Cambridge IGCSE · Thinka-original Practice Paper

2024 Cambridge IGCSE Biology (0610) Practice Paper with Answers

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

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

Section A

Answer all questions. Write your answers in the spaces provided on the question paper.
6 Question · 80 marks
Question 1 · structured
11 marks
Fig. 1.1 shows a diagram representing a bacterium (prokaryotic) and a plant cell (eukaryotic).

(a) State three structural features present in a eukaryotic plant cell that are absent from a prokaryotic bacterial cell. [3]

(b) A student placed a sample of plant epidermal cells and a sample of red blood cells into separate test tubes containing distilled water.

(i) Describe and explain the expected appearance of the plant cells after 30 minutes. Use the term water potential in your explanation. [4]

(ii) Describe and explain the expected appearance of the red blood cells after 30 minutes. [4]
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Worked solution

(a) Three structural differences:
1. Nucleus (or nuclear envelope/membrane)
2. Mitochondria
3. Chloroplasts / large permanent vacuole

(b)(i) Plant cell response:
- Description: The cells become turgid (firm/swollen).
- Osmosis: Water enters the cells by osmosis.
- Water potential: Water moves from a region of higher water potential (distilled water outside) to a region of lower water potential (inside the cell cytoplasm/vacuole).
- Role of cell wall: The rigid cellulose cell wall resists the turgor pressure and prevents the cell from bursting.

(b)(ii) Red blood cell response:
- Description: The cells will swell and burst (haemolysis / lysis).
- Explanation: Water enters the red blood cells by osmosis down a water potential gradient.
- Contrast: Unlike plant cells, red blood cells do not have a cell wall. The cell membrane is flexible and weak, so it ruptures under osmotic pressure.

Marking scheme

Part (a) [Max 3 marks]:
- Award 1 mark for each correct structure (up to 3):
- Nucleus / nuclear membrane [1]
- Mitochondria [1]
- Chloroplasts [1]
- Large permanent vacuole [1]
- (Note: Do not accept 'cell wall' unless qualified as 'cellulose cell wall', as bacteria also have cell walls but of different composition).

Part (b)(i) [Max 4 marks]:
- Plant cells become turgid / firm / swollen [1]
- Water enters the cell by osmosis [1]
- From a higher water potential outside to a lower water potential inside the cell [1]
- Across a partially permeable membrane [1]
- Cell wall prevents the cell from bursting / exerts turgor pressure [1]

Part (b)(ii) [Max 4 marks]:
- Red blood cells swell and burst / lyse / undergo haemolysis [1]
- Water enters by osmosis [1]
- Red blood cells lack a cell wall [1]
- Cell membrane is fragile / cannot withstand the turgor pressure [1]
Question 2 · structured
13 marks
A student investigated the reproductive structures of two different wild plant species, Species A and Species B.

The student recorded the following observations about the flowers:
* Species A: Large, bright pink petals; sweet scent; nectar present at the base of the carpel; small, sticky stigma enclosed inside the flower.
* Species B: Tiny green petals; no scent; no nectar; long, feathery stigmas hanging outside the flower; long filaments with dangling anthers.

(a) (i) Identify the method of pollination for:
- Species A
- Species B
[2 marks]

(ii) Explain how two features of Species B, described above, adapt it for its method of pollination.
[4 marks]

(b) (i) Describe the pathway and process from the moment a pollen grain lands on a compatible stigma of Species A until fertilization occurs.
[4 marks]

(ii) State the precise definition of fertilization in flowering plants.
[1 mark]

(c) State one advantage and one disadvantage of self-pollination to a plant species compared to cross-pollination.
[2 marks]
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Worked solution

(a) (i)
* Species A: Insect-pollinated (or animal-pollinated)
* Species B: Wind-pollinated

(ii)
* Feature 1: Long, feathery stigmas hanging outside the flower.
* Explanation 1: Provides a large surface area to trap pollen grains drifting in the air currents.
* Feature 2: Long filaments with dangling/exposed anthers.
* Explanation 2: Allows the anthers to be easily shaken by the wind, releasing pollen directly into the air currents.

(b) (i)
1. The pollen grain absorbs nutrients/moisture from the stigma and germinates.
2. A pollen tube grows downwards through the tissues of the style.
3. This growth is directed towards the ovary, guided by chemical signals.
4. The pollen tube enters the ovule through a tiny opening called the micropyle.
5. The male gamete nucleus travels down the pollen tube.
6. The male gamete nucleus fuses with the female gamete (ovule) nucleus inside the ovule.

(ii)
* Fertilization is defined as the fusion of a male gamete (pollen) nucleus with a female gamete (ovule) nucleus.

(c)
* Advantage: Does not rely on external pollinators or wind, meaning pollination is highly reliable; useful for colonizing isolated habitats.
* Disadvantage: Reduces genetic variation in the offspring, making the population more vulnerable to environmental changes or disease outbreaks.

Marking scheme

(a) (i) [Total: 2 marks]
* Species A: Insect/animal-pollinated [1]
* Species B: Wind-pollinated [1]

(ii) [Total: 4 marks]
Award 1 mark for identifying an appropriate adaptation and 1 mark for its explanation, up to a maximum of two features.
* Feature: Long / feathery / sticky stigma [1]
* Explanation: Large surface area to trap wind-borne pollen [1]
* Feature: Stamens / anthers / filaments hanging outside the flower [1]
* Explanation: Allows pollen to be easily swept away by the wind [1]
* Feature: Tiny/dull petals OR lack of scent/nectar [1]
* Explanation: Avoids wasting energy/resources on structures not needed for wind pollination [1]

(b) (i) [Total: 4 marks]
Award 1 mark for each correct point up to a maximum of 4:
* Pollen grain germinates on the stigma [1]
* Pollen tube grows down style [1]
* Growth of the tube is guided towards the ovary/micropyle [1]
* Pollen tube enters the ovule (via the micropyle) [1]
* Male (gamete) nucleus travels down the pollen tube [1]
* Fusion of male nucleus and female/ovule nucleus [1]

(ii) [Total: 1 mark]
* Fusion of a male gamete nucleus and a female gamete nucleus [1]
(Note: Reject 'fusion of pollen and egg' without the explicit mention of nuclei)

(c) [Total: 2 marks]
* One mark for any valid advantage: e.g., higher chance of successful pollination / does not rely on pollinators / preserves adapted gene combinations / useful in isolated habitats [1]
* One mark for any valid disadvantage: e.g., results in lower genetic diversity / inbreeding depression / limited ability to adapt to changing environmental conditions or disease [1]
Question 3 · structured
10 marks
Alcohol dehydrogenase is an enzyme produced in the liver that catalyzes the breakdown of alcohols.

(a) The production of alcohol dehydrogenase is controlled by a specific gene in the DNA. Describe the process of protein synthesis that leads to the production of this enzyme in a cell. [4]

(b) Explain why enzymes, such as alcohol dehydrogenase, are specific to only one type of substrate. Use the term "active site" in your answer. [2]

(c) Ethylene glycol is a toxic substance found in antifreeze. In the body, alcohol dehydrogenase converts ethylene glycol into a highly dangerous chemical called glycolic acid. Ethanol (drinking alcohol) can be administered as an antidote because it acts as a competitive inhibitor.

Using the lock and key hypothesis, suggest and explain how administering ethanol prevents the conversion of ethylene glycol to glycolic acid. [4]
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Worked solution

(a) The gene coding for alcohol dehydrogenase in the DNA is transcribed to make a complementary mRNA molecule in the nucleus. This mRNA molecule leaves the nucleus through a nuclear pore and enters the cytoplasm, where it binds to a ribosome. As the mRNA passes through the ribosome, the ribosome assembles amino acids brought by tRNA molecules in a specific sequence determined by the codons on the mRNA. This chain of amino acids folds to form the active alcohol dehydrogenase enzyme.

(b) Every enzyme has an active site with a specific and unique three-dimensional shape. This active site is complementary in shape only to its specific substrate. Other molecules with different shapes cannot fit into the active site, meaning no reaction can be catalyzed.

(c) Ethanol has a very similar shape to ethylene glycol and is also complementary to the active site of the alcohol dehydrogenase enzyme. When ethanol is administered, it competes with ethylene glycol and binds to the enzyme's active site. This blocks the active site, preventing ethylene glycol from entering and binding to form enzyme-substrate complexes. Consequently, the rate of conversion of ethylene glycol to toxic glycolic acid is greatly reduced, allowing the body to safely excrete the unmetabolized ethylene glycol.

Marking scheme

(a) Max [4 marks]:
- mRNA is a copy of the gene / DNA [1]
- mRNA travels from nucleus to cytoplasm / to ribosome [1]
- mRNA passes through a ribosome [1]
- Ribosome assembles / joins amino acids together [1]
- Amino acids are assembled in a specific sequence [1]
- tRNA molecules carry amino acids to ribosome [1]

(b) Max [2 marks]:
- Active site has a specific / unique shape [1]
- Active site is complementary to the shape of the substrate [1]
- Only the substrate fits into the active site / reference to lock and key model [1]

(c) Max [4 marks]:
- Ethanol has a similar shape to ethylene glycol / is complementary to the active site [1]
- Ethanol binds to / fits into the active site of alcohol dehydrogenase [1]
- Ethanol blocks the active site / acts as a competitive inhibitor [1]
- Prevents ethylene glycol from binding to the active site / forming enzyme-substrate complexes [1]
- Reduces / stops the production of toxic glycolic acid [1]
Question 4 · structured
12 marks
Many modern crop plants, such as bread wheat, have been developed from wild ancestors through selective breeding.

(a) Define the term *selective breeding*. [2]

(b) Contrast natural selection with artificial selection. In your answer, refer to:
* what exerts the selection pressure
* who benefits from the selected traits
* the relative time scale of each process [4]

(c)
(i) State two desirable features, other than disease resistance, that a plant breeder might select for in a cereal crop like wheat. [2]
(ii) Outline how a breeder would develop a new variety of wheat that is both disease-resistant and produces a high yield of grain. [4]
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Worked solution

### Model Solution

**(a)**
Selective breeding is the selection by humans of individuals with desirable features, crossing these individuals to produce the next generation, and selecting the offspring that show the desirable features to breed again.

**(b)**
* **Selection pressure:** In natural selection, selection pressure is exerted by environmental factors (such as predators, climate, or pathogens). In artificial selection, the selection pressure is exerted by human choice.
* **Beneficiary:** In natural selection, the selected traits improve the adaptation and survival of the organism in its natural habitat. In artificial selection, the selected traits are of economic, agricultural, or aesthetic value to humans, and may actually reduce the survival fitness of the organism in the wild.
* **Time scale:** Natural selection occurs over very long, geological timescales (many thousands or millions of years). Artificial selection occurs much more rapidly over a relatively small number of generations.

**(c)(i)**
Any two of:
* Short, stiff stems (to prevent lodging/falling over in high winds)
* Larger grain size
* Faster growth rate or early ripening
* Drought tolerance / salinity tolerance
* Improved nutritional quality (e.g., higher protein content)

**(c)(ii)**
1. Cross-pollinate a high-yielding wheat plant with a disease-resistant wheat plant.
2. Collect and plant the resulting seeds, and grow the offspring.
3. Test the offspring and select only those individuals that show both high yield and disease resistance.
4. Cross-breed these selected offspring together (or self-pollinate them).
5. Repeat this process of selection and breeding over many successive generations until a stable variety that consistently breeds true for both traits is established.

Marking scheme

**(a)** [Max 2 marks]
* selection by humans of individuals with desirable/useful features; [1]
* crossing/breeding these individuals to produce the next generation (and selecting offspring with desirable features); [1]

**(b)** [Max 4 marks]
* (agent of selection) in natural selection, selection pressure is exerted by environmental factors / named environmental factor, whereas in artificial selection, it is exerted by humans; [1]
* (beneficiary) in natural selection, traits increase survival/fitness/adaptation of the organism, whereas in artificial selection, traits benefit humans / have economic or aesthetic value; [1]
* (time scale) natural selection is slow / occurs over geological time scales, whereas artificial selection is fast / occurs over fewer generations; [1]
* (mating) natural selection involves random mating / competition for mates, whereas artificial selection involves controlled/deliberate breeding; [1]

**(c)(i)** [Max 2 marks]
Two of the following:
* short / stiff / strong stems; [1]
* large grain size / large seeds; [1]
* fast growth rate / early maturation; [1]
* tolerance to environmental stress (e.g. drought / salt / frost); [1]
* uniform growth / ripening; [1]
* high protein/gluten content / high baking quality; [1]
* *Reject: disease resistance (given in prompt)*

**(c)(ii)** [Max 4 marks]
* cross-pollinate / breed high-yield plant with disease-resistant plant; [1]
* grow offspring and identify / select individuals showing both high yield and disease resistance; [1]
* breed / self-pollinate these selected offspring; [1]
* repeat selection and breeding process over many / several generations; [1]
* until the offspring consistently show both traits / breed true; [1]
Question 5 · structured
19 marks
A marine ecosystem is studied by researchers. In this ecosystem, kelp (a large multicellular alga) and phytoplankton are the primary producers. Kelp is grazed upon by sea urchins and crabs. Sea otters feed on both sea urchins and crabs. Killer whales are apex predators that hunt sea otters. (a)(i) Define the term ecosystem. [2 marks] (a)(ii) State the principal source of energy input into this marine food web. [1 mark] (a)(iii) Identify one primary consumer from the described food web. [1 mark] (b)(i) In a specific area of this marine ecosystem, the energy stored in the biomass of kelp is \(1,200,000\text{ kJ m}^{-2}\text{ y}^{-1}\). The energy transferred and stored in the biomass of sea urchins is \(108,000\text{ kJ m}^{-2}\text{ y}^{-1}\). Calculate the efficiency of energy transfer from kelp to sea urchins. Show your working. [2 marks] (b)(ii) Explain why only a small percentage of the energy in the kelp is transferred to the sea urchins. [4 marks] (c)(i) Sea otters consume large quantities of crabs and sea urchins, which contain high concentrations of lipids (fats). Describe the chemical digestion of fats in the mammal's alimentary canal, including the names of the enzyme and products of this digestion. [3 marks] (c)(ii) State where bile is produced and describe its role in fat digestion. [2 marks] (d) Describe how the products of fat digestion are absorbed into the lymphatic system through the structure of a villus in the ileum. [4 marks]
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Worked solution

(a)(i) An ecosystem is a unit containing all of the organisms and their environment, interacting together, in a given area. (a)(ii) The principal source of energy is sunlight (or light from the Sun). (a)(iii) Sea urchins or crabs are primary consumers. (b)(i) To calculate efficiency: \((\text{Energy in consumer} / \text{Energy in producer}) \times 100 = (108,000 / 1,200,000) \times 100 = 9\%\). (b)(ii) Energy transfer efficiency is low because: 1. Not all of the kelp is eaten (some parts are left behind). 2. Some eaten parts are indigestible and egested as feces. 3. Kelp loses energy as heat from respiration. 4. Kelp uses energy for its own metabolic processes and growth. (c)(i) Fats are chemically digested by lipase (produced by the pancreas and active in the small intestine). Lipase breaks down fats into fatty acids and glycerol. (c)(ii) Bile is produced in the liver and stored in the gallbladder. It emulsifies fats, breaking large fat droplets into smaller droplets to increase the surface area for lipase action, and neutralizes acidic food from the stomach to create an alkaline optimum pH for lipase. (d) The inner wall of the ileum contains villi to increase surface area. Fatty acids and glycerol diffuse into the epithelial cells of the villi. Inside these cells, they are recombined into lipids/fats. These fats then enter the lacteals, which are specialized capillaries of the lymphatic system.

Marking scheme

(a)(i) Max 2 marks: 1 mark for stating it is a unit containing all organisms and their environment interacting together, 1 mark for specifying "in a given area". (a)(ii) 1 mark for light / sunlight / solar energy. (a)(iii) 1 mark for sea urchins / crabs. (b)(i) Max 2 marks: 1 mark for correct working formula/substitution \((108,000 / 1,200,000) \times 100\), 1 mark for correct final value of \(9\%\). (b)(ii) Max 4 marks: 1 mark for energy lost as heat during respiration of the producer, 1 mark for parts of the kelp not eaten/consumed, 1 mark for undigested food egested as feces, 1 mark for energy lost in excretory products (e.g., urea/ammonia). (c)(i) Max 3 marks: 1 mark for identifying lipase as the digestive enzyme, 1 mark for identifying fatty acids and glycerol as the products of digestion, 1 mark for stating lipase is secreted by the pancreas (or active in the small intestine). (c)(ii) Max 2 marks: 1 mark for stating bile is produced in the liver, 1 mark for stating bile emulsifies fats (increases surface area for lipase) or neutralizes stomach acid. (d) Max 4 marks: 1 mark for stating villi / microvilli increase the surface area for absorption, 1 mark for fatty acids and glycerol diffusing into epithelial cells, 1 mark for recombining into fats within the epithelial cells, 1 mark for entering the lacteal, 1 mark for stating that lacteals lead into the lymphatic system.
Question 6 · structured
15 marks
The Mountain Bongo (Tragelaphus eurycerus isaaci) is a critically endangered subspecies of antelope. Captive breeding programs are used to conserve this species. (a) (i) Explain why maintaining high genetic diversity is crucial for the long-term survival of the Mountain Bongo. [3] (ii) State what is meant by the term biodiversity. [2] (b) Conservationists often use artificial insemination (AI) in captive breeding programs. (i) Describe how artificial insemination is carried out in mammals. [3] (ii) Suggest two advantages of using artificial insemination instead of natural mating to conserve endangered mammals globally. [2] (c) Table 1.1 shows the genetic diversity (measured as average heterozygosity) of three different populations of Mountain Bongos over five generations. Population A: Captive population managed using a global computerized studbook and artificial insemination. Population B: Captive population kept in a single small zoo, breeding only by natural mating within the group. Population C: Wild population living in a fragmented forest reserve. Table 1.1: [Population A | Gen 1: 0.82 | Gen 5: 0.79] [Population B | Gen 1: 0.80 | Gen 5: 0.44] [Population C | Gen 1: 0.78 | Gen 5: 0.52]. (i) Calculate the percentage decrease in genetic diversity for Population B from Generation 1 to Generation 5. Show your working and give your answer to two significant figures. [2] (ii) Explain the differences in genetic diversity between Population A and Population B at Generation 5. [3]
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Worked solution

a) i) High genetic diversity is essential because it avoids inbreeding depression (the expression of harmful recessive alleles) and ensures the population retains a large gene pool. This genetic variation allows the population to adapt to future environmental changes such as new pathogens, climate shift, or habitat alterations, protecting them from extinction. ii) Biodiversity is defined as the variation of life forms, encompassing species diversity (species richness and abundance), genetic variation within species, and the variety of ecosystems. b) i) Semen containing sperm is collected from a selected male, evaluated for viability, and often frozen in liquid nitrogen for preservation. It is then mechanically introduced using specialized tools into the reproductive tract (uterus or cervix) of a female when she is in estrus (fertile period). ii) Advantages include: no need to transport heavy, stressed, or aggressive animals across global sites (only semen is shipped); reduced risk of spreading sexually transmitted diseases; and the ability to use frozen sperm from deceased or distant males. c) i) Percentage decrease = ((0.80 - 0.44) / 0.80) * 100 = (0.36 / 0.80) * 100 = 45%. ii) Population A is managed with a global computerized studbook and AI, which allows controlled breeding between genetically distant individuals across different locations, maintaining high genetic diversity. In contrast, Population B has a very small, closed gene pool with no introduction of new alleles, leading to high levels of inbreeding and genetic drift, which significantly decreases genetic diversity over five generations.

Marking scheme

a) i) Award 1 mark for each point up to 3 max: - reduces/prevents inbreeding depression / expression of harmful recessive alleles; - provides a larger gene pool / wider range of alleles; - increases ability of the population to adapt to environmental changes (e.g. disease / climate change); - reduces risk of extinction / increases long-term viability. ii) Award 1 mark for each point up to 2 max: - variety of different species; - genetic variation within species; - range of different habitats/ecosystems. b) i) Award 1 mark for each point up to 3 max: - semen/sperm collected from male; - semen is frozen / stored / evaluated; - semen is mechanically introduced/syringed/inserted into female's reproductive tract / uterus / cervix. ii) Award 1 mark for each point up to 2 max: - avoids the stress/risk of transporting live animals; - reduces the risk of sexually transmitted diseases; - allows sperm from deceased/distant males to be used; - overcomes physical mating incompatibilities. c) i) Award 2 marks for correct answer (45). Award 1 mark for correct working if final answer is incorrect: - working: ((0.80 - 0.44)/0.80) * 100. ii) Award 1 mark for each point up to 3 max: - Population A is managed using a studbook to breed unrelated/genetically distant individuals; - AI in Population A allows genes to be introduced from other populations (increasing gene flow); - Population B is closed/isolated with a small gene pool; - Population B suffers from high inbreeding / genetic drift / loss of alleles.

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