Cambridge IGCSE · thinka-original Practice Paper

2024 Cambridge IGCSE Biology (0610) Practice Paper with Answers

Thinka Nov 2024 (V3) Cambridge IGCSE-Style Mock — Biology (0610)

80 marks75 mins2024
An original Thinka practice paper modelled on the structure and difficulty of the Nov 2024 (V3) Cambridge IGCSE Biology (0610) paper. Not affiliated with or reproduced from Cambridge.

Extended Theory Paper

Answer all questions. Use a dark blue or black pen. You may use a calculator. Show all your workings.
6 Question · 80 marks
Question 1 · structured
13 marks
1 (a) Define physical digestion and explain how it differs from chemical digestion. [2]

(b) Describe how mechanical digestion by teeth assists the chemical digestion of food in the human body. [3]

(c) Complete Table 1.1 by writing the correct enzyme, its main substrate, or the product(s) formed in the human alimentary canal.

Table 1.1
| Enzyme | Secreted by | Substrate | Product(s) |
| :--- | :--- | :--- | :--- |
| Amylase | Salivary glands / Pancreas | ................................................ | Maltose |
| Lipase | Pancreas | Fats and oils | ................................................ |
| ................................................ | Gastric glands | Protein | Peptides / amino acids |
| Maltase | Small intestine epithelium | Maltose | ................................................ |
[4]

(d) Explain how hydrochloric acid in gastric juice ensures efficient protein digestion by pepsin and protects the body against infection. [4]
Show answer & marking scheme

Worked solution

(a) Physical digestion is the breakdown of food into smaller pieces without any chemical change to the food molecules. In contrast, chemical digestion is the chemical breakdown of large, insoluble molecules into small, soluble molecules, typically mediated by enzymes.

(b) Chewing and grinding food with teeth breaks it down into physically smaller pieces, which significantly increases the total surface area of the food. This larger surface area allows digestive enzymes, such as salivary amylase, to make more contact and digest the food molecules more rapidly and efficiently.

(c) The completed rows of Table 1.1 are:
- Substrate for Amylase: Starch
- Product(s) for Lipase: Fatty acids and glycerol
- Enzyme for Protein: Pepsin (or protease)
- Product(s) for Maltase: Glucose

(d) Hydrochloric acid in the stomach provides a highly acidic environment (optimum pH around 1.5 - 2.0). This low pH activates pepsinogen into the active enzyme pepsin and denatures proteins, exposing their peptide bonds for easier enzymatic cleavage. Furthermore, this extreme acidity destroys the cell membranes and enzymes of ingested pathogens / bacteria, protecting the body against infection.

Marking scheme

(a) Max 2 marks:
- Definition of physical digestion: breakdown of food into smaller pieces without changing chemical composition / chemical structure [1]
- Distinction: chemical digestion involves chemical change / breaking of chemical bonds / enzyme action [1]

(b) Max 3 marks:
- Chewing / grinding food breaks it into smaller pieces / mechanical breakdown [1]
- Increases the surface area of the food [1]
- Increases rate of enzyme action / more frequent successful collisions between enzymes and food substrates [1]
- Mixes food with saliva / lubricates food to form a bolus [1]

(c) 4 marks total:
- 1 mark for each correct entry: Starch [1]; Fatty acids and glycerol (both required) [1]; Pepsin (accept protease) [1]; Glucose [1]

(d) Max 4 marks:
- HCl provides acidic conditions / optimum pH (1.5 - 2.0) for pepsin [1]
- Denatures dietary proteins to expose peptide bonds for digestion [1]
- Kills bacteria / pathogens present in food [1]
- Disrupts cell membranes / denatures enzymes of pathogens [1]
Question 2 · structured
13 marks
2 (a) Define translocation and state the plant tissue responsible for this process. [2]

(b) With reference to the terms source and sink, explain the movement of sucrose in a plant during:

(i) early spring, when new shoots are growing from an underground storage tuber. [3]

(ii) late summer, when the plant is actively photosynthesising and storing nutrients in its tuber. [3]

(c) Describe how the structure of phloem sieve tube elements is adapted to their function. [3]

(d) Distinguish between the transport of water in the xylem and the translocation of sucrose in the phloem. [2]
Show answer & marking scheme

Worked solution

(a) Translocation is the movement of sucrose and amino acids in phloem from a source to a sink. The plant tissue responsible for this process is the phloem.

(b) (i) In early spring, the underground storage tuber acts as the source because starch stored inside it is converted into sucrose and released. The growing shoots act as the sink because they actively consume sucrose for cell division, respiration, and growth. Thus, sucrose is translocated upwards from the tuber to the shoots.

(ii) In late summer, the mature green leaves act as the source because they photosynthesise and produce a surplus of sucrose. The storage tuber acts as the sink because it imports sucrose and converts it to starch for long-term storage. Thus, sucrose is translocated downwards from the leaves to the tuber.

(c) Sieve tube elements are adapted by having sieve plates with large pores, allowing continuous flow of cytoplasmic sap. They also lack a nucleus, ribosomes, and a large vacuole, leaving the lumen clear to reduce resistance to flow. Furthermore, they are closely associated with companion cells, which contain many mitochondria to generate the ATP required for active loading of sucrose.

(d) Xylem transport is unidirectional (upwards from roots to leaves only), whereas phloem translocation is bidirectional (from source to sink, upwards or downwards). Xylem transports water and dissolved mineral ions passively (driven by transpiration pull), whereas phloem transports organic solutes (sucrose and amino acids) actively (requiring metabolic energy).

Marking scheme

(a) Max 2 marks:
- Definition: transport of sucrose and amino acids from a source to a sink [1]
- Tissue: phloem [1]

(b) (i) Max 3 marks:
- Underground tuber is the source AND growing shoots are the sink [1]
- Starch in tuber is converted to sucrose [1]
- Sucrose is transported upwards to support shoot respiration / growth [1]

(b) (ii) Max 3 marks:
- Leaves are the source AND tuber is the sink [1]
- Leaves photosynthesise to produce sucrose [1]
- Sucrose is transported downwards to the tuber to be stored as starch [1]

(c) Max 3 marks:
- Sieve plates with pores allow easy flow of cell sap [1]
- Absence of nucleus / ribosomes / vacuole / minimal cytoplasm reduces resistance to flow [1]
- Companion cells contain many mitochondria to provide ATP for active loading [1]

(d) Max 2 marks:
- Xylem is unidirectional (upwards only) vs phloem is bidirectional (upwards/downwards) [1]
- Xylem transports water and minerals vs phloem transports sucrose and amino acids [1]
- Xylem transport is passive (transpiration pull) vs phloem transport is active [1]
Question 3 · structured
13 marks
3 (a) Distinguish between active immunity and passive immunity. Include one example of how each type of immunity can be acquired in your answer. [4]

(b) Describe how the body's physical and chemical barriers defend against pathogens. [3]

(c) Explain the role of the following white blood cells in the human immune response:

(i) phagocytes [3]

(ii) lymphocytes [3]
Show answer & marking scheme

Worked solution

(a) Active immunity is the body's defense against a pathogen by antibody production within the organism itself. It is long-term and leads to the production of memory cells. It can be acquired naturally through infection or artificially through vaccination. Passive immunity is the short-term defense against a pathogen by ready-made antibodies acquired from another individual. It does not produce memory cells. It can be acquired naturally across the placenta or through breast milk, or artificially by injection of an antibody serum.

(b) Physical barriers, such as the skin, form a continuous outer layer that prevents the entry of pathogens, while nose hairs trap dust and airborne microbes. Chemical barriers, such as mucus, trap inhaled microbes, and hydrochloric acid in the stomach kills ingested bacteria. Saliva and tears also contain antimicrobial enzymes (lysozyme) that destroy bacterial cell walls.

(c) (i) Phagocytes carry out phagocytosis. They detect foreign pathogens, move towards them, and engulf them by surrounding them with pseudopodia. Once enclosed in a phagocytic vacuole, lysosomes fuse with the vacuole to release digestive enzymes, breaking down and killing the pathogen.

(ii) Lymphocytes recognize specific foreign antigens on pathogens. They produce complementary antibodies that bind specifically to these antigens. This causes agglutination (clumping of pathogens) to make it easier for phagocytes to engulf them, or neutralizes toxins. Some lymphocytes also differentiate into memory cells, which persist in the body to launch a rapid, massive immune response upon re-infection.

Marking scheme

(a) Max 4 marks:
- Active immunity: defense by antibody production by the host organism / long-term defense [1]
- Example of active: natural infection / vaccination [1]
- Passive immunity: short-term defense by ready-made antibodies from another source / does not produce memory cells [1]
- Example of passive: antibodies across placenta / breast milk / injection of antiserum [1]

(b) Max 3 marks:
- Skin as a physical barrier preventing entry [1]
- Hairs in nose trap dust / pathogens [1]
- Mucus traps microbes [1]
- Stomach acid / HCl kills ingested pathogens [1]
- Lysozyme in tears / saliva kills bacteria [1]

(c) (i) Max 3 marks:
- Phagocytosis / engulfing pathogens [1]
- Formation of vacuole / vesicle [1]
- Release of lysosomal / digestive enzymes to break down pathogen [1]

(c) (ii) Max 3 marks:
- Production of specific / complementary antibodies [1]
- Antibodies bind to antigens on pathogens [1]
- Agglutination / neutralisation of toxins [1]
- Production of memory cells for long-term immunity [1]
Question 4 · structured
14 marks
4 (a) State the word equation for anaerobic respiration in:

(i) human muscle cells. [1]

(ii) yeast cells. [1]

(b) Compare anaerobic respiration in human muscle cells with aerobic respiration in human muscle cells. [4]

(c) A runner completed a 200 m sprint race.

(i) Explain why the runner's breathing rate and heart rate remained high for several minutes after the race finished. Use the term oxygen debt in your explanation. [4]

(ii) Describe how the lactic acid produced during the sprint is removed from the body. [4]
Show answer & marking scheme

Worked solution

(a) (i) glucose \(\rightarrow\) lactic acid

(ii) glucose \(\rightarrow\) carbon dioxide + ethanol (alcohol)

(b) Anaerobic respiration occurs without oxygen, whereas aerobic respiration requires oxygen. Anaerobic respiration produces only lactic acid as a product, while aerobic respiration produces carbon dioxide and water. Anaerobic respiration releases a much smaller amount of energy per glucose molecule compared to aerobic respiration. Additionally, anaerobic respiration occurs entirely in the cytoplasm, whereas aerobic respiration involves mitochondria.

(c) (i) During the sprint, the runner's muscles required energy very rapidly, exceeding the rate at which oxygen could be delivered. Consequently, the muscles respired anaerobically, causing lactic acid to build up and creating an oxygen debt. After the race, the heart and breathing rates must stay high to supply the extra oxygen required to oxidize and break down this accumulated lactic acid, thereby repaying the oxygen debt.

(ii) Lactic acid is carried in the bloodstream from the contracting muscle tissues to the liver. In the liver, some of the lactic acid is aerobically respired / oxidized using oxygen, breaking it down into carbon dioxide and water. The remaining lactic acid is converted back into glucose and stored as glycogen in the liver cells.

Marking scheme

(a) (i) 1 mark:
- glucose -> lactic acid [1] (reject if oxygen or energy is shown on left; ignore energy on right)

(a) (ii) 1 mark:
- glucose -> carbon dioxide + ethanol / alcohol [1]

(b) Max 4 marks:
- Anaerobic does not use oxygen vs aerobic does [1]
- Anaerobic produces lactic acid vs aerobic produces carbon dioxide and water [1]
- Anaerobic releases less energy (per glucose molecule) vs aerobic releases much more [1]
- Anaerobic occurs only in cytoplasm vs aerobic involves mitochondria [1]
- Both use glucose as the starting substrate [1]

(c) (i) Max 4 marks:
- Anaerobic respiration occurs during sprint due to insufficient oxygen delivery [1]
- Lactic acid builds up in muscles / blood [1]
- Creating an oxygen debt [1]
- Extra oxygen is needed post-race to oxidize / break down lactic acid [1]
- Elevated breathing and heart rates deliver extra oxygen to target tissues [1]

(c) (ii) Max 4 marks:
- Lactic acid is transported in blood [1]
- From muscles to liver [1]
- Oxidized using oxygen to produce carbon dioxide and water [1]
- Some lactic acid converted back to glucose / glycogen [1]
Question 5 · structured
13 marks
5 (a) Explain the difference between continuous variation and discontinuous variation, providing one example of each in humans. [4]

(b) Flower colour in a plant species is controlled by a single gene with two codominant alleles: \(C^R\) (red) and \(C^W\) (white). Heterozygous plants have pink flowers.

(i) State the genotype of a plant with pink flowers. [1]

(ii) Draw a genetic diagram to show the expected phenotypic ratio of the offspring when two plants with pink flowers are crossed. [5]

(c) Explain why the inheritability of flower colour in this species is classified as discontinuous variation. [3]
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Worked solution

(a) Continuous variation has a complete range of intermediate phenotypes between two extremes (e.g., height or mass), whereas discontinuous variation has distinct, separate categories with no intermediates (e.g., ABO blood groups). Continuous variation is typically influenced by both genetics and environmental factors, while discontinuous variation is determined solely by genes.

(b) (i) \(C^R C^W\)

(ii)
- Parental Phenotypes: Pink flower \(\times\) Pink flower
- Parental Genotypes: \(C^R C^W \times C^R C^W\)
- Gametes: \(C^R\), \(C^W\) and \(C^R\), \(C^W\)
- Offspring Genotypes: \(C^R C^R\), \(C^R C^W\), \(C^R C^W\), \(C^W C^W\)
- Offspring Phenotypes: 1 Red (\(C^R C^R\)) : 2 Pink (\(C^R C^W\)) : 1 White (\(C^W C^W\))
- Phenotypic Ratio: 1 Red : 2 Pink : 1 White

(c) The flower colour is determined solely by the alleles inherited, which means the plants fall into three distinct, non-overlapping phenotypic categories (red, pink, or white). There are no intermediate shades on a continuous scale (e.g., no light-pinkish-red on a spectrum), and the environment has no influence on the phenotypic outcome.

Marking scheme

(a) Max 4 marks:
- Continuous variation: range of intermediate phenotypes / no distinct categories [1]
- Example of continuous: height / mass / skin color [1]
- Discontinuous variation: distinct categories / no intermediates [1]
- Example of discontinuous: ABO blood groups / gender / tongue rolling [1]
- Cause: continuous is genetic and environmental vs discontinuous is genetic only [1]

(b) (i) 1 mark:
- \(C^R C^W\) [1]

(b) (ii) Max 5 marks:
- Parental genotypes: \(C^R C^W \times C^R C^W\) [1]
- Gametes shown: \(C^R\) and \(C^W\) from both parents [1]
- Offspring genotypes correctly determined: \(C^R C^R\), \(C^R C^W\), \(C^R C^W\), \(C^W C^W\) [1]
- Phenotypes linked to genotypes: red, pink, white [1]
- Phenotypic ratio: 1 Red : 2 Pink : 1 White [1]

(c) Max 3 marks:
- Results in distinct, separate categories with no intermediates [1]
- Determined solely by genes / alleles / genotypes [1]
- Not affected / influenced by environmental factors [1]
Question 6 · structured
14 marks
6 (a) Human insulin is produced commercially using genetically modified bacteria. Describe the process of producing insulin using recombinant DNA technology, naming the key enzymes involved in cutting and joining DNA. [6]

(b) Explain why bacteria are useful organisms for biotechnology and genetic modification. [4]

(c) People with haemophilia cannot clot their blood effectively when injured.

(i) State the role of platelets in normal blood clotting. [2]

(ii) Explain the physiological importance of blood clotting. [2]
Show answer & marking scheme

Worked solution

(a) The human gene that codes for insulin is first identified and isolated from pancreatic cells. The gene is cut out using a specific restriction enzyme, which leaves complementary single-stranded DNA ends called sticky ends. At the same time, a plasmid vector is extracted from a bacterium and cut open using the exact same restriction enzyme to ensure matching sticky ends. The human insulin gene and the cut plasmid are joined together using the enzyme DNA ligase, forming a recombinant plasmid. This recombinant plasmid is inserted back into a bacterial cell. These transformed bacteria are grown under controlled conditions in industrial fermenters, where they replicate rapidly and express the human gene to produce insulin, which is then harvested and purified.

(b) Bacteria are highly advantageous in biotechnology because they have a rapid rate of reproduction, allowing massive yields of a protein in a short time. They contain plasmids, which are easily extracted and serve as ideal vectors for recombinant DNA. Additionally, they share the universal genetic code, enabling them to read and translate human genes. Finally, growing bacteria in fermenters has minimal space requirements and few ethical concerns compared to using animals.

(c) (i) When a blood vessel is damaged, platelets clump together at the site of the injury to form a temporary plug. They also release chemical substances that trigger a cascade of reactions, converting the soluble plasma protein fibrinogen into insoluble fibrin, which forms a mesh to trap red blood cells.

(ii) Blood clotting is essential because it seals broken blood vessels to prevent excessive loss of blood (haemorrhage). Moreover, the clot forms a barrier that prevents the entry of harmful pathogens into the bloodstream, minimizing the risk of infection.

Marking scheme

(a) Max 6 marks:
- Human insulin gene identified and isolated [1]
- Cut using restriction enzymes [1]
- Creating sticky ends [1]
- Plasmid vector cut using the same restriction enzyme [1]
- DNA ligase used to join gene and plasmid [1]
- Forms a recombinant plasmid [1]
- Recombinant plasmid inserted into bacteria [1]
- Bacteria cultured in fermenter to express insulin [1]

(b) Max 4 marks:
- Rapid reproduction rate [1]
- Plasmids are easy to extract / manipulate [1]
- Genetic code is universal (so human genes can be translated) [1]
- Simple nutritional requirements / grown easily in fermenters [1]
- No / few ethical concerns compared to using animals [1]

(c) (i) Max 2 marks:
- Form a temporary plug at site of injury [1]
- Release chemicals to trigger conversion of soluble fibrinogen to insoluble fibrin [1]
- Fibrin forms a mesh that traps red blood cells [1]

(c) (ii) Max 2 marks:
- Prevents excessive blood loss / haemorrhage [1]
- Prevents entry of bacteria / pathogens to reduce infection risk [1]

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