Syllabus Edition

First teaching 2023

First exams 2025

|

Variation (CIE A Level Biology)

Exam Questions

2 hours11 questions
1a
Sme Calculator
2 marks

A group of 1000 people were chosen at random and surveyed as part of a population study. The participants were asked about their characteristics. 

One characteristic that was surveyed was the participant's height measurements. 

Sketch a graph in the space below to predict the distribution of individuals' height values. 

Axes for height normal distribution curve SQ

1b
Sme Calculator
1 mark

Name the type of variation shown in the example in part (a). 

1c
Sme Calculator
2 marks

Another characteristic that was surveyed was hair colour. 

It was found that most individuals had black, brown, blonde or ginger hair, but a small number of individuals had hair colours like pink, blue and green. 

Describe the factors that can cause variation in hair colour.

1d
Sme Calculator
1 mark

Some characteristics that are more likely to be examples of continuous variation are those that are coded for by several genes that work in combination to produce the phenotype. 

State the scientific term used for this type of characteristic. 

Did this page help you?

2a
Sme Calculator
4 marks

Table 1 shows four characteristics displayed by various organisms.

Table 1

Organism Characteristic Phenotypic characteristic? (✓/)
 human  brown eyes  
 human  expresses the CFTR protein*  
 plant  contains gene for two-tone leaves  
 fungi  red and white spotted fruiting bodies (mushrooms)  

* the CFTR protein is an important ion channel protein associated with clearing mucus in the airways; cystic fibrosis sufferers produce a faulty copy of this protein

Place a tick () in the box(es) in the final column that describe phenotypic characteristics and a cross (✘) in the other characteristic(s). 

2b
Sme Calculator
2 marks

State two environmental factors that can affect an organism's phenotype.

Give one example from the animal kingdom and one from the plant kingdom.

2c
Sme Calculator
4 marks

Fig. 1 lists the four main sources of genetic variation within a population.luc3Q7BV_cie-ial-17-1-q2c-e---sq

Fig. 1

Link each source of genetic variation to its description with a straight line. 

2d
Sme Calculator
1 mark

Explain how a mutation may not necessarily bring about a change in an organism's phenotype.

Did this page help you?

3a
Sme Calculator
2 marks

Before performing a statistical test such as the t-test, a null hypothesis must be formed.

State what is meant by a null hypothesis.

3b
Sme Calculator
2 marks

State the two statistical calculations that must be performed on each set of raw data before calculating the t-value between two sets of data.

3c
Sme Calculator
2 marks

State the relationship between the size of the calculated t-value and the probability of a difference between two data sets having occurred by chance.

3d
Sme Calculator
2 marks

Table 1 shows a table of critical values for the t-test which compared the number of leaves of shrubs of the same species growing in different habitats.

The null hypothesis states that the habitat has no significant effect on the number of leaves per shrub. 

Table 1

Degrees of freedom Value of t
1 6.31 12.7 63.7 63.6
2 2.92 4.30 9.93 31.6
3 2.35 3.18 5.84 12.9
4 2.13 2.78 4.60 8.61
5 2.02 2.57 4.03 6.87
6 1.94 2.45 3.71 5.96
7 1.90 2.37 3.50 5.41
8 1.86 2.31 3.36 5.04
9 1.83 2.26 3.25 4.78
10 1.81 2.23 3.17 4.59
Probability that chance could have produced this value of t 0.10 0.05 0.01 0.001
Confidence level 10% 5% 1% 0.1%

For a value of t calculated to be 3.15 at 9 degrees of freedom, state whether you would accept or reject the null hypothesis at a 5% confidence level.

Give a reason for your answer. 

Did this page help you?

4a
Sme Calculator
1 mark

Explain what is meant by the term phenotypic variation.

4b
Sme Calculator
3 marks

Gregor Mendel famously carried out experiments on pea plants. Fig. 1 shows the range in height of some pea plants. 

cie-ial-17-1-e-q4b

Fig. 1

Describe the variation in height shown by pea plants in Fig. 1

4c
Sme Calculator
2 marks

Describe the cause of the variation shown by the pea plants in Fig. 1.

4d
Sme Calculator
2 marks

A student investigated the height of pea plants in the classroom. They grew 10 pea plants in red light and 10 pea plants in blue light. The plants were left to grow for 3 weeks and height was measured for each plant and a mean height for each condition was calculated. 

State, with a reason, which statistical test should be carried out to determine if there is a significant difference in the heights of the pea plants from each condition. 

Did this page help you?

1a3 marks

Cacti are plants that are adapted to live in arid conditions. They store water in thick stems which are protected from grazing by herbivores through the presence of multiple spines. The number of spines that are present on cacti may vary greatly, depending on the frequency of grazing that they experience.

Fig. 1 shows the number of spines present in a species of cacti.

17-1-fig-2-1
Fig. 1

State the type of variation that is represented by Fig. 1 and provide two reasons for your answer.

1b2 marks

Suggest the possible impact of environmental factors on the variation observed in Fig. 1, with regards to the number of spines on cacti.

1c3 marks

Contrast the genetic basis of continuous and discontinuous variation with each other.

Did this page help you?

2a
Sme Calculator
2 marks

Students investigated the effect of light intensity on leaf length in the Mexican sword plant (Echinodorus palaefolius). They had two groups (A and B) consisting of ten plants each, with group A grown in a laboratory where the lights were kept on, while group B was grown under similar conditions to group A but kept under low light conditions. The plants were grown at these different light intensities for 14 days, and the average leaf length of each plant was calculated for each group.

The results of the investigation is shown in Table 1.

Table 1

Plant number Leaf length of group A Leaf length of group B
1 38 57
2 39 60
3 40 56
4 40 59
5 37 58
6 36 61
7 37 60
8 37 57
9 39 57
10 39 59
Mean 38.2  

Calculate the mean leaf length for group B

Show your working.

2b2 marks

The students used the t-test to compare the means of groups A and B.

State two features of the data that allow for the use of the t-test.

2c1 mark

State a null hypothesis for this investigation.

2d3 marks

After performing the t-test, the students calculated a t-value of 29.7. Table 2 shows different t-values and the probability that the differences between the data sets are due to chance.

Table 217-1-table-3-2

Using the information in Table 2, discuss the conclusions that the students can draw from their results.

Did this page help you?

3a
Sme Calculator
2 marks

Scientists investigated the effect of light on the germination of begonia seeds. They had three groups consisting of 30 seeds each, for both light and dark conditions. The seeds were exposed to light or darkness for 72 hours before scientists calculated the mean number of seeds that germinated out of 30 seeds for each light condition. All other conditions were kept constant between the seeds.

Fig. 1 shows the results of this investigation.

17-1-fig-4-1
Fig. 1

Calculate the mean number of seeds germinating in dark conditions as a percentage of the mean number of seeds that germinated in light conditions.

Show your working and give your answer to one decimal place.

3b2 marks

Explain why the t-test would not be a suitable statistical test to use on the data represented in Fig.1.

3c
Sme Calculator
3 marks

The house sparrow (Passer domesticus) is a small bird commonly found in most parts of the world. They feed mainly on seeds and they show variation in beak size. Two small populations of sparrows, consisting of 20 birds each, were studied to investigate the difference in beak size displayed by each population. 

The biologists conducting the study calculated the mean beak size for each population, as well as the standard deviation for each. The results of these calculations are represented in Table 1

Table 1

  Population 1 Population 2
Mean 13 14
Standard deviation (S) 1.80 1.74

Fig. 2 represents the formula for calculating the t-value.

17-1-fig-4-2
Fig. 2

Using the information in Table 1 and Fig. 2, calculate the t-value for this data set.

Show your working and give your answer to two decimal places.

3d2 marks

The null hypothesis for this investigation states that there is no significant difference between the beak sizes of the two populations. The t-value at a probability of 0.05 is 1.96 at 38 degrees of freedom. 

With reference to the t-value calculated in part c), state the conclusion that can be made regarding the difference in beak size between the two populations.

Did this page help you?

4a
Sme Calculator
1 mark

A scientist investigated whether a high-protein diet improves muscle density in rats. Rats were given either a diet high in protein or a diet deficient in protein, and their muscle density was measured weekly over 10 weeks. 

Identify the independent variable in this investigation. 

4b
Sme Calculator
3 marks

The results of the investigation are shown in Table 1. 

Table 1

  With high protein diet Without high protein diet
Mean muscle density / arbitrary units 125 117
Standard deviation 9.97 10.68

Describe what the mean and standard deviation values in Table 1 suggest about the rats' muscle density with and without a high protein diet.

4c
Sme Calculator
3 marks

The starting muscle mass of mean muscle density with a high protein diet was 98 arbitrary units. 

Calculate the percentage change in mean muscle mass for a high protein diet for the data shown in Table 1. Give your answer to three significant figures. 

4d
Sme Calculator
3 marks

Explain why a t-test would be an appropriate statistical test for this data set.

Did this page help you?

1a5 marks

Fig. 1 shows a type of cell division occurring in the reproductive organs of humans.

17-1-fig-1-1Fig. 1

(i)
Identify the type of cell division illustrated in Fig. 1.

[1]

(ii)
Explain how the type of cell division identified in part (i) contributes to variation in the phenotype of the resulting offspring.

[4]

1b2 marks

Tall tomato plants are produced by the presence of a dominant allele T. Two homozygous tall plants were crossed and seeds from this cross were planted in soils with different nutrient compositions. The mean height of each group of tomato plants was calculated and the results are shown in Table 1.

Table 1

Soil type Mean height / cm
A 35
B 12
C 26
D 40

Suggest reasons for the variation in height that was observed in the tomato plants.

1c3 marks

Seeds from the tomato plants grown in soil type B were taken and planted in soil type where they reached an average height of 42 cm.

Explain this observation.

Did this page help you?

2a
Sme Calculator
5 marks

Meiosis is one process that contributes to genetic variation.

(i)

State precisely the stage of meiosis where single chromosomes line up on the equator.

[1]

(ii)

Outline the events taking place during anaphase I of meiosis.

[2]

(iii)

Describe how crossing over during meiosis leads to genetic variation.

[2]

2b
Sme Calculator
5 marks

Mutation also causes genetic variation. Some populations of water hemp, Amaranthus tuberculatus, have evolved herbicide resistance as a result of a mutation. This is a problem for farmers as water hemp grows in crop fields, lowering productivity.

Two populations of water hemp were tested for resistance to the herbicide mesotrione. One was a population known to be resistant (control) and the other was a test population, whose resistance was unknown.

  • Leaves were removed and immersed in a radioactively labelled solution of mesotrione.
  • The leaves absorbed some mesotrione and became radioactive.
  • Resistant leaves are able to degrade mesotrione by metabolism.
  • The time for 50 % of absorbed mesotrione to degrade was calculated by measuring the radioactivity of the leaves.

The results are shown in Table 1.

Table 1

population of water
hemp

mean time for 50 % of
absorbed mesotrione to
degrade / hours
standard
deviation
test 27.5 4.75
control 10.1 2.34

(i)

Explain how the results in Table 1 show that the two populations differ in their resistance to mesotrione.

[2]

(ii)

Explain why this example of genetic variation is important for natural selection in water hemp populations.

[2]

(iii)

Farmers can send in a sample of leaves of water hemp from their fields to a laboratory to be tested for resistance to mesotrione or other herbicides.

   Suggest the benefit of this to a farmer.

[1]

2c
Sme Calculator
4 marks

The null hypothesis states there is no significant difference between the mean times for 50 % of absorbed mesotrione to degrade in the two populations.

A t‐test can be carried out to compare these two means. The critical value for t at the p = 0.05 significance level is 2.23.

(i)

Use the formula in Fig. 1 to calculate the value of t.

Show your working.

bold italic t bold space equals space fraction numerator open vertical bar x with bar on top subscript 1 minus space x with bar on top subscript 2 close vertical bar over denominator square root of open parentheses fraction numerator s subscript 1 superscript 2 over denominator n subscript 1 end fraction plus fraction numerator s subscript 2 superscript 2 over denominator n subscript 2 end fraction close parentheses end root end fraction         

Key
x with bar on top = mean
s = standard deviation
n1 = 6 (number of readings for test population)
n2 = 6 (number of readings for control population)

Fig. 1

 [2]

(ii)

Use your calculated value of t to explain whether the null hypothesis should be accepted or rejected.

[2]

Did this page help you?

3a
Sme Calculator
2 marks

The t-test is a statistical test used to compare two sets of biological data.

State the purpose of t-test.

3b
Sme Calculator
7 marks

A number of boy students in a school measured their maximum handspan using a ruler as shown in Fig. 1.


3LoasJ6R_cie-ial-17-1-q3b-e---sq

Fig. 1

Table 1 shows their class data:

Table 1

Student Hand span / cm
 Ali 20.3
 Billy 21.1
 Caleb 22.2
 Dylan 19.5
 Eli 20.0
 Faaz 21.3
 Gabe 18.6
 Harry 23.3
 Ismail 19.8
 Joe 19.9
(i)

Calculate the mean value x̄ of the students' handspans.

[2]

(ii)

Use Table 2 below to calculate and sum the values of (x-x̄)2

Student Hand span / cm x-x̄ (x-x̄)2
 Ali 20.3    
 Billy 21.1    
 Caleb 22.2    
 Dylan 19.5    
 Eli 20.0    
 Faaz 21.3    
 Gabe 18.6    
 Harry 23.3    
 Ismail 19.8    
 Joe 19.9    
 Mean x̄ (from part (i)      

[3]

(iii)

From the values you have found in Table 2, calculate the standard deviation S using the formula below.

sd-formula---sq

[2]

3c
Sme Calculator
2 marks

In a linked experiment, groups of 10 male students from two different countries (Countries A and B) also took part in this experiment. 

All the boys taking part were of the same age group as the boys in Table 1 and also measured their maximum hand spans. 

Their data calculated out as follows:

Table 2

  Country A Country B
 Mean / cm 2 = 19.94 1 = 21.03
 Standard deviation S2 = 0.756 S1 = 1.687
 Sample size  n2 = 10 n1 = 10

State a null hypothesis for this investigation. 

3d
Sme Calculator
3 marks

From the data described in part (c), calculate the t value between the data from Country A and Country B.

The formula for calculating t is as follows:

bold italic t bold space equals space fraction numerator open vertical bar x with bar on top subscript 1 minus space x with bar on top subscript 2 close vertical bar over denominator square root of open parentheses fraction numerator s subscript 1 superscript 2 over denominator n subscript 1 end fraction plus fraction numerator s subscript 2 superscript 2 over denominator n subscript 2 end fraction close parentheses end root end fraction         

3e
Sme Calculator
3 marks
Table 3 shows a table of critical values for the t-test carried out in parts (c) and (d).

Table 3

Degrees of freedom Value of t
1 6.31 12.7 63.7 63.6
2 2.92 4.30 9.93 31.6
3 2.35 3.18 5.84 12.9
4 2.13 2.78 4.60 8.61
5 2.02 2.57 4.03 6.87
6 1.94 2.45 3.71 5.96
7 1.90 2.37 3.50 5.41
8 1.86 2.31 3.36 5.04
9 1.83 2.26 3.25 4.78
10 1.81 2.23 3.17 4.59
Probability that chance could have produced this value of t 0.10 0.05 0.01 0.001
Confidence level 10% 5% 1% 0.1%

Use the value of t that you calculated in part (d) to either reject or accept your null hypothesis from part (c).

Give a reason for your decision. 

Did this page help you?