AQA A-Level Biology Paper 1, November 2021: Question 8

10 marks · Medium difficulty · Extended Answer

Describe triglyceride formation, identify fatty acid structures, evaluate experimental root growth data in Helianthus annuus populations, and explain species definition.

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Question

Question text

08.1 Describe how a triglyceride molecule is formed.

[3 marks]

08.2 Table 5 shows some properties of four fatty acids.

Table 5

Number of Number of

Fatty acid carbon atoms in double bonds

the R group in the R group

Caprylic acid 8 0

Palmitoleic

16 1

acid

Stearic acid 18 0

Linoleic acid 18 2

Figure 10 shows diagrams of these fatty acids.

Figure 10

Put a tick ( ) in one box that contains correct information about one of these fatty

acids.

[1 mark]

Caprylic acid is an unsaturated fatty acid represented by diagram L.

Linoleic acid is a saturated fatty acid represented by diagram N.

Palmitoleic acid is an unsaturated fatty acid represented by diagram K.

Stearic acid is a saturated fatty acid represented by diagram26 M.

The percentage of saturated fatty acids compared with unsaturated fatty acids found

in lipid stores in seeds differs in different populations.

Scientists investigated two populations of the plant, Helianthus annuus.

The scientists grew young plants from seeds collected from each population. They

placed the seeds on wet tissue paper so that the root growth was visible.

They grew seeds from each population at two temperatures:

• warm temperature of 24 °C

• cool temperature of 10 °C

After 10 days, the scientists measured the length of each root.

Table 6 shows some of the properties of the two populations and the scientists’

results.

Table 6

In the seed – Mean length of

Mean length of

Mean root after 10

root after 10

Temperature percentage °

days at 24 °C / days at 10 C /

Population in natural of fatty

mm mm

environment acids that

are (± 2 x standard (± 2 x standard

saturated deviation) deviation)

1 Warm 10.9 8.2 (±1.0) 3.1 (±0.3)

2 Cool 6.1 5.5 (±0.9) 4.3 (±0.2)

The mean ±2 × standard deviation includes 95% of the data.

08.3 The scientists used a data logger to measure the length of the root rather than a ruler.

Suggest one reason why they used a data logger and explain why this was important

in this investigation.

[1 mark]

08.4 It is known that:

• during respiration saturated fatty acids yield more energy than unsaturated fatty

acids

• saturated fatty acids have higher melting points than unsaturated fatty acids

• lipases in seeds act more rapidly on liquid substrates.

Use this information and Table 6 to show how each population is better adapted for its

natural environment when compared with the other population.

*26* [4 marks]

08.5 Although these two populations are completely separate and show genetic variation,

they are both called Helianthus annuus.

Explain why they are both given this name.

[1 mark]

Mark scheme

Show the mark scheme

Question Marking Guidance Mark Comments

Accept all marks in

08.1 1. One glycerol and three fatty acids;

suitably labelled

diagram OR in a

2. Condensation (reactions) and removal of

3 balanced equation

three molecules of water;

3. Ester bond(s) (formed);

08.2 Palmitoleic acid is an unsaturated fatty acid 1

represented by diagram K;

08.3 1. To increase accuracy/resolution because Ignore 'precision'

differences/lengths are small;

2. To increase accuracy because reduces risk of

human error;

1 max

3. To increase accuracy because roots are less

(likely to be) damaged;

4. To reduce error/uncertainty because – A-LEVEL BIOLOGY – –

differences/lengths are small;

1. Population 1 grew longer roots in warm

08.4

temperatures and population 2 grew longer

roots in cool temperatures; 2. Accept: 'Standard

deviations do not

2. Standard deviations do not overlap so overlap showing

difference (in mean) unlikely to be/not due to difference (in

chance; mean likely to be)

significant'

3. Population 1 (is better adapted to warm 4

conditions because it) has more saturated fatty

acids so more energy available (and more

3. and 4. Accept for

growth);

‘fatty acids’, fat

4. Population 2 (is better adapted to cool

conditions because it) has more

unsaturated/liquid fatty acids so more lipase

activity (and more growth);

08.5 Same species

(If mated) can produce fertile offspring 1

(It is) genus and species name;

How to answer it

Biological Molecules & Adaptation Study Guide

AQA A-Level Biology

What this question tests

This question assesses your understanding of lipid biochemistry (triglyceride formation and fatty acid structure), practical methodology (measuring equipment and data loggers), data interpretation involving standard deviations, evolutionary adaptation, and binomial nomenclature (species definition).

Question 08.1: Triglyceride Formation

✅ Correct Answer

1 glycerol and 3 fatty acids.
Condensation reaction(s) removing 3 molecules of H₂O.
Ester bond(s) formed.

💡 Key Knowledge

Triglycerides are formed by esterification. Each hydroxyl group on glycerol reacts with the carboxyl group of a fatty acid. Three water molecules are released per triglyceride synthesized.

🧠 Exam Technique

Make sure to specify the exact quantities (one glycerol, three fatty acids, three water molecules). Vague statements like "glycerol joins to fatty acids" will not score full marks.

Marks available: 3 marks

Question 08.2: Identifying Fatty Acids

✅ Correct Answer

Palmitoleic acid is an unsaturated fatty acid represented by diagram K.

💡 Key Knowledge

Table 5 indicates palmitoleic acid has 16 carbon atoms in the R group and 1 double bond. Diagram K visually shows a chain with 16 carbons and a kink (double bond).

❌ Common Errors

Confusing carbon chain length with double bonds, or failing to look for the characteristic "kink" in the hydrocarbon chain caused by a cis double bond in unsaturated fatty acids.

Marks available: 1 mark

Question 08.3: Data Loggers vs. Rulers

✅ Correct Answer (1 max)

To increase accuracy/resolution because differences/lengths are small.
To reduce the risk of human error.
To prevent root damage (handling roots with physical measuring tools).
To reduce error/uncertainty.

🧠 Exam Technique

Note the examiner guidance: Ignore "precision". Always link the use of automated sensors or data loggers directly to minimizing human error or handling delicate biological specimens.

Marks available: 1 mark

Question 08.4: Data Analysis & Adaptation

✅ Correct Answer (4 marks)

Trend 1: Population 1 grew longer roots in warm temperatures, whereas Population 2 grew longer roots in cool temperatures.
Statistical Significance: Standard deviations do not overlap, showing the difference in the mean is likely significant / not due to chance.
Population 1 Adaptation: Better adapted to warm conditions because it has more saturated fatty acids, leading to more stored energy available for growth.
Population 2 Adaptation: Better adapted to cool conditions because it has more unsaturated/liquid fatty acids, allowing more rapid lipase activity for faster breakdown and growth at lower temperatures.

🧠 Exam Technique

When questions provide bulleted information ("It is known that..."), you must explicitly use those rules to connect the data in the table to the biological explanation. Mention standard deviation overlap to prove significance.

❌ Common Errors

Students often forget to mention standard deviations when discussing differences in means, or describe the physiological properties of lipids without linking them back to the specific temperature environments shown in Table 6.

Marks available: 4 marks

Question 08.5: Species Definition & Nomenclature

✅ Correct Answer

They are the same species / if mated they can produce fertile offspring / it consists of a genus and species name.

💡 Key Knowledge

The biological species concept defines a species as a group of similar organisms that can interbreed to produce living, fertile offspring. In binomial nomenclature, the second term denotes the specific epithet (species).

Marks available: 1 mark

Topics

Biology · Practical skills · 3.1 Biological molecules · 3.4 Genetic information, variation and relationships between organisms · 3.6 Organisms respond to changes in their internal and external environments (A-level only) · Data analysis · Uncertainty and evaluation

Question and mark scheme from the AQA A-Level Biology examination, Paper 1, November 2021. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.