(a) The diagram shows apparatus a student uses to compare inhaled and exhaled air.
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The student breathes into and out of the mouthpiece for one minute.
(i) Explain which flask exhaled air passes through.
(ii) Explain the changes that will happen in the limewater in flask A and in flask B.
(iii) The student uses limewater to compare the composition of exhaled and inhaled air. Suggest an alternative substance that they could use.
(b) Describe the role of the diaphragm and the intercostal muscles in inhalation.
Most-appropriate topic codes (Edexcel IGCSE Biology):
• 2(f): Respiration — part (a)(ii)
• Practical Investigation: Investigate breathing in humans — part (a)(iii)
▶️ Answer/Explanation
(a)(i) An explanation that makes reference to two of the following:
- (Flask B) as (mouthpiece) connected to long tube in flask B / (mouthpiece) connected to flask A is shorter (1)
- (Flask B) tube (from mouthpiece) in limewater / tube in liquid in flask B / causes bubbles (in limewater) in B / not in liquid / draws air in flask A (1)
- cannot inhale in flask B as limewater would be sucked in / would get mouthful of liquid / Flask A for inhalation as no liquid drawn up (1)
(2 marks)
(a)(ii) An explanation that makes reference to two of the following:
- (Limewater) in flask A stays clear / no change / less cloudy / goes cloudy slowly (1)
- (Limewater in) flask B goes (more) cloudy / milky / cloudy quicker (1)
- as less carbon dioxide \( \text{(CO}_2\text{)} \) in inhaled / atmospheric air / (more) carbon dioxide in exhaled (1)
- due to respiration (1)
(2 marks)
(a)(iii) • Sodium hydrogencarbonate / sodium bicarbonate (solution) / hydrogencarbonate / bicarbonate indicator (1)
(1 mark)
(b) An answer that makes reference to four of the following:
- diaphragm contracts (1)
- diaphragm flattens / moves down (1)
- (external) intercostal muscles contract (1)
- rib cage raised / moves out (1)
- volume (of chest cavity / thorax) increases (1)
- pressure in (chest cavity / thorax) decreases / reduces (1)
- air drawn into lungs / lungs inflate (1)
(4 marks)
The diagram shows an insect-pollinated flower called a lily.
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(a) (i) Which of these is the name of structure P?
A) anther
B) filament
C) stigma
D) style
(ii) Which of these is the name of structure Q?
A) leaf
B) petal
C) stem
D) style
(iii) On which structure do pollen grains germinate?
A) P
B) Q
C) R
D) S
(b) Describe how the structures of P, Q and S would differ in a wind-pollinated flower.
(c) Plants can also reproduce by asexual reproduction.
(i) Give one natural method that plants use to reproduce asexually.
(ii) Give one artificial method that a plant grower may use to reproduce a plant asexually.
(iii) Suggest why a plant grower may choose to reproduce a plant asexually rather than allowing the plant to reproduce sexually.
Most-appropriate topic codes (Edexcel IGCSE Biology):
• 4.3: Structures of insect-pollinated and wind-pollinated flowers — parts (a)(i), (a)(ii), (a)(iii), (b)
• 4.7: Asexual reproduction in plants (natural and artificial) — parts (c)(i), (c)(ii), (c)(iii)
▶️ Answer/Explanation
(a)(i) A (anther)
B is not the answer as it is the filament
C is not the answer as it is the stigma
D is not the answer as it is the style
(a)(ii) B (petal)
A is not the answer as it is the leaf
C is not the answer as it is the stem
D is not the answer as it is the style
(a)(iii) D (S, stigma)
A is not the answer as they do not germinate on P (anther)
B is not the answer as they do not germinate on Q (petal)
C is not the answer as they do not germinate on R (stem)
(b) An answer that makes reference to the following:
• P / anthers exposed / hanging out / outside / eq (1)
• S / stigma feathery / exposed / hanging out / longer / outside / eq (1)
• Q / petals smaller / absent / green / not coloured / not scented / eq (1)
(c)(i) One natural method, e.g.:
• runners / eq
allow corms / bulbs / rhizomes / tubers
(c)(ii) One artificial method, e.g.:
• cuttings / eq
allow grafting / layering / micropropagation / tissue culture
(c)(iii) An answer that makes reference to two of the following:
• no / (less) genetic variation / a clone / eq (1)
• maintain phenotype / colour / flavour / desired characteristic / same characteristics / eq (1)
• faster / eq (1)
• seeds not viable / produce rare plants / eq (1)
The diagram shows a cycle found in ecosystems.
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(a) Give the name of the cycle.
(b) (i) Which process is represented by the letter W?
A) combustion
B) decomposition
C) feeding
D) respiration
(ii) Which process is represented by the letter X?
A) combustion
B) decomposition
C) feeding
D) respiration
(c) Name a group of organisms that is responsible for decomposition.
(d) Carbon dioxide is released into the atmosphere by the decomposition of organic material. The rate of this decomposition depends on a number of factors.
Design an investigation to find out if changing the pH of organic material affects the rate of decomposition.
Include experimental details in your answer and write in full sentences.
Most-appropriate topic codes (Edexcel IGCSE Biology):
• 5(c): Describe the stages in the carbon cycle (specification point 5.10) — parts (a), (b)(i), (b)(ii)
• 5(b): Feeding relationships / decomposers — part (c)
• 5(c): Decomposition as part of the carbon cycle — part (d)
• Assessment Objective A03: Experimental skills, analysis and evaluation — part (d)
▶️ Answer/Explanation
(a) Carbon (cycle) (1)
(b)(i) The only correct answer is A (combustion)
B is not the answer as W is not decomposition
C is not the answer as W is not feeding
D is not the answer as W is not respiration
(b)(ii) The only correct answer is D (respiration)
A is not the answer as X is not combustion
B is not the answer as X is not decomposition
C is not the answer as X is not feeding
(c) bacteria / fungi / allow correct genus and species names (1)
(d) A description that makes reference to six of the following points:
• C Use different pH of organic material / add acid / alkali / add buffers / etc. (1)
• O Use the same organic material / plant / age / species / type / mass / volume of organic material (same state of decay) (1)
• R Repeat (at each different pH) (1)
• M1 Measure change / loss in mass of organic material (initial mass – final mass) / volume of carbon dioxide released / change in hydrogen-carbonate indicator / change in limewater / etc. (1)
• M2 After a stated time period (e.g., 1 hour+) / measure time taken for same mass to decompose (1)
• S1 Keep the same temperature / oxygen levels / etc. (1)
• S2 Keep the same water / mineral ions / humidity / volume of each acid/alkali/buffer / same bacteria/fungi/decomposer added / etc. (allow same number of bacteria/decomposers) (1)
Example full-sentence answer structure:
I would set up several containers with the same type, mass, and state of decay of organic material. I would adjust the pH in each container to different values using buffers or acids/alkalis, while keeping other factors like temperature, oxygen availability, and humidity the same. I would then measure the loss in mass of the organic material after a fixed period, such as one week, to determine the rate of decomposition. This experiment would be repeated at each pH level to ensure reliability.
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.
| Age of mother in years | Data for mothers who did not smoke during pregnancy | Data for mothers who did smoke during pregnancy | ||
|---|---|---|---|---|
| number of mothers | percentage of babies with low birth mass | number of mothers | percentage of babies with low birth mass | |
| 19 and under | 1331 | 11.5 | 356 | 16.0 |
| 20–24 | 11243 | 9.8 | 1677 | 13.2 |
| 25–29 | 24099 | 9.0 | 2211 | 13.3 |
| 30–34 | 28695 | 9.2 | 1847 | 14.5 |
| 35–39 | 16537 | 10.5 | 934 | 21.1 |
| 40 and over | 3242 | 12.3 | 181 | 22.1 |
(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):
• 4(a): Reproduction — spec. point 4.11: describe the role of the placenta in the nutrition of the developing embryo
• Mathematical Skills: Handling data (Appendix 3) — calculations, ratios, percentages, and interpreting data from tables
• AO2/AO3: Application, analysis, and evaluation of data and methods — interpreting study data, identifying correlation vs. causation, evaluating conclusions
▶️ Answer/Explanation
(a)(i)
Number of mothers aged ≤19 who smoked = 356
Total mothers aged ≤19 = \(1331 + 356 = 1687\)
Percentage = \( \frac{356}{1687} \times 100 \)
\( \approx 21.1\% \)
Answer: 21.1% (allow 21%)
(a)(ii)
Total non-smokers = \(1331 + 11243 + 24099 + 28695 + 16537 + 3242 = 85147\)
Total smokers = \(356 + 1677 + 2211 + 1847 + 934 + 181 = 7206\)
Ratio (non-smokers : smokers) = \( \frac{85147}{7206} : 1 \)
\( \approx 11.82 : 1 \)
Nearest whole number ratio = \(12 : 1\)
Answer: \(12 : 1\)
(b)
An answer that makes reference to points such as:
- Yes, smoking is important / correlated:
- At every age group, the percentage of low birth mass babies is higher for smokers than non-smokers.
- The difference is especially large in older age groups (e.g., 35–39: 10.5% vs 21.1%).
- The study involves large numbers, giving statistical weight.
- No, it is not the sole/main factor / correlation ≠ causation:
- There is also a clear age effect: for non-smokers, low birth mass % is higher in very young (≤19: 11.5%) and older mothers (≥40: 12.3%).
- The effect of smoking appears greater in older mothers, suggesting an interaction; older mothers may have smoked for longer.
- Other factors not recorded could be responsible (confounding variables): mother’s mass/weight, healthcare, diet, alcohol/drug use, prenatal care, genetics, premature birth.
- No information on frequency/duration of smoking.
- Biological knowledge: smoking reduces oxygen supply to fetus (via carbon monoxide binding haemoglobin, vasoconstriction), impairs placental function, but other factors also affect fetal growth.
- Conclusion: The data shows a strong association, but the student’s conclusion that smoking is the main cause is an oversimplification. Other factors, especially age and unmeasured variables, also play significant roles.
Scientists carry out an experiment to see if reducing the availability of oxygen affects the production of yoghurt.
They use increasing acidity as a measure of yoghurt production.
They record the acidity of two cultures, one with a reduced oxygen level and one with a normal oxygen level, over 210 minutes.
The table shows their results.
| Time in minutes | Acidity (%) reduced oxygen level | Acidity (%) normal oxygen level |
|---|---|---|
| 0 | 0.20 | 0.20 |
| 30 | 0.22 | 0.22 |
| 60 | 0.25 | 0.24 |
| 90 | 0.40 | 0.25 |
| 120 | 0.50 | 0.31 |
| 150 | 0.62 | 0.41 |
| 180 | 0.70 | 0.51 |
| 210 | 0.70 | 0.70 |
(a) Explain why increasing acidity can be used as a measure of yoghurt production.
(b) Give one abiotic variable that the scientists should control in their experiment.
(c)(i) Plot a line graph to show how the percentage acidity changes over the period of 210 minutes for the reduced oxygen level and for the normal oxygen level.
Use a ruler to join the points with straight lines.
(ii) Explain why the changes in percentage acidity are different in the reduced oxygen level and in the normal oxygen level cultures.
Most-appropriate topic codes (Edexcel IGCSE Biology):
• 2(f): Respiration — Aerobic vs. anaerobic respiration (2.36) — parts (a, c(ii))
• Appendix 3: Mathematical skills — Handling data, Graphs — part (c)(i)
• Experimental Skills: (Pages 30, 47) — Controlling variables, interpreting data — parts (b, c(ii))
▶️ Answer/Explanation
(a) An explanation that makes reference to two of the following:
- anaerobic respiration (1)
- (by) bacteria / named bacteria (e.g., Lactobacillus) (1)
- produces lactic acid / eq (1)
(b) • temperature / volume / mass of milk / lactose content of milk / type of milk / eq (1)
(c)(i) A graph that includes:
- scales linear and at least half page and axes the correct way around (time on \(x\)-axis) (1)
- (time in) minutes and acidity % labelled (1)
- straight lines joining points (1)
- points correctly plotted (1)
- key to identify normal and low oxygen (1)
(c)(ii) An explanation that makes reference to the following:
- reduced oxygen (more) anaerobic respiration (of milk) / less aerobic respiration / eq (1)
- acidity increases faster / sooner / more rapidly / eq (1)
Marking note for (c)(ii): Allow converse for more oxygen (i.e., normal oxygen level has more aerobic respiration, so acidity increases more slowly).
The diagram shows part of the gut of a rabbit. The rabbit is a primary consumer and eats mainly grass and other vegetable material.
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(a) Name the parts labelled A, B, C and D.
(b) The gut of a rabbit has a large caecum and appendix. These contain bacteria that are able to produce the enzyme cellulase. Explain how these bacteria help the rabbits with their diet of plant material.
(c) The human gut has a caecum and appendix but they are much smaller than those in the rabbit.
(i) Suggest why the human gut only has a small caecum and appendix.
(ii) In humans the appendix also acts as a store of useful bacteria. Scientists have discovered that patients who have had their appendix removed are more likely to develop infections of the colon.
Explain how having no appendix may increase the likelihood of bacterial infections of the colon.
Most-appropriate topic codes (Edexcel IGCSE Biology):
• 2(e): Nutrition — Humans (digestion and enzymes) — parts (b), (c)(i)
• 6(a): Use of biological resources — Food production (role of micro-organisms) — part (b)
• 3(i): Structure and functions in living organisms: Part 2 — Excretion — part (c)(ii) context
• 5(d): Ecology and the environment — Human influences on the environment — part (c)(ii) context
▶️ Answer/Explanation
(a)
A: oesophagus / gullet (1)
B: stomach (1)
C: small intestine / ileum / duodenum / jejunum (1)
D: large intestine / colon (1)
(b) An explanation that makes reference to three of the following:
• (plants contain) cellulose (1)
• (cellulose) digested by cellulase / enzyme (1)
• into glucose (1)
• energy released / respiration (1)
(c)(i)
• humans do not digest cellulose / do not eat only plant material / omnivores / eat fewer plants / vegetables (1)
(c)(ii) An answer that makes reference to two of the following:
• removes useful bacteria / fewer / no useful bacteria (1)
• reduces competition (1)
• pathogenic bacteria increase / survive / multiply / grow / more harmful bacteria (1)
(a) The diagram shows a food web from a woodland ecosystem.
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(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.
| Level | Energy (kJ m−2 year−1) |
|---|---|
| producers | \(8.7 \times 10^5\) |
| primary consumers | \(1.4 \times 10^4\) |
| secondary consumers | \(1.6 \times 10^3\) |
(i) The light energy reaching the producers is \(7.1 \times 10^6\) kJ per m2 per year.
Explain why the plants cannot absorb all of this energy.
(ii) 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 photograph shows how a woodlouse can curl up into a ball. This behaviour is an example of a reflex response.
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(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):
• 5(a): The organism in the environment — part (b)(i)
• 3(j): Co-ordination and response — parts (c)(i), (c)(ii)
• 4(b): Inheritance — part (c)(iii) (Natural selection)
▶️ Answer/Explanation
(a)(i) D (oak tree)
A is not correct as it is not the producer
B is not correct as it is not the producer
C is not correct as it is not the producer
(a)(ii) A (blackbird)
B is not correct as it is not a secondary and tertiary consumer
C is not correct as it is not a secondary and tertiary consumer
D is not correct as it is not a secondary and tertiary consumer
(b)(i) An explanation that makes reference to the following:
• some light / colours / wavelengths are reflected (1)
• light falls on flowers / not on leaves / some does not fall on chloroplasts / chlorophyll (1)
Ignore: shade. Allow: not enough chloroplasts.
(b)(ii)
• Producer to primary consumer: \( \frac{1.4 \times 10^4}{8.7 \times 10^5} \times 100 = 1.61\% \) (1) (allow 1.6%)
• Primary to secondary consumer: \( \frac{1.6 \times 10^3}{1.4 \times 10^4} \times 100 = 11.4\% \) (1) (allow 11%, 11.43%)
OR calculation of energy lost:
• Producer to primary: \(100 – 1.61 = 98.4\%\) lost (1)
• Primary to secondary: \(100 – 11.4 = 88.6\%\) lost (1)
• Conclusion: The student is incorrect; the transfer from primary to secondary consumer is more efficient (as 11.4% > 1.61%).
(c)(i) • involuntary / automatic / does not involve the brain / spontaneous / without thinking / unconscious (1)
(c)(ii) • protects from predators / only exposes hard shell (1)
Ignore: danger / harm / injury.
(c)(iii) An answer that makes reference to four of the following:
• (gene) mutation / mutated gene (1)
• variation (in rolling behaviour) (1)
• woodlice that roll up survive / are not eaten (1) (allow converse)
• (survivors) reproduce (1) (allow converse)
• pass on alleles / genes (for rolling behaviour) (1)
Do not accept: just “pass on mutation / behaviour”.
The rate of photosynthesis is affected by different factors. One factor is the concentration of carbon dioxide in the air.
(a) The percentage of oxygen in the air is 21%. This is equivalent to a concentration of 210000 parts per million.
The percentage of carbon dioxide in the air is 0.04%. Calculate this concentration in parts per million.
(b) The graph shows the effect of increasing the concentration of carbon dioxide in the air on the relative rate of photosynthesis at different temperatures.
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(i) Describe the effect of increasing the concentration of carbon dioxide on the relative rate of photosynthesis at 5°C.
(ii) Describe how the effect of increasing the concentration of carbon dioxide on the relative rate of photosynthesis changes when the temperature is increased.
(iii) Explain the effect of increasing the temperature from 5°C to 35°C on the relative rate of photosynthesis.
(c) The scientists who carried out this study concluded that the effect of increasing the concentration of carbon dioxide on the rate of growth of a plant is dependent on temperature and also on the minerals that the plants can absorb.
(i) Explain how lacking a named mineral might affect plant growth.
(ii) Explain how a named factor can affect the rate of photosynthesis, other than carbon dioxide concentration, temperature and minerals absorbed.
Most-appropriate topic codes (Edexcel IGCSE Biology):
• 2.20: Understand how varying carbon dioxide concentration, light intensity and temperature affect the rate of photosynthesis — parts (b), (c)(ii)
• 2.22: Understand that plants require mineral ions for growth, and that magnesium ions are needed for chlorophyll and nitrate ions are needed for amino acids — part (c)(i)
• Mathematical Skills (Appendix 3): Arithmetic and numerical computation — part (a)
▶️ Answer/Explanation
(a)
0.04% = 0.04 in 100
\(1\,000\,000 \div 100 = 10\,000\)
\(10\,000 \times 0.04 = 400\)
concentration = 400 ppm (2 marks)
Allow 1 mark for correct calculation of 10,000.
(b)(i)
An answer that makes reference to the following:
• (Increasing CO2) increases (relative) rate of photosynthesis (1)
• (Begins to) level off / reaches maximum (at 0.10% at approx. 1.12) / rate of increase steepest up to 0.06% / eq (1)
(b)(ii)
An answer that makes reference to two of the following:
• Steepest increase / most effect at 35°C / higher temperatures / eq (1)
• Increasing CO2 increases (relative) rate of photosynthesis at all temperatures (1)
• At CO2 of 0.03%, temperature is not the limiting factor / CO2 is limiting factor so increasing the temperature has little effect (1)
(b)(iii)
An explanation that makes reference to three of the following:
• Increasing temperature increases (relative) rate of photosynthesis (1)
• As more (kinetic) energy (supplied to molecules) / eq (1)
• More (frequent) collisions / faster collisions (between substrate and enzyme molecules) / more enzyme-substrate complexes formed / eq (1)
• Temperature becomes limiting factor at 5°C / increasing CO2 has less effect on rate / eq (1)
(c)(i)
An explanation that makes reference to one pair of the following:
• Nitrate (1) for / amino acids / proteins / stunted growth / eq (1)
OR
• Magnesium (1) for chlorophyll / yellow leaves / chlorosis / eq (1)
OR
• Other correct mineral (e.g., potassium, phosphate) (1) with correct effect (e.g., poor root growth, poor flowering) (1)
(c)(ii)
An explanation that makes reference to two of the following:
• (Increasing) light (intensity) (1) provides energy for / required for photosynthesis / absorbed / trapped / used by chloroplasts / chlorophyll / stomata open more / eq (1)
OR
• Changing wavelength / colour of light (1) some not absorbed / affect photosynthesis (1)
OR
• (Availability of) water (1) as reactant / substrate / used in photosynthesis (1)
Allow other valid factors such as pollutants or oxygen concentration.
Alkaptonuria is an inherited condition caused by the presence of recessive alleles.
(a) State what is meant by a recessive allele.
(b) Alkaptonuria is first diagnosed in children when it is noticed that they produce very dark urine that turns black when exposed to air.
A woman and a man do not have alkaptonuria. They have a child who has the condition.
The woman and the man are expecting a second child.
(i) Draw a genetic diagram to show the genotypes of the woman and the man, the gametes they produce and the possible phenotypes and genotypes of the second child.
(ii) Calculate the probability that the second child is male and does not have the condition.
(c) Alkaptonuria is caused by the body being unable to break down the amino acids tyrosine and phenylalanine. This leads to a build-up of a toxin that causes damage to joints and tendons and can also lead to heart valve damage in later life.
A new drug treatment is being tested that can slow the damage to the joints and tendons. Scientists selected 40 adults who all had alkaptonuria. They placed each patient into one of two groups. One group was given the drug treatment and the other group acted as a control. The scientists then compared the symptoms of the patients in each group after three years.
(i) Describe what is meant by the control group.
(ii) The table compares the control group with the drug treatment group. It shows the numbers starting and completing the trial and those showing harmful effects. It also compares improvements in two symptoms of alkaptonuria.
| Control group | Drug group | |
|---|---|---|
| number of patients starting trial | 20 | 20 |
| number of patients completing trial | 17 | 16 |
| number of patients showing adverse effects | 0 | 2 |
| number of patients that died | 0 | 1 |
| decrease in time taken to stand up and walk 3 m in seconds | 0.54 | 1.33 |
| increase in distance in metres walked in 6 minutes | 6.7 | 51.5 |
Evaluate whether the new drug should be recommended as an effective treatment for alkaptonuria.
(d) Other scientists have suggested that eating fewer proteins that contain tyrosine and phenylalanine would reduce the symptoms of alkaptonuria. Suggest why eating fewer of these proteins may be difficult.
Most-appropriate topic codes (Pearson Edexcel International GCSE in Biology (Modular)):
• 3: Biology content / Assessment Information (Experimental skills & Evaluation): — parts (c)(i), (c)(ii), (d)
• 2(c): Biological molecules (proteins) — part (d)
▶️ Answer/Explanation
(a) • only expressed / seen in phenotype of homozygous / requires two / both (copies) to be expressed in phenotype / only seen / expressed / shown if no dominant allele present / not expressed if dominant allele present / eq (1)
(b)(i) Correct genetic diagram:
- Parents: Aa and Aa (1)
- Gametes: A and a from each parent (1)
- Offspring genotypes: AA, Aa, Aa, aa (1)
- Offspring phenotypes: 3 no symptoms : 1 alkaptonuria (1)
(b)(ii) probability = \(0.375\) / \(37.5\%\) / \(3/8\) (1)
(c)(i) • did not receive the drug / no treatment / eq (1)
(c)(ii) An evaluation that makes reference to five of the following points: (5)
- Yes (the drug is effective) because:
- control group so valid study /eq (1)
- drug group showed improvement/reduced symptoms / eq (1)
- quicker to stand up (1.33 s vs 0.54 s decrease) / eq (1)
- improved distance walked (51.5 m vs 6.7 m increase) / eq (1)
- No (caution/reservations) because:
- small group size / should be tested on more people / repeat study / eq (1)
- not all finished study (16/20 vs 17/20) / eq (1)
- adverse / side effects (2 patients in drug group) / eq (1)
- one died in drug group /eq (1)
- no information on age / mass / sex / other health conditions / activity / eq (1)
(d) An answer that makes reference to two of the following points: (2)
- patients may not know what foods contain these (amino acids)/eq (1)
- (tyrosine and phenylalanine are) present in many / most foods / most proteins /eq (1)
- proteins (that contain them) are required for growth / repair / eq (1)
Farmers sometimes use biological control to reduce the damage to their crops caused by pests such as insects.
(a) Which of these is an advantage of using biological control over chemical control?
A) it lasts a short time
B) it leads to bioaccumulation
C) it is specific
D) it is quicker
(b) Aphids are tiny insects that have very sharp mouthparts. They push these mouthparts into the phloem found in stems. They then feed on the phloem contents.
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(i) Name two substances the aphids obtain from the phloem.
(ii) Explain how aphids feeding from the phloem of crop plants can lead to a reduction in yield.
(c) Silverflies and hoverflies are two species of insects whose larvae feed on aphids.
Scientists investigate the feeding behaviour of these species in a laboratory experiment.
This is the scientists’ method:
- place a single silverfly in a container
- place a single hoverfly in a separate container
- keep the containers at 12°C
- put 30 aphids in each container
- count the number of aphids consumed each day for several days
- determine the mean number of aphids consumed per day
The scientists repeat the method at two higher temperatures.
The graph shows the scientists’ results.
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The scientists conclude that the hoverfly is the most effective biological control agent for aphids.
Discuss the scientists’ conclusion, referring to information in the graph and the scientists’ method in your answer.
Most-appropriate topic codes (Edexcel IGCSE Biology):
• 2(e): Nutrition (in plants) / 3(h): Transport (Phloem function) – part (b)(i)
• 2(e): Nutrition (in plants) – part (b)(ii)
• 5(a): The organism in the environment (Investigating populations) / 6(a): Food production (Biological control) – part (c)
▶️ Answer/Explanation
(a) C (it is specific)
A is not correct because biological control is often longer-lasting, not short.
B is not correct because bioaccumulation is a disadvantage of some chemical pesticides.
D is not correct because biological control is generally slower acting than chemical control.
(b)(i) An answer that includes the following:
- sucrose / sugars (1)
- amino acids (1)
Additional guidance: ignore water, reject starch/glucose. Allow named amino acid.
(b)(ii) An explanation that makes reference to four of the following:
- less sucrose / sugars transported to storage organs / fruits / grains / tubers (1)
- less amino acids for protein synthesis / growth (1)
- less energy (ATP) from respiration for growth / active transport (1)
- less material for making new plant tissue (e.g., leaves, roots) (1)
- weakened plant more susceptible to disease (1)
- aphids may spread plant viruses (1)
Note: The answer must logically link aphid feeding (removing phloem contents) to reduced plant growth or storage, leading to lower crop yield.
(c) Discussion that includes the following points:
Support for the conclusion:
- From the graph, hoverflies eat/consume more aphids than silverflies at all three temperatures. (1)
- Therefore, fewer hoverflies might be needed to control an aphid population. (1)
- The difference is greatest (approx. ×3) at 12°C, and still significant (approx. ×1.9) at 18°C. (1)
- Both species consume more aphids at higher temperatures. (1)
Limitations / Points against being fully conclusive:
- The method used only one larva per container – should use more larvae / not repeated enough for reliable mean. (1)
- It was a laboratory experiment, not a natural habitat / field study – conditions are controlled and simplified. (1)
- Only two species (hoverfly and silverfly) were compared – other potential biological control agents exist. (1)
- Temperature range tested (12–18°C) may not represent all seasonal field conditions. (1)
A balanced answer would reference both the supporting data and the methodological limitations to ‘discuss’ the conclusion.
