Selection & evolution: A Level Biology questions with answers
Free Selection & evolution questions for A Level Biology (9700), part of Genetics & evolution. Each question has the answer and the marking points an examiner would look for.
In a population of 2000 animals, 9% show a recessive phenotype caused by allele a.
(a)
Use the Hardy–Weinberg equations to calculate the frequency of the dominant allele and the number of heterozygous animals.
[4]
(b)
State three conditions needed for the Hardy–Weinberg principle to apply.
[3]
(c)
Explain how allopatric speciation could occur in this species.
[4]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
p = 0.7, 840
q2 = 0.09 B1 q = 0.3 B1 2pq = 0.42 M1 840 A1
4
(b)
large population; random mating; no mutation, migration or selection
any three valid conditions B3
3
(c)
geographical isolation; different selection pressures; allele frequencies change by natural selection and genetic drift; reproductive isolation so populations can no longer interbreed to produce fertile offspring
geographical barrier B1 different selection pressures / drift B1 change in allele frequencies B1 reproductively isolated / cannot produce fertile offspring B1
4
Question 2Structured · 3 marksA* challenge
The Hardy–Weinberg principle estimates allele frequencies.
In a population, 16% of individuals show a recessive phenotype. Calculate the frequency of heterozygotes.
[3]
Show answers and mark scheme
Part
Answer and marking points
Marks
0.48
q² = 0.16 so q = 0.4 C1 p = 0.6 C1 2pq = 0.48 A1
3
Question 3Structured · 10 marks
A student counts the number of stomata in a field of view on the lower surface of leaves from plants growing in the shade (A) and in full sun (B). Table 3.1 shows the results.
leaf
1
2
3
4
5
6
7
8
shade (A)
36
40
38
41
37
39
42
35
sun (B)
48
50
48
47
47
47
52
43
Table 3.1
(a)
State a suitable null hypothesis for this investigation.
[1]
(b)
The standard deviation for A is 2.45 and for B is 2.60. Calculate the mean for each group and use the t-test formula t = |x̄₁ − x̄₂| ÷ √(s₁²/n₁ + s₂²/n₂) to calculate t.
[3]
(c)
For 14 degrees of freedom, the critical value of t at p = 0.05 is 2.145. Explain what your answer to (b) shows.
[2]
(d)
Explain why there are 14 degrees of freedom.
[1]
(e)
Suggest why leaves in full sun have more stomata.
[2]
(f)
Suggest one disadvantage to the plant of having many stomata.
[1]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
no significant difference in mean number of stomata
there is no significant difference between the mean number of stomata of leaves in shade and in sun B1
1
(b)
means 38.5, 47.8; t = 7.32
means 38.5 and 47.8 B1 substitution into formula M1 t = 7.32 A1
3
(c)
significant difference; reject null
t = 7.32 > 2.145 B1 the difference in means is significant (not due to chance), so reject the null hypothesis B1
B decreases if banding conspicuous; selection pressure from predation; fewer B passed on
frequency of B decreases (if banding is dominant and conspicuous in woodland) B1 predation is a selection pressure against banded snails B1 fewer B alleles passed on B1
3
(c)
natural selection acting; assumption of no selection broken
the principle assumes no selection B1 here predation selects against one phenotype B1
2
(d)
crossing over; independent assortment; random fertilisation; mutation
Any 2 of: crossing over in meiosis; independent assortment; random fertilisation; mutation B2
2
Question 5Structured · 13 marks
In a population, 9% of individuals show a recessive phenotype controlled by allele a.
(a)
Calculate the frequency of the recessive allele and of the dominant allele.
[2]
(b)
Calculate the percentage of heterozygous carriers in the population.
[2]
(c)
State the conditions needed for the Hardy–Weinberg principle to apply.
[3]
(d)
Explain how allopatric speciation can occur.
[4]
(e)
Explain how genetic drift can change allele frequencies in a small population.
[2]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
q = 0.30; p = 0.70
q = √0.09 = 0.30 M1 p = 1 − q = 0.70 A1
2
(b)
42.0%
2pq M1 42.0% A1
2
(c)
large population; random mating; no mutation, selection, migration
Any 3 of: large population; random mating; no mutation; no natural selection; no migration B3
3
(d)
geographical barrier; no gene flow; different selection pressures / mutations; reproductive isolation, cannot interbreed
a geographical barrier separates populations B1 no gene flow between them B1 different mutations and selection pressures change allele frequencies B1 eventually they can no longer interbreed to produce fertile offspring (reproductive isolation) B1
4
(e)
chance events; large effect in small population; alleles lost or fixed
random changes in allele frequency due to chance (which individuals reproduce) B1 has a greater effect in small populations, so alleles may be lost or fixed B1
2
Question 6Structured · 3 marks
In a population, 13% of individuals show a recessive phenotype. Use the Hardy–Weinberg principle to calculate the frequency of heterozygous individuals.
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Original Cambridge-style practice questions written for mentorriyaz.in. These are not Cambridge past paper questions, and this site is not affiliated with or endorsed by Cambridge University Press & Assessment. Updated 2026-11-04. Paper generator · mentorriyaz.in