0610/22 Oct/Nov 2025 Worked Answers (IGCSE Biology 0610 Extended)
40 questions · 40 marks · 45 minutes
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Worked answers for 27 questions
- Step 1: A neurone is a single specialised cell. Step 2: The brain and the liver are each made of several different tissues working together, so each is an organ. Step 3: This gives organ, organ and cell for the brain, liver and neurone.Method:Decide whether each structure is a single cell, a tissue or a whole organ.Examiner tips
- Cell to tissue to organ to organ system is the order of increasing size.
- Step 1: Pair each base with its partner: A with T and C with G. Step 2: A C T T A G A T pairs with T G A A T C T A. Step 3: So the complementary strand reads TGAATCTA.Method:Write the complementary base beneath each base using the pairing rules.Examiner tips
- Complementary base pairing: A-T and C-G.
- Step 1: Enzymes are proteins that act as biological catalysts. Step 2: Pepsin is a protease enzyme that digests protein in the stomach. Step 3: Fibrinogen is a clotting protein, mRNA is a nucleic acid, and saliva is a fluid containing the enzyme amylase, so none of these is itself an enzyme.Method:Identify which substance is a protein that catalyses a reaction.Examiner tips
- Many enzyme names end in -ase, but pepsin is a well-known exception.
- Step 1: In the duodenum, bile and digestive enzymes act on food, so digestion takes place there. Step 2: Ingestion happens at the mouth and egestion at the anus. Step 3: Assimilation happens in body cells after absorption, so the function in the duodenum is digestion.Method:Match each term to where it happens along the gut and pick the duodenum role.Examiner tips
- Digestion continues in the duodenum where enzymes and bile are added.
- Step 1: Water is a small molecule and is absorbed in both the small intestine and the colon. Step 2: Amino acids and glucose are small molecules but are absorbed in the small intestine, not the colon. Step 3: Maltose is a double sugar, so it is not a small molecule, which leaves water as the correct row.Method:Check both whether each substance is a small molecule and where it is absorbed.Examiner tips
- The colon mainly absorbs water; most nutrients are absorbed in the small intestine.
- Step 1: Water on the surfaces of the mesophyll cells turns from liquid into water vapour. Step 2: This change from liquid to vapour is evaporation. Step 3: The vapour then diffuses out of the leaf through the stomata, so the process moving water into the air spaces is evaporation.Method:Identify the change from liquid water to water vapour as evaporation.Examiner tips
- Water evaporates from mesophyll cell surfaces, then diffuses out through stomata.
- Step 1: Greasing the upper surface blocks water loss from it, so loss through the upper surface = (no grease) minus (upper greased) = 12.0 minus 8.7 = 3.3 g in seven days. Step 2: The mean daily rate = 3.3 g divided by 7 days. Step 3: 3.3 divided by 7 = 0.47 g per day.Method:Use the difference between the two treatments, then convert to a daily rate.Examiner tips
- Subtract to find the loss through one surface, then divide by the number of days.
- Step 1: Translocation is the transport of dissolved food in the phloem. Step 2: The substances moved are sucrose and amino acids. Step 3: They move from a source (where they are made or stored) to a sink (where they are used or stored), giving sucrose and amino acids from source to sink.Method:Recall the substances and direction of transport in the phloem.Examiner tips
- Phloem carries sucrose and amino acids from source to sink.
- Step 1: Urea is made in the liver, so blood leaving the liver in the hepatic vein has the highest urea concentration. Step 2: The kidneys remove urea from the blood, so blood leaving the kidney in the renal vein has the lowest urea concentration. Step 3: This gives the hepatic vein for the highest and the renal vein for the lowest.Method:Track where urea is added (liver) and removed (kidney) and name the vessels leaving each.Examiner tips
- Blood leaving the liver is highest in urea; blood leaving the kidney is lowest.
- Step 1: When breathing in, the diaphragm contracts and flattens, increasing the volume of the chest. Step 2: The external intercostal muscles also contract, pulling the ribs up and out. Step 3: Both muscles therefore contract, increasing chest volume and drawing air in.Method:Recall the muscle actions that increase chest volume during inhalation.Examiner tips
- Breathing in: diaphragm and external intercostals both contract.
- Step 1: During hard exercise muscles respire anaerobically and produce lactic acid, which diffuses into the blood, so statement 1 is correct. Step 2: The blood carries the lactic acid to the liver, so statement 2 is correct. Step 3: In the liver the lactic acid is broken down using oxygen (aerobically), so statement 3 is wrong, leaving 1 and 2 only.Method:Judge each statement and remember lactic acid is removed aerobically in the liver.Examiner tips
- Lactic acid is broken down in the liver using oxygen (the oxygen debt).
- Step 1: Moving into bright light makes the pupil get smaller, so the circular muscles contract and the radial muscles relax. Step 2: Looking at a near object makes the ciliary muscles contract, so the lens becomes thicker (more rounded). Step 3: The correct set is radial muscles relax, pupil constricts, ciliary muscles contract and the lens gets thicker.Method:Work out the pupil response to bright light and the lens response to a near object.Examiner tips
- Bright light: pupil constricts. Near object: ciliary muscles contract and lens thickens.
- Step 1: Adrenaline prepares the body for action ("fight or flight"). Step 2: It increases the blood glucose concentration to supply more energy and increases the pulse rate to deliver glucose and oxygen faster. Step 3: So both the blood glucose concentration and the pulse rate increase.Method:Recall adrenaline fight-or-flight effects on glucose and heart rate.Examiner tips
- Adrenaline raises blood glucose and increases heart rate.
- Step 1: Resistance spreads because bacteria exposed to antibiotics are selected for survival if they are resistant. Step 2: Reducing unnecessary exposure lowers this selection pressure. Step 3: Using antibiotics only when they are needed, to treat bacterial infections, limits the development of resistance.Method:Choose the action that reduces unnecessary exposure of bacteria to antibiotics.Examiner tips
- Only use antibiotics for bacterial infections, and finish the full course.
- Step 1: Asexual reproduction involves only one parent and produces genetically identical offspring. Step 2: A single bacterium dividing into two identical bacteria has one parent and identical offspring. Step 3: The other examples all involve gametes and fertilisation, so they are sexual reproduction.Method:Look for the example with one parent and identical offspring.Examiner tips
- No gametes and no fertilisation means asexual reproduction.
- Step 1: Male secondary sexual characteristics are controlled by the hormone testosterone. Step 2: Testosterone is made in the testes and causes changes such as a deeper voice and body hair. Step 3: Insulin and glucagon control blood glucose, and lactase is a digestive enzyme, so testosterone is the answer.Method:Recall the male sex hormone and rule out the glucose hormones and the enzyme.Examiner tips
- Testosterone is the male sex hormone made in the testes.
- Step 1: The fetus makes the waste product urea, which must be removed. Step 2: Urea passes across the placenta from the fetus blood into the mother blood to be excreted by the mother. Step 3: Glucose and oxygen move from mother to fetus, carbon dioxide moves from fetus to mother, and glycogen does not cross, so urea moving from fetus to mother is correct.Method:Decide whether each substance is a food, a gas or a waste, and set its net direction.Examiner tips
- Food and oxygen go to the fetus; wastes (carbon dioxide, urea) go to the mother.
- Step 1: HIV is carried in body fluids such as blood and is spread when these fluids pass between people. Step 2: Sharing a needle passes infected blood directly into another person, so action 2 transmits HIV. Step 3: Hugging and sharing a plate do not transfer body fluids, so only action 2 can transmit HIV.Method:Pick the action that transfers body fluids between people.Examiner tips
- HIV spreads through blood and other body fluids, not casual contact.
- Step 1: Males have the sex chromosomes XY in their body cells. Step 2: During meiosis the pair separates, so each sperm receives only one sex chromosome. Step 3: That single chromosome is either an X or a Y, so a single sperm contains X or Y.Method:Recall the male sex chromosomes and that meiosis gives one per sperm.Examiner tips
- Each sperm carries one sex chromosome: either X or Y.
- Step 1: Every body cell contains the same genes, including the genes for antibodies. Step 2: Cells only make a protein if the gene for it is switched on (expressed). Step 3: Antibody genes are expressed only in lymphocytes, so only lymphocytes make antibody proteins.Method:Recall that all cells share the same genes but express different ones.Examiner tips
- Different cell types express different genes from the same gene set.
- Step 1: During protein synthesis the ribosome is the site where the protein is built. Step 2: It joins amino acids together in the order set by the mRNA to form a chain. Step 3: The mRNA carries the code from the gene, while the ribosome assembles the amino acids into the protein.Method:Recall the ribosome role as the site of joining amino acids.Examiner tips
- Ribosomes assemble amino acids into proteins using the mRNA code.
- Step 1: The speckled chicken has one white allele and one black allele, while the white chicken has two white alleles. Step 2: The speckled parent passes on either a white or a black allele; the white parent always passes on a white allele. Step 3: This gives half the offspring with two white alleles (white) and half with one white and one black allele (speckled), a ratio of 1 white : 1 speckled.Method:Write the alleles for each parent and combine the gametes in a cross.Examiner tips
- A heterozygous-by-homozygous cross gives a 1:1 phenotype ratio.
- Step 1: Stem cells are unspecialised cells that have not yet developed into a particular type of cell. Step 2: They divide by mitosis to produce more cells, some of which can then specialise. Step 3: So stem cells are unspecialised cells that divide by mitosis.Method:Recall that stem cells are unspecialised and divide by mitosis.Examiner tips
- Stem cells are unspecialised and divide by mitosis.
- Step 1: A floating leaf has its upper surface in the air and its lower surface in the water. Step 2: Stomata must be on the upper surface so that gases can be exchanged with the air, not blocked by water below. Step 3: So a hydrophyte with floating leaves has many stomata on the upper surface.Method:Think about which leaf surface is in air and where stomata must be.Examiner tips
- Floating leaves have stomata on the upper surface and large air spaces to float.
- Step 1: A pyramid of energy shows the energy passing through each trophic level in a fixed area over a fixed time. Step 2: So the unit must include energy (kJ), an area (per m²) and a time (per year). Step 3: The unit kJ per m² per year includes all three, so it is correct.Method:Check that the unit includes energy, area and time.Examiner tips
- Energy flow units combine energy, area and time, for example kJ/m²/year.
- Step 1: Dead organisms contain proteins that need to be broken down. Step 2: Decomposer microorganisms break down these proteins and release ammonium ions. Step 3: Nitrification, denitrification and nitrogen fixation act on other nitrogen compounds, so decomposition is the process that makes ammonium ions from protein.Method:Match the breakdown of protein to ammonium with the work of decomposers.Examiner tips
- Decomposers break down proteins in dead material to release ammonium ions.
- Step 1: A monoculture grows one crop species over a large area. Step 2: This allows large amounts of the same crop to be planted, managed and harvested efficiently, giving an increased crop yield. Step 3: Lower biodiversity, more disease and less variation are disadvantages, so the advantage is the increased crop yield.Method:Separate the benefit (yield) from the drawbacks of monocultures.Examiner tips
- Monocultures increase yield but reduce biodiversity and increase disease risk.
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