Feversham College
A-level Biology (7401/7402)
Name:
Class:
Section 7 and 8 Practical Skills
Author:
Date:
Time:
224
Marks:
192
Comments:
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Q1.In cats, males are XY and females are XX. A gene on the X chromosome controls fur colour in cats. The
allele G codes for ginger fur and the allele B codes for black fur. These alleles are codominant. Heterozygous
females have ginger and black patches of fur and their phenotype is described as tortoiseshell.
(a)
Explain what is meant by codominant alleles.
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(1)
(b)
Male cats with a tortoiseshell phenotype do not usually occur. Explain why.
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(1)
(c)
A tortoiseshell female was crossed with a black male. Use a genetic diagram to show all
the possible genotypes and the ratio of phenotypes expected in the offspring of this cross.
Use XG to indicate the allele G on an X chromosome.
Use XB to indicate the allele B on an X chromosome.
Genotypes of offspring .................................................................................
Phenotypes of offspring ................................................................................
Ratio of phenotypes ......................................................................................
(3)
(d)
Polydactyly in cats is an inherited condition in which cats have extra toes. The allele for
polydactyly is dominant.
(i)
In a population, 19% of cats had extra toes. Use the Hardy-Weinberg equation to
calculate the frequency of the recessive allele for this gene in this population.
Show your working.
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Answer = ............................
(2)
(ii)
Some cat breeders select for polydactyly. Describe how this would affect the
frequencies of the homozygous genotypes for this gene in their breeding populations
over time.
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(1)
(Total 8 marks)
Q2.(a)
In fruit flies, the genes for body colour and wing length are linked. Explain what this means.
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(1)
A scientist investigated linkage between the genes for body colour and wing length.
He carried out crosses between fruit flies with grey bodies and long wings and fruit flies with
black bodies and short wings.
Figure 1 shows his crosses and the results.
•
•
G represents the dominant allele for grey body and g represents the recessive allele
for black body.
N represents the dominant allele for long wings and n represents the recessive allele
for short wings.
Figure 1
Phenotype of parents
Genotype of parents
grey body,
long wings
black
body,
short
wings
×
GGNN
ggnn
GgNn
Genotype of offspring
Phenotype of offspring
all grey body, long wings
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These offspring were crossed with flies homozygous for black body and short wings.
The scientist’s results are shown in Figure 2.
Figure 2
GgNn crossed with ggnn
Grey body, Black body, Grey body, Black body,
long wings short wings short wings long wings
Number of
offspring
(b)
975
963
186
194
Use your knowledge of gene linkage to explain these results.
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(4)
(c)
If these genes were not linked, what ratio of phenotypes would the scientist have
expected to obtain in the offspring?
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(1)
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(d)
Which statistical test could the scientist use to determine whether his observed results
were significantly different from the expected results?
Give the reason for your choice of statistical test.
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(2)
(Total 8 marks)
Q3.(a)
Explain what is meant by the term phenotype.
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(2)
(b)
Tay-Sachs disease is a human inherited disorder. Sufferers of this disease often die
during childhood. The allele for Tay-Sachs disease t, is recessive to allele T, present in
unaffected individuals. The diagram shows the inheritance of Tay-Sachs in one family.
(i)
Explain one piece of evidence from the diagram which proves that the allele for
Tay-Sachs disease is recessive.
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(ii)
Explain one piece of evidence from the diagram which proves that the allele for
Tay-Sachs disease is not on the X chromosome.
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(2)
(c)
(i)
In a human population, one in every 1000 children born had Tay-Sachs disease. Use
the Hardy-Weinberg equation to calculate the percentage of this population you
would expect to be heterozygous for this gene. Show your working.
Answer = ...................................... %
(3)
(ii)
The actual percentage of heterozygotes is likely to be lower in future generations
than the answer to part (c)(i). Explain why.
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(1)
(Total 10 marks)
Q4.
Colour blindness is controlled by a gene on the X chromosome. The allele for colour
blindness, Xb, is recessive to the allele for normal colour vision, XB . The gene controlling the
presence of a white streak in the hair is not sex linked, with the allele for the presence of a white
streak, H, being dominant to the allele for the absence of a white streak, h.
(a)
Explain why colour blindness is more common in men than in women.
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(2)
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(b)
The diagram shows a family tree in which some of the individuals have colour blindness or
have a white streak present in the hair.
(i)
What are the genotypes of individuals 5 and 6?
Individual 5
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Individual 6
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(2)
(ii)
Give the possible genotypes of the gametes produced by
individual 5;
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individual 6.
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(1)
(iii)
What is the probability that the first child of individuals 5 and 6 will be a colour
blind boy with a white streak in his hair? Show your working.
Answer ............................................
(2)
(Total 7 marks)
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Q5.Oestrogen is a substance produced by the enzyme aromatase. In females, the main source of
oestrogen is the ovaries but aromatase is produced by many other organs in the body, including
the lungs. Oestrogen can stimulate the development of some lung tumours. In these tumours,
binding of oestrogen to cell-surface receptors stimulates cell division.
Scientists investigated whether two drugs could prevent lung tumours in female mice. First, they
removed the ovaries from these mice. They then injected the mice with a tumour-causing
chemical found in tobacco twice a day for 4 weeks. The mice were then randomly allocated to one
of four groups. Each group contained 10 mice.
•
•
•
•
Group Q was given a placebo. This placebo did not contain either drug.
Group R was given the drug anastrozole. This inhibits the enzyme aromatase.
Group S was given the drug fulvestrant. This binds to oestrogen receptors.
Group T was given both anastrozole and fulvestrant.
The mice were given these drugs each week during weeks 5−15 of the investigation.
(a)
The scientists removed the ovaries from the mice for the investigation. They also gave the
mice injections of the substrate of aromatase each day.
Explain why these steps were necessary.
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(2)
(b)
The scientists predicted that fulvestrant would be more effective when given with
anastrozole than when given alone.
Use the information provided to suggest why they predicted this.
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(2)
At week 15, the lungs of the mice were removed and examined. The scientists then
determined the number of tumours present and the mean tumour area for each group.
Figure 1 and Figure 2 show the scientists’ results.
Figure 1
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Figure 2
(c)
The scientists concluded that both drugs should be used together to reduce the risk of
lung cancer in women exposed to tobacco products.
Do you agree? Explain your answer.
(5)
(d)
The scientists used tumour area as an indicator of tumour size.
Explain why tumour area may not be the best indicator of tumour size and suggest a more
reliable measurement.
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(2)
(e)
The scientists repeated the investigation but this time they did not give the drugs until
week 9.
Suggest why they gave the drugs at week 9, rather than at week 5.
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(2)
(f)
Another group of scientists is currently using these drugs in human trials. However, the
control group is not being given a placebo.
Suggest why a placebo is not being given and what is being given to this group instead.
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(2)
(Total 15 marks)
Q6.
(a)
Individuals in a population show phenotypic variation.
Give the two types of factor that cause this variation.
1 ...................................................................................................................
2 ...................................................................................................................
(2)
(b)
What is allopatric speciation?
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(2)
(Total 4 marks)
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Q7.
(a)
What is sympatric speciation?
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(2)
Littorina saxatilis is a snail found on rocky seashores. It has a shell and a muscular foot that it
uses to move and to attach to rocks. Crabs are predators of this snail. The crabs use their claws
to break open the snails’ shells, or pull the snails from their shells.
Two forms of this snail are common in the UK.
Form T lives near the top of the shore. It lives in cracks in rock. Wave action is greatest near the
top of the shore and there are very few crabs.
Form M lives on the middle shore. On the middle shore there are many crabs. Unlike form T, the
snails of form M live on the open rock and not in cracks.
Forms T and M were produced by natural selection. The drawings show both forms of the snail.
The table shows features of these forms.
Form of Littorina saxatilis
Feature
T
M
Size of shell
Small
Large
Thickness of shell
Thin
Thick
Large
Small
Size of opening of shell
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(b)
Use this information to answer the following question.
Give two differences between forms T and M.
For each difference suggest how one environmental factor may have caused differential
survival in the snail populations leading to this difference.
Difference 1 ................................................................................................
Suggestion ...................................................................................................
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Difference 2 .................................................................................................
Suggestion ...................................................................................................
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(4)
(c)
Scientists placed male and female snails of forms T and M into an aquarium. They
recorded how many form T males mated with form T females and how many mated with
form M females.
The scientists found that the probability of a form T male mating with a form T female was
greater than 90 %. They interpreted this result as evidence that speciation was taking place.
Explain why.
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(2)
(Total 8 marks)
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Q8.
Warfarin is a substance which inhibits blood clotting. Rats which eat warfarin are killed due to
internal bleeding. Some rats are resistant to warfarin as they have the allele WR.
Rats have three possible genotypes:
WRWR
WRWS
WSWS
resistant to warfarin
resistant to warfarin
susceptible (not resistant) to warfarin.
In addition, rats with the genotype WRWR require very large amounts of vitamin K in their diets. If
they do not receive this they will die within a few days due to internal bleeding.
(a)
How can resistance suddenly appear in an isolated population of rats which has never
before been exposed to warfarin?
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(1)
(b)
A population of 240 rats was reared in a laboratory. They were all fed on a diet containing
an adequate amount of vitamin K. In this population, 8 rats had the genotype WSWS, 176 had
the genotype WRWS and 56 had the genotype WRWR.
(i)
Use these figures to calculate the actual frequency of the allele WR in this population.
Show your working.
Answer ..................................
(2)
(ii)
The diet of the rats was then changed to include only a small amount of vitamin K.
The rats were also given warfarin. How many rats out of the population of 240 would
be likely to die within a few days?
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(1)
(c)
In a population of wild rats, 51% were resistant to warfarin.
(i)
Use the Hardy-Weinberg equation to estimate the percentage of rats in this
population which would be heterozygous for warfarin resistance. Show your working.
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Answer ....................................... %
(3)
(ii)
If all the susceptible rats in this population were killed by warfarin, more susceptible
rats would appear in the next generation. Use a genetic diagram to explain how.
(2)
(iii)
The graph shows the change in the frequency of the WS allele in an area in which
warfarin was regularly used. Describe and explain the shape of the curve.
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(iv)
Give two assumptions that must be made when using the Hardy-Weinberg equation.
1 …......................................................................................................
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2 …......................................................................................................
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(2)
(Total 15 marks)
Q9.
(a)
Explain the meaning of these ecological terms.
Population ....................................................................................................
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Community ...................................................................................................
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(2)
(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.
1 ……...................................................................................................
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2 ……...................................................................................................
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(2)
(ii)
Calculate the number of woodlice in the area under investigation. Show your
working.
Answer ......................................................
(2)
(c)
Explain how you would use a quadrat
to estimate the number of dandelion plants
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in a field measuring 100 m by 150 m.
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(3)
(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
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the niche of species A;
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the niche of species B;
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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? .............
(2)
(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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(4)
(Total 15 marks)
Q10.
Sea otters were close to extinction at the start of the 20th century. Following a ban on
hunting sea otters, the sizes of their populations began to increase. Scientists studied the
frequencies of two alleles of a gene in one population of sea otters. The dominant allele, T, codes
for an enzyme. The other allele, t, is recessive and does not produce a functional enzyme.
In a population of sea otters, the allele frequency for the recessive allele, t, was found to be 0.2.
(a)
(i)
Use the Hardy-Weinberg equation to calculate the percentage of homozygous
recessive sea otters in this population. Show your working.
Answer ..................................... %
(2)
(ii)
What does the Hardy-Weinberg principle predict about the frequency of the t allele
after another 10 generations?
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(1)
(b)
Several years later, scientists
Page 17repeated their study on this population.
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They found that the frequency of the recessive allele had decreased.
(i)
A statistical test showed that the difference between the two frequencies of the t
allele was significant at the P = 0.05 level.
Use the terms probability and chance to help explain what this means.
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(2)
(ii)
What type of natural selection appears to have occurred in this population of sea
otters? Explain how this type of selection led to a decrease in the frequency of the
recessive allele.
Type of selection ................................................................................
Explanation .........................................................................................
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(2)
(Total 7 marks)
Q11.Malaria is a disease that is spread by insects called mosquitoes. In Africa, DDT is a pesticide used
to kill mosquitoes, to try to control the spread of malaria.
Mosquitoes have a gene called KDR. Today, some mosquitoes have an allele of this gene, KDR
minus, that gives them resistance to DDT. The other allele, KDR plus, does not give resistance.
Scientists investigated the frequency of the KDR minus allele in a population of mosquitoes in an
African country over a period of 10 years.
The figure below shows the scientists’ results.
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Year
(a)
Use the Hardy–Weinberg equation to calculate the frequency of mosquitoes heterozygous
for the KDR gene in this population in 2003.
Show your working.
Frequency of heterozygotes in population in 2003 ...................................
(2)
(b)
Suggest an explanation for the results in the figure above.
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(4)
The KDR plus allele codes for the sodium ion channels found in neurones.
(c)
When DDT binds to a sodium ion channel, the channel remains open all the time.
Use this information to suggest how DDT kills insects.
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(2)
(d)
Suggest how the KDR minus allele gives resistance to DDT.
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(2)
(Total 10 marks)
Q12.
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.
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Length of band X =..................................... µm
(2)
(ii)
Explain the difference in appearance between transverse sections A and C in
Figure 1.
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(1)
(b)
Explain what leads to the differences in appearance between the relaxed myofibril and the
contracted myofibril.
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(4)
(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
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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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(1)
(ii)
Give the numbers of two people in Figure 2 who are definitely carriers of muscular
dystrophy.
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(1)
(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:
XD = normal X chromosome
Xd = X chromosome carrying the allele for muscular dystrophy
Y
= normal Y chromosome
6
7
Parental phenotypes
Unaffected
Unaffected
Parental genotypes
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Gametes
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Offspring genotypes
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Offspring phenotypes
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Probability of having a son with DMD ...................................................
(4)
(d)
DMD is caused by a deletion mutation in the gene for a muscle protein called dystrophin.
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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.
Number of copies of gene fragment per cell
Child
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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(1)
(ii)
The number of copies of gene fragments F and G shows that person 12 is male.
Explain how.
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(2)
(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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(Extra space) ......................................................................................
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(3)
(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
Left leg - injected
with cultured cells
suspended in salt
solution
Percentage of muscle
fibres labelled with
antibody
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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(1)
(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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(4)
(Total 25 marks)
Q13.
Cocaine is a highly addictive and illegal drug.
The release of the neurotransmitter dopamine in specific synapses in the brain leads to feelings
of pleasure. Dopamine is removed from synapses by dopamine transporter proteins in the
plasma membrane of neurones. Cocaine binds to the dopamine transporter protein.
Figure 1 shows a dopamine transporter protein and molecules of cocaine and dopamine.
Figure 1
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(a)
Using all of the information, suggest how cocaine leads to feelings of pleasure.
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(Extra space) ................................................................................................
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(3)
(b)
(i)
Scientists isolated a mutated gene for the dopamine transporter protein.
Name one method that the scientists could have used to produce many copies of the
mutated gene in the laboratory.
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(1)
(ii)
Copies of the gene were then inserted into early embryos of mice. When these mice
were born, samples of their DNA were tested using DNA probes to make sure that the
mutated gene was present in the mice.
What is a DNA probe?
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(c)
Figure 2 shows dopamine transporter proteins produced from the normal gene and from
the mutated gene.
Figure 2
Explain how the mutation leads to the production of a protein that transports dopamine but
is not affected by cocaine.
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(3)
(Total 9 marks)
Q14.
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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(1)
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The table shows the number of base pairs present in the fragments.
Fragment
Number of base pairs (× 103)
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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(2)
(c)
(i)
Write 6 above the appropriate box on the diagram to show the position you would
expect fragment 6 to have reached.
(1)
(ii)
Explain how you arrived at your answer.
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(1)
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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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(1)
(Total 6 marks)
Q15.
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
Mean volume of tumour / mm3
Number of days
of treatment
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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2 ...................................................................................................................
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(2)
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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
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Figure 3
Treatment
Mean volume of tumour
following 70 days treatment
/mm3 (± 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)
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Feversham College
(Total 15 marks)
Q16.(a)
(i)
A mutation of a tumour suppressor gene can result in the formation of a tumour.
Explain how.
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(2)
(ii)
Not all mutations result in a change to the amino acid sequence of the encoded
polypeptide.
Explain why.
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(1)
(b)
Some cancer cells have a receptor protein in their cell-surface membrane that binds to a
hormone called growth factor. This stimulates the cancer cells to divide.
Scientists have produced a monoclonal antibody that stops this stimulation.
Use your knowledge of monoclonal antibodies to suggest how this antibody stops the
growth of a tumour.
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[Extra space] ................................................................................................
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(3)
(Total 6 marks)
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Feversham College
Q17.
One hypothesis for the cause of cancer of the colon (large intestine) is that Clostridium
bacteria present in the gut can convert bile steroids into cancer-causing substances.
S
(a)
Explain the presence of bile in the colon.
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(2)
(b)
The concentrations of bile steroids and numbers of Clostridium bacteria were measured in
people with colon cancer and in controls without colon cancer. The table shows the results.
Concentration of
bile steroids
Number of
Clostridium
bacteria
Percentage of
cancer patients
Percentage of
controls
P
high
high
low
low
high
low
high
low
76
13
7
4
9
8
34
49
<0.01
<0.01
<0.01
<0.01
A statistical test showed there was a significant difference between the cancer patients and
the controls in each of the four categories.
(i)
Explain how the results could be used to support the hypothesis that Clostridium
bacteria convert bile steroids into substances which cause colon cancer.
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(2)
(ii)
Explain how the results indicate that other factors may be involved in causing colon
cancer.
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(1)
S
(c)
Human cells contain genes that control their growth and division. One of these genes
codes for a protein that prevents cell division. The substances formed from bile steroids by
Clostridium bacteria may cause gene mutation. Describe and explain how these
substances could cause colon cancer.
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(4)
(Total 9 marks)
Q18.Cyanide is a poisonous substance. Cyanogenic clover plants produce cyanide when their tissues
are damaged. The ability to produce cyanide is controlled by genes at loci on two different
chromosomes. The dominant allele, A, of one gene controls the production of an enzyme which
converts a precursor to linamarin. The dominant allele, L, of the second gene controls the
production of an enzyme which converts linamarin to cyanide. This is summarised in the diagram.
(a)
Acyanogenic clover plants cannot produce cyanide. Explain why a plant with the genotype
aaLl cannot produce cyanide.
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(1)
(b)
A clover plant has the genotype AaLl.
(i)
Give the genotypes of the male gametes which this plant can produce.
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(1)
(ii)
Explain how meiosis results in this plant producing gametes with these genotypes.
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(2)
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Feversham College
(c)
Two plants, heterozygous for both of these pairs of alleles, were crossed. What proportion
of the plants produced from this cross would you expect to be acyanogenic but able to
produce linamarin? Use a genetic diagram to explain your answer.
(3)
In an investigation, cyanogenic and acyanogenic plants were grown together in pots. Slugs
were placed in each pot and records were kept of the number of leaves damaged by the
feeding of the slugs over a period of 7 days. The results are shown in Table 1.
Table 1
(d)
Undamaged
Damaged
Cyanogenic plants
160
120
Acyanogenic plants
88
192
A x2 test was carried out on the results.
(i)
Suggest the null hypothesis that was tested.
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(1)
(ii)
x2 was calculated. When this value was looked up in a table, it was found to
correspond to a probability of less than 0.05. What conclusion can you draw from
this?
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(3)
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A second investigation was carried out in a field of grass which had been undisturbed
for many years. Table 2 shows the population density of slugs and the numbers of
cyanogenic and acyanogenic clover plants at various places in the field.
Table 2
Number of
acyanogenic clover
plants per m2
Number of
cyanogenic clover
plants per m2
Very low
26
10
Low
17
26
High
0
10
Very high
0
5
Population density of
slugs
(e)
Explain the proportions of the two types of clover plant in different parts of the field.
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(Extra space)..................................................................................................
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(4)
(Total 15 marks)
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