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Biology

CAIE2024

(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:44)

Question

Biology

CAIE2024

Spea multiplicata is one of several species of American spadefoot toad. (a) Young spadefoot toads are called tadpoles and live in water in ponds. S. multiplicata tadpoles show three different phenotypes due to genetic variation. The three phenotypes are: detritus feeder, intermediate and carnivore. Detritus feeders are small, and carnivores are large. Intermediates vary in size between the two extremes. A detritus feeder and a carnivore are shown in Fig. 7.1. carnivore detritus feeder fairy shrimp Fig. 7.1 Detritus feeders: • eat detritus (small pieces of dead organic matter) and algae (photosynthetic protoctists) • have smooth mouthparts, small jaw muscles and long intestines. Intermediates: • can eat all available food (detritus, algae and fairy shrimps) • have teeth‑like mouthparts, medium‑sized jaw muscles and medium‑sized intestines. Carnivores: • eat fairy shrimps and other small animals • have teeth‑like mouthparts, large jaw muscles and short intestines. Scientists counted the number of each type of tadpole in two different ponds: pond 1 and pond 2. (cid:300)(cid:205)(cid:266)(cid:190)(cid:288)(cid:180)(cid:237)(cid:200)(cid:245)(cid:207)(cid:298)(cid:197)(cid:266)(cid:224)(cid:249)(cid:182)(cid:258)(cid:215) © UCLES 2024 (cid:300)(cid:298)(cid:182)(cid:243)(cid:209)(cid:288)(cid:248)(cid:261)(cid:230)(cid:251)(cid:251)(cid:275)(cid:202)(cid:249)(cid:185)(cid:259)(cid:240)(cid:258) 9700/41/O/N/24 (cid:293)(cid:213)(cid:293)(cid:213)(cid:181)(cid:181)(cid:197)(cid:245)(cid:229)(cid:213)(cid:181)(cid:197)(cid:197)(cid:277)(cid:229)(cid:213)(cid:293)(cid:213) NIGRAM SIHT NI ETIRW TON OD NIGRAM SIHT NI ETIRW TON OD NIGRAM SIHT NI ETIRW TON OD NIGRAM SIHT NI ETIRW TON OD NIGRAM SIHT NI ETIRW TON OD PMT * 0000800000015 *

Question

Biology

CAIE2024

During aerobic respiration, cells respire substrates such as glucose to produce ATP. Some events that occur during aerobic respiration are: • The respiratory substrate breaks down into smaller and smaller molecules. These series of reactions are described as catabolism. • Coenzymes take part in various reactions. In some reactions, coenzymes are reduced or oxidised. • Carbon dioxide is released. (a) Aerobic respiration occurs in four successive stages: glycolysis (G), link reaction (LR), Krebs cycle (KC) and oxidative phosphorylation (OP). Complete Table 3.1 to show which events occur in each stage of aerobic respiration. Use a tick (✓) to show that the event does occur or a cross (✗) to show that the event does not occur. Table 3.1 stage event G LR KC OP catabolism coenzyme is reduced or oxidised a coenzyme forms a covalent bond with a respiratory intermediate carbon dioxide is released [4] (cid:300)(cid:209)(cid:266)(cid:190)(cid:288)(cid:180)(cid:237)(cid:200)(cid:245)(cid:207)(cid:298)(cid:197)(cid:266)(cid:222)(cid:252)(cid:181)(cid:254)(cid:215) © UCLES 2024 (cid:300)(cid:298)(cid:183)(cid:242)(cid:209)(cid:294)(cid:255)(cid:283)(cid:223)(cid:260)(cid:266)(cid:211)(cid:206)(cid:218)(cid:253)(cid:267)(cid:240)(cid:258) 9700/41/O/N/24 (cid:293)(cid:229)(cid:245)(cid:277)(cid:181)(cid:181)(cid:197)(cid:181)(cid:245)(cid:261)(cid:293)(cid:197)(cid:261)(cid:277)(cid:293)(cid:213)(cid:277)(cid:213) NIGRAM SIHT NI ETIRW TON OD NIGRAM SIHT NI ETIRW TON OD NIGRAM SIHT NI ETIRW TON OD NIGRAM SIHT NI ETIRW TON OD NIGRAM SIHT NI ETIRW TON OD PMT * 0000800000007 *

Question

Biology

CAIE2024

(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:8)(cid:44) (b) A new hand‑held technological device shows the main type of respiratory substrate being used in the cells of a person. The device consists of a carbon dioxide sensor and air‑flow meter. The person inhales through the device for a fixed time and then exhales into it. The device calculates the respiratory quotient (RQ) value to show whether the cells are mainly respiring carbohydrates or lipids. (i) Explain how the device calculates the RQ value and how this shows whether the cells are mainly respiring carbohydrates or lipids. … ........................................................................................................................................ ........................................................................................................................................... … ........................................................................................................................................ … ........................................................................................................................................ … ........................................................................................................................................ … ........................................................................................................................................ … ........................................................................................................................................ ..................................................................................................................................... [4] (ii) State the difference in the relative energy values of carbohydrates and lipids as respiratory substrates, and explain the reasons for the difference. … ........................................................................................................................................ … ........................................................................................................................................ … ........................................................................................................................................ … ........................................................................................................................................ … ........................................................................................................................................ … .................................................................................................................................. [3] [Total: 11] © UCLES 2024 (cid:300) (cid:300) (cid:211) (cid:298) (cid:266) (cid:184) (cid:190) (cid:241) (cid:288) (cid:217) (cid:180) (cid:292) (cid:237) (cid:259) (cid:200) (cid:299) (cid:245) (cid:218) (cid:207) (cid:246) (cid:298) (cid:247) (cid:197) (cid:278) (cid:266) (cid:266) (cid:222) (cid:226) (cid:250) (cid:217) (cid:181) (cid:267) (cid:254) (cid:256) (cid:215) (cid:258) 9700/41/O/N/24 [Turn over (cid:293)(cid:229)(cid:261)(cid:213)(cid:245)(cid:213)(cid:229)(cid:213)(cid:293)(cid:245)(cid:181)(cid:197)(cid:261)(cid:245)(cid:261)(cid:277)(cid:261)(cid:213) NIGRAM SIHT NI ETIRW TON OD NIGRAM SIHT NI ETIRW TON OD NIGRAM SIHT NI ETIRW TON OD NIGRAM SIHT NI ETIRW TON OD NIGRAM SIHT NI ETIRW TON OD PMT * 0000800000008 *

Question

Biology

CAIE2024

The leaves of Mimosa pudica plants are made of a number of structures known as pinnae. The pinnae fold when the leaf is touched. This closes the leaf. Fig. 4.1 shows an open leaf of M. pudica before it is touched. Fig. 4.2 shows the same leaf that has closed after being touched. pinnae A open (not touched) B closed (touched) Fig. 4.1 Fig. 4.2 (a) A touch stimulus to an M. pudica leaf causes an action potential to be generated. The action potential results in changes in cells, which cause the leaf to close. Fig. 4.3 shows the mechanism in M. pudica cells that causes the leaf to close. action protons Cl − and K+ water moves cells potential after pumped into ions move out out of cells become touch stimulus extensor cells of cells by osmosis flaccid Fig. 4.3 The leaves of M. pudica and the leaves of Venus fly traps move in response to touch stimuli, but the mechanisms that cause the responses are different. Describe the differences between the mechanism shown in Fig. 4.3 and the mechanism that causes the closure of the modified leaves in Venus fly traps. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [3] (cid:300)(cid:209)(cid:266)(cid:190)(cid:288)(cid:180)(cid:237)(cid:200)(cid:245)(cid:207)(cid:298)(cid:197)(cid:266)(cid:224)(cid:252)(cid:181)(cid:256)(cid:215) © UCLES 2024 (cid:300)(cid:298)(cid:184)(cid:244)(cid:217)(cid:298)(cid:241)(cid:286)(cid:221)(cid:252)(cid:240)(cid:269)(cid:236)(cid:254)(cid:251)(cid:283)(cid:248)(cid:258) 9700/41/O/N/24 (cid:293)(cid:277)(cid:213)(cid:213)(cid:181)(cid:213)(cid:229)(cid:245)(cid:261)(cid:213)(cid:197)(cid:197)(cid:197)(cid:245)(cid:229)(cid:277)(cid:277)(cid:213) NIGRAM SIHT NI ETIRW TON OD NIGRAM SIHT NI ETIRW TON OD NIGRAM SIHT NI ETIRW TON OD NIGRAM SIHT NI ETIRW TON OD NIGRAM SIHT NI ETIRW TON OD PMT * 0000800000009 *

Question

Biology

CAIE2024

The mammalian kidney is responsible for: • the excretion of urea • osmoregulation (the homeostatic control of the water potential of the blood). (a) (i) Outline how and where the excretory product urea is made in the body. ........................................................................................................................................... ........................................................................................................................................... ..................................................................................................................................... [2] (ii) Homeostatic control of the water potential of blood includes receptors, effectors and target cells. Identify the names and locations of these components of homeostatic control in osmoregulation. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ..................................................................................................................................... [3] (b) The glomerulus and Bowman’s capsule of the nephron are important in the formation of urine. Outline the role of the glomerulus and Bowman’s capsule in the formation of urine. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [3] (cid:300)(cid:209)(cid:266)(cid:190)(cid:288)(cid:180)(cid:237)(cid:200)(cid:245)(cid:207)(cid:298)(cid:197)(cid:266)(cid:223)(cid:252)(cid:183)(cid:256)(cid:215) © UCLES 2024 (cid:300)(cid:298)(cid:182)(cid:242)(cid:220)(cid:296)(cid:263)(cid:253)(cid:225)(cid:267)(cid:257)(cid:249)(cid:178)(cid:220)(cid:297)(cid:179)(cid:264)(cid:258) 9700/41/O/N/24 (cid:293)(cid:245)(cid:277)(cid:213)(cid:181)(cid:181)(cid:293)(cid:181)(cid:213)(cid:181)(cid:293)(cid:261)(cid:261)(cid:181)(cid:293)(cid:213)(cid:229)(cid:213) NIGRAM SIHT NI ETIRW TON OD NIGRAM SIHT NI ETIRW TON OD NIGRAM SIHT NI ETIRW TON OD NIGRAM SIHT NI ETIRW TON OD NIGRAM SIHT NI ETIRW TON OD PMT * 0000800000013 *

Question

Biology

CAIE2024

Question

Biology

CAIE2024

(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:4)(cid:44) (b) Fig. 1.2 shows an action potential in a rat neurone and Fig. 1.3 shows an action potential in a snail neurone. rat snail

Question

Biology

CAIE2024

(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:5)(cid:44)

Question

Biology

CAIE2024

(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:3)(cid:44)

Question

Biology

CAIE2024

The leaves of many plants have stomata that show a regular daily rhythm of opening and closing over a period of 24 hours. (a) Explain why it is important for plants to open and close their stomata in a daily rhythm. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [3] (b) Fig. 2.1 shows the results of an experiment to monitor this rhythm over three days and nights for the plant Arabidopsis thaliana. The percentage of open stomata is shown. Each day consisted of 14 hours of light (white bar) and each night consisted of 10 hours of darkness (black bar). 100 Key day open night stomata 50 / %

Question

Biology

CAIE2024

(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:7)(cid:44)

Question

Biology

CAIE2024

Lunularia cruciata is a primitive plant. Its body consists of a flattened sheet of photosynthetic tissue called a thallus. Fig. 5.1 shows L. cruciata with two different types of reproductive structure, labelled A and B, on its surface. A B Fig. 5.1 (a) Structures A and B and the thallus of L. cruciata are haploid. Explain what is meant by haploid. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [2] (b) Structure A contains pale discs of tissue, C, that can germinate into new L. cruciata. The new plants that develop from C are genetically identical to the parent plant in Fig. 5.1. Structure B contains male sperm that are chemically attracted to swim to female eggs on a neighbouring parent plant. When the egg and sperm fuse, they form structure D. Structure D develops to produce spores that grow into new plants that are genetically different from the two parent plants. Identify which of the structures A, B, C and D are: associated with sexual reproduction ………………………………………… produced by mitosis ………………………………………… the site of meiosis ………………………………………… Each letter may be used once, more than once, or not at all. [3] (cid:300)(cid:209)(cid:266)(cid:190)(cid:288)(cid:180)(cid:237)(cid:200)(cid:245)(cid:207)(cid:298)(cid:197)(cid:266)(cid:221)(cid:252)(cid:183)(cid:254)(cid:215) © UCLES 2024 (cid:300)(cid:298)(cid:181)(cid:244)(cid:212)(cid:300)(cid:249)(cid:252)(cid:227)(cid:243)(cid:247)(cid:183)(cid:200)(cid:256)(cid:271)(cid:227)(cid:256)(cid:258) 9700/41/O/N/24 (cid:293)(cid:197)(cid:181)(cid:277)(cid:181)(cid:213)(cid:261)(cid:245)(cid:229)(cid:293)(cid:197)(cid:261)(cid:197)(cid:213)(cid:229)(cid:277)(cid:229)(cid:213) NIGRAM SIHT NI ETIRW TON OD NIGRAM SIHT NI ETIRW TON OD NIGRAM SIHT NI ETIRW TON OD NIGRAM SIHT NI ETIRW TON OD NIGRAM SIHT NI ETIRW TON OD PMT * 0000800000011 *

Question

Biology

CAIE2024

(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:6)(cid:44) ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [6] (c) The experiment was repeated with A. thaliana plants that were left in darkness from 14 to 96 hours. The results are shown in Fig. 2.2. 100 Key day open night stomata 50 / %

Question

Biology

CAIE2024

(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:7)(cid:44)

Question

Biology

CAIE2024

Lunularia cruciata is a primitive plant. Its body consists of a flattened sheet of photosynthetic tissue called a thallus. Fig. 5.1 shows L. cruciata with two different types of reproductive structure, labelled A and B, on its surface. A B Fig. 5.1 (a) Structures A and B and the thallus of L. cruciata are haploid. Explain what is meant by haploid. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [2] (b) Structure A contains pale discs of tissue, C, that can germinate into new L. cruciata. The new plants that develop from C are genetically identical to the parent plant in Fig. 5.1. Structure B contains male sperm that are chemically attracted to swim to female eggs on a neighbouring parent plant. When the egg and sperm fuse, they form structure D. Structure D develops to produce spores that grow into new plants that are genetically different from the two parent plants. Identify which of the structures A, B, C and D are: associated with sexual reproduction ………………………………………… produced by mitosis ………………………………………… the site of meiosis ………………………………………… Each letter may be used once, more than once, or not at all. [3] (cid:300)(cid:209)(cid:266)(cid:190)(cid:288)(cid:180)(cid:237)(cid:200)(cid:245)(cid:207)(cid:298)(cid:197)(cid:266)(cid:221)(cid:250)(cid:184)(cid:254)(cid:215) © UCLES 2024 (cid:300)(cid:228)(cid:197)(cid:250)(cid:212)(cid:293)(cid:282)(cid:264)(cid:227)(cid:243)(cid:268)(cid:194)(cid:220)(cid:177)(cid:270)(cid:299)(cid:285)(cid:258) 9700/43/O/N/24 (cid:293)(cid:213)(cid:213)(cid:213)(cid:181)(cid:181)(cid:197)(cid:245)(cid:197)(cid:261)(cid:245)(cid:261)(cid:197)(cid:213)(cid:293)(cid:245)(cid:197)(cid:213) NIGRAM SIHT NI ETIRW TON OD NIGRAM SIHT NI ETIRW TON OD NIGRAM SIHT NI ETIRW TON OD NIGRAM SIHT NI ETIRW TON OD NIGRAM SIHT NI ETIRW TON OD PMT * 0000800000011 *

Question

Biology

CAIE2024

(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:8)(cid:44) (b) A new hand‑held technological device shows the main type of respiratory substrate being used in the cells of a person. The device consists of a carbon dioxide sensor and air‑flow meter. The person inhales through the device for a fixed time and then exhales into it. The device calculates the respiratory quotient (RQ) value to show whether the cells are mainly respiring carbohydrates or lipids. (i) Explain how the device calculates the RQ value and how this shows whether the cells are mainly respiring carbohydrates or lipids. … ........................................................................................................................................ ........................................................................................................................................... … ........................................................................................................................................ … ........................................................................................................................................ … ........................................................................................................................................ … ........................................................................................................................................ … ........................................................................................................................................ ..................................................................................................................................... [4] (ii) State the difference in the relative energy values of carbohydrates and lipids as respiratory substrates, and explain the reasons for the difference. … ........................................................................................................................................ … ........................................................................................................................................ … ........................................................................................................................................ … ........................................................................................................................................ … ........................................................................................................................................ … .................................................................................................................................. [3] [Total: 11] © UCLES 2024 (cid:300) (cid:300) (cid:211) (cid:228) (cid:266) (cid:200) (cid:190) (cid:251) (cid:288) (cid:217) (cid:180) (cid:285) (cid:237) (cid:292) (cid:200) (cid:279) (cid:245) (cid:218) (cid:207) (cid:246) (cid:298) (cid:268) (cid:197) (cid:291) (cid:266) (cid:278) (cid:222) (cid:271) (cid:252) (cid:220) (cid:182) (cid:195) (cid:254) (cid:285) (cid:215) (cid:258) 9700/43/O/N/24 [Turn over (cid:293)(cid:181)(cid:293)(cid:277)(cid:245)(cid:181)(cid:293)(cid:213)(cid:261)(cid:277)(cid:261)(cid:197)(cid:261)(cid:245)(cid:197)(cid:245)(cid:293)(cid:213) NIGRAM SIHT NI ETIRW TON OD NIGRAM SIHT NI ETIRW TON OD NIGRAM SIHT NI ETIRW TON OD NIGRAM SIHT NI ETIRW TON OD NIGRAM SIHT NI ETIRW TON OD PMT * 0000800000008 *

Question

Biology

CAIE2024

(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:44)

Question

Biology

CAIE2024

The leaves of Mimosa pudica plants are made of a number of structures known as pinnae. The pinnae fold when the leaf is touched. This closes the leaf. Fig. 4.1 shows an open leaf of M. pudica before it is touched. Fig. 4.2 shows the same leaf that has closed after being touched. pinnae A open (not touched) B closed (touched) Fig. 4.1 Fig. 4.2 (a) A touch stimulus to an M. pudica leaf causes an action potential to be generated. The action potential results in changes in cells, which cause the leaf to close. Fig. 4.3 shows the mechanism in M. pudica cells that causes the leaf to close. action protons Cl − and K+ water moves cells potential after pumped into ions move out out of cells become touch stimulus extensor cells of cells by osmosis flaccid Fig. 4.3 The leaves of M. pudica and the leaves of Venus fly traps move in response to touch stimuli, but the mechanisms that cause the responses are different. Describe the differences between the mechanism shown in Fig. 4.3 and the mechanism that causes the closure of the modified leaves in Venus fly traps. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [3] (cid:300)(cid:209)(cid:266)(cid:190)(cid:288)(cid:180)(cid:237)(cid:200)(cid:245)(cid:207)(cid:298)(cid:197)(cid:266)(cid:224)(cid:250)(cid:182)(cid:256)(cid:215) © UCLES 2024 (cid:300)(cid:228)(cid:200)(cid:250)(cid:217)(cid:295)(cid:274)(cid:274)(cid:221)(cid:252)(cid:259)(cid:300)(cid:184)(cid:179)(cid:250)(cid:211)(cid:277)(cid:258) 9700/43/O/N/24 (cid:293)(cid:261)(cid:181)(cid:277)(cid:181)(cid:181)(cid:293)(cid:245)(cid:293)(cid:181)(cid:245)(cid:197)(cid:197)(cid:245)(cid:293)(cid:245)(cid:245)(cid:213) NIGRAM SIHT NI ETIRW TON OD NIGRAM SIHT NI ETIRW TON OD NIGRAM SIHT NI ETIRW TON OD NIGRAM SIHT NI ETIRW TON OD NIGRAM SIHT NI ETIRW TON OD PMT * 0000800000009 *

Question

Biology

CAIE2024

N = estimate of population size n = number of individuals captured in first sample

Question

Biology

CAIE2024

unmarked) captured in second sample m = number of marked individuals recaptured in

Question

Biology

CAIE2024

second sample population size = ......................................................... [2] [Total: 10] The boundaries and names shown, the designations used and the presentation of material on any maps contained in this question paper/insert do not imply official endorsement or acceptance by Cambridge Assessment International Education concerning the legal status of any country, territory, or area or any of its authorities, or of the delimitation of its frontiers or boundaries. Permission to reproduce items where third‑party owned material protected by copyright is included has been sought and cleared where possible. Every reasonable effort has been made by the publisher (UCLES) to trace copyright holders, but if any items requiring clearance have unwittingly been included, the publisher will be pleased to make amends at the earliest possible opportunity. To avoid the issue of disclosure of answer‑related information to candidates, all copyright acknowledgements are reproduced online in the Cambridge Assessment International Education Copyright Acknowledgements Booklet. This is produced for each series of examinations and is freely available to download at www.cambridgeinternational.org after the live examination series. Cambridge Assessment International Education is part of Cambridge Assessment. Cambridge Assessment is the brand name of the University of Cambridge Local Examinations Syndicate (UCLES), which is a department of the University of Cambridge. (cid:300)(cid:209)(cid:266)(cid:190)(cid:288)(cid:180)(cid:237)(cid:200)(cid:245)(cid:207)(cid:298)(cid:197)(cid:266)(cid:222)(cid:249)(cid:182)(cid:260)(cid:215) © UCLES 2024 (cid:300)(cid:228)(cid:200)(cid:252)(cid:215)(cid:291)(cid:261)(cid:292)(cid:215)(cid:249)(cid:245)(cid:174)(cid:252)(cid:274)(cid:271)(cid:203)(cid:285)(cid:258) 9700/43/O/N/24 (cid:293)(cid:245)(cid:277)(cid:213)(cid:245)(cid:245)(cid:229)(cid:245)(cid:197)(cid:197)(cid:213)(cid:261)(cid:197)(cid:181)(cid:293)(cid:245)(cid:245)(cid:213) NIGRAM SIHT NI ETIRW TON OD NIGRAM SIHT NI ETIRW TON OD NIGRAM SIHT NI ETIRW TON OD NIGRAM SIHT NI ETIRW TON OD NIGRAM SIHT NI ETIRW TON OD PMT

Question

Biology

CAIE2024

(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:4)(cid:44) (b) As the newly formed glomerular filtrate passes along the proximal convoluted tubule, selective reabsorption takes place. The fluid remaining in the proximal convoluted tubule eventually forms urine. Table 1.1 lists three of the components of blood plasma that enter the glomerulus in a healthy person. Complete Table 1.1 by writing: • increased, if the component is present and is higher in concentration than in blood plasma • decreased, if the component is present and is lower in concentration than in blood plasma • same, if the component is present and is of the same concentration as blood plasma • not present, if the component is absent. You may use each response once, more than once or not at all. Table 1.1 component of blood plasma component in newly component in urine entering glomerulus formed glomerular filtrate glucose large plasma proteins urea [3] (c) Sweating during exercise can lead to a response in the posterior pituitary gland and in the kidney. State the response to sweating that occurs in the posterior pituitary gland and the response that occurs in the kidney. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [2] [Total: 9] © UCLES 2024 (cid:300) (cid:300) (cid:207) (cid:224) (cid:266) (cid:202) (cid:190) (cid:241) (cid:288) (cid:217) (cid:180) (cid:290) (cid:237) (cid:293) (cid:200) (cid:291) (cid:245) (cid:226) (cid:207) (cid:260) (cid:298) (cid:254) (cid:197) (cid:182) (cid:266) (cid:215) (cid:221) (cid:179) (cid:252) (cid:261) (cid:183) (cid:235) (cid:254) (cid:285) (cid:215) (cid:258) 9700/42/O/N/24 [Turn over (cid:293)(cid:197)(cid:277)(cid:277)(cid:181)(cid:181)(cid:261)(cid:181)(cid:277)(cid:277)(cid:229)(cid:261)(cid:197)(cid:245)(cid:229)(cid:213)(cid:261)(cid:213) NIGRAM SIHT NI ETIRW TON OD NIGRAM SIHT NI ETIRW TON OD NIGRAM SIHT NI ETIRW TON OD NIGRAM SIHT NI ETIRW TON OD NIGRAM SIHT NI ETIRW TON OD PMT * 0000800000004 *

Question

Biology

CAIE2024

(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:5)(cid:44)

Question

Biology

CAIE2024

(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:6)(cid:44)

Question

Biology

CAIE2024

The treatment of many diseases has been improved by the use of genetic technology in medicine. (a) Explain what is meant by genetic engineering. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [3] (b) (i) Leber Congenital Amaurosis (LCA) is an inherited eye disease. In LCA, the photoreceptor cells in the retina die at an early age. This causes impaired vision (reduced eyesight) in children, which can progress to blindness. Mutations in different genes cause different forms of LCA. One form of this disease, LCA2, is caused by a mutation in the RPE65 gene. Gene therapy has been used to treat LCA2. Outline how an inherited eye disease, such as LCA2, is treated with gene therapy. ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ........................................................................................................................................... ..................................................................................................................................... [3] (ii) Suggest why the eye is a suitable organ for gene therapy. ........................................................................................................................................... ..................................................................................................................................... [1] © UCLES 2024 (cid:300) (cid:300) (cid:207) (cid:224) (cid:266) (cid:201) (cid:190) (cid:243) (cid:288) (cid:209) (cid:180) (cid:286) (cid:237) (cid:283) (cid:200) (cid:278) (cid:245) (cid:228) (cid:207) (cid:252) (cid:298) (cid:252) (cid:197) (cid:252) (cid:266) (cid:225) (cid:223) (cid:295) (cid:252) (cid:243) (cid:183) (cid:187) (cid:256) (cid:293) (cid:215) (cid:258) 9700/42/O/N/24 [Turn over (cid:293)(cid:245)(cid:181)(cid:213)(cid:181)(cid:213)(cid:293)(cid:245)(cid:293)(cid:197)(cid:261)(cid:261)(cid:261)(cid:277)(cid:293)(cid:277)(cid:261)(cid:213) NIGRAM SIHT NI ETIRW TON OD NIGRAM SIHT NI ETIRW TON OD NIGRAM SIHT NI ETIRW TON OD NIGRAM SIHT NI ETIRW TON OD NIGRAM SIHT NI ETIRW TON OD PMT * 0000800000006 *

Question

Biology

CAIE2024

(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:7)(cid:44) (c) LCA10 is a different form of LCA caused by a recessive mutation in the CEP290 gene. This gene codes for the protein CEP290, which is involved in the correct functioning of photoreceptor cells in the retina. The mutation in CEP290 causes an error to be made when the primary transcript is spliced to form messenger RNA (mRNA). The abnormal mRNA that is formed has an extra sequence of RNA nucleotides, known as exon X, between exon 26 and exon 27. Exon X contains a STOP codon. Fig. 3.1 compares the effect of the mutation in CEP290 with the normal gene expression. normal LCA10 exon intron exon section of DNA exon mutation exon of CEP290

Question

Biology

CAIE2024

(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:8)(cid:44) Key: eye not treated eye treated impaired 0.2 acuity

Question

Biology

CAIE2024

(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:2)(cid:1)(cid:44)

Question

Biology

CAIE2024

Genetic crosses can be used to investigate patterns of inheritance. (a) A mutation in a gene involved in fruit colour in tomato plants, Solanum lycopersicum, results in the production of yellow fruits instead of red fruits. A genetic cross was carried out between a pure breeding plant with a red fruit and a pure breeding plant with a yellow fruit to produce the F1 generation. All offspring plants produced red fruits. The F1 plants were then crossed with each other and the seeds produced were planted to obtain the F2 generation. Construct a genetic diagram to show the cross of the F1 generation that produced this F2 generation. Use the symbols R and r for the alleles. [3] (b) A theoretical dihybrid cross involves two genes located on different autosomes. Each gene has two alleles, one dominant and one recessive. A parent, homozygous dominant for both genes, is crossed with a parent that is homozygous recessive for both genes. This produces F1 individuals that are then crossed to produce the F2 generation. State the phenotypic ratio of this dihybrid F2 generation and explain why some of these offspring phenotypes are different from the original parental phenotypes. ratio ........................................................................................................................................... explanation ............................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [3] [Total: 6] (cid:300)(cid:209)(cid:266)(cid:190)(cid:288)(cid:180)(cid:237)(cid:200)(cid:245)(cid:207)(cid:298)(cid:197)(cid:266)(cid:221)(cid:252)(cid:184)(cid:254)(cid:215) © UCLES 2024 (cid:300)(cid:224)(cid:204)(cid:241)(cid:207)(cid:298)(cid:261)(cid:239)(cid:232)(cid:256)(cid:248)(cid:293)(cid:219)(cid:201)(cid:260)(cid:187)(cid:293)(cid:258) 9700/42/O/N/24 (cid:293)(cid:197)(cid:213)(cid:213)(cid:181)(cid:277)(cid:229)(cid:277)(cid:245)(cid:197)(cid:261)(cid:197)(cid:261)(cid:213)(cid:261)(cid:277)(cid:293)(cid:213) NIGRAM SIHT NI ETIRW TON OD NIGRAM SIHT NI ETIRW TON OD NIGRAM SIHT NI ETIRW TON OD NIGRAM SIHT NI ETIRW TON OD NIGRAM SIHT NI ETIRW TON OD PMT * 0000800000011 *

Question

Biology

CAIE2024

(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:2)(cid:2)(cid:44)

Question

Biology

CAIE2024

Phenotypic variation exists in natural populations. There are many causes of variation. Natural selection determines which phenotypes are advantageous. (a) Variation in a particular characteristic can be described as either discontinuouso r continuous. Table 5.1 contains a list of statements that apply to discontinuous variation, continuous variation or both. Complete both columns of Table 5.1. Put a tick (✓) in the box if the statement applies and leave the box empty if the statement does not apply. Table 5.1 statement discontinuous continuous variation variation often involves one gene only environmental factors may affect gene expression there is an additive effect of genes that contributes to the phenotype there are distinct differences between the various forms of a characteristic [4] © UCLES 2024 (cid:300) (cid:300) (cid:211) (cid:224) (cid:266) (cid:203) (cid:190) (cid:242) (cid:288) (cid:215) (cid:180) (cid:288) (cid:237) (cid:265) (cid:200) (cid:255) (cid:245) (cid:209) (cid:207) (cid:250) (cid:298) (cid:265) (cid:197) (cid:228) (cid:266) (cid:255) (cid:221) (cid:177) (cid:250) (cid:216) (cid:184) (cid:187) (cid:254) (cid:277) (cid:215) (cid:258) 9700/42/O/N/24 [Turn over (cid:293)(cid:197)(cid:229)(cid:277)(cid:245)(cid:245)(cid:197)(cid:245)(cid:293)(cid:181)(cid:213)(cid:197)(cid:261)(cid:181)(cid:293)(cid:213)(cid:245)(cid:213) NIGRAM SIHT NI ETIRW TON OD NIGRAM SIHT NI ETIRW TON OD NIGRAM SIHT NI ETIRW TON OD NIGRAM SIHT NI ETIRW TON OD NIGRAM SIHT NI ETIRW TON OD PMT * 0000800000012 *

Question

Biology

CAIE2024

(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:2)(cid:3)(cid:44) (b) There is variation in the quantity of vitamin D stored in the body. Vitamin D has an important role in keeping bones healthy. The main storage form of vitamin D in the body is serum 25‑hydroxyvitamin D (serum 25‑OHD). A study was carried out on 262 healthy women to investigate if the concentration of serum 25‑OHD varied between summer and winter. The women had taken no vitamin D supplements. The age range of the women in the sample was 40 to 72 years old. Table 5.2 shows the results of the study. Table 5.2 group mean concentration of standard serum 25‑OHD deviation / ng cm–3 sampled during summer n = 138 32.7 7.6 sampled during winter n = 124 28.5 8.3 whole sample n = 262 30.7 8.2 Additional analysis showed that there was no significant correlation between age and serum 25‑OHD concentration. (i) Explain what is meant by standard deviation. ........................................................................................................................................... ..................................................................................................................................... [1] (ii) The t‑test was used to compare the mean concentration of serum 25‑OHD when sampled during the summer with the mean concentration of serum 25‑OHD when sampled during the winter, as shown in Table 5.2. Calculate the value of t using the formula provided. x -x key to symbols: t = 1 2 ¯ x = mean s2 s2 s = sample standardf d1ev + iat2iopn n n n = sample size (number of observations) 1 2 Give your answer to four significant figures. There is space for your working. t‑test value ......................................................... [3] (cid:300)(cid:209)(cid:266)(cid:190)(cid:288)(cid:180)(cid:237)(cid:200)(cid:245)(cid:207)(cid:298)(cid:197)(cid:266)(cid:223)(cid:252)(cid:184)(cid:256)(cid:215) © UCLES 2024 (cid:300)(cid:224)(cid:203)(cid:243)(cid:215)(cid:294)(cid:251)(cid:266)(cid:230)(cid:248)(cid:258)(cid:235)(cid:221)(cid:269)(cid:246)(cid:235)(cid:285)(cid:258) 9700/42/O/N/24 (cid:293)(cid:245)(cid:245)(cid:277)(cid:181)(cid:245)(cid:197)(cid:213)(cid:261)(cid:277)(cid:229)(cid:197)(cid:197)(cid:181)(cid:197)(cid:213)(cid:293)(cid:213) NIGRAM SIHT NI ETIRW TON OD NIGRAM SIHT NI ETIRW TON OD NIGRAM SIHT NI ETIRW TON OD NIGRAM SIHT NI ETIRW TON OD NIGRAM SIHT NI ETIRW TON OD PMT * 0000800000013 *

Question

Biology

CAIE2024

(a) Paper chromatography is a technique that can be used to separate and identify different chloroplast pigments. Describe how the results of paper chromatography can be used to identify chloroplast pigments. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [3] (b) Carotene and xanthophyll are chloroplast pigments. Describe the role played by these pigments in photosynthesis. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [3] (cid:300)(cid:205)(cid:266)(cid:190)(cid:288)(cid:180)(cid:237)(cid:200)(cid:245)(cid:207)(cid:298)(cid:197)(cid:266)(cid:222)(cid:249)(cid:181)(cid:260)(cid:215) © UCLES 2024 (cid:300)(cid:224)(cid:204)(cid:244)(cid:214)(cid:290)(cid:246)(cid:247)(cid:227)(cid:240)(cid:254)(cid:255)(cid:227)(cid:300)(cid:212)(cid:267)(cid:269)(cid:258) 9700/42/O/N/24 (cid:293)(cid:293)(cid:229)(cid:213)(cid:181)(cid:277)(cid:261)(cid:213)(cid:261)(cid:229)(cid:213)(cid:261)(cid:197)(cid:245)(cid:197)(cid:277)(cid:229)(cid:213) NIGRAM SIHT NI ETIRW TON OD NIGRAM SIHT NI ETIRW TON OD NIGRAM SIHT NI ETIRW TON OD NIGRAM SIHT NI ETIRW TON OD NIGRAM SIHT NI ETIRW TON OD PMT * 0000800000017 *

Question

Biology

CAIE2024

.......

Question

Biology

CAIE2024

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Question

Biology

CAIE2024

Question

Biology

CAIE2024

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Question

Biology

CAIE2024

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Question

Biology

CAIE2024

D [4] (c) Lambert‑Eaton myasthenic syndrome (LEMS) is a rare disorder of the neuromuscular junction. A person with LEMS produces antibodies that bind to the voltage‑gated calcium channels on the presynaptic knob. One symptom of LEMS is weaker muscle contraction. Suggest and explain why LEMS leads to weaker muscle contraction. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [4] [Total: 11] © UCLES 2024 (cid:300) (cid:300) (cid:207) (cid:224) (cid:266) (cid:202) (cid:190) (cid:243) (cid:288) (cid:215) (cid:180) (cid:294) (cid:237) (cid:242) (cid:200) (cid:273) (cid:245) (cid:220) (cid:207) (cid:257) (cid:298) (cid:252) (cid:197) (cid:292) (cid:266) (cid:283) (cid:223) (cid:274) (cid:251) (cid:292) (cid:183) (cid:179) (cid:258) (cid:277) (cid:215) (cid:258) 9700/42/O/N/24 [Turn over (cid:293)(cid:181)(cid:181)(cid:213)(cid:245)(cid:245)(cid:197)(cid:181)(cid:181)(cid:293)(cid:261)(cid:197)(cid:197)(cid:181)(cid:229)(cid:213)(cid:261)(cid:213) NIGRAM SIHT NI ETIRW TON OD NIGRAM SIHT NI ETIRW TON OD NIGRAM SIHT NI ETIRW TON OD NIGRAM SIHT NI ETIRW TON OD NIGRAM SIHT NI ETIRW TON OD PMT * 0000800000020 *

Question

Biology

CAIE2024

(a) Blood glucose concentration is maintained around a set point by homeostasis. Explain the principles of homeostasis. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [4] (b) Glycogen phosphorylase catalyses the conversion of glycogen to glucose in liver cells. The production of glycogen phosphorylase is coded for by the gene PYGL. A mutation in PYGL leads to a condition called glycogen storage disease type VI (GSDVI), in which glycogen is not broken down efficiently. Suggest and explain why cell signalling by glucagon is likely to be affected in the liver cells of a person with GSDVI. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [3] © UCLES 2024 (cid:300) (cid:300) (cid:211) (cid:224) (cid:266) (cid:201) (cid:190) (cid:242) (cid:288) (cid:220) (cid:180) (cid:294) (cid:237) (cid:300) (cid:200) (cid:270) (cid:245) (cid:231) (cid:207) (cid:268) (cid:298) (cid:258) (cid:197) (cid:274) (cid:266) (cid:217) (cid:224) (cid:198) (cid:249) (cid:227) (cid:182) (cid:267) (cid:258) (cid:269) (cid:215) (cid:258) 9700/42/O/N/24 [Turn over (cid:293)(cid:213)(cid:197)(cid:213)(cid:181)(cid:213)(cid:197)(cid:181)(cid:277)(cid:229)(cid:213)(cid:197)(cid:197)(cid:181)(cid:229)(cid:277)(cid:197)(cid:213) NIGRAM SIHT NI ETIRW TON OD NIGRAM SIHT NI ETIRW TON OD NIGRAM SIHT NI ETIRW TON OD NIGRAM SIHT NI ETIRW TON OD NIGRAM SIHT NI ETIRW TON OD PMT * 0000800000024 *

Question

Biology

CAIE2024

During aerobic respiration, cells respire substrates such as glucose to produce ATP. Some events that occur during aerobic respiration are: • The respiratory substrate breaks down into smaller and smaller molecules. These series of reactions are described as catabolism. • Coenzymes take part in various reactions. In some reactions, coenzymes are reduced or oxidised. • Carbon dioxide is released. (a) Aerobic respiration occurs in four successive stages: glycolysis (G), link reaction (LR), Krebs cycle (KC) and oxidative phosphorylation (OP). Complete Table 3.1 to show which events occur in each stage of aerobic respiration. Use a tick (✓) to show that the event does occur or a cross (✗) to show that the event does not occur. Table 3.1 stage event G LR KC OP catabolism coenzyme is reduced or oxidised a coenzyme forms a covalent bond with a respiratory intermediate carbon dioxide is released [4] (cid:300)(cid:209)(cid:266)(cid:190)(cid:288)(cid:180)(cid:237)(cid:200)(cid:245)(cid:207)(cid:298)(cid:197)(cid:266)(cid:222)(cid:250)(cid:182)(cid:254)(cid:215) © UCLES 2024 (cid:300)(cid:228)(cid:199)(cid:252)(cid:209)(cid:299)(cid:288)(cid:295)(cid:223)(cid:260)(cid:245)(cid:230)(cid:194)(cid:295)(cid:256)(cid:195)(cid:269)(cid:258) 9700/43/O/N/24 (cid:293)(cid:181)(cid:277)(cid:213)(cid:181)(cid:213)(cid:261)(cid:181)(cid:277)(cid:293)(cid:213)(cid:197)(cid:261)(cid:277)(cid:229)(cid:181)(cid:245)(cid:213) NIGRAM SIHT NI ETIRW TON OD NIGRAM SIHT NI ETIRW TON OD NIGRAM SIHT NI ETIRW TON OD NIGRAM SIHT NI ETIRW TON OD NIGRAM SIHT NI ETIRW TON OD PMT * 0000800000007 *

Question

Biology

CAIE2024

Scientists use many different techniques in genetic engineering. (a) Sometimes the gene for genetic engineering cannot be extracted from the donor organism. Instead, the gene is synthesised using one of two different methods. Outline the two methods for synthesising a gene for use in genetic engineering. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [3] (b) DNA ligase and DNA polymerase are two enzymes that are used in genetic engineering. Complete Table 8.1 to show the roles of DNA ligase and DNA polymerase in genetic engineering. Use a tick (✓) if the enzyme has the role or a cross (✗) if the enzyme does not have the role. Table 8.1 role in genetic engineering DNA ligase DNA polymerase joins two sections of sugar phosphate backbone in DNA adds a gene to a plasmid adds free activated DNA nucleotides to a polynucleotide [3] (cid:300)(cid:205)(cid:266)(cid:190)(cid:288)(cid:180)(cid:237)(cid:200)(cid:245)(cid:207)(cid:298)(cid:197)(cid:266)(cid:223)(cid:251)(cid:183)(cid:258)(cid:215) © UCLES 2024 (cid:300)(cid:228)(cid:200)(cid:251)(cid:212)(cid:287)(cid:289)(cid:282)(cid:234)(cid:268)(cid:249)(cid:258)(cid:192)(cid:274)(cid:234)(cid:291)(cid:285)(cid:258) 9700/43/O/N/24 (cid:293)(cid:229)(cid:197)(cid:277)(cid:181)(cid:181)(cid:197)(cid:181)(cid:277)(cid:213)(cid:229)(cid:261)(cid:261)(cid:213)(cid:229)(cid:245)(cid:181)(cid:213) NIGRAM SIHT NI ETIRW TON OD NIGRAM SIHT NI ETIRW TON OD NIGRAM SIHT NI ETIRW TON OD NIGRAM SIHT NI ETIRW TON OD NIGRAM SIHT NI ETIRW TON OD PMT * 0000800000019 *

Question

Biology

CAIE2024

hours You must answer on the question paper. No additional materials are needed. INSTRUCTIONS ● Answer all questions. ● Use a black or dark blue pen. You may use an HB pencil for any diagrams or graphs. ● Write your name, centre number and candidate number in the boxes at the top of the page. ● Write your answer to each question in the space provided. ● Do not use an erasable pen or correction fluid. ● Do not write on any bar codes. ● You may use a calculator. ● You should show all your working and use appropriate units. INFORMATION ● The total mark for this paper is 100. ● The number of marks for each question or part question is shown in brackets [ ]. This document has 24 pages. DC (DE) 348495 © UCLES 2024 [Turn over *5489725721* PMT * 0000800000001 * (cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:2)(cid:44) (cid:172)(cid:79)(cid:138)(cid:62)(cid:160)(cid:52)(cid:109)(cid:72)(cid:117)(cid:79)(cid:170)(cid:69)(cid:138)(cid:96)(cid:124)(cid:54)(cid:128)(cid:87) (cid:172)(cid:170)(cid:54)(cid:116)(cid:79)(cid:162)(cid:145)(cid:114)(cid:107)(cid:120)(cid:134)(cid:62)(cid:104)(cid:120)(cid:110)(cid:139)(cid:112)(cid:130) (cid:165)(cid:149)(cid:149)(cid:149)(cid:53)(cid:117)(cid:133)(cid:85)(cid:165)(cid:133)(cid:165)(cid:133)(cid:133)(cid:85)(cid:133)(cid:85)(cid:117)(cid:85) * 0000800000002 *

Question

Biology

CAIE2024

(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:3)(cid:44)

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Biology

CAIE2024

Question

Biology

CAIE2024

(cid:44)(cid:1)(cid:1)(cid:1)(cid:1)(cid:9)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:1)(cid:5)(cid:44)

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Biology

CAIE2024

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Biology

CAIE2024

Question

Biology

CAIE2024

Scientists use many different techniques in genetic engineering. (a) Sometimes the gene for genetic engineering cannot be extracted from the donor organism. Instead, the gene is synthesised using one of two different methods. Outline the two methods for synthesising a gene for use in genetic engineering. ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ................................................................................................................................................... ............................................................................................................................................. [3] (b) DNA ligase and DNA polymerase are two enzymes that are used in genetic engineering. Complete Table 8.1 to show the roles of DNA ligase and DNA polymerase in genetic engineering. Use a tick (✓) if the enzyme has the role or a cross (✗) if the enzyme does not have the role. Table 8.1 role in genetic engineering DNA ligase DNA polymerase joins two sections of sugar phosphate backbone in DNA adds a gene to a plasmid adds free activated DNA nucleotides to a polynucleotide [3] (cid:300)(cid:205)(cid:266)(cid:190)(cid:288)(cid:180)(cid:237)(cid:200)(cid:245)(cid:207)(cid:298)(cid:197)(cid:266)(cid:223)(cid:249)(cid:184)(cid:258)(cid:215) © UCLES 2024 (cid:300)(cid:298)(cid:184)(cid:241)(cid:212)(cid:290)(cid:258)(cid:294)(cid:234)(cid:268)(cid:262)(cid:247)(cid:212)(cid:223)(cid:235)(cid:235)(cid:256)(cid:258) 9700/41/O/N/24 (cid:293)(cid:181)(cid:229)(cid:213)(cid:181)(cid:213)(cid:261)(cid:181)(cid:245)(cid:181)(cid:277)(cid:261)(cid:261)(cid:213)(cid:293)(cid:277)(cid:213)(cid:213) NIGRAM SIHT NI ETIRW TON OD NIGRAM SIHT NI ETIRW TON OD NIGRAM SIHT NI ETIRW TON OD NIGRAM SIHT NI ETIRW TON OD NIGRAM SIHT NI ETIRW TON OD PMT * 0000800000019 *

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