Topic: 2.1
Fig. 1.1 is a diagram of an animal cell.
(a) Complete Table 1.1 by writing the name and function of each of the labelled parts of the cell shown in Fig. 1.1.
(b) (i) Explain how water moves into and out of animal cells.
(ii) A sample of red blood cells was taken from a person. The red blood cells were put into three test‑tubes. Each test‑tube contained a different liquid:
- blood plasma
- pure water
- a very concentrated salt solution.
Fig. 1.2 shows the appearance of the red blood cells when examined using a microscope. Identify the liquids the red blood cells were immersed in. Write your answers in the spaces provided in Fig. 1.2.
(iii) State the name of the molecule in red blood cells that combines with oxygen.
(iv) Fig. 1.3 shows a drawing of a white blood cell.
Identify the type of white blood cell shown in Fig. 1.3.
▶️ Answer/Explanation
Ans:
(a) 
(b)(i) Water moves via osmosis—from high to low water potential through a partially permeable membrane, driven by the random motion of water molecules.
(b)(ii) From top to bottom in Fig. 1.2:
• Blood plasma (normal cell shape)
• Concentrated salt solution (shrunken cells, water leaves by osmosis)
• Pure water (swollen/burst cells, water enters by osmosis)
(b)(iii) Haemoglobin binds oxygen in red blood cells.
(b)(iv) The white blood cell is a phagocyte, identifiable by its lobed nucleus and ability to engulf pathogens.
Topic: 16.3
Fig. 2.1 is a photograph of a spider plant, Chlorophytum comosum. Spider plants can reproduce by producing an identical plantlet, which grows away from the single parent plant.
(a) (i) Identify one feature in Fig. 2.1 that shows that spider plants are monocotyledons.
(ii) Describe the advantages and disadvantages of asexual reproduction in a population of spider plants in the wild.
(b) Fig. 2.2 shows pollen that has been released from the flowers of a hazel tree, Corylus avellana. These flowers are wind‑pollinated.
(i) Explain why a hazel tree has to produce very large amounts of pollen.
(ii) Describe how a stigma from a wind‑pollinated plant is adapted for pollination.
(c) The hazel tree uses cross‑pollination rather than self‑pollination. Describe cross‑pollination.
(d) Outline the events that occur in the flower after pollination.
▶️ Answer/Explanation
(a)(i) Ans: Parallel veins in the leaves or strap-shaped leaves.
Monocotyledons like spider plants have leaves with parallel veins, unlike dicots with branched veins.
(a)(ii) Ans:
- Advantages: Rapid reproduction, no need for pollinators, offspring genetically identical to parent (well-adapted to stable environment).
- Disadvantages: Low genetic variation increases vulnerability to diseases/environmental changes; competition among identical offspring.
(b)(i) Ans: Wind pollination is inefficient, so large pollen quantities increase the chance of pollen reaching a stigma.
(b)(ii) Ans: Feathery/sticky stigmas protrude outside the flower to trap airborne pollen effectively.
(c) Ans: Cross-pollination transfers pollen from anthers of one plant to the stigma of another (same species), promoting genetic diversity.
(d) Ans: Post-pollination events:
- Pollen tube grows down the style.
- Male nucleus travels to the ovule via the pollen tube.
- Fertilization occurs (male + female nuclei fuse).
- Zygote forms, developing into a seed.
Topic: 17.1
(a) Proteins are biological molecules. Explain how proteins are made.
(b) Fig. 3.1 is a diagram of enzyme 1 catalysing the breakdown of molecule A into two smaller molecules, B and C.
Molecule D, shown in Fig. 3.2, is added to the mixture shown in Fig. 3.1. This causes the rate of production of molecules B and C to decrease.
(i) Using the information in Fig. 3.1 and Fig. 3.2, suggest why the rate of production of molecules B and C decreases after molecule D is added.
(ii) Fig. 3.3 shows enzyme 1 after a change in its environmental conditions.
Suggest two changes that could cause the effect on enzyme 1 shown in Fig. 3.3
▶️ Answer/Explanation
(a) Protein Synthesis:
- Transcription: mRNA is synthesized in the nucleus as a complementary copy of a DNA gene.
- mRNA Processing: The mRNA strand exits the nucleus and attaches to a ribosome in the cytoplasm.
- Translation: The ribosome reads the mRNA sequence in codons (3-base groups).
- tRNA Role: Transfer RNA (tRNA) molecules bring specific amino acids to the ribosome based on mRNA codons.
- Peptide Bond Formation: Amino acids are linked via peptide bonds to form a polypeptide chain.
- Folding & Modification: The polypeptide folds into a functional 3D protein structure, sometimes with additional modifications (e.g., glycosylation).
Key Point: The sequence of amino acids (primary structure) determines the protein’s final shape and function.
(b)(i) Decreased Reaction Rate:
Molecule D acts as a competitive inhibitor:
- D has a shape complementary to enzyme 1’s active site, competing with substrate A for binding.
- When D binds, it blocks A from entering the active site, reducing the breakdown of A into B and C.
Note: The inhibition is reversible; increasing substrate A concentration can overcome the effect of D.
(b)(ii) Enzyme Denaturation (Fig. 3.3):
Two environmental changes causing the distorted enzyme shape:
- High Temperature: Excessive heat breaks hydrogen bonds and other weak interactions, altering the active site.
- Extreme pH: Acids or bases disrupt ionic bonds and change the enzyme’s charge distribution, leading to misfolding.
Effect: Denaturation is irreversible—the enzyme permanently loses its catalytic function.
Topic: 18.3
Characteristics of plants are controlled by genes. Alleles are alternative forms of a gene.
(a) Explain how new alleles are formed.
(b) The number of hours of daylight affects when plants produce flowers. This ensures that flowers are produced at an appropriate time of year for each species. This response to the number of hours of daylight is controlled by genes.
(i) Wheat is a crop plant. A farmer wants to produce a type of wheat that flowers when there are fewer hours of daylight. Describe how the farmer could selectively breed this type of wheat.
(ii) Wheat is grown from seeds. State two conditions needed for the germination of seeds.
(iii) Different types of wheat have also evolved through natural selection. Explain how natural selection differs from artificial selection.
(c) Scientists can selectively breed for increased pest resistance in wheat. Explain the benefits of growing wheat with increased pest resistance.
▶️ Answer/Explanation
Ans
(a) New alleles are formed through mutations, which are changes in the DNA base sequence. These can occur spontaneously or be induced by mutagens like radiation or chemicals.
(b)(i) The farmer could selectively breed wheat by:
- Selecting plants that flower under shorter daylight conditions.
- Crossing these plants to produce offspring.
- Repeating the process over generations to stabilize the trait.
(b)(ii) Seed germination requires:
- Water (for metabolic reactions)
- Oxygen (for respiration)
- Suitable temperature (for enzyme activity)
(b)(iii) Natural selection differs from artificial selection:
- Natural selection is driven by environmental pressures, favoring traits that enhance survival and reproduction.
- Artificial selection is human-directed, targeting specific desirable traits (e.g., early flowering).
(c) Benefits of pest-resistant wheat:
- Reduces pesticide use, minimizing environmental pollution.
- Improves crop yield and quality by reducing pest damage.
- Protects non-target species (e.g., pollinators) and farm workers’ health.
Topic: 19.2
Phytoplankton are microorganisms that photosynthesise.
(a) State the chemical equation for photosynthesis.
(b) A scientist made notes about a marine food web. Fig. 5.1 shows the notes she made.
(ii) State the principal source of energy for this food web.
(iii) Table 5.1 describes features of the food web in Fig. 5.2. Complete Table 5.1, using the information in Fig. 5.2
(iv) Fig. 5.3 is a photograph of an orca.
Using the information in Fig. 5.2, predict and explain the most likely effect of a decrease in the orca population on the population size of:
(c) Krill and seals can be harvested to produce omega‑3 fatty acids. Many people take omega‑3 fatty acids as a dietary supplement as they are thought to have health benefits in humans.
(i) List the chemical elements found in all fats.
(ii) Using the information in Fig. 5.2, explain why it is more energy efficient to use krill as a source of omega‑3 fatty acids rather than seals.
(d) Describe how fats ingested by humans are digested and absorbed.
▶️ Answer/Explanation
Ans:
(a) The chemical equation for photosynthesis is:
\(6CO_2 + 6H_2O \xrightarrow{\text{light energy}} C_6H_{12}O_6 + 6O_2\)
(b)(ii) The principal energy source is sunlight, which phytoplankton convert into chemical energy via photosynthesis.
(b)(iii) 
(b)(iv)
Krill: Population would decrease due to increased predation by crabeater seals and Adélie penguins (orca’s prey competitors).
Leopard seals: Population would increase due to reduced predation pressure from orcas.
(c)(i) All fats contain carbon (C), hydrogen (H), and oxygen (O).
(c)(ii) Krill are more energy-efficient because:
1. They occupy a lower trophic level (less energy lost).
2. Energy is lost at each trophic level through metabolism, excretion, and other processes (~90% loss per level).
(d) Fat digestion and absorption:
1. Emulsification by bile salts (in small intestine).
2. Enzymatic breakdown by lipase (from pancreas) into fatty acids + glycerol.
3. Absorption via lacteals in villi (small intestine).
4. Transported as chylomicrons to lymphatic system.
Topic: 20.4
Fig. 6.1 shows a boreal toad, Anaxyrus boreas. This toad is listed as an endangered species.
(a) State the genus of the boreal toad.
(b) One reason for the boreal toad’s recent decrease in population is a disease caused by chytrid fungus. State two reasons other than disease that could cause the boreal toad to become endangered.
(c) Scientists investigated whether increasing the number of beneficial bacteria that live on the toads’ skin could provide protection against the fungus. The scientists took 42 toads and placed 21 in group 1 and 21 in group 2.
• Group 1 – the toads’ skin was treated with beneficial bacteria.
• Group 2 – the toads’ skin was not treated with beneficial bacteria.
The scientists measured the number of toads that became infected with the fungus.
Table 6.1 shows the results.
(i) Calculate the percentage of the 21 toads that have the fungal infection after 100 days in group 1 in Table 6.1. Give your answer to two significant figures. Space for working.
(ii) Using the information in Table 6.1, describe the effect of treating the toads with beneficial bacteria.
(d) There are conservation programmes to reintroduce boreal toads into their natural habitat.
(i) Describe why conservation programmes are set up for animals like the boreal toad.
(ii) Conservation programmes for other endangered animals, such as birds, use artificial insemination as part of a captive breeding programme. Describe how artificial insemination is carried out as part of a captive breeding programme.
▶️ Answer/Explanation
Ans
(a) Anaxyrus
Explanation: The genus is the first part of the binomial name Anaxyrus boreas.
(b) Any two from: climate change, habitat destruction, pollution, introduced species, overharvesting
Explanation: These factors can reduce populations by altering habitats or directly harming toads.
(c)(i) 62%
Calculation: \[ \text{Percentage infected} = \left( \frac{13}{21} \right) \times 100 = 61.9\% \approx 62\% \text{ (to 2 s.f.)} \]
(c)(ii) Bacterial treatment reduces fungal infections after 40 days. By day 100, only 62% of treated toads were infected vs. 100% of untreated toads.
Explanation: Beneficial bacteria compete with or inhibit the fungus, providing protection.
(d)(i) Conservation programs maintain biodiversity, prevent extinction, preserve ecosystems, and protect genetic diversity.
Explanation: These efforts ensure ecological balance and potential future benefits (e.g., medical research).
(d)(ii) Artificial insemination involves:
- Collecting and screening sperm from males
- Inducing ovulation in females
- Inserting sperm into the reproductive tract
- Maximizing genetic diversity by selecting donors
Explanation: This technique boosts breeding success in captive populations.