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    Biology (0610)

    0610/42 May/June 2025 Worked Answers (IGCSE Biology 0610 Extended)

    30 questions · 80 marks · 75 minutes

    Question papers and mark schemes are copyright Cambridge International. We do not reproduce them: the worked answers here are written by The Practice Book. Have the paper open alongside. Get the official paper from Cambridge

    Worked answers for 23 questions
    1. Question 1(a)(ii)

      1 marksFunction of cartilage in the trachea
      Step 1: The cartilage rings are stiff and form a support in the wall of the airway. Step 2: This support holds the airway open so that air can always pass through, even when the pressure inside falls. Step 3: Producing mucus is the job of goblet cells, moving mucus is done by cilia, and gas exchange happens in the alveoli.
      Method:
      Recall that the firm cartilage rings hold the airway open during breathing.
      Examiner tips
      • Cartilage rings keep the airway open so it does not collapse.
    2. Question 1(a)(iii)

      4 marksHow inspiration occurs
      Step 1: During inspiration the diaphragm contracts and flattens while the external intercostal muscles contract to pull the ribs up and out. Step 2: These movements increase the volume of the thorax, so the pressure inside the chest falls below atmospheric pressure. Step 3: Air then moves into the lungs from the higher pressure outside to equalise the pressure.
      Method:
      Link the muscle movements to a larger chest volume, a lower pressure and air moving in.
      Examiner tips
      • Inspiration: diaphragm flattens, ribs up and out, volume up, pressure down, air in.
    3. Step 1: Goblet cells in the lining of the airways secrete sticky mucus that traps pathogens and dust particles in the air. Step 2: Ciliated cells have tiny hair-like cilia that beat to sweep the trapped mucus up and out of the airways. Step 3: Red blood cells, palisade cells and root hair cells do not line the airways, so the other options are wrong.
      Method:
      Name the goblet and ciliated cells and match each to its protective role.
      Examiner tips
      • Goblet cells make mucus to trap; ciliated cells sweep the mucus away.
    4. Question 1(b)(ii)

      4 marksEffect of thick mucus on exercise
      Step 1: Thicker, stickier mucus narrows and blocks the airways, reducing the flow of air and the surface area available for gas exchange. Step 2: Less oxygen therefore reaches the blood and the muscles, so less aerobic respiration can take place. Step 3: With less respiration, less energy is released for muscle contraction, so the person cannot exercise as well.
      Method:
      Trace the chain from blocked airways to reduced oxygen, respiration and energy for exercise.
      Examiner tips
      • Blocked airways mean less oxygen, less respiration and less energy for muscle contraction.
    5. Step 1: Each Ff parent produces gametes carrying F or f, so the cross Ff x Ff gives offspring FF, Ff, Ff and ff. Step 2: Only the ff child has two recessive alleles and therefore has cystic fibrosis, which is 1 of the 4 possible offspring. Step 3: So the probability of a child with cystic fibrosis is 1 in 4, which is 25% or 0.25.
      Method:
      Set out the cross, identify the ff offspring and express its share as a probability.
      Examiner tips
      • For a recessive condition only the double-recessive (ff) offspring is affected.
    6. Step 1: HIV is carried in the blood and other body fluids of an infected person. Step 2: It can pass directly from one person to another through contact with infected blood, such as a shared needle. Step 3: Influenza is spread through the air, cholera through contaminated water, and athlete’s foot by contact with surfaces, so HIV is the blood-borne pathogen.
      Method:
      Recall which pathogen is carried in the blood and spread through blood contact.
      Examiner tips
      • HIV is a blood-borne virus spread by contact with infected blood.
    7. Question 2(b)(ii)

      2 marksIndirect transmission of pathogens
      Step 1: Indirect transmission means the pathogen passes through the environment, such as food, water or objects, before reaching a new person. Step 2: Contaminated food, water and surfaces are all environmental routes, so they are indirect. Step 3: Blood contact, sexual contact, touching an infected person and passing through the placenta or breast milk are all direct routes, so only the first pair is indirect.
      Method:
      Separate routes through the environment (indirect) from body-to-body routes (direct).
      Examiner tips
      • Indirect transmission goes through food, water, air, surfaces or animal vectors.
    8. Question 2(c)

      3 marksRole of antibodies in defence
      Step 1: Each antibody has a specific shape that is complementary to the antigen on a particular pathogen. Step 2: The antibody binds to that antigen, then marks, clumps together or neutralises the pathogen or its toxins. Step 3: This makes the pathogen easier to destroy, for example by phagocytosis, so the first option is correct.
      Method:
      Describe the specific binding of an antibody to an antigen and what it does to the pathogen.
      Examiner tips
      • Antibodies are specific, bind to antigens and mark pathogens for destruction.
    9. Question 2(d)

      3 marksActive versus passive immunity
      Step 1: In active immunity the body is stimulated by infection or vaccination to make its own antibodies and memory cells. Step 2: The memory cells allow a fast secondary response, so active immunity is long-term. Step 3: In passive immunity ready-made antibodies are received from outside, such as in breast milk, so no memory cells are made and the protection is only short-term.
      Method:
      Contrast where the antibodies come from and whether memory cells and long-term protection form.
      Examiner tips
      • Active: body makes antibodies, memory cells, long-term. Passive: ready-made antibodies, short-term.
    10. Question 3(b)(i)

      1 marksMolecule that carries genes
      Step 1: Chromosomes are made of a molecule called DNA. Step 2: A gene is a length of DNA at a particular position that codes for a protein. Step 3: Starch and glycogen are storage carbohydrates and haemoglobin is a protein, so none of these carries genes.
      Method:
      Recall that chromosomes and genes are made of DNA.
      Examiner tips
      • Chromosomes are made of DNA, and genes are sections of DNA.
    11. Step 1: Breeding the GM crop for several generations lets scientists check that the new gene is reliably passed on and expressed in every plant. Step 2: It also lets them confirm the plants give a good, consistent yield and quality and show no unexpected harmful effects. Step 3: Growing many generations also builds up a large stock of seed to sell to farmers, so the first option is correct.
      Method:
      Pick the reasons that show the crop is reliable, useful and safe before farmers grow it.
      Examiner tips
      • Many generations check the gene is stable, useful and safe, and build up seed stocks.
    12. Step 1: A restriction enzyme cuts the required gene out of the donor DNA, and the same enzyme is used to cut open a bacterial plasmid so that the cut ends are complementary. Step 2: The gene is joined into the plasmid using the enzyme ligase, forming recombinant DNA. Step 3: The recombinant plasmid is put into a bacterium, which then multiplies and expresses the gene to make the protein.
      Method:
      Order the steps: cut gene and plasmid, join with ligase, insert plasmid into bacterium.
      Examiner tips
      • Restriction enzyme cuts, ligase joins, plasmid carries the gene into the bacterium.
    13. Step 1: Pollination is prevented if the GM pollen cannot reach the wild plants. Step 2: Growing the crops in glasshouses or removing the anthers stops the pollen from being released or carried to the wild plants. Step 3: Planting the crops close to wild plants, adding more insects, or making more pollen would all increase pollination, so the first option is correct.
      Method:
      Choose the method that physically stops the GM pollen reaching wild plants.
      Examiner tips
      • Isolate the crop, grow it under cover, or remove the anthers to stop pollen spreading.
    14. Question 3(c)(ii)

      2 marksAdvantages of GM crops
      Step 1: GM crops can be engineered to contain extra nutrients, such as added vitamins, improving their nutritional value. Step 2: They can also be given a higher yield or a longer shelf life, which are clear advantages. Step 3: The other options describe impossible or untrue claims, so only the first pair are real advantages.
      Method:
      Select the pair describing real, useful traits added to GM crops.
      Examiner tips
      • GM advantages include better nutrition, higher yield, longer shelf life and disease resistance.
    15. Step 1: An adaptive feature is passed on from parents, so it is an inherited feature. Step 2: It is a feature that increases the chance an organism can survive in its environment. Step 3: It also helps the organism to reproduce, so the words are inherited, survive and reproduce.
      Method:
      Recall the standard definition of an adaptive feature and choose the matching words.
      Examiner tips
      • Adaptive feature = inherited feature that helps an organism survive and reproduce.
    16. Step 1: There are 1000 micrometres in 1 millimetre, so to change micrometres to millimetres you divide by 1000. Step 2: 1.276 ÷ 1000 = 0.001276. Step 3: So 1.276 µm is equal to 0.001276 mm.
      Method:
      Divide the micrometre value by 1000 to convert it to millimetres.
      Examiner tips
      • Micrometres to millimetres: divide by 1000.
    17. Question 4(c)(ii)

      1 marksCarbohydrate that builds cell walls
      Step 1: Plant cell walls are made of the carbohydrate cellulose. Step 2: Cellulose forms strong fibres that give the cell wall its strength and support. Step 3: Glucose is a small sugar, glycogen is an animal storage carbohydrate, and protein is not a carbohydrate, so cellulose is the answer.
      Method:
      Recall that the carbohydrate forming plant cell walls is cellulose.
      Examiner tips
      • Plant cell walls are made of cellulose, a carbohydrate.
    18. Question 4(c)(iii)

      1 marksStem adaptation of a xerophyte
      Step 1: In a dry habitat the plant must save as much water as possible. Step 2: A swollen, fleshy stem can store water inside it for use when there is no rain. Step 3: Thin draining stems, large open pores and no outer covering would all lose water, so they are not adaptations to dryness.
      Method:
      Choose the stem feature that helps store water rather than lose it.
      Examiner tips
      • A swollen, fleshy (succulent) stem stores water for dry periods.
    19. Question 5(a)(ii)

      1 marksWays nitrogen fixation occurs
      Step 1: Nitrogen fixation turns unreactive nitrogen gas into useful nitrogen compounds. Step 2: Lightning provides enough energy to combine nitrogen with oxygen, and nitrogen-fixing bacteria convert nitrogen gas into compounds in the soil and in root nodules. Step 3: Photosynthesis, respiration, evaporation and combustion do not fix nitrogen, so the first option is correct.
      Method:
      Recall the two natural routes that fix nitrogen: lightning and nitrogen-fixing bacteria.
      Examiner tips
      • Nitrogen is fixed by lightning and by nitrogen-fixing bacteria.
    20. Step 1: The movement of water across a partially permeable membrane is called osmosis. Step 2: Water moves from the soil, where there is a higher water concentration, into the cell, where there is a lower water concentration. Step 3: Active transport moves ions using energy, respiration releases energy, and transpiration is water loss from leaves, so osmosis is correct.
      Method:
      Recall that water movement across a partially permeable membrane is osmosis.
      Examiner tips
      • Water crossing a partially permeable membrane moves by osmosis.
    21. Step 1: Both a root hair cell and a bacterial cell have a cell wall, cell membrane, cytoplasm, ribosomes and DNA, so these are the similarities. Step 2: The root hair cell also has a nucleus, mitochondria and a large vacuole, which the bacterial cell does not have. Step 3: The bacterial cell has no nucleus and its DNA forms a loop, so the first option correctly gives the differences and similarities.
      Method:
      Separate the structures common to both cells from those found only in the plant cell.
      Examiner tips
      • Both have wall, membrane, cytoplasm, ribosomes, DNA; only the plant cell has a nucleus, mitochondria and vacuole.
    22. Step 1: Enzyme M works best in alkaline conditions, so a pH of 2 is far from its optimum. Step 2: The very acidic conditions denature the enzyme, changing the shape of its active site. Step 3: The protein substrate can no longer fit into the changed active site, so no jelly is broken down.
      Method:
      Link the wrong pH to denaturation, a changed active site and loss of activity.
      Examiner tips
      • Extreme pH denatures an enzyme, changing the active site so the substrate cannot bind.
    23. Question 6(c)

      2 marksIdentifying a stomach protease
      Step 1: Enzyme L is a protease that works best at the very low pH of about 2. Step 2: The stomach is the part of the digestive system with strongly acidic conditions, and the protease that works there is pepsin. Step 3: Amylase digests starch in the mouth, lipase digests fats in the small intestine, and trypsin works in the small intestine, so pepsin in the stomach is correct.
      Method:
      Match the acid-loving protease to pepsin and locate it in the stomach.
      Examiner tips
      • Pepsin is the protease of the acidic stomach.

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