AQA AS Level Biology Paper 1, November 2020: Question 9

10 marks · Medium difficulty · Short Answer

Explain the effects of genetic modification on aquaporin expression, membrane permeability, stomatal density, and calculate ion-to-water movement ratios.

Practise this question

Question

Exam question starting with number 09 about aquaporin channel proteins, genetically modified tobacco plants, and ion and water molecule movement across membranes, broken down into four parts (09.1 to 09.4) totaling 10 marks.
Question text

09 Channel proteins called aquaporins enable water to be transported across

membranes. Aquaporins are produced in cells when genes coding for the proteins

are expressed. One aquaporin gene is called PIP1b. The expression of PIP1b in

tobacco plant cells produces an aquaporin located in their cell membranes.

Scientists have produced genetically modified tobacco plants. The scientists 5

inserted a gene from a different species into the DNA of tobacco plant cells. This

gene causes an increase in the rate of transcription of the PIP1b gene.

The scientists found that the stomatal density of leaves from tobacco plants with

the inserted gene was greater than that of unmodified control plants.

In a different investigation, scientists measured the movement of potassium ions 10

and water molecules through cell-surface membranes and vacuole membranes.

They found 6 potassium ions moved for every 150 water molecules across vacuole

membranes. They found 3 potassium ions moved for every 1500 water molecules

across cell-surface membranes.

Use information from the passage and your own understanding to answer the

questions.

09.1 Explain how the proteome of a cell from a genetically modified tobacco plant

(lines 5–7) differs from that of a cell from an unmodified control tobacco plant.

[2 marks]

09.2 Explain how an increase in the rate of transcription of the PIP1b gene (lines 6–7) will

affect the permeability of tobacco plant cell membranes to water.

[2 marks]

09.3 Suggest and explain one advantage and one disadvantage of increased stomatal

density on the growth of tobacco plant leaves (lines 8–9).

[4 marks]

*22* Advantage

Disadvantage

09.4 How much greater is the ratio of movement of potassium ions to movement of water

molecules across a vacuole membrane than across a cell-surface membrane

(lines 10–14)? Show your working.

[2 marks]

Answer

Mark scheme

Show the mark scheme Mark scheme table detailing answers for questions 09.1 through 09.4, providing marking guidance, point allocations, and accepted alternatives for each sub-question.

Question Marking Guidance Mark Comments

09.1 1. Expression of gene from different species; 2

2. (So) a new/one more protein (in set/range of

proteins);

09.2 1. More/increased aquaporin/channel protein 2

(made)

2. Increased (water) permeability;

09.3 Advantage 4

1. More carbon dioxide uptake;

2. More photosynthesis so faster/more growth; 3. Accept plant wilts

for ‘more water

loss’

Disadvantage

4. Accept reduced

3. More water loss/transpiration

metabolic/chemical

4. Less photosynthesis so slower/less growth; reactions

09.4 Correct answer for 2 marks = x20;; 2

Accept for 1 mark, any correct simplification of a

correct ratio eg 6 : 150 = 1 : 25

TOTAL 10

How to answer it

Plant Transport, Gene Expression & Ratios Study Guide

What this question tests

This question assesses your understanding of molecular genetics (transcription and the proteome), membrane transport mechanisms (aquaporins and water permeability), gas exchange trade-offs in plants (stomata, photosynthesis, and transpiration), and quantitative data manipulation (calculating ratios and proportional differences).

Question 09.1 (2 marks)

Explain how the proteome of a cell from a genetically modified tobacco plant differs from that of a cell from an unmodified control tobacco plant.

✅ Correct Answer

  • The genetically modified plant expresses a gene from a different species (1 mark).
  • This results in a new or additional protein being produced within its set/range of proteins (1 mark).

💡 Key Knowledge

The proteome is defined as the full range of proteins that a cell is able to produce at a given time. Introducing foreign DNA into tobacco plants means instructions for a novel polypeptide are transcribed and translated, expanding the organism's total proteome.

🧠 Exam Technique

Be precise with biological terminology. Examiners look for explicit references to expression of a foreign gene and the resulting production of a new/additional protein. Do not just say "it has more DNA"—focus specifically on proteins.

❌ Common Errors

Students often lose marks by confusing the genome with the proteome, or vaguely stating that "genes make cells different" without explaining that translation creates a novel protein.

Question 09.2 (2 marks)

Explain how an increase in the rate of transcription of the PIP1b gene will affect the permeability of tobacco plant cell membranes to water.

✅ Correct Answer

  • More/increased aquaporins or channel proteins are made (1 mark).
  • This leads to increased water permeability across the membrane (1 mark).

💡 Key Knowledge

Aquaporins are specific channel proteins that facilitate the rapid diffusion of water molecules (osmosis) across biological membranes. Increased transcription leads to more mRNA, higher rates of translation, and a greater density of these channel proteins embedded in the phospholipid bilayer.

🧠 Exam Technique

Link the molecular event (transcription) directly to structural changes (more channel proteins) and functional consequences (permeability). Always explicitly mention water permeability.

❌ Common Errors

Candidates frequently forget to link transcription to translation and protein synthesis, jumping straight from gene expression to permeability without mentioning aquaporin channels.

Question 09.3 (4 marks)

Suggest and explain one advantage and one disadvantage of increased stomatal density on the growth of tobacco plant leaves.

✅ Correct Answer

Advantage (max 2 marks):

  • More carbon dioxide uptake (1 mark).
  • Leads to more photosynthesis, resulting in faster or greater growth (1 mark).

Disadvantage (max 2 marks):

  • More water loss/transpiration (1 mark).
  • Leads to less photosynthesis due to water stress/wilting/reduced metabolic reactions, or slower/less growth (1 mark).

💡 Key Knowledge

Stomata are the primary sites for gas exchange in leaves. They allow CO₂ to diffuse into the leaf for the Calvin cycle in photosynthesis, but simultaneously allow water vapour to escape via transpiration, creating a constant physiological trade-off for terrestrial plants.

🧠 Exam Technique

This is a "suggest and explain" question. You must pair a valid feature with a logical biological consequence affecting growth. Ensure your explanation connects the gas exchange point all the way through to sugar production or cellular hydration.

❌ Common Errors

Simply stating "more water loss" is insufficient for the second marking point of the disadvantage unless linked to a consequence such as plant wilting, reduced turgor, or slower growth/metabolism.

Question 09.4 (2 marks)

How much greater is the ratio of movement of potassium ions to movement of water molecules across a vacuole membrane than across a cell-surface membrane? Show your working.

✅ Correct Answer

  • Correct final answer = ×20 (2 marks).
  • Alternative / 1 mark awarded for any correct simplification of a correct ratio (e.g. vacuole ratio = 6 : 150 simplified to 1 : 25 , or equivalent working steps).

📐 Step-by-Step Calculation

  1. Extract data for vacuole membrane: 6 potassium ions moved for every 150 water molecules. Ratio = 6 : 150 (or simplified to 1 : 25 ).
  2. Extract data for cell-surface membrane: 3 potassium ions moved for every 1500 water molecules. Ratio = 3 : 1500 (or simplified to 1 : 500 ).
  3. Put both ratios over a common denominator or compare unit rates:
    Vacuole ratio per water molecule = 6 / 150 = 0.04
    Cell-surface ratio per water molecule = 3 / 1500 = 0.002
  4. Calculate how much greater: 0.04 / 0.002 = 20 -fold greater.

🧠 Exam Technique

Always show your intermediate ratios clearly. Even if your final calculation goes awry, examiners award 1 mark for accurately extracting and simplifying the ratio data from the text.

❌ Common Errors

Inverting the fractions or mixing up vacuole and cell-surface data values. Double-check your numbers against the stem text (lines 10–14) before performing division.

Topics

Biology · 3.2 Cells · 3.4 Genetic information, variation and relationships between organisms

Question and mark scheme from the AQA AS Level Biology examination, Paper 1, November 2020. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.