A Level Biology (9700) topical questions with answers
Every A Level Biology topic, with free past-paper-style questions and full mark schemes. Pick a chapter to practise its questions, or generate a complete 9700-format paper in seconds, from easy warm-ups to A*-level "brutal" papers.
Paper 4 · A Level Structured Questions (A Level · 100 marks · 2h); Paper 5 · Planning, Analysis & Evaluation (A Level · 30 marks · 1h 15m).
Sample A Level Biology questions
A mixed set from across the syllabus. Questions marked A* challenge come from the hardest bank.
Question 1Structured · 10 marksA* challenge
The numbers of individuals of four species found in a quadrat survey were 20, 12, 6, 4.
(a)
Calculate Simpson’s index of diversity, D = 1 − Σ(n/N)2.
[3]
(b)
In a mark–release–recapture study, 40 beetles were marked; later 45 were caught, of which 10 were marked. Estimate the population size and state two assumptions.
[4]
(c)
Explain why conserving biodiversity is important.
[3]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
0.662
N = 42 B1 Σ(n/N)2 calculated M1 0.662 A1
3
(b)
180; no migration/births/deaths, marks not lost, marking does not affect survival, random mixing
c1 × c2 ÷ m M1 180 A1 two valid assumptions B2
4
(c)
ecological stability/food webs; source of drugs/genes; economic (tourism) and ethical reasons
any three valid reasons B3
3
Question 2Structured · 3 marksA* challenge
The kidney regulates water content.
Explain how ADH increases water reabsorption in the collecting duct.
[3]
Show answers and mark scheme
Part
Answer and marking points
Marks
aquaporins inserted; more water reabsorbed by osmosis
ADH binds receptors on collecting duct cells B1 vesicles containing aquaporins fuse with the membrane B1 more water reabsorbed by osmosis B1
3
Question 3Structured · 13 marks
In a population, 16% 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 = 0.40; p = 0.60
q = √0.16 = 0.40 M1 p = 1 − q = 0.60 A1
2
(b)
48.0%
2pq M1 48.0% 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 4Structured · 9 marks
This question is about genetic screening and gene therapy.
(a)
Describe how DNA sequencing data can be used in genetic screening for an inherited disease.
[2]
(b)
Explain how a DNA probe locates a particular allele.
[3]
(c)
Discuss two ethical issues raised by genetic screening of embryos.
[2]
(d)
State one vector used to deliver genes in gene therapy and one risk associated with it.
[2]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
compare person’s sequence with known disease alleles; identify mutation / carriers
the person’s DNA is sequenced or probed for known disease-causing alleles B1 identifying affected individuals or carriers B1
2
(b)
single-stranded labelled DNA complementary to target; hybridises; label detected (fluorescent/radioactive)
a short single-stranded DNA sequence complementary to the target allele B1 labelled with a fluorescent or radioactive marker B1 it hybridises with the target and is detected B1
3
(c)
selection of embryos (designer babies); discarding embryos; insurance/discrimination
embryos may be selected for non-medical traits / discarded embryos B1 information could be used for discrimination (insurance, employment) B1
2
(d)
viral vector (adenovirus/retrovirus); immune response / insertion causes cancer
viruses (e.g. adeno-associated virus / retrovirus) or liposomes B1 immune reaction / insertional mutagenesis causing cancer B1
2
Question 5Structured · 4 marks
In a dihybrid cross, two genes are inherited together.
(a)
Explain why the observed ratio may differ from 9 : 3 : 3 : 1.
[2]
(b)
State the purpose of a chi-squared test in genetics.
[2]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
linkage or epistasis
genes may be linked on the same chromosome (or epistasis) B1 so they are not inherited independently B1
2
(b)
whether difference between observed and expected is significant
to test whether the difference between observed and expected results is significant or due to chance B2
2
Question 6Structured · 8 marks
This question is about meiosis and genetic variation.
(a)
Describe the behaviour of chromosomes in prophase I and metaphase I of meiosis.
[3]
(b)
Explain how crossing over and independent assortment produce genetic variation.
[3]
(c)
Calculate the number of different combinations of chromosomes possible in human gametes from independent assortment alone.
[2]
Show answers and mark scheme
Part
Answer and marking points
Marks
(a)
homologous chromosomes pair (bivalents); crossing over at chiasmata; pairs line up on equator randomly
homologous chromosomes pair to form bivalents B1 non-sister chromatids cross over at chiasmata B1 bivalents line up on the equator in random orientation B1
3
(b)
exchange of alleles between chromatids; new combinations; random orientation gives 2ⁿ combinations
crossing over exchanges sections between non-sister chromatids, forming new combinations of alleles B1 independent assortment of bivalents gives 2n possible chromosome combinations in gametes B2
3
(c)
2²³ = 8.39 × 10⁶
223M1 8.39 × 106A1
2
Frequently asked questions
Are these real 9700 past paper questions?
No. They are original questions written to match the A Level Biology 9700 syllabus and the style of real Cambridge papers, so you can practise beyond the past papers you have already done.
Do the Biology questions have answers?
Yes. Every question on these pages has its answer, and every generated paper comes with a full Cambridge-style mark scheme you can download as a PDF.
Can I make a full A Level Biology practice paper?
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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