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Question 1

Read the passage below. Use the information in the passage and your own knowledge to answer the questions that follow.

Toystory

Toystory is a bull who was born in 2001. The photograph shows Toystory.

In the world of dairy farming Toystory is a famous bull. He fathered 500 000 offspring but did not mate with any cows. He was able to father so many offspring because his semen was collected and then used to fertilise cows using artificial insemination.

Toystory’s mother was a high milk producer and his father was a popular bull. He was sold for $4 000 by his owner to a specialist breeding company called Genex.

Genex started to collect semen from Toystory when he was four years old. The semen is carefully collected, using a teaser animal and an artificial rubber vagina.

The semen is divided into many separate samples. These samples are put in small straws and frozen in liquid nitrogen. The straws can then be sold and sent to dairy farmers around the world to inseminate their cows. A total of 2.4 million samples of semen from Toystory was sold in more than 50 countries around the world.

Toystory’s reputation grew as the offspring he fathered went on to be high milk producers. His semen straws sold for over $60 each. Bull semen can now be sold as sexed or unsexed samples, with sexed samples being more expensive to purchase.

Toystory was highly valued because his offspring produced large quantities of milk of a desired composition. His semen was effective at getting cows pregnant, his daughters gave birth easily and were strong. He had a rare mix of fertility, genetics and appearance.

One of his daughters sold for $300 000 in 2009. His record number of offspring is unlikely to be beaten. This is because bulls are often retired earlier as new genetic advances are discovered.

(a) Suggest why Genex waited until Toystory was four years old before beginning to collect his semen (line 9).

(b) Explain how the semen from the bull is used to fertilise cows using artificial insemination.

(c) (i) Suggest why the semen is stored in liquid nitrogen (line 13).
(ii) Sexed semen is guaranteed to produce offspring of one sex. Suggest why dairy farmers would prefer to used sexed semen (line 19).

(d) Determine the percentage success of Toystory’s semen samples in producing offspring (line 2 and line 15).

(e) Describe how scientists could investigate which of two bulls is the best to use as a father in dairy farming.

(f) Explain why the composition of milk is important to consumers (line 22).

(g) (i) Scientists are now using cloning to produce animals. Describe the stages that are required to clone a bull.
(ii) Give two advantages of using cloning rather than selective breeding to produce offspring.

Most-appropriate topic codes (Edexcel IGCSE Biology (Modular)):

• 3(a): Reproduction — parts (a), (b), (c)(ii)
• 5(a): Food production — parts (e), (f)
• 5(b): Selective breeding — part (e)
• 5(d): Cloning — parts (g)(i), (g)(ii)
• 2(d): Movement of substances into and out of cells / Practical application — part (c)(i)
• Appendix 3: Mathematical skills — part (d)
▶️ Answer/Explanation
Solution

(a)
• so semen contained sperm / (bull is) (sexually) mature / sperm in semen / gone through puberty / fully developed.

(b)
An explanation that includes two of the following points:
• collect semen / sperm from penis of bull (1)
• insert straw into / inject semen (into cow) (1)
• put (it / semen / sperm) in vagina / uterus / womb / cervix (1)

(c)(i)
• preserve (sperm) / keep (sperm) alive / viable / prevent growth of microorganisms / slow down metabolism.

(c)(ii)
• provide females (produce milk) / will produce cows.

(d)
\( 500\,000 \div 2.4 \text{ million} = 0.2083 \)
Percentage = \( 0.2083 \times 100 = 20.83\% \) (allow 1 mark for ÷ 2.4 million).

(e)
A description that makes reference to three of the following points:
• use semen (from each bull) to fertilise (many / similar) cows (1)
• collect / measure milk yields (1)
• from each daughter / offspring of these cows / mother of bull (1)
• select bull with highest (average) milk yield (across all daughters) (1)

(f)
An explanation that makes reference to two of the following points:
• (milk that contains) (most) fat (1)
• (most) protein (1)
• (most) vitamins (1)
• (milk that contains) (most) calcium (1)

(g)(i)
A description that makes reference to four of the following points:
• nucleus from (body) cell of bull (1)
• insert this nucleus into enucleated egg cell (1)
• electric shock (1)
• mitosis / cell division (1)
• embryo into uterus / womb (1)
• surrogate mother (1)

(g)(ii)
An explanation that makes reference to two of the following points:
• genetically identical / no genetic variation / same (combination of) alleles (1)
• quicker process (1)

Question 2

Deforestation is the cutting down of trees.

After deforestation many dead leaves are left on the forest floor.

A student investigates the decomposition of these leaves.

This is her method.

  • collect four samples of dead leaves each with a mass of 6.0 kg
  • label the samples P, Q, R and S
  • cut sample P into small pieces and keep at \(10^\circ C\)
  • cut sample Q into small pieces and keep at \(20^\circ C\)
  • do not cut sample R and keep at \(10^\circ C\)
  • do not cut sample S and keep at \(20^\circ C\)
  • measure the mass of each sample after three months

The graph shows her results.

(a) Explain the results obtained by the student.

(b) Calculate the difference between the rate of decomposition in sample P and the rate of decomposition in sample Q.

Give your answer in kg per month.

(c) The student needs to control biotic variables in her investigation.

Give two biotic variables she should control.

Most-appropriate topic codes (Edexcel IGCSE Biology):

• 4(c): Cycles within ecosystems — part (a) Decomposition
• 2(c): Biological molecules — part (a) Role of enzymes in decomposition
• Appendix 3: Mathematical skills — part (b) Calculation
• 4(a): The organism in the environment — part (c) Biotic variables
▶️ Answer/Explanation
Solution

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

  • more (decomposition) / faster with warmer temperatures / eq (1)
    (allow mass remains high in low temp)
  • enzymes (1)
  • more (decomposition) / faster with cut material / eq (1)
    (allow mass remains high in uncut)
  • more surface area (1)
  • fungi / bacteria (1)
    (allow converse)

Example answer: Decomposition is faster at \(20^\circ C\) (samples Q and S) than at \(10^\circ C\) (samples P and R) because enzymes in decomposers like fungi and bacteria work more efficiently at higher temperatures. Cutting the leaves into small pieces (samples P and Q) increases the surface area available for decomposers and their enzymes to act on, leading to faster decomposition compared to uncut leaves (samples R and S) at the same temperature.

(b) Calculation:

  1. Mass loss for P: \(6.0 – 3.6 = 2.4 \text{ kg}\)
  2. Rate for P: \(2.4 \div 3 = 0.8 \text{ kg per month}\)
  3. Mass loss for Q: \(6.0 – 2.0 = 4.0 \text{ kg}\)
  4. Rate for Q: \(4.0 \div 3 = 1.333… \text{ kg per month}\)
  5. Difference: \(1.333… – 0.8 = 0.533… \text{ kg per month}\)

Answer: \(0.53\) (allow \(0.5\), \(0.53\), \(0.533\), etc.) kg per month

Award full marks for correct numerical answer without working.

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

  • species / type of leaves / plant (1)
  • age of plant / leaves (1)
    (ignore volume of leaves)
  • same (number of) / type of decomposers / eq (1)
  • insects or organisms that might consume leaf / eq (1)

Example answer: 1. The species/type of leaf used. 2. The presence/absence of specific decomposers (e.g., fungi, bacteria).

Question 3

Plant roots absorb water from soil.

This water is transported to the leaves and then moves into the air.

(a) Which of these processes is used to absorb water from the soil?

A) active transport
B) diffusion
C) evaporation
D) osmosis

(b) Name the tissue that transports water to the leaves.

(c) Name the process that moves water vapour into the air.

(d) Which of these reduces the movement of water from the leaves into the air?

A) high light intensity
B) low air humidity
C) low air temperature
D) windy conditions

(e) Give two uses of water in a plant.

Most-appropriate topic codes (Edexcel IGCSE Biology):

• 2(h): Transport — parts (a), (b), (c), (d), (e)
• 2(d): Movement of substances into and out of cells — part (a)
• 2(h): Transport – Flowering plants — parts (b), (c), (d), (e)
▶️ Answer/Explanation
Solution

(a) D osmosis
The only correct answer is D osmosis.
A is not correct as it is not how plants absorb water.
B is not correct as it is not how plants absorb water.
C is not correct as it is not how plants absorb water.

(b) xylem / xylem vessels

(c) transpiration / evaporation / diffusion / evapotranspiration

(d) C low air temperature
A is not correct as it does not reduce the movement of water from the leaves into the air.
B is not correct as it does not reduce the movement of water from the leaves into the air.
D is not correct as it does not reduce the movement of water from the leaves into the air.

(e) An answer that makes reference to two of the following points:
• support / turgor / eq (1)
• photosynthesis / eq (1)
• cooling (1)
• reactions / solvent / transport of mineral ions / named mineral ion / eq (1)

Question 4

A scientist investigates the effect of growth hormone (GH) on the body mass of rats.

This is his method.

  • give one rat a GH solution every day for 500 days
  • give another rat a control solution every day for 500 days
  • measure the mass of each rat each week for 500 days

The graph shows his results.

(a) Suggest how the control solution differs from the GH solution.

(b) Calculate the average rate of growth of the rat given GH solution from 100 days to 500 days. Give your answer in \(g\) per day.

(c) The scientist controlled all the variables in his investigation. Suggest two abiotic variables he controls.

(d) The scientist repeats his investigation using more rats. Explain why using more rats improves his investigation.

(e) GH increases transcription in cells. Explain why this affects the growth of rats.

Most-appropriate topic codes (Edexcel IGCSE Biology):

• 1.1, 3.44: Characteristics of living organisms, Hormones (growth) — context of the investigation
• 3.31: Homeostasis — controlled variables in (c)
• Appendix 3: Mathematical skills — calculation in (b)
• 3(b): Inheritance — transcription and growth in (e)
• Appendix 4: Command words (Suggest, Calculate, Explain) — across all parts
▶️ Answer/Explanation
Solution

(a) • no GH / water / saline / eq

(b) Increase in mass between 100 and 500 days: \(485 – 230 = 255g\)
Divide by the number of days: \(400\) days
Average rate = \(255 \div 400 = 0.6375\) \(g\)/day
Allow any answer between \(0.6375\) and \(0.65\) \(g\)/day.

(c) An answer that makes reference to two of the following points:
• temperature (1)
• (mass of) food / diet / type of food / eq (1)
• water (1)
• size of cage (1)
• time (of day measurements are taken) (1)
• volume of solution given / eq (1)

(d) An explanation that makes reference to two of the following points:
• avoids making wrong conclusion based on one / few result(s) / conclusion is valid / eq (1)
• can calculate mean / average (1)
• results are reliable / increase reliability (1)
• anomalous results recognised / eq (1)

(e) An answer that makes reference to three of the following points:
• (more) mRNA made (1)
• (more) translation (1)
• (more) proteins / polypeptides made (1)
• proteins are needed for growth / to make enzymes / muscle / tissue / eq (1)

Question 5

Human populations are at risk from infectious disease.

(a) Describe the different types of pathogen. Refer to a disease that each type of pathogen causes in your answer.

(b) Explain how vaccination protects humans from pathogens.

Most-appropriate topic codes (Edexcel IGCSE Biology):

• 1(b): Variety of living organisms / Pathogens — part (a) [Describing viruses, bacteria, fungi, protocysts and related diseases]
• 1.4: Understand the term pathogen — part (a) [Context]
• 2(h): Transport — part (b) [Vaccination and immune response, memory cells]
• 3.13B: Understand how vaccination results in the manufacture of memory cells — part (b) [Core explanation]
▶️ Answer/Explanation
Solution

(a) A description that makes reference to six of the following points (typically three pathogen types with their descriptions and matching diseases):

  • Virus: non-living organisms / small particles / protein coat (capsid) / relies on other organisms for reproduction / eq (1)
    Disease: AIDS / Influenza / HIV (1) [Disease mark must match pathogen type. Allow plant disease e.g., Tobacco Mosaic Virus (TMV).]
  • Bacteria: microscopic single-celled / prokaryotic / no nucleus / have nucleoid / plasmids (1)
    Disease: Pneumonia / Cholera / Tuberculosis (1)
  • Fungus: not able to carry out photosynthesis / saprotrophic nutrition / single-celled or hyphal / cell wall made of chitin / eq (1)
    Disease: Athlete’s foot / Ringworm (1)
  • Protocysts (Protozoa): microscopic single-celled / eukaryotic (1)
    Disease: Malaria (caused by Plasmodium) (1)

Scoring Note: The first mark for each pathogen is for its description, the second is for naming a correct associated disease. A pair (description + disease) for one pathogen type scores 2 marks. “HIV / AIDS” scores the disease mark for a virus but not the pathogen description mark if the description is incorrect. “Virus non-living causing cholera” scores the pathogen description mark but not the disease mark (cholera is bacterial).

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

  • dead / weakened / harmless / attenuated pathogen / or its antigens are introduced (into the body) / eq (1) [Allow ‘weakened strain’.]
  • this stimulates the production of memory cells / lymphocytes (1)
  • this leads to a (secondary) immune response upon future infection with the actual pathogen (1)
  • antibodies are produced faster / in greater quantity / sooner (1)
Question 6

Variation in a population can have different causes.

(a) Which of these will not lead to an increase in genetic variation in a population of plants?




(b) Explain how a change in the DNA of a microorganism can reduce its ability to digest a substance.

(c) Explain why a change in DNA may not affect the phenotype of an organism. 

Most-appropriate topic codes (Edexcel IGCSE Biology):

• 3(a): Reproduction — Sexual vs. asexual reproduction (part a)
• 3(b): Inheritance — Mutation, Protein synthesis (parts b, c)
• 2(c): Biological molecules — Enzymes (part b)
▶️ Answer/Explanation
Solution

(a) A (asexual reproduction)

  • B is not correct as it increases genetic variation
  • C is not correct as it increases genetic variation
  • D is not correct as it increases genetic variation

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

  1. Different (sequence of) bases in DNA / eq (1)
  2. Changes mRNA / codons (1)
  3. Transcription (1)
  4. Change tRNA / anticodons / (sequence of) amino acids (1)
  5. Translation (1)
  6. Changes structure / shape of protein / eq (1) (changes shape of active site = 2 marks)
  7. Changes active site (1)
  8. Enzyme not functional / no binding / no enzyme-substrate complex formed / eq (1)

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

  1. As some triplets / codons code for same amino acid / degenerative code / eq (1)
  2. No change in protein / polypeptide / enzyme produced (1)
  3. Active site not changed / affected (1)
  4. Mutation / allele may be recessive (1)
  5. So not expressed in phenotype / if heterozygous / dominant allele present / eq (1)
  6. Mutation may occur in a non-coding sequence of DNA / eq (1)
Question 7

Gas exchange in a flowering plant changes depending on conditions.

(a) Complete the passage by writing a suitable word or words in each blank space.

Plants carry out photosynthesis to produce ………………….. To enable this process to occur the leaf cells absorb carbon dioxide and release oxygen.

At the same time the cells in the leaves are respiring. This means that they are using …………………. and producing carbon dioxide. If the leaves are in bright sunlight, then the rate of photosynthesis will be …………………. than the rate of respiration. If the leaves are in dim light, then the rate of respiration will be greater than the rate of photosynthesis and there will be a net production of …………………..

In conditions when there is no net absorption or release of carbon dioxide the rate of photosynthesis and respiration are …………………. and the plant is at its compensation point.

(b) Describe how you could use hydrogen-carbonate indicator to investigate the effect of light intensity on net gas exchange in a leaf.

Most-appropriate topic codes (Edexcel IGCSE Biology):

• 2(e): Nutrition (Photosynthesis) — part (a)
• 2(g): Gas exchange (Flowering plants) — parts (a), (b)
• 2(f): Respiration — part (a)
• 2(g): Practical Investigation — part (b)
▶️ Answer/Explanation
Solution

(a) An answer that makes reference to the following points:

  • carbohydrate / named carbohydrate (e.g., glucose, starch) (1)
  • oxygen (1) allow carbohydrate / named carbohydrate
  • higher / greater / more (1)
  • carbon dioxide (1)
  • equal / the same / balanced (1)

Completed passage: Plants carry out photosynthesis to produce carbohydrate (e.g., glucose). To enable this process to occur the leaf cells absorb carbon dioxide and release oxygen. At the same time the cells in the leaves are respiring. This means that they are using oxygen and producing carbon dioxide. If the leaves are in bright sunlight, then the rate of photosynthesis will be higher than the rate of respiration. If the leaves are in dim light, then the rate of respiration will be greater than the rate of photosynthesis and there will be a net production of carbon dioxide. In conditions when there is no net absorption or release of carbon dioxide the rate of photosynthesis and respiration are equal and the plant is at its compensation point.

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

  • (how light intensity is varied) foil / muslin / move lamp / eq (1) allow light and dark
  • leaf in test tube with bung / use flask with delivery tube / eq (1)
  • (look for colour change after) same / stated time (1)
  • same size / species / type / surface area / eq (1)
  • same temperature / same volume of indicator (1)
  • correct colour change so goes yellow with increased CO\(_2\) in dark / goes dark red/ red/ purple with reduced CO\(_2\) in light / eq (1)

Suggested answer: Place a leaf in a test tube containing a set volume of hydrogen-carbonate indicator and seal it. Repeat with another leaf in a second test tube. Cover one tube with foil to provide dark conditions (low light intensity) and leave the other in bright light. Keep all other variables (e.g., temperature, leaf size/species, time) the same. After a set time, observe the colour change. The indicator goes yellow if there is a net production of CO\(_2\) (respiration > photosynthesis in dim/dark) and purple/red if there is a net uptake of CO\(_2\) (photosynthesis > respiration in bright light).

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