Cambridge IGCSE · Thinka 原創模擬試題

2024 Cambridge IGCSE Biology (0610) 模擬試題連答案詳解

Thinka Nov 2024 (V3) Cambridge International A Level-Style Mock — Biology (0610)

80 75 分鐘2024
An original Thinka practice paper modelled on the structure and difficulty of the Nov 2024 (V3) Cambridge International A Level 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 題目 · 79.98
題目 1 · structured
13.33
(a) Describe the steps involved in using genetic engineering to produce human insulin from bacteria. [6 marks]

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

(c) State one other medical product made by genetically modified bacteria and its purpose. [3.33 marks]
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解題

(a) The gene for human insulin is identified and cut out of human DNA using a restriction enzyme, leaving sticky ends. A plasmid is extracted from a bacterium and cut open using the same restriction enzyme, which ensures complementary sticky ends. The human insulin gene is inserted into the cut plasmid and joined using the enzyme DNA ligase to form a recombinant plasmid. This recombinant plasmid is inserted back into a bacterium (host cell). The transgenic bacteria are grown in a fermenter, where they multiply rapidly and produce human insulin, which is then extracted and purified.

(b) Bacteria are highly useful for genetic modification because they have rapid reproduction rates, meaning large quantities of the product can be made quickly. They possess plasmids which are easy to extract, manipulate, and re-insert. They have simple nutritional requirements, and there are fewer ethical concerns compared to using animals. Also, they share the same universal genetic code, allowing them to translate human genes.

(c) Human growth hormone (HGH) is another medical product made by genetically modified bacteria. It is used to treat growth disorders/deficiencies in children.

評分準則

Part (a): Max 6 marks
- Identify and cut out human insulin gene using a restriction enzyme (1)
- Leaves sticky ends (1)
- Cut bacterial plasmid open with the same restriction enzyme (1)
- Ensures complementary sticky ends (1)
- Join human insulin gene and plasmid using DNA ligase (1)
- Form recombinant plasmid (1)
- Insert plasmid back into bacterium (1)
- Grow transgenic bacteria in a fermenter to produce insulin (1)

Part (b): Max 4 marks
- Rapid reproduction rate (1)
- Presence of plasmids which are easy to manipulate/transfer (1)
- No/few ethical concerns (compared to animals) (1)
- Simple nutritional requirements / cheap to culture (1)
- Universal genetic code allows translation of human genes (1)

Part (c): Max 3.33 marks
- Named medical product (e.g., human growth hormone / factor VIII / hepatitis B vaccine) (1.33)
- Correct purpose (e.g., treatment of growth deficiency / haemophilia / prevention of hepatitis B infection) (2)
題目 2 · structured
13.33
(a) Define the term translocation. [2 marks]

(b) Distinguish between a 'source' and a 'sink' in a plant, providing examples of each during different times of a plant's life cycle. [5 marks]

(c) Describe how the structure of phloem sieve tube elements is adapted to their function. [4 marks]

(d) Explain how aphids can be used to study the contents of phloem tissue. [2.33 marks]
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解題

(a) Translocation is the movement of sucrose and amino acids in phloem from a source to a sink.

(b) A source is a part of a plant that releases sucrose or amino acids, such as photosynthesising leaves (in summer) or germinating seeds/storing tubers (in early spring). A sink is a part of a plant that receives or stores sucrose or amino acids, such as roots, growing buds, flowers, fruits, or tubers (during the active growing season in summer). The same organ can be a source at one time (e.g., a tuber releasing nutrients in spring) and a sink at another time (e.g., a tuber storing starch in autumn).

(c) Phloem sieve tube elements are adapted by: having little cytoplasm and no nucleus/vacuole to allow unimpeded flow of sap; having sieve plates with pores at the end walls to allow solutes to flow from cell to cell; having companion cells adjacent to them which contain many mitochondria to provide the energy (ATP) needed for active transport of sucrose into the sieve tubes.

(d) Aphids feed on phloem sap by inserting their specialized mouthparts, called stylets, directly into a single sieve tube element. The high pressure within the phloem forces the sap out through the stylet. Researchers can anesthetize the aphid, cut off its body leaving the stylet in place, and collect the exuded sap from the cut stylet to analyze its chemical composition (which consists mainly of sucrose and amino acids).

評分準則

Part (a): Max 2 marks
- Movement of sucrose and amino acids (1)
- From source to sink (1)

Part (b): Max 5 marks
- Source defined as an organ that exports sucrose/amino acids (1)
- Sink defined as an organ that imports/uses/stores sucrose/amino acids (1)
- Example of source: photosynthesising leaves in summer OR storing tuber in spring (1)
- Example of sink: growing roots/fruits in summer OR growing shoot in spring (1)
- Explanation that roles can change/reverse depending on the season/stage of life cycle (1)

Part (c): Max 4 marks
- Sieve tube cells have very little cytoplasm / no nucleus / no vacuole to reduce resistance to flow (1)
- Sieve plates with pores allow continuous flow from cell to cell (1)
- Companion cells contain many mitochondria to provide ATP (1)
- For active loading/transport of sucrose (1)

Part (d): Max 2.33 marks
- Aphid inserts mouthpart/stylet into phloem/sieve tube (1)
- High pressure forces sap out of stylet (1)
- Body of aphid is severed, sap is collected and analyzed (0.33)
題目 3 · structured
13.33
(a) Explain the difference between active immunity and passive immunity. [4 marks]

(b) Describe how a vaccine leads to long-term active immunity against a pathogen. [6 marks]

(c) Explain why passive immunity only provides short-term protection. [3.33 marks]
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解題

(a) Active immunity involves the host's own immune system producing antibodies in response to an antigen, resulting in the production of memory cells and long-term protection. Passive immunity is the temporary transfer of ready-made antibodies from another individual or source (such as through breast milk, placenta, or serum injection), which does not involve memory cells and offers immediate but temporary protection.

(b) A vaccine contains harmless, dead, or weakened pathogens (or their antigens). When injected or ingested, the antigens on the surface of these pathogens are recognized by lymphocytes. Lymphocytes undergo rapid cell division by mitosis to produce clones of plasma cells. These plasma cells secrete specific antibodies that bind to the antigens. Crucially, memory cells are also produced during this primary response. These memory cells persist in the blood for a long time. If the actual pathogen enters the body later, the memory cells recognize it immediately, dividing rapidly to produce a much larger concentration of antibodies much more quickly, neutralizing the pathogen before disease symptoms can develop.

(c) Passive immunity only provides short-term protection because the antibodies are foreign proteins and are gradually broken down by the body's metabolic processes and excreted. Furthermore, because the recipient's own lymphocytes were not activated, no memory cells are produced to manufacture new antibodies if the host is exposed to the pathogen again in the future.

評分準則

Part (a): Max 4 marks
- Active: antibodies made by own body VS passive: antibodies from external source (1)
- Active: results in memory cells VS passive: no memory cells produced (1)
- Active: slow to develop VS passive: immediate protection (1)
- Active: long-term protection VS passive: short-term/temporary protection (1)

Part (b): Max 6 marks
- Vaccine contains harmless/weakened/dead pathogen or antigens (1)
- Antigens trigger response by lymphocytes (1)
- Lymphocytes divide/clone (1)
- Produce specific antibodies (1)
- Memory cells are produced and remain in blood (1)
- On reinfection, memory cells recognize antigen and respond rapidly (1)
- Rapid and large-scale antibody production prevents illness (1)

Part (c): Max 3.33 marks
- Transferred/injected antibodies are broken down/destroyed/excreted (2)
- No memory cells are produced (1.33)
題目 4 · structured
13.33
(a) State the balanced chemical equation for anaerobic respiration in yeast. [3 marks]

(b) Describe the role of anaerobic respiration in bread-making. [4 marks]

(c) Describe the effects of lactic acid accumulation in human muscles during vigorous exercise and how the body subsequently removes it. [6.33 marks]
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解題

(a) The balanced chemical equation for anaerobic respiration in yeast is: \(C_6H_{12}O_6 \rightarrow 2C_2H_5OH + 2CO_2\).

(b) In bread-making, yeast is mixed with flour, water, and sugar to form dough. The yeast respires anaerobically, producing carbon dioxide gas and ethanol. The carbon dioxide gas becomes trapped in the dough, forming bubbles that cause the dough to rise. When the dough is baked, the heat causes the gas bubbles to expand further, making the bread light and airy. The ethanol produced evaporates during the baking process, leaving no alcohol in the final loaf.

(c) During vigorous exercise, muscle cells do not receive enough oxygen to meet energy demands solely through aerobic respiration, so they perform anaerobic respiration, producing lactic acid. Lactic acid accumulates in the muscles, lowering the pH. This causes muscle fatigue and pain/cramps. After exercise, the body has an 'oxygen debt'. The lactic acid is transported by the blood from the muscles to the liver. In the liver, aerobic respiration of a portion of the lactic acid provides the energy needed to convert the remaining lactic acid back into glucose/glycogen. This process requires extra oxygen, which is why heart rate and breathing rate remain high after exercise.

評分準則

Part (a): Max 3 marks
- Correct formulas for reactants and products: \(C_6H_{12}O_6\) and \(C_2H_5OH\) and \(CO_2\) (1)
- Balanced correctly: 1 glucose yields 2 ethanol and 2 carbon dioxide (1)
- No oxygen on the left side (1)

Part (b): Max 4 marks
- Yeast respires anaerobically / ferments (1)
- Produces carbon dioxide gas (1)
- Carbon dioxide bubbles trap in dough and cause it to rise (1)
- Baking evaporates the ethanol and expands the carbon dioxide bubbles (1)

Part (c): Max 6.33 marks
- Lactic acid builds up in muscles due to anaerobic respiration (1)
- Lowers pH and causes muscle fatigue / pain / cramps (1)
- Creates an oxygen debt (1)
- Lactic acid is transported by blood from muscles to the liver (1)
- Lactic acid is broken down / converted to glucose in the liver (1)
- Oxygen is required to break down/convert lactic acid (1)
- Breathing and heart rates remain high to supply this oxygen (0.33)
題目 5 · structured
13.33
(a) Describe the physical digestion of lipids by bile, and explain why this is important for subsequent chemical digestion. [4 marks]

(b) Identify the three main types of human teeth and describe their specific functions in physical digestion. [3 marks]

(c) Describe the chemical digestion of starch in the alimentary canal, naming the enzyme, the substrate, the intermediate product, and the final product. [6.33 marks]
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解題

(a) Bile is produced in the liver and stored in the gall bladder before being released into the duodenum. Bile contains bile salts which physically digest lipids by emulsifying them. This process breaks down large droplets of lipids into many tiny droplets. Emulsification does not break chemical bonds, but it significantly increases the surface area of lipids available for lipase enzymes to act upon, which increases the rate of chemical digestion into fatty acids and glycerol.

(b) The three main types of human teeth and their functions are:
1. Incisors: have sharp, chisel-shaped edges used for biting and cutting pieces of food.
2. Canines: are pointed teeth used for tearing or gripping food.
3. Premolars/Molars: have broad, flat surfaces with ridges (cusps) used for crushing and grinding food to increase its surface area.

(c) Starch is a large, insoluble polysaccharide. Chemical digestion of starch begins in the mouth, where the enzyme salivary amylase breaks down starch (substrate) into maltose (an intermediate product, which is a disaccharide). This process continues in the duodenum with pancreatic amylase. Finally, the enzyme maltase, which is located on the membranes of the epithelium lining the small intestine, breaks down the maltose into glucose (the final product, which is a soluble monosaccharide) that can be easily absorbed into the blood.

評分準則

Part (a): Max 4 marks
- Bile contains bile salts (1)
- Emulsifies lipids / breaks large droplets into small droplets (1)
- Physical digestion / no chemical bonds broken (1)
- Increases surface area of lipids (1)
- Lipase can chemically digest lipids faster (1)

Part (b): Max 3 marks
- Incisors: biting / cutting (1)
- Canines: gripping / tearing (1)
- Premolars / Molars: grinding / crushing (1)

Part (c): Max 6.33 marks
- Starch is a large, insoluble polysaccharide (1)
- Amylase is the enzyme that breaks down starch into maltose (1)
- Salivary amylase in mouth / pancreatic amylase in duodenum (1)
- Maltose is the intermediate product / disaccharide (1)
- Maltase breaks down maltose into glucose (1)
- Glucose is the final product / soluble monosaccharide (1)
- Maltase is located on the membranes of the epithelium lining the small intestine (0.33)
題目 6 · structured
13.33
(a) Sickle-cell anaemia is caused by a codominant allele. Use a genetic diagram to predict the possible genotypes and phenotypes of children born to two parents who both have the sickle-cell trait (\(Hb^A Hb^S\)). [5 marks]

(b) Explain how the mutation that causes sickle-cell anaemia affects the structure and function of red blood cells. [4.33 marks]

(c) Describe the difference between continuous and discontinuous variation, and state which type of variation sickle-cell anaemia represents. [4 marks]
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解題

(a) The genetic diagram is as follows:
- Parent Genotypes: \(Hb^A Hb^S \times Hb^A Hb^S\)
- Gametes: \(Hb^A\) and \(Hb^S\) from each parent
- Punnett Square fertilization outcomes:
- \(Hb^A Hb^A\) (1/4 or 25% chance)
- \(Hb^A Hb^S\) (2/4 or 50% chance)
- \(Hb^S Hb^S\) (1/4 or 25% chance)
- Phenotypes:
- \(Hb^A Hb^A\): Normal red blood cells (no sickle-cell trait)
- \(Hb^A Hb^S\): Sickle-cell trait (mostly healthy, mild symptoms under low oxygen, resistant to malaria)
- \(Hb^S Hb^S\): Sickle-cell anaemia (severe symptoms, abnormal haemoglobin)

(b) The mutation is a change in the DNA base sequence of the gene that codes for the beta-globin polypeptide of haemoglobin. This leads to a single incorrect amino acid in the polypeptide chain. As a result, the haemoglobin molecules polymerize into long, stiff fibers when oxygen levels are low. This distorts the red blood cells, forcing them to collapse from their flexible biconcave disc shape into a rigid sickle/crescent shape. These rigid sickle-shaped cells cannot bend to squeeze through small capillaries, causing blockages that stop blood flow and cut off oxygen supply to tissues. They are also fragile and break down easily, leading to a shortage of red blood cells (anaemia).

(c) Continuous variation is variation that results in a range of phenotypes between two extremes, with no distinct categories (e.g., height, body mass). It is usually caused by multiple genes (polygenic inheritance) and influenced by environmental factors. Discontinuous variation results in a limited number of distinct, non-overlapping categories (e.g., ABO blood groups). It is caused by a single gene or a small number of genes, with little or no environmental influence. Sickle-cell anaemia is an example of discontinuous variation.

評分準則

Part (a): Max 5 marks
- Correct parental genotypes shown: \(Hb^A Hb^S\) and \(Hb^A Hb^S\) (1)
- Correct gametes shown: \(Hb^A\) and \(Hb^S\) (1)
- Correct offspring genotypes in genetic diagram: \(Hb^A Hb^A\), \(Hb^A Hb^S\), \(Hb^S Hb^S\) (1)
- Correct ratio of genotypes (1:2:1) or percentages (25% : 50% : 25%) (1)
- Correctly linking each genotype to its corresponding phenotype (Normal, Sickle-cell trait, Sickle-cell anaemia) (1)

Part (b): Max 4.33 marks
- Mutation is a change in the DNA base sequence of the haemoglobin gene (1)
- Leads to abnormal haemoglobin that clumps/polymerizes in low oxygen (1)
- Red blood cells change shape from biconcave disc to rigid sickle/crescent shape (1)
- Rigid cells block capillaries, reducing oxygen delivery to tissues (1)
- Sickle cells are destroyed rapidly, causing anaemia (0.33)

Part (c): Max 4 marks
- Continuous: range of phenotypes / no distinct categories VS Discontinuous: distinct / separate / non-overlapping categories (1)
- Continuous: influenced by both genes and environment VS Discontinuous: genetic only / no environmental influence (1)
- Continuous: polygenic VS Discontinuous: single/few genes (1)
- Sickle-cell anaemia is discontinuous variation (1)

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