules do;
(accept semi-permeable)
2 max

(b)     (i)      diffusion
(reject facilitated)
1
(ii)     higher rate of exchange / diffusion;
prevents cooling of the blood / prevents increase in viscosity;
2

(iii)     concentration gradient maintained / equilibrium never achieved;
blood always meets fluid with lower concentration of urea;
diffusion / exchange along the whole length of surface;
2 max
(iv)    0.2 × 60 = 12 dm^3 h^-1;
(principle: volume per hour)
12 × 5 = 60 dm^3;
(correct answer 2 marks)
2
[10]

M6.          (a)     Population – organisms of one species in an ecosystem / habitat / area;
Community – organisms of all species / all populations in an
ecosystem / habitat / area;
2

(b)     (i)      No immigration / migration (Ignore references to emigration);
No reproduction (Ignore references to death);
Idea of mixing;
Marking does not influence behaviour / increase vulnerability
to predation;
Sample / population large enough;
max 2

(ii)     ; 672;
Correct answer (however derived) scores 2 marks
Incorrect answer with evidence of correct method scores 1 mark.
2
(c)     Principle of randomly placed quadrats and method of producing random
quadrats; (Reject ‘throwing’)
Valid method of obtaining no. dandelions in given area (mean per
quadrat / total no. in many quadrats);
Multiply to give estimate for total field area;
3

(d)     (i)      Niche of A – 1;
Niche of B – 3;
Too small for B / too hot for A – 4;
Too large for A / too cold for B – 2;
All four correct = 2 marks; any 2 correct = 1 mark
2

(ii)     Original population living in one area / 2 species evolved in
the area;
Idea of genetic variability;
Concept of reproductive isolation;
Possible mechanism;
Gene pools become increasingly different;
Until interbreeding does not produce fertile offspring;
max 4
[15]

M7.          (a)     Endonuclease / restriction enzyme;
1
(b)     DNA made of base pairs;
Each base pair is same length / occupies same distance
along backbone;
2

(c)     (i)      Second blank box from left labelled 6;
1
(ii)     Distance moved depends on length / number of base pairs / 
second longest fragment / second shortest distance identified;
1
(d)     5;
1
[6]

M8.          (a)     (i)      where mitosis / division / growing / occurs
(reject growing cells)
1
(ii)     to distinguish chromosomes / chromosomes not visible
without stain;
1
(iii)     to let light through / thin layer;
1

(b)     (i)      74 + 18 / 982;
= 9.4% / 9%;
2
(allow 1 mark for identifying prophase & metaphase i.e.92 or correct method using wrong figures)

(ii)     genetic differences / different types of garlic;
time of day;
chance;
age of root tip;
water availability;
temperature;
nutrient availability;
(environmental factors = 1 but cannot be awarded in addition to a named environmental factor)
2 max
[7]

 

M9.          (a)     Krebs cycle / link reaction / pyruvate to acetylcoenzyme A;
Q Accept valid alternative for any of these steps.
1

(b)     (Respiratory reactions controlled by) enzymes;
Rate decreases as less kinetic energy / fewer collisions (between substrate and active site) fewer E-S complexes formed;
2

(c)     Requires hydrogen / electrons / is reduction;
Hydrogens from reduced NAD / reduced NAD reduces (pyruvic acid) / reduced NAD oxidised;
Information may be on diagram
2

(d)     Respiring anaerobically;
(Anaerobic respiration / respiration with nitrogen) less efficient / produces less ATP;
More anaerobic respiration / more glucose / substrate must be respired to produce same amount of ATP (so more carbon dioxide produced);
3
[8]

M10.          (a)     (i)      Taxis;
Ignore references to positive and negative, and prefixes such as photo-
Accept taxes / tactic
Allow phonetic spelling
1
(ii)     Moves towards stimulus / towards light;
Direction must be correct.
1

(b)     Gravity;
Antennae involved;
Doesn’t show light is involved / doesn’t respond to light as they are
unable to see / as eyes are covered;
Accept geotaxis
3

(c)     Helps them to leave the soil / ground / reach the surface;
Disperse / produce new colonies;
Avoid competition;
2 max
[7]

M11.(a)     Light (intensity) / temperature / air movement / humidity;
1
(b)     Prevent air entering / continuous water column;
Allow answer in context of shoot, xylem or potometer.
1

(c)     Distance and time;
Reject ‘amount bubble moves’
1
Radius / diameter / area (of capillary tube);
1

(d)     (used to provide) turgidity / support / description of;
(used in) photosynthesis / (produced in) respiration;
Apparatus not sealed / ’leaks’;
2 max

(e)     (i)      Returns bubble (to start);
1
(ii)     Increases reliability (of results) / anomalous result can be identified;
Q Ignore references to validity / precision / accuracy etc.
1
[8]

M12.(a)     (i)      1.      Sex;
2.      Lifestyle;
Stress, smoking, diet etc are examples of lifestyle.
3.      Body mass;
3. Allow weight for mark point 3.
4.      Health;
Reject: height.
5.      Ethnicity;
6.      Genetic factors / family history;
2 max

(ii)     1.      Large sample / number / 410 000;
Reject: random
2.      Long time period / 8.5 / many years;
3.      Different countries / more than one country;
2

(b)     Correct answer of 209 / 209.1 = 2 marks;
Answer of 210 = one mark
Incorrect answer but multiplies by 8.5 = 1 mark;
2

(c)     Age affects risk of cancer;
Must relate to cancer not just to illness
1

(d)     1.      Correlation does not mean causal relationship;
1. Reject casual for point 1.
Reference to 'due to other factors' on its own is not enough for a mark
2.      Tea / coffee contains other substances / different amounts of caffeine / estimated intake (of tea / coffee);
3.      No control group;
4.      Only one type of cancer studied / further studies required / only one investigation / study / group;
4

(e)     (i)      1.      Treated the same;
2. Accept decaffeinated
2.      No caffeine;
2. Reject placebo.
2

(ii)     1.      Absorb different amounts;
Reject: Different body masses
2.      Broken down by enzymes / digested;
3.      Different blood volumes;
4.      Differences in metabolism;
5.      Caffeine from a different source;
1 max

(iii)    1.      Less oxygen / glucose to (cancer) cells;
'Reduces cell division' on its own should not be credited.
2.      Less carcinogens;
3.      Reduces spread of cancer (cells);
1 max
[15]

M13.(a)     1.      To show the effect of the inhibitor / drug;
2.      To show the effect of yoghurt (on its own does not affect blood
glucose);
2

(b)     1.      Food is a factor affecting blood glucose / different foods contain 
different amounts of starch / glucose / sugar / carbohydrate;
Accept converse
2.      To keep starch / fibre intake the same / similar;
Accept something in food which affects the inhibitor
2

(c)     1.      Fewer E-S complexes formed;
2.      (With inhibitor) less / no starch digested to maltose ;
Require knowledge that maltose comes from starch
3.      (So) less / no glucose from maltose;
Require knowledge that glucose comes from maltose
Accept no glucose
4.      (So) less absorption of glucose (from gut);
2 max

(d)     Suitable reason; with explanation;
Paired responses – do not mix and match
Ignore references to correlation does not prove causation, 
it could be due to other factors
Examples,
1.      Need larger sample / only 30 mice / only 15 mice in each group;
Accept small sample size
2.      Might not be representative / anomalies might have a bigger or smaller effect;
Accept mean not reliable
OR
3.      Investigation only lasted 20 days;
Experiment was not long enough
4.      Can’t see what longer term effects are;
OR
5.      Fall in blood glucose is small / numbers from graph;
6.      Mice with inhibitor still have a large rise in blood glucose / so don’t know if differences significant;
Accept differences are due to chance
OR
7.      No stats / SDs / SEs;
8.      So don’t know if differences significant;
OR
9.       Blood glucose could continue to fall;
10.     which could be harmful;
OR
11.     No group without yoghurt;
12.     So cannot compare to other groups;
2 max
[8]

M14.(a)     1.      (Some of the) light that passes through is absorbed
         by chlorophyll b;
2.      This is light of around 500 and / or around 640;
Accept any value or range between 460 and 540 and / 
or 600 and 670
2

(b)     (i)      Supports hypothesis 2 (no mark)
1.      Greater carotenoid found in sun leaves than shade leaves of beech tree;
2.      Sun leaves exposed to much brighter light than shade leaves;
OR
It supports hypothesis 2 because it does not support hypothesis 1 (no mark)
3.      Although carotenoids absorb wavelengths of light that pass through leaves;
4.      There are not more carotenoids in shade leaves;
2

(ii)     1.      Mass of pigments / carotenoids in sun and shade leaves of other trees;
2.      Position of carotenoids in leaf cells;
3.      Effect of bright light on (isolated) chlorophyll;
4.      Whether without carotenoids chlorophyll is damaged (supporting hypothesis 2) / photosynthesis is reduced (supporting hypothesis 1);
1 max
[5]

M15.(a)     1.      Thin slice/section;
2.      Put on slide in water / solution / stain;
3.      Add cover slip;
Accept: ‘between two slides’
Max 2
(b)     200 (μm);;
OR
1.      Divide image length by key length eg 64/16 = 4;
2.      Multiply by 50 eg 4 × 50;
Accept for 2 marks answers in the range of 185-217 (μm)
Max 1 mark for responses not within the range
Accept: measurements in the ranges 63-65mm and 15-17mm
2
(c)     1.      Select large number of cells / select cells at random;
Accept: > 3 for “large number”
Accept: many fields of view for ‘large number of cells’
Accept: all cells in field of view
2.      Count number of chloroplasts;
3.      Divide number of chloroplasts by number of cells;
Ignore: ‘calculate the mean’
3
[7]

M16.          (a)     Lactose intolerance in babies/babies don’t make/have lactase;
Lactose (in milk) causes colic/crying/discomfort;
Lactase breaks down lactose/milk sugar;
2 max

(b)     To avoid prejudice/bias from mother (when recording crying);
Accept for mother she/they/their
Accept to avoid mothers being concerned that their
child has lactose intolerance. Q - Reject vague ref. to fair test/avoid bias with no ref. to mothers.
1

(c)     One variable; with explanation;
Example,
Type of milk used;
So same concentration of lactose;
Same age of children;
Change in enzyme production with age;

Same age of children;
Change in milk consumption with age;
Same volume of milk (per kg baby);
So same dosage (of lactose);
Accept ref. to controlling other factors in diet
Reject time in the context of duration of investigation, given in stem
Accept e.g. time of feeding each day
Accept temperature of milk, related to action of lactase enzyme
2 max

(d)     There is a decrease in crying;
Could be other causes/symptoms of colic;
Babies might cry for reasons besides colic/might have colic and
not cry/can’t be sure they have colic, they can’t talk;
Don’t know number of babies in trial/need a larger study;
So don’t know how reliable mean is;
Standard deviation not given/spread of data;
So don’t know whether difference is significant;
Babies still crying (for 1.43 hours);
Recording by mothers might not be reliable;
Longer running study to make sure effect (of lactase) lasts;
3 max
[8]

M17.         (a)     A = envelope/membrane/phospholipid (bilayer);B = capsid / nucleocapsid / capsomere / protein;
2
(i)      (HIV is) invading cells which make new viruses;Cells release viruses into blood;
2
(ii)     Virus remains dormant/exists as provirus/exists as DNA in host DNA;
Accept virus stays in cells
1

(c)     HIV destroys T cells;More (free) viruses produced leads to fall in T-cells;(So fewer) T-cells activate B-cells/memory cells;
Reduced/no antibody production;Immune system not working properly/inability to fight infection;Opportunistic infections;
4 max
[9]

M18.          (a)     Correct answer: 1.25;
Ignore working
OR    (if wrong answer)
 / = 1 mark
125 but wrong order of magnitude = 1 mark
2

(ii)     C has myosin / thick (and actin / thin) filaments;
OR
A has only actin / thin (/ no myosin / no thick) filaments;
1 max

(b)     When contracted:
Thick & thin filaments/myosin & actin overlap more;
Interaction between myosin heads & actin / cross-links form;
Movement of myosin head;
Thin filaments / actin moved along thick filaments / myosin;
Movement of thin filaments / actin pulls Z-lines closer together;
Displacement of tropomyosin to allow interaction;
Role of Ca^2+;
Role of ATP;
Allow ref. to ‘sliding filament mechanism’ /
described if no other marks awarded
4 max

(c)     (i)      8 has DMD but 3 and 4 do not / 12 has DMD but 6 and 7
do not / neither parent has the condition but their child has;
Allow parents 3 and 4 give 8, parents 6 and 7 give 12
1
(ii)     4 AND 7;
1

(iii)     Parental genotypes:  6 = X^DY AND 7 = X^DX^d
AND
Gametes correct for candidate’s P genotypes ‒ e.g.
X^D and Y + X^D   and  X^d;
Offspring genotypes correctly derived from gametes e.g.
X^DX^D + X^DX^d + X^DY + X^dY;
Male offspring with MD correctly identified: X^dY;
Probability = 0.25 / correct for candidates offsprings genotypes;
Accept ¼ / 1 in 4 / 1:3 / 25%
NOT ‘3:1’ / ‘1:4’
4

(d)     (i)      No gene fragment G;
1
(ii)     Only one copy of gene fragment F;
Male has only one X-chromosome / is XY
(c.f. female has two / is XX);
2

(iii)     10 has only one copy of gene fragment G;
10 has only one normal X-chromosome / has one abnormal /
has only one normal allele / has one X^d / is X^DX^d / is heterozygous;
11 has two normal X-chromosomes / has 2 normal alleles /
is X^DX^D / has not got X^d / has 2 copies of (F and) G;
3

(e)     (i)      To prevent rejection / prevent antibody production vs. injected cells /
injected cells have (foreign) antigen (on surface);
1
(ii)     Shows effect of cells / not just effect of injection / not just effect of
salt solution;
1

(iii)     Only one person tested so far ‒ need more to see if similar results /
need more to see if reliable;
Need to assess if new (dystrophin positive) muscle fibres are
functional / if muscle becomes functional;
Can’t tell how widespread effect is in the muscle / sample taken
near injection site;
Need to test for harmful side effects;
Need to test if successful for other mutations of dystrophin gene;
Need to assess permanence / longevity of result/insufficient time
allowed in investigation;
(In this patient) only small response / %;
Further sensible suggestion;
4 max
[25]

M19.          (a)     less muscle / thin(ner) wall in left atrium;
1
(b)     (i)      pressure of left ventricle higher than pressure of left atrium;
1
(ii)     closing of the semi-lunar valves / pocket valves;
pressure in artery / aorta is higher than ventricle;
2
[4]

M20.          (a)     47 213;
1
(b)     (i)      there is no difference in the proportion / number of influenza cases
between the 5 vaccines;
(reject vaccinated versus no vaccinated)
1
(ii)     significant difference in proportion / number of cases of influenza
between the vaccines / the null hypothesis should be rejected;
1

(c)     sample size small;
possible differences in exposure to infection;
exposure to different strains / mutants;
possible differences in existing immunity;
possible differences in sex / age;
possible differences in socio-economic status;
2 max
[5]

M21.          (i)      for correct use of sigma;
numerator = 380 and denominator = 132;
2
2.87 to 2.9 gains 2 marks
(do not allow 2.8 or denominator = 135)
(ii)      more types of prey found on strawberries;
1
[3]

M22.         (a)     chloroplast, so cell photosynthesises and moves to optimum / best light intensity for photosynthesis;
avoids damage due to bright light;
2

(b)     (i)      2700
1
(ii)      = 27 225 000 / 27 × 10^6 = 2 marks
(allow 1 mark for principle: )
2

(c)     (i)      rate slightly slower / not affected in first 20 / 30 minutes / lower
peak than control;
then decreases / much lower (than control);
(allow 1 mark for increase in first 20 / 30 minutes, then decreased, if not compared with control / normal)
(disqualify flagellum grows longer)
2

(ii)     1.       actinomycin has no effect (on growth of flagella);
          even though mRNA production / transcription prevented;
(accept references to ‘expt 1’)
2.       (re)growth little affected by puromycin at first;
protein synthesis inhibited, so likely to be using proteins
present;
4
[11]

M23.          (a)     used in (aerobic) respiration / to provide energy / ATP
(and not replaced by breathing) / used up by muscle and not replaced;
(reject used up and used up and not replaced)
1
(b)     36;
1

(c)     converted back to pyruvate / glycogen / glucose / CO_2 and H_2O;
reacted with oxygen / oxidised;
(reject “breaking down” with respect to glycogen and glucose)
2

(d)     (i)      vasoconstriction / contraction of muscles in arteries / arterioles / 
arteries / arterioles close;
(reject contraction of arteries / arterioles / capillaries)
1
(ii)     supplies oxygen / glucose or removal of carbon dioxide / lactate;
so cells can respire when not contracting / breathing;
2
[7]

M24.          (a)     left ventricle;
1
(b)     (i)      (left) ventricle / heart relaxes / diastole / filling / not contracting;
1
(ii)     elastic tissue / wall;
recoils / springs back (to maintain pressure);
(“contraction / muscle causing recoil” negates second point)
2

(c)     correct answer, 666 to 667 gains 2 marks; allow 1 mark for principle;
correct time for 1 heartbeat as 90 (ms) or 630÷7 / 
60 ÷ incorrect time identified from graph;
2
(d)     correct answer, 0.03, gains 2 marks;
(allow 1 mark for correct working, 16.6 ÷ 550, if answer wrong)
2
[8]

M25.          (a)     electrons;
from chlorophyll / photolysis;
2
(b)     (i)      RuBP combines with carbon dioxide to produce 2 x GP;
1
(ii)     less used to combine with carbon dioxide / 
less used to form glycerate 3-phosphate;
1

(c)     (i)      used in photosynthesis allows detection of products;
1
(ii)     ATP and reduced NADP not formed;
GP is not being used to form RuBP / is being formed from RuBP;
2
(iii)     used in respiration / formation of starch / cellulose;
1
[8]

M26.          (a)     Tapes / string / axes laid out at right angles / grid area;
Method of obtaining random co-ordinates;
Do not allow “Use random number generator”
2

(b)     (i)      Decrease then remain constant;
From 200 cm / over 150 cm;
2
(ii)     Oxygen decreasing because soil becomes more compacted / not
replaced;
Decrease in oxygen leads to fewer aerobes surviving;
2

(c)     Anaerobic bacteria replace aerobic as oxygen decreased by aerobic bacteria;
Remove competition;
Aerobic bacteria no longer able to survive in these conditions;
3

(d)     (i)      Near the surface / in top 50 cm;
Table shows decrease with time at greater depths;
2
(ii)     Decrease;
Fewer aerobic bacteria with depth;
Oxygen concentration decreases / less oxygen at depth;
3

(e)     Probability greater than 95% / 0.95;
Results are not due to chance / results are significant;
Because bars do not overlap;
3
(f)      Plot as graph;
Draw line of best fit;
Read off appropriate value;
3
[20]

M27.          (a)     It is a measure of the concentration of a gas
(in a mixture of gases or a liquid);
1
(b)     37-38%
Accept 36 – 39

(c)     muscle contraction causes increased respiration;
increased CO_2 production lowering blood pH / lactate released
lowering blood pH;
increased heat released therefore increased temperature;
increased O_2 consumption lowering tissue PO_2;
4

(d)     haemoglobin has a lower affinity for oxygen;
more O_2 for respiration;
2
(e)     3.4 times = 2 marks
(incorrect answer in which candidate shows amount of oxygen removed at rest is 4.6 and amount removed during exercise is 15.8 = 1 mark)
2

(f)      Nearly all O_2 is transported by haemoglobin / v. little transported in plasma;
EITHER
Haemoglobin is (nearly) fully saturated with O_2 at the alveoli both at
rest and when exercising;
Therefore no (very little) further increase is possible;
OR
Haemoglobin is only 95% saturated with oxygen at the alveoli;
Therefore enriching inspired / air with oxygen will raise this to 100%;
3

(g)     increased depth / rate / pulmonary ventilation;
increase stroke volume / heart rate / Q increases blood flow rate;
arterioles [Accept artery] supplying the muscles
dilate / vasodilation / greater proportion of blood flow to the muscles;
max 3
[15]

M28.          (a)     (i)      Pattern described as constant / decrease to 04.00 / 06.00
then rising;
1
(ii)     Corresponds to ventricles contracting / systole;
1
(iii)     Less / little difference between maximum and
minimum / less variation / constant / not pulsed / smoother;
pressure in vein lower
2

(b)     (i)      The larger the molecule, the less permeable;
Over 68 000 walls not permeable;
2
(ii)     Plasma proteins / albumin and globulin too large to leave capillary;
Water lost / Increase in concentration of proteins in blood / plasma;
2
(iii)     Haemoglobin in red blood cells / 
Haemoglobin too large to pass through membrane of RBC / 
Red blood cells (containing haemoglobin) too large to pass
through wall;
1
[9]

M29.          (a)     Publicity about vaccination / better health education / risks of ‘flu epidemics;
(Accept: now free on NHS (though only since 2000) / better awareness / more commonly available)
1

(b)     (i)      1990: 26% of 7.4million = 1.92million and 2000: 64% of 7.8 million = 4.99million;
increase = 3.07 million;
2
(Correct reading of all 4 figures from graph = 1)
(Correct answer but no ‘millions’ = 1)
(Correct method resulting from wrong graph reading = 1)

(ii)     Over 50% of population being vaccinated;
But only from 2000 onwards;
(Principle of more people being vaccinated each year = 1)
2

(iii)     Different strain / type of virus each year / virus mutates;
With different antigens;
Influenza antibodies / memory cells (rapidly) destroyed / need replacing;
max 2

(c)     (Protein coat) carries antigens which stimulates B-cells / production of antibodies;
Production of memory cells;
2
[9]

M30.          (a)     (i)      Selecting the nettle plant:
Random number table avoids bias in placing of quadrat;
‘Nearest centre’ avoids bias in choosing plant to measure;
1 mark for “method avoids bias”
2
(ii)     Measuring the sixth leaf:
To allow valid comparison / so as not to introduce another variable;
Reduces / avoids influence of growth / age / light / shading;
2

(b)     Definition of range + SD / effect of outliers on range + SD;
Ranges are similar in both areas;
Suggests that variation within populations is similar;
SD smaller in area of high light intensity;
Shows that area of high light intensity is a more uniform population;
4
(c)     1.164 / 1.16 / 1.2 , however derived = 2 marks
0.83 – 0.86 / 1.1, however derived = 1 mark
Evidence of correct use of both sets of figures, but inappropriate answer = 1 mark
2
[10]

M31.          (a)     0.1 / 0.9 (s);
1
(b)     Two marks for correct answer of 75 (beats per minute);
One mark for incorrect answer based on cardiac cycle taking 0.8 seconds;
2

(c)     (i)      Pressure in ventricle higher than pressure in atria;
1
(ii)     Prevents backflow of blood / prevents flow from ventricles to atria;
1

(d)     Increase (in stroke volume) as blood pressure increases, remains constant / plateaus; after 3 kPa / when stroke volume = 82cm^3
2
(e)     Two marks for correct answer of 80;
One mark for incorrect answer recognising that ventricle contracts once every cardiac cycle / stroke volume = 70 cm^3
2

(f)      1       Muscles (surrounding veins) contract and press on (walls of) vein and squeezes blood along veins;
2       Valves prevent backflow / ensure flow in one direction;
3       Systole / contraction of heart pumps blood (through arteries) into veins / residual arterial pressure / negative pressure in chest due to inspiration;
4       Recoil of heart muscle during diastole / after contraction;
5       Draws blood from veins into atria;
Accept sucks
6       Wide lumen little resistance / friction
6
[15]

M32.          (a)     In table:
 
D | All 3 correct = 2 marks;;
B | 2 correct = 1 mark;
C | 0 or 1 correct = 0 marks

2

(b)     (i)      myelin insulates / prevents ion movement; saltation / described 
re leaping node to node;
2
(ii)     cat has higher body temperature;
ignore references to homoiothermy’ / warm-blooded
faster diffusion of ions / faster opening of ion pores / gates / channels;
2

(c)     1       increasing stimulus (potential) causes decrease in potential
difference / rise in potential at P;
2       1 or 2 is sub-threshold / 1 or 2 does not give action potential / 3 or 4 is above threshold / 3 or 4 does give an action potential;
3       influx of Na^+ ions; (not just Na / sodium)
4       voltage-gated channels (in axon membrane) opens / opens Na^+
channels / membrane more permeable to Na^+
(NOT just Na / sodium);
5       sufficient for stimulation of adjacent region of axon therefore impulse propagated (from P to Q);
5

(d)     1       X / Acetylcholine → opening of Na^+ channels / increases
Na^+ permeability and Na^+ ion entry into Z;
2       Y / Cl^– entry - lowers potential / increases potential
difference / makes potential more negative;
3       X stimulates and Y inhibits (Z);
4       balance of impulses from X and Y determines whether Z
fires action potential / determines whether potential rises
above threshold;
4
[15]

M33.          (a)     Given only saline;
Otherwise treated exactly the same way;
2

(b)     Ethical consideration, e.g., leads to death / suffering of mice;
Large number to improve reliability / reduce sampling error;
Number of mice related to cost / space available / animal husbandry;
2 max

(c)     Vary in shape / do not grow uniformly;
Q Allow descriptions of variation in shape.
1

(d)     7.44 and 1.74;;
7.42 and 1.72;;
(Ratio) 4.28 : 1;;
(Ratio) 4.31 : 1;;
(Percentage decrease) 76.6%;;
(Percentage decrease) 76.8%;;
Any of the answers shown gain two marks.
An answer of 23.4% or 23.2%
Percentage decrease gains one mark.
Correct method of calculating rate / ratio / percentage increase with an incorrect answer gains one mark.
2 max

(e)     Reference to Mitosis;
As chromosomes cannot attach (to spindle) / chromatids cannot separate (on spindle);
Q Do not penalise confusion between chromosomes and chromatids in second marking point
Cell division / cell cycle slows down;
Q Mitosis slows down = 2 marks
Q Mitosis stopped = 1 mark
Q Mitosis must be spelt correctly
3

(f)      (i)      (Degree of) spread / variation from the mean;
1
(ii)     Both chemicals (on their own) slow down growth / are effective;
Taxol is more effective than OGF;
Combined treatment (seems) most effective;
SD overlap for OGF with taxol and taxol (on its own) so not conclusive / could be chance / both treatments could be equally effective;
Q Ignore all references to significance
4
[15]

M34.(a)     Straight lines point to point as not possible to predict intermediate values / values between points;
1

(b)     Increases then levels / falls;
Maximum antibody production 180 units / at dose of 0.25 g per kg;
2

(c)     Two marks for correct answer of 57.14 / 57.1;;
One mark for incorrect answer in which candidate clearly divides difference in antibody production / 60 by 105;
2

(d)     Takes into account different masses of mice / allows comparison;
Accept different weights of mice.
Do not accept different size.
1

(e)     Sheep red blood cells have antigens (on their surface);
Antigens are proteins foreign to mice / are non-self;
Stimulate B cells to produce antibodies;
3

(f)     Response only observed in mice;
Disease organisms not investigated;
Not all disease caused by pathogens / cured by antibodies;
i.e. not tested on humans
2 max
[11]

M35.(a)     1.      No oxygen can enter;
2.      Ethanol produced during anaerobic respiration;
OR
3.      No ethanol / carbon dioxide can escape;
4.      Allows accuracy of measuring;

OR
5.      To prevent entry of / contamination with microorganisms;
6.      Prevent competition with yeast;
Any two pairs of answers
Second mark of each pair must be related to the first point of the pair.
4 max

(b)     1.      Yeast respiring aerobically;
2.      Oxygen used equal to carbon dioxide produced;
2

(c)     1.      7.0 / 7;
2.      Ethanol production starts;
2

(d)     (i)      1.      Repeat;
2.      Identify anomalies / see if results are similar / enough results for statistical test / give more reliable mean;
3.      Carry out statistical test / statistical analysis;
4.      Ensure results are significant / find probability of results being due to chance;
5.      Peer review;
6.      Allows procedure to be checked / see if other scientists get similar results;
Two pairs of linked points, each pair a suggestion and an explanation. The explanation must relate to the suggestion to gain the second point of the pair.
4

(ii)     1.      Curve levelling off / rate of increase is decreasing / very little extra ethanol produced;
2.      Becomes less cost effective / less profit;
2. Accept a description of cost effectiveness
2

(iii)    1.      (Funding agency) might want particular results;
2.      Results may be withheld / results may not be published / results may be confidential;
2 max
[16]