Cambridge IGCSE · thinka-original Practice Paper

2026 Cambridge IGCSE Biology (0610) Practice Paper with Answers

Thinka Jun 2026 (V3) Cambridge IGCSE-Style Mock — Biology (0610)

320 marks300 mins2026
An original Thinka practice paper modelled on the structure and difficulty of the Jun 2026 (V3) Cambridge IGCSE Biology (0610) paper. Not affiliated with or reproduced from Cambridge.

Paper 1 / Paper 2: Multiple Choice Objective Test

Answer all 40 four-option multiple choice questions on the separate answer sheet within 45 minutes.
40 Question · 40 marks
Question 1 · Multiple Choice
1 marks
A student investigated the effect of temperature on the rate of an enzyme-catalysed reaction. Which statement explains why the rate of reaction decreases rapidly at temperatures above the optimum temperature?
  1. A.Substrate molecules lose kinetic energy and collide less frequently with active sites.
  2. B.The active site changes shape so substrate molecules no longer fit into it.
  3. C.The activation energy required for the reaction increases significantly.
  4. D.The enzyme molecules are permanently converted into product molecules.
Show answer & marking scheme

Worked solution

Above the optimum temperature, high thermal energy causes the enzyme molecules to vibrate excessively, breaking bonds holding their three-dimensional structure together. This denatures the enzyme, altering the shape of the active site so that the substrate is no longer complementary and cannot bind to form enzyme-substrate complexes.

Marking scheme

B [1 mark]

A is incorrect as kinetic energy increases with temperature.
C is incorrect as denaturation does not raise the reaction's baseline activation energy in this manner.
D is incorrect as enzymes act as biological catalysts and are not converted into products.
Question 2 · Multiple Choice
1 marks
A piece of dialysis tubing containing a 10% sucrose solution is placed into a beaker of distilled water. Which row correctly describes the net movement of water molecules and the resulting change in the mass of the dialysis tubing?
  1. A.into the dialysis tubing; mass increases
  2. B.into the dialysis tubing; mass decreases
  3. C.out of the dialysis tubing; mass increases
  4. D.out of the dialysis tubing; mass decreases
Show answer & marking scheme

Worked solution

Distilled water has a higher water potential than the 10% sucrose solution inside the dialysis tubing. Therefore, water moves into the dialysis tubing down a water potential gradient by osmosis, causing the mass of the dialysis tubing to increase.

Marking scheme

A [1 mark]

B is incorrect because water entry causes mass to increase, not decrease.
C and D are incorrect because the net movement of water is into the tubing, not out of it.
Question 3 · Multiple Choice
1 marks
Which hormone stimulates ovulation during the human menstrual cycle and where is it secreted?
  1. A.FSH; secreted by the ovaries
  2. B.LH; secreted by the pituitary gland
  3. C.oestrogen; secreted by the pituitary gland
  4. D.progesterone; secreted by the ovaries
Show answer & marking scheme

Worked solution

Luteinising hormone (LH) is produced and secreted by the pituitary gland. A surge in LH mid-cycle triggers ovulation (the release of an egg from an ovary).

Marking scheme

B [1 mark]

A is incorrect because FSH stimulates follicle development, not ovulation, and is secreted by the pituitary gland.
C is incorrect because oestrogen is secreted by the ovaries.
D is incorrect because progesterone maintains the uterine lining and does not stimulate ovulation.
Question 4 · Multiple Choice
1 marks
Which process in the nitrogen cycle converts ammonium ions into nitrate ions in the soil?
  1. A.deamination
  2. B.denitrification
  3. C.nitrification
  4. D.nitrogen fixation
Show answer & marking scheme

Worked solution

Nitrification is the biological oxidation of ammonium ions first into nitrite ions and then into nitrate ions by nitrifying bacteria in the soil.

Marking scheme

C [1 mark]

A is incorrect because deamination occurs in the liver to remove amino groups from excess amino acids.
B is incorrect because denitrification converts nitrate ions into nitrogen gas.
D is incorrect because nitrogen fixation converts nitrogen gas into ammonia or ammonium ions.
Question 5 · Multiple Choice
1 marks
A photomicrograph shows a palisade mesophyll cell. The measured width of the cell in the image is \(48\text{ mm}\). The actual width of the cell is \(80\ \mu\text{m}\). What is the magnification of the image?
  1. A.\(\times 0.6\)
  2. B.\(\times 60\)
  3. C.\(\times 600\)
  4. D.\(\times 6000\)
Show answer & marking scheme

Worked solution

First, convert both values to the same unit (millimetres):
\(\text{Actual size} = 80\ \mu\text{m} = \frac{80}{1000}\text{ mm} = 0.08\text{ mm}\).
Using the formula:
\(\text{Magnification} = \frac{\text{Image size}}{\text{Actual size}} = \frac{48\text{ mm}}{0.08\text{ mm}} = \times 600\).

Marking scheme

C [1 mark]

A is incorrect due to dividing actual size by image size without proper conversion.
B is incorrect by a factor of 10 from arithmetic conversion error.
D is incorrect by a factor of 10 from an incorrect unit conversion.
Question 6 · Multiple Choice
1 marks
Which sequence correctly shows the transfer of genetic information during the synthesis of a specific protein?
  1. A.amino acid sequence \(\rightarrow\) mRNA \(\rightarrow\) DNA base sequence \(\rightarrow\) protein
  2. B.DNA base sequence \(\rightarrow\) mRNA \(\rightarrow\) amino acid sequence \(\rightarrow\) protein
  3. C.mRNA \(\rightarrow\) DNA base sequence \(\rightarrow\) protein \(\rightarrow\) amino acid sequence
  4. D.protein \(\rightarrow\) amino acid sequence \(\rightarrow\) mRNA \(\rightarrow\) DNA base sequence
Show answer & marking scheme

Worked solution

The sequence of bases in a gene (DNA) determines the sequence of bases in mRNA, which carries the code to the ribosome, where it determines the specific sequence of amino acids joined together to make a protein.

Marking scheme

B [1 mark]

A, C, and D are incorrect as they do not follow the correct flow of genetic information: DNA \(\rightarrow\) mRNA \(\rightarrow\) amino acid sequence \(\rightarrow\) protein.
Question 7 · Multiple Choice
1 marks
In a healthy human, which substance is completely reabsorbed from the renal filtrate back into the blood in the nephron?
  1. A.glucose
  2. B.mineral ions
  3. C.urea
  4. D.water
Show answer & marking scheme

Worked solution

In the nephron, all glucose filtered into the Bowman's capsule is reabsorbed back into the blood capillaries from the proximal convoluted tubule via active transport, so that none is normally present in the urine.

Marking scheme

A [1 mark]

B is incorrect because only some mineral ions are reabsorbed.
C is incorrect because urea is mostly excreted as waste.
D is incorrect because most, but not all, water is reabsorbed.
Question 8 · Multiple Choice
1 marks
Which chamber of the mammalian heart has the thickest muscular wall and generates the highest pressure to pump blood into the aorta?
  1. A.left atrium
  2. B.left ventricle
  3. C.right atrium
  4. D.right ventricle
Show answer & marking scheme

Worked solution

The left ventricle has the thickest muscular wall because it must generate enough pressure to pump oxygenated blood through the aorta to the entire systemic circulation throughout the body.

Marking scheme

B [1 mark]

A and C are incorrect because atria have thin walls as they only pump blood into the ventricles.
D is incorrect because the right ventricle pumps deoxygenated blood at lower pressure to the lungs via the pulmonary artery.
Question 9 · Multiple Choice
1 marks
A student investigated the rate of an enzyme-catalysed reaction at temperatures ranging from \(10\text{ }^\circ\text{C}\) to \(70\text{ }^\circ\text{C}\). At \(65\text{ }^\circ\text{C}\), the reaction ceased entirely. Which statement correctly explains why the reaction stopped?
  1. A.The substrate molecules had lost all of their kinetic energy.
  2. B.The enzyme had denatured so the substrate no longer fit into the active site.
  3. C.The activation energy of the reaction increased to a level that prevented successful collisions.
  4. D.The enzyme had been fully consumed during the earlier stages of the reaction.
Show answer & marking scheme

Worked solution

At high temperatures above the optimum (such as \(65\text{ }^\circ\text{C}\)), the enzyme molecule gains excessive kinetic energy, causing the bonds maintaining its tertiary structure to break. This alters the three-dimensional shape of the active site (denaturation). Consequently, the substrate can no longer bind to form enzyme-substrate complexes.

Marking scheme

[1] B — The enzyme had denatured so the substrate no longer fit into the active site; [0] for incorrect options.
Question 10 · Multiple Choice
1 marks
Cylinders of potato tissue of equal initial mass were immersed in sucrose solutions of different concentrations (\(0.0\text{ mol/dm}^3\), \(0.2\text{ mol/dm}^3\), \(0.4\text{ mol/dm}^3\), and \(0.8\text{ mol/dm}^3\)) for 60 minutes. The cylinder in the \(0.8\text{ mol/dm}^3\) sucrose solution showed a significant decrease in mass. What caused this loss of mass?
  1. A.Sucrose molecules diffused out of the plant cells down a concentration gradient.
  2. B.Sucrose molecules were actively transported into the plant cells.
  3. C.Water moved into the plant cells from a region of lower water potential to higher water potential.
  4. D.Water moved out of the plant cells by osmosis down a water potential gradient.
Show answer & marking scheme

Worked solution

A \(0.8\text{ mol/dm}^3\) sucrose solution has a lower water potential than the cytoplasm/cell sap of potato cells. Water moves out of the potato cells down a water potential gradient by osmosis, resulting in a loss of cellular mass.

Marking scheme

[1] D — Water moved out of the plant cells by osmosis down a water potential gradient; [0] for incorrect options.
Question 11 · Multiple Choice
1 marks
Which row correctly identifies the hormone responsible for stimulating the repair and thickening of the uterine lining following menstruation, along with its primary site of secretion?
  1. A.hormone: LH; site of secretion: ovary
  2. B.hormone: oestrogen; site of secretion: ovary
  3. C.hormone: progesterone; site of secretion: pituitary gland
  4. D.hormone: FSH; site of secretion: pituitary gland
Show answer & marking scheme

Worked solution

Oestrogen is secreted by the developing ovarian follicles in the ovary. One of its key roles during the follicular phase of the menstrual cycle is to stimulate the repair and thickening of the endometrium (uterine lining).

Marking scheme

[1] B — hormone: oestrogen; site of secretion: ovary; [0] for incorrect options.
Question 12 · Multiple Choice
1 marks
A micrograph shows a mammalian red blood cell with a measured image diameter of \(14\text{ mm}\). The magnification of the image is \(\times 2000\). What is the actual diameter of the red blood cell in micrometres (\(\mu\text{m}\))?
  1. A.\(0.007\text{ }\mu\text{m}\)
  2. B.\(0.07\text{ }\mu\text{m}\)
  3. C.\(7.0\text{ }\mu\text{m}\)
  4. D.\(70\text{ }\mu\text{m}\)
Show answer & marking scheme

Worked solution

Convert the image size to micrometres: \(14\text{ mm} = 14\times 1000\text{ }\mu\text{m} = 14\,000\text{ }\mu\text{m}\). Use the formula \(\text{Actual size} = \frac{\text{Image size}}{\text{Magnification}} = \frac{14\,000\text{ }\mu\text{m}}{2000} = 7.0\text{ }\mu\text{m}\).

Marking scheme

[1] C — 7.0 µm; [0] for incorrect numerical conversions or formula applications.
Question 13 · Multiple Choice
1 marks
In a healthy human kidney, which substance passes from the blood into the renal capsule during ultrafiltration and is subsequently entirely reabsorbed into the capillaries from the nephron tubule?
  1. A.glucose
  2. B.protein
  3. C.urea
  4. D.water
Show answer & marking scheme

Worked solution

Glucose is small enough to be filtered out of the glomerulus into the Bowman's (renal) capsule during ultrafiltration. In a healthy individual, 100% of the filtered glucose is reabsorbed back into the blood by active transport along the proximal convoluted tubule.

Marking scheme

[1] A — glucose; [0] for incorrect substances.
Question 14 · Multiple Choice
1 marks
Which statement correctly describes the action of the human heart during systemic circulation?
  1. A.Deoxygenated blood enters the left atrium via the pulmonary veins.
  2. B.Oxygenated blood is pumped from the left ventricle into the aorta at high pressure.
  3. C.Deoxygenated blood is pumped from the left ventricle into the pulmonary artery.
  4. D.Oxygenated blood enters the right atrium via the vena cava.
Show answer & marking scheme

Worked solution

During ventricular systole, the muscular left ventricle contracts forcefully to pump oxygenated blood through the aortic semilunar valve into the aorta at high pressure, supplying the body tissues.

Marking scheme

[1] B — Oxygenated blood is pumped from the left ventricle into the aorta at high pressure; [0] for incorrect statements.
Question 15 · Multiple Choice
1 marks
A double-stranded DNA molecule contains 120 base pairs. On one strand, there are 28 adenine (A) bases and 32 thymine (T) bases. What is the total number of cytosine (C) bases present in this entire double-stranded DNA molecule?
  1. A.30
  2. B.60
  3. C.120
  4. D.180
Show answer & marking scheme

Worked solution

120 base pairs means there are \(120 \times 2 = 240\) total bases. Since strand 1 has 28 A and 32 T, strand 2 must have 28 T and 32 A to pair with them. Total A = \(28 + 32 = 60\), and total T = \(32 + 28 = 60\). Together, A + T = 120 bases. The remaining bases are C and G: \(240 - 120 = 120\) bases. Because C pairs with G in equal quantities, total C = \(120 / 2 = 60\).

Marking scheme

[1] B — 60; [0] for incorrect base calculation.
Question 16 · Multiple Choice
1 marks
Why do natural food chains rarely contain more than five trophic levels?
  1. A.Top carnivores produce too many offspring to be supported by primary consumers.
  2. B.Decomposers intercept all available chemical energy before it reaches secondary consumers.
  3. C.Most of the energy is lost as heat from respiration and in excretory products at each trophic level.
  4. D.Primary producers cannot absorb sufficient solar radiation to fuel more than four consumers.
Show answer & marking scheme

Worked solution

Energy transfer between trophic levels is inefficient; approximately 90% of energy is lost at each level due to metabolic heat from respiration, movement, excretion, and uneaten body parts. By the fifth trophic level, insufficient energy remains to sustain another viable population.

Marking scheme

[1] C — Most of the energy is lost as heat from respiration and in excretory products at each trophic level; [0] for incorrect explanations.
Question 17 · multiple_choice
1 marks
An investigation was carried out into the effect of temperature on the rate of an enzyme-catalysed reaction. Which statement explains why the rate of reaction decreases rapidly above the optimum temperature?
  1. A.The substrate molecules lose kinetic energy and collide less frequently with the active site.
  2. B.The active site changes shape so that the substrate molecule can no longer bind.
  3. C.The activation energy required for the reaction increases significantly.
  4. D.The bonds holding the substrate molecule together break before collisions occur.
Show answer & marking scheme

Worked solution

At temperatures above the optimum, high thermal energy causes the enzyme molecule to vibrate excessively, breaking the internal bonds that maintain its three-dimensional structure. This causes the active site to change shape (denature), meaning the substrate is no longer complementary and cannot bind to form enzyme-substrate complexes.

Marking scheme

B ; [1 mark]
Question 18 · multiple_choice
1 marks
Fresh plant tissue with a water potential equivalent to a \(0.3\text{ mol/dm}^3\) sucrose solution is placed into a concentrated \(0.8\text{ mol/dm}^3\) sucrose solution. Which row correctly describes the net movement of water and the resulting condition of the plant cells?
  1. A.Net movement of water: into the cells; Condition of plant cells: turgid
  2. B.Net movement of water: into the cells; Condition of plant cells: plasmolysed
  3. C.Net movement of water: out of the cells; Condition of plant cells: turgid
  4. D.Net movement of water: out of the cells; Condition of plant cells: plasmolysed
Show answer & marking scheme

Worked solution

A \(0.8\text{ mol/dm}^3\) solution is more concentrated and therefore has a lower water potential than the plant tissue (\(0.3\text{ mol/dm}^3\)). Water moves down the water potential gradient out of the cells by osmosis, causing the cytoplasm and cell membrane to pull away from the cell wall, leaving the cells plasmolysed.

Marking scheme

D ; [1 mark]
Question 19 · multiple_choice
1 marks
Which row correctly matches a hormone involved in the human menstrual cycle with its site of production and its primary role?
  1. A.Hormone: FSH; Site of production: ovaries; Role: triggers the release of an egg cell
  2. B.Hormone: LH; Site of production: pituitary gland; Role: triggers ovulation
  3. C.Hormone: oestrogen; Site of production: pituitary gland; Role: stimulates repair and thickening of the uterus lining
  4. D.Hormone: progesterone; Site of production: ovaries; Role: causes development of follicles in the ovary
Show answer & marking scheme

Worked solution

Luteinising hormone (LH) is secreted by the pituitary gland and causes the release of a mature egg from an ovary (ovulation). FSH is also secreted by the pituitary gland (not ovaries). Oestrogen is produced by the ovaries (not pituitary). Progesterone maintains the uterine lining and is secreted by the ovaries/corpus luteum.

Marking scheme

B ; [1 mark]
Question 20 · multiple_choice
1 marks
Why do food chains in natural ecosystems rarely consist of more than five trophic levels?
  1. A.A large proportion of energy is lost as heat and waste at each successive level.
  2. B.Top carnivores have lower metabolic rates and reproduce much less frequently.
  3. C.Decomposers return all consumer energy directly to the primary producers.
  4. D.Total biomass increases substantially from one trophic level to the next.
Show answer & marking scheme

Worked solution

Energy is transferred inefficiently between trophic levels (typically only about 10% is passed on), with the rest being lost as heat through respiration, in excretion, or as undigested material. Consequently, after four or five trophic levels, there is insufficient energy remaining to sustain a viable population of top predators.

Marking scheme

A ; [1 mark]
Question 21 · multiple_choice
1 marks
A photomicrograph shows a specialised human cell with an image diameter of \(28\text{ mm}\). The magnification of the image is \(\times 3500\). What is the actual diameter of the cell in micrometres (\(\mu\text{m}\))?
  1. A.\(0.008\text{ }\mu\text{m}\)
  2. B.\(0.80\text{ }\mu\text{m}\)
  3. C.\(8.0\text{ }\mu\text{m}\)
  4. D.\(98\text{ }\mu\text{m}\)
Show answer & marking scheme

Worked solution

\(\text{Actual size} = \frac{\text{Image size}}{\text{Magnification}} = \frac{28\text{ mm}}{3500} = 0.008\text{ mm}\). To convert millimetres to micrometres: \(0.008\text{ mm} \times 1000 = 8.0\text{ }\mu\text{m}\).

Marking scheme

C ; [1 mark]
Question 22 · multiple_choice
1 marks
In a species of plant, petal colour is controlled by a single gene. The allele for purple petals (\(P\)) is dominant over the allele for white petals (\(p\)). A purple-flowered plant was crossed with a white-flowered plant, producing 49 purple-flowered offspring and 51 white-flowered offspring. What are the genotypes of the parent plants?
  1. A.\(PP\) and \(pp\)
  2. B.\(Pp\) and \(pp\)
  3. C.\(Pp\) and \(Pp\)
  4. D.\(PP\) and \(Pp\)
Show answer & marking scheme

Worked solution

The phenotypic ratio of the offspring is approximately \(1:1\) (49 purple to 51 white). A white-flowered plant must have the homozygous recessive genotype (\(pp\)). For white offspring (\(pp\)) to be produced, the purple parent must contribute a recessive allele \(p\), so its genotype must be heterozygous (\(Pp\)). A cross of \(Pp \times pp\) yields \(1\,Pp : 1\,pp\).

Marking scheme

B ; [1 mark]
Question 23 · multiple_choice
1 marks
Which row correctly identifies substances filtered from the blood at the renal capsule (glomerulus) and a substance that is completely reabsorbed into the blood along the nephron in a healthy human?
  1. A.Filtered into renal capsule: glucose, mineral ions, and urea; Completely reabsorbed: glucose
  2. B.Filtered into renal capsule: proteins, glucose, and urea; Completely reabsorbed: urea
  3. C.Filtered into renal capsule: red blood cells, salts, and water; Completely reabsorbed: salts
  4. D.Filtered into renal capsule: proteins, mineral ions, and glucose; Completely reabsorbed: water
Show answer & marking scheme

Worked solution

During ultrafiltration at the glomerulus, small molecules (glucose, mineral ions, water, and urea) pass into the renal capsule, while large molecules such as proteins and blood cells remain in the capillaries. In a healthy human, all of the filtered glucose is reabsorbed back into the blood.

Marking scheme

A ; [1 mark]
Question 24 · multiple_choice
1 marks
Which heart chamber has the thickest muscular wall, and into which vessel does it directly pump oxygenated blood?
  1. A.Left ventricle, which pumps blood directly into the aorta
  2. B.Left ventricle, which pumps blood directly into the pulmonary artery
  3. C.Right ventricle, which pumps blood directly into the aorta
  4. D.Right ventricle, which pumps blood directly into the pulmonary artery
Show answer & marking scheme

Worked solution

The left ventricle has the thickest muscular wall because it must generate high pressure to pump oxygenated blood through the systemic circulation via the aorta to the rest of the body.

Marking scheme

A ; [1 mark]
Question 25 · Multiple Choice
1 marks
An experiment investigated the rate of an enzyme-catalysed reaction at different temperatures from \(10\text{ }^\circ\text{C}\) to \(60\text{ }^\circ\text{C}\).

Which statement explains why the rate of reaction decreases rapidly at temperatures above the optimum temperature?
  1. A.The kinetic energy of substrate molecules becomes too low for successful collisions.
  2. B.The active site changes shape so substrate molecules no longer fit.
  3. C.Substrate molecules are decomposed by excess heat before collisions occur.
  4. D.All enzyme molecules are permanently converted into product molecules.
Show answer & marking scheme

Worked solution

At temperatures above the optimum, high thermal energy causes the enzyme molecules to vibrate excessively, breaking bonds that maintain their tertiary structure. This alters the shape of the active site (denaturation), meaning the substrate molecule can no longer fit or form enzyme-substrate complexes.

Marking scheme

B [1 mark]; A is incorrect because kinetic energy increases with temperature; C and D describe incorrect biological mechanisms.
Question 26 · Multiple Choice
1 marks
Cylinders of potato tissue of identical length and mass were placed in test-tubes containing sucrose solutions of different concentrations: \(0.0\text{ mol}/\text{dm}^3\), \(0.4\text{ mol}/\text{dm}^3\), \(0.8\text{ mol}/\text{dm}^3\), and \(1.2\text{ mol}/\text{dm}^3\).

After two hours, in which concentration of sucrose solution will the potato cells have the highest turgor pressure?
  1. A.\(0.0\text{ mol}/\text{dm}^3\)
  2. B.\(0.4\text{ mol}/\text{dm}^3\)
  3. C.\(0.8\text{ mol}/\text{dm}^3\)
  4. D.\(1.2\text{ mol}/\text{dm}^3\)
Show answer & marking scheme

Worked solution

A solution of \(0.0\text{ mol}/\text{dm}^3\) (pure water) has the highest water potential compared to the potato cells. Water enters the cells down a water potential gradient by osmosis, causing the vacuoles to swell and push firmly against the cell walls, resulting in the highest turgor pressure.

Marking scheme

A [1 mark]; B, C, and D are higher sucrose concentrations where less water enters or water is lost, resulting in lower turgor pressure or flaccidity/plasmolysis.
Question 27 · Multiple Choice
1 marks
Which row correctly identifies the site of secretion of luteinising hormone (LH) and its main function during the human menstrual cycle?
  1. A.site of secretion: ovary; main function: stimulates repair of the uterus lining
  2. B.site of secretion: pituitary gland; main function: stimulates ovulation
  3. C.site of secretion: ovary; main function: maintains the thickness of the uterus lining
  4. D.site of secretion: pituitary gland; main function: stimulates the development of follicles
Show answer & marking scheme

Worked solution

Luteinising hormone (LH) is produced and secreted by the pituitary gland. A surge in LH concentration around day 14 of the menstrual cycle triggers ovulation (the release of a mature egg from an ovarian follicle).

Marking scheme

B [1 mark]; A, C, and D state incorrect secretion organs or incorrect primary hormonal roles.
Question 28 · Multiple Choice
1 marks
A woodland food chain is shown:

oak tree \(\rightarrow\) caterpillars \(\rightarrow\) blue tits \(\rightarrow\) sparrowhawk

The oak tree absorbs \(4\,000\,000\text{ kJ}\) of light energy, but only \(1\%\) is incorporated into biomass during photosynthesis.

Assuming a \(10\%\) efficiency of energy transfer between all subsequent trophic levels, how much energy is transferred to the sparrowhawk?
  1. A.\(40\text{ kJ}\)
  2. B.\(400\text{ kJ}\)
  3. C.\(4\,000\text{ kJ}\)
  4. D.\(40\,000\text{ kJ}\)
Show answer & marking scheme

Worked solution

Step 1: Energy fixed by oak tree (producer) = \(1\%\text{ of } 4\,000\,000\text{ kJ} = 40\,000\text{ kJ}\).
Step 2: Energy to caterpillars (primary consumer) = \(10\%\text{ of } 40\,000\text{ kJ} = 4\,000\text{ kJ}\).
Step 3: Energy to blue tits (secondary consumer) = \(10\%\text{ of } 4\,000\text{ kJ} = 400\text{ kJ}\).
Step 4: Energy to sparrowhawk (tertiary consumer) = \(10\%\text{ of } 400\text{ kJ} = 40\text{ kJ}\).

Marking scheme

A [1 mark]; B is the energy reaching blue tits; C is the energy reaching caterpillars; D is the energy fixed by the oak tree.
Question 29 · Multiple Choice
1 marks
A photomicrograph shows a mammalian red blood cell with an observed image diameter of \(35\text{ mm}\).

The magnification of the image is \(\times 5\,000\).

What is the actual diameter of this red blood cell in micrometres (\(\mu\text{m}\))?
  1. A.\(0.007\text{ }\mu\text{m}\)
  2. B.\(0.70\text{ }\mu\text{m}\)
  3. C.\(7.0\text{ }\mu\text{m}\)
  4. D.\(70\text{ }\mu\text{m}\)
Show answer & marking scheme

Worked solution

Actual size \(= \text{Image size} / \text{Magnification}\).
Image size \(= 35\text{ mm} = 35\,000\text{ }\mu\text{m}\).
Actual diameter \(= 35\,000\text{ }\mu\text{m} / 5\,000 = 7.0\text{ }\mu\text{m}\).

Marking scheme

C [1 mark]; A, B, and D result from incorrect unit conversion between millimetres and micrometres.
Question 30 · Multiple Choice
1 marks
A mother with blood group A and a father with blood group B have a child who has blood group O.

What is the probability that their next child will have blood group AB?
  1. A.\(0.00\)
  2. B.\(0.25\)
  3. C.\(0.50\)
  4. D.\(0.75\)
Show answer & marking scheme

Worked solution

Because blood group O has the genotype \(I^O I^O\), the child must receive one \(I^O\) allele from each parent. Therefore, the mother's genotype is \(I^A I^O\) and the father's genotype is \(I^B I^O\).
A cross between \(I^A I^O\) and \(I^B I^O\) yields offspring genotypes \(I^A I^B\) (group AB), \(I^A I^O\) (group A), \(I^B I^O\) (group B), and \(I^O I^O\) (group O) in a \(1:1:1:1\) ratio. Thus, the probability of blood group AB is \(1/4 = 0.25\).

Marking scheme

B [1 mark]; A, C, and D are incorrect probabilities resulting from incorrect parental genotype deduction.
Question 31 · Multiple Choice
1 marks
Samples of fluid were extracted from three distinct regions of a healthy human kidney nephron:

- Region 1: blood in the afferent arteriole entering the glomerulus
- Region 2: filtrate inside the Bowman's capsule
- Region 3: urine exiting the collecting duct

Which row correctly shows the presence (+) or absence (-) of glucose and protein molecules in these samples?
  1. A.Region 1: glucose +, protein +; Region 2: glucose +, protein -; Region 3: glucose -, protein -
  2. B.Region 1: glucose +, protein +; Region 2: glucose -, protein -; Region 3: glucose -, protein -
  3. C.Region 1: glucose +, protein -; Region 2: glucose +, protein +; Region 3: glucose +, protein -
  4. D.Region 1: glucose -, protein +; Region 2: glucose +, protein -; Region 3: glucose -, protein +
Show answer & marking scheme

Worked solution

Region 1 (blood) contains both glucose (+) and large protein molecules (+). Region 2 (filtrate) contains glucose (+) because small molecules pass through the basement membrane during ultrafiltration, but proteins are too large to pass (-). Region 3 (urine of a healthy person) contains neither glucose (-) nor protein (-) because all glucose is selectively reabsorbed in the proximal convoluted tubule.

Marking scheme

A [1 mark]; B, C, and D misidentify the selectivity of ultrafiltration or complete glucose reabsorption.
Question 32 · Multiple Choice
1 marks
Which set of events occurs during ventricular systole in the human cardiac cycle?
  1. A.Atrioventricular valves close, semilunar valves open, and ventricular pressure rises.
  2. B.Atrioventricular valves open, semilunar valves close, and ventricular pressure falls.
  3. C.Atrioventricular valves close, semilunar valves close, and ventricular pressure falls.
  4. D.Atrioventricular valves open, semilunar valves open, and ventricular pressure rises.
Show answer & marking scheme

Worked solution

During ventricular systole, the ventricular myocardium contracts, causing ventricular pressure to increase. This elevated pressure pushes the atrioventricular (bicuspid and tricuspid) valves shut to prevent backflow into the atria, and forces open the semilunar valves so blood is pumped into the aorta and pulmonary artery.

Marking scheme

A [1 mark]; B describes ventricular diastole; C and D describe incorrect combinations of valve states during contraction.
Question 33 · Multiple Choice
1 marks
The effect of temperature on the rate of an amylase-catalysed reaction was investigated. The rate increased between \(10\,^{\circ}\text{C}\) and \(40\,^{\circ}\text{C}\), but decreased sharply above \(45\,^{\circ}\text{C}\).

Which statement explains why the rate of reaction decreases above the optimum temperature?
  1. A.Enzyme molecules lose kinetic energy and collide less frequently with substrate molecules.
  2. B.The active site denatures so that the substrate molecule no longer fits.
  3. C.Substrate molecules are converted into product molecules at a faster rate.
  4. D.The activation energy required for the catalysed reaction is significantly reduced.
Show answer & marking scheme

Worked solution

Above the optimum temperature, excessive thermal energy disrupts the bonds maintaining the three-dimensional structure of the enzyme's active site. As the active site denatures, its shape is no longer complementary to the substrate, preventing the formation of enzyme-substrate complexes.

Marking scheme

B ; [1 mark]
[A is incorrect: kinetic energy increases with temperature, not decreases]
[C is incorrect: rate of product formation decreases due to denaturation]
[D is incorrect: enzymes lower activation energy, but temperature does not reduce it further]
Question 34 · Multiple Choice
1 marks
Four potato cylinders of equal mass (\(5.0\text{ g}\)) were placed into four different sucrose solutions, \(\text{P}\), \(\text{Q}\), \(\text{R}\), and \(\text{S}\). After two hours, their masses were measured again:

- Solution \(\text{P}\): \(5.8\text{ g}\)
- Solution \(\text{Q}\): \(5.0\text{ g}\)
- Solution \(\text{R}\): \(4.2\text{ g}\)
- Solution \(\text{S}\): \(3.6\text{ g}\)

In which solution was the initial water potential higher than that of the potato cells?
  1. A.Solution \(\text{P}\) only
  2. B.Solution \(\text{Q}\) only
  3. C.Solutions \(\text{R}\) and \(\text{S}\) only
  4. D.Solutions \(\text{P}\), \(\text{Q}\) and \(\text{R}\)
Show answer & marking scheme

Worked solution

Water moves by osmosis from a region of higher water potential to a region of lower water potential down a water potential gradient. When placed in solution P, the potato cylinder gained mass (from \(5.0\text{ g}\) to \(5.8\text{ g}\)), showing a net movement of water into the cells. Therefore, solution P had a higher water potential than the potato cells.

Marking scheme

A ; [1 mark]
[B is incorrect: mass remained unchanged, meaning water potentials were equal]
[C is incorrect: mass decreased in R and S, meaning external water potential was lower]
[D is incorrect: combines gained, unchanged, and lost mass solutions]
Question 35 · Multiple Choice
1 marks
Which hormone is secreted by the developing follicle in the ovary to stimulate the repair and thickening of the endometrium during the human menstrual cycle?
  1. A.Follicle-stimulating hormone (FSH)
  2. B.Luteinising hormone (LH)
  3. C.Oestrogen
  4. D.Progesterone
Show answer & marking scheme

Worked solution

Oestrogen is secreted by the developing follicle in the ovaries. It promotes the repair and thickening of the uterine lining (endometrium) following menstruation and stimulates the surge of LH required for ovulation.

Marking scheme

C ; [1 mark]
[A is incorrect: FSH is produced by the pituitary gland]
[B is incorrect: LH is produced by the pituitary gland]
[D is incorrect: progesterone is secreted mainly by the corpus luteum after ovulation]
Question 36 · Multiple Choice
1 marks
The table compares the concentrations of three substances in blood plasma entering the glomerulus, filtrate in the renal capsule, and urine in a healthy human:

| Substance | Concentration in plasma / \(\text{g dm}^{-3}\) | Concentration in renal filtrate / \(\text{g dm}^{-3}\) | Concentration in urine / \(\text{g dm}^{-3}\) |
|---|---|---|---|
| 1 | 80.0 | 0.0 | 0.0 |
| 2 | 1.0 | 1.0 | 0.0 |
| 3 | 0.3 | 0.3 | 18.0 |

Which row correctly identifies substances 1, 2, and 3?

| | substance 1 | substance 2 | substance 3 |
|---|---|---|---|
| A | glucose | protein | urea |
| B | protein | glucose | urea |
| C | protein | urea | glucose |
| D | urea | glucose | protein |
  1. A.substance 1 = glucose, substance 2 = protein, substance 3 = urea
  2. B.substance 1 = protein, substance 2 = glucose, substance 3 = urea
  3. C.substance 1 = protein, substance 2 = urea, substance 3 = glucose
  4. D.substance 1 = urea, substance 2 = glucose, substance 3 = protein
Show answer & marking scheme

Worked solution

Substance 1 is protein (too large to pass through the glomerulus into the renal capsule, so \(0.0\text{ g dm}^{-3}\) in filtrate). Substance 2 is glucose (filters freely but is entirely reabsorbed along the nephron, so \(0.0\text{ g dm}^{-3}\) in urine). Substance 3 is urea (filters freely and is concentrated in the urine as water is reabsorbed).

Marking scheme

B ; [1 mark]
[A is incorrect: misidentifies protein and glucose]
[C is incorrect: urea is not completely reabsorbed and glucose does not concentrate in urine]
[D is incorrect: inverted substance roles]
Question 37 · Multiple Choice
1 marks
Which sequence correctly traces the pathway of an oxygenated red blood cell leaving the lungs and travelling to the kidneys?
  1. A.pulmonary vein \(\rightarrow\) left atrium \(\rightarrow\) left ventricle \(\rightarrow\) aorta \(\rightarrow\) renal artery
  2. B.pulmonary artery \(\rightarrow\) right atrium \(\rightarrow\) right ventricle \(\rightarrow\) aorta \(\rightarrow\) renal vein
  3. C.pulmonary vein \(\rightarrow\) right atrium \(\rightarrow\) right ventricle \(\rightarrow\) pulmonary artery \(\rightarrow\) renal artery
  4. D.vena cava \(\rightarrow\) left atrium \(\rightarrow\) left ventricle \(\rightarrow\) aorta \(\rightarrow\) renal artery
Show answer & marking scheme

Worked solution

Oxygenated blood returns from the lungs via the pulmonary vein into the left atrium, moves through the bicuspid valve into the left ventricle, is pumped out via the aorta, and travels through systemic branches including the renal artery to reach the kidneys.

Marking scheme

A ; [1 mark]
[B is incorrect: pulmonary artery carries deoxygenated blood to the lungs]
[C is incorrect: blood enters left atrium from pulmonary vein, not right atrium]
[D is incorrect: vena cava carries deoxygenated blood from the body, not lungs]
Question 38 · Multiple Choice
1 marks
A strand of mRNA contains the triplet base sequence:

\(\text{AUG - CCA - UUC}\)

What was the base sequence on the template strand of the DNA molecule from which this mRNA sequence was transcribed?
  1. A.\(\text{TAC - GGT - AAG}\)
  2. B.\(\text{TAC - GGU - AAG}\)
  3. C.\(\text{UAC - GGU - AAG}\)
  4. D.\(\text{ATC - CCA - TTG}\)
Show answer & marking scheme

Worked solution

In transcription, mRNA bases pair complementarily with DNA template bases: A in mRNA pairs with T in DNA, U in mRNA pairs with A in DNA, G pairs with C, and C pairs with G.
- A \(\rightarrow\) T, U \(\rightarrow\) A, G \(\rightarrow\) C (AUG \(\rightarrow\) TAC)
- C \(\rightarrow\) G, C \(\rightarrow\) G, A \(\rightarrow\) T (CCA \(\rightarrow\) GGT)
- U \(\rightarrow\) A, U \(\rightarrow\) A, C \(\rightarrow\) G (UUC \(\rightarrow\) AAG)
Thus, the DNA template sequence is TAC - GGT - AAG.

Marking scheme

A ; [1 mark]
[B is incorrect: contains uracil (U), which is not found in DNA]
[C is incorrect: RNA sequence, not DNA]
[D is incorrect: incorrect base complementarity]
Question 39 · Multiple Choice
1 marks
A culture of bacteria is grown in a closed flask with a fixed volume of nutrient broth. After a period of exponential growth, the population enters the stationary phase.

Which statement correctly describes the population during the stationary phase?
  1. A.The rate of reproduction is greater than the rate of cell death.
  2. B.The rate of reproduction equals the rate of cell death.
  3. C.The rate of reproduction stops completely and all cells die immediately.
  4. D.The rate of reproduction increases rapidly because waste products have been eliminated.
Show answer & marking scheme

Worked solution

During the stationary phase of a sigmoid population growth curve, the rate of cell division (reproduction) equals the rate of cell death because factors such as food supply, accumulation of waste products, or space become limiting.

Marking scheme

B ; [1 mark]
[A is incorrect: describes the exponential / log phase]
[C is incorrect: describes the death phase]
[D is incorrect: waste products accumulate in a closed flask, they do not decrease to zero]
Question 40 · Multiple Choice
1 marks
A photomicrograph shows a palisade mesophyll cell. The length of the cell in the image is \(48\text{ mm}\). The actual length of the cell is \(60\,\mu\text{m}\).

What is the magnification of the photomicrograph?
  1. A.\(\times 1.25\)
  2. B.\(\times 80\)
  3. C.\(\times 800\)
  4. D.\(\times 8000\)
Show answer & marking scheme

Worked solution

First convert units so that both values are in the same unit:
Image length \(= 48\text{ mm} = 48\,000\,\mu\text{m}\).
Actual length \(= 60\,\mu\text{m}\).

\(\text{Magnification} = \frac{\text{image size}}{\text{actual size}} = \frac{48\,000}{60} = \times 800\).

Marking scheme

C ; [1 mark]
[A is incorrect: dividing 60 by 48 without unit conversion]
[B is incorrect: arithmetic error by factor of 10]
[D is incorrect: arithmetic error by factor of 10]

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Practise This Topic

Paper 3 / Paper 4: Structured Theory Examination

Answer all questions in the spaces provided on the question paper in 1 hour 15 minutes.
15 Question · 46 marks
Question 1 · Short Answer & Definition
2 marks
Define the term enzyme.
Show answer & marking scheme

Worked solution

An enzyme is defined as a biological catalyst that speeds up the rate of a chemical reaction and is made of protein.

Marking scheme

a protein; [1]
(biological) catalyst / substance that speeds up a chemical reaction (without being changed/used up); [1]
[Total: 2]
Question 2 · Short Answer & Definition
2 marks
Define the term osmosis.
Show answer & marking scheme

Worked solution

Osmosis is the net movement of water molecules from an area of higher water potential (dilute solution) to an area of lower water potential (concentrated solution) across a partially permeable membrane.

Marking scheme

net movement / diffusion of water molecules from a region of higher water potential to a region of lower water potential / down a water potential gradient; [1]
through a partially permeable membrane; [1]
[Total: 2]
Question 3 · Short Answer & Definition
2 marks
Define the term homeostasis.
Show answer & marking scheme

Worked solution

Homeostasis refers to the maintenance of a constant internal environment within an organism, such as regulating body temperature or blood glucose concentration.

Marking scheme

maintenance of a constant / stable; [1]
internal environment (within the body); [1]
[Total: 2]
Question 4 · Short Answer & Definition
2 marks
Define the term deamination and state the organ in which it occurs.
Show answer & marking scheme

Worked solution

Deamination is the removal of the nitrogen-containing amino group from excess amino acids to form urea, and this chemical process takes place in the liver.

Marking scheme

definition: removal of the nitrogen-containing part of amino acids (to form urea); [1]
organ: liver; [1]
[Total: 2]
Question 5 · Short Answer & Definition
2 marks
Define the term population.
Show answer & marking scheme

Worked solution

A population is defined as a group of organisms of one species that live together in the same geographical area at the same time.

Marking scheme

group of organisms of one / the same species; [1]
living in the same area at the same time; [1]
[Total: 2]
Question 6 · Short Answer & Definition
2 marks
Define the term gene.
Show answer & marking scheme

Worked solution

A gene is a specific length of DNA that contains the genetic code for synthesising a particular protein.

Marking scheme

length / sequence / section of DNA; [1]
that codes for a protein; [1] (A polypeptide)
[Total: 2]
Question 7 · Short Answer & Definition
2 marks
Define the term pathogen.
Show answer & marking scheme

Worked solution

A pathogen is a disease-causing organism, such as a bacterium, virus, or fungus.

Marking scheme

disease-causing; [1]
organism / microorganism / agent; [1]
[Total: 2]
Question 8 · Short Answer & Definition
2 marks
Define the term translocation.
Show answer & marking scheme

Worked solution

Translocation is the movement of sucrose and amino acids through the phloem from regions of production or release (sources) to regions of storage or utilisation (sinks).

Marking scheme

movement of sucrose / amino acids in phloem; [1]
from a source to a sink / from regions of production to regions of storage or use; [1]
[Total: 2]
Question 9 · Data Analysis & Table Completion
4 marks
A student investigated the effect of \(\text{pH}\) on the activity of amylase.

Equal volumes of amylase and starch suspension were mixed at different \(\text{pH}\) values at a constant temperature of \(35\,^{\circ}\text{C}\). The time taken for starch to be completely broken down was recorded.

The results are shown in Table 1.1.

Table 1.1

| \(\text{pH}\) | Time taken to completely break down starch / s | Rate of reaction / \(\text{s}^{-1}\) |
| :--- | :--- | :--- |
| 4.0 | 250 | 0.004 |
| 5.0 | 100 | 0.010 |
| 6.0 | 40 | 0.025 |
| 7.0 | 25 | [A] |
| 8.0 | 50 | 0.020 |
| 9.0 | 200 | 0.005 |

(a) Calculate the rate of reaction at \(\text{pH}\) 7.0 to complete space [A] in Table 1.1.
$$\text{Rate} = \frac{1}{\text{time taken}}$$
Give your answer to two significant figures. [2]

(b) Explain why the rate of reaction was significantly lower at \(\text{pH}\) 9.0 than at \(\text{pH}\) 7.0. [2]
Show answer & marking scheme

Worked solution

(a)
$$\text{Rate} = \frac{1}{25\text{ s}} = 0.04\text{ s}^{-1}$$
Expressed to two significant figures: \(0.040\text{ s}^{-1}\).

(b)
At \(\text{pH}\) 9.0, the \(\text{pH}\) is far above the optimum for amylase (which is \(\text{pH}\) 7.0). Extreme \(\text{pH}\) disrupts the ionic and hydrogen bonds maintaining the three-dimensional structure of the enzyme, altering the conformation of the active site (denaturation). Consequently, the substrate (starch) is no longer complementary in shape and cannot bind to form enzyme-substrate complexes, reducing the rate of hydrolysis.

Marking scheme

(a)
- Correct working: \(\frac{1}{25}\) [1];
- \(0.040\) (must be given to 2 sig figs) [1];

(b) Any two from:
- Enzyme / amylase is denatured [1];
- Shape of active site changes / active site no longer complementary to substrate [1];
- Substrate / starch cannot bind / fewer or no enzyme-substrate complexes formed [1];

[Total: 4]
Question 10 · Data Analysis & Table Completion
4 marks
A student investigated osmosis using cylinders cut from a fresh potato tuber.

Cylinders of equal dimensions were immersed in sucrose solutions of various concentrations for 60 minutes.

The results are shown in Table 2.1.

Table 2.1

| Concentration of sucrose solution / \(\text{mol/dm}^3\) | Initial mass / g | Final mass / g | Change in mass / g | Percentage change in mass / \(\%\) |
| :--- | :--- | :--- | :--- | :--- |
| 0.0 | 3.20 | 3.68 | +0.48 | +15.0 |
| 0.2 | 3.20 | 3.44 | +0.24 | +7.5 |
| 0.4 | 3.20 | 3.20 | 0.00 | 0.0 |
| 0.6 | 3.20 | 2.96 | -0.24 | [B] |
| 0.8 | 3.20 | 2.72 | -0.48 | -15.0 |

(a) Calculate the percentage change in mass for the potato cylinder placed in the \(0.6\,\text{mol/dm}^3\) sucrose solution to complete space [B] in Table 2.1. Include the sign. [2]

(b) Using the data in Table 2.1, state the estimated concentration of sucrose inside the potato tuber cells and explain what occurs to the cells in the \(0.8\,\text{mol/dm}^3\) solution. [2]
Show answer & marking scheme

Worked solution

(a)
$$\text{Percentage change in mass} = \left(\frac{\text{Change in mass}}{\text{Initial mass}}\right) \times 100$$
$$\text{Percentage change in mass} = \left(\frac{-0.24}{3.20}\right) \times 100 = -7.5\,\%$$

(b)
At \(0.4\,\text{mol/dm}^3\), there is \(0.0\,\%\) change in mass, indicating no net movement of water because the water potential of the solution equals the internal water potential of the cells. In the \(0.8\,\text{mol/dm}^3\) solution, the external sucrose solution has a lower water potential than the cell sap. Water moves out of the cytoplasm and vacuole via osmosis down the water potential gradient, leading to cell plasmolysis and loss of turgor (flaccidity).

Marking scheme

(a)
- Correct working: \(\frac{-0.24}{3.20} \times 100\) or \(\frac{2.96 - 3.20}{3.20} \times 100\) [1];
- \(-7.5\) (accept minus sign or 'decrease of 7.5') [1];

(b)
- Concentration inside cells = \(0.4\text{ mol/dm}^3\) [1];
- Water moves out (of potato cells) by osmosis / down a water potential gradient causing cells to become flaccid / plasmolysed [1];

[Total: 4]
Question 11 · Data Analysis & Table Completion
4 marks
Samples of blood plasma, glomerular filtrate, and urine were collected from a healthy individual and analysed for the concentration of various chemical components.

The findings are summarised in Table 3.1.

Table 3.1

| Substance | Concentration in blood plasma / \(\text{g/dm}^3\) | Concentration in glomerular filtrate / \(\text{g/dm}^3\) | Concentration in urine / \(\text{g/dm}^3\) |
| :--- | :--- | :--- | :--- |
| Protein | 75.0 | 0.0 | 0.0 |
| Glucose | 1.0 | 1.0 | [C] |
| Urea | 0.3 | 0.3 | 18.0 |
| Mineral ions | 8.0 | 8.0 | 16.0 |

(a) State the concentration value that should be recorded in space [C] for glucose in urine and explain why this value is observed in a healthy person. [2]

(b) Calculate how many times more concentrated urea is in the urine compared to the glomerular filtrate. [2]
Show answer & marking scheme

Worked solution

(a)
In a healthy person, glucose is filtered across the glomerulus into Bowman's capsule, but \(100\%\) of it is selectively reabsorbed back into the bloodstream via active transport along the nephron tubule (proximal convoluted tubule). Therefore, the concentration in urine is \(0.0\,\text{g/dm}^3\).

(b)
$$\text{Concentration factor} = \frac{\text{Concentration in urine}}{\text{Concentration in glomerular filtrate}} = \frac{18.0\,\text{g/dm}^3}{0.3\,\text{g/dm}^3} = 60$$

Marking scheme

(a)
- \(0.0\) / \(0\) (g/dm\(^3\)) [1];
- All / 100% of glucose is (selectively) reabsorbed (back into the blood / tubule capillaries) [1];

(b)
- Correct working: \(\frac{18.0}{0.3}\) [1];
- \(60\) / \(60\times\) (times) [1];

[Total: 4]
Question 12 · Data Analysis & Table Completion
4 marks
A culture of yeast cells was grown in a nutrient broth in a closed conical flask over a 24-hour period. The population density was recorded every 4 hours.

The results are presented in Table 4.1.

Table 4.1

| Time / \(\text{hours}\) | Population density / \(\times 10^4\text{ cells/cm}^3\) |
| :--- | :--- |
| 0 | 4.0 |
| 4 | 6.0 |
| 8 | 18.0 |
| 12 | 54.0 |
| 16 | 90.0 |
| 20 | 96.0 |
| 24 | 96.0 |

(a) Calculate the percentage increase in yeast population density between hour 4 and hour 12. Show your working. [2]

(b) Identify the growth phase occurring between 20 and 24 hours and explain why the yeast population stops increasing during this phase. [2]
Show answer & marking scheme

Worked solution

(a)
$$\text{Increase in density} = 54.0 - 6.0 = 48.0 \times 10^4\text{ cells/cm}^3$$
$$\text{Percentage increase} = \left(\frac{48.0}{6.0}\right) \times 100 = 800\,\%$$

(b)
Between 20 and 24 hours, the population reaches the stationary phase (growth rate equals death rate). In a closed system, nutrients (e.g. glucose/sugar) become depleted and toxic waste products (e.g. ethanol/carbon dioxide lowering pH) accumulate, limiting cell division and increasing the mortality rate.

Marking scheme

(a)
- Correct working: \(\frac{54.0 - 6.0}{6.0} \times 100\) or \(\frac{48.0}{6.0} \times 100\) [1];
- \(800\) (\%)[1];

(b)
- Stationary phase [1];
- (Depletion of) nutrients / food supply OR accumulation of toxic waste / ethanol / acid [1];

[Total: 4]
Question 13 · Data Analysis & Table Completion
4 marks
A student investigated their ventilation during different levels of physical activity.

Minute ventilation is the total volume of air inspired per minute and is calculated using the equation:
$$\text{Minute ventilation} = \text{Breathing rate} \times \text{Tidal volume}$$

Table 5.1 displays the collected data.

Table 5.1

| Activity state | Breathing rate / \(\text{breaths/min}\) | Tidal volume / \(\text{dm}^3\) | Minute ventilation / \(\text{dm}^3\text{/min}\) |
| :--- | :--- | :--- | :--- |
| At rest | 14 | 0.50 | 7.0 |
| Running on treadmill | 28 | 2.25 | [D] |
| 10 min post-exercise recovery | 18 | 0.75 | 13.5 |

(a) Calculate the minute ventilation when running on the treadmill to complete space [D] in Table 5.1. [1]

(b) Explain why both breathing rate and tidal volume remain elevated at '10 min post-exercise recovery' compared to 'At rest'. [3]
Show answer & marking scheme

Worked solution

(a)
$$\text{Minute ventilation} = 28\text{ breaths/min} \times 2.25\text{ dm}^3 = 63.0\text{ dm}^3\text{/min}$$

(b)
During strenuous running, the demand for energy outstrips oxygen supply to muscles, causing anaerobic respiration to occur alongside aerobic respiration. This results in the buildup of lactic acid and builds an oxygen debt. During recovery, breathing rate and tidal volume remain elevated to supply additional oxygen to the blood to transport lactic acid to the liver, where it is oxidised/broken down into carbon dioxide and water, and to repay the oxygen debt.

Marking scheme

(a)
- \(63\) / \(63.0\) (dm\(^3\)/min) [1];

(b) Any three from:
- Anaerobic respiration occurred (during exercise) [1];
- Lactic acid was produced / built up (in muscles) [1];
- Oxygen debt is created [1];
- Extra / more oxygen is needed to break down / oxidise lactic acid (in the liver) / repay oxygen debt [1];

[Total: 4]
Question 14 · Data Analysis & Table Completion
4 marks
The blood glucose concentration of a healthy human volunteer was measured over a period of 120 minutes following the consumption of a meal rich in starch.

The results are shown in Table 6.1.

Table 6.1

| Time after meal / \(\text{min}\) | Blood glucose concentration / \(\text{mmol/dm}^3\) |
| :--- | :--- |
| 0 | 4.8 |
| 30 | 8.4 |
| 60 | 6.6 |
| 90 | 5.2 |
| 120 | 4.8 |

(a) Calculate the rate of decrease in blood glucose concentration between 30 minutes and 90 minutes. State the correct units in your answer. [2]

(b) Describe the biological mechanism that brings about the decrease in blood glucose concentration between 30 minutes and 120 minutes. [2]
Show answer & marking scheme

Worked solution

(a)
$$\text{Decrease in concentration} = 8.4 - 5.2 = 3.2\,\text{mmol/dm}^3$$
$$\text{Time elapsed} = 90 - 30 = 60\,\text{min}$$
$$\text{Rate of decrease} = \frac{3.2}{60} \approx 0.0533\,\text{mmol/(dm}^3\cdot\text{min)}$$
(Accept \(0.053\,\text{mmol/dm}^3\text{/min}\) or \(0.0533\,\text{mmol/dm}^3\text{ per min}\)).

(b)
The high blood glucose level is detected by the pancreas, which releases the hormone insulin into the bloodstream. Insulin binds to liver and muscle cells, increasing their uptake of glucose and promoting the enzyme-controlled condensation of glucose into glycogen for storage, thus lowering the glucose concentration back to normal levels.

Marking scheme

(a)
- Correct calculation: \(\frac{8.4 - 5.2}{60} = \frac{3.2}{60}\) [1];
- \(0.053\) (or \(0.0533\)) WITH correct unit: \(\text{mmol}/\text{dm}^3\text{/min}\) (or \(\text{mmol}\cdot\text{dm}^{-3}\cdot\text{min}^{-1}\)) [1];

(b)
- Pancreas secretes / releases insulin [1];
- (Insulin stimulates) liver / muscle cells to take up glucose AND convert it to glycogen [1];

[Total: 4]
Question 15 · theory
6 marks
(a) Define the term gene. [1]

(b) State the role of mRNA in protein synthesis. [2]

(c) Explain how the sequence of bases in a gene determines the specific 3D shape and biological activity of an enzyme. [3]
Show answer & marking scheme

Worked solution

(a) A gene is defined as a length of DNA that codes for a protein.
(b) mRNA is formed in the nucleus as a complementary copy of the gene's DNA code. It passes out of the nucleus and travels to the ribosomes in the cytoplasm, where the code is translated into a polypeptide chain.
(c) The triplet code of bases in the DNA determines the sequence of amino acids assembled during translation. This primary sequence of amino acids causes the chain to fold in a unique, precise three-dimensional structure. In enzymes, this folding creates a catalytic active site whose shape is complementary to a specific substrate molecule, allowing the enzyme to function.

Marking scheme

(a) length / sequence of DNA that codes for a protein; [1]
(b) any two from:
- carries a copy of the genetic code / base sequence; [1]
- from the nucleus / DNA; [1]
- to the ribosome / site of protein synthesis (in the cytoplasm); [1]
(c) any three from:
- base sequence determines the sequence / order of amino acids; [1]
- sequence of amino acids determines the specific folding / 3D shape of the protein; [1]
- active site has a specific / unique shape; [1]
- active site is complementary (in shape) to the substrate; [1]

Paper 5 / Paper 6: Practical Assessment & Alternative to Practical

Answer all practical and planning questions in 1 hour to 1 hour 15 minutes.
5 Question · 46 marks
Question 1 · Investigation Table & Data Evaluation
15 marks
A student investigated the effect of different concentrations of sucrose solution on cylinders cut from a sweet potato (Ipomoea batatas).

Method:
1. A cork borer was used to cut six cylinders of tissue from the same sweet potato.
2. A scalpel and ruler were used to cut each cylinder to an initial length of \(40\text{ mm}\).
3. Each cylinder was gently blotted with a paper towel and its initial mass was measured and recorded using an electronic balance.
4. Six boiling tubes were set up containing \(25\text{ cm}^3\) of sucrose solution of concentrations: \(0.0\), \(0.2\), \(0.4\), \(0.6\), \(0.8\), and \(1.0\text{ mol/dm}^3\).
5. One sweet potato cylinder was placed into each boiling tube and left for \(60\text{ minutes}\).
6. After \(60\text{ minutes}\), each cylinder was removed, blotted gently with a paper towel, and its final mass was measured and recorded.

The student recorded their results in Table 1.

Table 1

| Sucrose concentration / \(\text{mol/dm}^3\) | Initial mass / \(\text{g}\) | Final mass / \(\text{g}\) | Change in mass / \(\text{g}\) | Percentage change in mass / \(\%\) |
| :--- | :--- | :--- | :--- | :--- |
| \(0.0\) | \(3.20\) | \(3.68\) | \(+0.48\) | \(+15.0\) |
| \(0.2\) | \(3.15\) | \(3.40\) | \(+0.25\) | \(+7.9\) |
| \(0.4\) | \(3.25\) | \(3.28\) | \(+0.03\) | [A] |
| \(0.6\) | \(3.10\) | \(2.91\) | \(-0.19\) | \(-6.1\) |
| \(0.8\) | \(3.30\) | \(2.87\) | \(-0.43\) | [B] |
| \(1.0\) | \(3.20\) | \(2.62\) | \(-0.58\) | \(-18.1\) |

(a) (i) Calculate the percentage change in mass for the cylinders placed in the \(0.4\text{ mol/dm}^3\) and \(0.8\text{ mol/dm}^3\) sucrose solutions. Give your answers to one decimal place. [2]

(ii) State why calculating the percentage change in mass allows a more valid comparison than using the raw change in mass. [1]

(b) On the grid, plot a line graph of percentage change in mass against sucrose concentration. Draw a smooth curve or join the points with ruled straight lines. [4]

(c) (i) Use your graph to estimate the sucrose concentration that has an equivalent water potential to the sweet potato tissue. Indicate clearly on your graph how you determined this value. [2]

(ii) Explain the change in mass of the cylinder placed in the \(0.0\text{ mol/dm}^3\) sucrose solution. [2]

(d) (i) State two variables that were kept constant in this investigation. [2]

(ii) Identify one potential source of error in step 3 or step 6 of the method and suggest an improvement to minimize this error. [2]
Show answer & marking scheme

Worked solution

(a) (i)
Percentage change for \(0.4\text{ mol/dm}^3\):
$$\text{Percentage change} = \left(\frac{+0.03}{3.25}\right) \times 100 = +0.923\% \approx +0.9\%$$
Percentage change for \(0.8\text{ mol/dm}^3\):
$$\text{Percentage change} = \left(\frac{-0.43}{3.30}\right) \times 100 = -13.030\% \approx -13.0\%$$

(a) (ii)
The initial masses of the plant cylinders were not identical (varied between \(3.10\text{ g}\) and \(3.30\text{ g}\)); calculating percentage change takes into account differences in starting mass to allow a fair comparison.

(b)
- Axes: \(x\)-axis labelled as *Sucrose concentration / \(\text{mol/dm}^3\)* and \(y\)-axis labelled as *Percentage change in mass / \(\%\)*.
- Scale: Linear scale on both axes spanning more than \(50\%\) of the grid in both directions; \(y\)-axis requires positive and negative values (from \(-20\%\) to \(+20\%\)).
- Plotting: All 6 points plotted accurately to within \(\pm 0.5\) small square with neat small crosses or circled dots.
- Line: Smooth line of best fit or straight points connected cleanly with a ruler without double lines/feathers.

(c) (i)
The concentration where there is no net movement of water (percentage change in mass \(= 0\%\)) is read from the point where the line intersects the \(x\)-axis (expected range: \(0.42\text{ to } 0.44\text{ mol/dm}^3\)). Working shown with intercept markings/dashed lines to the axis.

(c) (ii)
Water entered the cells/tissue by osmosis; moving down a water potential gradient from a higher water potential in the pure water/hypotonic solution (\(0.0\text{ mol/dm}^3\)) to a lower water potential inside the sweet potato cells through a partially permeable cell membrane.

(d) (i)
Any two from:
- Volume of sucrose solution (\(25\text{ cm}^3\))
- Immersion time (\(60\text{ minutes}\))
- Initial length of cylinder (\(40\text{ mm}\))
- Source / variety of sweet potato
- Temperature of room / water bath

(d) (ii)
- Source of error: Blotting was subjective / inconsistent force applied when drying cylinders with paper towels, leading to different amounts of surface liquid being removed / liquid squeezed out.
- Improvement: Standardise blotting by rolling each cylinder a fixed number of times (e.g. twice) with uniform light pressure / using a mechanical press or fixed drying time.

Marking scheme

(a)(i) 1 mark for each correct value:
- \(+0.9\%\) (accept \(0.9\) or \(+0.92\));
- \(-13.0\%\) (accept \(-13\) or \(-13.03\));
[2 marks]

(a)(ii) Cylinders had different initial masses / starting masses were not the same;
[1 mark]

(b) 4 marks for graph plotting:
- A (Axes): both axes fully labelled with quantity and units (Sucrose concentration / \(\text{mol/dm}^3\) and Percentage change in mass / \(\%\));
- S (Scale): linear, sensible scale using at least half the grid in both dimensions, with \(0\) on y-axis placed appropriately to include positive and negative values;
- P (Plotting): all 6 data points plotted accurately to within half a small square;
- L (Line): neat, continuous thin line drawn connecting points with a ruler OR single smooth curve passing through points with no extrapolation beyond \(1.0\text{ mol/dm}^3\);
[4 marks]

(c)(i)
- Correct reading of \(x\)-intercept where \(y = 0\%\) from candidate's graph (\(0.42\text{ to } 0.44\text{ mol/dm}^3\));
- Clear intercept line/indication shown on graph;
[2 marks]

(c)(ii)
- Water moved into cells / tissue by osmosis;
- Down a water potential gradient / from higher water potential outside to lower water potential inside cells across a partially permeable membrane;
[2 marks]

(d)(i) Any two from:
- Volume of solution (\(25\text{ cm}^3\));
- Incubation / immersion time (\(60\text{ minutes}\));
- Initial length / diameter of cylinder;
- Same sweet potato;
- Temperature;
[2 marks]

(d)(ii)
- Error: Variable blotting / removal of differing volumes of surface liquid OR moisture lost to evaporation during weighing;
- Improvement: Standardise blotting procedure (e.g. roll each cylinder fixed number of times under a set weight) OR weigh in closed container / weighing boat;
[2 marks]
Question 2 · short-answer
8 marks
Fig. 1.1 shows a photomicrograph of a transverse section through the root of a young dicotyledonous plant.

[Fig. 1.1: A transverse section of a dicotyledonous root showing an outer layer of epidermis, a wide cortex of parenchyma tissue, an inner endodermis ring, and a central vascular cylinder (stele) containing a 4-armed (x-shaped) core of xylem surrounded by patches of phloem. A straight line labelled XY runs horizontally across the total width of the root section, measuring 75 mm.]

(a) Make a large, labelled outline drawing of the root cross-section shown in Fig. 1.1.
Do not draw individual cells.

(b) (i) Measure the length of line XY on Fig. 1.1.

Length of line XY = .................................................... mm

(ii) The magnification of the image in Fig. 1.1 is \(\times 25\).
Calculate the actual diameter of the root along line XY.
Give your answer in micrometres (\(\mu\text{m}\)).

Show your working.
Show answer & marking scheme

Worked solution

(a) Drawing criteria:
- Size: Drawing occupies more than 50% of the provided space.
- Outline: Single, clean, continuous unbroken pencil lines with no sketching, feathering, or shading.
- Proportion and detail: Correct concentric arrangement of tissue layers drawn (outer single epidermal boundary, wide cortex region, distinct inner central cylinder with 4-lobed/x-shaped central xylem).
- Label: One correct label (e.g. epidermis, cortex, xylem, phloem, or vascular cylinder/stele) indicated by a ruled label line touching the intended tissue layer without crossing any other label lines.

(b) (i)
Measurement of line XY = \(75\text{ mm}\) (acceptable range: \(74\text{--}76\text{ mm}\)).

(ii)
Step 1: Use the magnification formula to find actual size in mm:
\[\text{Actual size (mm)} = \frac{\text{Image size (mm)}}{\text{Magnification}} = \frac{75\text{ mm}}{25} = 3.0\text{ mm}\]

Step 2: Convert millimetres to micrometres (\(1\text{ mm} = 1000\ \mu\text{m}\)):
\[3.0\text{ mm} \times 1000 = 3000\ \mu\text{m}\]
(Allow ecf if line XY measurement is between \(74\) and \(76\text{ mm}\), giving \(2960\) to \(3040\ \mu\text{m}\)).

Marking scheme

(a)
1. Size: drawing occupies at least half of the available box/space; [1]
2. Outline: single, sharp, unbroken continuous lines with no shading / cross-hatching / overlapping sketchy strokes; [1]
3. Detail: shows outer layer (epidermis), broad middle region (cortex), and central vascular core with a distinct lobed/x-shaped pattern (xylem); [1]
4. Label: at least one valid anatomical label (cortex / epidermis / xylem / phloem / stele / vascular cylinder) with a straight, ruled label line touching the structure; [1]

(b)(i)
75 (mm); [1]
[A: 74 to 76 mm]

(b)(ii)
1. Correct substitution into formula: image length ÷ 25 (e.g. \(75 \div 25\) or \(3.0\text{ mm}\)); [1]
2. Correct conversion of mm to \(\mu\text{m}\) (multiplying by 1000); [1]
3. Final calculated value: 3000 (\(\mu\text{m}\)); [1]
[ecf allowed for incorrect measurement from (b)(i): (measured length / 25) × 1000]
[R: if incorrect or omitted units prevent validation of the conversion]
Question 3 · Experimental Planning & Design
6 marks
Pectinase is an enzyme widely used in the food industry to break down cell wall components in fruit pulp and increase the yield of extracted juice.

Plan an investigation to determine the effect of temperature on the volume of fruit juice produced from apple purée using pectinase.

You are provided with apple purée, a 1% solution of pectinase enzyme, standard laboratory glassware, and access to thermostatically controlled water-baths.

Your plan should include:
- the independent variable and how you will change it, including the range and values you will use
- the dependent variable and the method you will use to measure it
- the variables that must be kept constant and how you will control them
- a safety precaution
- how you will ensure the reliability of your results.
Show answer & marking scheme

Worked solution

1. Independent Variable: Use at least five different temperatures across a suitable range (e.g., 20 °C, 30 °C, 40 °C, 50 °C, and 60 °C), maintained using thermostatically controlled water-baths.
2. Equilibration: Place separate tubes of a known mass of apple purée and a known volume of 1% pectinase in each water-bath and leave them for 5 minutes so they reach the test temperature before mixing.
3. Dependent Variable: Mix the apple purée and pectinase thoroughly, place the mixture into a funnel lined with filter paper resting over a measuring cylinder, and record the volume of clear juice collected (in \(\text{cm}^3\)) after a specified time period (e.g., 15 minutes) using a measuring cylinder read at eye level.
4. Controlled Variables:
- Mass of apple purée (e.g., 20 g measured on an electronic balance) from the same batch/variety of apple.
- Volume of pectinase solution (e.g., 2 \(\text{cm}^3\) measured with a syringe) and enzyme concentration (1%).
- Filtration/incubation time (e.g., exactly 15 minutes timed with a stopwatch).
- pH of the mixture (kept constant, or monitored using a buffer).
5. Repetitions / Reliability: Repeat the entire procedure at least 3 times at each temperature, identify any anomalous results, and calculate a mean volume of juice for each temperature.
6. Safety: Wear safety goggles to prevent eye irritation from accidental splashes of enzyme solution, and use heat-resistant gloves or tongs when handling hot water-baths to prevent burns.

Marking scheme

Award one mark for each valid point up to a maximum of 6 marks:

- MP1 (Independent variable): at least five named temperatures across a sensible range (e.g. 20 °C to 60 °C / minimum 5 temperatures) AND method of maintaining temperature stated (water-bath / thermostatically controlled water-bath);
- MP2 (Dependent variable): measure the volume of juice collected AND specified apparatus (measuring cylinder / graduated syringe / balance to measure mass);
- MP3 (Timing / Rate): state a specific incubation / filtration time (e.g. 10 to 30 minutes) using a timer / stopwatch;
- MP4 (Controlled variable 1): keep volume / concentration of pectinase constant (e.g. 2 \(\text{cm}^3\) of 1% pectinase);
- MP5 (Controlled variable 2): keep mass / volume of apple purée constant (e.g. 20 g) / use same type / batch of apple;
- MP6 (Experimental procedure): equilibrate the pectinase and apple purée separately in the water-bath before mixing / ensure consistent stirring;
- MP7 (Reliability): repeat the experiment at least three times at each temperature and calculate a mean (reject: 'repeat to get an average' without stating at least 3 repeats / repeat at each temperature);
- MP8 (Safety): appropriate safety precaution with valid reason (e.g. wear safety goggles to protect eyes from enzyme splashes / use heat-resistant gloves or tongs to avoid burns or scalds from hot water-baths).
Question 4 · Experimental Planning & Design
6 marks
Pectinase is an enzyme widely used in the food industry to break down cell wall components in fruit pulp and increase the yield of extracted juice.

Plan an investigation to determine the effect of temperature on the volume of fruit juice produced from apple purée using pectinase.

You are provided with apple purée, a 1% solution of pectinase enzyme, standard laboratory glassware, and access to thermostatically controlled water-baths.

Your plan should include:
- the independent variable and how you will change it, including the range and values you will use
- the dependent variable and the method you will use to measure it
- the variables that must be kept constant and how you will control them
- a safety precaution
- how you will ensure the reliability of your results.
Show answer & marking scheme

Worked solution

1. Independent Variable: Use at least five different temperatures across a suitable range (e.g., 20 °C, 30 °C, 40 °C, 50 °C, and 60 °C), maintained using thermostatically controlled water-baths.
2. Equilibration: Place separate tubes of a known mass of apple purée and a known volume of 1% pectinase in each water-bath and leave them for 5 minutes so they reach the test temperature before mixing.
3. Dependent Variable: Mix the apple purée and pectinase thoroughly, place the mixture into a funnel lined with filter paper resting over a measuring cylinder, and record the volume of clear juice collected (in \(\text{cm}^3\)) after a specified time period (e.g., 15 minutes) using a measuring cylinder read at eye level.
4. Controlled Variables:
- Mass of apple purée (e.g., 20 g measured on an electronic balance) from the same batch/variety of apple.
- Volume of pectinase solution (e.g., 2 \(\text{cm}^3\) measured with a syringe) and enzyme concentration (1%).
- Filtration/incubation time (e.g., exactly 15 minutes timed with a stopwatch).
- pH of the mixture (kept constant, or monitored using a buffer).
5. Repetitions / Reliability: Repeat the entire procedure at least 3 times at each temperature, identify any anomalous results, and calculate a mean volume of juice for each temperature.
6. Safety: Wear safety goggles to prevent eye irritation from accidental splashes of enzyme solution, and use heat-resistant gloves or tongs when handling hot water-baths to prevent burns.

Marking scheme

Award one mark for each valid point up to a maximum of 6 marks:

- MP1 (Independent variable): at least five named temperatures across a sensible range (e.g. 20 °C to 60 °C / minimum 5 temperatures) AND method of maintaining temperature stated (water-bath / thermostatically controlled water-bath);
- MP2 (Dependent variable): measure the volume of juice collected AND specified apparatus (measuring cylinder / graduated syringe / balance to measure mass);
- MP3 (Timing / Rate): state a specific incubation / filtration time (e.g. 10 to 30 minutes) using a timer / stopwatch;
- MP4 (Controlled variable 1): keep volume / concentration of pectinase constant (e.g. 2 \(\text{cm}^3\) of 1% pectinase);
- MP5 (Controlled variable 2): keep mass / volume of apple purée constant (e.g. 20 g) / use same type / batch of apple;
- MP6 (Experimental procedure): equilibrate the pectinase and apple purée separately in the water-bath before mixing / ensure consistent stirring;
- MP7 (Reliability): repeat the experiment at least three times at each temperature and calculate a mean (reject: 'repeat to get an average' without stating at least 3 repeats / repeat at each temperature);
- MP8 (Safety): appropriate safety precaution with valid reason (e.g. wear safety goggles to protect eyes from enzyme splashes / use heat-resistant gloves or tongs to avoid burns or scalds from hot water-baths).
Question 5 · structured
11 marks
A student investigated the distribution and population size of a plant species, hoary plantain (Plantago media), growing across a meadow bordering a dense woodland.

A 50 m tape measure was laid out as a line transect perpendicular to the woodland edge into the meadow.
A $0.5\text{ m} \times 0.5\text{ m}$ ($0.25\text{ m}^2$) quadrat was placed at 10 m intervals along the transect line, and the number of individual hoary plantain plants inside each quadrat was counted.

The results are shown in Table 1.1.

Table 1.1

| Distance from woodland edge / m | Number of plants per $0.25\text{ m}^2$ quadrat |
| :--- | :--- |
| 0 | 2 |
| 10 | 6 |
| 20 | 11 |
| 30 | 17 |
| 40 | 24 |
| 50 | 32 |

(a) Plot a line graph on the grid of the number of plants per $0.25\text{ m}^2$ quadrat against the distance from the woodland edge. [4]

(b) (i) Describe the relationship between the distance from the woodland edge and the number of plants per quadrat. [1]

(ii) Use your graph to estimate the number of plants per quadrat at a distance of 25 m from the woodland edge. Show clearly on your graph how you obtained your answer. [2]

(c) The open meadow at the far end of the transect has a total area of $1200\text{ m}^2$.
To estimate the total population of hoary plantain in this open meadow, the student threw ten $0.25\text{ m}^2$ quadrats at random across the meadow.
The total number of hoary plantain plants counted across all 10 quadrats combined was 310.

Calculate the estimated total population of hoary plantain plants in the $1200\text{ m}^2$ meadow.

Show your working. [2]

(d) State two ways the student could improve the accuracy or reliability of the population estimate in (c). [2]
Show answer & marking scheme

Worked solution

(a) Graph requirements:
- A (Axes): Distance from woodland edge / m on x-axis, Number of plants per $0.25\text{ m}^2$ quadrat on y-axis (both with units).
- S (Scale): Linear, occupying $\ge 50\%$ of grid along both axes (e.g. x-axis: 0 to 50, y-axis: 0 to 35).
- P (Plotting): All 6 points accurately plotted within $\pm 0.5$ small square.
- L (Line): Smooth line of best fit OR points joined with sharp, straight lines with no extrapolation beyond 0 or 50.

(b) (i) As the distance from the woodland edge increases, the number of hoary plantain plants per quadrat increases.
(ii) Reading from graph at 25 m: line drawn up from 25 m on x-axis to the curve and across to the y-axis gives approximately 13.5 to 14.5 plants (acceptable integer: 14).

(c) Total area sampled by 10 quadrats $= 10 \times 0.25\text{ m}^2 = 2.5\text{ m}^2$.
Mean plant density $= \frac{310\text{ plants}}{2.5\text{ m}^2} = 124\text{ plants per m}^2$.
Estimated total population in $1200\text{ m}^2 = 124 \times 1200 = 148\text{ }800$ plants.

(d) 1. Use a random coordinate method (e.g. using a grid and random number table/generator) rather than throwing quadrats to eliminate placement bias.
2. Sample a larger number of quadrats (e.g. 20 or more) to increase sample size and reliability.

Marking scheme

(a)
1. A — Axes labelled with quantity and units (Distance from woodland edge / m and *Number of plants per $0.25\text{ m}^2$ quadrat*); [1]
2. S — Sensible linear scale using more than half the grid in both dimensions; [1]
3. P — All six points plotted accurately to within $\pm 0.5$ small square; [1]
4. L — Clean, thin line drawn either point-to-point with a ruler or smooth best-fit curve without feathery lines / not extrapolated beyond 0 or 50 m; [1]

(b) (i)
- As distance from woodland edge increases, number of plants increases / positive correlation; [1]
(A: greater distance gives more plants / ora)
(R: directly proportional)

(ii)
- Clear construction lines shown on the graph at 25 m on the x-axis to the line and intercept on the y-axis; [1]
- Correct reading from candidate's plotted graph in the range $13.5$ to $14.5$ (or whole number 14); [1]
(ecf applies if graph is drawn incorrectly)

(c)
- Correct calculation of total area sampled ($2.5\text{ m}^2$) OR mean number per $\text{m}^2$ ($124$) OR mean per quadrat ($31$); [1]
- $148\text{ }800$ (or $1.488 \times 10^5$); [1]
(Allow 2 marks for correct final answer without working shown)

(d) Any two from:
- Use a grid system with random coordinates / random number generator (to avoid bias from throwing); [1]
- Increase the number of quadrats used / increase sample size; [1]
- Use larger quadrats / measure over a longer period / different seasons; [1]
- Count plants that overlap boundaries using a standardised rule (e.g., top/right rule); [1]
(Maximum 2 marks)

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