AQA GCSE Biology Biology Paper 1 (Higher), June 2022: Question 6

14 marks · High Demand difficulty · Extended Answer

Calculate surface area to volume ratio from standard form data and explain adaptations of organisms, gas exchange surfaces, and metabolic rates relating to size.

Practise this question

Question

Table 6 displays data for organisms A to E: surface area, volume, and surface area to volume ratio, with organism C's ratio given as X:1. Question 06.1 asks to calculate value X. Question 06.2 asks for the relationship between organism size and surface area to volume ratio. Question 06.3 asks why organism D needs a respiratory system but organism B does not. Table 7 gives metabolic rates for organisms D (890) and E (75). Question 06.4 asks why the metabolic rate of warm-blooded organism D is greater than organism E. Question 06.5 presents Figure 8 showing cross-sectional diagrams of alveoli and villi, asking how they are adapted to increase absorption.
Question text

06 Table 6 shows information about five different organisms.

Table 6

Surface area Volume Surface area to

Organism 2 3

in m in m volume ratio

A 6.04 × 10−8 1.65 × 10−12 36606:1

B 3.21 × 10−3 1.25 × 10−6 2568:1

C 9.96 × 10−3 1.35 × 10−4 X:1

D 4.61 × 10−1 1.57 × 10−2 29:1

E 1.99 × 101 6.12 × 100 3:1

06.1 Calculate value X in Table 6.

Give your answer to the nearest whole number.

[3 marks]

X (nearest whole number) =

06.2 What is the relationship between the size of an organism and its surface area

to volume ratio?

Use Table 6.

[1 mark]

06.3 Organism B exchanges gases with the environment directly through its skin.

Organism D exchanges gases with the environment using its respiratory system.

Explain why organism D requires a respiratory system, but organism B does not

require a respiratory system.

[2 marks]

Table 6 is repeated below.

Table 6

Surface area Volume Surface area to

Organism 2 3

in m in m volume ratio

A 6.04 × 10−8 1.65 × 10−12 36606:1

B 3.21 × 10−3 1.25 × 10−6 2568:1

C 9.96 × 10−3 1.35 × 10−4 X:1

D 4.61 × 10−1 1.57 × 10−2 29:1

E 1.99 × 101 6.12 × 100 3:1

Table 7 shows information about organism D and organism E.

Table 7

Metabolic rate in

Organism

arbitrary units

D 890

E 29 75

06.4 Organisms D and E both keep a constant body temperature (warm-blooded).

Explain why the metabolic rate of organism D is greater than the metabolic rate of

organism E.

Use information from Table 6 and Table 7.

[4 marks]

06.5 Organism D and organism E both have alveoli in the lungs and villi in the

small intestine.

Figure 8 shows some alveoli and some villi.

Figure 8

Describe how the alveoli and the villi are adapted to increase absorption.

[4 marks]

Mark scheme

Show the mark scheme The mark scheme details marks across parts 06.1 to 06.5: 06.1 awards up to 3 marks for dividing 9.96 × 10^-3 by 1.35 × 10^-4 to reach 74. 06.2 awards 1 mark for identifying the inverse relationship between size and surface area to volume ratio. 06.3 awards 2 marks for organism D having a smaller SA:V ratio meaning diffusion distance is too large. 06.4 awards 4 marks for linking higher SA:V ratio in D to faster heat loss, requiring a higher respiration rate to generate heat. 06.5 provides a 4-mark levels-of-response grid detailing adaptations of alveoli and villi such as large surface area, thin walls, and close capillary blood supply.

Question 6

AO /

Question Answers Extra information Mark

Spec. Ref.

if no answer in answer space

06.1 allow answer in Table 6 AO2

4.1.3.1

9.96 ×10-3 0.00996 1

allow

1.35 ×10-4 0.000135

73.77…

74 (:1) allow a correctly derived whole

number from an incorrect

calculation

do not accept if unit is given

AO /

Spec. Ref.

06.2 as size increases, (surface area allow they are inversely 1 AO2

to volume) ratio decreases proportional or they are 4.1.3.1

negatively correlated

allow as one increases, the

other decreases

allow as size decreases,

(surface area to volume) ratio

increases

AO /

Spec. Ref.

06.3 allow converse for B throughout

D has a smaller surface area to 1 AO3

volume ratio (than B)

(so) diffusion distance is too allow (so) diffusion takes too 1 AO2

large (to meet demands of cells long (to meet demands of cells /

/ organism) organism) 4.1.3.1

AO / 23

Spec. Ref.

06.4 allow converse for E throughout

AO3

D has a larger surface area to allow D has a larger surface 1

volume ratio and so will lose area to volume ratio and so

heat more quickly (per unit temperature of D will drop more

volume than E) quickly AO2

ignore E loses more heat

(overall)

(D) requires greater rate of 1 AO2

respiration

(as) respiration is a (large) part 1 AO2

of metabolism

(so) need to generate more heat 1 4.1.3.1

(to keep itself warm) 4.4.2.1

4.4.2.3

allow (so) needs to release

more heat (to keep itself warm)

do not accept energy produced

/ made / created

AO /

Question Answers Mark

Spec. Ref.

06.5 Level 2: Scientifically relevant facts, events or processes are 3−4 AO1

identified and given in detail to form an accurate account. 4.1.3.1

4.1.3.3

Level 1: Facts, events or processes are identified and simply 1−2 4.2.2.2

stated but their relevance is not clear.

No relevant content 0

Indicative content

• both have a large surface area

o to maximise diffusion

• both have thin walls or have walls that are one cell thick

o to reduce diffusion distance / time

• both are in close proximity to blood supply

o to reduce diffusion distance / time

• both have a good blood supply or both have a capillary network

o to maintain concentration gradient

• villi have microvilli

o to (further) increase surface area

• cells of villi contain many mitochondria

o for active transport

For Level 2 reference to functions of structural details of both

alveoli and villi is required.

Total Question 6 14

How to answer it

Surface Area to Volume Ratio, Metabolism & Exchange Surfaces

What this question tests

Calculating ratios in standard form, interpreting how organism size alters surface area to volume (SA:V) ratios, explaining why multicellular organisms need specialised exchange surfaces, linking heat loss and metabolic rate in endotherms, and detailing structural adaptations of alveoli and villi for efficient absorption.

Part 06.1 • 3 Marks

Calculating Surface Area to Volume Ratio

Calculate value X for Organism C (to the nearest whole number)

📐 Step-by-Step Calculation

Step 1: Formula setup
Ratio = Surface Area ÷ Volume
9.96 × 10⁻³ ÷ 1.35 × 10⁻⁴
(or 0.00996 ÷ 0.000135)
Step 2: Unrounded value
73.777...
Step 3: Rounding
Rounding to nearest whole number: 74
Mark 1: Correct fraction setup • Mark 2: 73.77... • Mark 3: 74

❌ Common Errors

  • Adding units: The mark scheme states: "do not accept if unit is given". Ratios do not have units!
  • Power of 10 input errors: Forgetting brackets around negative indices on calculators (e.g. dividing by 1.35 then multiplying by 10⁻⁴ ).
  • Incorrect rounding: Writing 73 instead of rounding up to 74.
Part 06.2 • 1 Mark

Size vs Surface Area to Volume Ratio

What is the relationship between organism size and SA:V ratio?

✅ Correct Answer

As the size (volume) of an organism increases, its surface area to volume ratio decreases.

1 mark: Clear inverse relationship stated.

🧠 Exam Technique

Always check both extremes of Table 6:

  • Organism A (tiny volume: 1.65 × 10⁻¹² m³) → Huge SA:V (36606:1)
  • Organism E (large volume: 6.12 × 10⁰ m³) → Tiny SA:V (3:1)

Acceptable terms: "inversely proportional" or "negatively correlated".

Part 06.3 • 2 Marks

Requirement for a Specialised Respiratory System

Explain why organism D needs a respiratory system, but organism B does not

✅ Marking Points

  • Point 1: Organism D has a smaller surface area to volume ratio than organism B. [1 mark]
  • Point 2: Therefore, the diffusion distance is too large (or diffusion across skin is too slow) to supply oxygen to inner cells / meet metabolic demands. [1 mark]

💡 Key Knowledge

Small organisms (like B) have a high SA:V and short diffusion distance, so simple diffusion through their outer surface suffices.

Larger organisms (like D) have interior cells situated far from the surface; without a specialised exchange system and transport network, oxygen cannot reach them fast enough.

Part 06.4 • 4 Marks

Surface Area, Heat Loss & Metabolic Rate

Explain why the metabolic rate of D (890) is greater than E (75)

✅ Model Answer (4 Marks)

  1. Organism D has a larger surface area to volume ratio (29:1 vs 3:1) and so loses heat faster per unit volume than E. [1 mark]
  2. To maintain a constant body temperature, D requires a greater rate of respiration. [1 mark]
  3. Respiration is a major component of an organism's metabolism. [1 mark]
  4. Increased respiration generates/releases more heat to keep D warm. [1 mark]

❌ Critical Biological Traps

  • Fatal Phrasing: Never write "energy is created" or "energy is produced" — energy cannot be created! Say "heat is released / generated" or "energy transferred".
  • Confusion over total vs relative heat loss: A larger animal loses more total heat, but a smaller animal loses heat much faster per unit of body mass/volume.
Part 06.5 • 4 Marks (Level of Response)

Structural Adaptations of Alveoli and Villi

Describe how alveoli and villi are adapted to increase absorption

🧠 How to Secure Level 2 (3–4 Marks)

You must link structural features to their exact functional benefits for both alveoli and villi. Mentioning features without their mechanism limits you to Level 1 (1–2 marks).

💡 Common Shared Adaptations

  • Large Surface Area: Both are folded/numerous, which maximises the rate of diffusion.
  • Extremely Thin Walls: Both have single-cell-thick epithelial walls, providing a very short diffusion distance.
  • Rich Capillary Network: Extensive blood supply removes absorbed molecules rapidly, maintaining a steep concentration gradient.
  • Close Proximity: Capillaries lie immediately beneath the surface to minimise transport distance.

💡 Specific Adaptations for Villi

  • Microvilli: The epithelial cells have microscopic folds on their membrane, further multiplying the surface area available for absorption.
  • Many Mitochondria: Cells contain abundant mitochondria to supply ATP via aerobic respiration for the active transport of nutrients (like glucose) against concentration gradients.

Topics

Biology · B1: Cell Biology · B2: Organisation · B4: Bioenergetics

Question and mark scheme from the AQA GCSE Biology examination, Biology Paper 1 (Higher), June 2022. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.