a) Fig. 1.1 is a diagram of the digestive system.
Each letter may be used once, more than once or not at all.
State the letter of the part shown in Fig. 1.1:
- that produces bile
- that produces gastric juice
- that produces urea
- where maltose is digested
- where trypsin acts
(b) A student investigated the effect of bile on the digestion of fat in milk.
They set up three different test-tubes:
• test-tube A contained milk and bile
• test-tube B contained milk and lipase
• test-tube C contained milk, lipase and bile.
They used an indicator that is pink in alkaline solutions and colourless in acidic solutions.
They added the same volume of indicator to each test-tube.
The student observed and recorded the colour of the contents of each test-tube at 0 minutes, 20 minutes and 40 minutes.
Table 1.1 shows the results of the investigation
(i) Explain the results for test-tubes B and C in Table 1.1.
(ii) Explain the purpose of test-tube A in Table 1.1.
(c) The action of lipase is affected by temperature.
Fig. 1.2 shows the axes for a graph of the effect of temperature on the activity of lipase.
Complete the graph by:
• drawing a line to show the expected effect of temperature on the activity of lipase
• adding a label line and a label to show the point at which all the lipase has been denatured.
(d) Explain why lipase cannot be used to catalyse the breakdown of proteins.
▶️ Answer/Explanation
(a)
- that produces bile: K (Liver produces bile, stored in the gallbladder)
- that produces gastric juice: C (Stomach secretes gastric juice containing HCl and pepsin)
- that produces urea: K (Liver converts ammonia to urea in the ornithine cycle)
- where maltose is digested: H (Small intestine contains maltase for maltose breakdown)
- where trypsin acts: H (Trypsin is active in the small intestine)
(b)(i)
Test-tube B turns acidic due to lipase breaking down fats into fatty acids. Test-tube C turns acidic faster because bile emulsifies fats, increasing surface area for lipase action. The indicator changes from pink (alkaline) to colorless (acidic) as pH drops.
(b)(ii)
Test-tube A acts as a control to confirm that bile alone does not digest fats or change pH. It ensures observed changes in B and C are due to lipase activity.
(c)
The graph should show an increase in lipase activity up to an optimum temperature (~37°C), then a sharp decline as enzymes denature. Label the point where the curve meets the x-axis as “All lipase denatured.”
(d)
Lipase is specific to fats due to its active site’s complementary shape. Proteins cannot bind to lipase’s active site, preventing enzyme-substrate complex formation. Thus, lipase cannot catalyze protein breakdown.
(a) A student investigated osmosis in potato plant cells.
He immersed cubes of potato tissue in water and different concentrations of sucrose solution for 30 minutes.
The masses of the potato cubes were measured before and after immersion.
The percentage changes in mass were calculated.
Table 2.1 shows the results
(i) Using the information in Table 2.1, calculate the percentage change in mass at 1.00mol\(dm^{–3}\).
Give your answer to two decimal places
(ii) Using the information in Table 2.1, explain the difference in the results between the 0.6mol\(dm^{–3} and the 0.8mol\(dm^{–3}\) sucrose solutions.
Use the term water potential in your answer.
(iii) Describe the expected appearance of a cell from a potato cube that has been immersed in distilled water for 30 minutes.
(b) Describe how the process of active transport differs from the process of osmosis
(c) State the type of plant cells that use active transport to absorb mineral ions from the
environment.
(d) Explain the effect of a lack of magnesium ions on the colour of plant leaves
▶️ Answer/Explanation
(a)(i) –13.28 (%)
Explanation: The percentage change in mass is calculated using the formula: \[ \text{Percentage Change} = \left( \frac{\text{Final Mass} – \text{Initial Mass}}{\text{Initial Mass}} \right) \times 100 \] From Table 2.1, at \(1.00 \, \text{mol}\,dm^{-3}\), the mass decreases, resulting in a negative percentage change of –13.28%.
(a)(ii) The potato in \(0.8 \, \text{mol}\,dm^{-3}\) loses more mass than in \(0.6 \, \text{mol}\,dm^{-3}\) due to a steeper water potential gradient. Water moves from high water potential (potato) to low water potential (sucrose solution). The \(0.8 \, \text{mol}\,dm^{-3}\) solution has a lower water potential, causing more water to leave the potato cells.
(a)(iii) The cell becomes turgid (swollen) as water enters by osmosis. The vacuole enlarges, and the cell membrane presses against the cell wall.
(b) Active transport requires energy (ATP) to move substances against their concentration gradient, whereas osmosis is passive and involves water movement down its water potential gradient.
(c) Root hair cells use active transport to absorb mineral ions from the soil.
(d) Magnesium is essential for chlorophyll synthesis. A deficiency leads to chlorosis (yellowing of leaves) due to reduced chlorophyll production.
Topic: 11.1
(a) Fig. 3.1 is a photomicrograph of some cells lining the trachea
(i) Describe the role of goblet cells.
(ii) Explain how the cell labelled X in Fig. 3.1 is adapted for its function.
(iii) State the name of one other part of the body where the type of cell labelled X in Fig. 3.1 is found.
(b) Table 3.1 contains some features of the breathing system.
Complete Table 3.1 to show the actions of each feature of the breathing system that occur to cause inspiration.
(c) State the name of the gas that is excreted by the breathing system.
(d) Good ventilation is one feature of gas exchange surfaces.
State two other features.
(e) State the name of the gas exchange surface in humans.
▶️ Answer/Explanation
(a)(i) Goblet cells produce mucus, which traps pathogens and particulates, preventing them from entering the lungs. The mucus also moistens the airways.
(a)(ii) The cell labelled X is a ciliated cell. It has hair-like cilia that move rhythmically to sweep mucus (with trapped pathogens) upwards towards the throat, where it can be swallowed or expelled.
(a)(iii) Ciliated cells are also found in the bronchi and bronchioles.
(c) The gas excreted by the breathing system is carbon dioxide (and water vapour).
(d) Two other features of gas exchange surfaces are: large surface area (for efficient diffusion) and thin surface (short diffusion distance). A good blood supply is also essential.
(e) The gas exchange surface in humans is the alveoli (tiny air sacs in the lungs).
(a) Fig. 4.1 shows the effect of light intensity on the rate of photosynthesis at different temperatures and concentrations of carbon dioxide.
Describe and explain the reasons for the shape of lines B and C in Fig. 4.1.
(b) \(C_6H_{12}O_6\) is one of the products of photosynthesis.
State the chemical formula of the other product.
(c) Outline how the carbohydrates made during photosynthesis are used in plants.
▶️ Answer/Explanation
(a) The shape of lines B and C is determined by limiting factors in photosynthesis. Initially, both lines rise as light intensity increases because light provides energy for the reaction. Line B (higher CO2 concentration) levels off at a higher rate than C because CO2 becomes limiting for C at lower light intensities. Temperature also affects B, limiting its maximum rate.
(b) The other product is oxygen, with the chemical formula \(O_2\). This is derived from the water molecules split during the light-dependent reactions.
(c) Carbohydrates from photosynthesis are used in respiration for energy, stored as starch, converted to sucrose for transport, or used to build cellulose for cell walls. They may also form nectar, fruits, or other compounds like amino acids.
(a) Fig. 5.1 shows the stages involved in protein synthesis.
(i) State the names of the parts labelled X, Y and Z in Fig. 5.1.
(ii) State what determines the sequence of the amino acids in the protein that is produced.
(iii) Explain why the sequence of amino acids is important in the production of receptor molecules for neurotransmitters.
(b) Explain why body cells can have different specialised functions even though they contain the same genes.
(c) Allele frequency in a population can be changed by natural selection and artificial selection. State the meaning of the term allele.
(d) Describe how artificial selection differs from natural selection.
(e) Mutation causes formation of new alleles which increases genetic variation. State two other sources of genetic variation in populations.
▶️ Answer/Explanation
(a)(i)
X – nucleus
Y – mRNA
Z – ribosome
Explanation: X is the nucleus where DNA is transcribed into mRNA (Y), which then travels to the ribosome (Z) for translation into proteins.
(a)(ii) sequence / order, of bases in, mRNA / DNA / gene
Explanation: The sequence of amino acids is determined by the triplet codons in mRNA, which are transcribed from DNA. This ensures the correct protein structure.
(a)(iii) (the sequence of amino acids) determines the shape of the, protein / receptor ,complementary / specific, shape is required for the receptor molecule to, bind / attach / fit, to neurotransmitter
Explanation: The precise folding of the protein (due to amino acid sequence) ensures the receptor’s binding site matches the neurotransmitter, enabling proper signaling.
(b) not all genes are, expressed / switched on / activated
cells only produce the (specific) proteins they need
Explanation: Cell differentiation occurs via selective gene expression. For example, muscle cells express actin/myosin genes, while neurons express neurotransmitter-related genes.
(c) either
an alternative form of a gene
Explanation: Alleles are variants of a gene (e.g., blue/brown eye color alleles). They arise from mutations and contribute to genetic diversity.
(d) humans choose, specific feature(s) / desired features
humans choose, individuals / offspring, to reproduce
selection not influenced by environment / humans manipulate environment
decreases (genetic) variation / decreases size of gene pool
faster / shorter
for economic / aesthetic, reasons
no / less, evolution
e.g. decrease in fitness
Explanation: Artificial selection is intentional (e.g., breeding crops for yield), while natural selection is environment-driven (e.g., camouflage in prey).
(e) Any two from:
meiosis
random mating / cross pollination / crossbreeding , random fertilisation
Explanation: Meiosis shuffles alleles via crossing over, and random fertilization combines gametes unpredictably, increasing variation beyond mutations.
(a) A scientist monitored the changes in the pH in muscles before, during and after two minutes of vigorous exercise.
The changes in pH are caused by the production of lactic acid.
Complete the sentences to describe and explain the results in Fig. 6.1.
The pH decreases from ……………………………………….. to ……………………………………….. during vigorous exercise.
There is not enough ……………………………………….. supplied to the muscles.
The body respires anaerobically. The lactic acid produced builds up in the muscles causing an ……………………………………….. debt.
It takes ……………………………………….. minutes for the muscle pH to return to its initial level after exercise.
The pH value increases after vigorous exercise has ended, as lactic acid is transported in the ……………………………………….. to the ……………………………………….. .
During this time the breathing rate and ……………………………………….. rate remain high.
(b) Yeast can respire anaerobically.
(i) Complete the balanced chemical equation for anaerobic respiration in yeast.
\(C_6H_{12}O_6\)
(ii) Yeast belongs to the kingdom fungus.
State one cell component that is present in yeast cells but is absent in animal cells.
▶️ Answer/Explanation
(a) The pH decreases from 7.07 to 6.55 during vigorous exercise. This occurs due to lactic acid buildup from anaerobic respiration when oxygen is insufficient (oxygen / oxygenated blood). The resulting oxygen debt is repaid post-exercise, taking 31–32 minutes for pH recovery. Lactic acid is transported via the blood(stream) / plasma to the liver, while heart/pulse rate remains elevated.
(b)(i) The balanced equation for yeast anaerobic respiration is: \[ C_6H_{12}O_6 \rightarrow 2C_2H_5OH + 2CO_2 \]
(b)(ii) Yeast cells contain cell wall / (large) vacuole / plasmid(s), which animal cells lack.