Cambridge IGCSE · thinka-original Practice Paper

2023 Cambridge IGCSE Biology (0610) Practice Paper with Answers

Thinka Jun 2023 (V3) Cambridge IGCSE-Style Mock — Biology (0610)

160 marks180 mins2023
An original Thinka practice paper modelled on the structure and difficulty of the Jun 2023 (V3) Cambridge IGCSE Biology (0610) paper. Not affiliated with or reproduced from Cambridge.

Paper 23

Answer all 40 multiple-choice questions.
40 Question · 40 marks
Question 1 · multiple-choice
1 marks
The table shows the initial and final mass of four potato chips placed in four sucrose solutions of different concentrations.

$$\begin{array}{|c|c|c|} \hline \text{potato chip} & \text{initial mass / g} & \text{final mass / g} \\ \hline 1 & 2.5 & 2.8 \\ \hline 2 & 2.5 & 2.5 \\ \hline 3 & 2.5 & 2.2 \\ \hline 4 & 2.5 & 2.0 \\ \hline \end{array}$$

Which potato chip was placed in a solution with a water potential equal to that of the potato cells?
  1. A.1
  2. B.2
  3. C.3
  4. D.4
Show answer & marking scheme

Worked solution

Potato chip 2 did not change in mass (initial mass = 2.5 g, final mass = 2.5 g). This indicates that there was no net movement of water into or out of the potato cells by osmosis, meaning the water potential of the surrounding solution was equal to that of the potato cells.

Marking scheme

Award 1 mark for selecting B.
Question 2 · multiple-choice
1 marks
A student uses a potometer to measure the rate of water uptake of a leafy shoot. Under which combination of environmental conditions will the rate of water uptake be the slowest?
  1. A.temperature: low | humidity: high | wind speed: low
  2. B.temperature: low | humidity: low | wind speed: high
  3. C.temperature: high | humidity: high | wind speed: high
  4. D.temperature: high | humidity: low | wind speed: low
Show answer & marking scheme

Worked solution

Transpiration rate decreases when temperature is low (less kinetic energy of water molecules), humidity is high (gentler water potential gradient between the leaf and the air), and wind speed is low (water vapour is not blown away from the stomata, maintaining a high humidity around the leaf). Thus, the slowest rate of water uptake occurs under low temperature, high humidity, and low wind speed.

Marking scheme

Award 1 mark for selecting A.
Question 3 · multiple-choice
1 marks
Which row correctly describes the state of the heart valves and the direction of blood flow when the ventricles contract?
  1. A.atrioventricular valves: closed | semilunar valves: open | direction of blood flow: ventricles to arteries
  2. B.atrioventricular valves: closed | semilunar valves: open | direction of blood flow: atria to ventricles
  3. C.atrioventricular valves: open | semilunar valves: closed | direction of blood flow: ventricles to arteries
  4. D.atrioventricular valves: open | semilunar valves: closed | direction of blood flow: veins to atria
Show answer & marking scheme

Worked solution

When the ventricles contract, pressure inside the ventricles increases. This causes the atrioventricular valves to close, preventing blood from flowing back into the atria, and forces the semilunar valves to open, allowing blood to flow from the ventricles into the arteries (pulmonary artery and aorta).

Marking scheme

Award 1 mark for selecting A.
Question 4 · multiple-choice
1 marks
A healthy person's glomerular filtrate contains glucose, but their urine does not. Which process is responsible for this difference?
  1. A.active transport in the ureter
  2. B.filtration in the glomerulus
  3. C.selective reabsorption in the kidney tubule
  4. D.deamination in the liver
Show answer & marking scheme

Worked solution

Glomerular filtration allows small molecules like glucose to pass from the blood into the kidney tubule. In a healthy kidney, all of this glucose is selectively reabsorbed back into the blood capillaries from the kidney tubule by active transport, so none is left in the urine.

Marking scheme

Award 1 mark for selecting C.
Question 5 · multiple-choice
1 marks
Which row correctly matches the biological molecule with its monomer subunits and the positive result for its diagnostic test?
  1. A.biological molecule: protein | monomer subunits: amino acids | test reagent and positive result: biuret test turns purple
  2. B.biological molecule: starch | monomer subunits: glucose | test reagent and positive result: Benedict's solution turns blue
  3. C.biological molecule: glycogen | monomer subunits: fatty acids and glycerol | test reagent and positive result: iodine solution turns blue-black
  4. D.biological molecule: DNA | monomer subunits: nucleotides | test reagent and positive result: ethanol emulsion test turns cloudy white
Show answer & marking scheme

Worked solution

Proteins are large biological molecules made up of amino acid subunits. The diagnostic test for proteins is the biuret test, which changes colour from blue to purple/violet in the presence of protein.

Marking scheme

Award 1 mark for selecting A.
Question 6 · multiple-choice
1 marks
In humans, the allele for wet earwax ($$W$$) is dominant to the allele for dry earwax ($$w$$). A heterozygous father and a homozygous recessive mother have a child. What is the probability that their child will have dry earwax?
  1. A.0%
  2. B.25%
  3. C.50%
  4. D.75%
Show answer & marking scheme

Worked solution

Identify the genotypes of the parents:
- Heterozygous father: $$Ww$$
- Homozygous recessive mother: $$ww$$

Perform the genetic cross:
- Father's gametes: $$W$$ and $$w$$
- Mother's gametes: all $$w$$
- Offspring genotypes: $50\%$ $$Ww$$ (wet earwax) and $50\%$ $$ww$$ (dry earwax).

Therefore, the probability of the child having dry earwax is $50\%$.

Marking scheme

Award 1 mark for selecting C.
Question 7 · multiple-choice
1 marks
Consider the following food chain:

$$\text{phytoplankton} \rightarrow \text{zooplankton} \rightarrow \text{small fish} \rightarrow \text{herring} \rightarrow \text{seal}$$

Which organism represents both a secondary consumer and is at the third trophic level?
  1. A.zooplankton
  2. B.small fish
  3. C.herring
  4. D.seal
Show answer & marking scheme

Worked solution

Let's analyze the level of each organism:
- Phytoplankton: Producer (1st trophic level)
- Zooplankton: Primary consumer (2nd trophic level)
- Small fish: Secondary consumer (3rd trophic level)
- Herring: Tertiary consumer (4th trophic level)
- Seal: Quaternary consumer (5th trophic level)

Therefore, the small fish is the secondary consumer and is at the third trophic level.

Marking scheme

Award 1 mark for selecting B.
Question 8 · multiple-choice
1 marks
Which statement correctly describes the effect of pH on enzyme activity?
  1. A.Increasing pH always increases the rate of reaction.
  2. B.Extreme pH values alter the shape of the active site, denaturing the enzyme.
  3. C.All human enzymes have an optimum pH of 7.
  4. D.Low pH values increase the activation energy of the reaction.
Show answer & marking scheme

Worked solution

Enzymes are proteins with a specific three-dimensional shape, including an active site. Extreme pH values (both too acidic or too alkaline relative to the enzyme's optimum pH) can disrupt the chemical bonds holding the protein's shape, changing the shape of the active site so that the substrate can no longer bind. This process is called denaturation.

Marking scheme

Award 1 mark for selecting B.
Question 9 · multiple-choice
1 marks
Which row correctly describes the features of active transport?
  1. A.Requires energy: Yes, Direction of movement: Down a concentration gradient, Uses carrier proteins: No
  2. B.Requires energy: Yes, Direction of movement: Against a concentration gradient, Uses carrier proteins: Yes
  3. C.Requires energy: No, Direction of movement: Against a concentration gradient, Uses carrier proteins: Yes
  4. D.Requires energy: No, Direction of movement: Down a concentration gradient, Uses carrier proteins: No
Show answer & marking scheme

Worked solution

Active transport requires energy released from respiration to move molecules or ions against a concentration gradient (from a region of lower concentration to a region of higher concentration) across a cell membrane, utilizing specific carrier proteins.

Marking scheme

Award 1 mark for the correct option B.
Question 10 · multiple-choice
1 marks
Which statement correctly explains why the wall of the left ventricle of the human heart is thicker than the wall of the right ventricle?
  1. A.The left ventricle contains a larger volume of blood than the right ventricle.
  2. B.The left ventricle must contract with more force to pump blood a longer distance around the body.
  3. C.The left ventricle receives blood under very high pressure directly from the pulmonary veins.
  4. D.The left ventricle must pump blood under higher pressure to the lungs to maximize gas exchange.
Show answer & marking scheme

Worked solution

The left ventricle pumps blood through the aorta to the rest of the body (systemic circulation), which is a much longer distance than the distance to the lungs. Therefore, it requires a thicker muscular wall to contract with greater force and generate higher pressure.

Marking scheme

Award 1 mark for the correct option B.
Question 11 · multiple-choice
1 marks
Under which combination of environmental conditions will the rate of transpiration in a plant be lowest?
  1. A.high humidity, low wind speed, low temperature
  2. B.high humidity, high wind speed, high temperature
  3. C.low humidity, low wind speed, high temperature
  4. D.low humidity, high wind speed, low temperature
Show answer & marking scheme

Worked solution

High humidity reduces the concentration gradient of water vapour between the inside and outside of the leaf. Low wind speed allows water vapour to accumulate near the stomata, further reducing diffusion. Low temperature decreases the kinetic energy of water molecules, slowing evaporation.

Marking scheme

Award 1 mark for the correct option A.
Question 12 · multiple-choice
1 marks
Where in the human body is urea produced, and what is the source material from which it is formed?
  1. A.produced in the kidney, source material is excess glucose
  2. B.produced in the kidney, source material is excess amino acids
  3. C.produced in the liver, source material is excess glucose
  4. D.produced in the liver, source material is excess amino acids
Show answer & marking scheme

Worked solution

Urea is produced in the liver during the process of deamination, which involves the removal of the nitrogen-containing part of excess amino acids.

Marking scheme

Award 1 mark for the correct option D.
Question 13 · multiple-choice
1 marks
A mother has blood group A and genotype \(I^A I^O\). The father has blood group B and genotype \(I^B I^O\). What is the probability that their first child will have blood group O?
  1. A.0%
  2. B.25%
  3. C.50%
  4. D.75%
Show answer & marking scheme

Worked solution

The possible alleles from the mother are \(I^A\) and \(I^O\), and from the father are \(I^B\) and \(I^O\). Crossing these genotypes results in the 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). The probability of genotype \(I^O I^O\) is 1 out of 4, which is 25%.

Marking scheme

Award 1 mark for the correct option B.
Question 14 · multiple-choice
1 marks
Which statement describes a major advantage of conserving plant species in a seed bank rather than as grown plants in a botanical garden?
  1. A.Seeds are kept active so they can continuously photosynthesise.
  2. B.Large quantities of genetic material can be stored in a very small space.
  3. C.Plants are kept in their natural ecological habitats to allow natural selection.
  4. D.Pest species can be introduced to test the seeds' natural resistance.
Show answer & marking scheme

Worked solution

Seed banks keep seeds dormant under dry and freezing conditions. This allows a vast number of individuals and species to be preserved compactly in a very small space, conserving significant genetic diversity cost-effectively.

Marking scheme

Award 1 mark for the correct option B.
Question 15 · multiple-choice
1 marks
How does increasing the temperature from 20 °C to 35 °C affect an enzyme-catalysed reaction?
  1. A.The kinetic energy of the enzyme and substrate molecules decreases, reducing the rate of reaction.
  2. B.The shape of the active site changes permanently, preventing substrate binding.
  3. C.The frequency of effective collisions between enzyme and substrate molecules increases, increasing the rate of reaction.
  4. D.The enzyme molecules are completely denatured, stopping the reaction.
Show answer & marking scheme

Worked solution

Increasing the temperature increases the kinetic energy of the enzyme and substrate molecules. They move faster and collide more frequently and with more energy, leading to an increased rate of reaction up to the enzyme's optimum temperature.

Marking scheme

Award 1 mark for the correct option C.
Question 16 · multiple-choice
1 marks
In the genetic modification of bacteria to produce human insulin, what is inserted into the bacterial cells to produce the hormone?
  1. A.recombinant plasmids containing the human insulin gene
  2. B.human ribosomes that translate insulin mRNA
  3. C.the entire nucleus of a human pancreatic cell
  4. D.active insulin enzymes extracted from human blood
Show answer & marking scheme

Worked solution

During genetic modification, the human insulin gene is joined with a bacterial plasmid to form a recombinant plasmid. This recombinant plasmid is then inserted into the bacterial cells, which then express the gene to produce human insulin.

Marking scheme

Award 1 mark for the correct option A.
Question 17 · multiple-choice
1 marks
An epidermal strip from a red onion leaf is placed in a concentrated sucrose solution. Which observation correctly describes the appearance of the cells after 30 minutes?
  1. A.The vacuole increases in volume and the cell membrane is pressed firmly against the cell wall.
  2. B.The vacuole decreases in volume and the cell membrane pulls away from the cell wall.
  3. C.Water moves into the vacuole by active transport, causing the cell to burst.
  4. D.Solute particles move into the cell, causing the cytoplasm to swell .
Show answer & marking scheme

Worked solution

Water leaves the onion cells by osmosis from a region of higher water potential (inside the cell) to a region of lower water potential (the concentrated sucrose solution). This causes plasmolysis, where the vacuole shrinks and the cell membrane pulls away from the cell wall.

Marking scheme

Correct option B is awarded 1 mark.
Question 18 · multiple-choice
1 marks
The table shows the time taken for a starch-amylase mixture to turn completely yellow-brown when tested with iodine solution at different temperatures. At which temperature is the amylase activity the highest? (10 degrees C: 15 minutes, 25 degrees C: 8 minutes, 40 degrees C: 2 minutes, 55 degrees C: no change after 30 minutes)
  1. A.10 degrees C
  2. B.25 degrees C
  3. C.40 degrees C
  4. D.55 degrees C
Show answer & marking scheme

Worked solution

At 40 degrees C, the time taken for starch to be completely digested (so that it no longer gives a blue-black color with iodine) is the shortest (2 minutes), which indicates the highest rate of amylase activity. At 55 degrees C, the enzyme has denatured, so starch remains undigested.

Marking scheme

Correct option C is awarded 1 mark.
Question 19 · multiple-choice
1 marks
A potometer is used to measure the rate of water uptake in a leafy shoot. Which set of environmental conditions will produce the lowest rate of transpiration?
  1. A.high wind speed, low humidity, high temperature
  2. B.low wind speed, high humidity, low temperature
  3. C.low wind speed, low humidity, high temperature
  4. D.high wind speed, high humidity, low temperature
Show answer & marking scheme

Worked solution

High humidity decreases the water potential gradient between the air spaces inside the leaf and the surrounding atmosphere. Low wind speed allows water vapor to accumulate near the stomata, further reducing this gradient, and low temperature reduces the kinetic energy of water molecules, slowing down evaporation.

Marking scheme

Correct option B is awarded 1 mark.
Question 20 · multiple-choice
1 marks
Which statement correctly explains why the muscle wall of the left ventricle is thicker than the muscle wall of the right ventricle?
  1. A.The left ventricle must pump a larger volume of blood with each heartbeat.
  2. B.The left ventricle pumps blood directly to the lungs under high pressure.
  3. C.The left ventricle pumps blood to the systemic circulation, requiring greater pressure to overcome higher resistance.
  4. D.The left ventricle receives deoxygenated blood that is more viscous than oxygenated blood.
Show answer & marking scheme

Worked solution

The left ventricle pumps blood to the systemic circulation (around the entire body), which is a much longer distance with higher resistance than the pulmonary circulation (to the lungs) served by the right ventricle. Therefore, the left ventricle requires a thicker muscle wall to generate the higher pressures needed.

Marking scheme

Correct option C is awarded 1 mark.
Question 21 · multiple-choice
1 marks
Which row correctly identifies the process that produces urea, the organ where it is produced, and the organ where it is excreted from the body?
  1. A.assimilation / kidney / liver
  2. B.deamination / liver / kidney
  3. C.deamination / kidney / bladder
  4. D.filtration / liver / kidney
Show answer & marking scheme

Worked solution

Urea is produced during the process of deamination, which occurs in the liver when excess amino acids are broken down. The urea is then transported in the blood plasma to the kidneys, where it is filtered out and excreted in urine.

Marking scheme

Correct option B is awarded 1 mark.
Question 22 · multiple-choice
1 marks
A liquid food sample is tested. It gives a blue color with Benedict’s reagent after heating, a blue-black color with iodine solution, and a purple color with Biuret reagent. What does the sample contain?
  1. A.reducing sugar and protein, but no starch
  2. B.starch and protein, but no reducing sugar
  3. C.reducing sugar and starch, but no protein
  4. D.reducing sugar only
Show answer & marking scheme

Worked solution

The Benedict’s test remains blue, meaning no reducing sugar is present. The iodine test turns blue-black, indicating the presence of starch. The Biuret test turns purple, indicating the presence of protein. Thus, the sample contains starch and protein, but no reducing sugar.

Marking scheme

Correct option B is awarded 1 mark.
Question 23 · multiple-choice
1 marks
In pea plants, the allele for tall height (T) is dominant to the allele for dwarf height (t). If a heterozygous tall plant is crossed with a dwarf plant, what is the probability that an offspring will be dwarf?
  1. A.0%
  2. B.25%
  3. C.50%
  4. D.75%
Show answer & marking scheme

Worked solution

The heterozygous tall plant has the genotype Tt. The dwarf plant has the genotype tt. Crossing Tt with tt produces offspring with genotypes Tt (tall) and tt (dwarf) in a 1:1 ratio. Therefore, the probability of obtaining a dwarf offspring is 50% (0.5).

Marking scheme

Correct option C is awarded 1 mark.
Question 24 · multiple-choice
1 marks
Consider this food chain: microscopic algae -> water flea -> small fish -> kingfisher. Which organism represents the secondary consumer?
  1. A.microscopic algae
  2. B.water flea
  3. C.small fish
  4. D.kingfisher
Show answer & marking scheme

Worked solution

Microscopic algae are the producers. The water flea is the primary consumer because it feeds on the producer. The small fish is the secondary consumer because it feeds on the primary consumer.

Marking scheme

Correct option C is awarded 1 mark.
Question 25 · multiple_choice
1 marks
The table shows some features of three processes by which substances move across cell membranes.

$$\begin{array}{|c|c|c|c|} \hline \text{process} & \begin{array}{c} \text{requires energy} \\ \text{from respiration} \end{array} & \begin{array}{c} \text{moves molecules against a} \\ \text{concentration gradient} \end{array} & \begin{array}{c} \text{movement of water} \\ \text{molecules only} \end{array} \\ \hline 1 & \text{no} & \text{no} & \text{no} \\ \hline 2 & \text{yes} & \text{yes} & \text{no} \\ \hline 3 & \text{no} & \text{no} & \text{yes} \\ \hline \end{array}$$

Which row correctly identifies processes 1, 2 and 3?
  1. A.1 = active transport, 2 = diffusion, 3 = osmosis
  2. B.1 = diffusion, 2 = active transport, 3 = osmosis
  3. C.1 = osmosis, 2 = active transport, 3 = diffusion
  4. D.1 = diffusion, 2 = osmosis, 3 = active transport
Show answer & marking scheme

Worked solution

Process 1 does not require energy, does not move against a concentration gradient, and is not limited to water. This is diffusion.
Process 2 requires energy and moves molecules against a concentration gradient. This is active transport.
Process 3 does not require energy, does not move against a concentration gradient, and involves only water molecules moving. This is osmosis.

Marking scheme

Award 1 mark for the correct option B.
Question 26 · multiple_choice
1 marks
Which statement about enzymes is correct?
  1. A.They are carbohydrate molecules that act as biological catalysts.
  2. B.They are used up in the reactions that they catalyse.
  3. C.Their rate of reaction decreases as the kinetic energy of the reactants increases up to the optimum temperature.
  4. D.Their active site has a shape that is complementary to a specific substrate.
Show answer & marking scheme

Worked solution

A is incorrect because enzymes are protein molecules, not carbohydrates.
B is incorrect because enzymes act as catalysts and are not used up in the reactions they catalyse.
C is incorrect because the rate of reaction increases (not decreases) as kinetic energy increases up to the optimum temperature.
D is correct because the active site of an enzyme is complementary in shape to a specific substrate molecule.

Marking scheme

Award 1 mark for the correct option D.
Question 27 · multiple_choice
1 marks
Why is the muscular wall of the left ventricle thicker than the muscular wall of the right ventricle in the human heart?
  1. A.The left ventricle contains more blood than the right ventricle.
  2. B.The left ventricle has to pump blood at a higher pressure to the lungs.
  3. C.The left ventricle has to pump blood a longer distance around the body.
  4. D.The left ventricle receives oxygenated blood from the vena cava.
Show answer & marking scheme

Worked solution

The left ventricle pumps blood around the entire body (systemic circulation), which is a much longer distance and requires higher pressure than pumping blood to the lungs (pulmonary circulation), which is done by the right ventricle.

Marking scheme

Award 1 mark for the correct option C.
Question 28 · multiple_choice
1 marks
Which row correctly describes active and passive immunity?

$$\begin{array}{|c|c|c|} \hline & \text{active immunity} & \text{passive immunity} \\ \hline \text{A} & \begin{array}{c} \text{produced by the injection of} \\ \text{ready-made antibodies, providing} \\ \text{long-term protection} \end{array} & \begin{array}{c} \text{produced by infection with a} \\ \text{pathogen, providing} \\ \text{short-term protection} \end{array} \\ \hline \text{B} & \begin{array}{c} \text{produced by infection with a} \\ \text{pathogen, providing} \\ \text{short-term protection} \end{array} & \begin{array}{c} \text{produced by the transfer of} \\ \text{antibodies from mother to baby,} \\ \text{providing long-term protection} \end{array} \\ \hline \text{C} & \begin{array}{c} \text{produced by a vaccine containing} \\ \text{weakened pathogens, providing} \\ \text{long-term protection} \end{array} & \begin{array}{c} \text{produced by the injection of} \\ \text{ready-made antibodies, providing} \\ \text{short-term protection} \end{array} \\ \hline \text{D} & \begin{array}{c} \text{produced by the transfer of} \\ \text{antibodies from mother to baby,} \\ \text{providing short-term protection} \end{array} & \begin{array}{c} \text{produced by a vaccine containing} \\ \text{weakened pathogens, providing} \\ \text{long-term protection} \end{array} \\ \hline \end{array}$$
  1. A.A
  2. B.B
  3. C.C
  4. D.D
Show answer & marking scheme

Worked solution

Active immunity is gained after an antigen enters the body (either via infection or vaccination), triggering an immune response where the body produces its own antibodies and memory cells, which results in long-term protection. Passive immunity is the short-term defense acquired by the transfer of ready-made antibodies into the body (such as from mother to fetus, or by injection), and it does not result in the production of memory cells.

Marking scheme

Award 1 mark for the correct option C.
Question 29 · multiple_choice
1 marks
What is the correct pathway taken by a urea molecule from the site of its production until it is excreted from the human body?
  1. A.kidney $\rightarrow$ renal vein $\rightarrow$ liver $\rightarrow$ ureter $\rightarrow$ bladder $\rightarrow$ urethra
  2. B.liver $\rightarrow$ renal artery $\rightarrow$ kidney $\rightarrow$ ureter $\rightarrow$ bladder $\rightarrow$ urethra
  3. C.liver $\rightarrow$ renal vein $\rightarrow$ kidney $\rightarrow$ urethra $\rightarrow$ bladder $\rightarrow$ ureter
  4. D.kidney $\rightarrow$ renal artery $\rightarrow$ liver $\rightarrow$ ureter $\rightarrow$ bladder $\rightarrow$ urethra
Show answer & marking scheme

Worked solution

Urea is produced in the liver by deamination of excess amino acids. It is transported in the blood through the circulatory system to the renal artery, entering the kidney where it is filtered. It then travels down the ureter to the bladder for storage, and is excreted from the body via the urethra.

Marking scheme

Award 1 mark for the correct option B.
Question 30 · multiple_choice
1 marks
A cross between a heterozygous tall pea plant ($Tt$) and a homozygous short pea plant ($tt$) is carried out. What is the expected ratio of phenotypes in the offspring?
  1. A.all tall plants
  2. B.1 tall plant : 1 short plant
  3. C.3 tall plants : 1 short plant
  4. D.1 tall plant : 3 short plants
Show answer & marking scheme

Worked solution

The gametes produced by the heterozygous parent are $T$ and $t$. The gametes produced by the homozygous recessive parent are only $t$.
Crossing these gives offspring genotypes: $Tt$ (tall) and $tt$ (short) in a $1:1$ ratio. Therefore, the phenotype ratio is $1\text{ tall plant} : 1\text{ short plant}$.

Marking scheme

Award 1 mark for the correct option B.
Question 31 · multiple_choice
1 marks
Which sequence of events is correct during the genetic modification of bacteria to produce human insulin?
  1. A.insert the human insulin gene into a plasmid $\rightarrow$ insert the recombinant plasmid into a bacterium $\rightarrow$ extract the human insulin gene from human cells $\rightarrow$ grow the bacteria in a fermenter
  2. B.extract the human insulin gene from human cells $\rightarrow$ insert the human insulin gene into a plasmid $\rightarrow$ insert the recombinant plasmid into a bacterium $\rightarrow$ grow the bacteria in a fermenter
  3. C.grow the bacteria in a fermenter $\rightarrow$ extract the human insulin gene from human cells $\rightarrow$ insert the human insulin gene into a plasmid $\rightarrow$ insert the recombinant plasmid into a bacterium
  4. D.extract the human insulin gene from human cells $\rightarrow$ insert the human insulin gene into a bacterium $\rightarrow$ insert the bacterium into a plasmid $\rightarrow$ grow the plasmid in a fermenter
Show answer & marking scheme

Worked solution

The correct sequence starts with isolating/extracting the human gene for insulin from human cells. Next, this gene is inserted into a vector, which is a bacterial plasmid. The recombinant plasmid is then introduced into bacterial cells. Finally, these genetically modified bacteria are grown in large quantities in a fermenter to produce insulin.

Marking scheme

Award 1 mark for the correct option B.
Question 32 · multiple_choice
1 marks
Which combination of environmental conditions would result in the highest rate of transpiration in a plant?

$$\begin{array}{|c|c|c|c|} \hline & \text{humidity} & \text{temperature} & \text{wind speed} \\ \hline \text{A} & \text{high} & \text{high} & \text{low} \\ \hline \text{B} & \text{low} & \text{low} & \text{high} \\ \hline \text{C} & \text{low} & \text{high} & \text{high} \\ \hline \text{D} & \text{high} & \text{low} & \text{low} \\ \hline \end{array}$$
  1. A.A
  2. B.B
  3. C.C
  4. D.D
Show answer & marking scheme

Worked solution

Low humidity increases the concentration gradient of water vapour between the inside of the leaf and the surrounding air. High temperature increases the kinetic energy of water molecules, increasing the rate of evaporation. High wind speed sweeps away water vapour from the leaf surface, maintaining a steep concentration gradient. Together, these conditions maximize the rate of transpiration.

Marking scheme

Award 1 mark for the correct option C.
Question 33 · multiple-choice
1 marks
Four equal-sized cylinders of potato tissue were placed in four sucrose solutions of different concentrations: 0.1 mol/dm³, 0.3 mol/dm³, 0.5 mol/dm³, and 0.7 mol/dm³. After two hours, the percentage change in mass of each cylinder was calculated. Which cylinder was placed in the most concentrated sucrose solution?
  1. A.the cylinder that gained the most mass
  2. B.the cylinder that had no change in mass
  3. C.the cylinder that lost the least mass
  4. D.the cylinder that lost the most mass
Show answer & marking scheme

Worked solution

In a highly concentrated sucrose solution (0.7 mol/dm³), the water potential of the solution is much lower than the water potential of the potato cells. Consequently, water moves out of the potato cells by osmosis down a water potential gradient, causing the potato cylinder to lose the most mass.

Marking scheme

Award 1 mark for the correct option (D).
Question 34 · multiple-choice
1 marks
The rate of transpiration in a plant was measured under different environmental conditions. Which combination of factors will result in the lowest rate of transpiration?
  1. A.low humidity, high wind speed, high light intensity
  2. B.high humidity, low wind speed, low light intensity
  3. C.high humidity, high wind speed, low light intensity
  4. D.low humidity, low wind speed, high light intensity
Show answer & marking scheme

Worked solution

Transpiration is the evaporation of water from mesophyll cell surfaces followed by the diffusion of water vapour through the stomata. High humidity decreases the water potential gradient between the leaf interior and the outside air. Low wind speed prevents the removal of water vapour around the stomata, and low light intensity causes stomatal closure. Together, these factors minimize transpiration.

Marking scheme

Award 1 mark for the correct option (B).
Question 35 · multiple-choice
1 marks
An enzyme-catalysed reaction was carried out at different pH values while keeping all other variables constant. The rate of reaction was measured. At pH 2, the rate was very high. At pH 7, the rate was zero. Which enzyme is most likely responsible for this reaction?
  1. A.amylase
  2. B.lipase
  3. C.pepsin
  4. D.trypsin
Show answer & marking scheme

Worked solution

Pepsin is an acidic protease found in the stomach that functions optimally at a very low pH (pH 1.5 to 2). Neutral pH conditions (pH 7) denature pepsin, rendering it completely inactive.

Marking scheme

Award 1 mark for the correct option (C).
Question 36 · multiple-choice
1 marks
Which chamber of the human heart has the thickest muscular wall, and into which blood vessel does it directly pump blood?
  1. A.left atrium into the pulmonary vein
  2. B.left ventricle into the aorta
  3. C.right ventricle into the pulmonary artery
  4. D.right atrium into the vena cava
Show answer & marking scheme

Worked solution

The left ventricle has the thickest muscular wall of all cardiac chambers because it must generate high enough pressure to force blood throughout the entire systemic circulation of the body. It pumps oxygenated blood directly into the aorta.

Marking scheme

Award 1 mark for the correct option (B).
Question 37 · multiple-choice
1 marks
Which substances are filtered from the blood in the glomerulus into the Bowman's capsule of a healthy human kidney?
  1. A.glucose, urea, mineral ions, and water
  2. B.glucose, proteins, red blood cells, and urea
  3. C.urea, mineral ions, proteins, and water
  4. D.mineral ions, red blood cells, water, and starch
Show answer & marking scheme

Worked solution

Ultrafiltration in the glomerulus forces small molecules such as glucose, urea, mineral ions, and water into the Bowman's capsule. Large proteins and red blood cells are too large to pass through the filtration barrier, and starch is not present in human blood.

Marking scheme

Award 1 mark for the correct option (A).
Question 38 · multiple-choice
1 marks
In a species of plant, the allele for red flowers (R) is dominant to the allele for white flowers (r). A heterozygous red-flowered plant is crossed with a white-flowered plant. What is the expected ratio of phenotypes in the offspring?
  1. A.1 red : 1 white
  2. B.3 red : 1 white
  3. C.all red
  4. D.1 red : 3 white
Show answer & marking scheme

Worked solution

A heterozygous red-flowered plant has the genotype Rr, and a white-flowered plant is homozygous recessive (rr). The cross (Rr x rr) yields gametes R and r from one parent, and r from the other. The resulting offspring genotypes are 50% Rr (red-flowered) and 50% rr (white-flowered), giving a 1:1 phenotypic ratio.

Marking scheme

Award 1 mark for the correct option (A).
Question 39 · multiple-choice
1 marks
The list shows two food chains in an ecosystem:

1. Oak tree → Caterpillar → Blue tit → Hawk
2. Oak tree → Aphid → Ladybird → Blue tit

Which organism occupies the trophic levels of both secondary consumer and tertiary consumer?
  1. A.aphid
  2. B.blue tit
  3. C.caterpillar
  4. D.hawk
Show answer & marking scheme

Worked solution

In food chain 1, the blue tit is a secondary consumer because it feeds on a primary consumer (caterpillar). In food chain 2, the blue tit is a tertiary consumer because it feeds on a secondary consumer (ladybird). Therefore, the blue tit occupies both consumer levels.

Marking scheme

Award 1 mark for the correct option (B).
Question 40 · multiple-choice
1 marks
During the production of human insulin using genetically modified bacteria, which structure is cut using restriction enzymes to allow the insertion of the human insulin gene?
  1. A.the bacterial cell wall
  2. B.a bacterial plasmid
  3. C.the bacterial chromosome
  4. D.a human ribosome
Show answer & marking scheme

Worked solution

Bacterial plasmids are small, circular rings of DNA. They are used as vectors in genetic engineering. Restriction enzymes are used to cut the plasmid DNA open at specific sites, allowing the human insulin gene to be inserted.

Marking scheme

Award 1 mark for the correct option (B).

Ready to test yourself?

Turn these notes into exam-style practice. Get unlimited AI questions on this topic with instant marking and explanations.

Practice This Topic

Paper 43

Answer all structured questions on the question paper.
7 Question · 80 marks
Question 1 · structured
11 marks
1 (a) Active transport and diffusion are two methods by which substances move across cell membranes. Table 1.1 lists several features of these processes. Complete Table 1.1 by placing a tick (✓) in the correct column(s) for each feature. Some features may apply to one process, both processes, or neither process.

Table 1.1
Feature | Active transport | Diffusion
- movement of particles can occur down a concentration gradient | |
- requires energy released from respiration | |
- requires transport proteins in cell membranes | |
- occurs in non-living systems | |

(b) Explain why active transport is important in root hair cells.

(c) Describe how the rate of diffusion of oxygen into an animal cell is affected by:
(i) the concentration gradient of oxygen
(ii) the surface area of the cell membrane.
Show answer & marking scheme

Worked solution

(a)
Table 1.1 completed as follows:
- 'movement of particles can occur down a concentration gradient': Diffusion only (tick under Diffusion)
- 'requires energy released from respiration': Active transport only (tick under Active transport)
- 'requires transport proteins in cell membranes': Active transport only (tick under Active transport)
- 'occurs in non-living systems': Diffusion only (tick under Diffusion)

(b) Root hair cells use active transport to absorb mineral ions (such as nitrate or magnesium ions) from the soil. The concentration of these ions in the soil is lower than their concentration inside the vacuole/cytoplasm of the root hair cell. Therefore, the cells must move these ions against their concentration gradient, which requires energy from respiration and specific carrier proteins.

(c)
(i) A steeper concentration gradient (greater difference in oxygen concentration between the outside and inside of the cell) increases the rate of diffusion, as more oxygen molecules collide with and cross the membrane per unit time.
(ii) A larger surface area of the cell membrane increases the rate of diffusion, as there is more space/membrane available for oxygen molecules to pass through simultaneously.

Marking scheme

(a) [4 marks total, 1 mark per correct row]:
- Row 1: Tick under Diffusion only
- Row 2: Tick under Active transport only
- Row 3: Tick under Active transport only
- Row 4: Tick under Diffusion only

(b) [3 marks maximum]:
- to absorb mineral ions / named example (e.g. nitrates/magnesium) ;
- because concentration of ions is lower in soil than inside root hair cells / movement against concentration gradient ;
- requires energy (released from respiration) / uses carrier proteins ;

(c)
(i) [2 marks maximum]:
- steeper gradient increases rate / directly proportional ;
- because of greater net movement of particles / more collisions ;
(ii) [2 marks maximum]:
- larger surface area increases rate ;
- because more space/area is available for diffusion to occur ;
Question 2 · structured
12 marks
2 (a) Define the term transpiration.

(b) Explain how water moves upwards in xylem vessels, with reference to cohesion, adhesion, and transpiration pull.

(c) A student investigated the effect of wind speed on the rate of water uptake (transpiration rate) in a leafy shoot. Table 2.1 shows the results of this investigation.

Table 2.1
Wind speed / m s^{-1} | Rate of transpiration / g h^{-1}
0.5 | 1.2
1.5 | 2.6
2.5 | 4.1
3.5 | 5.2

Explain why the rate of transpiration increases as the wind speed increases from 0.5 m s^{-1} to 3.5 m s^{-1}.

(d) State two environmental factors, other than wind speed, that affect the rate of transpiration.
Show answer & marking scheme

Worked solution

(a) Transpiration is the loss of water vapour from plant leaves by evaporation of water at the surfaces of the mesophyll cells followed by the diffusion of water vapour through the stomata into the atmosphere.

(b) Water evaporates from the leaves, creating a transpiration pull that draws water upwards. Water molecules are polar and are held together by forces of attraction called cohesion, forming a continuous, unbroken column of water inside the xylem vessel. Water molecules also attract to the cellulose walls of the xylem vessels by adhesion, which helps support the column of water against gravity.

(c) Moving air/wind carries away water vapour that has accumulated outside the stomata on the leaf surface. This maintains a steep water potential gradient between the inside of the leaf (air spaces) and the outside atmosphere, increasing the rate of diffusion of water vapour out of the leaf.

(d) Temperature and light intensity (or humidity).

Marking scheme

(a) [3 marks maximum]:
- loss of water vapour from leaves/aerial parts of plant ;
- evaporation (of water) at surface of mesophyll cells ;
- diffusion of water vapour through stomata ;

(b) [4 marks maximum]:
- evaporation/loss of water from leaves creates transpiration pull/tension ;
- draws up a continuous column of water ;
- cohesion is the attraction between water molecules (holding them together) ;
- adhesion is the attraction between water molecules and xylem walls ;

(c) [3 marks maximum]:
- wind moves/removes water vapour away from the leaf surface/stomata ;
- maintains / increases the water potential / concentration gradient (between inside and outside of leaf) ;
- increases rate of diffusion of water vapour (out of stomata) ;

(d) [2 marks maximum]:
- temperature ;
- light intensity / humidity ;
[Reject: wind speed]
Question 3 · structured
12 marks
3 (a) Fig. 3.1 represents a diagrammatic cross-section through a human heart.
(i) State the name of the muscular wall that separates the left and right sides of the heart and describe its function.
(ii) Explain why the muscular wall of the left ventricle is much thicker than the muscular wall of the right ventricle.

(b) Describe the pathway taken by a red blood cell as it travels from the vena cava, through the heart chambers, to the lungs, and out through the aorta.

(c) Suggest two ways a doctor can monitor the activity of a patient's heart.
Show answer & marking scheme

Worked solution

(a)
(i) The septum. Its function is to separate oxygenated blood on the left side of the heart from deoxygenated blood on the right side of the heart, preventing them from mixing.
(ii) The left ventricle has a thicker muscle wall because it must pump blood under high pressure a much longer distance to the entire body (systemic circulation). The right ventricle only pumps blood under lower pressure a short distance to the lungs (pulmonary circulation).

(b) Deoxygenated blood enters the right atrium via the vena cava. The right atrium contracts, pumping blood through the tricuspid valve into the right ventricle. The right ventricle contracts, pumping blood through the semi-lunar valve into the pulmonary artery, which carries it to the lungs for gas exchange. Oxygenated blood returns from the lungs to the left atrium via the pulmonary vein. The left atrium contracts, passing blood through the bicuspid valve into the left ventricle. Finally, the left ventricle contracts, pumping oxygenated blood through the semi-lunar valve into the aorta to be distributed to the body.

(c) Listening to the heart sounds (using a stethoscope) and measuring the pulse rate (or recording an electrocardiogram / ECG).

Marking scheme

(a)
(i) [2 marks total]:
- septum ;
- prevents mixing of oxygenated and deoxygenated blood ;
(ii) [3 marks maximum]:
- left ventricle pumps blood to the whole body / longer distance / systemic circulation ;
- requires higher pressure / more force ;
- right ventricle only pumps to the lungs / shorter distance / pulmonary circulation ;

(b) [5 marks maximum]:
- vena cava to right atrium to right ventricle ;
- right ventricle to pulmonary artery to lungs ;
- pulmonary vein to left atrium to left ventricle ;
- left ventricle to aorta to body ;
- ref. to contraction of atrium/ventricle ;
- ref. to valves (atrioventricular / semi-lunar) preventing backflow ;

(c) [2 marks maximum]:
- listen to heart valves closing/sounds using a stethoscope ;
- measure pulse rate / heart rate ;
- electrocardiogram / ECG ;
Question 4 · structured
11 marks
4 (a) Urea is a nitrogenous waste product excreted by the kidneys. Describe how urea is formed in the human body.

(b) State the name of:
(i) the blood vessel that carries deoxygenated blood away from the kidneys
(ii) the component of blood that transports dissolved urea to the kidneys.

(c) Describe the process of ultrafiltration and selective reabsorption in a kidney nephron.

(d) State why glucose is present in the fluid entering the nephron but is not normally found in the urine of a healthy person.
Show answer & marking scheme

Worked solution

(a) Excess amino acids cannot be stored in the body. They are transported to the liver, where they undergo deamination. During deamination, the nitrogen-containing amine group is removed from the amino acid and converted into ammonia, which is highly toxic. The liver quickly converts ammonia into urea, which is less toxic and can be safely transported in the blood to the kidneys for excretion.

(b)
(i) Renal vein.
(ii) Plasma.

(c) Blood enters the glomerulus under high pressure. Small molecules such as water, glucose, urea, and salts are forced out of the capillary network through the basement membrane into the Bowman's capsule (ultrafiltration). Large molecules like proteins and red blood cells remain in the blood. As the filtrate moves along the nephron, useful substances such as all glucose, some water, and some salts are actively transported/reabsorbed back into the surrounding capillaries (selective reabsorption).

(d) Glucose is small enough to pass through the filter in the glomerulus during ultrafiltration, so it enters the nephron. However, in a healthy person, 100% of the filtered glucose is selectively reabsorbed back into the blood capillaries at the proximal convoluted tubule, leaving none to be excreted in the urine.

Marking scheme

(a) [3 marks maximum]:
- excess amino acids transported to liver ;
- amine group removed / deamination ;
- ammonia formed and converted to urea ;

(b)
(i) [1 mark]:
- renal vein ;
(ii) [1 mark]:
- plasma ;

(c) [4 marks maximum]:
- high pressure in glomerulus forces small molecules/filtrate out ;
- water / glucose / urea / ions enter Bowman's capsule / ultrafiltration ;
- proteins / blood cells too large to be filtered ;
- useful substances / all glucose / some water / some ions reabsorbed into blood ;
- active transport (of glucose/ions) / osmosis (of water) ;

(d) [2 marks maximum]:
- glucose molecule is small enough to pass through glomerulus/filter ;
- all/100% of glucose is selectively reabsorbed back into the blood ;
Question 5 · structured
11 marks
5 (a) State the chemical elements present in:
(i) carbohydrates
(ii) proteins.

(b) Table 5.1 shows three large biological molecules, the simpler molecules they are broken down into during digestion, and the enzymes that catalyse these reactions. Complete Table 5.1 by filling in the missing information.

Table 5.1
Large biological molecule | Simpler product molecules | Digestive enzyme
Starch | Maltose | (i)
Lipids (fats and oils) | (ii) | Lipase
Proteins | (iii) | Protease

(c) Explain why the rate of an enzyme-catalysed reaction decreases rapidly at temperatures above the optimum temperature.
Show answer & marking scheme

Worked solution

(a)
(i) Carbon, hydrogen, and oxygen.
(ii) Carbon, hydrogen, oxygen, nitrogen (and sometimes sulfur).

(b)
(i) Amylase
(ii) Fatty acids and glycerol
(iii) Amino acids

(c) Enzymes are proteins. At high temperatures (above the optimum), the increased kinetic energy causes the atoms within the enzyme molecule to vibrate violently, breaking the weak hydrogen/ionic bonds that maintain its three-dimensional shape. This changes the specific shape of the active site. As a result, the substrate molecule can no longer fit into the active site, and no enzyme-substrate complexes can be formed. The enzyme has been denatured.

Marking scheme

(a)
(i) [1 mark]:
- carbon, hydrogen, oxygen (all three required) ;
(ii) [2 marks]:
- carbon, hydrogen, oxygen, nitrogen (all four required for 1 mark) ;
- sulfur (1 mark) ;

(b) [3 marks total, 1 mark per correct answer]:
- (i) amylase ;
- (ii) fatty acids and glycerol (both required) ;
- (iii) amino acids ;

(c) [4 marks maximum]:
- enzymes are proteins ;
- high kinetic energy breaks bonds in enzyme structure ;
- changes the shape of the active site ;
- substrate no longer fits active site / no longer complementary ;
- no enzyme-substrate complexes formed ;
- enzyme is denatured ;
[Reject: enzyme is killed / dead]
Question 6 · structured
12 marks
6 (a) Outline the role of the placenta in supporting the growth and development of a fetus.

(b) The Rhesus blood group system is determined by a single gene. The allele for Rhesus-positive (D) is dominant to the allele for Rhesus-negative (d).
A Rhesus-negative woman has a child with a heterozygous Rhesus-positive man.
(i) State the genotype of the Rhesus-negative woman.
(ii) Complete a Punnett square to show the possible genotypes and phenotypes of the offspring of this couple.
(iii) State the probability that the child will be Rhesus-positive.

(c) Explain the difference between homozygous and heterozygous genotypes.
Show answer & marking scheme

Worked solution

(a) The placenta allows the exchange of materials between the mother's blood and the fetus's blood. It delivers useful substances such as oxygen, glucose, amino acids, and antibodies from the mother to the fetus, and removes waste products such as carbon dioxide and urea from the fetal blood. It also secretes hormones (like progesterone) to maintain the uterine lining and support pregnancy.

(b)
(i) dd
(ii) The Punnett square is completed with gametes: woman (d, d) and man (D, d).
Offspring genotypes are: Dd, dd, Dd, dd.
Offspring phenotypes are:
- Dd: Rhesus-positive
- dd: Rhesus-negative
(iii) 0.5 (or 50% or 1 in 2).

(c) A homozygous genotype has two identical alleles of a particular gene (e.g. DD or dd), whereas a heterozygous genotype has two different alleles of that gene (e.g. Dd). Homozygous individuals breed true for that trait, while heterozygous individuals do not. In a heterozygous genotype, the dominant allele is expressed in the phenotype, while the recessive allele remains unexpressed.

Marking scheme

(a) [3 marks maximum]:
- provides oxygen / glucose / amino acids / useful nutrients to fetus ;
- removes carbon dioxide / urea / wastes from fetus ;
- allows transfer of antibodies (passive immunity) ;
- produces progesterone / hormones to maintain pregnancy ;
- prevents maternal and fetal blood mixing / protects against pressure/pathogens ;

(b)
(i) [1 mark]:
- dd ;
(ii) [3 marks total]:
- parent gametes correct: d and d on one side, D and d on the other ;
- offspring genotypes correct in grid: Dd and dd ;
- phenotypes correctly matched to genotypes (Dd = Rhesus-positive, dd = Rhesus-negative) ;
(iii) [1 mark]:
- 0.5 / 50% / 1/2 / 1 in 2 ;

(c) [4 marks maximum]:
- homozygous has two identical alleles ;
- heterozygous has two different alleles ;
- homozygous breeds true ;
- in heterozygous, only dominant allele is expressed in phenotype / recessive is masked ;
Question 7 · structured
11 marks
7 (a) State what is meant by the term endangered species.

(b) Farmers sometimes apply excessive amounts of chemical fertilisers to crop fields. Explain how the runoff of these fertilisers into a nearby lake can lead to eutrophication and the death of fish.

(c) Describe the role of seed banks in conserving endangered plant species.
Show answer & marking scheme

Worked solution

(a) An endangered species is a species of animal or plant that is seriously at risk of extinction (dying out completely).

(b) When excessive fertilisers run off into a lake, the high concentration of nitrates and phosphates causes rapid growth of algae on the surface of the water, forming an algal bloom. This layer of algae blocks sunlight from reaching aquatic plants deeper in the lake, preventing them from photosynthesising, so they die. Decomposers, such as aerobic bacteria, feed on the dead plants and reproduce rapidly, using up dissolved oxygen in the water for respiration. The oxygen concentration in the lake drops severely, causing fish and other aquatic organisms to suffocate and die.

(c) Seed banks collect, dry, and store seeds from endangered plant species at very low temperatures (below freezing). This preserves their viability for decades or centuries. Seed banks take up very little space, allow genetic diversity to be maintained, and protect species from extinction in the wild. If the wild population is destroyed, these seeds can be germinated to reintroduce the species into its natural habitat.

Marking scheme

(a) [2 marks total]:
- species at risk of extinction / dying out ;
- numbers have fallen to low levels ;

(b) [5 marks maximum]:
- fertiliser runoff contains nitrates/phosphates, causing algal bloom ;
- algae block light from reaching submerged plants ;
- submerged plants cannot photosynthesise and die ;
- bacteria / decomposers decay / feed on dead plants ;
- bacteria multiply rapidly and use up dissolved oxygen (for aerobic respiration) ;
- fish die due to lack of oxygen / suffocation ;

(c) [4 marks maximum]:
- seeds are dried and stored at low temperatures / frozen ;
- keeps seeds viable / dormant for a long time ;
- small space required to conserve many species ;
- maintains genetic diversity ;
- seeds can be germinated to reintroduce plants to wild if extinct ;

Paper 63

Answer all practical-style alternative questions.
2 Question · 40 marks
Question 1 · practical
20 marks
### 1 A student investigated the nutrient content of three plant extracts:
- extract **X**
- extract **Y**
- extract **Z**

The student performed three food tests on each extract:

**Test 1: Testing for reducing sugars**
- Benedict's reagent was added to each extract and the tubes were placed in a hot water-bath for five minutes.
- *Observations:* Extract **X** and **Y** remained blue. Extract **Z** turned brick-red.

**Test 2: Testing for starch**
- Iodine solution was added to each extract on a white tile.
- *Observations:* Extract **X** turned blue-black. Extract **Y** and **Z** remained brown.

**Test 3: Testing for protein**
- Biuret reagent was added to each extract.
- *Observations:* Extract **Y** and **Z** turned purple. Extract **X** remained blue.

(a) (i) Prepare a table to record the observations for all three tests on each extract. Do not include conclusions in your table. [5]

(a) (ii) Using the observations, state which nutrients are present in each extract. [3]
- extract **X**:
- extract **Y**:
- extract **Z**:

(a) (iii) Identify one safety hazard associated with Test 1. [1]

(b) The concentration of vitamin C and fat in three different fruit juices, **P**, **Q**, and **R**, was determined.

It was found that:
- Juice **P** contained vitamin C only.
- Juice **Q** contained fat only.
- Juice **R** contained both vitamin C and fat.

(b) (i) State the reagent used when testing for vitamin C. [1]

(b) (ii) Describe the method for the emulsion test for fats. [2]

(b) (iii) The results for one of the juices are shown in Table 1.1.

**Table 1.1**
| test | observation |
|---|---|
| vitamin C | the solution remains blue / does not decolourise |
| fat | a cloudy white emulsion has formed |

Identify the juice from the results provided in Table 1.1. [1]
- juice:

(b) (iv) Explain how you identified the juice from the results provided in Table 1.1. [1]

(c) Three types of fresh fruit juices (pineapple, papaya, and kiwi) contain different concentrations of the enzyme protease. Protease catalyses the breakdown of protein (such as in gelatin or milk protein) to form soluble, clear products.

Plan an investigation to compare the concentrations of protease in the three types of fruit juice. [6]
Show answer & marking scheme

Worked solution

(a) (i) A suitable table should be drawn with clear columns and rows, showing observations without conclusions:

| Extract | Test 1: Benedict's test / reducing sugars | Test 2: Iodine test / starch | Test 3: Biuret test / protein |
|---|---|---|---|
| **X** | blue | blue-black | blue |
| **Y** | blue | brown | purple |
| **Z** | brick-red | brown | purple |

(a) (ii)
- Extract **X**: starch
- Extract **Y**: protein
- Extract **Z**: reducing sugars and protein

(a) (iii) Use of a hot water-bath / hot liquids can cause burns to skin; or Benedict's reagent is corrosive / irritant.

(b) (i) DCPIP (solution).

(b) (ii) Add ethanol to the sample, shake/mix thoroughly, then pour/add water. Look for the formation of a white / cloudy emulsion.

(b) (iii) Juice **Q**

(b) (iv) The positive emulsion test (cloudy white emulsion) shows fat is present, and the negative DCPIP test (solution remains blue) shows vitamin C is absent, which matches the description for juice **Q**.

(c) Suggested plan:
- Independent variable: The three different fruit juices (pineapple, papaya, kiwi).
- Dependent variable: Measure the time taken for a protein suspension (e.g., milk or gelatin) to go clear, or measure the volume of protein digested in a set time.
- Controlled variables: Keep volume of fruit juice constant, volume and concentration of protein suspension constant, temperature constant (using a water-bath), and pH constant.
- Replicates: Repeat the procedure at least three times for each juice and calculate a mean.
- Control: Use boiled fruit juice as a control to show that the reaction is enzyme-catalysed.
- Safety: Wear eye protection / gloves when handling juices.

Marking scheme

**(a)(i) Prepare table [5]**
- 1. Table drawn with clear borders and lines;
- 2. Headings including 'Extract / Sample' and 'Test name / Reagent used';
- 3. All nine observations correctly recorded;
- 4. No conclusions (nutrient names) included in the table;
- 5. Correct colours associated with positive/negative tests (e.g. blue for negative Benedict's/Biuret, brown for negative Iodine).

**(a)(ii) Identify nutrients [3]**
- Extract **X**: starch;
- Extract **Y**: protein;
- Extract **Z**: reducing sugar AND protein; (Reject: 'sugar' alone, must specify reducing sugar / simple sugar)

**(a)(iii) Safety hazard [1]**
- Hot water / hot water-bath / heat source can cause burns; OR Benedict's reagent is an irritant / chemicals in eyes;

**(b)(i) Reagent for vitamin C [1]**
- DCPIP / dichlorophenolindophenol;

**(b)(ii) Emulsion test method [2]**
- 1. Add ethanol to the sample and shake / mix;
- 2. Add water and observe for a white / cloudy emulsion;

**(b)(iii) Identify juice [1]**
- Juice **Q**;

**(b)(iv) Explain identification [1]**
- The white emulsion indicates fat is present, and the failure of DCPIP to decolourise indicates vitamin C is absent, which is characteristic of juice **Q**;

**(c) Planning investigation [6]**
- *Max 6 marks from any of the following categories:*
- 1. **Independent Variable**: Use of the three types of fresh fruit juices (pineapple, papaya, kiwi);
- 2. **Dependent Variable**: Measure the time taken for the protein suspension (milk / gelatin) to clear / go from cloudy to transparent;
- 3. **Method**: Detail of measuring a set volume of protein (e.g. 5 cm³) and adding a set volume of juice (e.g. 1 cm³);
- 4. **Controlled variable 1**: Keep temperature constant (using a water-bath);
- 5. **Controlled variable 2**: Keep concentration/volume of protein suspension constant;
- 6. **Controlled variable 3**: Keep pH constant (using a buffer);
- 7. **Reliability**: Repeat each test at least twice / do three trials in total and calculate the average/mean;
- 8. **Safety**: Wear safety goggles / gloves to prevent irritation from fruit acids/proteases;
- 9. **Control**: Use boiled/denatured fruit juice to show protease is responsible for the clearing.
Question 2 · practical
20 marks
### 2 (a) The nutrient content of diet can affect hemoglobin levels in the blood.

In a study, the dietary iron intake and hemoglobin levels of 100 male athletes were monitored over one year. The athletes were all between 18 and 30 years of age.

The scientists:
- calculated the mean daily iron intake for each athlete
- measured the change in hemoglobin concentration in their blood using standard lab assays.

The results for five of the athletes are shown in Table 2.1.

**Table 2.1**
| mean daily iron intake for each athlete / mg per day | mean change in hemoglobin concentration / g per dm³ per year |
|---|---|
| 5 | -8.5 |
| 10 | -4.2 |
| 15 | -0.5 |
| 20 | +1.5 |
| 25 | +3.8 |

(a) (i) Plot a line graph on a grid of the data in Table 2.1. [4]
*(Note: For this online question, outline the scales you would choose for both axes and describe the shape of the resulting curve.)*

(a) (ii) State two conclusions for the data in Table 2.1. [2]

(a) (iii) Identify the independent variable in this investigation. [1]

(a) (iv) Describe two variables that the scientists should have considered when selecting athletes for the study. [2]

(a) (v) Suggest a reason for a large number of athletes (100) being included in the study. [1]

(a) (vi) State one way this study is not representative of the general population. [1]

(a) (vii) A student stated that more athletes were gaining hemoglobin than were losing hemoglobin. Explain why this statement may not be correct for the data in this study. [1]

(b) Fig. 2.1 is a photograph of a broad bean seed.

$$\text{Line PQ runs from the top-left tip to the bottom-right tip across the seed. (Magnification } \times 1.5\text{)}$$

(b) (i) Make a large drawing of the seed shown in Fig. 2.1. [4]

(b) (ii) The length of line PQ represents the length of the seed in Fig. 2.1.
- Measure the length of line PQ on the photograph. Assume the measured length of line PQ is 90 mm.
- Use your measurement and the formula to calculate the actual length of the seed.
$$\text{magnification} = \frac{\text{length of line PQ on Fig. 2.1}}{\text{actual length of the seed}}$$
Give your answer to three significant figures. Show your working. [3]

(b) (iii) Fig. 2.2 shows a seed of the same species grown under nitrogen-deficient conditions.
State one way the seed in Fig. 2.2 would be different from the healthy seed in Fig. 2.1. [1]
Show answer & marking scheme

Worked solution

(a) (i) Axes choice:
- x-axis: Mean daily iron intake / mg per day (scale: 0 to 30, with intervals of 5).
- y-axis: Mean change in hemoglobin concentration / g per dm³ per year (scale: -10 to +5, with intervals of 2).
- Shape: A straight or slightly curved line showing a positive upward trend from (-8.5 at 5 mg) to (+3.8 at 25 mg).

(a) (ii)
- 1. As mean daily iron intake increases, the change in hemoglobin concentration increases / becomes more positive.
- 2. Athletes with less than 15-16 mg per day of iron intake experience a decrease in hemoglobin concentration, while those with more than 16 mg per day show an increase.

(a) (iii) Mean daily iron intake.

(a) (iv) Any two of: age, starting hemoglobin levels, training intensity, overall health status / medical history, or baseline diet.

(a) (v) To reduce the effect of anomalies / to make the study more reliable and representative.

(a) (vi) The study only included male athletes aged 18 to 30 (not females, older adults, sedentary individuals, or children).

(a) (vii) We do not know the exact number of athletes in each iron intake category; we only have the average changes for these specific daily intakes.

(b) (i) A large, clean drawing of the broad bean seed showing its overall bean shape, hilum, and seed coat outline without shading.

(b) (ii) Measurement: 90 mm.
Formula: actual length = PQ / magnification = 90 / 1.5 = 60.0 mm.

(b) (iii) The seed is smaller / shriveled / wrinkled / has irregular shape.

Marking scheme

**(a)(i) Plot graph [4]**
- 1. Axes labelled with units: x-axis 'mean daily iron intake / mg per day', y-axis 'mean change in hemoglobin concentration / g per dm³ per year';
- 2. Scales: Linear, with data occupying at least half of the grid in both directions (x-axis: 0 to 30, y-axis: -10 to +5);
- 3. Points plotted accurately to within half a small square;
- 4. A clean, smooth, single line of best fit drawn through the points.

**(a)(ii) Conclusions [2]**
- 1. Increasing iron intake increases hemoglobin change / concentration (or reverse argument);
- 2. Below ~16 mg per day, hemoglobin levels decrease, whereas above ~16 mg per day, hemoglobin levels increase;

**(a)(iii) Independent variable [1]**
- Mean daily iron intake (per athlete);

**(a)(iv) Controlled selection variables [2]**
- *Any two from:*
- Age (must be restricted, e.g., 18-30);
- Initial health status / no pre-existing blood disorders;
- Training load / exercise levels;
- Gender / sex;
- Weight / body mass;

**(a)(v) Purpose of large sample size [1]**
- To identify/remove anomalies OR to increase reliability OR to get a representative sample;

**(a)(vi) Non-representative nature [1]**
- *Any one from:*
- Only males were studied / no females;
- Only athletes were studied / physical activity levels are not representative of sedentary people;
- Only a narrow age range (18-30) was studied;

**(a)(vii) Explaining statement flaw [1]**
- The distribution of the 100 athletes across the different daily iron intake groups is unknown / there might be very few athletes in the high-intake groups;

**(b)(i) Broad bean seed drawing [4]**
- 1. Outline: Clear, single, continuous line without shading or stippling;
- 2. Size: Larger than or equal to the size of the photograph / PQ line;
- 3. Detail 1: Showing the oval/kidney shape of the seed accurately;
- 4. Detail 2: Showing the hilum (scar) clearly on one side;

**(b)(ii) Calculation [3]**
- 1. Measured value of PQ = 90 mm (allow 89-91 mm);
- 2. Actual length calculation = 90 / 1.5 = 60.0 mm;
- 3. Correct units (mm) and rounded to 3 significant figures (60.0);

**(b)(iii) Seed difference [1]**
- Shriveled / wrinkled / smaller size / irregular shape;

Wondering how well you actually know this?

thinka is an AI practice app for IGCSE & IB students: unlimited questions, instant auto-marking, and detailed step-by-step solutions. 100,000+ students use it to confirm they actually know it, not just think they do.

Want more questions like this? Practice unlimited on thinka, instant answers included.

Start Practicing Free