AQA A-Level · Thinka-original Practice Paper

2024 AQA A-Level Biology 7402 Practice Paper with Answers

Thinka Jun 2024 AQA A Level-Style Mock — Biology 7402

260 marks360 mins2024
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An original Thinka practice paper modelled on the structure and difficulty of the Jun 2024 AQA A Level Biology 7402 paper. Not affiliated with or reproduced from AQA.

7402/1 Paper 1 Section A

Answer all questions in the spaces provided.
10 Question · 90.8 marks
Question 1 · Short Answer & Calculation
9.1 marks
Ecologists investigated a population of violet ground beetles (\(Carabus\ violaceus\)) in a woodland with an area of \(2.5\text{ hectares}\) (where \(1\text{ hectare} = 10,000\text{ m}^2\)). They captured \(120\) beetles, marked them with a small spot of non-toxic, weather-resistant paint, and released them. Five days later, they captured \(150\) beetles in the same area, of which \(18\) were marked.

(a) Calculate the estimated population density of these beetles per \(100\text{ m}^2\). Show your working.

(b) Explain two precautions the ecologists must take when marking the beetles to ensure the accuracy of the mark-release-recapture method.

(c) Suggest how a high rate of emigration during the five-day period would affect the calculated population size estimate.
Show answer & marking scheme

Worked solution

(a)
First, calculate the total estimated population size (\(N\)) using the Lincoln Index formula:
\(N = \frac{M \times C}{R}\)
where:
- \(M\) (number marked initially) = \(120\)
- \(C\) (total captured in second sample) = \(150\)
- \(R\) (number recaptured with marks) = \(18\)

\(N = \frac{120 \times 150}{18} = \frac{18,000}{18} = 1000\) beetles.

Next, convert the woodland area into square metres:
\(2.5\text{ hectares} \times 10,000\text{ m}^2\text{ ha}^{-1} = 25,000\text{ m}^2\).

Finally, calculate the density per \(100\text{ m}^2\):
\(\text{Density} = \left(\frac{1000}{25,000}\right) \times 100 = 4\text{ beetles per } 100\text{ m}^2\).

(b)
1. The paint must be non-toxic so it does not increase the mortality rate of marked beetles compared to unmarked ones.
2. The paint must not make the beetles more visible to predators, which would also artificially increase their mortality rate.
3. The paint must be waterproof/durable so it does not rub off before the second capture.

(c)
If marked beetles emigrate (leave the study area), fewer marked beetles will be available for recapture. This will decrease the value of \(R\) (recaptures), which acts as the denominator. A smaller denominator results in an overestimation of the total population size.

Marking scheme

Part (a): 4 Marks
- Correct formula or substitution to find total population: \(\frac{120 \times 150}{18}\) (1 mark)
- Correct total population of 1000 beetles (1 mark)
- Conversion of area to \(25,000\text{ m}^2\) (1 mark)
- Correct final density of 4 (1 mark)

Part (b): 3 Marks (Max 3)
- Non-toxic paint / does not affect survival (1 mark)
- Disguised/does not increase predation risk (1 mark)
- Does not hinder movement or mating behaviour (1 mark)
- Paint is permanent/does not rub off (1 mark)

Part (c): 2.1 Marks
- Explains that emigration leads to fewer marked individuals in the area/less chance of recapture (1 mark)
- Correctly identifies that the calculated population size will be an overestimate (1.1 marks)
Question 2 · Short Answer & Calculation
9.1 marks
An ecological study tracked primary succession on a sand dune system over 150 years. The percentage cover of marram grass (\(Ammophila\ arenaria\)) and sessile oak trees (\(Quercus\ petraea\)) was measured at regular intervals. At Year 60, the percentage cover of sessile oak was \(5\%\). At Year 150, the percentage cover of sessile oak had reached \(75\%\).

(a) Calculate the mean rate of increase in percentage cover of sessile oak per year between Year 60 and Year 150. Give your answer to 2 decimal places.

(b) Describe how pioneering species like marram grass facilitate the colonization of the sand dunes by later successional species like sessile oak.

(c) Explain why the population of marram grass eventually declines to zero as the oak woodland climax community is established.
Show answer & marking scheme

Worked solution

(a)
- Change in percentage cover: \(75\% - 5\% = 70\%\)
- Change in time: \(150\text{ years} - 60\text{ years} = 90\text{ years}\)
- Mean rate of increase: \(\frac{70}{90} = 0.7777...\%\text{ per year}\)
- Rounded to 2 decimal places: \(0.78\%\text{ per year}\)

(b)
Pioneer species stabilize the loose sand with their extensive root networks. When they die and decay, they are decomposed by microorganisms, which adds organic matter (humus) to the sand. This improves soil nutrient availability and increases water retention, making the abiotic conditions less hostile and allowing later successional species to germinate and survive.

(c)
Sessile oak trees grow taller and form a dense canopy. This shades out the ground level, reducing the light intensity reaching the marram grass. Marram grass is adapted to open, high-light environments and cannot photosynthesize sufficiently in the shade, leading to a competitive exclusion by dominant woody species.

Marking scheme

Part (a): 3 Marks
- Correct difference in percentage cover (70) and time (90) (1 mark)
- Correct calculation: \(70 / 90\) (1 mark)
- Correct rounding to 2 decimal places: 0.78 (1 mark)

Part (b): 3.1 Marks
- Stabilisation of soil/dunes by roots (1 mark)
- Decomposition of organic matter / formation of humus (1 mark)
- Makes environment less hostile by increasing nutrients/water retention (1.1 marks)

Part (c): 3 Marks
- Climax community species (oak trees) grow taller and block light / form a canopy (1 mark)
- Light becomes a limiting factor for photosynthesis in marram grass (1 mark)
- Outcompeted for resources (light/space/nutrients) leading to competitive exclusion (1 mark)
Question 3 · Short Answer & Calculation
9.1 marks
A student used cell fractionation and ultracentrifugation to isolate mitochondria from a sample of fresh chicken liver.

(a) Before homogenizing the liver tissue, the student placed it in an ice-cold, isotonic, and buffered solution. Explain the importance of each of these three conditions.

(b) The isolated mitochondria were viewed using a transmission electron microscope (TEM). In one micrograph, the measured length of a single mitochondrion was \(4.8\text{ cm}\). The magnification of the image was \(\times 15,000\). Calculate the actual length of the mitochondrion in micrometres (\(\mu\text{m}\)). Show your working.
Show answer & marking scheme

Worked solution

(a)
- **Ice-cold**: Reduces the activity of hydrolytic enzymes (such as lysosomal enzymes) that might be released during homogenization, preventing them from digesting organelles.
- **Isotonic**: Has the same water potential as the liver cells/organelles. This prevents water from entering the mitochondria by osmosis (which would cause them to burst) or leaving them (which would cause them to shrink/damage).
- **Buffered**: Maintains a constant pH to prevent the denaturation of membrane proteins and enzymes within the organelles.

(b)
- Formula: \(\text{Actual Size} = \frac{\text{Image Size}}{\text{Magnification}}\)
- Convert image size to micrometres (\(\mu\text{m}\)):
\(4.8\text{ cm} = 48\text{ mm} = 48,000\text{ }\mu\text{m}\).
- Calculate actual length:
\(\text{Actual Length} = \frac{48,000}{15,000} = 3.2\text{ }\mu\text{m}\).

Marking scheme

Part (a): 5 Marks
- Ice-cold reduces enzyme/protease activity (1 mark)
- Prevents digestion of organelles/mitochondria (1 mark)
- Isotonic prevents osmotic movement of water / maintains constant water potential (1 mark)
- Organelles do not burst/shrink (1 mark)
- Buffer prevents pH changes to stop protein/enzyme denaturation (1 mark)

Part (b): 4.1 Marks
- Correct conversion of cm to mm (48 mm) or directly to \(\mu\text{m}\) (\(48,000\text{ }\mu\text{m}\)) (2 marks)
- Correct calculation: \(48,000 / 15,000\) (1 mark)
- Correct final answer: 3.2 (1.1 marks)
Question 4 · Short Answer & Calculation
9.1 marks
A student prepared a root tip squash of garlic (\(Allium\ sativum\)) to observe the stages of mitosis. Out of \(450\) cells examined under the light microscope, \(18\) were identified as being in metaphase.

(a) Calculate the percentage of cells that were in metaphase.

(b) The complete cell cycle for these cells takes \(24\text{ hours}\). Assuming that the proportion of cells in any stage is directly proportional to the time spent in that stage, calculate how many minutes a single cell spends in metaphase. Show your working.

(c) Describe the distinct behavior of chromosomes during metaphase and contrast this with their behavior during anaphase.
Show answer & marking scheme

Worked solution

(a)
Percentage of cells in metaphase:
\(\frac{18}{450} \times 100 = 4\%\)

(b)
First, convert the total duration of the cell cycle into minutes:
\(24\text{ hours} \times 60\text{ minutes/hour} = 1440\text{ minutes}\).

Multiply the total minutes by the proportion of cells in metaphase:
\(1440\text{ minutes} \times 0.04 = 57.6\text{ minutes}\).

(c)
- **Metaphase**: Chromosomes line up along the equator (metaphase plate) of the cell. Spindle fibres attach to the centromeres of each chromosome.
- **Anaphase**: The centromeres split. Spindle fibres contract, pulling sister chromatids apart to opposite poles of the spindle. The chromatids are now referred to as individual chromosomes.

Marking scheme

Part (a): 2 Marks
- Correct fraction: \(18/450\) (1 mark)
- Correct percentage: 4% (1 mark)

Part (b): 3.1 Marks
- Conversion of 24 hours to 1440 minutes (1 mark)
- Correct multiplication of fraction by 1440: \(0.04 \times 1440\) (1 mark)
- Correct calculation of 57.6 minutes (1.1 marks)

Part (c): 4 Marks
- (Metaphase) Chromosomes align at the equator/middle of the cell (1 mark)
- (Metaphase) Spindle fibres attach to centromeres (1 mark)
- (Anaphase) Centromeres divide / chromatids split (1 mark)
- (Anaphase) Chromatids/chromosomes pulled to opposite poles by contracting spindle fibres (1 mark)
Question 5 · Short Answer & Calculation
9.1 marks
A student investigated the response of woodlice to moisture using a choice chamber containing four distinct quadrants: Damp/Dark, Damp/Light, Dry/Dark, and Dry/Light. At the start, \(40\) woodlice were introduced into the center of the apparatus. After 15 minutes, the number of woodlice in each quadrant was recorded:

- Damp/Dark: \(22\)
- Damp/Light: \(10\)
- Dry/Dark: \(6\)
- Dry/Light: \(2\)

(a) Perform a Chi-squared (\(\chi^2\)) test to determine whether the distribution of woodlice differs significantly from a random (equal) distribution. Show your working and state the calculated \(\chi^2\) value to 1 decimal place.

(b) Using the critical value of \(7.81\) for \(3\) degrees of freedom at the \(p = 0.05\) significance level, state your statistical conclusion.

(c) Explain the survival value of the woodlice moving towards and remaining in damp, dark environments.
Show answer & marking scheme

Worked solution

(a)
- Null hypothesis (\(H_0\)): There is no preference; woodlice are distributed equally among the 4 quadrants.
- Expected number (\(E\)) in each quadrant if distributed equally: \(\frac{40}{4} = 10\).
- Formula: \(\chi^2 = \sum \frac{(O - E)^2}{E}\)

Let's calculate the term for each quadrant:
- Damp/Dark: \(\frac{(22 - 10)^2}{10} = \frac{12^2}{10} = 14.4\)
- Damp/Light: \(\frac{(10 - 10)^2}{10} = 0\)
- Dry/Dark: \(\frac{(6 - 10)^2}{10} = \frac{(-4)^2}{10} = 1.6\)
- Dry/Light: \(\frac{(2 - 10)^2}{10} = \frac{(-8)^2}{10} = 6.4\)

Sum of terms:
\(\chi^2 = 14.4 + 0 + 1.6 + 6.4 = 22.4\).

(b)
Since the calculated \(\chi^2\) value of \(22.4\) is greater than the critical value of \(7.81\), we reject the null hypothesis. The probability (\(p\)) that the observed distribution occurred by chance is less than 0.05. The woodlice show a significant preference for certain conditions.

(c)
Damp environments prevent desiccation (water loss) because woodlice have permeable exoskeletons/gills for gas exchange. Dark environments reduce the risk of detection by predators and are also typically damper, which enhances survival and reproductive success.

Marking scheme

Part (a): 4 Marks
- Correctly identifies expected value for each category is 10 (1 mark)
- Correct calculation of individual \((O-E)^2/E\) terms (at least two correct) (1 mark)
- Correct formula application (1 mark)
- Correct final calculated \(\chi^2\) value of 22.4 (1 mark)

Part (b): 2.1 Marks
- States calculated value is greater than critical value (1 mark)
- Rejects the null hypothesis / states result is highly significant (1.1 marks)

Part (c): 3 Marks
- Dampness prevents water loss / desiccation (1 mark)
- Woodlice have gas exchange surfaces/gills that must remain moist (1 mark)
- Darkness reduces risk of predation (1 mark)
Question 6 · Short Answer & Calculation
9.1 marks
The effect of indole-3-acetic acid (IAA) concentration on the growth of oat coleoptile segments was investigated. The mean elongation of segments incubated in different concentrations of IAA was recorded.

- Control (\(0\text{ mol dm}^{-3}\) IAA): \(1.5\text{ mm}\)
- Treated (\(10^{-6}\text{ mol dm}^{-3}\) IAA): \(8.4\text{ mm}\)

(a) Calculate the percentage increase in coleoptile elongation of the treated segments compared to the control. Show your working.

(b) Explain how the distribution of IAA within a shoot changes when the shoot is exposed to unilateral light from one side.

(c) Describe the mechanism by which IAA causes cell elongation on the shaded side of a plant shoot.
Show answer & marking scheme

Worked solution

(a)
- Absolute increase: \(8.4\text{ mm} - 1.5\text{ mm} = 6.9\text{ mm}\)
- Percentage increase: \(\frac{6.9}{1.5} \times 100 = 460\%\)

(b)
When unilateral light shines on a shoot, IAA is actively transported laterally from the illuminated side to the shaded side of the shoot tip. This results in a higher concentration of IAA on the shaded side of the shoot than on the light-exposed side.

(c)
IAA stimulates the active transport of hydrogen ions (\(\text{H}^+\)) from the cytoplasm into the cell wall matrix. This lowers the pH of the cell wall, which activates enzymes called expansins. Expansins break the hydrogen bonds between cellulose microfibrils, loosening the cell wall. Water enters the cell by osmosis, increasing turgor pressure, which stretches the loosened cell wall and causes cell elongation.

Marking scheme

Part (a): 3 Marks
- Correct subtraction to find growth increase: \(6.9\text{ mm}\) (1 mark)
- Correct expression: \((6.9 / 1.5) \times 100\) (1 mark)
- Correct final percentage: 460% (1 mark)

Part (b): 2.1 Marks
- IAA moves/transports laterally (1 mark)
- From light side to shaded side of the shoot tip (1.1 marks)

Part (c): 4 Marks
- IAA stimulates active transport of hydrogen ions (\(\text{H}^+\)) into cell walls (1 mark)
- Acidification of cell wall activates expansin proteins (1 mark)
- Cellulose microfibrils loosen / bonds break (1 mark)
- Osmotic water uptake / turgor pressure stretches the wall (1 mark)
Question 7 · Short Answer & Calculation
9.1 marks
An enzyme-controlled reaction was monitored in the presence and absence of a chemical inhibitor.

- Initial rate of reaction without inhibitor: \(4.5\text{ mmol dm}^{-3}\text{ s}^{-1}\)
- Initial rate of reaction with a non-competitive inhibitor: \(1.8\text{ mmol dm}^{-3}\text{ s}^{-1}\)

(a) Calculate the percentage inhibition of the enzyme activity caused by the non-competitive inhibitor.

(b) Explain how a non-competitive inhibitor reduces the rate of an enzyme-controlled reaction.

(c) State how the effect of a competitive inhibitor on the rate of reaction differs from that of a non-competitive inhibitor when substrate concentration is significantly increased.
Show answer & marking scheme

Worked solution

(a)
- Decrease in rate: \(4.5 - 1.8 = 2.7\text{ mmol dm}^{-3}\text{ s}^{-1}\)
- Percentage inhibition: \(\frac{2.7}{4.5} \times 100 = 60\%\)

(b)
A non-competitive inhibitor binds to the enzyme at an allosteric site (a site other than the active site). This binding alters the tertiary structure of the enzyme, changing the shape of its active site. Consequently, the substrate can no longer bind to the active site to form enzyme-substrate complexes, even if the substrate concentration is extremely high.

(c)
Increasing the substrate concentration will overcome the effect of a competitive inhibitor, eventually reaching the maximum rate of reaction (\(V_{\max}\)), because the substrate outcompetes the inhibitor for the active site. In contrast, increasing substrate concentration has no effect on a non-competitive inhibitor, and the maximum rate of reaction (\(V_{\max}\)) cannot be reached.

Marking scheme

Part (a): 3.1 Marks
- Correct subtraction: \(4.5 - 1.8 = 2.7\) (1 mark)
- Calculation setup: \((2.7 / 4.5) \times 100\) (1 mark)
- Correct final answer: 60% (1.1 marks)

Part (b): 3 Marks
- Binds to allosteric site / site other than active site (1 mark)
- Changes tertiary structure/shape of the active site (1 mark)
- Substrate can no longer bind / no enzyme-substrate complexes form (1 mark)

Part (c): 3 Marks
- High substrate concentration overcomes competitive inhibition / reaches \(V_{\max}\) (1 mark)
- Competitive inhibitor competes for the active site (1 mark)
- High substrate concentration does not overcome non-competitive inhibition / cannot reach \(V_{\max}\) (1 mark)
Question 8 · Short Answer & Calculation
9.1 marks
A student carried out paper chromatography to separate a mixture of amino acids. The solvent was allowed to migrate until the solvent front reached \(12.5\text{ cm}\) from the start line. One of the amino acids, tyrosine, migrated a distance of \(7.5\text{ cm}\) from the start line.

(a) Calculate the retardation factor (\(R_f\)) value of tyrosine in this solvent system.

(b) Draw or describe the general structural formula of an amino acid, identifying the functional groups present.

(c) Explain how a change in the primary structure of a protein can lead to a completely non-functional enzyme.
Show answer & marking scheme

Worked solution

(a)
- Formula: \(R_f = \frac{\text{Distance travelled by substance}}{\text{Distance travelled by solvent front}}\)
- \(R_f = \frac{7.5}{12.5} = 0.60\) (or \(0.6\))

(b)
An amino acid consists of a central carbon atom (\(\text{C}_\alpha\)) bonded to:
1. An amine/amino group (\(-\text{NH}_2\))
2. A carboxyl group (\(-\text{COOH}\))
3. A hydrogen atom (\(-\text{H}\))
4. A variable side chain (\(-\text{R}\) group)

(c)
The primary structure is the specific sequence of amino acids in a polypeptide chain. A change in this sequence alters the position of the R-groups. This changes where hydrogen, ionic, and disulfide bonds form during folding, which alters the final tertiary structure of the protein. If the tertiary structure is altered, the shape of the active site changes, meaning the substrate is no longer complementary, no enzyme-substrate complexes can form, and the enzyme becomes non-functional.

Marking scheme

Part (a): 2 Marks
- Correct division setup: \(7.5 / 12.5\) (1 mark)
- Correct final answer: 0.6 or 0.60 (1 mark)

Part (b): 3.1 Marks
- Describes/draws central carbon with H (1 mark)
- Identifies amine/amino group (\(-\text{NH}_2\)) (1 mark)
- Identifies carboxyl group (\(-\text{COOH}\)) and variable R group (1.1 marks)

Part (c): 4 Marks
- Primary structure determines the sequence of amino acids (1 mark)
- Dictates the position of hydrogen, ionic, and disulfide bonds (1 mark)
- Alters the tertiary structure / 3D shape of the protein (1 mark)
- Active site changes shape so it is no longer complementary to the substrate / no ES complexes form (1 mark)
Question 9 · Short Answer & Calculation
9 marks
### Question 1

(a) Describe how a peptide bond is formed between two amino acids. **[2 marks]**

(b) A scientist investigated the effect of an inhibitor on the rate of reaction of an amylase enzyme. They measured the mass of maltose produced over time. The results are shown in **Table 1**.

**Table 1**

| Time / s | Mass of maltose with no inhibitor / mg | Mass of maltose with inhibitor / mg |
| :--- | :--- | :--- |
| 0 | 0.0 | 0.0 |
| 20 | 3.2 | 1.1 |
| 40 | 4.8 | 1.8 |
| 60 | 5.0 | 2.1 |

Calculate the percentage decrease in the initial rate of reaction (from 0 to 20 seconds) caused by the inhibitor. Show your working. **[3 marks]**

(c) The inhibitor used in this investigation was a non-competitive inhibitor. Explain how a non-competitive inhibitor reduces the rate of an enzyme-controlled reaction. **[4 marks]**
Show answer & marking scheme

Worked solution

### Part (a)
1. A condensation reaction occurs between the carboxyl group (\(-\text{COOH}\)) of one amino acid and the amino group (\(-\text{NH}_2\)) of another amino acid.
2. This reaction releases a molecule of water (\(\text{H}_2\text{O}\)) and forms a peptide bond.

### Part (b)
1. Calculate the initial rate for the control (no inhibitor):
\(\text{Rate}_{\text{control}} = \frac{3.2\text{ mg} - 0.0\text{ mg}}{20\text{ s}} = 0.16\text{ mg s}^{-1}\)
2. Calculate the initial rate with the inhibitor:
\(\text{Rate}_{\text{inhibitor}} = \frac{1.1\text{ mg} - 0.0\text{ mg}}{20\text{ s}} = 0.055\text{ mg s}^{-1}\)
3. Calculate the percentage decrease:
\(\text{Percentage decrease} = \frac{0.16 - 0.055}{0.16} \times 100 = \frac{0.105}{0.16} \times 100 = 65.625\%\)
Rounding to 3 significant figures gives **65.6%** (accept 66%).

### Part (c)
1. The non-competitive inhibitor binds to the enzyme at an alternative site (allosteric site) rather than the active site.
2. This binding alters the tertiary structure of the enzyme.
3. Consequently, this changes the specific shape of the active site.
4. Therefore, the substrate is no longer complementary, preventing the formation of enzyme-substrate (E-S) complexes.

Marking scheme

**Part (a): [2 marks total]**
* **1 mark** for stating it is a condensation reaction between the carboxyl group of one amino acid and the amino group of another.
* **1 mark** for stating that a water molecule is released.

**Part (b): [3 marks total]**
* **1 mark** for correct calculation of control rate (\(0.16\)) AND inhibitor rate (\(0.055\)).
* **1 mark** for correct subtraction and division to find the percentage change: \(\frac{0.16 - 0.055}{0.16} \times 100\) (or equivalent ratio method).
* **1 mark** for correct final answer of **65.6%** (accept **66%** or **65.63%**).
* *Note: If the final answer is correct with no working shown, award 3 marks.*

**Part (c): [4 marks total]**
* **1 mark** for binding to the allosteric site / a site other than the active site.
* **1 mark** for changing the tertiary structure of the enzyme.
* **1 mark** for changing the shape of the active site (so it is no longer complementary).
* **1 mark** for preventing substrate binding / preventing enzyme-substrate (E-S) complexes from forming.
Question 10 · Short Answer & Calculation
9 marks
### Question 2

(a) Explain why a homogenizing solution used in cell fractionation must be cold, isotonic, and buffered. **[3 marks]**

(b) A student used an optical microscope with an eyepiece graticule to measure the size of chloroplasts in plant cells. The stage micrometer had divisions spaced exactly 10 \(\mu\)m apart. At a magnification of \(\times 400\), 40 divisions of the eyepiece graticule lined up exactly with 16 divisions of the stage micrometer.

Calculate the actual length of one eyepiece graticule unit at this magnification. Show your working. **[2 marks]**

(c) The student measured a single chloroplast and found its length was 2.5 eyepiece graticule units.

Calculate the actual length of this chloroplast in micrometers (\(\mu\)m) and determine the magnification of a drawing of this chloroplast where its length is drawn as 35 mm. Show your working. **[4 marks]**
Show answer & marking scheme

Worked solution

### Part (a)
1. **Cold:** To reduce enzyme activity so that hydrolytic enzymes (e.g., from lysosomes) do not digest or damage the organelles.
2. **Isotonic:** To maintain the same water potential as the organelles, preventing the net movement of water by osmosis so that organelles do not burst or shrivel.
3. **Buffered:** To maintain a constant pH, preventing denaturation of essential proteins and enzymes within the organelles.

### Part (b)
1. Determine the actual distance represented by 16 stage micrometer divisions:
\(16 \text{ divisions} \times 10 \text{ }\mu\text{m} = 160 \text{ }\mu\text{m}\)
2. Determine the value of 1 eyepiece graticule unit:
\(1 \text{ eyepiece unit} = \frac{160 \text{ }\mu\text{m}}{40} = 4.0 \text{ }\mu\text{m}\)

### Part (c)
1. Calculate actual length of chloroplast:
\(\text{Actual length} = 2.5 \text{ eyepiece units} \times 4.0 \text{ }\mu\text{m/unit} = 10.0 \text{ }\mu\text{m}\)
2. Convert the drawing size to micrometers:
\(35 \text{ mm} = 35 \times 1000 = 35,000 \text{ }\mu\text{m}\)
3. Calculate magnification:
\(\text{Magnification} = \frac{\text{Drawing size}}{\text{Actual size}} = \frac{35,000 \text{ }\mu\text{m}}{10 \text{ }\mu\text{m}} = \times 3500\) (or \(3500\))

Marking scheme

**Part (a): [3 marks total]**
* **1 mark** for Cold: reduces/prevents enzyme activity (to prevent organelle damage).
* **1 mark** for Isotonic: prevents osmosis / net water movement (so organelles do not burst/shrivel).
* **1 mark** for Buffered: maintains pH to prevent denaturation of proteins/enzymes.

**Part (b): [2 marks total]**
* **1 mark** for showing the step: \(16 \times 10 = 160 \text{ }\mu\text{m}\).
* **1 mark** for correct answer: **4.0** (or **4**) \(\mu\)m.

**Part (c): [4 marks total]**
* **1 mark** for calculating actual length: **10.0** (or **10**) \(\mu\)m.
* **1 mark** for correct conversion of mm to \(\mu\)m (e.g., \(35 \text{ mm} = 35,000 \text{ }\mu\text{m}\)).
* **1 mark** for correct formula: \(\text{Magnification} = \frac{\text{Drawing size}}{\text{Actual size}}\).
* **1 mark** for correct magnification: **\(\times 3500\)** (or **3500**).
* *Note: If a student uses an incorrect actual length from part (b), allow full follow-through marks for part (c) if calculation is correct.*

7402/2 Paper 2 Section A

Answer all questions in the spaces provided.
10 Question · 90.99999999999999 marks
Question 1 · Short Answer & Calculation
9.1 marks
A biologist investigated the population of a ground beetle species in a deciduous woodland using the mark-release-recapture method. In the first sample, 120 beetles were captured, marked with non-toxic, weather-resistant paint, and released. Five days later, a second sample of 150 beetles was captured. Of these, 18 were found to have been marked.

(a) Calculate the estimated population size of the ground beetles in this woodland. Show your working.

(b) State two assumptions of the mark-release-recapture method that must be met to ensure the calculated estimate is reliable.

(c) Describe how the biologist could use random sampling with a grid and coordinates to select sampling locations within the woodland, rather than selecting them subjectively.
Show answer & marking scheme

Worked solution

(a) Population size is calculated using the formula: \(N = \frac{M \times C}{R}\), where \(M\) is the number marked in the first sample (120), \(C\) is the total captured in the second sample (150), and \(R\) is the number of marked recaptures (18).
\(N = \frac{120 \times 150}{18} = \frac{18000}{18} = 1000\).

(b) Key assumptions include:
1. No births, deaths, immigration, or emigration have occurred during the 5-day period.
2. The mark does not affect the survival or behaviour of the beetles (e.g., does not make them more visible to predators).
3. The marked beetles mix fully and randomly back into the population before the second capture.

(c) To conduct random sampling: Set up a grid over the woodland area using two long measuring tapes at right angles to create x and y axes. Use a random number generator to obtain pairs of coordinates. Place the quadrat at the intersection of these coordinates to sample the area without subjective bias.

Marking scheme

Total Marks: 9.1
- (a) [3 marks total] 1 mark for correct formula or substituting correct values into formula: \(\frac{120 \times 150}{18}\). 1 mark for correct calculation step: \(18000 / 18\). 1 mark for correct final answer: 1000.
- (b) [3 marks total] 1.5 marks for each valid assumption stated (max 2): e.g., no birth/death/migration (accept: population is closed) (1.5 marks); marking does not affect survival/predation (1.5 marks); marked individuals distribute evenly (1.5 marks).
- (c) [3.1 marks total] 1 mark for establishing a grid system using measuring tapes at right angles. 1 mark for using a random number generator (accept: random number table, computer program) to get coordinates. 1.1 marks for placing the quadrats at the intersection of these coordinates to remove bias.
Question 2 · Short Answer & Calculation
9.1 marks
During the primary succession of a sand dune ecosystem, the gross primary productivity (GPP) of a pioneer grass species was measured as \(8.4 \times 10^3\text{ kJ m}^{-2}\text{ yr}^{-1}\). The respiratory loss (R) for this species was determined to be \(4.8 \times 10^3\text{ kJ m}^{-2}\text{ yr}^{-1}\).

(a) Calculate the net primary productivity (NPP) of this pioneer grass species. Include the correct unit in your final answer.

(b) Explain how the establishment of pioneer species alters the abiotic environment, making it more suitable for other species to colonise.

(c) Suggest why biomass accumulation eventually stabilizes as succession reaches a climax community.
Show answer & marking scheme

Worked solution

(a) Net Primary Productivity is calculated using the equation: \(NPP = GPP - R\).
\(NPP = 8.4 \times 10^3 - 4.8 \times 10^3 = 3.6 \times 10^3\text{ kJ m}^{-2}\text{ yr}^{-1}\) (or \(3600\text{ kJ m}^{-2}\text{ yr}^{-1}\)).

(b) Pioneer species are adapted to harsh conditions. When they die and decompose, they add organic matter (humus) to the soil. This improves soil structure, increases water retention, and adds nutrients (like nitrates), making the abiotic environment less hostile for other species to establish.

(c) In a climax community, dominant species (such as mature trees) reach a stable equilibrium. Large trees have high respiratory demands to maintain their large non-photosynthetic biomass (trunks, roots). As a result, the rate of photosynthesis (GPP) is balanced by respiration (R), leading to near-zero net biomass accumulation.

Marking scheme

Total Marks: 9.1
- (a) [3 marks total] 1 mark for correct formula \(NPP = GPP - R\). 1 mark for correct numerical answer: \(3.6 \times 10^3\) or \(3600\). 1 mark for correct unit: \(\text{kJ m}^{-2}\text{ yr}^{-1}\).
- (b) [3 marks total] 1 mark for pioneer species dying/decomposing to form humus. 1 mark for increasing soil nutrient content / water retention. 1 mark for making the environment less hostile/more suitable for subsequent species.
- (c) [3.1 marks total] 1.1 marks for stating that gross productivity balances respiratory losses (GPP \(\approx\) R). 1 mark for mentioning the high respiratory demand of non-photosynthetic biomass in mature trees. 1 mark for stating that species composition remains constant and dominant species prevent further new colonisers.
Question 3 · Short Answer & Calculation
9.1 marks
An electron micrograph of a mitochondrion from a liver cell shows its length as \(4.5\text{ cm}\). The magnification of the image is \(\times 30,000\).

(a) Calculate the actual length of the mitochondrion in micrometres (\(\mu\text{m}\)). Show your working.

(b) During cell fractionation, the liver tissue was homogenised in an ice-cold, isotonic, and buffered solution. Explain the biological reason for each of these three conditions.

(c) Name the organelle that would pellet out first during differential centrifugation, and state which pellet would contain the enzymes responsible for aerobic respiration.
Show answer & marking scheme

Worked solution

(a) Actual size = Image size / Magnification.
Convert image size to micrometres: \(4.5\text{ cm} = 45\text{ mm} = 45,000\text{ }\mu\text{m}\).
Actual length = \(\frac{45,000}{30,000} = 1.5\text{ }\mu\text{m}\).

(b) The conditions are essential for the following reasons:
- Ice-cold: Reduces the activity of hydrolytic enzymes (like lysosomal enzymes) to prevent them from digesting or damaging organelles.
- Isotonic: Maintains the same water potential as the organelles to prevent water from entering or leaving them by osmosis, which could cause them to burst or shrink.
- Buffered: Maintains a constant pH to prevent the denaturation of structural proteins and enzymes within the organelles.

(c) The heaviest organelle, the nucleus, pellets out first (at low speed). The pellet containing mitochondria (which contains the enzymes for aerobic respiration) is obtained at the second, higher centrifugation speed.

Marking scheme

Total Marks: 9.1
- (a) [3 marks total] 1 mark for converting cm to \(\mu\text{m}\) (\(45,000\text{ }\mu\text{m}\)). 1 mark for dividing by magnification (\(45,000 / 30,000\)). 1 mark for correct final answer: \(1.5\text{ }\mu\text{m}\) (accept 1.5).
- (b) [3.1 marks total] 1 mark for Ice-cold: prevents/slows enzyme activity to stop organelle degradation. 1 mark for Isotonic: prevents osmotic movement of water to prevent bursting/shrinkage of organelles. 1.1 marks for Buffered: maintains pH to prevent enzyme/protein denaturation.
- (c) [3 marks total] 1.5 marks for identifying the nucleus/nuclei as pelleting out first. 1.5 marks for identifying the second pellet (mitochondria) as containing the enzymes for aerobic respiration.
Question 4 · Short Answer & Calculation
9.1 marks
A spherical bacterium has a diameter of \(1.2\text{ }\mu\text{m}\), while a spherical eukaryotic human cell has a diameter of \(12.0\text{ }\mu\text{m}\).

(a) Calculate the ratio of the surface area to volume ratio (SA:V) of the bacterium to that of the eukaryotic cell. Show your working.
(Note: Surface area of a sphere = \(4\pi r^2\); Volume of a sphere = \(\frac{4}{3}\pi r^3\))

(b) Describe two structural differences between a prokaryotic cell wall and a plant cell wall.

(c) Explain why eukaryotic cells benefit from having membrane-bound organelles, whereas bacteria do not require them.
Show answer & marking scheme

Worked solution

(a) Let us simplify the SA:V ratio formula for a sphere:
\(SA:V = \frac{4\pi r^2}{\frac{4}{3}\pi r^3} = \frac{3}{r}\).
For the bacterium (diameter = \(1.2\text{ }\mu\text{m}\), radius \(r = 0.6\text{ }\mu\text{m}\)):
\(SA:V = \frac{3}{0.6} = 5\text{ }\mu\text{m}^{-1}\).
For the eukaryotic cell (diameter = \(12.0\text{ }\mu\text{m}\), radius \(r = 6.0\text{ }\mu\text{m}\)):
\(SA:V = \frac{3}{6.0} = 0.5\text{ }\mu\text{m}^{-1}\).
Ratio of bacterium SA:V to eukaryotic cell SA:V = \(\frac{5}{0.5} = 10\).
Therefore, the ratio is 10:1.

(b) 1. A prokaryotic cell wall contains murein (peptidoglycan), whereas a plant cell wall is composed of cellulose.
2. Prokaryotic walls lack microfibril structures and pectin, which are present in plant cell walls.

(c) Eukaryotic cells are much larger, so their SA:V ratio is too low for simple diffusion to meet cellular demands. Organelles compartmentise biochemical processes, keeping enzymes and substrates concentrated. Bacteria have a very large SA:V ratio, allowing rapid diffusion of nutrients and wastes across the plasma membrane to sustain metabolic functions without internal membrane compartments.

Marking scheme

Total Marks: 9.1
- (a) [3.1 marks total] 1 mark for correctly determining the radius of both cells (\(0.6\text{ }\mu\text{m}\) and \(6.0\text{ }\mu\text{m}\)). 1 mark for calculating both SA:V values (\(5\) and \(0.5\)). 1.1 marks for the correct final simplified ratio of 10:1 (or 10).
- (b) [3 marks total] 1.5 marks for stating prokaryotic wall is murein/peptidoglycan vs plant cell wall is cellulose. 1.5 marks for describing structural differences in composition/linkages (e.g., cellulose microfibrils vs cross-linked glycan chains).
- (c) [3 marks total] 1 mark for noting the larger size and smaller SA:V ratio of eukaryotic cells. 1 mark for stating that organelles allow compartmentalisation / concentrated reaction microenvironments. 1 mark for relating bacterial small size to rapid diffusion pathways that negate the need for internal membranes.
Question 5 · Short Answer & Calculation
9.1 marks
The refractory period of a mammalian neurone axon lasts for \(1.5\text{ ms}\).

(a) Calculate the maximum theoretical frequency of action potentials (in \(\text{s}^{-1}\)) that can pass along this axon during this period. Show your working.

(b) Explain how the refractory period ensures that action potentials are unidirectional and discrete.

(c) Explain how myelination increases the speed of conduction of action potentials along an axon.
Show answer & marking scheme

Worked solution

(a) Convert the refractory period to seconds:
\(1.5\text{ ms} = 1.5 \times 10^{-3}\text{ s} = 0.0015\text{ s}\).
Frequency = \(\frac{1}{\text{Refractory period}}\).
Frequency = \(\frac{1}{0.0015} \approx 666.67\text{ s}^{-1}\) (or \(667\text{ s}^{-1}\)).

(b) During the refractory period, sodium ion voltage-gated channels are closed and inactivated. This prevents another action potential from being generated immediately behind the active region, ensuring the impulse only moves forward (unidirectional). It also ensures that individual action potentials are separated from one another (discrete).

(c) Myelin acts as an electrical insulator. It prevents ion flow across the axon membrane except at the gaps, known as the Nodes of Ranvier. Consequently, local currents cannot flow continuously through the insulated sheath; instead, depolarisation 'jumps' from one node to the next. This process is called saltatory conduction, which greatly increases impulse speed compared to unmyelinated fibres.

Marking scheme

Total Marks: 9.1
- (a) [3 marks total] 1 mark for converting ms to seconds (\(0.0015\text{ s}\)). 1 mark for setting up formula \(1 / t\). 1 mark for correct calculation: 667 (accept 666.7 or 666.67).
- (b) [3 marks total] 1.5 marks for explaining unidirectionality: sodium channels behind the action potential are inactive/closed, so depolarisation cannot spread backwards. 1.5 marks for explaining discrete nature: a second action potential cannot merge with the first because there must be a recovery delay before channels can reopen.
- (c) [3.1 marks total] 1 mark for explaining myelin acts as an electrical insulator. 1 mark for stating that action potentials/depolarisation only occur at the Nodes of Ranvier. 1.1 marks for describing saltatory conduction (depolarisation 'jumping' from node to node).
Question 6 · Short Answer & Calculation
9.1 marks
A patient with diabetes has a blood glucose concentration of \(9.2\text{ mmol dm}^{-3}\). After receiving an injection of insulin, their blood glucose concentration drops to \(5.6\text{ mmol dm}^{-3}\) over a period of 45 minutes.

(a) Calculate the average rate of blood glucose clearance in \(\mu\text{mol dm}^{-3}\text{ min}^{-1}\). Show your working.

(b) Describe how insulin binds to target cells and stimulates the rapid uptake of glucose from the blood.

(c) Explain the role of negative feedback in restoring blood glucose levels when they fall below the normal range.
Show answer & marking scheme

Worked solution

(a) Calculate the total change in concentration:
\(\Delta C = 9.2 - 5.6 = 3.6\text{ mmol dm}^{-3}\).
Convert \(3.6\text{ mmol dm}^{-3}\) to \(\mu\text{mol dm}^{-3}\):
\(3.6 \times 1000 = 3600\text{ }\mu\text{mol dm}^{-3}\).
Calculate the rate by dividing by time (45 minutes):
\(\text{Rate} = \frac{3600\text{ }\mu\text{mol dm}^{-3}}{45\text{ min}} = 80\text{ }\mu\text{mol dm}^{-3}\text{ min}^{-1}\).

(b) Insulin binds to specific glycoprotein receptors on the cell surface membrane of target cells (such as liver and muscle cells). This binding induces a conformational change that causes vesicles containing glucose transporter proteins (GLUT4) to fuse with the cell surface membrane, increasing the membrane's permeability to glucose. It also activates intracellular enzymes that convert glucose to glycogen (glycogenesis).

(c) When blood glucose falls below normal, alpha cells in the islets of Langerhans in the pancreas detect the change and secrete glucagon (while beta cells reduce insulin secretion). Glucagon binds to receptors on liver cells, activating enzymes for glycogenolysis (glycogen to glucose) and gluconeogenesis (non-carbohydrates to glucose). The glucose is released into the blood, raising levels back to normal and terminating the glucagon secretion loop.

Marking scheme

Total Marks: 9.1
- (a) [3 marks total] 1 mark for calculating the difference: \(3.6\text{ mmol dm}^{-3}\). 1 mark for converting millimoles to micromoles: \(3600\text{ }\mu\text{mol}\). 1 mark for dividing by 45 to obtain the correct final answer: 80.
- (b) [3.1 marks total] 1 mark for stating insulin binds to specific receptors on target cell membranes. 1.1 marks for describing the fusion of GLUT4/glucose transporter vesicles with the cell membrane. 1 mark for mentioning the activation of enzymes for glycogenesis (converting glucose to glycogen).
- (c) [3 marks total] 1 mark for detection of low glucose by pancreas alpha cells and secretion of glucagon. 1 mark for glucagon triggering glycogenolysis/gluconeogenesis in the liver. 1 mark for stating that returning glucose to normal levels inhibits the initial corrective response (negative feedback).
Question 7 · Short Answer & Calculation
9.1 marks
An enzyme-controlled reaction was set up using a protease enzyme. In the first 3 minutes of the reaction, \(120\text{ mg}\) of protein substrate was fully hydrolysed.

(a) Calculate the initial rate of this reaction in \(\mu\text{g s}^{-1}\). Show your working.

(b) Contrast how competitive and non-competitive inhibitors affect the rate of an enzyme-controlled reaction.

(c) Describe how the tertiary structure of a protein enzyme is maintained and explain how high temperatures can cause denaturing.
Show answer & marking scheme

Worked solution

(a) Convert mass from milligrams to micrograms:
\(120\text{ mg} = 120 \times 1000 = 120,000\text{ }\mu\text{g}\).
Convert time from minutes to seconds:
\(3\text{ minutes} = 3 \times 60 = 180\text{ seconds}\).
Calculate the rate:
\(\text{Rate} = \frac{120,000\text{ }\mu\text{g}}{180\text{ s}} \approx 666.67\text{ }\mu\text{g s}^{-1}\) (or \(667\text{ }\mu\text{g s}^{-1}\)).

(b) Competitive inhibitors have a similar shape to the substrate and bind to the active site. They can be overcome by increasing substrate concentration. Non-competitive inhibitors bind to an allosteric site (away from the active site), changing the tertiary structure and shape of the active site so the substrate can no longer bind. Their effect cannot be overcome by adding more substrate.

(c) The tertiary structure is maintained by interactions between the R-groups of amino acids, including hydrogen bonds, ionic bonds, disulfide bridges, and hydrophobic interactions. High temperatures provide kinetic energy that causes molecules to vibrate excessively. This breaks weak hydrogen and ionic bonds, causing the protein to lose its specific 3D shape (denature), and changing the shape of the active site.

Marking scheme

Total Marks: 9.1
- (a) [3 marks total] 1 mark for converting mg to \(\mu\text{g}\) (\(120,000\)). 1 mark for converting minutes to seconds (\(180\)). 1 mark for correct calculation: 666.7 (accept 667 or 666.67).
- (b) [3 marks total] 1 mark for competitive inhibitors binding to the active site vs non-competitive binding to an allosteric site. 1 mark for describing the change in active site shape by non-competitive inhibitors. 1 mark for explaining that competitive inhibition can be overcome by excess substrate, but non-competitive cannot.
- (c) [3.1 marks total] 1 mark for listing at least two bonds maintaining tertiary structure (hydrogen, ionic, disulfide). 1 mark for stating high kinetic energy breaks hydrogen/ionic bonds at high temperatures. 1.1 marks for explaining that this alters the specific three-dimensional shape of the active site so substrate is no longer complementary.
Question 8 · Short Answer & Calculation
9.1 marks
A student performed paper chromatography to separate a mixture of amino acids. The solvent front migrated a total distance of \(14.2\text{ cm}\) from the origin line. Amino acid X migrated a distance of \(6.4\text{ cm}\).

(a) Calculate the \(R_f\) value of amino acid X. Give your answer to 2 decimal places. Show your working.

(b) Describe how a peptide bond is formed between two amino acids, including the type of reaction and any other molecules produced.

(c) Explain how the sequence of amino acids in a polypeptide chain determines the final three-dimensional structure of a globular protein.
Show answer & marking scheme

Worked solution

(a) The \(R_f\) value is calculated using the formula:
\(R_f = \frac{\text{Distance moved by amino acid}}{\text{Distance moved by solvent front}}\).
\(R_f = \frac{6.4\text{ cm}}{14.2\text{ cm}} \approx 0.4507\).
To 2 decimal places, the answer is 0.45.

(b) A peptide bond is formed by a condensation reaction between the amine group (\(-\text{NH}_2\)) of one amino acid and the carboxyl group (\(-\text{COOH}\)) of another. This reaction releases one molecule of water (\(\text{H}_2\text{O}\)).

(c) The specific sequence of amino acids represents the primary structure of the protein. The positions of specific R-groups (side chains) along the chain determine where hydrogen bonds, ionic bonds, hydrophobic interactions, and disulfide bridges form during folding. These interactions fold the polypeptide into a highly specific, unique tertiary shape essential for its function.

Marking scheme

Total Marks: 9.1
- (a) [3 marks total] 1 mark for using the correct formula: \(\frac{6.4}{14.2}\). 1 mark for performing the division. 1 mark for rounding correctly to two decimal places: 0.45 (reject other values like 0.46 or unrounded 0.451).
- (b) [3 marks total] 1 mark for identifying it as a condensation reaction. 1 mark for explaining the bond forms between the amine group of one amino acid and the carboxyl group of another. 1 mark for stating that water is produced.
- (c) [3.1 marks total] 1 mark for stating the sequence is the primary structure. 1 mark for explaining that the R-group properties determine the types of bonds formed (ionic, hydrogen, disulfide). 1.1 marks for stating these bonds lock the protein into its final, specific three-dimensional conformation.
Question 9 · Short Answer & Calculation
9.1 marks
An ecologist investigated the population size of the land snail *Cepaea nemoralis* on a coastal sand dune system.

First, they collected 112 snails, marked their shells with non-toxic, weather-resistant paint, and released them back into the habitat.

Four days later, they collected a second sample of 95 snails from the same area, of which 14 were marked.

**(a)** Calculate the estimated population size of the snails on this sand dune. Show your working. [2 marks]

**(b)** The ecologist wanted to ensure the marking technique did not affect the survival rate of the snails. Explain why using a bright, highly visible fluorescent paint would be inappropriate for this study. [2 marks]

**(c)** The sand dune system undergoes primary succession. Describe how the pioneer species modify the abiotic environment to allow other species to colonise. [3 marks]

**(d)** Give two assumptions of the mark-release-recapture method, other than the marking not affecting survival. [2 marks]
Show answer & marking scheme

Worked solution

**(a)** Using the Lincoln Index formula:
\( N = \frac{M \times C}{R} \)
where \( M = 112 \) (number marked in first sample), \( C = 95 \) (total recaptured in second sample), and \( R = 14 \) (number marked in second sample).
\( N = \frac{112 \times 95}{14} = 8 \times 95 = 760 \) snails.

**(b)** Highly visible paint makes the snails more conspicuous to predators (such as birds), significantly increasing their predation rate. This violates the key assumption that marked and unmarked individuals have equal chances of survival, leading to an artificially lower number of recaptured marked snails and an overestimation of the population size.

**(c)** Pioneer species (e.g., marram grass) stabilise the shifting sand dunes with their extensive root networks. When these pioneer plants die and decompose, they add organic matter (humus) to the sand, which increases the soil's nutrient content (such as nitrogen) and significantly improves its water retention capacity. This makes the environment less hostile and allows other species to colonise.

**(d)** Any two of the following assumptions:
1. There is no immigration or emigration (the population is closed).
2. There are no births or deaths during the period of the investigation.
3. Marked individuals distribute themselves randomly and mix completely back into the population.
4. The marks do not rub off, fade, or become lost during the study.

Marking scheme

**(a)**
• 1 Mark: Correct working shown: \( \frac{112 \times 95}{14} \).
• 1 Mark: Correct final answer of 760 (accept without working).

**(b)**
• 1 Mark: Paint makes marked snails more visible/conspicuous to predators (e.g., birds), increasing predation.
• 1 Mark: This decreases the survival rate of marked individuals relative to unmarked individuals / violates assumption of equal survival rates (resulting in overestimation).

**(c)**
• 1 Mark: Pioneer species stabilise the sand/soil using roots/rhizomes.
• 1 Mark: Pioneer species die and are decomposed by saprobionts to form organic matter/humus.
• 1 Mark: This increases soil nutrient content (e.g., nitrates/minerals) OR improves water-holding capacity of the soil.

**(d)**
• 1 Mark: Any one valid assumption (e.g., no migration / no births or deaths).
• 1 Mark: A second valid assumption (e.g., marked individuals mix randomly / mark is not lost).
Question 10 · Short Answer & Calculation
9.1 marks
A student investigated the effect of exercise intensity on heart rate and blood pH. Immediately after 10 minutes of moderate exercise, the student's blood pH decreased from 7.41 to 7.25.

**(a)** Explain how a decrease in blood pH during exercise leads to an increase in heart rate. [4 marks]

**(b)** Before exercise, the student's resting heart rate was 68 beats per minute (bpm). Immediately after exercise, it was 145 bpm. Calculate the percentage increase in heart rate. Show your working and give your answer to 3 significant figures. [2 marks]

**(c)** Parasympathetic stimulation affects heart rate. Describe how the parasympathetic nervous system decreases heart rate from its peak back to resting level. [3 marks]
Show answer & marking scheme

Worked solution

**(a)** 1. Increased cellular respiration during exercise produces more carbon dioxide, which dissolves in the blood to form carbonic acid, lowering blood pH.
2. This decrease in pH is detected by chemoreceptors located in the carotid arteries, aorta, and medulla oblongata.
3. These chemoreceptors send more frequent nerve impulses to the cardioacceleratory centre in the medulla oblongata.
4. The medulla oblongata sends more frequent impulses along the sympathetic nervous system to the sinoatrial node (SAN).
5. This causes the SAN to increase the frequency of electrical impulses it generates, leading to an increase in heart rate.

**(b)** 1. Calculate the increase in heart rate: \( 145 - 68 = 77 \) bpm.
2. Calculate the percentage increase: \( \frac{77}{68} \times 100 = 113.235... \% \).
3. Rounding to 3 significant figures gives 113%.

**(c)** 1. The cardioinhibitory centre in the medulla oblongata sends nerve impulses down the parasympathetic (vagus) nerve to the SAN.
2. Acetylcholine is released by the parasympathetic nerve endings at the SAN.
3. Acetylcholine binds to receptors on the SAN, decreasing the frequency of electrical impulses generated by the SAN, which slows down the heart rate back to resting levels.

Marking scheme

**(a)**
• 1 Mark: Chemoreceptors (in the carotid bodies / aortic bodies / medulla oblongata) detect the change/decrease in blood pH.
• 1 Mark: Chemoreceptors send more frequent nerve impulses to the cardioacceleratory centre in the medulla oblongata.
• 1 Mark: Medulla oblongata / cardioacceleratory centre sends more frequent impulses down the sympathetic nervous system/nerves.
• 1 Mark: Sympathetic nerves stimulate the sinoatrial node (SAN) to increase the frequency of electrical impulses/waves of depolarisation.

**(b)**
• 1 Mark: Correct working shown: \( \frac{145 - 68}{68} \times 100 \) (or \( \frac{77}{68} \times 100 \)).
• 1 Mark: Correct answer of 113% (must be given to 3 significant figures; accept 113).

**(c)**
• 1 Mark: Cardioinhibitory centre (in the medulla) sends impulses down the parasympathetic (or vagus) nerve.
• 1 Mark: Acetylcholine is released at the sinoatrial node (SAN).
• 1 Mark: This decreases/reduces the frequency of electrical impulses (waves of depolarisation) sent out by the SAN.

7402/3 Paper 3 Section A

Answer all questions in Section A.
6 Question · 52.98 marks
Question 1 · Short Answer & Comprehensive Calculation
8.83 marks
An ecological study was conducted to estimate the population size of the meadow grasshopper, *Chorthippus parallelus*, in a nature reserve using the mark-release-recapture method.

In the first sample, 142 grasshoppers were captured, marked with a non-toxic fluorescent dye, and released.
After 5 days, a second sample of 125 grasshoppers was captured from the same area. Of these, 18 were found to be marked.

(a) Calculate the estimated population size of *Chorthippus parallelus* in the nature reserve. Show your working. (2 marks)
(b) The ecologist calculated the 95% confidence interval for this population estimate to be \(986 \pm 145\). State what is meant by a "95% confidence interval". (2 marks)
(c) State three assumptions that must be met for the mark-release-recapture method to provide a valid estimate of population size. (3 marks)
(d) Explain how marking the grasshoppers with too much fluorescent dye could lead to an overestimation of the population size. (2 marks)
Show answer & marking scheme

Worked solution

(a)
Using the Lincoln-Petersen index formula:
\(N = \frac{M \times C}{R}\)
where:
\(M\) (number of individuals marked in the first sample) = 142
\(C\) (total number of individuals captured in the second sample) = 125
\(R\) (number of marked individuals recaptured) = 18

\(N = \frac{142 \times 125}{18} = 986.11\)
Rounded to the nearest whole number: 986.

(b)
A 95% confidence interval means that there is a 95% probability (or confidence) that the true population size lies within the calculated range of 841 to 1131 (i.e., \(986 \pm 145\)).

(c)
Any three of the following:
1. No immigration or emigration during the study period.
2. No births or deaths (the population size is stable) during the study period.
3. Marked individuals redistribute themselves randomly within the population.
4. The mark does not affect survival rates (e.g., does not make them more visible to predators).
5. The mark does not rub off or fade.

(d)
- Excessive dye could make the grasshoppers more visible to predators, increasing their mortality rate (or making them less mobile).
- This decreases the number of marked individuals surviving to be recaptured (so \(R\) decreases).
- Since \(R\) is in the denominator of the equation, a smaller value of \(R\) results in a larger, overestimated population size \(N\).

Marking scheme

Part (a):
- 1 mark for correct working showing numbers substituted into formula: \(\frac{142 \times 125}{18}\)
- 1 mark for correct final calculation of 986 (allow 986.11)

Part (b):
- 1 mark for stating there is a 95% chance/probability that the true population size lies in the interval.
- 1 mark for correctly identifying the range (841 to 1131).

Part (c):
- 1 mark per correct assumption listed (up to 3 marks max).

Part (d):
- 1 mark for identifying that the dye increases predation or mortality of marked individuals, reducing the number of recaptured marked individuals (\(R\)).
- 1 mark for explaining that a lower \(R\) mathematically results in a higher/overestimated population size estimate.
Question 2 · Short Answer & Comprehensive Calculation
8.83 marks
An ecologist investigated the diversity of plant species in two different areas of a sand dune system: the pioneer community on the foredunes (Area A) and a climax community further inland (Area B).

The table below shows the number of individuals of each plant species found in a \(10\text{ m} \times 10\text{ m}\) quadrat in each area.

| Species | Area A (Foredunes) | Area B (Inland Climax) |
| :--- | :--- | :--- |
| *Ammophila arenaria* (Marram grass) | 84 | 5 |
| *Leymus arenarius* (Sand wild rye) | 16 | 0 |
| *Festuca rubra* (Red fescue) | 8 | 42 |
| *Lotus corniculatus* (Bird's-foot trefoil) | 2 | 28 |
| *Calluna vulgaris* (Heather) | 0 | 55 |
| *Betula pendula* (Silver birch) | 0 | 12 |

The index of diversity (\(d\)) is calculated using the formula:

\(d = \frac{N(N - 1)}{\sum n(n - 1)}\)

where \(N\) = total number of organisms of all species and \(n\) = total number of organisms of each species.

(a) Calculate the index of diversity (\(d\)) for Area A and Area B. Give your answers to 2 decimal places. Show your working. (4 marks)
(b) Using the calculated indices of diversity, compare the two communities and explain how abiotic and biotic factors change as succession progresses from Area A to Area B. (5 marks)
Show answer & marking scheme

Worked solution

(a)
**For Area A:**
\(N = 84 + 16 + 8 + 2 = 110\)
\(N(N - 1) = 110 \times 109 = 11990\)
\(\sum n(n - 1) = (84 \times 83) + (16 \times 15) + (8 \times 7) + (2 \times 1)\)
\(\sum n(n - 1) = 6972 + 240 + 56 + 2 = 7270\)
\(d = \frac{11990}{7270} \approx 1.649 \approx 1.65\)

**For Area B:**
\(N = 5 + 42 + 28 + 55 + 12 = 142\)
\(N(N - 1) = 142 \times 141 = 20022\)
\(\sum n(n - 1) = (5 \times 4) + (42 \times 41) + (28 \times 27) + (55 \times 54) + (12 \times 11)\)
\(\sum n(n - 1) = 20 + 1722 + 756 + 2970 + 132 = 5600\)
\(d = \frac{20022}{5600} \approx 3.575 \approx 3.58\)

(b)
- Area B has a significantly higher index of diversity than Area A (3.58 vs 1.65).
- In Area A (pioneer stage), abiotic conditions are extreme/harsh (e.g., high wind, dry sand, high salinity), so only highly adapted species like Marram grass can survive.
- Over time, pioneer species colonize and die, adding organic matter/humus to the sand. This improves soil nutrient levels and water retention.
- This modification of the environment makes conditions less hostile, allowing new species (e.g., heather, fescue, birch trees) to colonize and outcompete pioneer species.
- In Area B (climax community), there are more complex niches, food webs, and biotic interactions (such as competition and mutualism), leading to high stability and diversity.

Marking scheme

Part (a):
- 1 mark for correct \(N(N-1)\) values (11990 for Area A, 20022 for Area B).
- 1 mark for correct \(\sum n(n-1)\) values (7270 for Area A, 5600 for Area B).
- 1 mark for correct Area A diversity index of 1.65 (accept 1.6).
- 1 mark for correct Area B diversity index of 3.58 (accept 3.6).

Part (b):
- 1 mark for stating Area B has higher species diversity than Area A.
- 1 mark for mentioning harsh/hostile abiotic conditions in the pioneer stage (Area A).
- 1 mark for explaining that pioneer species decompose to form humus/organic matter, changing soil quality/nutrient levels.
- 1 mark for stating that environmental modification makes conditions less hostile, allowing colonization by other species.
- 1 mark for mentioning climax community (Area B) stability, increased niches, or biotic interactions.
Question 3 · Short Answer & Comprehensive Calculation
8.83 marks
A student used an eyepiece graticule and a stage micrometer to measure the size of chloroplasts in a sample of palisade mesophyll cells.

At a magnification of \(\times 400\), 11 divisions on the stage micrometer aligned exactly with 100 divisions on the eyepiece graticule.
Each division on the stage micrometer is \(0.01\text{ mm}\) wide.

(a) Calculate the actual size of one eyepiece graticule division in micrometres (\(\mu\text{m}\)). Show your working. (2 marks)
(b) The student measured a single chloroplast and found it to be 4 eyepiece graticule divisions long. Calculate the actual length of the chloroplast in micrometres (\(\mu\text{m}\)). Show your working. (2 marks)
(c) Describe the principles and limitations of using a transmission electron microscope (TEM) to study the detailed internal structure of a chloroplast. (5 marks)
Show answer & marking scheme

Worked solution

(a)
- 11 divisions on the stage micrometer = \(11 \times 0.01\text{ mm} = 0.11\text{ mm}\).
- Convert mm to \(\mu\text{m}\): \(0.11\text{ mm} \times 1000 = 110\ \mu\text{m}\).
- Since 100 eyepiece divisions align with 110 \(\mu\text{m}\):
1 eyepiece division = \(\frac{110\ \mu\text{m}}{100} = 1.1\ \mu\text{m}\).

(b)
- Chloroplast length = \(4\text{ eyepiece divisions} \times 1.1\ \mu\text{m per division} = 4.4\ \mu\text{m}\).

(c)
**Principles of TEM:**
- A beam of electrons is transmitted through a very thin section of the specimen.
- Denser areas of the specimen absorb/scatter more electrons and appear darker, while less dense areas allow electrons to pass through and appear brighter.
- This yields a 2D image with extremely high resolution due to the very short wavelength of the electron beam.

**Limitations of TEM:**
- Specimen must be placed in a vacuum, which means living cells/specimens cannot be observed.
- The preparation of specimens (ultrathin slicing and heavy metal staining) is complex and can introduce artifacts (structures that are not naturally present).
- It does not produce a 3D image (unlike SEM) and only produces black-and-white images.

Marking scheme

Part (a):
- 1 mark for calculating the total distance of stage micrometer divisions in micrometres (110 \(\mu\text{m}\)).
- 1 mark for dividing by 100 to get 1.1 \(\mu\text{m}\).

Part (b):
- 1 mark for setting up calculation: \(4 \times 1.1\).
- 1 mark for correct final answer of 4.4 \(\mu\text{m}\).

Part (c):
- 1 mark for stating that an electron beam passes through thin sections of the specimen.
- 1 mark for explaining that differential absorption/scattering of electrons creates contrast (dark vs light regions).
- 1 mark for mentioning high resolution due to the short wavelength of electrons.
- 1 mark for mentioning that living specimens cannot be viewed because of the vacuum requirement.
- 1 mark for mentioning that preparation steps can create artifacts or that images are 2D/black-and-white.
Question 4 · Short Answer & Comprehensive Calculation
8.83 marks
The rate of diffusion of substance X across the cell membrane of a model cell is determined by Fick's Law:

\(\text{Rate of Diffusion} \propto \frac{\text{Surface Area} \times \text{Difference in Concentration}}{\text{Diffusion Path Length}}\)

A student compared two cell types:
- Cell Type 1: A spherical lymphocyte with a radius of \(5\ \mu\text{m}\).
- Cell Type 2: A cuboidal epithelial cell with microvilli. The flat base of the cell is \(10\ \mu\text{m} \times 10\ \mu\text{m}\), but the microvilli on its apical (top) surface increase the surface area of that face by a factor of 6.5.

(a) Calculate the total surface area of Cell Type 1. Use the formula for the surface area of a sphere: \(A = 4\pi r^2\). Use \(\pi = 3.14\). Show your working. (2 marks)
(b) Calculate the total surface area of Cell Type 2, assuming it is a cube of side length \(10\ \mu\text{m}\) where only the top face has microvilli (and the other 5 faces are flat square surfaces). Show your working. (3 marks)
(c) Explain two ways in which a cell can be adapted to increase the rate of active transport of a substance across its membrane, other than by increasing surface area. (4 marks)
Show answer & marking scheme

Worked solution

(a)
- Surface area of Cell Type 1: \(A = 4 \times 3.14 \times (5)^2\)
- \(A = 4 \times 3.14 \times 25 = 314\ \mu\text{m}^2\).

(b)
- A standard cube with side length \(10\ \mu\text{m}\) has 6 faces.
- Area of each flat face = \(10\ \mu\text{m} \times 10\ \mu\text{m} = 100\ \mu\text{m}^2\).
- There are 5 flat faces on Cell Type 2:
\(5 \times 100 = 500\ \mu\text{m}^2\).
- The apical face is increased by a factor of 6.5 due to microvilli:
\(100 \times 6.5 = 650\ \mu\text{m}^2\).
- Total surface area of Cell Type 2 = \(500 + 650 = 1150\ \mu\text{m}^2\).

(c)
- **Adaptation 1:** Increased density of specific carrier proteins (or channel proteins) embedded in the cell membrane, allowing more molecules to be transported across the membrane simultaneously.
- **Adaptation 2:** Increased abundance of mitochondria in the cell cytoplasm. This increases the rate of aerobic respiration, which generates more ATP needed to power active transport against the concentration gradient.

Marking scheme

Part (a):
- 1 mark for substitution: \(4 \times 3.14 \times 25\).
- 1 mark for correct calculation of 314 \(\mu\text{m}^2\).

Part (b):
- 1 mark for calculating the area of 5 flat faces: \(5 \times 100 = 500\ \mu\text{m}^2\).
- 1 mark for calculating the area of the top face with microvilli: \(100 \times 6.5 = 650\ \mu\text{m}^2\).
- 1 mark for correct total surface area of 1150 \(\mu\text{m}^2\).

Part (c):
- 1 mark for stating: more carrier proteins in the membrane.
- 1 mark for explaining: increases transport capacity/rate of movement.
- 1 mark for stating: more mitochondria in the cell.
- 1 mark for explaining: provides more ATP/energy from aerobic respiration for active transport.
Question 5 · Short Answer & Comprehensive Calculation
8.83 marks
The speed of conduction of an action potential along an axon depends on the diameter of the axon and whether it is myelinated.

An investigator measured the conduction velocity in two different neurones at \(20^\circ\text{C}\):
- Neurone A (myelinated, diameter = \(12\ \mu\text{m}\)): Conduction velocity = \(72\text{ m s}^{-1}\)
- Neurone B (unmyelinated, diameter = \(600\ \mu\text{m}\)): Conduction velocity = \(24\text{ m s}^{-1}\)

(a) Calculate the ratio of the conduction velocity of Neurone A to Neurone B per micrometre (\(\mu\text{m}\)) of axon diameter. Show your working. (3 marks)
(b) Explain how myelination increases the rate of conduction of an action potential along an axon. (3 marks)
(c) Explain why a reflex arc is rapid and how this benefits the organism. (3 marks)
Show answer & marking scheme

Worked solution

(a)
- Conduction velocity per \(\mu\text{m}\) for Neurone A = \(\frac{72\text{ m s}^{-1}}{12\ \mu\text{m}} = 6\text{ m s}^{-1}\ \mu\text{m}^{-1}\).
- Conduction velocity per \(\mu\text{m}\) for Neurone B = \(\frac{24\text{ m s}^{-1}}{600\ \mu\text{m}} = 0.04\text{ m s}^{-1}\ \mu\text{m}^{-1}\).
- Ratio of Neurone A to Neurone B = \(\frac{6}{0.04} = 150\).
- Therefore, the ratio is \(150:1\) (or the conduction velocity per unit diameter is 150 times faster in the myelinated neurone).

(b)
- Myelin acts as an electrical insulator, which prevents ion leakage and depolarization along the myelinated regions of the axon.
- Action potentials/depolarization can only occur at the gaps in the myelin sheath, known as the nodes of Ranvier (where there is a high concentration of voltage-gated sodium channels).
- This causes the action potential to jump from node to node in a process called saltatory conduction, which is much faster than continuous wave depolarization along the entire length of the axon.

(c)
- Reflex arcs are rapid because they only involve a small number of neurones (sensory, relay, and motor neurones) and therefore have very few synapses, which minimizes synaptic delay.
- They bypass the conscious processing centres of the brain, making the pathway shorter and fully involuntary.
- Benefit: This allows extremely rapid, automatic responses to harmful stimuli (e.g., pulling away from heat), minimizing tissue damage or danger.

Marking scheme

Part (a):
- 1 mark for calculating the velocity per \(\mu\text{m}\) for Neurone A (6).
- 1 mark for calculating the velocity per \(\mu\text{m}\) for Neurone B (0.04).
- 1 mark for the correct simplified ratio (150:1 or 150).

Part (b):
- 1 mark for stating that myelin acts as an electrical insulator.
- 1 mark for stating that depolarization/action potentials can only occur at the nodes of Ranvier.
- 1 mark for explaining that the impulse jumps from node to node / saltatory conduction.

Part (c):
- 1 mark for mentioning that there are few synapses/neurones in a reflex arc (minimising synaptic delay).
- 1 mark for stating that it bypasses conscious brain areas / is involuntary.
- 1 mark for stating that it prevents or minimises damage to body tissues.
Question 6 · Short Answer & Comprehensive Calculation
8.83 marks
The enzyme lactate dehydrogenase (LDH) catalyses the conversion of pyruvate to lactate.

In an experiment, the initial rate of reaction of LDH was measured at different substrate concentrations in the presence and absence of a newly developed drug, Compound Z.

The results are shown below:
- Maximum rate of reaction (\(V_{\max}\)) without Compound Z = \(180\ \mu\text{mol dm}^{-3}\text{ s}^{-1}\)
- Maximum rate of reaction (\(V_{\max}\)) with \(5\text{ mmol dm}^{-3}\) Compound Z = \(180\ \mu\text{mol dm}^{-3}\text{ s}^{-1}\)
- Substrate concentration needed to reach half \(V_{\max}\) (\(K_m\)) without Compound Z = \(2.4\text{ mmol dm}^{-3}\)
- Substrate concentration needed to reach half \(V_{\max}\) (\(K_m\)) with Compound Z = \(7.2\text{ mmol dm}^{-3}\)

(a) State the type of inhibition exhibited by Compound Z. Explain your answer using the data provided. (3 marks)
(b) Calculate the percentage increase in the substrate concentration required to reach half \(V_{\max}\) when Compound Z is added. Show your working. (2 marks)
(c) Describe how the structure of a protein, such as an enzyme, is determined by its primary structure. (4 marks)
Show answer & marking scheme

Worked solution

(a)
- Compound Z is a competitive inhibitor.
- The maximum rate of reaction (\(V_{\max}\)) is unaffected and remains at \(180\ \mu\text{mol dm}^{-3}\text{ s}^{-1}\) because high substrate concentrations can overcome the inhibition.
- The substrate concentration required to reach half \(V_{\max}\) (\(K_m\)) increases from 2.4 to 7.2, indicating that more substrate is needed to successfully compete with the inhibitor for the active site.

(b)
- Increase in substrate concentration (\(K_m\)) = \(7.2 - 2.4 = 4.8\text{ mmol dm}^{-3}\).
- Percentage increase = \(\frac{\text{Increase}}{\text{Original Value}} \times 100\%
= \frac{4.8}{2.4} \times 100\%
= 200\%\).

(c)
- The primary structure is the specific sequence of amino acids in the polypeptide chain, which is coded for by DNA.
- The unique sequence and properties of the R-groups (side chains) of these amino acids determine where chemical bonds form when the protein folds.
- These bonds include hydrogen bonds, ionic bonds, and disulfide bridges.
- The precise folding creates a specific 3D shape (tertiary structure), which determines the formation and specific shape of the active site that is complementary to its substrate.

Marking scheme

Part (a):
- 1 mark for stating 'competitive inhibitor'.
- 1 mark for explaining that \(V_{\max}\) remains unchanged because high substrate concentration overcomes the inhibitor.
- 1 mark for explaining that \(K_m\) increases because the inhibitor competing for the active site reduces the enzyme's affinity for the substrate.

Part (b):
- 1 mark for calculating the absolute increase (4.8).
- 1 mark for the correct final answer of 200%.

Part (c):
- 1 mark for stating primary structure is the specific sequence of amino acids.
- 1 mark for stating that the sequence/type of R-groups determines the position of bonds.
- 1 mark for naming hydrogen bonds, ionic bonds, or disulfide bridges.
- 1 mark for explaining that folding produces a specific tertiary structure / complementary active site.

7402/3 Paper 3 Section B (Synoptic Essay)

Answer one question from Section B.
1 Question · 25 marks
Question 1 · Synoptic Essay
25 marks
Answer one question from Section B.

Write an essay on:
The roles and importance of cycles in living organisms and ecosystems.
Show answer & marking scheme

Worked solution

Exemplar Content and Synoptic Links

An outstanding essay should integrate knowledge from multiple areas of the specification. Below are the key cycles that can be discussed, along with their molecular/cellular details and biological importance:

1. The Mitotic Cell Cycle:
• Details: Consists of Interphase (G1, S, G2 phases) and M-phase (Mitosis: prophase, metaphase, anaphase, telophase, followed by cytokinesis). Includes DNA replication during S phase, chromosome condensation, spindle attachment, separation of sister chromatids to opposite poles, and nuclear division.
• Importance: Ensures genetic consistency by producing two genetically identical daughter cells. Vital for growth of multicellular organisms, replacement/repair of damaged tissues, and asexual reproduction.

2. The Cardiac Cycle:
• Details: Comprises atrial systole, ventricular systole, and diastole. Coordinated by pressure changes and the opening/closing of atrioventricular and semi-lunar valves.
• Importance: Establishes and maintains continuous blood flow and blood pressure. Delivers vital substances (oxygen, glucose, amino acids) to respiring tissues and removes toxic waste products (carbon dioxide, lactate) to maintain homeostasis.

3. The Krebs Cycle (Aerobic Respiration):
• Details: Occurs in the mitochondrial matrix. Acetylcoenzyme A combines with a 4C compound to form a 6C compound. Through decarboxylation and dehydrogenation, the 6C compound is converted back to the 4C compound, yielding reduced NAD, reduced FAD, carbon dioxide, and ATP via substrate-level phosphorylation.
• Importance: Acts as a vital link in cellular respiration. Regenerates electron carriers (NADH, FADH2) to power the electron transport chain, which produces the majority of ATP needed to drive active transport, muscle contraction, and anabolic synthesis.

4. The Calvin Cycle (Light-Independent Reaction of Photosynthesis):
• Details: Occurs in the stroma of chloroplasts. Carbon dioxide is fixed to ribulose bisphosphate (RuBP) by the enzyme rubisco to form glycerate 3-phosphate (GP). GP is then reduced to triose phosphate (TP) using ATP and reduced NADP from the light-dependent reaction. TP is used to regenerate RuBP and synthesize organic macromolecules.
• Importance: Converts inorganic carbon dioxide into organic matter (hexose sugars, starch, cellulose, lipids, and amino acids). This chemical energy forms the basis of biomass transfer through almost all terrestrial and aquatic food webs.

5. Nutrient Cycles (Nitrogen and Phosphorus):
• Details: Nitrogen cycle involves nitrogen fixation (by mutualistic or free-living bacteria), nitrification (ammonium ions converted to nitrites then nitrates by nitrifying bacteria), denitrification, and ammonification (decomposition of saprobionts). Phosphorus cycle involves weathering, plant uptake, assimilation, and saprobiontic decay.
• Importance: Essential elements like nitrogen and phosphorus are present in finite amounts. Recycling these nutrients ensures they are perpetually available to producers. Nitrogen is essential for amino acids, proteins, and nucleic acids; phosphorus is essential for ATP, phospholipids, and nucleic acids.

Marking scheme

AQA Essay Marking Criteria (25 Marks Total)

1. Scientific Content (Max 16 marks):
• 13-16 marks (Outstanding/Excellent): Covers a broad range of biological cycles (at least 4 distinct cycles). The biological principles are explained with a high degree of technical accuracy, including precise molecular, cellular, or physiological terminology.
• 9-12 marks (Good): Covers several cycles (3-4) with good detail, or many cycles with average detail. Minor inaccuracies may be present, but the core biological mechanisms are sound.
• 5-8 marks (Poor/Below Average): Only covers 1 or 2 cycles in detail, or many cycles but with very superficial or highly descriptive information lacking explanatory depth.
• 1-4 marks (Very Poor): Shows little understanding. Only mentions a single cycle briefly with major errors.

2. Breadth of Knowledge (Max 3 marks):
• 3 marks: The essay addresses at least four distinct areas from different parts of the specification (e.g., cell division, physiology, biochemistry, ecology).
• 2 marks: The essay addresses three distinct areas.
• 1 mark: The essay addresses only two distinct areas.

3. Relevance (Max 3 marks):
• 3 marks: All material is directly relevant to 'cycles' and their roles/importance. No significant digressions or unnecessary detailed information on non-cyclic processes.
• 2 marks: Most material is relevant, with small sections of irrelevant background information.
• 1 mark: A substantial portion of the essay consists of irrelevant material.

4. Quality of Written Communication (Max 3 marks):
• 3 marks: Highly coherent structure, logical sequence, fluent use of scientific terminology, and clear paragraphs.
• 2 marks: Generally well-structured, but may lack smooth transitions or contains minor grammatical lapses.
• 1 mark: Poorly structured, disjointed points, or difficult to read.

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