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Edexcel iGCSE Biology 4BI1 - Paper 1B -Co-ordination and response- Exam Style Questions- New Syllabus

Question

Fly larvae will continuously move around the edge of a Petri dish when placed under a lamp.

The diagram shows a fly larva in a Petri dish.

Design an investigation to find out if changing the colour of light from the lamp affects the speed that the larvae move.

Include experimental details in your answer and write in full sentences.

Most-appropriate topic codes (Edexcel IGCSE Biology):

2(j): Co-ordination and response — investigating response to light stimulus
3: Assessment information: Experimental skills — designing investigations and practical skills
Appendix 5: Suggested practical investigations — investigating responses to external stimuli
▶️ Answer/Explanation
Solution

An answer that makes reference to the following:

C – change colour of light / wavelength of light / use different filters (1)

O – same species / type / age / size / food (1) Ignore same larva unqualified

R – use several larvae with each light colour / repeat (1) Ignore repeat with different colours

M1 – measure distance larvae move / measure arc of dish / number of circuits (1) Accept measure how long it takes to move set distance for M1 and M2

M2 – time / stated time (1) Accept time taken to do a number of laps for two marks (M1 and M2)

S1 & S2 (two from) – same humidity / food / light intensity / temperature / of dish / surface of Petri / same dish (2) Accept clean dish between tests

Full sentence example: “I would change the colour of light using different filters while keeping the same species, age, and size of larvae, using several larvae for each colour and repeating the experiment. I would measure the distance moved along the Petri dish edge and time this movement, while controlling variables like temperature, humidity, and light intensity.”

Question

The diagram shows a section through a human eye with some structures labelled.

(a) (i) Which structures refract light onto the retina?

A) E and J
B) F and H
C) G and K
D) I and J

(ii) Which structure controls the amount of light reaching the retina?

A) E
B) F
C) G
D) H

(iii) Which structure contains light sensitive cells?

A) H
B) I
C) J
D) K

(b) The eye can focus on near objects and distant objects.

Describe the changes that take place in the eye when it focuses on a near object.

(c) Some people develop cataracts in their lenses as they get older. The diagram shows how a cataract changes the appearance of a person’s eye.

(i) Explain how cataracts affect a person’s vision.
(ii) The treatment for a person with cataracts is to remove the affected lenses.
Suggest what additional treatment is needed for the person.

Most-appropriate topic codes (Edexcel IGCSE Biology):

3(j): Co-ordination and response — Structure and function of the eye — parts (a)(i), (a)(ii), (a)(iii), (b), (c)(i), (c)(ii)
▶️ Answer/Explanation
Solution

(a)(i) B (F and H)
A is not the answer as E and J do not refract light.
C is not the answer as G and K do not refract light.
D is not the answer as I and J do not refract light.

(a)(ii) C (G)
A is not the answer as E does not control the amount of light.
B is not the answer as F does not control the amount of light.
D is not the answer as H does not control the amount of light.

(a)(iii) D (K)
A is not the answer as H does not contain light sensitive cells.
B is not the answer as I does not contain light sensitive cells.
C is not the answer as J does not contain light sensitive cells.

(b) A description that makes reference to four of the following points:

  1. the ciliary muscle(s) contract / structure I contract(s) / eq (1)
  2. the tension in the suspensory ligament(s) is reduced / loosened / slackens / eq (1)
  3. the lens becomes thick(er) / wide(r) / fat(ter) / round(er) / more convex / eq (1)
  4. the light is refracted more / bent more / eq (1)
  5. accommodation (1)

Ignore relax.

(c)(i) An explanation that makes reference to two of the following points:

  • lens becomes cloudy / opaque / misty / milky / eq (1)
  • less light enters eye / eq (1)
  • image dimmer / not formed / vision blurred / distorted / not clear / blindness / eq (1)

Ignore poor vision.

(c)(ii)

  • glasses / contact lens / new lens inside eye / eye drops / eq (1)

Ignore laser eye surgery.

Question

The body has a hormonal control system that coordinates some of the body’s responses.

(a) The table shows the effects of some of the hormones and the gland that produces each hormone. Complete the table by giving the missing information.

EffectName of hormoneName of gland
converts blood glucose into glycogen pancreas
stimulates the development of male secondary sexual characteristics  
increases heart rateadrenaline 
maintains the uterus lining  

(b) Plants also respond to changes in their environment.

(i) Explain how plants benefit from the responses of their roots and stems to the direction of light they receive.
(ii) In many plants, flowering is stimulated by the number of hours of daylight. Suggest why flowering in many plants is stimulated by the number of hours of daylight rather than by temperature. 

Most-appropriate topic codes (Edexcel IGCSE Biology):

2(j): Co-ordination and response (Humans) — part (a)
2(j): Co-ordination and response (Flowering plants) — parts (b)(i), (b)(ii)
3(a): Reproduction (Humans) — part (a)
▶️ Answer/Explanation
Solution

(a)

EffectName of hormoneName of gland
converts blood glucose into glycogeninsulinpancreas
stimulates the development of male secondary sexual characteristicstestosteronetestis
increases heart rateadrenalineadrenal gland
maintains the uterus liningprogesteroneovary / corpus luteum / placenta

(b)(i) An explanation that makes reference to four of the following points:

  1. Stems (grow) towards light / are positively phototropic.
  2. This increases photosynthesis / light absorption.
  3. It allows the plant to grow away from competition / other plants.
  4. Roots (grow) away from light / are negatively phototropic.
  5. This directs them into the soil to absorb water / minerals / for anchorage.
  6. This anchors the plant firmly.

(b)(ii) An explanation that makes reference to two of the following points:

  1. Hours of daylight are a better / more reliable indicator of season / time of year.
  2. Daylight length does not vary (day to day) / is constant for a given date, whereas temperature can vary (day to day) / is less predictable.
  3. More hours of daylight allow insects to see flowers (for pollination).
  4. More insect pollination occurs in daylight / insects pollinate in daylight.
Question

The diagram shows the human female reproductive system with some structures labelled.

(a) (i) In which structure does ovulation occur?

A) W
B) X
C) Y
D) Z

(ii) Explain the importance of progesterone in the menstrual cycle.
(iii) Explain how the placenta enables a developing embryo to obtain nutrients.

(b) For fertilisation to occur, a sperm needs to reach an ovum.

(i) Which of these describes the number of chromosomes in the nucleus of a human sperm?

A) diploid number of 23 chromosomes
B) diploid number of 46 chromosomes
C) haploid number of 23 chromosomes
D) haploid number of 46 chromosomes

(ii) Scientists are developing a contraceptive drug to reduce fertility. The drug is tested on 25 male mice.

Graph 1 shows the effect that different concentrations of the drug have on the mean concentration of sperm in semen.

Graph 2 shows the effect that different concentrations of the drug have on the percentage of sperm that can move.

The scientists conclude that the drug could be used as a contraceptive to reduce fertility in humans.

Discuss the scientists’ conclusion. Use information in the graphs and your own knowledge to support your answer.

Most-appropriate topic codes (Edexcel IGCSE Biology):

3(a): Reproduction — parts (a)(i), (a)(ii), (a)(iii), (b)(i)
2(j): Co-ordination and response — part (a)(ii) – hormonal control
3(b): Inheritance — part (b)(i) – chromosome numbers
5(a): Use of biological resources — part (b)(ii) – drug development and testing
Assessment Objective A02: Application of knowledge and understanding — part (b)(ii)
▶️ Answer/Explanation
Solution

(a)(i) B (X)
A is incorrect as ovulation occurs in the ovary
C is incorrect as ovulation occurs in the ovary
D is incorrect as ovulation occurs in the ovary

(a)(ii) An answer that makes reference to two of the following:
• (progesterone is released) after ovulation / in second half of cycle / eq (1)
• maintains uterus lining / maintain endometrium / prevents menstruation / eq (1)
• so embryo can implant in the lining / to support an embryo / to support fetus / eq (1)

(a)(iii) An answer that makes reference to three of following:
• villi / villus (1)
• large surface (area) (1)
• (blood / capillaries) maintains gradient / makes steep gradient / eq (1)
• thin / short distance / maternal and fetal blood are close / eq (1)
• diffusion / active transport (1)

(b)(i) C (haploid number of 23 chromosomes)
A is incorrect as sperm are not diploid
B is incorrect as sperm are not diploid
D is incorrect as sperm have 23 chromosomes

(b)(ii) An answer that makes reference to five of the following:
• sperm movement decreases with concentration / eq (1)
• no further effect between 0.4 and 0.8 / levels off after 0.4 / eq (1)
• no effect on sperm concentration / small reduction / slight effect / unclear effect / fluctuates / eq (1)
• fertilisation would not occur (if sperm cannot swim) / eq (1)
• (sperm) needs to swim to oviducts / Fallopian tubes / fertilisation occurs in oviducts / Fallopian tubes / eq (1)
• (may not work as) some sperm can (always) move / reach eggs / reach oviducts (with any concentration) / eq (1)
• sample size is small / no repeats (so not reliable) / eq (1)
• may not have same effect in humans / may have side effects / health impacts / humans would needs bigger doses / eq (1)
• no mention of age of mice / eq (1)

Question

Scientists can investigate the effect of exercise on breathing rate.

An athlete wears a face mask that covers their nose and mouth. The mask contains electronic sensors that measure and record the athlete’s breathing rate as they exercise.

An investigation using this mask produces these results.

Time since start of exercise in minutesBreathing rate in breaths per minute
020
228
435
645
848
1050
1250

(a) (i) Calculate the percentage change in the breathing rate from the start of exercise to the breathing rate at 10 minutes. 
(ii) Plot a line graph to show how breathing rate changes during exercise. Join your points with straight lines. 
(iii) Explain the change in breathing rate during the 12 minutes of exercise. 

(b) Explain why the breathing rate would remain high for a few minutes after the exercise has finished. 

(c) The difficulty with measuring breathing rate during exercise is that wearing a mask may affect breathing rate and performance.

A different method of recording breathing rate is to wear a shirt that contains sensors that record chest movements.

Suggest one advantage and one disadvantage of using a shirt that measures breathing rate by recording chest movements. 

Most-appropriate topic codes (Edexcel IGCSE Biology):

2(g): Gas exchange — parts (a)(i), (a)(ii), (a)(iii), (b)
2(f): Respiration — part (b)
Appendix 3: Mathematical skills — part (a)(i)
Appendix 4: Command word taxonomy — parts (a)(i) (Calculate), (a)(iii) (Explain), (b) (Explain), (c) (Suggest)
3(j): Co-ordination and response (Homeostasis) — part (a)(iii)
▶️ Answer/Explanation
Solution

(a)(i)

Calculation:
Change in rate = 50 – 20 = 30 breaths per minute.
Percentage change = \( \frac{30}{20} \times 100 = 150\% \)

Answer: 150%

(a)(ii)

A correctly plotted line graph should have:
• Time (minutes) on the x-axis.
• Breathing rate (breaths per minute) on the y-axis.
• A linear, appropriate scale (e.g., y-axis starting at 0 or 10).
• All points correctly plotted within half a small square of their true position.
• Points joined with straight lines.
• Both axes labelled with correct units.

(Refer to the mark scheme: 5 marks for correct scale, lines, axis orientation, units, and accurate plotting.)

(a)(iii)

An explanation that makes reference to three of the following points:
1. (The increase is due to the need for) increased intake of oxygen / eq.
2. To release more carbon dioxide / eq.
3. (To provide more oxygen) to red blood cells / for muscles / for (aerobic) respiration / eq.
4. The breathing rate reaches a maximum / plateaus / stops increasing at 10-12 minutes / eq.
5. The chest cannot inflate and deflate any quicker / there is a physical limit / eq.

(b)

An explanation that makes reference to two of the following points:
• (There is a build-up of) lactic acid / lactate (in muscles) / eq.
• Due to anaerobic respiration occurring during intense exercise / eq.
• An oxygen debt / EPOC (Excess Post-exercise Oxygen Consumption) needs to be repaid / eq.
• Oxygen is required to break down / oxidise the lactic acid.

(c)

Advantage:
• Mouth/nose/face is not covered / does not interfere with airflow / allows the subject to breathe normally / more freely / eq.

Disadvantage:
• Muscle movement during exercise may not be due to breathing / other movements may be recorded / it is an indirect measurement / eq.
• The shirt may be heavy/uncomfortable and affect performance / eq.
• Sweat may affect the sensors / eq.

Question

A reflex is a rapid response to a stimulus.

(a) The withdrawal of a hand when a finger touches a hot object is an example of a reflex.

Describe the pathway of a nerve impulse in the reflex arc involved in this response.

(b) The photograph shows an African land snail.

When an African land snail is touched on its eyestalk, the snail goes back into its shell. The snail then remains in its shell for a period of time.

A student investigates the effect that repeatedly touching the eyestalk has on the time the snail remains in its shell.

This is the student’s method.

  • gently touch the eyestalk of a snail with a cottonwool bud so that the snail goes back into its shell
  • measure the time the snail remains in its shell
  • after the snail has come back out of its shell, touch the eyestalk again so that the snail goes back into its shell again
  • measure the time the snail remains in its shell
  • repeat the stimulus six more times

The student’s results are shown in the table.

Stimulus numberTime snail remains in its shell (seconds)
112
212
310
411
510
67
75
83

(i) Plot a graph to show the effect of stimulus number on the time the snail remains in its shell. Join your points with ruled, straight lines.

(ii) Describe the effect of increasing stimulus number on the time the snail remains in its shell.

(iii) Suggest a benefit to the snail of the difference in response to the repeated stimulus.

Most-appropriate topic codes (Edexcel IGCSE Biology):

3(j): Co-ordination and response (Humans) — part (a)
3(j): Co-ordination and response (General) — parts (b)(ii), (b)(iii)
Appendix 4: Command word taxonomy — parts (a), (b)(ii), (b)(iii)
Appendix 5: Suggested practical investigations — part (b) investigation context
▶️ Answer/Explanation
Solution

(a) A description that makes reference to four of the following points:

  1. receptor (generates impulse) / eq (1)
  2. (impulse) passes along sensory neurone (to relay / motor neurone / to CNS / spinal cord) (1)
  3. (sensory / relay / CNS / spinal cord) to motor neurone (1)
  4. to muscle / effector (1)
  5. crosses synapses (1)
  6. (using) neurotransmitters (1)

(b)(i) Graph marking points:

  • S: linear scales for both axes (1)
  • L: straight line joining points (1)
  • A: axes correct way round (stimulus number on x-axis, time on y-axis) (1)
  • A: axes both labelled, and with units on y axis (Time/seconds) (1)
  • P: points plotted correctly (1)

(Graph should show a general decrease in time with increasing stimulus number, with a steeper decline after stimulus 5/6).

(b)(ii) A description that makes reference to the following points:

  • (time) decreases / gets less / becomes shorter / eq (1)
  • falls (more steeply) after 5 / from 6 touches / small decrease until 5 or 6 / eq (1)

(b)(iii) An explanation that makes reference to two of the following points:

  • (few stimuli) could represent danger / allow escape / escapes predator / snail not eaten / helps snail survive / eq (1)
  • (repeated stimuli) suggests no danger / no longer represents danger / snail recognises there is no threat / has learnt that there is no danger / eq (1)
  • (when not staying in shell) snail can eat for longer / can continue moving / does not waste energy / eq (1)
Question

The diagram shows two samples of blood seen using a high power microscope. One sample is of normal blood and the other sample is from a patient with a blood condition.

(a) State two differences between the normal blood sample and the sample from the patient with the blood condition.

(b) The line P-Q shown on the diagram has an actual length of \( 25 \; \mu m \).
\( [1000 \; \mu m = 1 \; mm] \)
Calculate the magnification of the diagram.

(c) In a healthy person \( 1 \; cm^3 \) of blood contains \( 5.0 \times 10^9 \) red blood cells.
An adult has \( 5.0 \) litres (\( 5.0 \; dm^3 \)) of blood in their body.
Calculate the number of red blood cells in the body of a healthy adult. Give your answer in standard form.
\( [1 \; dm^3 = 1000 \; cm^3] \)

(d) Using the information from the diagram and your own knowledge, comment on the likely effect of the blood condition on the patient.

Most-appropriate topic codes (Edexcel IGCSE Biology):

2(h): Transport — parts (a), (c), (d)
Appendix 3: Mathematical skills — parts (b), (c)
2(j): Co-ordination and response (Homeostasis) — part (d)
▶️ Answer/Explanation
Solution

(a) An answer that makes reference to two of the following:
• More red cells / erythrocytes in the blood condition (1).
• Fewer lymphocytes in the blood condition (1).
• Fewer monocytes in the blood condition (1).
• Fewer neutrophils in the blood condition (1).
If no reference to a named white blood cell, allow “fewer white blood cells” for 1 mark.

(b)
• Measure the line P-Q from the diagram (e.g., 14 mm) (1).
• Convert actual length to mm: \( 25 \; \mu m = 25 / 1000 = 0.025 \; mm \).
• Magnification = Image size / Actual size = \( \frac{14}{0.025} \).
• Magnification = \( \times 560 \) (allow range 520–600) (1).
Full working required for full marks.

(c)
• Volume of blood in \( cm^3 \): \( 5.0 \; dm^3 = 5.0 \times 1000 = 5000 \; cm^3 \).
• Number of red blood cells = \( (5.0 \times 10^9) \times 5000 \).
• \( = 2.5 \times 10^{13} \) (2).
Answer must be in standard form.

(d) An answer that makes reference to four of the following logical points, showing cause and effect:
1. Condition shows more red cells, but they may contain less haemoglobin (e.g., in some anaemias) (1).
2. This leads to less oxygen transport (1).
3. Resulting in less energy / ATP production via respiration (1).
4. Causing symptoms like tiredness, shortness of breath, or lactic acid buildup (1).
5. Diagram shows fewer white blood cells (lymphocytes, monocytes, neutrophils), which could mean a reduced immune response (1).
6. Leading to fewer antibodies produced or less phagocytosis (1).
7. Therefore, the patient may be more susceptible to infections (1).
Answers should link the observed differences to physiological effects on the patient.

Question

The diagram shows a human reflex arc with some structures labelled. It is the withdrawal reflex, which allows a person to remove their hand from a hot object quickly.

(a) (i) Which structure is the sensory receptor?

A) L
B) P
C) Q
D) S

(ii) Which structure is the cell body of the relay neurone?

A) L
B) N
C) M
D) P

(iii) Which structure is the motor neurone?

A) L
B) M
C) R
D) S

(iv) Which structure is the effector?

A) L
B) M
C) Q
D) S

(b) A nerve impulse travels at 50 metres per second.

(i) Calculate the time taken for an impulse to travel 180 cm.

(ii) The length of the nerve pathway in the withdrawal reflex arc is 180 cm.

The time a person takes to remove their hand away from a hot object is called the response time.

Explain why the actual response time is different from the time taken for a nerve impulse to travel 180 cm.

Most-appropriate topic codes (Edexcel IGCSE Biology):

2(j): Co-ordination and response — parts (a)(i), (a)(ii), (a)(iii), (a)(iv), (b)(ii)
Appendix 4: Mathematical skills — part (b)(i)
▶️ Answer/Explanation
Solution

(a)(i) A (L)
Explanation: In a reflex arc, the sensory receptor (e.g., a pain receptor in the skin) detects the stimulus. In the diagram, L is located at the point of stimulus (hand near heat).
B is not the answer as P is the cell body of the relay neurone.
C is not the answer as Q is the sensory neurone.
D is not the answer as S is the effector (muscle).

(a)(ii) D (P)
Explanation: The relay (or intermediate) neurone’s cell body is located in the spinal cord’s grey matter. P is shown within the central nervous system region.
A is not the answer as L is the sensory receptor.
B is not the answer as N is likely part of the motor pathway or synapse.
C is not the answer as M is the motor neurone’s cell body.

(a)(iii) C (R)
Explanation: The motor neurone carries the impulse from the central nervous system to the effector. R is shown connecting to the effector (muscle).
A is not the answer as L is the sensory receptor.
B is not the answer as M is the cell body of the motor neurone, not the entire neurone.
D is not the answer as S is the effector.

(a)(iv) D (S)
Explanation: The effector is the muscle (or gland) that carries out the response. S is labelled on the muscle in the arm.
A is not the answer as L is the sensory receptor.
B is not the answer as M is the cell body of the motor neurone.
C is not the answer as Q is the sensory neurone.

(b)(i)
Calculation:
180 cm = 1.8 m
Speed = 50 m/s
Time = Distance ÷ Speed = \( \frac{1.8}{50} \) = 0.036 s
Answer: 0.036 seconds or \( 3.6 \times 10^{-2} \) s

(b)(ii)
Explanation: The actual response time is longer than the simple impulse travel time because:
1. There is a delay at synapses between neurones (e.g., between sensory and relay, and relay and motor neurones).
2. This delay is due to chemical transmission involving the diffusion of neurotransmitters across the synaptic cleft.
3. Time is required for the sensory receptor to generate the initial nerve impulse in response to the stimulus.
4. Time is also needed for the effector (muscle) to contract after receiving the impulse.

Therefore, the total response time includes synaptic delays, impulse generation time, and muscle contraction time, not just the conduction speed along the neurones.

Question

Farmers and scientists have worked together for the last 50 years to develop crop plants and farm animals with desired characteristics.

(a) Explain why selective breeding with crop plants is easier and quicker than selective breeding with farm animals. 

(b) Selective breeding in cattle has been used to improve features such as higher milk yield, longer life and increased muscle mass.
Describe how selective breeding can be used to improve milk yield in dairy cattle.

(c) Scientists and farmers are also selecting animals based on behaviour and temperament.

(i) Suggest why an animal with a calmer temperament may be more suitable for the farmer on a dairy farm. 

(ii) Cattle behaviour is also affected by the hormone adrenaline. When cattle are more stressed, they have higher levels of adrenaline.
The effect of adrenaline on the cattle nervous system is similar to the effect of adrenaline in humans.
Describe the production of adrenaline and its effects on the body. 

(iii) Stress in cattle can also affect their reproductive system, leading to damage to the placenta.
Describe the role of the placenta in reproduction.

Most-appropriate topic codes (Edexcel IGCSE Biology):

5(b): Selective breeding — parts (a) and (b)
2(j): Co-ordination and response — part (c)(ii)
3(a): Reproduction (Humans) — part (c)(iii)
5(a): Food production (Animal behaviour) — part (c)(i)
▶️ Answer/Explanation
Solution

(a) An explanation that makes reference to three of the following:

  • Plants produce many more seeds/offspring (higher yield) per generation, providing a larger pool for selection.
  • Plants have shorter life cycles/ reach maturity quicker, allowing more generations to be bred in a given time.
  • Plants can be self-pollinated, allowing easier control of breeding and inheritance of desired traits.
  • The environment for plants (e.g., glasshouses) is easier to control than for large farm animals.

Additional detail: For example, a wheat plant can produce hundreds of seeds in one season, while a cow typically produces one calf per year.

(b) A description that makes reference to four of the following:

  1. Select (mate/breed) bulls (fathers) that come from female relatives (daughters/mothers/sisters) with the highest milk yield.
  2. Select (mate/breed) cows (mothers) that themselves have the highest milk yield or come from high-yielding families.
  3. Select the female offspring from these matings that show the highest milk yield.
  4. Repeat this process over many generations to accumulate the genes for high milk yield.
  5. Artificial insemination is often used to spread the genes of the best bulls widely.

(c)(i) An answer that makes reference to two of the following:

  • Calmer cows are easier and safer to manage, handle, and milk.
  • They are less likely to fight, injure other animals, or harm farm workers.
  • They are less likely to run around, which wastes energy that could be used for growth or milk production.
  • Reduced stress can lead to better overall health and higher milk yields.

(c)(ii) A description that makes reference to four of the following:

  1. Production: Adrenaline is produced and secreted by the adrenal glands (located above the kidneys).
  2. Heart Rate: It causes an increase in heart rate.
  3. Purpose: It prepares the body for ‘fight or flight’ by diverting resources to muscles.
  4. Blood Flow: Increases blood flow to muscles; blood is diverted away from the gut/intestines.
  5. Breathing: Increases breathing rate and depth to get more oxygen.
  6. Energy: Converts glycogen in the liver to glucose, increasing blood sugar levels for rapid energy.
  7. Other: Causes pupil dilation and speeds up reaction times.

(c)(iii) A description that makes reference to three of the following:

  • Diffusion Surface: It provides a large surface area for diffusion between maternal and fetal blood.
  • Supply of Nutrients: It supplies the fetus with oxygen, glucose, amino acids, vitamins, and minerals from the mother’s blood.
  • Supply of Antibodies: It allows the passage of antibodies from mother to fetus, providing passive immunity.
  • Removal of Waste: It removes fetal waste products (carbon dioxide, urea) into the mother’s blood for excretion.
  • Hormone Secretion: It secretes hormones (e.g., progesterone) to help maintain the pregnancy.

Note: Answers must indicate the direction of substance transfer (e.g., from mother to fetus).

Question

The diagram shows the structure of a human eye.

(a) (i) Give the name of structure Z.

(ii) Describe how structures X and Y control the shape of a lens when focusing on a near object.

(b) A cataract occurs when the lens becomes cloudy.
Severe cataracts are the main cause of blindness around the world.

(i) Explain why people with cataracts are unable to see clearly.

(ii) Exposure to ultraviolet light is thought to be a risk factor for the development of cataracts.
The table shows the results of a 25-year investigation into the effect of different mean hours of sunlight on agricultural workers in one south Asian country.

Comment on the results of the investigation.

Most-appropriate topic codes (Edexcel IGCSE Biology):

2(j): Co-ordination and response — parts (a)(i), (a)(ii), (b)(i)
4(d): Human influences on the environment — part (b)(ii) context
▶️ Answer/Explanation
Solution

(a) (i) iris

Explanation: Structure Z is the iris. The iris is the colored part of the eye that controls the size of the pupil, thereby regulating the amount of light that enters the eye.

(a) (ii) A description that makes reference to the following three points:

  • X / ciliary muscle, contracts
  • Y / suspensory ligaments go slack / loosen
  • lens becomes thicker / more spherical / more convex

Detailed Explanation: When focusing on a near object, the ciliary muscles (X) contract. This contraction reduces the tension on the suspensory ligaments (Y), causing them to become slack or loosen. Because these ligaments are no longer pulling the lens taut, the natural elasticity of the lens causes it to become thicker, more spherical, and more convex. This change in shape increases the refractive power of the lens, allowing it to bend light rays more sharply and focus the image of the close object correctly onto the retina.

(b) (i) An explanation that makes reference to two of the following:

  • less light can pass through / refraction affected
  • to retina / fovea / rods / cones / photoreceptors
  • so fewer impulses (to brain)

Detailed Explanation: A cataract causes the normally clear lens to become cloudy or opaque. This cloudiness scatters and blocks light, preventing a clear, sharp image from being formed. Less light can pass through the lens to reach the light-sensitive cells (rods and cones) in the retina at the back of the eye. Furthermore, the cloudiness disrupts the precise refraction (bending) of light rays that is necessary for focusing. As a result, the light rays are not properly focused onto the retina, leading to a blurred or dim image. This faulty image results in fewer or confused nerve impulses being sent to the brain via the optic nerve, causing unclear vision.

(b) (ii) An answer that makes reference to four of the following points based on the data:

  • numbers (of cataracts) increase as exposure to sunlight increases (e.g., 2 at 7 hrs vs 35 at 12 hrs).
  • group sizes are different (e.g., 100 at 7 hrs vs 540 at 12 hrs), which must be considered when interpreting raw numbers.
  • the proportion/percentage of cases fluctuates/does not change much between 7 and 10 hours (approx 2%, 1.9%, 2.0%, 1.6%).
  • the proportion/percentage rises more noticeably after 10/11 hours (3.2% at 11 hrs, 6.5% at 12 hrs).
  • the large number of people in the study overall makes it somewhat reliable, OR it is less reliable as data is only from one country/climate.
  • other factors like age, genetics, diet, or protective gear (sunglasses, hats) could affect the results and are not accounted for.

Detailed Comment: The data suggests a general trend where the number of people developing cataracts increases with higher mean daily hours of sunlight exposure, from 2 cases at 7 hours to 35 cases at 12 hours. However, because the group sizes vary significantly (e.g., 100 people vs 540 people), it’s more accurate to look at the percentage of each group affected. Calculating these percentages (7h: 2%, 8h: ~1.9%, 9h: 2%, 10h: ~1.6%, 11h: ~3.2%, 12h: ~6.5%) reveals that the risk appears relatively stable and low (around 2%) for exposures up to 10 hours. A more substantial increase in risk is observed beyond 10 hours, jumping to 3.2% at 11 hours and 6.5% at 12 hours, indicating a potential threshold effect. The 25-year duration and large total sample size add weight to the findings. However, the study is limited to agricultural workers in one South Asian country, so the results may not be directly applicable to populations in different climates, with different genetics, or with different sun protection behaviors. Other unmeasured factors, such as the exact intensity of UV exposure, age of the workers, dietary intake of antioxidants, and use of protective eyewear, could also influence cataract development and are not considered in this data table.

Question

The diagram shows a human reflex response to touching a hot object. Some of the structures in the reflex arc are labelled.

Reflex Arc Diagram

(a) (i) Give the name of this reflex response.

(a) (ii) Draw an arrow on the diagram to show the direction of the nerve impulse at point X.

(a) (iii) Which structure detects the stimulus?

  • A. E
  • B. F
  • C. H
  • D. K

(a) (iv) Which structure is the motor neurone?

  • A. F
  • B. G
  • C. I
  • D. K

(b) (i) A neurone is 1.10 m in length. The speed of the nerve impulse in this neurone is 120 metres per second. Calculate the time, in seconds, for the impulse to pass along the neurone. Give your answer in standard form.

(b) (ii) Describe how the impulse passes from neurone to neurone in the reflex arc.

(c) (i) Pain has a survival function. If a person damages their ankle, they sense pain. Explain how sensing pain benefits an organism.

(c) (ii) Some medicines are used to reduce pain. These medicines work by preventing communication between the injured ankle and the brain. Explain which components of the nervous system may be affected by these medicines.

Most-appropriate topic codes (Edexcel IGCSE Biology):

2(j): Co-ordination and response — parts (a)(i), (a)(ii), (a)(iii), (a)(iv), (b)(ii), (c)(i), (c)(ii)
Appendix 4: Mathematical skills — part (b)(i)
▶️ Answer/Explanation
Solution

(a)(i) Withdrawal reflex / Somatic reflex / Simple reflex / Involuntary reflex / Automatic reflex / Unconscious reflex.

(a)(ii) An arrow pointing away from the spinal cord (towards the muscle).

(a)(iii) A (E) – sensory receptor in the skin.

(a)(iv) C (I) – motor neurone.

(b)(i) \( 9.2 \times 10^{-3} \) s (or \( 9.17 \times 10^{-3} \) s).

(b)(ii) The impulse passes across a synapse via neurotransmitters released from the presynaptic neurone, which diffuse across the synaptic gap and bind to receptors on the postsynaptic neurone, triggering a new electrical impulse.

(c)(i) Pain prevents further movement or use of the injured part, reducing further damage and allowing time for healing.

(c)(ii) Medicines may affect pain receptors, sensory neurones, relay neurones in the spinal cord/brain, or synaptic transmission by blocking neurotransmitter release or receptor binding.

Question

Genetically modified bacteria are used to produce the hormone insulin.

(a) Explain how these genetically modified bacteria are produced.

(b) Explain the role of insulin in the human body.

(c) Some people are unable to produce insulin. This condition is called diabetes mellitus. People with diabetes mellitus control the condition by using insulin injections, controlling their diet, and monitoring how much they exercise.

(i) Explain why the insulin is injected rather than taken by mouth.

(ii) State why people with diabetes mellitus need to monitor how much they exercise.

(iii) State how people with diabetes mellitus may need to modify their diet compared with people who do not have diabetes mellitus.

Most-appropriate topic codes (Edexcel IGCSE Biology):

5(c): Genetic modification (genetic engineering) — part (a)
2(j): Co-ordination and response — part (b)
2(e): Nutrition — part (c)(i), (c)(iii)
2(f): Respiration — part (c)(ii)
▶️ Answer/Explanation
Solution

(a)

Explanation: Genetically modified bacteria are produced using a process that involves isolating the human insulin gene and inserting it into a bacterial plasmid. First, the restriction enzyme is used to cut the gene/DNA that codes for the production of human insulin. The same restriction enzyme is then used to cut open a bacterial plasmid. This creates complementary pairings or ‘sticky ends’ on both the insulin gene and the plasmid. The ligase enzyme is then used to join or insert the insulin gene into the plasmid. Finally, this modified plasmid (now a vector) is inserted into, or taken up by, a bacterium. The bacterium then uses this new genetic information to produce human insulin.

(b)

Explanation: Insulin is a hormone that plays a crucial role in regulating blood glucose levels. When blood glucose concentration increases, for example after a meal, insulin is released. It causes the liver and muscles to take up glucose from the blood and convert it into glycogen for storage. This process reduces the high blood glucose level back to normal.

(c)(i)

Explanation: Insulin is a protein. If it were taken by mouth, it would be digested and broken down in the stomach and small intestine by protease enzymes like pepsin and trypsin. These enzymes would break the insulin down into its constituent amino acids, destroying its structure and function before it could ever be absorbed into the bloodstream to lower blood sugar. Injection delivers the insulin directly into the subcutaneous tissue, allowing it to be absorbed into the bloodstream intact.

(c)(ii)

Explanation: People with diabetes need to monitor their exercise because physical activity uses glucose as a source of energy. This increased glucose use can cause their blood glucose level to become too low, a dangerous condition known as hypoglycaemia.

(c)(iii)

Explanation: People with diabetes mellitus often need to control or limit the amount of carbohydrates, sugars, and glucose in their diet to help manage their blood sugar levels. They might also replace simple sugars with complex carbohydrates like starch.

Question

The following passage is about chemical coordination in animals and plants. Complete the passage by writing a suitable word in the blank space.

Animals and plants use chemicals to coordinate responses. In animals, some glands produce hormones which are transported in the ______ of the blood. A high glucose concentration in the blood stimulates the release of a hormone called ______ from the ______ . This causes an organ called the ______ to remove glucose from the blood and store it as a substance called ______ .

Plant shoots grow towards light. This response is called ______ . The movement of a chemical called ______ to the shaded side of the shoot causes the shoot to grow towards the light.

Most-appropriate topic codes (Edexcel IGCSE Biology):

2(h): Transport — “transported in the plasma of the blood”
2(j): Co-ordination and response — “high glucose concentration… stimulates release of insulin”, “plant shoots grow towards light”, “auxin movement”
2(e): Nutrition — “store it as glycogen” (carbohydrate storage in humans)
2(f): Respiration — “glucose” as a substrate for respiration
▶️ Answer/Explanation
Solution

Completed Passage:

Animals and plants use chemicals to coordinate responses. In animals, some glands produce hormones which are transported in the plasma of the blood. A high glucose concentration in the blood stimulates the release of a hormone called insulin from the pancreas. This causes an organ called the liver to remove glucose from the blood and store it as a substance called glycogen.

Plant shoots grow towards light. This response is called phototropism. The movement of a chemical called auxin to the shaded side of the shoot causes the shoot to grow towards the light.

Detailed Explanation:

Animal Hormonal Control (Blood Glucose):

  • Hormones are chemical messengers secreted by glands directly into the bloodstream. The liquid part of the blood that carries these hormones, along with other substances, is called the plasma.
  • When the concentration of glucose in the blood becomes too high (e.g., after eating a carbohydrate-rich meal), it is detected by specialized cells in the pancreas.
  • In response, the pancreas secretes the hormone insulin.
  • Insulin travels in the plasma to its target organ, the liver.
  • The liver responds by converting the excess glucose into an insoluble storage carbohydrate called glycogen. This process lowers the blood glucose concentration back to a normal level.

Plant Hormonal Control (Growth Response to Light):

  • Plants also coordinate their growth in response to environmental stimuli using chemicals. The growth of a plant shoot towards a light source is a classic example of a directional growth response called phototropism (specifically positive phototropism, as growth is towards the stimulus).
  • This response is controlled by a plant hormone (or plant growth regulator) called auxin (Indoleacetic Acid or IAA is a specific example).
  • When light is directional (coming mostly from one side), auxin produced at the shoot tip tends to move or accumulate on the shaded side of the shoot.
  • Auxin stimulates cell elongation. The higher concentration of auxin on the shaded side causes those cells to elongate more than the cells on the lit side.
  • This uneven growth on opposite sides of the shoot causes it to bend towards the light source.
Question

Reflexes are responses that protect the body and involve the central nervous system (CNS).

(a) (i) Give the name of one part of the CNS.

(a) (ii) If an animal steps on a sharp object, a reflex arc occurs so that it picks up its foot to prevent further damage. Describe how named neurones bring about a withdrawal reflex to protect the animal from damage.

(b) The photograph shows a breed of dog called a border collie.

Some border collies have a genetic condition called sensory neuropathy. Border collies with sensory neuropathy may be injured as their reflexes do not work well. Sensory neuropathy is caused by a recessive allele, n. The dominant allele for not having sensory neuropathy is N.

The diagram shows a family pedigree for some border collies.

(b) (i) Sex determination in dogs is controlled in the same way as in humans. Which combination of sex chromosomes is present in the body cells of a male dog?

A. XX
B. XY
C. Y
D. YY

(b) (ii) How many individuals in this family have the genotype Nn?

A. 2
B. 4
C. 5
D. 6

(b) (iii) Determine the probability that the next offspring produced by individuals P and Q is a male with sensory neuropathy. Include a genetic diagram in your answer.

(b) (iv) Dog breeders use selective breeding to try to remove harmful alleles from dog breeds. Explain how selective breeding could be used to remove the allele for sensory neuropathy from a population of border collies.

Most-appropriate topic codes (Edexcel IGCSE Biology):

2(j): Co-ordination and response — part (a)(i), (a)(ii)
3(b): Inheritance — parts (b)(i), (b)(ii), (b)(iii)
5(b): Selective breeding — part (b)(iv)
▶️ Answer/Explanation
Solution

(a)(i) Brain or Spinal Cord.

Explanation: The Central Nervous System (CNS) is composed of two main parts: the brain and the spinal cord. These structures are responsible for processing information and coordinating responses throughout the body.

(a)(ii) A description including four of the following points:

  1. A receptor in the skin detects the sharp object/stimulus.
  2. This generates an electrical impulse.
  3. The impulse travels along a sensory neurone to the CNS (spinal cord).
  4. In the spinal cord, the impulse may pass via a relay neurone across a synapse (using neurotransmitters).
  5. The impulse is then passed to a motor neurone.
  6. The motor neurone carries the impulse to an effector (a muscle in the leg).
  7. The muscle contracts, causing the foot to be withdrawn.

Explanation: This is an automatic, rapid, protective response called a reflex arc. It bypasses the brain to save time. The sensory neurone carries information towards the CNS, and the motor neurone carries the instruction away from the CNS to the muscle (the effector) which performs the action. Synapses ensure the impulse travels in one direction only.

(b)(i) B (XY)

Explanation: In mammals like dogs and humans, sex is determined by the X and Y chromosomes. Females have two X chromosomes (XX), and males have one X and one Y chromosome (XY). The Y chromosome carries genes that trigger male development.

(b)(ii) D (6)

Explanation: Sensory neuropathy is recessive (nn). Unaffected individuals must have at least one dominant allele (N). Any unaffected individual that has an affected offspring (nn) must be a carrier (heterozygous, Nn) because they had to pass on an ‘n’ allele to their affected child. By analyzing the pedigree, we can identify 6 such carriers (Nn).

(b)(iii) Probability = 1/8 or 0.125 or 12.5%

Genetic Diagram:

Parents: Nn (P) × Nn (Q)

Gametes: P produces N or n; Q produces N or n.

 Q Gametes ↓ P → | N | n | +-------+-------+ N | NN | Nn | +-------+-------+ n | Nn | nn | +-------+-------+ 

Explanation: The Punnett square shows the cross between two heterozygous parents (Nn × Nn). The genotypic ratio of the offspring is 1 NN : 2 Nn : 1 nn. Therefore, the probability of an offspring having sensory neuropathy (nn) is 1/4. Sex determination is independent of this autosomal gene. The probability of an offspring being male is 1/2. To find the combined probability of being male AND having nn, we multiply the independent probabilities: (1/4) × (1/2) = 1/8.

(b)(iv) Selective breeding, also known as artificial selection, could be used to remove the harmful ‘n’ allele by carefully choosing which dogs breed together.

The process would involve:

  1. Testing: First, all potential breeding dogs would be genetically tested to determine their genotype for the sensory neuropathy gene (NN, Nn, or nn).
  2. Selection of Parents: Only dogs that are confirmed to be homozygous dominant (NN) – meaning they do not carry the harmful allele – would be selected as breeding stock. Dogs with the genotype Nn (carriers) or nn (affected) would be prevented from breeding.
  3. Repetition over Generations: This selective process would be repeated over many generations. By never allowing the ‘n’ allele to be passed on, its frequency in the population’s gene pool would decrease.
  4. Goal: Over time, the aim would be to eliminate the ‘n’ allele entirely from the breeding population, resulting in a population where all individuals are NN and therefore cannot have or pass on sensory neuropathy.

Potential Challenge: A significant challenge is that carriers (Nn) are phenotypically normal (unaffected). Without genetic testing, breeders might accidentally select carriers for breeding, which would perpetuate the allele in the population. Therefore, reliable genetic testing is crucial for this strategy to be successful.

Question

The chromosomes in a human cell can be photographed and then arranged in pairs to show the karyotype.

White blood cells are often used to show the chromosomes in the karyotype.

The chromosomes in white blood cells are larger and easier to see when the white blood cells divide.

Diagram 1 shows a karyotype.

(a) (i) Explain why a red blood cell cannot be used to show a karyotype.

(ii) State the type of cell division that occurs in white blood cells.

(iii) The karyotype in diagram 1 is from a male. State how this can be deduced from the diagram.

(b) Diagram 2 shows a karyotype from a white blood cell of another person. The karyotype is from a female, and the person has a condition called Turner syndrome. This condition affects the development of the ovaries so they may not produce normal quantities of sex hormones.

(i) Comment on the differences between the karyotypes shown in diagram 1 and diagram 2 and the effects Turner syndrome will have on the person. Use information from the question and your own knowledge in your answer.

(ii) Suggest how the difference in the chromosomes of people with Turner syndrome may have been produced.

Most-appropriate topic codes (Edexcel IGCSE Biology):

2(b): Cell structure — part (a)(i)
3(a): Reproduction — part (a)(ii)
3(b): Inheritance — parts (a)(iii), (b)(i), (b)(ii)
2(h): Transport — part (a)(i)
2(j): Co-ordination and response — part (b)(i)
▶️ Answer/Explanation
Solution

(a)(i) A red blood cell cannot be used to show a karyotype because it lacks a nucleus. The nucleus is the organelle that contains the chromosomes. Since red blood cells in mammals are anucleated (they lose their nucleus during development to make more space for hemoglobin), they do not contain any chromosomal material that can be photographed and arranged into a karyotype.

(a)(ii) The type of cell division that occurs in white blood cells is mitosis. Mitosis is the process of nuclear division in eukaryotic cells that results in two daughter cells each having the same number and kind of chromosomes as the parent nucleus. It is the division responsible for general growth and repair in the body.

(a)(iii) The karyotype in Diagram 1 can be identified as male because the 23rd pair of chromosomes consists of one X chromosome and one Y chromosome. In humans, females have two X chromosomes (XX), while males have one X and one Y chromosome (XY). The presence of the Y chromosome is the definitive indicator of a male karyotype.

(b)(i) The key difference between the two karyotypes is the number of chromosomes. Diagram 1 (the male) shows the normal human diploid number of 46 chromosomes. Diagram 2 (the female with Turner syndrome) shows only 45 chromosomes; specifically, there is only one sex chromosome present (a single X chromosome), denoted as 45,X.

The effects of Turner syndrome on the person are significant. Due to the missing X chromosome and the resulting hormonal deficiencies (particularly oestrogen), the individual will likely not undergo normal puberty. This leads to a lack of development of secondary sexual characteristics (such as breast development). Furthermore, the ovaries are typically underdeveloped (streak ovaries) and do not produce viable eggs, making the person infertile. There can also be other health implications, such as short stature and potential heart defects.

(b)(ii) The chromosomal difference in Turner syndrome (45,X) is most commonly produced by a process called non-disjunction. This is an error that can occur during the formation of the gametes (eggs or sperm) in one of the parents. Specifically, during meiosis, the paired sex chromosomes (X and X in a female, or X and Y in a male) fail to separate properly. If an egg or sperm that lacks a sex chromosome (is nullisomic) fuses with a normal gamete containing one X chromosome, the resulting zygote will have only one X chromosome, leading to Turner syndrome.

Question

Yeast can be used in food production.

(a) (i) Which group of organisms does yeast belong to?

  • A. bacteria
  • B. fungi
  • C. plants
  • D. protoctists

(ii) Which substance is the cell wall of a yeast cell made of?

  • A. cellulose
  • B. chitin
  • C. sucrose
  • D. starch

(b) A teacher sets up an experiment to investigate the effect of temperature on the rate of respiration in yeast. The diagram shows part of the apparatus the teacher uses.

(i) Explain what additional apparatus the teacher will need to investigate the effect of temperature on the rate of respiration in this experiment.

(ii) State the purpose of the liquid paraffin on the surface of the glucose solution.

(iii) Give the name of a suitable chemical that could be used as indicator X to show that the yeast is respiring.

(iv) Diazine green changes colour from blue to pink as the oxygen levels in the yeast and glucose solution reduce. Explain how this colour change gives information about the respiration in the yeast.

(c) Explain why the rate of respiration in the yeast will change as the temperature is increased.

Most-appropriate topic codes (Edexcel IGCSE Biology):

1(b): Variety of living organisms — part (a)(i)
2(b): Cell structure — part (a)(ii)
5(a): Food production (Micro-organisms) — part (a)(i), (a)(ii), (b) context
2(f): Respiration — parts (b)(ii), (b)(iii), (b)(iv), (c)
2(j): Co-ordination and response (Homeostasis) — part (c) linked to enzyme temperature effects
5.6 practical: Investigate the role of anaerobic respiration by yeast — part (b) experimental context

▶️ Answer/Explanation
Solution

(a)(i) B (fungi)
Explanation: Yeast is a single-celled organism that belongs to the kingdom Fungi. Unlike bacteria (prokaryotes), yeast are eukaryotic cells. They are not plants as they do not perform photosynthesis and lack chloroplasts, and they are not protoctists, which is a kingdom typically containing diverse, often unicellular, eukaryotes that don’t fit into the other kingdoms.

(a)(ii) B (chitin)
Explanation: The cell wall of a yeast cell is primarily made of a complex carbohydrate called chitin, which is also found in the exoskeletons of insects and the cell walls of other fungi. This is different from plant cell walls, which are made of cellulose (A). Sucrose (C) is a sugar and starch (D) is a storage polysaccharide; neither are structural components of cell walls.

(b)(i) The teacher would need additional apparatus to control and measure temperature. This includes a water bath (or a beaker of water heated by a Bunsen burner) to contain the test tube and allow for precise temperature variation. A thermometer is essential to accurately measure the temperature of the yeast and glucose solution. A stopwatch or timer is also needed to measure the rate of the reaction, for example, by timing how long it takes for a colour change to occur or measuring the rate of gas production if that was being monitored.

(b)(ii) The layer of liquid paraffin acts as a seal on the surface of the glucose solution. Its purpose is to prevent oxygen from the air dissolving into the solution. This creates anaerobic conditions, ensuring that any respiration detected is anaerobic respiration (fermentation) rather than aerobic respiration. This is crucial for investigating this specific type of respiratory pathway.

(b)(iii) A suitable chemical indicator would be limewater (calcium hydroxide solution). Alternatively, hydrogencarbonate indicator could be used.
Explanation: Limewater is used to test for the product carbon dioxide (CO₂) of respiration. CO₂ turns limewater cloudy. Hydrogencarbonate indicator changes colour (from red to yellow) in the presence of increased carbon dioxide levels.

(b)(iv) Diazine green is an indicator that is blue when oxygen is present and pink when it is absent (reduced). The initial blue colour indicates there is oxygen dissolved in the solution, and the yeast is likely respiring aerobically. As the yeast cells respire, they use up the oxygen in the solution for aerobic respiration. The change in colour from blue to pink indicates that the oxygen concentration is decreasing. Once the oxygen is depleted and the solution turns pink, it shows that the yeast has switched to anaerobic respiration (fermentation) to continue producing energy.

(c) The rate of respiration in yeast changes with temperature due to the effect on enzymes, which control the metabolic reactions of respiration.
As temperature increases from a low value, the kinetic energy of the enzyme and substrate molecules increases. This causes them to move faster and collide more frequently. More collisions lead to the formation of more enzyme-substrate complexes, which increases the rate of respiration.
This increase continues up to an optimum temperature, where the respiration rate is at its maximum because enzymes are working most efficiently.
Beyond this optimum temperature, the high energy causes the hydrogen and ionic bonds that hold the enzyme’s shape to break. This alters the specific 3D shape of the enzyme’s active site. The substrate can no longer fit into the denatured active site, meaning fewer enzyme-substrate complexes can form. This causes the rate of respiration to decrease rapidly and eventually stop.

Question

(a) The diagram shows the structure of a human eye.

(i) Which of these is the structure labelled X?

  • A. conjunctiva
  • B. cornea
  • C. lens
  • D. retina

(ii) When looking at a close object, which row of the table shows the state of the ciliary muscles and suspensory ligaments? (1)

(iii) Explain how structure Y controls the light entering the eye when someone walks into a dark room. (2)

(b) Multiple sclerosis is a condition that can slow down the speed at which electrical impulses travel along neurones.

The time taken for the blink reflex to occur can be used to help diagnose if someone has multiple sclerosis.

The blink reflex causes the eyelid to close.

Air is blown on to the eye and the time taken for the eyelid to close is recorded.

The diagram shows the reflex pathway.

The speed the impulse moves along the reflex arc consisting of all three neurones in a person without multiple sclerosis is 77 metres per second.

The time taken for the blink reflex to occur in a person with multiple sclerosis is 0.0050 s.

The total length of the neurones in the reflex arc for the person with multiple sclerosis is 25 cm.

(i) Calculate the difference between the speed of impulse for the person with multiple sclerosis and for the person without multiple sclerosis, in metres per second. (3)

(ii) The speed of an impulse along the axon of the motor neurone for someone without multiple sclerosis is 120 metres per second.

Suggest why the speed of the impulse calculated along all three neurones is less than the speed of the impulse along only the motor neurone. (2)

Most-appropriate topic codes (Edexcel IGCSE Biology):

2(j): Co-ordination and response — parts (a)(i), (a)(ii), (a)(iii), (b)(ii)
Appendix 4: Mathematical skills — part (b)(i)
2(b): Cell structure — part (b)(ii) [neurone structure]
▶️ Answer/Explanation
Solution

(a)(i) B – cornea

Explanation: The cornea is the transparent layer at the very front of the eye that helps to focus light. The conjunctiva is a thin membrane covering the front of the eye and inner eyelids, the lens is behind the pupil, and the retina is at the back of the eye. Structure X is at the front, identifying it as the cornea.

(a)(ii) A – contracted, loose

Explanation: When looking at a close object, the ciliary muscles contract. This contraction reduces the tension on the suspensory ligaments, causing them to become slack or loose. This allows the lens to become more rounded and thicker, increasing its refractive power to focus on near objects.

(a)(iii)

Explanation: Structure Y is the iris, which contains muscles that control the size of the pupil. When someone walks into a dark room, the light intensity decreases. In response, the radial muscles in the iris contract. Simultaneously, the circular muscles in the iris relax. This combined action causes the pupil to dilate (widen), allowing more light to enter the eye so that you can see better in the dim conditions.

(b)(i) 27 m/s

Explanation: First, we need to find the speed of the impulse for the person with multiple sclerosis (MS). Speed is calculated using the formula: speed = distance / time.

The total length of the neurones (distance) is 25 cm. We need to convert this to metres to match the units of the answer (m/s). 25 cm = 0.25 m.

The time taken for the impulse is 0.0050 s.

So, speedMS = distance / time = 0.25 m / 0.0050 s = 50 m/s.

The speed for a person without MS is given as 77 m/s.

The difference in speed is: 77 m/s – 50 m/s = 27 m/s.

(b)(ii)

Explanation: The overall speed along the entire reflex arc (77 m/s) is slower than the speed along just the motor neurone (120 m/s) because the reflex arc includes synapses. Synapses are the junctions between neurones where impulses are transmitted using chemical messengers called neurotransmitters. This chemical diffusion process across the synaptic cleft is much slower than the electrical transmission of the impulse along a single neurone. Therefore, the presence of multiple synapses in the pathway (sensory to relay to motor) introduces delays, reducing the overall calculated speed of the reflex.

Question

The photograph shows a small animal called a woodlouse.

Woodlice often live under pieces of dead wood in dark, humid conditions.

Design an investigation to find out if light intensity affects the speed at which woodlice move.

Include experimental details in your answer and write in full sentences.

Most-appropriate topic codes (Edexcel IGCSE Biology):

4(a): The organism in the environment — population and habitat study, abiotic factors
2(j): Co-ordination and response — organisms responding to environmental stimuli
Appendix 6: Suggested practical investigations — investigating organism responses to environmental factors
2.80–2.85: Responding to stimuli, tropisms — experimental design and organism behaviour
▶️ Answer/Explanation
Solution

Investigation Design:

To investigate whether light intensity affects the speed at which woodlice move, I would set up the following experiment:

Apparatus: I would need a choice chamber with two compartments, a lamp to create different light intensities, a ruler, a stopwatch, woodlice of the same species and similar size, and materials to control other variables like humidity.

Method:

  1. I would place several woodlice (e.g., 10) in the choice chamber and allow them to acclimatize for 5 minutes.
  2. I would set up the lamp at different distances (e.g., 10 cm, 20 cm, 30 cm) from one side of the chamber to create varying light intensities, while keeping the other side dark.
  3. For each light intensity, I would measure how quickly the woodlice move from one point to another. I could do this by marking a start line and finish line in the chamber and timing how long it takes each woodlouse to cross a certain distance (e.g., 10 cm).
  4. I would repeat each test at least three times to ensure reliability and calculate mean speeds.
  5. I would control other variables by maintaining constant temperature, humidity, and using woodlice of the same species and size.
  6. I would record my results in a table showing light intensity (distance from lamp) and average speed of movement.

Expected Analysis: If light intensity affects movement speed, I would expect to see faster movement in darker conditions (as woodlice prefer dark environments) and slower movement in brighter light. I could plot a graph of light intensity vs. speed to visualize any relationship.

Safety Considerations: I would handle the woodlice gently to avoid harming them and ensure the lamp doesn’t overheat the chamber.

Question

The data in the table was collected in Japan during a seven-year study.

Scientists collected data on the age of mothers and whether they smoked during pregnancy.

They also recorded the percentage of the babies that had a low birth mass.

(a) (i) Calculate the percentage of mothers aged 19 years and under who smoked during pregnancy.

(ii) Determine the ratio of non-smokers to smokers used in the study.

Give the ratio as the nearest whole number (n) in the form n:1

(b) A student examines this data and concludes that smoking is the main factor that causes low birth mass.

Use the data and your own biological knowledge to comment on this conclusion.

Most-appropriate topic codes (Edexcel IGCSE Biology):

2(g): Gas exchange — biological effects of smoking (part b)
2(j): Co-ordination and response — hormonal regulation, pregnancy (part b)
3(a): Reproduction — pregnancy and fetal development (part b)
Appendix 4: Mathematical skills — percentage and ratio calculations (parts a(i), a(ii))
▶️ Answer/Explanation
Solution

(a)(i) 21.1%

Explanation: To calculate the percentage of mothers aged 19 and under who smoked during pregnancy, we need to find the proportion of smokers in that age group. The number of smokers is 356, and the total number of mothers (smokers + non-smokers) is 1331 + 356 = 1687. The percentage is calculated as (356 ÷ 1687) × 100 = 21.1% (rounded to one decimal place).

(a)(ii) 12:1

Explanation: To determine the ratio of non-smokers to smokers, we first need to calculate the total number of non-smokers and smokers across all age groups. Total non-smokers = 1331 + 11243 + 24099 + 28695 + 16537 + 3242 = 85147. Total smokers = 356 + 1677 + 2211 + 1847 + 934 + 181 = 7206. The ratio of non-smokers to smokers is 85147:7206. To express this as n:1, we divide both sides by 7206: 85147 ÷ 7206 ≈ 11.82, which rounds to the nearest whole number as 12. So the ratio is 12:1.

(b)

Explanation: While the data shows a correlation between smoking and low birth mass, it would be premature to conclude that smoking is the main factor based solely on this data. Here’s a balanced analysis:

Evidence supporting the student’s conclusion:

  • At every age group, the percentage of babies with low birth mass is higher for mothers who smoked compared to those who didn’t. For example, in the 35-39 age group, smokers had 21.1% low birth mass babies compared to 10.5% for non-smokers.
  • The study involves a large sample size (over 92,000 mothers total), which adds credibility to the findings.
  • Biologically, smoking reduces oxygen supply to the fetus due to carbon monoxide binding to hemoglobin more strongly than oxygen. This can impair fetal development and lead to lower birth weight.

Evidence challenging the student’s conclusion:

  • The data shows that age is also a significant factor. For non-smokers, the percentage of low birth mass babies is higher at both extremes of age (11.5% for ≤19 and 12.3% for ≥40) compared to the middle age groups (9.0-10.5%).
  • The effect of smoking appears to be more pronounced in older mothers. For mothers ≥40, smokers have 22.1% low birth mass compared to 12.3% for non-smokers – a difference of 9.8 percentage points, while for mothers 25-29, the difference is only 4.3 percentage points (13.3% vs 9.0%).
  • The study doesn’t account for other potential factors that could influence birth mass, such as:
    • Maternal nutrition and diet
    • Maternal weight and overall health
    • Alcohol consumption
    • Quality of prenatal care
    • Genetic factors
    • Number of years the mother has been smoking
  • The data is correlational, not experimental, so we cannot establish causation definitively.

In conclusion, while smoking appears to be a significant factor associated with low birth mass, the data suggests that it is not necessarily the main factor, as maternal age and other unmeasured variables also play important roles. A more comprehensive study controlling for these other factors would be needed to determine the primary cause of low birth mass.

Question

The diagram shows a food web from a woodland ecosystem.

(a) (i) Which organism in the food web is the producer?

  • A. blackbird
  • B. centipede
  • C. earthworm
  • D. oak tree

(ii) Which organism acts as both a secondary consumer and a tertiary consumer in the food web?

  • A. blackbird
  • B. earthworm
  • C. ground beetle
  • D. sparrowhawk

(b) The amount of energy transferred changes as you move along a food chain.

The data comes from an ecosystem containing producers, primary consumers and secondary consumers.

(i) The light energy reaching the producers is 7.1 × 10⁶ kJ per m² per year. Explain why the plants cannot absorb all of this energy.

(ii) The table shows that energy is transferred between producer and primary consumer and between primary consumer and secondary consumer. A student states that the energy transfer between producer and primary consumer is the most efficient. Determine whether the student’s statement is correct.

(c) Woodlice feed on dead and decaying plant material in the soil. The photographs show how a woodlouse can curl up into a ball. This behaviour is an example of a reflex response.

(i) State what is meant by a reflex response.

(ii) Give a reason why this reflex response benefits the woodlouse.

(iii) Describe how this reflex response could have evolved by natural selection.

Most-appropriate topic codes (Edexcel IGCSE Biology):

4(b): Feeding relationships — parts (a)(i), (a)(ii), (b)(i), (b)(ii)
2(j): Co-ordination and response — parts (c)(i), (c)(ii)
3(b): Inheritance — part (c)(iii) — evolution by natural selection
▶️ Answer/Explanation
Solution

(a)(i) D oak tree

Explanation: Producers are organisms that can make their own food through photosynthesis. In this food web, the oak tree is the only organism that produces its own food using sunlight, making it the producer. The other options (blackbird, centipede, and earthworm) are all consumers that rely on eating other organisms for energy.

(a)(ii) A blackbird

Explanation: A secondary consumer eats primary consumers, while a tertiary consumer eats secondary consumers. The blackbird acts as both because it eats primary consumers like earthworms (making it a secondary consumer) and may also eat other secondary consumers like ground beetles (making it a tertiary consumer).

(b)(i)

Explanation: Plants cannot absorb all the light energy that reaches them because some light is reflected off their surfaces, particularly light of certain wavelengths that chlorophyll cannot absorb efficiently. Additionally, some light falls on non-photosynthetic parts of the plant like flowers or stems, or may not hit chloroplasts directly, reducing the overall energy absorption efficiency.

(b)(ii)

Explanation: To determine the efficiency of energy transfer, we calculate the percentage of energy transferred between trophic levels:

Producer to primary consumer: \( \frac{1.4 \times 10^4}{8.7 \times 10^5} \times 100 = 1.61\% \)

Primary to secondary consumer: \( \frac{1.6 \times 10^3}{1.4 \times 10^4} \times 100 = 11.43\% \)

The energy transfer between primary and secondary consumers (11.43%) is significantly more efficient than between producers and primary consumers (1.61%). Therefore, the student’s statement is incorrect.

(c)(i)

Explanation: A reflex response is an automatic, involuntary reaction to a stimulus that does not require conscious thought or involvement of the brain. It is a rapid, protective mechanism that helps organisms respond quickly to potential threats.

(c)(ii)

Explanation: This reflex response benefits the woodlouse by protecting it from predators. When curled into a ball, the woodlouse exposes only its hard exoskeleton, making it difficult for predators to attack or consume its vulnerable body parts.

(c)(iii)

Explanation: This reflex response could have evolved through natural selection via the following process:
1. Initially, genetic mutation created variation in the woodlouse population, with some individuals having a slight tendency to curl when threatened.
2. Woodlice that could curl into a ball had a survival advantage as they were better protected from predators.
3. These surviving woodlice reproduced and passed on the alleles responsible for this curling behavior to their offspring.
4. Over many generations, this reflex response became more common in the population as individuals with this trait continued to survive and reproduce more successfully than those without it.

Question

The skin is an organ involved in temperature regulation.

(a) State why the skin is described as an organ.

(b) The diagram shows a section of human skin.

(i) Explain how the skin capillaries regulate body temperature when the body temperature increases.

(ii) Some people have suggested that drinking warm drinks may cool the body because it increases the rate of sweat production.

Design an investigation to determine whether drinking warm drinks increases the rate of sweat production.

Include experimental details in your answer and write in full sentences.

Most-appropriate topic codes (Edexcel IGCSE Biology):

2(a): Level of organisation — part (a)
2(j): Co-ordination and response — parts (b)(i), (b)(ii)
2.93: Role of skin in temperature regulation — part (b)(i)
Appendix 6: Suggested practical investigations — part (b)(ii)
▶️ Answer/Explanation
Solution

(a) The skin is described as an organ because it is made up of different tissues working together to perform specific functions.

Explanation: An organ is defined as a structure composed of at least two different types of tissues that work together to perform a particular function. The skin includes epithelial tissue (epidermis), connective tissue (dermis), nervous tissue (sensory receptors), and muscular tissue (arrector pili muscles), all collaborating in roles like protection, sensation, and temperature regulation.

(b)(i) When body temperature increases, the skin capillaries help regulate temperature through a process called vasodilation.

Explanation: Here’s a step-by-step explanation of the process:

  1. The arterioles leading to the skin capillaries widen (vasodilation).
  2. This allows more blood to flow through the capillaries near the skin surface.
  3. As blood is warmer than the external environment, heat is transferred from the blood to the environment.
  4. This heat loss occurs primarily through radiation (and some convection).
  5. The result is a lowering of the blood temperature, which helps cool the entire body.

This mechanism is crucial for maintaining a constant internal body temperature, especially during exercise or in hot environments.

(b)(ii) A suitable investigation would be carefully controlled to test the hypothesis.

Explanation: Here is a detailed experimental design:

Aim: To determine if drinking a warm drink increases the rate of sweat production compared to drinking a cold drink.

Variables:

  • Independent Variable: The temperature of the drink (e.g., warm water at 50°C vs. cold water at 5°C).
  • Dependent Variable: The rate of sweat production.
  • Control Variables: These must be kept constant to ensure a fair test.
    • Volume of drink consumed (e.g., 250 cm³).
    • Type of drink (e.g., plain water).
    • The individual’s activity level before and during the test (e.g., resting).
    • Room temperature and humidity.
    • Clothing worn by the individual.
    • The time of day.
    • The initial body temperature of the individual.

Method:

  1. Select a group of healthy participants of similar age, fitness, and body mass.
  2. Ensure they rest in a room with a constant, comfortable temperature for 20 minutes before starting.
  3. For the test, one participant will drink the warm water, and another (or the same participant on a different day) will drink the cold water.
  4. Immediately after drinking, a pre-weighed piece of absorbent cotton wool or a small, dry cloth is placed on a specific part of the skin (e.g., the forearm).
  5. After a fixed time period (e.g., 10 minutes), the cotton wool/cloth is removed and immediately weighed.
  6. The increase in mass is calculated, which represents the mass of sweat produced in that area over that time.
  7. The rate of sweat production can then be calculated (e.g., mass of sweat per minute).
  8. The experiment is repeated with the other temperature of drink.
  9. The entire investigation should be repeated with multiple participants to improve reliability.

Conclusion: The average rate of sweat production after drinking the warm drink is compared to the average rate after drinking the cold drink. If the warm drink causes a significantly higher rate, it would support the suggestion.

Question

The human body has different hormones that are produced by endocrine glands.

(a) The diagram labels some of the endocrine glands in the body.

Diagram of endocrine glands

(i) Which gland produces insulin?

  • A   
  • B   
  • C   
  • D

(ii) Which gland produces progesterone?

  • A   
  • B   
  • C   
  • D

(b) The human body has two systems of communication, nervous and hormonal.

(i) Students research the speed of nervous and hormonal communication. They find this data:

  • hormones travel at a speed of 420 centimetres per minute
  • nerve impulses travel at a speed of 55 metres per second

Determine the ratio of the speed of nervous communication to the speed of hormonal communication. Give your answer in the form n:1

(ii) Describe three other differences between the nervous system and the hormonal system.

Most-appropriate topic codes (Edexcel IGCSE Biology):

2(j): Co-ordination and response — parts (a), (b)(i), (b)(ii)
Appendix 4: Mathematical skills (unit conversion, ratio) — part (b)(i)
▶️ Answer/Explanation
Solution

(a)(i) A

Explanation: The pancreas, labeled A in the diagram, is the gland that produces insulin. Insulin is a crucial hormone that regulates blood glucose levels by facilitating the uptake of glucose into cells. The other glands (B, C, D) have different primary functions and do not produce insulin.

(a)(ii) D

Explanation: The gland labeled D (which would be the ovary in a female) produces progesterone. Progesterone is a key hormone involved in the menstrual cycle and in maintaining pregnancy. Glands A, B, and C have different roles and do not produce progesterone.

(b)(i) 786:1

Explanation: To find the ratio of nervous communication speed to hormonal communication speed, we first need both speeds in the same units. Let’s convert the hormonal speed from cm/min to m/s. Hormonal speed is 420 cm/min. There are 100 cm in a meter and 60 seconds in a minute, so we calculate: \( \frac{420 , \text{cm/min}}{100 , \text{cm/m} \times 60 , \text{s/min}} = \frac{420}{6000} , \text{m/s} = 0.07 , \text{m/s} \).

The nervous speed is given as 55 m/s. The ratio of nervous to hormonal speed is therefore: \( \frac{55}{0.07} = 785.7 \), which rounds to approximately 786. So, the ratio is 786:1.

(b)(ii)

Explanation: Three key differences between the nervous and hormonal systems are:

  1. Mode of Transmission: The nervous system uses electrical impulses that travel along neurons (nerve cells). In contrast, the hormonal system uses chemical messengers (hormones) that are secreted into the bloodstream and travel in the plasma.
  2. Speed and Target: Nervous responses are very fast and are directed to specific, localized target cells (like a specific muscle). Hormonal responses are generally slower and the hormones can have widespread effects on many different target cells or organs throughout the body.
  3. Duration of Effect: The effects of nervous system communication are typically short-lived. For example, a nerve impulse causes a muscle to contract and then relax quickly. The effects of the hormonal system are often longer-lasting. For instance, the hormone insulin regulates blood sugar over a period of hours.
Question

(a) Vital capacity is the maximum volume of air that a person can force out of their lungs in one breath.

The graph shows the relationship between vital capacity and age for a large number of people.

The vital capacity plotted is the median value for each age.

(i) State what is meant by the term median.

(ii) Give a reason why the median is used rather than the mean.

(iii) Describe the relationship between vital capacity and age shown by the graph.

(iv) Explain why vital capacity changes with age.

(v) Age is not the only variable that can change vital capacity. Give two other variables that can affect a person’s vital capacity.

(b) Describe a method you could use to demonstrate the effect of exercise on breathing rate in students.

Most-appropriate topic codes (Edexcel IGCSE Biology):

Appendix 4: Mathematical skills — Handling datapart (a)(i), (a)(ii), (a)(iii)
2(g): Gas exchange — Humansparts (a)(iv), (a)(v), (b)
2(j): Co-ordination and response — Humanspart (b)
Appendix 6: Suggested practical investigationspart (b)
▶️ Answer/Explanation
Solution

(a)(i) The middle value.

Explanation: The median is a measure of central tendency that represents the middle number in a sorted, ascending or descending, list of numbers. If there is an odd number of observations, the median is the middle one. If there is an even number, it is the average of the two middle numbers. It effectively splits the data set into two halves.

(a)(ii) Median used as each age has wide/high range / not (influenced by) affected by / any reference to / reduces effect of extreme values / data is skewed / less affected by anomalies.

Explanation: The median is preferred over the mean in this context because the data for each age group likely has a wide range or might be skewed by extreme values (outliers). The median is not affected by very high or very low values, making it a more reliable measure of the typical vital capacity for each age group than the mean, which can be pulled up or down by outliers.

(a)(iii) A description that makes reference to two of the following:

  • Increases (steeply) up to 20-25 years
  • Levels off / peaks 20-25 years
  • Decreases (steadily at constant rate) from 20-25 years

Explanation: The graph shows that vital capacity increases quite steeply as age increases from childhood up to the age range of 20-25 years. At this point, it reaches its peak and levels off, meaning it stays relatively constant for a short period. After this peak, from around 25 years onwards, the vital capacity begins to decrease steadily at a fairly constant rate as age continues to increase.

(a)(iv) An explanation that makes reference to two of the following:

  • (Increases) (up to 20 years) as body / size / mass / chest cavity / lungs / muscle increases / grows / develops
  • No / little change (from 20) no further growth / stop growing
  • (Decreases) (from 25 years) as diaphragm / intercostal muscle weaker / lungs less elastic / less recoil

Explanation: The increase in vital capacity during childhood and adolescence (up to around 20 years) is due to physical growth. As the body grows, the chest cavity enlarges, and the lungs and associated muscles (like the diaphragm and intercostal muscles) develop and increase in size, allowing for a greater maximum volume of air to be expelled. From around 20 to 25 years, growth stops, so vital capacity remains relatively stable. The gradual decrease after 25 years is associated with the natural aging process, where the respiratory muscles can become weaker, and the lung tissue loses some of its elasticity, reducing the efficiency and maximum capacity of the lungs.

(a)(v) An answer that makes reference to two of the following:

  • Smoking
  • Lung disease / infection / condition / asthma / bronchitis / emphysema
  • Fitness / how active you are
  • Body size / height / mass / genetics
  • Sex
  • Altitude at which you live
  • Pregnancy
  • Pollution

Explanation: Many factors other than age can influence a person’s vital capacity. Lifestyle choices like smoking can damage lung tissue and reduce capacity. Pre-existing lung diseases or conditions like asthma also have a significant impact. A person’s fitness level is crucial; more active individuals typically have stronger respiratory muscles and better lung function. Physical attributes such as body size, height, and sex are also determining factors, as larger body size and male sex are generally associated with a higher vital capacity. Environmental factors like the altitude you live at or exposure to pollution can also affect lung function and development.

(b) A description that makes reference to three of the following:

  • Count number of breaths per minute / number of breaths in stated time (at rest)
  • Breathing rate count per minute / number of breaths in stated time after / during exercise
  • Repeat / using more participants / use group
  • Control age / sex / fitness of subjects / amount / period of exercise

Explanation: To investigate the effect of exercise on breathing rate, you would first need to measure the resting breathing rate. This is done by counting the number of breaths a student takes in one minute while they are at rest. One breath consists of one inhalation and one exhalation. Then, the student would perform a standardised exercise, such as running on the spot or stepping onto a bench for a set amount of time (e.g., 5 minutes). Immediately after the exercise, you would count their breathing rate again for one minute. To make the results more reliable, you should repeat the experiment with the same student on different days and also use a group of students to collect more data. It is important to control variables that could affect the results, such as the age, sex, and fitness level of the participants, as well as the type, duration, and intensity of the exercise they perform.

Question

The diagram shows a wind-pollinated flower with some structures labelled W, X and Y.

(a) (i) Describe how structures W, X and Y are adapted for wind pollination.

(ii) Structures W, X and Y are adapted for wind pollination.

Give two other differences between wind-pollinated flowers and insect-pollinated flowers.

(b) Wind-pollinated flowers often cause an allergic response in people. This is known as hay fever.

Most people in the United Kingdom who get hay fever have the symptoms from April to September.

The graph shows the changes in total pollen count for three different plant types from March to September during one year in the United Kingdom.

As part of an investigation into pollen allergy, five people keep a diary of their hay fever symptoms. They do this for the same year as the pollen count.

The table gives their results.

Using the data in the table and the information from the graph, discuss the likely causes of the allergic responses in each person.

(c) The allergic response to pollen is part of the body’s immune response.

Explain what is meant by the term immune response.

Most-appropriate topic codes (Edexcel IGCSE Biology):

3(a): Reproduction — flowering plants, pollination adaptations — parts (a)(i), (a)(ii)
4(a): The organism in the environment — populations and sampling — part (b)
2(j): Co-ordination and response — immune response — part (c)
▶️ Answer/Explanation
Solution

(a)(i)

Answer:

  • W or stigma / feathery / large surface area / W stigma outside flower / exposed (to catch pollen)
  • X or anther outside flower / exposed (to disperse pollen)
  • Y or filament long / hinged / not rigid / can move (to disperse pollen)

Explanation: Wind-pollinated flowers have specific adaptations to ensure successful pollination without relying on insects. Structure W is the stigma, which is feathery and has a large surface area to effectively catch pollen grains floating in the air. It is positioned outside the flower to maximize exposure to wind-borne pollen. Structure X is the anther, which is also exposed outside the flower to allow pollen to be easily released and carried away by the wind. Structure Y is the filament, which is long and often hinged or flexible, allowing the anther to move freely in the wind, thereby enhancing pollen dispersal.

(a)(ii)

Answer: Two differences could be:

  • Wind-pollinated flowers have smaller / dull / green flowers / petals / no petals
  • Wind-pollinated flowers have no nectar / nectary
  • Wind-pollinated flowers have no scent
  • Wind-pollinated flowers have smaller / lighter / smooth pollen grains / more pollen produced

Explanation: Wind-pollinated flowers and insect-pollinated flowers have evolved different strategies. Wind-pollinated flowers typically have smaller, duller, often green petals, or may even lack petals altogether since they don’t need to attract insects visually. They do not produce nectar because there are no insects to reward, and they lack scent as aroma is unnecessary for attracting pollinators. Their pollen grains are smaller, lighter, and smooth to be easily carried by wind currents, and they produce pollen in much larger quantities to increase the chance of some grains reaching another flower.

(b)

Answer: The discussion should include references such as:

  • Link between pollen number and symptoms
  • Person A allergic to tree pollen (only)
  • Person B allergic to (mainly) grass pollen
  • Person B some / mild allergy to tree and weed
  • Person C allergic to all pollen / tree and grass and weed
  • Person D no pollen allergy
  • Person E allergic to (mainly) grass and weed
  • Person E some / mild allergy to tree
  • No species level data
  • Only one year of data

Explanation: By correlating the pollen count graph with the symptom diaries, we can identify likely pollen allergies for each person. Person A experiences severe symptoms in April and May, which coincides with high tree pollen counts, suggesting a specific allergy to tree pollen. Person B has severe symptoms in June and July when grass pollen is highest, indicating a primary allergy to grass pollen, with mild symptoms in other months possibly due to lower levels of tree and weed pollen. Person C suffers from April to September, covering the peaks of all three pollen types, suggesting allergies to tree, grass, and weed pollen. Person D reports no symptoms, indicating no pollen allergies. Person E has severe symptoms from June to September, aligning with high grass and weed pollen, and mild symptoms earlier in the year possibly due to tree pollen. It’s important to note that this data is from only one year and doesn’t identify specific plant species, so conclusions should be considered preliminary.

(c)

Answer: An explanation that includes:

  • Response to antigens / pathogen / allergen / bacteria / virus
  • By white blood cells / phagocytes / lymphocytes
  • Antibodies produced / phagocytosis / engulf

Explanation: An immune response is the body’s sophisticated defense mechanism against foreign substances called antigens, which can include pathogens like bacteria and viruses, or allergens like pollen. This process is primarily carried out by specialized white blood cells. Phagocytes are cells that engulf and destroy foreign particles through a process called phagocytosis. Lymphocytes are another type of white blood cell that produces antibodies – specific proteins that recognize and bind to particular antigens, marking them for destruction. In the case of hay fever, the immune system mistakenly identifies harmless pollen as a threat, triggering this response and causing the uncomfortable symptoms associated with allergies.

Question

Fur colour in rats is controlled by a gene with two alleles.

One allele codes for black fur colour. The other allele codes for agouti fur colour.

Several female rats with agouti coloured fur are mated with several male rats with black coloured fur.

All of the offspring have agouti coloured fur.

(a) (i) Explain which allele is dominant.

(ii) A male and female rat from these offspring are then mated together in a second cross.

Some of the offspring of this second cross have agouti coloured fur and some have black coloured fur.

Draw a genetic diagram to show this second cross. Include the genotypes of the parents, the gametes they produce, and the genotypes and the phenotypes of the offspring.

(iii) Calculate the probability of any one offspring from this second cross being male with agouti coloured fur.

(b) Scientists observed that genes that control fur colour in rats can affect rat behaviour, such as how tame they are.

Other genes that control the size of the adrenal glands and the production of neurotransmitters also affect rat behaviour.

Scientists also noticed that coat colour is associated with differences in anatomy and physiology, such as the size of the adrenal glands and the production of neurotransmitters.

(i) State the name of the type of genetic control where many genes control one phenotype.

(ii) Explain why the size of the adrenal glands and the production of neurotransmitters would affect rat behaviour.

(c) Some rats with white fur also have pink eyes.

These rats have pink eyes because they do not have pigment in their irises.

This means that their irises let light pass through, unlike the coloured irises found in other rats.

Explain how this difference in the iris affects vision in the rats with pink eyes.

Most-appropriate topic codes (Edexcel IGCSE Biology):

3(b): Inheritance — parts (a)(i), (a)(ii), (a)(iii), (b)(i)
2(j): Co-ordination and response — part (b)(ii)
2(j): Structure and function of the eye — part (c)
▶️ Answer/Explanation
Solution

(a) (i) The agouti allele is dominant.

Explanation: When agouti females are crossed with black males, all offspring are agouti. This shows that the agouti allele is expressed in the offspring even when combined with a black allele. In genetics, the allele that is expressed in the heterozygous condition (one copy) is the dominant allele. Therefore, the agouti allele must be dominant over the recessive black allele.

(a) (ii)

Genetic Diagram:

Parental genotypes: Aa × Aa (both heterozygous from the first cross)

Gametes: A or a from each parent

Offspring genotypes:

  • AA (homozygous agouti)
  • Aa (heterozygous agouti)
  • Aa (heterozygous agouti)
  • aa (homozygous black)

Phenotypes: 3 Agouti : 1 Black

Explanation: The parents from the first cross are both heterozygous (Aa) because they received one agouti allele from their agouti mother and one black allele from their black father. When two heterozygous individuals are crossed, the expected genotypic ratio is 1 AA : 2 Aa : 1 aa, and the phenotypic ratio is 3 agouti : 1 black, which matches the observed results.

(a) (iii) Probability = 37.5% or 0.375

Explanation: The probability of an offspring being agouti is ¾ (from the 3:1 phenotypic ratio). The probability of an offspring being male is ½ (assuming equal chance of male and female). Since these are independent events, we multiply the probabilities: (¾) × (½) = 3/8 = 0.375 = 37.5%.

(b) (i) Polygenic inheritance.

Explanation: When multiple genes (such as those for fur color, adrenal gland size, and neurotransmitter production) collectively influence a single phenotypic trait (like behavior), this is known as polygenic inheritance.

(b) (ii)

Explanation: The size of the adrenal glands affects the amount of adrenaline produced. Adrenaline is a key hormone in the “fight or flight” response, influencing heart rate and fear responses. Neurotransmitters are chemical messengers that transmit signals across synapses in the nervous system. Different levels of neurotransmitters can affect how quickly nerve impulses are transmitted, influencing reflex speed, reaction time, and overall responsiveness. Therefore, variations in adrenal gland size and neurotransmitter production can directly impact how a rat reacts to its environment, making it more or less tame, fearful, or aggressive.

(c)

Explanation: The iris normally contains pigment that blocks light from entering the eye except through the pupil. In rats with pink eyes, the lack of pigment in the iris means it cannot effectively regulate the amount of light entering the eye. Light passes through the iris itself, not just the pupil, resulting in too much light reaching the retina. This overexposure can cause poor vision, difficulty focusing, and potential damage to the light-sensitive cells of the retina, potentially leading to blindness over time.

Question

(a) The diagram shows a plant cell.

(i) Which part of this cell contains chlorophyll?

  • A. P
  • B. Q
  • C. R
  • D. S

(ii) Which of these is found in chlorophyll?

  • A. calcium
  • B. iron
  • C. magnesium
  • D. water

(iii) Describe the role of chlorophyll.

(b) Which of these is an example of positive phototropism?

  • A. a plant root growing away from light
  • B. a plant root growing downwards due to gravity
  • C. a plant stem growing towards light
  • D. a plant stem growing upwards due to gravity

(c) The table lists the roles of some substances found in living organisms. Complete the table by naming each substance. The first one has been done for you.

Most-appropriate topic codes (Edexcel IGCSE Biology):

2(b): Cell structure — part (a)(i)
2(e): Nutrition (Flowering plants) — part (a)(ii), (a)(iii)
2(j): Co-ordination and response (Flowering plants) — part (b)
2(e): Nutrition (Humans) — part (c)
2(j): Co-ordination and response (Humans) — part (c)
▶️ Answer/Explanation
Solution

1 (a)(i) D S

Explanation: In plant cells, chlorophyll is found in the chloroplasts, which are represented by structure S in the diagram. Chloroplasts are the organelles responsible for photosynthesis and contain the green pigment chlorophyll that captures light energy.

1 (a)(ii) C magnesium

Explanation: Chlorophyll molecules contain magnesium ions at their center. The magnesium atom is coordinated to four nitrogen atoms in the porphyrin ring structure of chlorophyll, which is essential for its light-absorbing properties during photosynthesis.

1 (a)(iii) Chlorophyll absorbs/traps light energy for photosynthesis to produce carbohydrates.

Detailed Explanation: Chlorophyll plays a crucial role in photosynthesis by absorbing light energy, primarily from the blue and red regions of the visible spectrum. This absorbed light energy is then converted into chemical energy through the process of photosynthesis. During photosynthesis, carbon dioxide and water are converted into glucose and oxygen using this captured energy. The glucose produced can then be used by the plant for energy or converted into starch for storage.

1 (b) C a plant stem growing towards light

Explanation: Positive phototropism refers to the growth movement of a plant part toward a light source. Plant stems exhibit positive phototropism as they grow toward light, which maximizes their exposure to sunlight for photosynthesis. This directional growth is controlled by the hormone auxin, which accumulates on the shaded side of the stem, causing those cells to elongate more rapidly.

1 (c)

Detailed Explanation:

Lipase is an enzyme that breaks down fats (lipids) into fatty acids and glycerol during digestion. It’s produced by the pancreas and works in the small intestine.

Neurotransmitter is a chemical messenger that diffuses across the synaptic cleft (the gap between neurons) to transmit nerve impulses from one neuron to another. Examples include acetylcholine and noradrenaline.

Vitamin C (ascorbic acid) is essential for preventing scurvy, a disease characterized by bleeding gums, joint pain, and fatigue. Vitamin C is necessary for collagen synthesis, which is important for maintaining healthy connective tissues.

Question

The diagram shows the position of some hormone producing glands in the female body.

(a) Which of these structures is the adrenal gland?

  • A. W
  • B. X
  • C. Y
  • D. Z

(b) The adrenal gland is an organ that secretes adrenaline.
State what is meant by the term organ.

(c) Adrenaline is released into the blood when there is danger.

The list gives the effects of adrenaline on different parts of the body.

  • dilates the pupil in the eye
  • increases heart rate
  • narrows small arteries in the intestine
  • converts glycogen into glucose in the liver

Explain the advantages of these effects to a person in danger.

Most-appropriate topic codes (Edexcel IGCSE Biology):

2(j): Co-ordination and response — parts (a), (b), (c)
2.94: Hormones — adrenaline source, role and effect
2.80–2.82: Response to environment and homeostasis — adrenaline as part of the fight-or-flight response
2(a): Level of organisation — definition of an organ (part b)
▶️ Answer/Explanation
Solution

(a) C Y

Explanation: The adrenal gland is correctly identified as structure Y in the diagram. The other options are incorrect because:

  • A (W) is not the adrenal gland
  • B (X) is not the adrenal gland
  • D (Z) is not the adrenal gland

The adrenal glands are located above the kidneys and are responsible for producing adrenaline and other hormones.

(b) An organ is several/different tissues carrying out a particular function/purpose.

Explanation: An organ is defined as a structure composed of multiple different types of tissues that work together to perform a specific function or purpose in the body. In this case, the adrenal gland contains various tissues that collectively produce and secrete adrenaline and other hormones.

(c) The advantages include:

  • Dilation of the pupil allows more light to enter the eye, improving vision and making the person more aware of their surroundings to better detect danger.
  • Increased heart rate pumps more blood to vital organs, particularly the lungs and muscles, ensuring they receive adequate oxygen and nutrients.
  • Narrowing small arteries in the intestine redirects blood flow away from digestion and toward the muscles, which are needed for physical response.
  • Conversion of glycogen to glucose in the liver increases blood sugar levels, providing more immediate energy for muscle cells through respiration.
  • These combined effects increase oxygen delivery to muscles, enhance glucose availability for energy production, support higher rates of respiration, reduce reliance on anaerobic respiration (which produces lactic acid), and ultimately help the person run faster or escape danger more effectively.

Detailed Explanation: When adrenaline is released during a dangerous situation, it triggers the “fight or flight” response. The pupil dilation enhances visual awareness, allowing the person to better assess threats. The increased heart rate ensures that oxygenated blood reaches critical areas like the brain and muscles more quickly. By diverting blood from the digestive system to the muscles, the body prioritizes immediate physical action over digestion. The conversion of stored glycogen into glucose provides a rapid energy source for muscle contraction. All these physiological changes work together to prepare the body for intense physical activity, whether that means fighting the threat or fleeing from it.

Question

Farmers can keep their animals indoors or outdoors.

A student uses this apparatus to compare the heat loss from animals kept indoors and outdoors.

He uses a covered beaker to represent animals indoors and an uncovered beaker to represent animals outdoors.

This is the student’s method:

  • pour 200 cm³ of hot water into each beaker
  • measure the temperature of the water in each beaker
  • cover one beaker with a cardboard box
  • measure the temperature of the water in each beaker after 30 minutes

The student repeats the investigation six times and calculates the mean temperature of the water for each beaker.

The table shows the student’s results:

(a) Give the dependent variable in this investigation.

(b) Give one reason why the student uses the same volume of water in each beaker.

(c) Calculate the difference between the percentage decrease in temperature for the uncovered beaker and the percentage decrease in temperature for the covered beaker.

(d) The student concludes that it is better for farmers to keep their animals indoors. Discuss this conclusion.

(e) Describe how the student could modify his investigation to find out if an animal’s body size affects heat loss.

Most-appropriate topic codes (Edexcel IGCSE Biology):

2(j): Co-ordination and response – homeostasis and temperature regulation — parts (d), (e)
2(h): Transport – surface area to volume ratio and need for transport systems — part (e)
5(a): Use of biological resources – food production and animal husbandry — parts (a), (b), (d)
Assessment Information: Experimental skills – planning investigations, variables, data analysis — parts (a), (b), (c), (e)
▶️ Answer/Explanation
Solution

(a) Temperature

Explanation: The dependent variable is what is being measured in the experiment. In this case, the student is measuring how the temperature changes over time in both the covered and uncovered beakers. The temperature depends on whether the beaker is covered or not, making it the dependent variable in this investigation.

(b) To ensure a fair test by keeping the surface area and volume constant.

Explanation: Using the same volume of water (200 cm³) in each beaker ensures that the surface area to volume ratio remains constant. This is important because heat loss is affected by both volume and surface area. If different volumes were used, it wouldn’t be a fair comparison as the rate of heat loss would be influenced by the different volumes rather than just the covering condition.

(c) 5%

Explanation: Let’s calculate this step by step:

For the uncovered beaker:

Temperature decrease = 80°C – 40°C = 40°C

Percentage decrease = (40 ÷ 80) × 100 = 50%

For the covered beaker:

Temperature decrease = 80°C – 44°C = 36°C

Percentage decrease = (36 ÷ 80) × 100 = 45%

Difference = 50% – 45% = 5%

The uncovered beaker lost 5% more heat than the covered beaker over the 30-minute period.

(d) The conclusion has both supporting and limiting factors.

Explanation: While the results show that covering reduces heat loss (with the covered beaker maintaining 4°C higher temperature), suggesting animals might lose less heat indoors, there are several limitations to this conclusion:

  • The difference is relatively small (only 5%), which may not be significant in real farming conditions
  • This model oversimplifies real animals – animals can move, generate their own heat, and have fur or feathers for insulation
  • Climate factors matter – in hot countries, keeping animals outdoors might actually be better to prevent overheating
  • Other factors like disease spread (higher indoors), animal welfare, and feeding costs aren’t considered
  • Animals kept indoors may use less energy maintaining body temperature, potentially allowing more energy for growth and production

(e) Use beakers/containers of different sizes with the same covering conditions.

Explanation: To investigate how body size affects heat loss, the student could modify the experiment by using beakers of different volumes (representing different body sizes) while keeping all other variables constant. For example, they could use 100 cm³, 200 cm³, and 300 cm³ beakers, all either covered or all uncovered, and measure the temperature decrease over the same time period. This would show how the surface area to volume ratio (which changes with size) affects the rate of heat loss, with smaller volumes (representing smaller animals) having a larger surface area to volume ratio and thus losing heat faster.

Question

Neurones transmit nerve impulses.
The diagram shows a neurone.

(a) This neurone is involved in the withdrawal reflex of a finger from a hot object.
The neurone transmits electrical impulses from the central nervous system to the effector.

(i) Draw a circle on the diagram to show the part of the neurone where the impulses are transferred to the effector.

(ii) What is the name of this type of neurone?

  • A. association
  • B. motor
  • C. relay
  • D. sensory

(iii) Explain the advantage of a withdrawal reflex when a finger touches a hot object.

(b) Some neurones do not have a myelin sheath.
A scientist investigates the impulse speed for neurones with different diameters.
She does this for neurones with a myelin sheath and neurones without a myelin sheath.
The graph shows the scientist’s results.

(i) Which conclusion is supported by the graph?

  • A. impulses are always faster in neurones with a myelin sheath
  • B. neurones always have a myelin sheath
  • C. wider neurones have a myelin sheath
  • D. wider neurones have faster impulses

(ii) A neurone with a myelin sheath has a diameter of 1.0 µm.
Use the graph to determine the speed of an impulse in this neurone.

(iii) The neurone is 90 cm long.
Calculate the time taken for an impulse to travel along this neurone.
Give your answer in standard form.

Most-appropriate topic codes (Edexcel IGCSE Biology):

2(j): Co-ordination and response — parts (a)(i), (a)(ii), (a)(iii)
2(b): Cell structure — part (b)(i), (b)(ii), (b)(iii)
Appendix 4: Mathematical skills — part (b)(iii)
▶️ Answer/Explanation
Solution

(a)(i) A circle should be drawn around the axon terminals.

Explanation: The axon terminals are the end parts of the neurone that form synapses with the effector (e.g., a muscle). When the nerve impulse reaches the axon terminals, it triggers the release of neurotransmitters, which cross the synapse and stimulate the effector to respond.

(a)(ii) B motor

Explanation: A motor neurone carries electrical impulses from the central nervous system (CNS) to an effector, such as a muscle or gland. In this case, the neurone is part of a reflex arc that causes the finger to withdraw from the hot object, so it must be a motor neurone. Association neurones are found within the CNS, relay neurones connect sensory and motor neurones (also within the CNS), and sensory neurones carry impulses from receptors to the CNS.

(a)(iii) An explanation that makes reference to two of the following points:

  • fast (1)
  • no brain involvement / no thought / automatic / involuntary / eq (1)
  • less damage / harm / eq (1)

Detailed Explanation: The withdrawal reflex is an involuntary and rapid response. Its main advantage is speed. Because the reflex arc bypasses the brain and involves only the spinal cord, the response happens much faster than if the brain had to process the information and decide on a response. This speed helps to minimize tissue damage by removing the finger from the harmful stimulus (the heat) as quickly as possible.

(b)(i) D wider neurones have faster impulses

Explanation: The graph shows that for both myelinated and unmyelinated neurones, as the diameter increases, the impulse speed also increases. Therefore, the conclusion that wider neurones have faster impulses is directly supported by the data. Option A is incorrect because the graph does not show that myelinated neurones are always faster (at very small diameters, the lines are close). Options B and C are not supported by the information in the graph.

(b)(ii) 4.4 m per s

Explanation: To find the speed, you locate 1.0 µm on the x-axis (Diameter of neurone). Then, you follow a vertical line up until it meets the line for “myelin sheath.” From that point, you move horizontally to the left to read the speed value on the y-axis (Impulse speed in m per s), which is 4.4 m/s.

(b)(iii) \(2.0 \times 10^{-1}\) s

Explanation: To calculate the time taken, we use the formula: \(\text{time} = \frac{\text{distance}}{\text{speed}}\).

First, ensure all units are consistent. The length is 90 cm, which is \(0.9\) m. The speed from part (b)(ii) is \(4.4\) m/s.

So, \(\text{time} = \frac{0.9}{4.4}\).

Performing the division: \(0.9 \div 4.4 = 0.204545…\) seconds.

Rounding to two significant figures (as appropriate for the data given) gives \(0.20\) seconds.

Finally, expressing \(0.20\) seconds in standard form is \(2.0 \times 10^{-1}\) s.

Question

Scientists investigate the effect of salt (NaCl) concentration on the germination of maize seeds.

Batches of seeds are watered with different concentrations of salt solution.

They count the number of seeds that germinate in each batch 12 days after sowing. They then determined the percentage germination.

(a) (i) Describe how the scientists could tell if a maize seed had germinated.

(ii) The scientists measured the germination after 12 days.

State two other abiotic variables that the scientists need to control in their investigation.

(b) The table shows the scientists’ results.

(i) Plot a line graph to show the effect of concentration of salt solution on germination.

Use a ruler to join your points with straight lines.

(ii) Explain the effect of increasing the concentration of salt solution on germination.

(c) As plants grow, they produce roots and stems.

(i) Compare the responses of roots and stems to gravity.

(ii) Compare the responses of roots and stems to light.

Most-appropriate topic codes (Edexcel IGCSE Biology):

3(a): Reproduction — part (a)(i), (b)(ii)
2(d): Movement of substances into and out of cells — part (b)(ii)
2(j): Co-ordination and response — part (c)(i), (c)(ii)
Appendix 4: Mathematical skills (Handling data & Graphs) — part (b)(i)
Appendix 5: Command word taxonomy — throughout
▶️ Answer/Explanation
Solution

(a)(i) The scientists could tell a maize seed had germinated by observing the emergence and growth of the radicle (the first root) or the plumule (the first shoot) from the seed. The seed coat may also split open as these structures begin to grow outwards.

(a)(ii) Two other abiotic variables that need to be controlled are:

  1. Temperature: Germination rates are highly dependent on temperature. All batches of seeds should be kept at the same, constant temperature throughout the investigation to ensure that any differences in germination are due to the salt concentration and not temperature fluctuations.
  2. Light: While some seeds require light to germinate and others don’t, consistency is key. The light intensity, duration (photoperiod), and quality should be the same for all seed batches to prevent light from being a confounding variable.

Other acceptable answers include: volume of solution, humidity, oxygen concentration, pH, or carbon dioxide levels.

(b)(i) To plot the line graph:

  • The independent variable (salt concentration in mmol) should be plotted on the x-axis.
  • The dependent variable (percentage germination) should be plotted on the y-axis.
  • The scale should be linear and appropriate for the data range (0-320 mmol for x-axis, 0-100% for y-axis).
  • Axes must be clearly labelled with the variable and its units (“Concentration of salt solution (mmol)” and “Percentage germination (%)”).
  • Each data point from the table should be accurately plotted.
  • Straight lines should be drawn with a ruler to connect each point sequentially.

(b)(ii) Increasing the concentration of the salt solution decreases the percentage germination of maize seeds. This happens because the salt solution creates a lower (more negative) water potential in the soil surrounding the seed compared to the water potential inside the seed. Water moves by osmosis from an area of higher water potential (inside the seed) to an area of lower water potential (the salty soil). As the salt concentration increases, this outward osmotic gradient strengthens, or the inward gradient weakens, resulting in less water being absorbed by the seed. Water is essential for germination as it rehydrates the seed tissues, activates enzymes that break down stored food reserves (like starch), and allows for the growth of the embryo. With insufficient water uptake, these metabolic processes are inhibited, leading to reduced or prevented germination.

(c)(i) Roots and stems respond differently to gravity. Roots are positively gravitropic (or positively geotropic), meaning they grow in the direction of the gravitational pull, downwards into the soil. Stems, on the other hand, are negatively gravitropic, meaning they grow against the direction of gravity, upwards away from the soil.

(c)(ii) Roots and stems also respond differently to light. Roots are negatively phototropic, meaning they grow away from a light source, which helps them remain buried in the soil. Stems are positively phototropic, meaning they grow towards a light source, which enables the leaves to maximize their exposure to sunlight for photosynthesis.

Question

Bacteria are genetically modified to make human proteins.

(a) Which part of a bacterium is used in genetic modification?

  • A) cell wall
  • B) nucleoid
  • C) plasmid
  • D) RNA

(b) The passage describes the use of an industrial fermenter to grow genetically modified bacteria.

Complete the passage by writing a suitable word in each blank space.

The hormone called ______ is needed to control the blood glucose levels in humans.

Bacteria have been genetically modified to produce this hormone.

The fermenter is cleaned with ______ before adding a nutrient broth that contains the genetically modified bacteria.

This cleaning provides aseptic conditions that prevent ______ from other bacteria.

Paddles are used to ______ the contents.

A gas called ______ is bubbled into the fermenter.

The ______ is controlled by using a cooling jacket.

Most-appropriate topic codes (Edexcel IGCSE Biology):

6(c): Genetic modification (genetic engineering) — part (a)
6(a): Use of an industrial fermenter — part (b)
3(j): Co-ordination and response (hormones) — part (b), first blank
▶️ Answer/Explanation
Solution

(a) C (plasmid)
A is incorrect because it is not the cell wall
B is incorrect because it is not the nucleoid
D is incorrect because it is not RNA

(b)
The hormone called insulin is needed to control the blood glucose levels in humans.
Bacteria have been genetically modified to produce this hormone.
The fermenter is cleaned with steam / hot water before adding a nutrient broth that contains the genetically modified bacteria.
This cleaning provides aseptic conditions that prevent contamination / competition from other bacteria.
Paddles are used to mix / stir / agitate the contents.
A gas called oxygen / \(O_2\) is bubbled into the fermenter.
The temperature is controlled by using a cooling jacket.

Acceptable alternatives for blank 2: disinfectant, bleach, sterilising fluid, alcohol, ethanol.
Acceptable alternatives for blank 3: infection.

Question

The passage describes reproduction in flowering plants.

Complete the passage by writing a suitable word in each blank space.

Flowers are organs that allow plants to carry out ______ reproduction.

The male gamete is contained within the ______ grains. These grains are released from the ______, which is found on top of the filament.

These grains need to land on the ______, the female part of the flower.

Grains can be transferred by wind or by animals. These animals are often insects such as _____ or butterflies.

The petals of insect-pollinated plants are often ______ and brightly coloured.

After pollination, the grains germinate and a tube grows down the ______ to reach the ovary.

In the ovary, the gametes fuse. This process is known as ______ .

Most-appropriate topic codes (Edexcel IGCSE Biology):

4(a): Reproduction in flowering plants — entire question
3(j): Co-ordination and response in flowering plants — reference to pollen tube growth (tropism)
▶️ Answer/Explanation
Solution

Completed Passage:

Flowers are organs that allow plants to carry out sexual reproduction.

The male gamete is contained within the pollen grains. These grains

are released from the anther, which is found on top of the filament.

These grains need to land on the stigma, the female part of the flower.

Grains can be transferred by wind or by animals. These animals are often insects such as

bees or butterflies. The petals of insect-pollinated plants are often

large and brightly coloured. After pollination, the grains

germinate and a tube grows down the style to reach the ovary.

In the ovary, the gametes fuse. This process is known as fertilisation.

Detailed Explanation:

1. sexual – Flowers are the reproductive structures of flowering plants (angiosperms) that enable sexual reproduction. This process involves the fusion of male and female gametes to produce seeds.

2. pollen – Pollen grains are the structures that contain the male gametes (sperm cells). They are produced in the anthers and are essential for fertilizing the female ovule.

3. anther – The anther is the part of the stamen (male reproductive organ) where pollen grains are produced and stored. It sits atop the filament and releases pollen during pollination.

4. stigma – The stigma is the receptive tip of the carpel (female reproductive organ). It’s sticky or feathery to capture pollen grains during pollination.

5. bees – Bees are common insect pollinators that transfer pollen from one flower to another as they collect nectar. Other insects like moths and flies can also serve as pollinators.

6. large – Insect-pollinated flowers typically have large, brightly colored petals and often produce scent or nectar to attract pollinators. This contrasts with wind-pollinated flowers which are usually small and inconspicuous.

7. style – After pollen lands on the stigma, it germinates and grows a pollen tube down through the style. The style is the slender stalk connecting the stigma to the ovary.

8. fertilisation – Fertilisation occurs when the male gamete from the pollen tube fuses with the female gamete (egg cell) in the ovule within the ovary. This fusion creates a zygote that develops into an embryo within the seed.

This complete process from pollination to fertilisation ensures genetic diversity in flowering plants through the combination of genetic material from two different parents.

Question

(a) The diagram shows three types of neurone.

(i) Give the names of these neurones.

(ii) Explain the role of these neurones in the withdrawal reflex of a finger from a hot object.

(b) A teacher uses this method to estimate the speed of a nerve impulse.

  • students stand in a circle and hold hands
  • student A in the circle starts a timer and at the same time squeezes the hand of student B on his left
  • when student B feels his hand being squeezed, he immediately squeezes the hand of the student on his left
  • this process continues around the circle of students until student A feels his hand being squeezed and he stops the timer

(i) Explain what other measurements the teacher would need to make in order to calculate the speed of a nerve impulse.

(ii) Describe whether the teacher’s method is likely to give an accurate estimate of the speed of a nerve impulse.

Most-appropriate topic codes (Edexcel IGCSE Biology):

3(j): Co-ordination and response in humans — parts (a)(i), (a)(ii), (b)(i), (b)(ii)
Practical Investigation: Investigate reaction times — part (b)
▶️ Answer/Explanation
Solution

(a)(i)

A: Sensory/afferent neurone

B: Relay/association neurone

C: Motor/efferent neurone

Explanation: Neurones are specialized nerve cells that transmit electrical impulses. Sensory neurones (A) carry impulses from receptors to the central nervous system. Relay neurones (B) are found within the central nervous system and connect sensory and motor neurones. Motor neurones (C) carry impulses from the central nervous system to effectors like muscles.

(a)(ii)

When a finger touches a hot object, sensory receptors in the skin detect the heat and generate a nerve impulse. The sensory neurone (A) carries this impulse from the finger to the spinal cord. In the spinal cord, the sensory neurone connects with a relay neurone (B), which then passes the impulse to a motor neurone (C). The motor neurone carries the impulse back to the muscles in the arm and hand, causing them to contract and pull the finger away from the hot object.

Explanation: This is an example of a reflex arc, which is an automatic, rapid response that doesn’t involve conscious thought. The pathway goes: receptor → sensory neurone → relay neurone (in spinal cord) → motor neurone → effector (muscle). This three-neurone system allows for quick protective responses to potentially harmful stimuli.

(b)(i)

The teacher would need to measure the total distance around the circle and the total time taken for the impulse to travel around the circle. To calculate speed using the formula speed = distance/time, both distance and time measurements are essential.

Explanation: Specifically, the teacher should measure the circumference of the circle (either by measuring the diameter and calculating, or by measuring the distance between students and multiplying by the number of students). The time is already measured from when student A starts the timer to when they stop it after feeling the squeeze return to them.

(b)(ii)

The teacher’s method is unlikely to give an accurate estimate of the speed of a nerve impulse. The main issue is that it measures reaction time rather than just nerve impulse speed, as there are delays at each synapse and in cognitive processing.

Explanation: There are several reasons why this method is inaccurate. First, there’s synaptic delay at each connection between students (approximately 0.5ms per synapse). Second, each student has processing time to recognize the squeeze and decide to squeeze the next hand. Third, the method involves conscious reaction rather than a true reflex. In a real nerve pathway, impulses travel much faster (up to 120 m/s in myelinated neurones) without these additional delays. A more accurate method would involve measuring reflex times in a controlled laboratory setting using specialized equipment.

Question

The diagram shows a section through the human eye.

(a) (i) Describe the role of structures A, C and D in focusing light from a near object onto the retina.

(a) (ii) Bright light can damage the retina.
Explain how a reflex involving structure B protects the retina from this damage.

(b) (i) There are tear glands that release liquid into the eye.
This liquid contains an enzyme called lysozyme that breaks down cell walls.
Suggest why the liquid contains lysozyme.

(b) (ii) Suggest why the liquid is maintained at pH 7.

Most-appropriate topic codes (Edexcel IGCSE Biology):

2(j): Co-ordination and response – the eye, accommodation and pupil reflex — parts (a)(i), (a)(ii)
5(c): Enzymes and their role as biological catalysts — part (b)(ii)
5(f): Defence against disease – the non-specific immune system — part (b)(i)
▶️ Answer/Explanation
Solution

(a)(i) The cornea (C) refracts/bends light rays. The ciliary muscles (D) contract. This causes the suspensory ligaments to slacken. The lens (A) becomes more rounded/thicker, increasing its refractive power to focus light from near objects onto the retina.

Explanation: Focusing on a near object requires the eye to increase its focusing power. This process is called accommodation. First, the cornea (structure C) performs the initial bending (refraction) of light rays entering the eye. To fine-tune the focus for a nearby object, the ciliary muscles (structure D) contract. This contraction reduces the tension on the suspensory ligaments, which are like tiny strings holding the lens. When these ligaments slacken, the natural elasticity of the lens (structure A) causes it to become more convex, meaning it becomes thicker and more rounded. This change in shape increases the lens’s refractive power, bending the light rays more sharply so that they converge precisely onto the retina, forming a clear image.

(a)(ii) In bright light, the circular muscles in the iris (B) contract and the radial muscles relax. This causes the pupil to constrict (become smaller), reducing the amount of light entering the eye and protecting the retina from damage.

Explanation: This is a protective reflex arc. When very bright light is detected by the retina, a nerve impulse is sent to the brain. The brain then sends an impulse to the iris (structure B). The iris contains two sets of muscles: circular muscles and radial muscles. In response to bright light, the circular muscles contract (like tightening a drawstring) while the radial muscles simultaneously relax. This combined action makes the pupil—the central hole in the iris—smaller (a process called constriction). By making the pupil smaller, the eye limits the quantity of intense light that can pass through to the sensitive retina, thereby preventing overstimulation and potential damage to the light-sensitive cells.

(b)(i) Lysozyme kills bacteria / microorganisms. This helps to prevent eye infections.

Explanation: The tear fluid is not just for lubrication; it’s also a part of the body’s first line of defense. Lysozyme is an enzyme that specifically targets and breaks down the cell walls of many types of bacteria. Bacterial cell walls are essential for their structure and survival. By breaking these walls down, lysozyme effectively kills the bacteria or prevents them from growing and multiplying. This antiseptic action helps to protect the surface of the eye from potential infections that could be caused by dust, debris, or other airborne pathogens.

(b)(ii) pH 7 is the optimum pH for lysozyme activity. It also prevents irritation or damage to the sensitive eye tissues.

Explanation: Enzymes are very sensitive to pH, and lysozyme is no exception. Its three-dimensional shape, particularly the shape of its active site, is ideal for breaking down bacterial cell walls at a neutral pH of 7. If the pH were too acidic or too alkaline, the enzyme could become denatured—its shape would change, and it would no longer function effectively, leaving the eye more vulnerable to infection. Furthermore, a neutral pH is gentle and non-irritating to the delicate cells on the surface of the eye (the cornea and conjunctiva). An acidic or alkaline fluid would cause significant discomfort, stinging, and could potentially damage these sensitive tissues.

Question

Inherited conditions may be caused by a dominant allele (D) or by a recessive allele (d).

The diagram shows a family pedigree for an inherited condition.

The shaded circle shows a female with the condition.

(a) Complete the table by giving the genotype of each individual.

One has been done for you.

(b) The parents have a third child.

Calculate the probability that this third child is female and has the condition.

(c) The graphs show changes in the concentrations of the hormones testosterone and oestrogen in child R and in child S between the ages of 11 to 15.

Explain how the changes in hormone concentrations affect the development of each child.

Most-appropriate topic codes (Edexcel IGCSE Biology):

3(b): Inheritance — parts (a), (b)
3.20: Genetic terms (dominant, recessive, genotype) — part (a)
3.24: Interpreting family pedigrees — part (a)
3.23/3.25: Monohybrid inheritance and probability calculations — part (b)
2(j): Co-ordination and response — part (c)
2.94/2.95B: Hormones (oestrogen, testosterone) and their roles — part (c)
3.13: Role of oestrogen and testosterone in secondary sexual characteristics — part (c)
▶️ Answer/Explanation
Solution

(a)

IndividualGenotype
PDd
QDd
RDd
Sdd

Explanation: The condition is caused by a recessive allele (d) because both parents (P and Q) are unaffected but have an affected child (S, genotype dd). This means both parents must be heterozygous carriers (Dd). Child R is unaffected, so their genotype must be Dd (as given). Child S is affected, so their genotype is dd.

(b) \( \frac{1}{8} \) / 0.125 / 12.5%

Explanation: To find the probability that the third child is female AND has the condition, we consider two independent events:

1. Probability of being female: This is \( \frac{1}{2} \) (assuming equal chance of male or female).

2. Probability of having the condition (genotype dd): Both parents are Dd. A Punnett square for Dd × Dd gives: 25% DD (unaffected), 50% Dd (unaffected carriers), 25% dd (affected). So, the probability of being affected is \( \frac{1}{4} \).

To find the combined probability, we multiply the probabilities of the two independent events: \( \frac{1}{2} \times \frac{1}{4} = \frac{1}{8} \).

So, the probability is \( \frac{1}{8} \), 0.125, or 12.5%.

(c)

Explanation: The graphs show distinct hormonal profiles for Child R and Child S, indicating their different sexes and the onset of puberty.

Child R: Shows a significant increase in testosterone concentration with age, while oestrogen levels remain low. This pattern is characteristic of male puberty. Testosterone promotes the development of male secondary sexual characteristics, such as increased muscle mass, deepening of the voice, growth of facial and body hair, and the production of sperm.

Child S: Shows a significant increase in oestrogen concentration with age, while testosterone levels remain low. This pattern is characteristic of female puberty. Oestrogen promotes the development of female secondary sexual characteristics, such as the development of breasts, widening of the hips, the onset of menstruation, and the distribution of body fat.

In summary, the differing hormone profiles drive the development of sex-specific physical characteristics during adolescence for each child.

Question

Humans can control their internal environment.

(a) State the term used to describe the control of an organism’s internal environment.

(b) Coordination uses hormones and nerves.

Some responses are simple reflex arcs.

Describe the structure and functioning of the withdrawal reflex of a finger from a hot object.

(c) Humans use their skin to regulate their body temperature. The diagram shows a section through the skin with two structures labelled A and B.

Changes take place in the skin when a person moves from a warm environment to a cold environment.

(i) Explain the changes that take place in structure A as a person enters a cold environment.

(ii) Explain the changes that take place in structure B as the person enters a cold environment.

(d) Modern humans have less body hair than their ancestors.

Explain how this evolutionary change was brought about by natural selection.

Most-appropriate topic codes (Edexcel IGCSE Biology):

2(j): Co-ordination and response — parts (a), (b), (c)(i), (c)(ii)
3(b): Inheritance — part (d)
▶️ Answer/Explanation
Solution

(a) Homeostasis

Explanation: Homeostasis is the biological term that describes the process by which organisms maintain a relatively stable internal environment despite changes in external conditions. This involves regulating various factors such as body temperature, blood glucose levels, and water balance through feedback mechanisms.

(b)

Explanation: The withdrawal reflex is an automatic, rapid response that protects the body from harm. When a finger touches a hot object, here’s how the reflex arc functions:

First, specialized heat receptors in the skin of the finger detect the damaging stimulus (the heat). These receptors generate an electrical impulse.

This impulse travels along a sensory neurone towards the spinal cord (part of the central nervous system). At the junction (synapse) between the sensory neurone and a relay neurone in the spinal cord, the impulse triggers the release of chemical messengers called neurotransmitters, which diffuse across the gap.

The relay neurone then passes the signal to a motor neurone. The impulse travels along the motor neurone to the effector, which in this case is the muscles in the arm and hand.

Finally, the muscles contract, pulling the hand away from the hot object. This entire process happens incredibly quickly, without involving the brain for conscious thought, which helps to minimize tissue damage.

(c)(i)

Explanation: Structure A represents a sweat gland. As a person enters a cold environment, the activity of the sweat glands is significantly reduced. This means less sweat is produced and released onto the skin’s surface. Since the evaporation of sweat is a primary method the body uses to lose heat, reducing sweat production helps to conserve body heat and prevent the body from cooling down too much in the cold.

(c)(ii)

Explanation: Structure B represents a blood vessel near the skin’s surface. In response to cold, these blood vessels undergo vasoconstriction. This means the smooth muscles in the vessel walls contract, causing the diameter of the blood vessels to narrow. As a result, less warm blood flows from the body’s core to the skin’s surface. This reduces the amount of heat lost from the blood to the external environment through radiation and convection. By diverting blood away from the skin, more heat is retained in the core of the body, which is essential for maintaining the temperature of vital organs.

(d)

Explanation: The reduction in body hair in modern humans compared to their ancestors is a classic example of evolution by natural selection. Initially, there would have been genetic variation within the ancestral population regarding the amount of body hair, caused by random mutations.

In the warm climates where early humans likely lived and became more active (e.g., through persistence hunting), having less body hair would have been advantageous. It would have allowed for more efficient cooling through sweating, preventing overheating during physical exertion. Individuals with less body hair were therefore more likely to survive the heat and have the energy to hunt and gather food.

These individuals, being better adapted to their environment, had a higher chance of surviving to reproductive age and producing offspring. They passed on the alleles (genes) for reduced body hair to their children.

Over many generations, the allele for less body hair became more common in the population, while the allele for thick body hair became less common, as those individuals were at a selective disadvantage. This gradual change in the inherited characteristics of the population is evolution by natural selection.

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