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OCR A Level Biology

Topic questions Human physiology Surface area to volume ratio

Surface area to volume ratio

8 questions · 76 marks · about 1.5 hours

Surface area to volume ratio questions

OCR H420 Surface area to volume ratio 8 questions

Unified Biology June 2025, Question 1 9 marks

The roots of plants contain a variety of cells and tissues.

(a)

The photomicrograph shows a transverse section through the root of a herbaceous dicotyledonous plant.

Figure for question 1

Name the cells labelled A and B in the photomicrograph.

[2]
Reveal mark scheme Mark scheme for question 1(a)
(b)

The outer surface of a root is made up of root epidermal cells.

A root hair cell is a type of root epidermal cell that has a hair‑like extension called a root hair.

A typical root epidermal cell has a surface area to volume ratio (SA: V) of 0.3: 1.

A root hair cell has a higher SA: V than a typical root epidermal cell.

The diagram shows a root hair cell. The diagram is not to scale.

Figure for question 1

The root hair has:

• a diameter of 10 maths symbolm

• a length of 800 maths symbolm.

(b)(i)

Calculate the SA: V of the root hair.

Assume the root hair is cylindrical.

Use the equations:

Surface area of cylinder = 2maths symbolr(r + l)

Volume of cylinder = maths symbolr2l

[3]
Reveal mark scheme Mark scheme for question 1(b)(i)
(b)(ii)

State how a high SA: V helps a root hair cell perform its function.

[1]
Reveal mark scheme Mark scheme for question 1(b)(ii)
(c)

Carbohydrates produced in palisade cells are transported in phloem to sinks, such as roots.

Describe how carbohydrates move from palisade cells into companion cells in phloem.

[3]
Reveal mark scheme Mark scheme for question 1(c)

Unified Biology June 2023, Question 1 10 marks

Cats are mammals.

(a)

A light micrograph of a section through the trachea of a cat is shown in Fig. 1.1.

Figure for question 1
(a)(i)

Name the tissues labelled A and B in Fig. 1.1.

[2]
Reveal mark scheme Mark scheme for question 1(a)(i)
(a)(ii)

Name the two types of cell in the box labelled C in Fig. 1.1 and outline their functions.

[2]
Reveal mark scheme Mark scheme for question 1(a)(ii)
(b)

Alveoli increase the surface area to volume ratio (SA:V) in the lungs of cats and other mammals.

(b)(i)

An alveolus in the lung of a cat has:

• a spherical shape

• a diameter of 0.13 mm

• a surface area of 0.053 mm2.

Calculate the SA:V of this alveolus. Use the formula: volume of sphere = fraction r3

[3]
Reveal mark scheme Mark scheme for question 1(b)(i)
(b)(ii)

Explain why the large SA:V of alveoli is an advantage to mammals.

[1]
Reveal mark scheme Mark scheme for question 1(b)(ii)
(c)

The nervous system of a cat coordinates a response to stop objects becoming stuck in their throat.

Sensory receptors at the back of the throat detect the stimulus of a large piece of food or a ball of hair pressing on the tissue. The response is shown in Fig. 1.2.

Fig. 1.2

Sensory neurone

Question 1, shown as an image

in the throat
medulla oblongata

Figure for question 1
Question 1, shown as an image
Figure for question 1

Explain two conclusions that can be drawn from Fig. 1.2 about the type of nervous response shown.

[2]
Reveal mark scheme Mark scheme for question 1(c)

Biological Processes June 2022, Question 16 13 marks

Also involves Root hair cells · Mineral ions in plants

Dicotyledons are a large and diverse group of multicellular plants that includes oak trees and dandelions.

(a)

Explain, using examples of transported substances, why multicellular plants need transport systems.

[3]
Reveal mark scheme Mark scheme for question 16(a)
(b)

Fig. 16.1 is a light micrograph of a transverse section through the leaf of a dicotyledon.

Figure for question 16
(b)(i)

Identify the tissue labelled B in Fig. 16.1.

[1]
Reveal mark scheme Mark scheme for question 16(b)(i)
(b)(ii)

State two functions of structure A.

[2]
Reveal mark scheme Mark scheme for question 16(b)(ii)
(c)

Fig. 16.2 on the insert is a coloured scanning electron micrograph (SEM) of a longitudinal section through vascular tissue in the stem of a dicotyledon.

Explain how two features, visible in Fig. 16.2, show that this tissue is adapted to its function.

[4]
Reveal mark scheme Mark scheme for question 16(c)
(d)

A mycorrhiza is mutualistic association between a fungus and a plant. The fungus penetrates root cells and assists with uptake of water and nutrients.

A group of scientists investigated the role of mycorrhiza in the uptake of phosphates by plants.

Wild type barley plants produce root hairs. Mutant (brb) barley plants do not produce root hairs.

brb mutant and wild type plants were grown in pots and half of each group had mycorrhiza fungus added. After 28 days, the dry mass and phosphate content of the plants was measured.

The results are shown in the table.

Plant typefungus addedMycorrhiza Dry mass (g pot–1) Phosphate content (mg plant–1)
brb mutantno
0.84 ± 0.10
1.22 ± 0.13
brb mutantyes
1.23 ± 0.16
2.15 ± 0.25
Wild typeno
3.57 ± 0.16
4.72 ± 0.23
Wild typeyes
2.97 ± 0.36
3.91 ± 0.52

The scientists concluded that the mycorrhiza compensated for the absence of root hairs in the mutant (brb) barley plants.

Evaluate this conclusion.

[3]
Reveal mark scheme Mark scheme for question 16(d)

Unified Biology June 2022, Question 4 12 marks

The sea sponge, Aplysina aerophoba, and the zebra shark, Stegostoma fasciatum, are both animals.

(a)

A. aerophoba does not have an internal circulatory system. Instead, it filters food and oxygen from the surrounding water, as shown in Fig. 4.1.

Figure for question 4
(a)(i)

Suggest why A. aerophoba does not need a circulatory system.

[1]
Reveal mark scheme Mark scheme for question 4(a)(i)
(a)(ii)

A diagram of the circulatory system of S. fasciatum is shown in Fig. 4.2.

Figure for question 4

Describe the type of circulatory system that S. fasciatum has.

[2]
Reveal mark scheme Mark scheme for question 4(a)(ii)
(b)

Both A. aerophoba and S. fasciatum reproduce sexually, but under particular conditions they are both able to reproduce asexually.

In asexual reproduction in A. aerophoba:

• clumps of diploid cells detach from the body of the sponge

• the cells reattach to a surface and grow into new, adult sponges.

In asexual reproduction in S. fasciatum:

• meiosis occurs in a female

• two of the haploid cells produced by meiosis fuse to form a diploid cell

• the diploid cell develops into a new shark.

A student stated, ‘When they reproduce asexually, both animals produce clones of themselves.’.

Evaluate the student’s statement.

[3]
Reveal mark scheme Mark scheme for question 4(b)
(c)

* Humans can produce artificial clones of animals.

Describe two methods for producing artificial clones of animals.

Additional answer space if required.

[6]
Reveal mark scheme Mark scheme for question 4(c)

Biological Processes November 2020, Question 5 1 mark

Large multicellular animals need a transport system for oxygen and carbon dioxide.

Large multicellular plants do not need a transport system for oxygen and carbon dioxide.

Which of the following statements, A to D, correctly explains these observations?

A Large plants have a low surface area to volume ratio.

B Plant cells have a low metabolic rate.

C Plants generate ATP during photosynthesis, so they do not need to respire.

D Plants generate oxygen during photosynthesis.

Your answer

[1]
Reveal mark scheme Mark scheme for question 5

Biological Processes November 2020, Question 21 18 marks

(a)

The table compares the features of airways in the lungs.

Complete the table by putting a tick (✓) in each box if the feature is present and a cross (8 the feature is absent in each structure.

The first row has been completed for you.

Figure for question 21

Question 21(b) begins on page 30

[3]
Reveal mark scheme Mark scheme for question 21(a)
(b)

Ventilation of the lungs creates air movements that can be measured and recorded.

(b)(i)

Name the apparatus used to measure and record these air movements.

[1]
Reveal mark scheme Mark scheme for question 21(b)(i)
(b)(ii)

The trace below shows a recording of ventilation movements from an individual subject.

Figure for question 21

Use the trace to estimate the maximum value for tidal volume during the first minute.

[1]
Reveal mark scheme Mark scheme for question 21(b)(ii)
(b)(iii)

After 60 seconds, the subject was told to breathe in as deeply as possible and then breathe out fully.

Use the trace to calculate the vital capacity of the subject.

[2]
Reveal mark scheme Mark scheme for question 21(b)(iii)
(c)

* Compare and contrast the mechanism of expiration during the first 60 seconds of the trace with the mechanism of expiration when the subject was told to breathe out fully.

Additional answer space if required.

[6]
Reveal mark scheme Mark scheme for question 21(c)
(d)

Complete the following statements about exchange surfaces.

Use the correct terms selected from the list below.

circulatory system concentration gradient diffusion pathway

flow of air lung capacity surface area

surface area to volume ratio ventilation

Large organisms have a large........ but they

have a small........ This means they need

a specialised exchange surface and a........

Two features of an efficient exchange surface are:

1. A good blood supply to maintain the........

2. A short........

[5]
Reveal mark scheme Mark scheme for question 21(d)

Unified Biology June 2019, Question 5 6 marks

All organisms exchange gases with their environment.

(a)
Figure for question 5 Figure for question 5
[2]
Reveal mark scheme Mark scheme for question 5(a)
(b)

Beet armyworm larvae eat a variety of plants, including tomato plants.

Scientists wanted to investigate how effective a chemical called methyl jasmonate was in stopping beet armyworm larvae from eating plants. They sprayed tomato plants with different concentrations of methyl jasmonate and recorded the final biomass of the plants.

The results are shown in Fig. 5.1.

Figure for question 5

applied to plants (mmol dm–3)

Fig. 5.1

The scientists wrote the following hypothesis:

Plants use methyl jasmonate as a defence against herbivory.

(b)(i)

What additional information do the scientists need to confirm their hypothesis?

[2]
Reveal mark scheme Mark scheme for question 5(b)(i)
(b)(ii)

Suggest one valid conclusion it is possible for the scientists to draw from the results in Fig. 5.1.

[1]
Reveal mark scheme Mark scheme for question 5(b)(ii)
(b)(iii)
Figure for question 5
[1]
Reveal mark scheme Mark scheme for question 5(b)(iii)

Unified Biology June 2017, Question 7 7 marks

Many organisms have evolved specialised gas exchange surfaces. One feature of these structures is their large surface area to volume ratio.

(a)(i)

Describe how the structures of the insect tracheal system and fish gills provide a large surface area for gas exchange.

insect tracheal system........

[2]
Reveal mark scheme Mark scheme for question 7(a)(i)
(a)(ii)

The lugworm, Arenicola marina, is a species of segmented worm that lives in burrows in damp sand. They have hair-like external gills that increase the surface area available for gas exchange.

Many other species of segmented worm do not have external gills.

Suggest why lugworms have evolved external gills

[1]
Reveal mark scheme Mark scheme for question 7(a)(ii)
(b)
Figure for question 7
[4]
Reveal mark scheme Mark scheme for question 7(b)

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