AQA GCSE Biology Biology Paper 2 (Higher), June 2023: Question 7

14 marks · High Demand difficulty · Extended Answer

Analyze hormone interactions during follicle development, calculate follicle numbers using volume of a sphere and graph data, and explain the inheritance of an FSH-deficiency allele.

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Question text

07 Hormones are important for regulating the menstrual cycle.

During the menstrual cycle, eggs mature inside follicles in the ovaries.

A 27-year-old woman was infertile.

A doctor tested a sample of the woman’s blood.

The test did not detect any follicle stimulating hormone (FSH) in the woman’s blood.

The doctor gave the woman daily injections of FSH for 7 days.

The doctor measured:

• the concentration of FSH in the woman’s blood

• the concentration of oestrogen in the woman’s blood

• the volumes of developing follicles in the ovaries.

Figure 12 shows the results.

Figure 12

07.1 Give evidence from Figure 12 that the follicles in the ovaries release oestrogen.

[1 mark]

07.2 Injection of FSH caused the development of a number of follicles.

The mean diameter of the follicles on day 11 was 22 millimetres.

Calculate the number of follicles in the woman’s ovaries on day 11.

Assume each follicle is a sphere.

Volume of a sphere = πr

r = radius

π = 3.14

Give your answer to the nearest whole number.

[5 marks]

Number of follicles (to the nearest whole number) =32

07.3 Before treatment with FSH, the woman had underdeveloped breasts.

Explain why the lack of FSH in the woman’s blood caused underdeveloped breasts.

[2 marks]

07.4 Usually males and females both produce FSH.

The woman had inherited a faulty gene for FSH production from each of her parents.

The woman’s parents both produce FSH.

Show how the woman’s parents could have a child that does not produce FSH.

You should:

• draw a Punnett square diagram

• identify the phenotype of each offspring genotype

• use the symbols below:

H = allele for making FSH

h = allele for not making FSH

[3 marks]

07.5 The woman continues to have injections of FSH.

The woman has a child with a man who is heterozygous for the FSH gene.

Explain why the probability that the child will be able to produce FSH is 0.5.

[3 marks]

Mark scheme

Show the mark scheme

Question 7

AO /

Question Answers Extra information Mark

Spec Ref.

07.1 as volume of follicles rises allow (positive) correlation 1 AO3

oestrogen concentration (in between oestrogen 4.5.3.4

4.5.3.6

blood) rises (for 7 / 8 days) concentration and volume of

follicles (for 7 / 8 days)

or oestrogen concentration is in

proportion to follicle volume (for

7 / 8 days)

do not accept an increase of

oestrogen concentration causes

an increase of follicle volume

AO /

Spec Ref.

07.2 (volume of one follicle)

43 4 π 3

= × 3.14 × 11 allow = 3 × × 11 1 AO2

= 5572.(4533) allow 5575(.279…) 1 AO2

(total volume of follicles) 1 AO3

= 39 000 (mm3)

39 000 allow use of an incorrect volume 1 AO2

= 6.99… (from Figure 12) and / or an

5572

incorrectly calculated volume of

one follicle

7 do not accept 7.0 1 AO2

4.5.3.6

AO /

Spec Ref.

07.3 (lack of FSH causes) lack of allow lack of FSH causes lack of 1 AO2

oestrogen (production) follicle development / growth / 4.5.3.4

maturation

breast development is allow (female) secondary sexual 1

dependent on oestrogen (from characteristics are dependent

follicles) on oestrogen (from follicles)

AO /

Spec Ref.

07.4 gametes correct: 1 AO2

H + h and H + h 4.6.1.6

4.6.1.7

correct derivation of offspring allow correct for gametes stated 1

genotypes: in mp1

HH Hh Hh hh

correct phenotype for each allow correct for genotypes 1

genotype stated in mp2

do not accept if no hh offspring

AO /

Spec Ref.

07.5 allow annotated genetic diagram AO3

for all marks 4.6.1.6

mother (has hh so) passes on h 1

father (has Hh so) passes on H 1

or h with equal probability

(so) child will be Hh / hetero- 1

zygous with 0.5 probability and

produces FSH

Total Question 7 14

How to answer it

Hormonal Control of the Menstrual Cycle & Inheritance

📋 WHAT THIS QUESTION TESTS

Essential GCSE Biology skills across human reproduction, graph analysis, and genetics:

Graph interpretation: Reading complex graphs with dual y-axes to identify positive correlation between follicle growth and oestrogen secretion.
Multi-step mathematical calculation: Reading a value from a graph, finding the radius from a diameter, calculating the volume of a 3D sphere ( V = 4/3 π r³ ), and calculating the total number of follicles.
Endocrine knowledge: Linking pituitary hormones (FSH) to follicle development, ovarian hormone production (oestrogen), and female secondary sexual characteristics.
Monohybrid inheritance & genetic crosses: Completing a Punnett square for carrier parents, matching genotypes to phenotypes, and explaining genetic probabilities ( 0.5 or 50%).

QUESTION 07.1

Evidence of Follicle Oestrogen Release

1 Mark • AO3 (Data Analysis)

✅ Mark Scheme Answer

As the volume of follicles rises, the oestrogen concentration in the blood rises (for the first 7 to 8 days).

Also accepted: Positive correlation between oestrogen concentration and follicle volume (up to day 7/8) OR oestrogen concentration is directly proportional to follicle volume (for 7–8 days).

🧠 Exam Technique

Look at both curves from Day 0 to Day 7/8. As one increases, the other increases. State this relationship explicitly using words like as [X] increases, [Y] increases.
Note the time frame: after Day 8, oestrogen drops while volume continues to peak at Day 11, so specify the initial rising phase.

❌ Common Misconception

Do NOT write: "An increase of oestrogen concentration causes an increase of follicle volume." The question asks for evidence that the follicles release oestrogen, not that oestrogen makes the follicles grow. Mention the correlation clearly!

QUESTION 07.2

Calculating the Number of Follicles

5 Marks • AO2 & AO3 (Applied Maths & Graph Reading)

📐 Step-by-Step Calculation

Step 1 — Find the radius:
Diameter = 22 mm → Radius r = 22 ÷ 2 = 11 mm
Step 2 — Calculate the volume of one follicle:
Volume = 4/3 × π × r³
Volume = (4 ÷ 3) × 3.14 × (11)³ = (4 ÷ 3) × 3.14 × 1331 = 5572.4533 mm³
(Allow 5575.28 mm³ if using full π on a calculator)
Step 3 — Read the total volume from Figure 12 on Day 11:
Locate Day 11 on the x-axis, follow up to the Volume curve, and read across to the right-hand y-axis:
Total volume = 39 000 mm³
Step 4 — Calculate the number of follicles:
Number = Total Volume ÷ Volume of one follicle
Number = 39 000 ÷ 5572.45... = 6.998...
Step 5 — Round to the nearest whole number:
6.998... ≈ 7

Mark Breakdown (5 marks total):
• [1 mark] Correct substitution of radius ( r = 11 ) into formula.
• [1 mark] Calculating volume of one follicle: 5572.45 (or 5575 ).
• [1 mark] Correctly reading 39 000 mm³ from Figure 12.
• [1 mark] Dividing total volume by single follicle volume ( 39 000 ÷ 5572 = 6.99... ).
• [1 mark] Correct final whole number: 7.

❌ Common Errors & Pitfalls

Diameter trap: Using r = 22 instead of halving it to get r = 11 .
Wrong axis: Reading the left-hand axis (hormone concentration) instead of the right-hand axis (total volume in mm³).
Incorrect rounding: Writing 7.0 instead of 7 . The question asks for the nearest whole number. 7.0 is given to one decimal place and loses the final mark!

💡 Examiner Tip

Always write down every single working step. If you misread the graph (e.g. reading 38 500 instead of 39 000), error-carried-forward (ECF) still allows you to pick up 4 out of the 5 marks!

QUESTION 07.3

FSH Deficiency and Breast Development

2 Marks • AO2 (Application of Knowledge)

✅ Mark Scheme Answer

Mark 1: Lack of FSH results in a lack of oestrogen production (because follicles do not develop/mature).
Mark 2: Breast development (female secondary sexual characteristics) depends on oestrogen.

💡 Key Biological Link

Understand the hormonal pathway:

Pituitary releases FSH → Follicles grow in ovary → Follicles release Oestrogen → Causes secondary sexual characteristics (breast growth)

If FSH is missing, follicles don't grow, so oestrogen isn't released, preventing breast development.

QUESTION 07.4

Inheritance of FSH Deficiency: Punnett Square

3 Marks • AO2 (Genetics Application)

✅ Punnett Square & Phenotypes

Both parents produce FSH but had a child who does not, meaning both parents must be heterozygous carriers (Hh).

Gametes	H	h
H	HH Produces FSH	Hh Produces FSH
h	Hh Produces FSH	hh Does NOT produce FSH

• Mark 1: Correct parental gametes: H and h for both parents.
• Mark 2: Correct offspring genotypes: HH, Hh, Hh, hh.
• Mark 3: Correct phenotypes identified: HH and Hh produce FSH; hh does not produce FSH.

❌ Common Errors

Forgetting to identify phenotypes: Many students draw the grid correctly but forget to state which offspring produce FSH and which do not.
Incorrect allele symbols: The question specified H and h. Do not invent other letters like F/f or A/a.
Missing the recessive homozygous state: You must show hh to prove how a child can inherit the condition.

QUESTION 07.5

Explaining the 0.5 Probability

3 Marks • AO3 (Genetic Reasoning)

✅ Mark Scheme Explanation

Mark 1: The mother is homozygous recessive ( hh ), so she always passes on the recessive allele h .
Mark 2: The father is heterozygous ( Hh ), so he passes on either H or h with equal probability (50% chance each).
Mark 3: The child will have genotype Hh with a probability of 0.5 (or 50%) and therefore will be able to produce FSH.

🧠 Quick Check via Cross

	H (Father)	h (Father)
h (Mother)	Hh (FSH)	hh (No FSH)

Out of 2 possible outcomes, exactly 1 is Hh (produces FSH).
Probability = 1 ÷ 2 = 0.5.

Note from Mark Scheme: You can gain all 3 marks by drawing and annotating a clear Punnett square diagram showing the mother as hh and the father as Hh .

Topics

Biology · B5: Homeostasis and Response · B6: Inheritance, Variation and Evolution

Question and mark scheme from the AQA GCSE Biology examination, Biology Paper 2 (Higher), June 2023. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.