Cambridge IGCSE · thinka-original Practice Paper

2025 Cambridge IGCSE Biology (0610) Practice Paper with Answers

Thinka Jun 2025 (V2) Cambridge IGCSE-Style Mock — Biology (0610)

160 marks180 mins2025
An original Thinka practice paper modelled on the structure and difficulty of the Jun 2025 (V2) Cambridge IGCSE Biology (0610) paper. Not affiliated with or reproduced from Cambridge.

Paper 2 (Extended Multiple Choice)

There are forty questions on this paper. Answer all questions. Choose the one you consider correct and record your choice on the multiple choice answer sheet.
40 Question · 40 marks
Question 1 · multiple-choice
1 marks
The table shows two examples of immunity. Which row correctly classifies these types of immunity?

| Row | Receiving antibodies through breast milk | Producing memory cells after an infection |
| :--- | :--- | :--- |
| A | active | active |
| B | active | passive |
| C | passive | active |
| D | passive | passive |
  1. A.A
  2. B.B
  3. C.C
  4. D.D
Show answer & marking scheme

Worked solution

Receiving antibodies through breast milk is an example of passive immunity because the body is receiving pre-made antibodies from another individual. Producing memory cells after an infection is active immunity because the body's own immune system is stimulated to make its own antibodies and memory cells.

Marking scheme

1 mark for the correct option (C).
Question 2 · multiple-choice
1 marks
The list shows some changes that occur in the thorax during ventilation:

1. The diaphragm contracts and flattens.
2. The internal intercostal muscles contract.
3. The volume of the thorax increases.
4. The pressure inside the lungs increases.

Which changes occur during inspiration (breathing in)?
  1. A.1 and 3 only
  2. B.1 and 4 only
  3. C.2 and 3 only
  4. D.2 and 4 only
Show answer & marking scheme

Worked solution

During inspiration, the diaphragm contracts and flattens (1), and the external intercostal muscles contract, causing the internal intercostal muscles to relax (meaning statement 2 is incorrect). These movements increase the volume of the thorax (3), which decreases the pressure inside the lungs so that air is drawn in down a pressure gradient.

Marking scheme

1 mark for the correct option (A).
Question 3 · multiple-choice
1 marks
During genetic modification, a human gene is inserted into a bacterial plasmid. Which enzymes are used to cut the DNA and to join the DNA fragments together?

| Row | Enzyme used to cut DNA | Enzyme used to join DNA fragments |
| :--- | :--- | :--- |
| A | protease | lipase |
| B | restriction enzyme | DNA ligase |
| C | DNA ligase | restriction enzyme |
| D | restriction enzyme | protease |
  1. A.A
  2. B.B
  3. C.C
  4. D.D
Show answer & marking scheme

Worked solution

Restriction enzymes are used to cut both the human DNA and plasmid DNA at specific sequences, leaving complementary sticky ends. DNA ligase is then used to join the sticky ends of the human gene and the plasmid together.

Marking scheme

1 mark for the correct option (B).
Question 4 · multiple-choice
1 marks
What is the effect of a temperature of 60 °C on the activity of a human digestive enzyme?
  1. A.The rate of reaction increases because of increased kinetic energy of the molecules.
  2. B.The enzyme becomes denatured, changing the shape of its active site so the substrate no longer fits.
  3. C.The enzyme's active site changes shape to fit the substrate better, increasing the rate of reaction.
  4. D.The substrate molecules change shape and can no longer bind to the active site.
Show answer & marking scheme

Worked solution

At high temperatures (such as 60 °C), enzymes are denatured. This means the active site loses its specific shape, so the substrate can no longer fit, and the rate of reaction drops to zero.

Marking scheme

1 mark for the correct option (B).
Question 5 · multiple-choice
1 marks
A plant cell with a water potential of -300 kPa is placed in a concentrated sucrose solution with a water potential of -600 kPa. Which statement describes what happens to the cell?
  1. A.Water enters the cell by osmosis and the cell becomes turgid.
  2. B.Water enters the cell by osmosis and the cell swells and bursts.
  3. C.Water leaves the cell by osmosis and the cell membrane pulls away from the cell wall.
  4. D.Water leaves the cell by osmosis and the cell wall collapses completely.
Show answer & marking scheme

Worked solution

Water moves from a region of higher water potential (-300 kPa) to a region of lower water potential (-600 kPa). Therefore, water leaves the plant cell by osmosis, causing the cytoplasm to shrink and the cell membrane to pull away from the cell wall (plasmolysis).

Marking scheme

1 mark for the correct option (C).
Question 6 · multiple-choice
1 marks
Which adaptive feature is correctly matched with its function in a xerophyte?
  1. A.large flat leaves to maximize the absorption of sunlight
  2. B.sunken stomata to trap moist air and reduce the rate of transpiration
  3. C.thin waxy cuticle to allow rapid diffusion of carbon dioxide
  4. D.small, shallow root systems because water is always abundant near the soil surface
Show answer & marking scheme

Worked solution

Xerophytes are adapted to survive in dry habitats. Sunken stomata trap moist air outside the stomatal pore, reducing the concentration gradient of water vapor between the leaf and the atmosphere, thereby reducing water loss by transpiration.

Marking scheme

1 mark for the correct option (B).
Question 7 · multiple-choice
1 marks
Which process in the nitrogen cycle converts ammonium ions into nitrate ions?
  1. A.active transport
  2. B.denitrification
  3. C.nitrification
  4. D.nitrogen fixation
Show answer & marking scheme

Worked solution

Nitrification is the biological oxidation of ammonia or ammonium ions to nitrite ions, and then to nitrate ions, by nitrifying bacteria in the soil.

Marking scheme

1 mark for the correct option (C).
Question 8 · multiple-choice
1 marks
Albinism in humans is inherited as an autosomal recessive trait. Two parents with normal skin pigmentation have an albino child. What is the probability that their next child will also be albino?
  1. A.0%
  2. B.25%
  3. C.50%
  4. D.75%
Show answer & marking scheme

Worked solution

Since both parents have normal skin pigmentation but have an albino child (aa), both parents must be heterozygous carriers (Aa). The cross Aa x Aa yields a 1 in 4 (25%) chance of producing an offspring with the homozygous recessive genotype (aa), which results in albinism.

Marking scheme

1 mark for the correct option (B).
Question 9 · multiple-choice
1 marks
Plant cells are placed in a concentrated sucrose solution. Which row correctly describes the changes to the cell’s state?
  1. A.water potential of cytoplasm decreases | turgor pressure decreases | vacuole volume decreases
  2. B.water potential of cytoplasm decreases | turgor pressure increases | vacuole volume increases
  3. C.water potential of cytoplasm increases | turgor pressure decreases | vacuole volume decreases
  4. D.water potential of cytoplasm increases | turgor pressure increases | vacuole volume increases
Show answer & marking scheme

Worked solution

When plant cells are placed in a concentrated sucrose solution (which has a lower water potential than the cytoplasm), water leaves the vacuole and cytoplasm by osmosis. As water leaves, the vacuole volume decreases, turgor pressure decreases, and the remaining cytoplasm becomes more concentrated, meaning its water potential decreases.

Marking scheme

Award 1 mark for the correct option (A).
Question 10 · multiple-choice
1 marks
The rate of an enzyme-catalysed reaction was measured at different substrate concentrations. At high substrate concentrations, the rate of reaction remains constant even if more substrate is added. What is the reason for this?
  1. A.All of the active sites of the enzyme molecules are occupied by substrate.
  2. B.The enzyme molecules have been completely denatured by high substrate concentration.
  3. C.Substrate molecules are repelling each other, preventing them from binding.
  4. D.The temperature of the reaction mixture has decreased due to the reaction.
Show answer & marking scheme

Worked solution

At very high substrate concentrations, the concentration of substrate molecules exceeds the availability of enzyme molecules. Thus, all the active sites of the enzyme are fully occupied (saturated) with substrate, meaning any additional substrate must wait for an active site to become free.

Marking scheme

Award 1 mark for the correct option (A).
Question 11 · multiple-choice
1 marks
Which statement correctly describes passive immunity?
  1. A.It is short-term because no memory cells are produced by the body.
  2. B.It is long-term because antibodies are actively made by the body's lymphocytes.
  3. C.It is acquired only through vaccination with a harmless form of a pathogen.
  4. D.It stimulates the production of memory cells to defend against future infections.
Show answer & marking scheme

Worked solution

Passive immunity is temporary (short-term) because the antibodies are introduced from an external source (such as across the placenta or through breast milk) rather than being made by the individual's own lymphocytes, and no memory cells are produced.

Marking scheme

Award 1 mark for the correct option (A).
Question 12 · multiple-choice
1 marks
How does the thin layer of water lining the alveoli assist in gas exchange?
  1. A.It dissolves oxygen so that it can diffuse across the gas exchange surface.
  2. B.It increases turgor pressure to keep the alveolus fully expanded.
  3. C.It actively pumps carbon dioxide out of the blood plasma.
  4. D.It acts as a barrier to prevent pathogens from reaching the blood capillary.
Show answer & marking scheme

Worked solution

Gases must dissolve in a liquid medium before they can diffuse across gas exchange membranes. The thin layer of water dissolves oxygen, allowing it to diffuse efficiently across the alveolar and capillary walls.

Marking scheme

Award 1 mark for the correct option (A).
Question 13 · multiple-choice
1 marks
During the genetic modification of bacteria to produce human proteins, which enzyme is used to join the useful human gene into a cut bacterial plasmid?
  1. A.DNA ligase
  2. B.amylase
  3. C.restriction enzyme
  4. D.protease
Show answer & marking scheme

Worked solution

DNA ligase is the enzyme responsible for joining the sugar-phosphate backbones of the human gene insert and the bacterial plasmid DNA, forming a continuous recombinant DNA molecule.

Marking scheme

Award 1 mark for the correct option (A).
Question 14 · multiple-choice
1 marks
In a species of plant, allele \(F^R\) produces red flowers and allele \(F^W\) produces white flowers. The heterozygous genotype \(F^R F^W\) produces pink flowers. If two pink-flowered plants are crossed, what is the expected ratio of phenotypes in the offspring?
  1. A.1 red : 2 pink : 1 white
  2. B.3 red : 1 white
  3. C.1 red : 1 white
  4. D.all pink
Show answer & marking scheme

Worked solution

This is an example of codominance. Crossing two pink heterozygotes (\(F^R F^W \times F^R F^W\)) produces offspring with genotypes in a ratio of 1 \(F^R F^R\) : 2 \(F^R F^W\) : 1 \(F^W F^W\). This directly corresponds to a phenotypic ratio of 1 red : 2 pink : 1 white.

Marking scheme

Award 1 mark for the correct option (A).
Question 15 · multiple-choice
1 marks
Which feature is an adaptive feature of a hydrophyte?
  1. A.large air spaces in the tissues to provide buoyancy
  2. B.stomata sunken in pits on the lower surface of the leaf
  3. C.a thick waxy cuticle covering the upper surface of the leaf
  4. D.an extensive, deep root system to anchor the plant
Show answer & marking scheme

Worked solution

Hydrophytes are adapted to living in aquatic environments. Large air spaces in their leaves and stems provide buoyancy, keeping them near the water surface where light levels are highest, and facilitating the diffusion of gases within the submerged tissues.

Marking scheme

Award 1 mark for the correct option (A).
Question 16 · multiple-choice
1 marks
Which process in the nitrogen cycle converts ammonium ions into nitrate ions in the soil?
  1. A.nitrification
  2. B.denitrification
  3. C.nitrogen fixation
  4. D.active transport
Show answer & marking scheme

Worked solution

Nitrification is the biological process carried out by nitrifying bacteria in the soil that oxidises ammonium ions first into nitrites, and then into nitrates, making nitrogen accessible to plants.

Marking scheme

Award 1 mark for the correct option (A).
Question 17 · multiple-choice
1 marks
The table shows the changes in mass of four different plant tissues, P, Q, R and S, after being placed in a beaker containing a 0.5 mol/dm³ sucrose solution for two hours.

$$\begin{array}{|c|c|c|} \hline \text{Tissue} & \text{Initial mass / g} & \text{Final mass / g} \\ \hline P & 2.5 & 2.1 \\ \hline Q & 2.5 & 2.5 \\ \hline R & 2.5 & 2.9 \\ \hline S & 2.5 & 3.2 \\ \hline \end{array}$$

Which tissue had the highest water potential before being placed in the sucrose solution?
  1. A.P
  2. B.Q
  3. C.R
  4. D.S
Show answer & marking scheme

Worked solution

A plant tissue with a higher water potential than the surrounding sucrose solution will lose water by osmosis, resulting in a net decrease in mass. Tissue P went from 2.5 g to 2.1 g, showing a net loss of water. Tissues R and S gained mass, indicating they had a lower water potential than the solution. Tissue Q showed no change, meaning its water potential was equal to the solution. Therefore, Tissue P had the highest initial water potential.

Marking scheme

Award 1 mark for the correct option A.
Question 18 · multiple-choice
1 marks
In humans, the allele for a certain genetic condition is recessive. A couple, both of whom do not have the condition, have three children. Their first child has the condition, while the next two do not. What is the probability that their fourth child will be heterozygous for the condition?
  1. A.25%
  2. B.50%
  3. C.75%
  4. D.100%
Show answer & marking scheme

Worked solution

Since the parents do not have the condition but have produced a child with the recessive condition (genotype aa), both parents must be heterozygous carriers (genotype Aa). For each pregnancy, the cross Aa × Aa has a constant probability of producing offspring genotypes: 25% AA, 50% Aa (heterozygous), and 25% aa. The genotypes of previous children do not affect the probability of future independent events. Thus, the probability that the fourth child is heterozygous is 50%.

Marking scheme

Award 1 mark for the correct option B.
Question 19 · multiple-choice
1 marks
Which statement correctly describes passive immunity?
  1. A.It is achieved by the active production of antibodies by the body’s own lymphocytes.
  2. B.It results in the long-term production of memory cells.
  3. C.It can be transferred from a mother to her baby through breast milk.
  4. D.It is only gained after a vaccination containing weakened pathogens.
Show answer & marking scheme

Worked solution

Passive immunity is the short-term defense against a pathogen by antibodies acquired from another individual. This includes the transfer of antibodies from a mother to her baby across the placenta or via breast milk (colostrum). Active immunity, on the other hand, involves the body's own production of antibodies and memory cells, often stimulated by infection or vaccination.

Marking scheme

Award 1 mark for the correct option C.
Question 20 · multiple-choice
1 marks
An enzyme-catalysed reaction was investigated at different temperatures. The table shows the rate of reaction at each temperature.

$$\begin{array}{|c|c|} \hline \text{Temperature / } ^\circ\text{C} & \text{Rate of reaction / arbitrary units} \\ \hline 10 & 12 \\ \hline 20 & 25 \\ \hline 30 & 50 \\ \hline 40 & 85 \\ \hline 50 & 10 \\ \hline 60 & 0 \\ \hline \end{array}$$

Which statement explains the change in the rate of reaction between 40 °C and 50 °C?
  1. A.The kinetic energy of the substrate and enzyme molecules decreased.
  2. B.The active sites of the enzyme molecules changed shape.
  3. C.The enzyme molecules were completely broken down into individual amino acids.
  4. D.The activation energy of the reaction increased significantly.
Show answer & marking scheme

Worked solution

Between 40 °C and 50 °C, the temperature exceeds the optimum range for the enzyme, causing denaturation. During denaturation, the specific three-dimensional shape of the enzyme's active site is permanently altered so that it no longer fits the substrate. This causes a dramatic drop in the rate of reaction. Note that denaturation does not break peptide bonds to destroy the enzyme back into individual amino acids.

Marking scheme

Award 1 mark for the correct option B.
Question 21 · multiple-choice
1 marks
Which row in the table correctly describes the state of the diaphragm and external intercostal muscles, as well as the movement of the ribcage, during quiet breathing out (expiration)?

$$\begin{array}{|c|c|c|c|} \hline \text{Row} & \text{Diaphragm} & \text{External intercostal muscles} & \text{Ribcage movement} \\ \hline \text{A} & \text{contracts} & \text{contract} & \text{upwards and outwards} \\ \hline \text{B} & \text{contracts} & \text{relax} & \text{downwards and inwards} \\ \hline \text{C} & \text{relaxes} & \text{contract} & \text{upwards and outwards} \\ \hline \text{D} & \text{relaxes} & \text{relax} & \text{downwards and inwards} \\ \hline \end{array}$$

Choose the correct row (A, B, C or D).
  1. A.A
  2. B.B
  3. C.C
  4. D.D
Show answer & marking scheme

Worked solution

During expiration (breathing out), the diaphragm relaxes and dome-shapes upwards, while the external intercostal muscles relax, allowing the ribcage to move downwards and inwards under the influence of gravity. This decreases the volume of the thorax, increasing the pressure and forcing air out of the lungs.

Marking scheme

Award 1 mark for the correct option D.
Question 22 · multiple-choice
1 marks
Some features of plants are listed below:

1. Thick waxy cuticle
2. Stomata on the upper epidermis of leaves
3. Large air spaces in tissues
4. Deep, extensive root systems

Which of these features are typical adaptations of hydrophytes (plants living in wet or aquatic environments)?
  1. A.1 and 3
  2. B.1 and 4
  3. C.2 and 3
  4. D.2 and 4
Show answer & marking scheme

Worked solution

Hydrophytes are adapted to living in aquatic environments. They have stomata on the upper epidermis of their leaves (feature 2) so they can exchange gases when floating on water. They also contain large air spaces in their tissues (feature 3) to provide buoyancy and store gases. A thick waxy cuticle (feature 1) and deep, extensive root systems (feature 4) are adaptations typical of xerophytes (desert plants) to conserve and absorb scarce water.

Marking scheme

Award 1 mark for the correct option C.
Question 23 · multiple-choice
1 marks
What are the roles of restriction enzymes and DNA ligase in the production of genetically modified bacteria?
  1. A.Restriction enzymes cut DNA at specific sequences; DNA ligase joins the cut DNA fragments together.
  2. B.Restriction enzymes join the cut DNA fragments together; DNA ligase cuts DNA at specific sequences.
  3. C.Restriction enzymes replicate the gene of interest; DNA ligase inserts the plasmid into the bacterium.
  4. D.Restriction enzymes insert the plasmid into the bacterium; DNA ligase replicates the gene of interest.
Show answer & marking scheme

Worked solution

In genetic engineering, restriction enzymes are used to cut DNA (both the gene of interest and the plasmid vector) at specific base sequences, leaving matching sticky ends. DNA ligase is then used to join the complementary DNA fragments together by forming phosphodiester bonds, sealing the gene of interest into the plasmid.

Marking scheme

Award 1 mark for the correct option A.
Question 24 · multiple-choice
1 marks
The sequence of transitions below represents part of the nitrogen cycle:

$$\text{Organic nitrogen compounds in dead plants} \xrightarrow{1} \text{Ammonium ions} \xrightarrow{2} \text{Nitrate ions} \xrightarrow{3} \text{Nitrogen gas}$$

Which row correctly identifies the biological processes represented by 1, 2 and 3?

$$\begin{array}{|c|c|c|c|} \hline \text{Row} & \text{Process 1} & \text{Process 2} & \text{Process 3} \\ \hline \text{A} & \text{decomposition} & \text{nitrification} & \text{denitrification} \\ \hline \text{B} & \text{nitrification} & \text{denitrification} & \text{nitrogen fixation} \\ \hline \text{C} & \text{decomposition} & \text{denitrification} & \text{nitrification} \\ \hline \text{D} & \text{nitrogen fixation} & \text{nitrification} & \text{denitrification} \\ \hline \end{array}$$

Choose the correct row (A, B, C or D).
  1. A.A
  2. B.B
  3. C.C
  4. D.D
Show answer & marking scheme

Worked solution

Transition 1 is the breakdown of organic nitrogen compounds in dead matter into ammonium ions, which is carried out by decomposers (decomposition). Transition 2 is the conversion of ammonium ions to nitrate ions by nitrifying bacteria (nitrification). Transition 3 is the conversion of nitrate ions back into nitrogen gas by denitrifying bacteria in anaerobic conditions (denitrification).

Marking scheme

Award 1 mark for the correct option A.
Question 25 · multiple-choice
1 marks
A plant cell with a water potential of -300 kPa is placed in a solution with a water potential of -700 kPa. Which row correctly describes the net movement of water and the state of the cell after a few hours?
  1. A.net movement of water: out of the cell | state of cell: plasmolysed
  2. B.net movement of water: out of the cell | state of cell: turgid
  3. C.net movement of water: into the cell | state of cell: plasmolysed
  4. D.net movement of water: into the cell | state of cell: turgid digital state of cell after equilibration is turgid when water enters the cell, which is not the case here.
Show answer & marking scheme

Worked solution

Water moves by osmosis down a water potential gradient from a region of higher water potential (less negative, -300 kPa) to a region of lower water potential (more negative, -700 kPa). Therefore, the net movement of water is out of the cell. This loss of water causes the cell membrane to pull away from the cell wall, resulting in the cell becoming plasmolysed.

Marking scheme

1 mark for the correct answer A.
Question 26 · multiple-choice
1 marks
A student set up four test-tubes containing starch solution and amylase at different temperatures. The time taken for the starch to be completely broken down was recorded: Tube 1 (15 degrees C) took 15 minutes, Tube 2 (25 degrees C) took 8 minutes, Tube 3 (35 degrees C) took 3 minutes, and Tube 4 (45 degrees C) took 12 minutes. At which temperature did the amylase have the highest rate of activity?
  1. A.15 degrees C
  2. B.25 degrees C
  3. C.35 degrees C
  4. D.45 degrees C
Show answer & marking scheme

Worked solution

The rate of enzyme activity is inversely proportional to the time taken for the substrate to be broken down. Amylase broke down all the starch in the shortest time (3 minutes) in Tube 3 at 35 degrees C, indicating this temperature gave the highest rate of activity.

Marking scheme

1 mark for the correct answer C.
Question 27 · multiple-choice
1 marks
Which statement correctly describes passive immunity?
  1. A.It is gained after an infection, leading to the production of memory cells by the host.
  2. B.It provides long-term protection through the injection of weakened pathogens.
  3. C.It involves the transfer of antibodies from another organism, providing immediate but temporary protection.
  4. D.It is stimulated by vaccination with antigens, causing the host's lymphocytes to produce antibodies.
Show answer & marking scheme

Worked solution

Passive immunity involves the transfer of ready-made antibodies from one individual to another (e.g., from mother to infant via breast milk, or through antibody injections). This provides immediate, short-term protection, but does not stimulate the recipient's immune system to produce memory cells.

Marking scheme

1 mark for the correct answer C.
Question 28 · multiple-choice
1 marks
The composition of four different gases in inspired and expired air is compared. Which gas has a percentage of approximately 0.04% in inspired air and 4.0% in expired air?
  1. A.Carbon dioxide
  2. B.Nitrogen
  3. C.Oxygen
  4. D.Water vapour
Show answer & marking scheme

Worked solution

Carbon dioxide is produced by aerobic respiration in body cells and excreted via the lungs. Consequently, expired air contains a significantly higher concentration of carbon dioxide (about 4.0%) compared to the concentration in inspired atmospheric air (about 0.04%).

Marking scheme

1 mark for the correct answer A.
Question 29 · multiple-choice
1 marks
In genetic modification, which enzymes are used to cut a specific gene from human DNA and to join it into a bacterial plasmid?
  1. A.Amylase and lipase
  2. B.Protease and ligase
  3. C.Restriction enzymes and ligase
  4. D.Restriction enzymes and protease
Show answer & marking scheme

Worked solution

Restriction enzymes are used to cut specific DNA sequences (leaving complementary sticky ends), while DNA ligase is used to join the target gene and plasmid vector together by forming covalent phosphodiester bonds.

Marking scheme

1 mark for the correct answer C.
Question 30 · multiple-choice
1 marks
Albinism is an inherited condition caused by a recessive allele, a. The dominant allele, A, results in normal skin pigmentation. Two parents with normal skin pigmentation have an albino child. What is the probability that their next child will also be albino?
  1. A.0%
  2. B.25%
  3. C.50%
  4. D.75%
Show answer & marking scheme

Worked solution

Because both parents have normal pigmentation but produced an albino child (genotype aa), they must both be carriers of the recessive allele, making their genotypes Aa. A cross between Aa and Aa parents yields a 1 in 4 (25%) chance of producing an offspring with the homozygous recessive aa genotype.

Marking scheme

1 mark for the correct answer B.
Question 31 · multiple-choice
1 marks
Which leaf feature is an adaptation of a xerophytic plant to reduce the rate of transpiration?
  1. A.Broad leaves with a thin cuticle
  2. B.Sunken stomata that trap moist air
  3. C.High density of stomata on the upper epidermis
  4. D.Reduced root system to decrease water absorption
Show answer & marking scheme

Worked solution

Sunken stomata trap moist air in the stomatal pits, reducing the concentration gradient of water vapour between the inside of the leaf and the surrounding atmosphere, which significantly slows down transpiration.

Marking scheme

1 mark for the correct answer B.
Question 32 · multiple-choice
1 marks
Which biological process in the nitrogen cycle converts ammonium ions into nitrate ions?
  1. A.Active transport
  2. B.Denitrification
  3. C.Nitrification
  4. D.Nitrogen fixation
Show answer & marking scheme

Worked solution

Nitrification is the biological process where nitrifying bacteria in the soil convert ammonium ions into nitrites, and then into nitrates, which plants can easily absorb and use to make proteins.

Marking scheme

1 mark for the correct answer C.
Question 33 · multiple_choice
1 marks
A patient is bitten by a venomous snake and is immediately injected with an antivenom containing antibodies. Three months later, they are bitten by the same species of snake. Why is the patient not protected from the venom during the second bite?
  1. A.The antivenom provided passive immunity, so no memory cells were produced.
  2. B.The patient's body had developed active immunity which had already faded.
  3. C.The antibodies from the first injection had converted into antigens.
  4. D.The venom caused a rapid mutation in the lymphocytes of the patient.
Show answer & marking scheme

Worked solution

Antivenom contains ready-made antibodies, which provides immediate but temporary protection. This is an example of passive immunity. Because the patient's own lymphocytes did not produce these antibodies, no memory cells were formed, meaning there is no long-term protection or secondary immune response upon subsequent exposure.

Marking scheme

1 mark: correct option selected (A). Reject all other options.
Question 34 · multiple_choice
1 marks
Which physiological changes occur in the human thorax during inspiration?
  1. A.Internal intercostal muscles contract, diaphragm relaxes, pressure in thorax increases.
  2. B.External intercostal muscles contract, diaphragm contracts, volume of thorax increases.
  3. C.External intercostal muscles contract, diaphragm relaxes, pressure in thorax decreases.
  4. D.Internal intercostal muscles contract, diaphragm contracts, volume of thorax decreases.
Show answer & marking scheme

Worked solution

During inhalation (inspiration), the external intercostal muscles contract to lift the ribcage up and out, while the diaphragm contracts and flattens downwards. These actions increase the volume of the thoracic cavity, reducing internal pressure and drawing air in.

Marking scheme

1 mark: correct option selected (B). Reject all other options.
Question 35 · multiple_choice
1 marks
Which row correctly identifies the roles of restriction enzymes and DNA ligase in genetic modification?
  1. A.restriction enzyme: joins DNA molecules together; DNA ligase: cuts DNA to leave sticky ends
  2. B.restriction enzyme: isolates the plasmid from a bacterium; DNA ligase: replicates the target gene
  3. C.restriction enzyme: cuts DNA to leave sticky ends; DNA ligase: joins DNA molecules together
  4. D.restriction enzyme: replicates the target gene; DNA ligase: isolates the plasmid from a bacterium
Show answer & marking scheme

Worked solution

Restriction enzymes are utilized to cut target DNA and bacterial plasmids at specific sites, often leaving single-stranded sticky ends. DNA ligase is then used to join the sticky ends of the target gene and plasmid together to form a recombinant plasmid.

Marking scheme

1 mark: correct option selected (C). Reject all other options.
Question 36 · multiple_choice
1 marks
Which statement describes what happens to an enzyme when it is heated slightly above its optimum temperature, but not completely denatured?
  1. A.The enzyme molecules lose all kinetic energy, causing the rate of reaction to drop immediately to zero.
  2. B.The active site changes shape permanently, allowing a wider variety of different substrates to bind to it.
  3. C.The activation energy of the reaction increases, making the substrate molecules more stable.
  4. D.The kinetic energy of the molecules increases, causing more frequent collisions, but the active site begins to lose its complementary shape.
Show answer & marking scheme

Worked solution

Heating slightly above the optimum temperature increases the kinetic energy of the molecules, which increases collision frequency. However, the heat begins to disrupt the delicate bonds holding the enzyme's three-dimensional structure together, starting to deform the active site so that it is no longer fully complementary to the substrate. This results in a decreased rate of reaction.

Marking scheme

1 mark: correct option selected (D). Reject all other options.
Question 37 · multiple_choice
1 marks
Plant cells are placed in a concentrated salt solution. Which description of the state of the cells and the direction of water movement is correct?
  1. A.Water moves into the vacuoles, and the cells become turgid.
  2. B.Water moves out of the vacuoles, and the cells become plasmolysed.
  3. C.Water moves out of the vacuoles, and the cells become turgid.
  4. D.Water moves into the vacuoles, and the cells become plasmolysed.
Show answer & marking scheme

Worked solution

A concentrated salt solution has a lower water potential than the cell sap. Water therefore moves out of the plant cells' vacuoles by osmosis, down a water potential gradient, through the partially permeable cell membrane. This loss of turgor pressure causes the cells to become flaccid and eventually plasmolysed (where the cell membrane pulls away from the cell wall).

Marking scheme

1 mark: correct option selected (B). Reject all other options.
Question 38 · multiple_choice
1 marks
Which adaptation of a xerophytic plant is correctly matched with how it helps the plant survive in dry conditions?
  1. A.reduced root system — prevents the absorption of toxic mineral salts
  2. B.thin waxy cuticle — allows rapid entry of carbon dioxide for photosynthesis
  3. C.sunken stomata — trap moist air to reduce the diffusion gradient of water vapour
  4. D.flat, wide leaves — increase the surface area for rapid transpiration
Show answer & marking scheme

Worked solution

Sunken stomata trap a pocket of humid (moist) air near the surface of the leaf. This moist air reduces the concentration gradient of water vapour between the inside and the outside of the leaf, significantly reducing water loss via transpiration.

Marking scheme

1 mark: correct option selected (C). Reject all other options.
Question 39 · multiple_choice
1 marks
In the nitrogen cycle, which process converts ammonium ions into nitrate ions?
  1. A.decomposition by decomposers
  2. B.nitrogen fixation by lightning
  3. C.denitrification by denitrifying bacteria
  4. D.nitrification by nitrifying bacteria
Show answer & marking scheme

Worked solution

Nitrification is the process carried out by nitrifying bacteria in the soil that converts ammonium ions first into nitrites, and then into nitrates, which plants can easily absorb.

Marking scheme

1 mark: correct option selected (D). Reject all other options.
Question 40 · multiple_choice
1 marks
A father has blood group AB and a mother has blood group O. What is the probability that their first child will have blood group A?
  1. A.50%
  2. B.25%
  3. C.75%
  4. D.0%
Show answer & marking scheme

Worked solution

The father with blood group AB has the genotype \(I^A I^B\). The mother with blood group O has the genotype \(I^O I^O\). A genetic cross yields offspring with genotypes \(I^A I^O\) (phenotype A) and \(I^B I^O\) (phenotype B) in a 1:1 ratio. Therefore, the probability of having a child with blood group A is 50%.

Marking scheme

1 mark: correct option selected (A). Reject all other options.

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Paper 4 (Extended Theory)

Answer all questions. Write your answers in the spaces provided on the question paper. You may use a calculator.
6 Question · 80 marks
Question 1 · structured
13 marks
Table 1.1 shows the ventilation rate and tidal volume of an athlete at rest and during two different intensities of running.

Table 1.1
| State | Ventilation rate / breaths per minute | Tidal volume / \(dm^3\) |
| :--- | :---: | :---: |
| Rest | 12 | 0.5 |
| Moderate running | 24 | 1.8 |
| Maximum sprinting | 40 | 2.5 |

(a) (i) Calculate the minute ventilation (volume of air breathed in one minute) during moderate running. Show your working and state the unit. [3]

(a) (ii) State and explain how the change in tidal volume during exercise supports the increased rate of aerobic respiration in muscle cells. [3]

(b) Describe how ciliated cells and goblet cells work together to protect the gas exchange system from pathogens during exercise. [4]

(c) Explain the role of the internal and external intercostal muscles during expiration during intense exercise. [3]
Show answer & marking scheme

Worked solution

(a) (i) Minute ventilation = ventilation rate \(\times\) tidal volume = \(24 \times 1.8 = 43.2\) \(dm^3\) \(min^{-1}\).
(a) (ii) Increased tidal volume delivers a larger volume of oxygen to the alveoli per breath. This maintains a steep concentration gradient between the alveoli and blood, increasing the rate of oxygen diffusion into the blood to meet the higher demand of respiring muscle cells.
(b) Goblet cells produce and secrete sticky mucus, which traps inhaled pathogens and dust particles. Ciliated cells have tiny hair-like cilia that beat in a coordinated wave-like motion to move the trapped pathogens in mucus up and out of the trachea towards the throat, where it is swallowed and destroyed by stomach acid.
(c) During intense exercise, expiration is active. The internal intercostal muscles contract, while the external intercostal muscles relax. This action actively pulls the ribcage downwards and inwards, rapidly decreasing the volume of the thorax and increasing thoracic pressure to force air out of the lungs quickly.

Marking scheme

(a) (i)
- Formula/working: \(24 \times 1.8\) [1]
- Correct calculation: 43.2 [1]
- Correct unit: \(dm^3\) \(min^{-1}\) or \(dm^3\)/min (accept litres per minute) [1]

(a) (ii)
- Tidal volume increases (by \(1.3\) \(dm^3\) / more air per breath) [1]
- Delivers more oxygen to the alveoli / maintains a steep concentration gradient [1]
- Increases rate of diffusion of oxygen into blood (to supply respiring muscles) [1]

(b)
- Goblet cells produce/secrete mucus [1]
- Mucus traps pathogens/dust/bacteria [1]
- Ciliated cells have cilia that beat/sweep [1]
- Move mucus upwards/away from lungs/towards throat [1]

(c)
- Internal intercostal muscles contract [1]
- External intercostal muscles relax [1]
- Ribcage is pulled down and in (active/forced expiration) [1]
Question 2 · structured
13 marks
Cholera is a transmissible disease caused by a bacterium. It results in severe diarrhoea.

(a) Explain how the cholera bacterium causes diarrhoea. [4]

(b) Explain why treatment with oral rehydration therapy (ORT) is highly effective, and state what it consists of. [3]

(c) Compare active immunity, such as that gained from a cholera vaccine, with passive immunity, such as that a newborn receives from breast milk. [6]
Show answer & marking scheme

Worked solution

(a) The cholera bacteria multiply in the small intestine and secrete a toxin. This toxin stimulates the cells lining the intestine to actively secrete chloride ions into the lumen of the gut. This lowers the water potential inside the gut lumen, creating a water potential gradient. Water moves out of the blood and intestinal cells into the lumen by osmosis, resulting in watery, loose feces (diarrhoea).
(b) ORT is highly effective because it directly rehydrates the body and replaces lost essential minerals, preventing dehydration-induced organ failure. It consists of a balanced mixture of clean water, salts (electrolytes such as sodium), and glucose.
(c) Active immunity involves the production of antibodies by the host's own immune system (lymphocytes) in response to antigens. It leads to the production of memory cells, which means it provides long-term, lasting protection, though the response is delayed. Passive immunity involves the transfer of ready-made antibodies from an external source (e.g., breast milk or injection). It does not produce memory cells and only provides short-term, temporary protection, but the defense is immediate.

Marking scheme

(a)
- Bacteria produce/release a toxin [1]
- Toxin stimulates secretion of chloride ions (\(Cl^-\)) into the gut/small intestine lumen [1]
- This lowers the water potential of the gut lumen / creates a water potential gradient [1]
- Water moves into the gut by osmosis, leading to watery feces/diarrhoea [1]

(b)
- ORT consists of water, salts (sodium), and glucose [1]
- Replaces lost water/electrolytes to prevent dehydration [1]
- Glucose aids active absorption of sodium/water in the gut [1]

(c)
Any six comparison points:
- Active: antibodies produced by own body/lymphocytes vs Passive: antibodies from external source [1]
- Active: triggered by antigen/pathogen exposure vs Passive: no antigen exposure required [1]
- Active: memory cells produced vs Passive: no memory cells produced [1]
- Active: long-term protection vs Passive: short-term/temporary protection [1]
- Active: delayed response (days to weeks) vs Passive: immediate protection [1]
- Active example: vaccine/infection vs Passive example: breast milk/placenta/antitoxin injection [1]
Question 3 · structured
13 marks
A student investigated the effect of temperature on the rate of starch breakdown by the enzyme amylase. The results are shown in Table 3.1.

Table 3.1
| Temperature / \(^\circ C\) | Rate of starch breakdown / arbitrary units |
| :---: | :---: |
| 10 | 1.2 |
| 20 | 3.5 |
| 30 | 7.8 |
| 40 | 12.4 |
| 50 | 4.1 |
| 60 | 0.0 |

(a) Describe and explain the trend shown by the data from \(10^\circ C\) to \(40^\circ C\). [4]

(b) Explain the results obtained at \(50^\circ C\) and \(60^\circ C\). [4]

(c) Suggest how the student could modify the experiment to obtain a more precise estimate of the optimum temperature for this amylase. [2]

(d) State the names of the substrate and product of the reaction catalysed by amylase, and name the reagent used to test for the presence of the product. [3]
Show answer & marking scheme

Worked solution

(a) As the temperature increases from \(10^\circ C\) to \(40^\circ C\), the rate of starch breakdown increases from \(1.2\) to \(12.4\) arbitrary units. This is because higher temperatures give the enzyme and starch molecules more kinetic energy. They move faster, leading to more frequent successful collisions, which results in a higher rate of enzyme-substrate complex formation.
(b) Above \(40^\circ C\), the rate of reaction drops sharply to \(4.1\) at \(50^\circ C\) and reaches \(0.0\) at \(60^\circ C\). This is due to denaturation of the amylase enzyme. High temperatures break the bonds holding the enzyme's three-dimensional structure together, changing the shape of the active site. The starch substrate can no longer fit into the active site. At \(60^\circ C\), all enzyme molecules are completely denatured, preventing any reaction from occurring.
(c) To find a more precise optimum temperature, the student should test at smaller temperature intervals (e.g., every \(2^\circ C\)) in the range between \(30^\circ C\) and \(50^\circ C\).
(d) Amylase catalyses the breakdown of starch (substrate) into maltose (product). The reagent used to test for maltose (a reducing sugar) is Benedict's reagent.

Marking scheme

(a)
- Description: Rate of starch breakdown increases as temperature increases (from 10 to 40 °C) [1]
- Explanation: Kinetic energy of molecules increases / molecules move faster [1]
- More frequent collisions (between amylase and starch) [1]
- More enzyme-substrate complexes formed per unit time [1]

(b)
- At 50/60 °C, the rate decreases / drops to zero [1]
- High temperature denatures the enzyme / amylase [1]
- Shape of the active site changes / is destroyed [1]
- Substrate (starch) can no longer fit / bind [1]

(c)
- Test more temperatures / smaller temperature intervals [1]
- Between 30 °C and 50 °C [1]

(d)
- Substrate: starch [1]
- Product: maltose (accept glucose/reducing sugar) [1]
- Reagent: Benedict's (reagent/solution) [1]
Question 4 · structured
14 marks
Albinism is an inherited condition in rabbits where they lack pigment in their fur and eyes. It is caused by a recessive allele, \(a\). The dominant allele, \(A\), results in normal pigmentation.

(a) Define the terms:
(i) genotype [1]
(ii) homozygous [1]

(b) Two rabbits with normal pigmentation were mated. They produced a litter of seven offspring: five had normal pigmentation and two were albino.
(i) State the genotypes of the parents. Explain your choice. [3]
(ii) Complete a genetic diagram to show the possible genotypes and phenotypes of the offspring from this cross. [5]
(iii) Calculate the probability that the next offspring produced by these parents will be a male rabbit with normal pigmentation. Show your working. [4]
Show answer & marking scheme

Worked solution

(a) (i) Genotype is defined as the genetic makeup of an organism in terms of the alleles present.
(a) (ii) Homozygous means having two identical alleles of a particular gene.
(b) (i) Both parents must have the genotype \(Aa\). Since they have normal pigmentation, they must possess at least one dominant allele (\(A\)). However, because they produced albino offspring (which must be homozygous recessive, \(aa\)), each parent must have contributed a recessive allele (\(a\)) to those offspring, meaning both parents must carry the recessive allele.
(b) (ii)
- Parental Phenotypes: Normal \(\times\) Normal
- Parental Genotypes: \(Aa\) \(\times\) \(Aa\)
- Gametes: \(A\) or \(a\) from each parent
- Offspring Genotypes: \(AA\), \(Aa\), \(Aa\), \(aa\)
- Offspring Phenotypes: Normal pigmentation (for \(AA\) and \(Aa\)) and Albino (for \(aa\))
- Phenotypic Ratio: \(3\) normal : \(1\) albino
(b) (iii) The probability of an offspring having normal pigmentation is \(\frac{3}{4}\) (or \(0.75\)). The probability of an offspring being male is \(\frac{1}{2}\) (or \(0.5\)). The combined probability is calculated by multiplying these independent probabilities: \(\frac{3}{4} \times \frac{1}{2} = \frac{3}{8}\) (or \(0.375\) / \(37.5\%\)).

Marking scheme

(a) (i)
- Genetic makeup of an organism in terms of the alleles present [1]

(a) (ii)
- Having two identical alleles of a particular gene [1]

(b) (i)
- Parental genotypes: \(Aa\) and \(Aa\) (or heterozygous) [1]
- Explanation: Parents must have normal phenotype so they carry the dominant \(A\) allele [1]
- Explanation: They produced albino offspring (\(aa\)), so each parent must have passed on a recessive \(a\) allele [1]

(b) (ii)
- Correct parental genotypes: \(Aa \times Aa\) [1]
- Correct gametes shown: \(A\) and \(a\) [1]
- Correct offspring genotypes: \(AA\), \(Aa\), \(Aa\), \(aa\) [1]
- Correct offspring phenotypes matched to genotypes: \(AA/Aa\) = normal, \(aa\) = albino [1]
- Correct ratio: \(3\) normal : \(1\) albino (or \(75\%\) to \(25\%\)) [1]

(b) (iii)
- Probability of normal pigmentation: \(\frac{3}{4}\) / \(0.75\) [1]
- Probability of being male: \(\frac{1}{2}\) / \(0.5\) [1]
- Multiplication of probabilities: \(\frac{3}{4} \times \frac{1}{2}\) [1]
- Final correct answer: \(\frac{3}{8}\) / \(0.375\) / \(37.5\%\) [1]
Question 5 · structured
14 marks
Human growth hormone (HGH) can be produced on an industrial scale using genetically modified bacteria grown in fermenters.

(a) Describe how a gene, such as the one for HGH, is isolated from a human cell and inserted into a bacterial plasmid. [6]

(b) State three reasons why bacteria are useful organisms for biotechnology and genetic modification. [3]

(c) Fermenters are used to grow the genetically modified bacteria. Explain the importance of maintaining the following conditions inside a fermenter:
(i) a constant temperature [2]
(ii) constant agitation (stirring) of the mixture [3]
Show answer & marking scheme

Worked solution

(a) To isolate the human gene for HGH, DNA is extracted from a human cell, and restriction enzymes are used to cut the DNA at specific base sequences, leaving single-stranded overhangs called 'sticky ends'. A bacterial plasmid is cut using the exact same restriction enzyme to produce complementary sticky ends. The human HGH gene and the cut plasmid are mixed together, and the enzyme DNA ligase is used to join their sticky ends together, forming a recombinant plasmid. This recombinant plasmid is then inserted back into a bacterial cell.
(b) Bacteria are useful in biotechnology because: they have a very rapid reproduction rate, allowing fast production of molecules; they contain small loops of DNA called plasmids which are easy to extract and manipulate; there are fewer ethical concerns compared to using animals; and they share the same genetic code as humans, allowing human proteins to be made correctly.
(c) (i) Maintaining a constant temperature is vital because bacteria grow best at their optimum temperature. If the temperature is too low, growth/metabolism is slow; if it is too high, the bacterial enzymes will denature, which kills the bacteria.
(c) (ii) Agitation (stirring) is necessary because it ensures an even distribution of nutrients (like glucose and amino acids), oxygen, and temperature throughout the fermenter. It also prevents the bacterial cells from settling at the bottom, maximizing their access to resources for growth.

Marking scheme

(a)
- Identify/isolate HGH gene [1]
- Cut human DNA using restriction enzymes [1]
- Leaves sticky ends (single-stranded overhangs) [1]
- Cut plasmid DNA with the same restriction enzyme [1]
- Sticky ends of plasmid and HGH gene are complementary [1]
- Join HGH gene and plasmid using DNA ligase [1]
- Forms a recombinant plasmid [1]
(Max 6 marks)

(b)
Any three from:
- Rapid reproduction rate [1]
- Presence of plasmids (easy to manipulate) [1]
- Few/no ethical concerns [1]
- Share same genetic code as humans [1]
- Ability to make complex molecules [1]

(c) (i)
- Maintain optimum temperature for bacterial enzymes/growth [1]
- Prevent denaturation of enzymes at high temperatures / slow metabolic rates at low temperatures [1]

(c) (ii)
- Even distribution of nutrients / oxygen / heat [1]
- Prevents settling of bacteria at the bottom [1]
- Maintains constant contact between bacteria and nutrients/oxygen [1]
Question 6 · structured
13 marks
Nitrogen is an essential element for all living organisms. Four major processes in the soil nitrogen cycle are described below:
- Process P: Atmospheric nitrogen gas is converted directly to ammonium ions in the soil.
- Process Q: Ammonium ions are converted to nitrate ions.
- Process R: Nitrate ions are absorbed by plant roots.
- Process S: Nitrate ions are converted back to atmospheric nitrogen gas.

(a) (i) Identify the processes P, Q, and S. [3]
(a) (ii) State the group of organisms responsible for carrying out process Q. [1]

(b) Explain how the conditions in waterlogged soils affect process S, and explain the consequences of this on plant growth. [4]

(c) Describe how nitrogen from plant proteins is recycled back into ammonium ions in the soil when a plant dies. [5]
Show answer & marking scheme

Worked solution

(a) (i) Process P is nitrogen fixation. Process Q is nitrification. Process S is denitrification.
(a) (ii) The organisms responsible for process Q are nitrifying bacteria.
(b) Waterlogged soils lack air spaces and therefore have anaerobic conditions (low oxygen concentration). These anaerobic conditions favor the activity of denitrifying bacteria, which carry out process S (denitrification). This converts soil nitrates back into nitrogen gas, leaving less nitrate in the soil for plants to absorb. Since plants need nitrates to make amino acids and proteins, a lack of soil nitrates leads to stunted plant growth.
(c) When a plant dies, decomposers (bacteria and fungi in the soil) break down its organic matter. They secrete extracellular enzymes (such as proteases) to digest plant proteins into amino acids. The decomposers absorb these amino acids. During their metabolism, they carry out deamination/breakdown of excess amino acids, releasing nitrogen as ammonia. This ammonia reacts with water in the soil to form ammonium ions, returning the nitrogen to the soil.

Marking scheme

(a) (i)
- P: nitrogen fixation [1]
- Q: nitrification [1]
- S: denitrification [1]

(a) (ii)
- Nitrifying bacteria [1]

(b)
- Waterlogged soil is anaerobic / lacks oxygen [1]
- Favors/increases denitrification (process S) / denitrifying bacteria [1]
- Decreases nitrate concentration in the soil [1]
- Plants have fewer nitrates to absorb, leading to reduced protein synthesis / stunted growth [1]

(c)
- Decomposers / saprotrophs / bacteria / fungi break down dead plant matter [1]
- Secrete proteases / extracellular digestive enzymes [1]
- Proteins are digested/broken down to amino acids [1]
- Amino acids are absorbed and broken down / deaminated [1]
- Ammonia / ammonium ions are released into the soil [1]

Paper 6 (Alternative to Practical)

Answer all questions. Write your answers in the spaces provided. Show all your working and use appropriate units.
3 Question · 39.900000000000006 marks
Question 1 · practical
13.3 marks
A student investigated the effect of temperature on the rate of lipase activity. Lipase breaks down fats in milk into fatty acids and glycerol. Phenolphthalein indicator was added to a mixture of milk and sodium carbonate, turning the solution pink (alkaline pH). As fatty acids are produced, the pH decreases, and the mixture turns from pink to colorless.

Six test-tubes were set up with equal volumes of milk, sodium carbonate solution, and phenolphthalein. Lipase solution was heated separately to the required temperatures before being mixed with the milk solutions. The time taken for the pink color to completely disappear was recorded.

The results are shown in Table 1.1.

**Table 1.1**
| Temperature / °C | Time taken for pink color to disappear / seconds |
| :--- | :--- |
| 10 | 480 |
| 20 | 240 |
| 30 | 120 |
| 40 | 60 |
| 50 | 180 |
| 60 | No change after 600 |

(a) (i) State the independent variable and the dependent variable in this investigation. [2]

(ii) Identify two variables that should be controlled in this investigation. [2]

(b) (i) Describe how the student could maintain the test-tubes at the required temperatures of 10°C, 30°C, and 50°C during the experiment. [1.3]

(ii) Suggest why the student kept the lipase solution and the milk solution in the water baths for 5 minutes *before* mixing them. [1]

(c) (i) Plot a line graph of the data in Table 1.1 on the grid. (Note: For the 60°C data point, assume no reaction occurred and do not plot it or plot it as a point above 600 s). [4]

(ii) Use your graph to estimate the time taken for the pink color to disappear at 35°C. Show on your graph how you obtained your answer. [2]

(d) State the conclusion for this experiment based on the results from 10°C to 40°C. [1]
Show answer & marking scheme

Worked solution

(a) (i) Independent variable: Temperature. Dependent variable: Time taken for the pink color to disappear.
(ii) Concentration of lipase, volume of lipase, volume of milk, volume of sodium carbonate solution.
(b) (i) Use thermostatically controlled water baths at each temperature, or beakers of water with ice (for 10°C) and hot water added periodically to maintain temperature.
(ii) To ensure that both solutions reached the desired testing temperature before the enzyme-catalyzed reaction started.
(c) (i) A line graph with Temperature / °C on the x-axis (scale 0 to 60) and Time taken / seconds on the y-axis (scale 0 to 500). Points plotted accurately: (10, 480), (20, 240), (30, 120), (40, 60), (50, 180).
(ii) Reading from the graph at 35°C should yield approximately 90 seconds (allow 85–95 depending on curve construction). A horizontal and vertical construction line must be shown starting from 35°C on the x-axis to the curve, and then to the y-axis.
(d) As temperature increases from 10°C to 40°C, the rate of lipase activity increases (or the time taken for the pink color to disappear decreases).

Marking scheme

**(a) (i)**
- Independent variable: Temperature [1]
- Dependent variable: Time taken / rate of reaction / color change time [1]

**(a) (ii)**
- Any two from: volume of milk / volume of sodium carbonate / volume of lipase / concentration of lipase [2]

**(b) (i)**
- Use of thermostatically controlled water baths / water baths with ice or hot water monitored with a thermometer [1.3]

**(b) (ii)**
- To equilibrate the solutions / ensure solutions reach the correct temperature before mixing [1]

**(c) (i)**
- Axes labeled correctly with units: Temperature / °C on x-axis, Time / s on y-axis [1]
- Suitable linear scale, with data occupying at least half the grid [1]
- Points plotted accurately within half a small square [1]
- Smooth curve of best fit or ruled straight lines between points [1]

**(c) (ii)**
- Reading of approximately 85–95 seconds (or matching candidate's graph) [1]
- Construction lines shown on the graph [1]

**(d)**
- As temperature increases (up to 40°C), rate of activity increases / time taken decreases [1]
Question 2 · practical
13.3 marks
A student investigated the effect of different concentrations of sodium chloride (salt) solution on the mass of sweet potato cylinders.

Five sweet potato cylinders of identical length were cut using a cork borer. The initial mass of each cylinder was recorded. Each cylinder was then placed into a test-tube containing a different concentration of sodium chloride solution. After 45 minutes, the cylinders were removed, gently dried, and weighed again.

The results are shown in Table 2.1.

**Table 2.1**
| Concentration of sodium chloride solution / % | Initial mass / g | Final mass / g | Change in mass / g | Percentage change in mass / % |
| :--- | :--- | :--- | :--- | :--- |
| 0.0 | 3.50 | 3.85 | +0.35 | +10.0 |
| 1.0 | 3.48 | 3.48 | 0.00 | 0.0 |
| 2.0 | 3.52 | 3.31 | -0.21 | -6.0 |
| 3.0 | 3.51 | 3.12 | -0.39 | **[Calculation 1]** |
| 4.0 | 3.49 | 2.97 | -0.52 | **[Calculation 2]** |

(a) (i) Calculate the missing percentage change in mass for the 3.0% and 4.0% sodium chloride solutions. Show your working and write your answers to one decimal place. [2]

(ii) Explain why calculating the percentage change in mass is more useful than just comparing the change in mass in grams. [1.3]

(b) Describe the trend shown by the sweet potato cylinders in the different concentrations of sodium chloride solution. [2]

(c) (i) Identify one potential source of error in the step where the cylinders are dried, and suggest an improvement to reduce this error. [2]

(ii) State two safety precautions the student should take when cutting the sweet potato cylinders. [2]

(d) Suggest why using 5 sweet potato cylinders in each solution instead of 1 would improve the reliability of the results. [1]

(e) Use the results in Table 2.1 to estimate the concentration of sodium chloride inside the sweet potato cells. Explain your reasoning. [3]
Show answer & marking scheme

Worked solution

(a) (i) For 3.0%: \((-0.39 / 3.51) \times 100 = -11.11\%\), rounded to -11.1%.
For 4.0%: \((-0.52 / 3.49) \times 100 = -14.90\%\), rounded to -14.9%.
(ii) Because the sweet potato cylinders did not all have the same starting mass, calculating percentage change allows for a fair comparison.
(b) As the concentration of sodium chloride solution increases, the percentage change in mass decreases (becomes more negative / changes from a gain in mass to a loss in mass).
(c) (i) Error: Cylinders may be dried inconsistently / too much water left on or too much water squeezed out. Improvement: Use a standardized drying method, such as rolling each cylinder on a paper towel exactly three times with minimal pressure.
(ii) Always cut away from your hands / body on a flat chopping tile; wear safety goggles to prevent injury from flying debris.
(d) It allows the calculation of a mean, which reduces the impact of anomalous results.
(e) The concentration inside the cells is approximately 1.0%. Reason: At 1.0% sodium chloride solution, there is no change in mass (0.0%), meaning there is no net movement of water by osmosis because the solution is isotonic to the cell sap.

Marking scheme

**(a) (i)**
- Calculation 1 (3.0%): -11.1 (%) [1]
- Calculation 2 (4.0%): -14.9 (%) [1]
*(Note: deduct 1 mark if negative sign is missing)*

**(a) (ii)**
- Cylinders had different initial masses / starting masses [1]
- Allows a fair comparison to be made [0.3]

**(b)**
- As salt concentration increases, mass change decreases / becomes more negative [1]
- Gains mass at 0.0% but loses mass at concentrations above 1.0% [1]

**(c) (i)**
- Source of error: Drying is inconsistent / varies between cylinders [1]
- Improvement: Standardize the drying method (e.g., roll a set number of times / use identical blotting paper) [1]

**(c) (ii)**
- Cut on a hard tile / flat surface (not in the hand) [1]
- Cut away from fingers / body [1]

**(d)**
- Minimizes the effect of anomalous results / increases reliability / allows a mean to be calculated [1]

**(e)**
- 1.0% [1]
- There is no net movement of water / no osmosis [1]
- Because water potential inside and outside the cell is equal / isotonic [1]
Question 3 · practical
13.3 marks
A student investigated the antibacterial properties of garlic extract using the agar well diffusion method.

An agar plate was inoculated with a non-pathogenic bacterium. Five wells of equal diameter were cut into the agar. Different concentrations of garlic extract were placed into wells 1 to 5. After incubation, the diameter of the zone of inhibition (clear zone where bacteria did not grow) was measured around each well.

The experiment was repeated three times, and the results are shown in Table 3.1.

**Table 3.1**
| Concentration of garlic extract / % | Zone of inhibition diameter / mm | | | Mean diameter / mm |
| :--- | :--- | :--- | :--- | :--- |
| | Trial 1 | Trial 2 | Trial 3 | |
| 10 | 12 | 11 | 13 | 12.0 |
| 20 | 15 | 14 | 16 | 15.0 |
| 40 | 18 | 19 | 17 | **[Calculation 1]** |
| 60 | 22 | 21 | 23 | 22.0 |
| 80 | 25 | 24 | 26 | 25.0 |

(a) Calculate the missing mean diameter for the 40% concentration of garlic extract. [1]

(b) Plot a bar chart on the grid to show the mean diameter of the zone of inhibition for each concentration of garlic extract. [4]

(c) State one reason why the agar plates were incubated at 25°C rather than at human body temperature (37°C). [1.3]

(d) Explain why the student should include a control well containing only sterile distilled water, and what result they would expect to observe in this well. [2]

(e) Ginger is also thought to have antibacterial properties. Plan an investigation to determine whether ginger extract has a greater antibacterial effect than garlic extract. [5]
Show answer & marking scheme

Worked solution

(a) Mean for 40% = \((18 + 19 + 17) / 3 = 54 / 3 = 18.0\) mm.
(b) A bar chart with 'Concentration of garlic extract / %' on the x-axis and 'Mean diameter of zone of inhibition / mm' on the y-axis. The y-axis scale must be linear and appropriate. Five bars must be drawn with equal width, separated by equal gaps, and plotted at the correct heights: 12.0, 15.0, 18.0, 22.0, 25.0.
(c) To prevent the growth of potential human pathogens, which grow best at 37°C.
(d) The distilled water well acts as a negative control to show that the water itself (or the physical cutting of the well) does not inhibit bacterial growth. The expected result is a zone of 0 mm (no clear zone around the well).
(e) Plan:
1. Prepare extracts of ginger and garlic at the same concentration (e.g., 50%).
2. Inoculate an agar plate with the same bacterial species.
3. Cut wells of equal diameter into the agar.
4. Add equal volumes of garlic extract into one well and ginger extract into another.
5. Include a control well with distilled water.
6. Incubate plates at the same temperature (e.g., 25°C) for the same duration (e.g., 24–48 hours).
7. Measure the diameter of the zone of inhibition for both extracts.
8. Repeat the experiment at least three times to calculate a mean.

Marking scheme

**(a)**
- 18.0 (mm) [1]

**(b)**
- Axes labeled correctly with units: Concentration of garlic extract / % on x-axis, Mean diameter / mm on y-axis [1]
- Linear scale on y-axis, occupying at least half the grid [1]
- Five separate bars of equal width, plotted accurately with gaps between them [1]
- All bars plotted accurately to within half a small square [1]

**(c)**
- To prevent / minimize growth of pathogens / harmful microorganisms (which grow best at human body temperature / 37°C) [1.3]

**(d)**
- To prove that the solvent / water does not cause inhibition of bacterial growth [1]
- Expected result: No zone of inhibition / 0 mm / bacteria grow right up to the edge of the well [1]

**(e)**
- Any five planning points from:
1. Use of same bacterial species to seed the agar [1]
2. Equal concentration of both ginger and garlic extracts [1]
3. Equal volumes of each extract placed into wells of equal size [1]
4. Constant incubation temperature (e.g., 25°C) and time (e.g., 24 hours) [1]
5. Measure and compare the zone of inhibition diameter [1]
6. Repeat experiment at least 3 times (and calculate mean) [1]
7. Relevant safety point (e.g., aseptic technique / washing hands / wearing gloves) [1]
*(max 5 marks)*

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