Topic – 9.1
(a) Fig. 1.1 is a photograph of a fish. Fig. 1.2 is a photograph of an amphibian.
State two visible features that distinguish the fish in Fig. 1.1 from the amphibian in Fig. 1.2.
(b) Fish, amphibians and mammals are all vertebrate groups. State the name of one other vertebrate group.
(c) Fig. 1.3 shows the circulatory system of a fish. Fig. 1.4 shows the circulatory system of an amphibian.
Describe the similarities and the differences between the circulatory systems of the fish and the amphibian in Fig. 1.3 and Fig. 1.4.
(d) Explain the advantages of the type of circulatory system in mammals compared with the type of circulatory system in fish.
(e) Explain how the structure of arteries and veins relates to the difference in the pressure of the blood transported by these vessels.
(f) Table 1.1 shows the names of some organs and the name of the main artery that brings blood to the organ. Complete Table 1.1.
▶️ Answer/Explanation
Ans:
(a) Fish have fins/scales, while amphibians have limbs/moist skin. These are distinguishing features visible in the images.
(b) Other vertebrate groups include reptiles and birds, which are also part of the five main vertebrate classes.
(c) Both fish and amphibians have a heart with valves, but fish have a 2-chambered heart and a single circulatory system, while amphibians have a 3-chambered heart and an incomplete double circulatory system.
(d) Mammals have a double circulatory system, which maintains higher blood pressure for efficient oxygen delivery, unlike fish where blood pressure drops after passing through gills.
(e) Arteries have thick, muscular walls to withstand high pressure, while veins have thinner walls and valves to prevent backflow due to low pressure.
(f) 
Topic – 14.4
(a) Fig. 2.1 shows the internal body temperature of a human and the external environmental temperature during six hours in one day.
(i) The internal body temperature range is from 36.4°C to 37.0°C. State the range of the external environmental temperature shown in Fig. 2.1.
(ii) Explain the results for the internal body temperature shown in Fig. 2.1.
(b) Fig. 2.2 shows a cross‑section through human skin.
Table 2.1 shows the names of some parts of the skin, the letter identifying the part in Fig. 2.2 and its role in maintaining internal body temperature. Complete Table 2.1.
▶️ Answer/Explanation
(a)(i) Ans: 28.0 (°C) to 39.0 (°C)
The graph shows the external temperature fluctuating between 28.0°C and 39.0°C over the six-hour period, while the internal body temperature remains stable.
(a)(ii) Ans:
The internal body temperature is maintained within a narrow range (36.4°C–37.0°C) due to homeostasis. When external temperature rises, vasodilation and sweating occur to release heat. When external temperature drops, vasoconstriction and shivering help retain heat. The hypothalamus regulates these responses.
(b) Ans:
The table is completed by identifying structures and their roles: sweat glands (cooling via evaporation), hair erector muscles (insulation by raising hairs), and blood vessels (vasodilation/vasoconstriction to regulate heat loss).
Topic – 6.2
(a) Fig. 3.1 shows a drawing of a root hair cell and Fig. 3.2 shows a drawing of a palisade cell.
Explain the reasons for the difference in the numbers of mitochondria and chloroplasts between the root hair cell and the palisade cell, shown in Fig. 3.1 and Fig. 3.2.
(b) Fig. 3.3 is a photomicrograph of a cross‑section of part of a xerophyte leaf.
(i) Explain why the part labelled A in Fig. 3.3 is a tissue.
(ii) Describe two ways the structure labelled B in Fig. 3.3 is adapted for its function.
(iii) Describe one way the leaves of xerophytes are adapted to their environment.
(iv) Describe one way the roots of xerophytes are adapted to their environment.
▶️ Answer/Explanation
(a)
Mitochondria: Root hair cells have more mitochondria because they require energy for active transport of minerals/ions against a concentration gradient. Palisade cells have fewer mitochondria as their primary function is photosynthesis.
Chloroplasts: Palisade cells contain chloroplasts for photosynthesis, while root hair cells lack chloroplasts because they are underground and do not receive light.
(b)(i)
The part labelled A is a tissue because it consists of a group of similar cells working together to perform a specific function (e.g., photosynthesis or structural support).
(b)(ii)
Adaptations of structure B (xylem vessel):
- Thick lignified walls for mechanical strength.
- Hollow lumen for efficient water transport.
(b)(iii)
Xerophyte leaf adaptation: Thick waxy cuticle to reduce water loss through transpiration.
(b)(iv)
Xerophyte root adaptation: Deep or widespread roots to maximize water absorption from a large soil area.
Topic – 5.1
(a) A student investigated the effect of lactase on three different liquids:
- milk
- lactose‑free milk
- sucrose solution.
The student used an indicator to test for the presence of glucose. A sample of each liquid was tested before and after treatment with lactase. The indicator turned brown in the presence of glucose. The indicator remained blue in the absence of glucose. Table 4.1 shows the results of the tests.
(i) Explain the results for the three liquids shown in Table 4.1.
(ii) The student kept the solutions at a temperature that was close to the optimum during the investigation. Using your knowledge of the effect of temperature on enzyme activity, explain why this was important.
(b) As part of a balanced diet, some governments recommend that children drink milk that has vitamin D added to it.
(i) Suggest the dietary reasons for this advice.
(ii) Describe what is meant by a balanced diet.
▶️ Answer/Explanation
Ans
(a)(i) The results show that:
- Milk: Lactase breaks down lactose into glucose (and galactose), causing the indicator to turn brown after treatment.
- Lactose-free milk: Already contains glucose due to prior lactase treatment, so the indicator remains brown before and after.
- Sucrose solution: Lactase does not act on sucrose (enzyme specificity), so no glucose is produced, and the indicator stays blue.
(a)(ii) Maintaining the optimum temperature ensures maximum enzyme activity. At this temperature, kinetic energy is ideal for enzyme-substrate collisions. Deviations (too high or too low) reduce activity—either by denaturing the enzyme or slowing molecular motion.
(b)(i) Vitamin D enhances calcium absorption, which is crucial for developing strong bones and teeth in children. It also prevents deficiencies like rickets.
(b)(ii) A balanced diet provides all essential nutrients (carbohydrates, proteins, fats, vitamins, minerals) in appropriate proportions to meet energy needs and maintain health.
Topic – 8.3
Fig. 5.1 is a graph showing the effect of temperature on the rate of transpiration from the upper and lower surfaces of a leaf that is provided with a constant supply of water
(a) Describe the results shown in Fig. 5.1.
(b) Explain reasons for the shape of the graph for the upper surface of the leaf at X and at Y in Fig. 5.1.
(c) Suggest how the structure of the lower surface differs from the upper surface of the leaf used in this investigation.
▶️ Answer/Explanation
Ans:
(a) The graph shows that the transpiration rate increases with temperature before plateauing. The lower surface has a higher transpiration rate than the upper surface, and both stabilize at nearly the same temperature.
(b) At X, the rate increases because higher temperature boosts evaporation and diffusion. At Y, the rate stabilizes as factors like humidity or stomatal limits restrict further increase.
(c) The lower surface likely has more stomata or a thinner cuticle, facilitating higher transpiration compared to the upper surface.
Topic – 10.1
(a) Polio is a viral disease that can cause nerve damage in humans. In one area, polio vaccination began in 1957. Fig. 6.1 shows the number of cases of polio in this area between 1950 and 1970.
(i) Calculate the percentage change in the number of cases of polio between 1950 and 1952 in Fig. 6.1. Give your answer to two significant figures. Space for working.
(ii) Explain how vaccination causes the results shown between 1958 and 1970 in Fig. 6.1.
(iii) Explain why the polio vaccine does not protect you from other diseases.
(b) Blood clotting helps to prevent some infections. Outline how a blood clot is formed and how it can prevent infections.
(c) State the name of the component of blood responsible for transporting blood cells.
▶️ Answer/Explanation
Ans
(a)(i) – 64 (%) / 64 (%) decrease
Explanation: The percentage decrease is calculated using the formula: \[ \text{Percentage Change} = \left( \frac{\text{Final Value} – \text{Initial Value}}{\text{Initial Value}} \right) \times 100 \] From the graph, the number of cases in 1950 is ~1100 and in 1952 is ~400. Substituting: \[ \left( \frac{400 – 1100}{1100} \right) \times 100 = -63.63\% \approx -64\% \text{ (to 2 s.f.)} \]
(a)(ii) Vaccination introduces weakened/inactivated polio virus, stimulating an immune response. Lymphocytes produce antibodies and memory cells, providing long-term immunity. Herd immunity reduces transmission, leading to fewer cases.
(a)(iii) The polio vaccine is specific—antibodies produced only recognize polio antigens. Memory cells do not respond to other pathogens, so protection is limited to polio.
(b) Blood clotting involves fibrinogen conversion to fibrin, forming a mesh that traps platelets and seals wounds. The clot acts as a barrier, preventing pathogen entry.
(c) plasma
Topic – 20.3
Fig. 7.1 is a flowchart showing the stages of eutrophication.
(a) Complete Fig. 7.1.
(b) A scientist obtained a sample of the bacterial decomposers and grew them in a flask. The resources available for bacterial growth in the flask became limiting. The size of the bacterial population was estimated during the investigation and these data were plotted on a graph.
(i) State the name of the expected shape of the population growth curve that would be drawn on the graph.
(ii) State the name of the initial phase of bacterial growth.
(iii) State one factor, other than a lack of resources, that would cause bacteria to die during the death phase.
▶️ Answer/Explanation
Ans:
(a) The missing labels in the eutrophication flowchart are:
- Nitrate/ammonium/phosphate ions (nutrient input).
- Algae/plants/producers (growth due to excess nutrients).
- Light (required for photosynthesis).
- Photosynthesise (process by which algae grow).
- Aerobic respiration (by decomposers consuming oxygen).
- Dissolved oxygen depletion (leading to aquatic organism death).
(b)(i) The population growth curve is sigmoid (S-shaped), showing lag, exponential, stationary, and death phases.
(b)(ii) The initial phase is the lag phase, where bacteria adapt to the environment before rapid growth.
(b)(iii) Bacteria die due to factors like toxin buildup, pH changes, or overcrowding, not just resource scarcity.