Edexcel iGCSE Biology 4BI1 - Paper 2B -Biological molecules- Exam Style Questions- New Syllabus

Question

Apples contain an enzyme called phenol oxidase.

When apple tissue is exposed to oxygen in the air, this enzyme turns the apple tissue brown.

(a) A student uses this method to investigate the effect of pH on the time taken for apple tissue to turn brown.

mash an apple to produce a pulp
place 25 g of this pulp into a beaker
mix a pH 7 buffer with the pulp to maintain the pH
check the colour of the pulp every five minutes
record the time when the pulp turns brown

The student repeats this method using pH buffers of 3, 5, 9, and 11.

The graph shows the student’s results.

(i) Calculate the percentage increase in time taken for the apple pulp to turn brown at pH 11 compared with the time taken at pH 7.
(ii) Explain why the change in pH affects the time taken for the apple pulp to turn brown.
(iii) Explain how the student could modify this method to give a more accurate measure of the time taken for the apple pulp to turn brown.

(b) A genetically modified (GM) variety of apple has been produced that does not turn brown when exposed to air.

These apples have a gene in their DNA that produces a section of mRNA with a complementary sequence to the mRNA for the phenol oxidase gene.

(i) State two differences between DNA and mRNA.
(ii) Explain why these GM apples do not turn brown when the apple tissue is exposed to air. Use your knowledge of protein synthesis to support your answer.

Most-appropriate topic codes (Edexcel IGCSE Biology):

• 2(c): Biological molecules — parts (a)(ii), (b)(ii)
• 2(c): Enzymes as biological catalysts — parts (a)(ii), (a)(iii)
• 3(b): Inheritance — parts (b)(i), (b)(ii)
• 5(c): Genetic modification (genetic engineering) — part (b)(ii)
• Appendix 3: Mathematical skills — part (a)(i)

▶️ Answer/Explanation

Solution

(a)(i)

• 1400 (%) (2)

One mark for 75 ÷ 5 = 15 OR 70 ÷ 5 = 14

Calculation: Time at pH 11 = 75 minutes, Time at pH 7 = 5 minutes

Percentage increase = \(\frac{75 – 5}{5} \times 100 = \frac{70}{5} \times 100 = 14 \times 100 = 1400\%\)

(a)(ii)

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

1. optimum pH is 7 (1)

2. (pH causes) enzyme to denature (1)

3. shape of enzyme changes / shape of active site changes (1)

4. substrate does not fit active site / enzyme no longer complementary / cannot form enzyme-substrate complexes (1)

(a)(iii)

An explanation that makes reference to two of the following points:

1. use a colour matching chart / light sensor / colorimeter (1)

2. (because) colour change is subjective / to standardise the end colour / so colour is always same (1)

3. use intervals of shorter than 5 minutes / shorter intervals / check more often / check colour constantly (1)

(b)(i)

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

1. DNA is double stranded / RNA is single stranded (1)

2. DNA has T (thymine) / RNA has U (uracil) (1)

3. DNA is a helix / RNA is not a helix (1)

4. DNA has deoxyribose / RNA has ribose (1)

(b)(ii)

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

1. (complementary) RNA binds to the phenol oxidase RNA (1)

2. translation cannot occur (1)

3. so enzyme / phenol oxidase not made (1)

AND

Maximum three from:

4. transcription makes mRNA (1)

5. RNA leaves nucleus and enters cytoplasm / RNA moves to ribosome (1)

6. tRNA brings / carries / transports amino acids (1)

7. (during translation) tRNA binds mRNA / anticodons bind codons (1)

8. amino acids join / amino acid chain / makes polypeptide / peptide bonds form (1)

Question

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

Pollination in fruit trees

Plants need to be pollinated to produce fruits and seeds. Some plants are self-pollinated and others are cross-pollinated.

Self-pollination is usually the transfer of pollen from flowers on the same tree. In fruit trees, self-pollination also includes the transfer of pollen from another tree of the same cultivar. A cultivar is a genetically identical group of trees produced by selective breeding. Self-pollinating plants need bees or other insects to transfer pollen. Examples of self-pollinated fruit trees are plum, apricot and peach that have sweet-tasting, soft fruits.

Other fruit trees require cross-pollination and can only produce fruit by fertilisation from a different cultivar. Cross-pollination is the transfer of pollen from one cultivar to the flower of a different cultivar. Self-incompatibility prevents the same plant or cultivar from fertilising its own flowers. For example, the pollen from one cultivar of an apple tree will not fertilise trees of the same cultivar. Most hard fruit trees such as pear and apple require the presence of two different compatible cultivars for pollination to result in fruit production.

Pollen can be transferred by wind, insects or birds. Many fruit trees are pollinated by insects. Nut trees are usually pollinated by wind.

After pollination the pollen grain must germinate to lead to fertilisation. The success of pollination and fertilisation depends on favourable environmental conditions. It also requires the right pollen grain reaching the right flower, as pollen can only fertilise specific, compatible flowers. For example, pollen from a peach flower will not pollinate apple flowers.

Cross-pollinating fruit trees should be planted no more than 60 m apart. This will ensure that bees visit the trees often enough to ensure pollination. Nut trees need to be planted no more than 15 m apart.

A further complication is that a few apple and pear cultivars, known as triploids, produce sterile pollen. Triploid cultivars have three sets of chromosomes within their cells. A triploid cultivar will require another cultivar for pollination, and the trees must flower at the same time.

(a) Explain the differences between self-pollination and cross-pollination. (lines 3 to 5 and 9 to 11)

(b) Describe the sequence of events from pollen landing on a flower to fruit production.

(c) Fruit trees often produce sweet-tasting fruit containing sugars.

(i) Sugars are carbohydrates. Name the three elements in carbohydrates.
(ii) Suggest how production of sweet-tasting fruit may help spread the seeds of fruit trees. (lines 7 to 8)

(d) Explain how self-incompatibility prevents self-fertilisation. (lines 11 to 13)

(e) Describe how the structure of insect-pollinated flowers helps to achieve pollination. (line 16)

(f) Give a reason why nut trees need to be planted closer together than fruit trees. (lines 23 to 25)

(g) Explain why triploid cultivars produce sterile pollen. (lines 26 to 29)

Most-appropriate topic codes (Edexcel IGCSE Biology):

• 3(a): Reproduction — parts (a), (b), (c)(ii), (d), (e), (f), (g)
• 2(c): Biological molecules — part (c)(i)
• 3(b): Inheritance — part (g)

▶️ Answer/Explanation

Solution

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

self-pollination within same plant / flower / cultivar / tree / uses one parent / eq (1)
cross-pollination from different plant / flower / cultivar / tree / uses two parents / eq (1)

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

germinates / eq (1)
pollen tube grows down style / goes down style / eq (1)
enters ovule (via micropyle) / eq (1)
male nucleus / gamete travels down pollen tube / eq (1)
fuses with ovum / fuses with female gamete / fertilises ovum / fertilises female gamete / eq (1)
ovule becomes seed / eq (1)
ovary becomes fruit / eq (1)

(c)(i) carbon hydrogen oxygen / C H O (1)

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

fruit eaten by animals / insects / birds / eq (1)
egested / lost in faeces / eq (1)
seeds dispersed / taken to new areas / eq (1)

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

pollen from same plant / flower / cultivar eq (1)
cannot germinate / no pollen tube growth / eq (1)
no fusion / fertilisation / eq (1)

(e) A description that makes reference to the following:

flowers are large / petals are large (so seen by insects) / eq (1)
flowers / petals are coloured (so seen by insects) / eq (1)
flowers are scented / have scented petals / eq (1)
have nectary / produce nectar / eq (1)
anthers within flower / stigma within flower so insect brushes pollen / transfers pollen to stigma / eq (1)

(f) An answer that includes:

wind pollinated / no insects to carry pollen long distance / eq (1)

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

triploid cells cannot divide by meiosis / cannot be divided by two / equally / eq (1)
cannot produce haploid (gametes) / produce one set / n chromosomes / eq (1)
fertilisation / fusion not possible / eq (1)

Question

The photograph shows a fish called tilapia. This type of fish is often grown in fish farms.

(a) Tilapia are a good source of nutrients for humans.

The table shows the percentage of the daily requirements for humans of three nutrients provided by a portion of tilapia.

Nutrient group	Percentage of daily requirement provided by a portion of tilapia
lipids	16
protein	38
carbohydrate	16

(i) Use the table to determine the ratio of lipid to protein to carbohydrate.

Give your answer in the form \(1 : n : 1\)

(ii) Describe how to test a sample of tilapia to show the presence of lipid.

(iii) State one function of protein in the human body.

(b) Scientists have produced genetically modified (GM) tilapia that grow faster.

Scientists investigate the growth of these GM tilapia compared with non-genetically modified (non-GM) tilapia.

This is the scientists’ method.

set up two equally sized tanks containing water
put non-GM tilapia into one tank
put an equal mass of GM tilapia into the second tank
feed the fish in each tank the same mass of protein pellets

The mass of each type of fish was measured at the start of the investigation and after seven months.

A measure called the feed conversion index was also calculated for each type of fish.

The table shows the scientists’ results.

Type of fish	Total starting mass of fish in g	Total mass of fish after seven months in g	Feed conversion index
non-GM	1250	2830	1.9
GM	1250	3750	1.2

(i) The mean rate of increase in mass of the non-GM tilapia during the seven months is \(226 \, \text{g}\) per month.

Calculate the mean rate of increase in mass, in g per month, of the GM tilapia.

Give your answer to three significant figures.

(ii) The feed conversion index is a measure of the mass of protein pellets used compared with the increase in mass of tilapia.

It is calculated using this formula.

\(\text{feed conversion index} = \frac{\text{total mass of protein pellets used}}{\text{increase in mass of tilapia}}\)

Use the information in the table to calculate the mass of protein pellets given to the non-GM tilapia.

(iii) Suggest why the GM tilapia have a lower feed conversion index than the non-GM tilapia.

(iv) Both groups of fish were fed the same mass of pellets and placed into the same sized tanks of water.

Give one other abiotic factor that the scientists should keep constant.

(v) The scientists conclude that farming GM tilapia is better for the environment than farming non-GM tilapia. This is because the GM tilapia would result in less nitrate in the water and cause less spread of disease. Evaluate this conclusion.

Use the information in the table and your own knowledge in your answer.

Most-appropriate topic codes (Edexcel IGCSE Biology):

• 2(c): Biological molecules — parts (a)(ii), (a)(iii)
• 2(e): Nutrition — part (a)(i), (a)(iii)
• 5(a): Food production (Fish farming) — parts (b)(i), (b)(ii), (b)(iii), (b)(iv), (b)(v)
• 5(c): Genetic modification (genetic engineering) — parts (b)(i), (b)(iii), (b)(v)
• 4(a): The organism in the environment (Abiotic factors) — part (b)(iv)

▶️ Answer/Explanation

Solution

(a)(i) \(1 : 2.4 : 1\)
Accept \(1: 2.375: 1\)

(a)(ii) A description that makes reference to the following points:
• ethanol / alcohol and add water (1)
• goes white / cloudy / white emulsion forms / milky / eq (1)
OR
• add Sudan III (1)
• red colour (in top layer) (1)
Accept alternative method: rub on paper / eq (1); paper goes transparent / clear / eq (1)

(a)(iii) • growth / repair / enzymes / build muscle / antibodies / eq (1)

(b)(i) • \(357 \, \text{g per month}\) (2)
Correct answer gains both marks.
Accept one mark for:
\(2500\) (increase) or division by \(7\) or \((3750-1250)\) or \(357.14…\)

(b)(ii) • \(3002 / 3000 \, \text{g}\) (1)

(b)(iii) An answer that makes reference to two of the following points (1 mark each):
• (GM fish) grow more / bigger / faster / eq
• produce less faeces / can digest more efficiently / absorb more / eq
• less respiration / lower metabolic rate / eq
• move less / slower swimming / eq
• eat more food / less food wasted / eq

(b)(iv) • temperature / light (intensity) / oxygen / salinity / pH / eq (1)

(b)(v) An evaluation that makes reference to the following points (up to 5 marks):
1. GM fish grow bigger/faster/harvest sooner / eq (1)
2. less food wasted / more food eaten / eq (1)
3. less faeces / urea / excretion / ammonia / eq (1)
4. less decomposition / fewer decomposers / eq (1)
5. less nitrification / fewer nitrifying bacteria / eq (1)
6. less eutrophication / algal growth / plant growth / eq (1)
7. less oxygen loss / more oxygen in water / eq (1)
8. due to less (bacterial) respiration / (more oxygen for) respiration of animals / eq (1)
9. no information about disease spread / pathogen spread / eq (1)
10. GM fish are (genetically) similar so may be more likely to catch/spread disease / eq (1)
11. not repeated / need more experiments / no idea of sample size / eq (1)
12. (GM) fish that escape may harm food chains / outcompete wild fish / may breed with wild fish / eq (1)
13. did not measure nitrates / eq (1)

Question

Cells use genetic information when carrying out protein synthesis.

(a) Give the difference between a gene and a genome.

(b) Describe the process of transcription.

(c) Explain why a change in the sequence of bases in the DNA does not always result in a change in enzyme function.

(d) Some characteristics are determined only by the genes, and some are determined by a combination of the genes and the environment. Give an example of each type of characteristic in humans.

determined only by the genes
determined by the genes and the environment

Most-appropriate topic codes (Edexcel IGCSE Biology):

• 3(b): Inheritance — parts (a), (c), (d) [genes, DNA, variation, mutations]
• 3(b): Protein synthesis — part (b) [transcription]
• 2(c): Biological molecules (Enzymes) — part (c) [effect on enzyme function]
• 3(a): Reproduction and inheritance — implicit link to genetic determination of characteristics

▶️ Answer/Explanation

Solution

(a) A gene is a section of DNA that codes for a single polypeptide/protein, whereas a genome includes all of the DNA/all of the genes in an organism. (Requires both parts for the mark.)

(b) A description that makes reference to four of the following points (1 mark each):

DNA unzips/unwinds.
Involves mRNA (messenger RNA).
mRNA copies the code/sequence of the DNA strand (DNA acts as a template).
Process occurs in the nucleus / mRNA leaves the nucleus.
mRNA goes to a ribosome.

(c) An explanation that makes reference to three of the following points (1 mark each):

Some different base sequences/codons may code for the same amino acid (degeneracy of genetic code).
No change in the protein/enzyme is produced.
The mutation may not affect the bonding/3D shape/active site of the enzyme.
The allele may be recessive and therefore not expressed.
The change may be in non-coding DNA.

(d) An answer that makes reference to the following (1 mark each):

Determined only by genes: Blood group, eye colour, iris colour. (Examples of single-gene traits.)
Determined by genes and environment: Mass, height, skin colour, body shape, hair colour, behavioural traits.

Question

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

Tissue culture and plants

In micropropagation, plant tissues are grown in vitro on plant tissue culture media, under aseptic conditions in a controlled environment. This technique is possible because plant cells can differentiate and become specialised cells. This allows them to change their metabolism, growth and development to form a whole plant.

Plant tissue culture media contain substances to support the normal growth and development of plants. The media are mainly composed of minerals, vitamins, and plant hormones. The pH of the media is kept constant.

Plant hormones play an essential role in determining how cells and tissues develop in culture media. Plant cells can differentiate into different tissues and cell types. The concentration of hormones can determine the tissue that develops. Auxins and cytokinins are the most widely used. A balance of both auxin and cytokinin leads to the development of a mass of undifferentiated cells known as a callus.

In vitro cell culture offers an alternative method for conserving endangered species and varieties. Tissue culture can be used when the plant species produce seeds that do not germinate or have seeds that cannot be stored for a long period of time. These can be successfully preserved using in vitro techniques for the maintenance of gene banks.

Embryo culture is a type of plant tissue culture that is used to grow embryos from seeds in nutrient media. In embryo culture, the plant develops directly from the embryo or indirectly through the formation of a callus and then subsequent formation of shoots and roots. The technique has been developed to break dormancy in seeds, and to reproduce rare species and haploid plants.

Scientists are also using cell suspension culture systems from which products can be extracted. A suspension culture is developed by transferring a portion of the callus into liquid media. The media are maintained under suitable conditions of agitation, light and temperature. This system can provide a continuous, reliable source of natural products independent of climate and soil conditions. The first commercial application of large-scale suspension cultivation of plant cells was carried out to produce shikonin. Shikonin is used in traditional Chinese medicine and is a potential anti-cancer treatment.

(a) State what is meant by the term in vitro. (line 1)

(b) Explain how plant cells differ from human cells in their ability to differentiate and specialise. (lines 2 and 3)

(c) Give the function of two named minerals included in the culture media. (line 6)

(d) Explain why the pH of the media needs to be kept constant. (line 7)

(e) Auxin also controls the response of plants to light. Describe a simple experiment you could do to show the phototropic response of plant stems to light.

(f) Explain why scientists want to conserve endangered plant species and varieties. (lines 13 and 14)

(g) Explain why plant cell suspension culture media are maintained under suitable conditions of agitation, light and temperature. (lines 24 and 25)

Most-appropriate topic codes (Edexcel IGCSE Biology):

• 5(d): Cloning — parts (a), (b), (g)
• 2(b): Cell structure — part (b)
• 2(e): Nutrition — part (c)
• 2(c): Biological molecules — part (d)
• 2(j): Co-ordination and response — part (e)
• 4(a): The organism in the environment — part (f)

▶️ Answer/Explanation

Solution

(a) In a test tube / culture dish / jar / glass / petri dish / container / in culture solution / in a lab / outside a living organism.

Explanation: The term “in vitro” literally means “in glass” in Latin, referring to biological processes that are conducted outside of a living organism in an artificial laboratory environment, such as in test tubes or petri dishes. This contrasts with “in vivo” experiments which are conducted within living organisms.

(b) Plant cells can differentiate into all/different types of tissues or specialized cells throughout the plant’s life and can form/regenerate a whole new plant.

Explanation: Plant cells exhibit totipotency, meaning that even mature, differentiated plant cells retain the ability to dedifferentiate and then redifferentiate into any cell type needed to regenerate an entire plant. This is why you can grow a new plant from a cutting. In contrast, human cells have much more limited differentiation capabilities. While stem cells can differentiate into various cell types, most human cells become permanently specialized during development and cannot revert back or form entirely new organisms.

(c)
1. Nitrate – for making amino acids/proteins/DNA/nucleic acids
2. Magnesium – for making chlorophyll/chloroplasts/photosynthesis

Explanation: Plant tissue culture media must contain essential minerals that support plant growth and development. Nitrate is crucial as it provides nitrogen, which is a fundamental component of amino acids, proteins, and nucleic acids (DNA and RNA). Without adequate nitrogen, plants cannot synthesize these essential biomolecules. Magnesium is a central component of the chlorophyll molecule, which is vital for photosynthesis as it captures light energy. Without magnesium, plants cannot produce chlorophyll effectively, leading to chlorosis (yellowing of leaves) and impaired photosynthesis.

(d) Enzymes are affected by pH/ work best at optimum pH. If pH changes, the shape of the active site can change/be denatured so substrates can no longer bind.

Explanation: Maintaining a constant pH is critical because enzymes, which catalyze all biochemical reactions in plant cells, are highly sensitive to pH changes. Each enzyme has an optimal pH range where it functions most efficiently. If the pH deviates from this range, the enzyme’s three-dimensional structure can be altered, changing the shape of its active site. This prevents substrates from binding properly, effectively denaturing the enzyme and halting the metabolic reactions it catalyzes. This would severely disrupt plant growth and development in the culture media.

(e) Place a shoot in light from one side/unidirectional light and another shoot in darkness/light all around. Leave both for a stated time/use shoots of same type/same temperature/other control variable. Observe/measure bending or growing towards light.

Explanation: To demonstrate phototropism (growth response to light), you would set up two identical young plant shoots. One would be placed in a location with light coming from only one direction (e.g., near a window), while the control would be placed in either complete darkness or with light evenly distributed from all sides. Both plants should be kept under the same temperature and watering conditions to ensure any differences are due to light direction only. After a few days, you would observe that the shoot exposed to unilateral light has bent toward the light source. This bending occurs because auxin hormone accumulates on the shaded side of the stem, promoting more cell elongation on that side and causing the stem to curve toward the light.

(f) To maintain biodiversity/reduce damage to ecosystems and to prevent extinction/keep species for future generations/for medicinal properties.

Explanation: Conserving endangered plant species is crucial for several reasons. Firstly, it maintains biodiversity, which ensures ecosystem stability and resilience. Each plant species plays a unique role in its ecosystem, and losing one can disrupt food webs and ecological balance. Secondly, it prevents extinction, preserving genetic diversity that might be valuable for future breeding programs, especially as climate changes. Many plants contain compounds with medicinal properties; for example, aspirin originated from willow bark. By conserving endangered species, we preserve potential future medicines and genetic resources that could be vital for human well-being.

(g) Agitation mixes contents/mixes oxygen/with plant cells. Light is for photosynthesis. Suitable temperature is for enzyme action.

Explanation: Suspension cultures require specific conditions to mimic optimal natural environments. Agitation (shaking or stirring) ensures that cells and nutrients are evenly distributed throughout the liquid media, preventing sedimentation. It also promotes gas exchange, ensuring oxygen (needed for respiration) is available and carbon dioxide (a product of respiration) is removed. Light is essential for photosynthetic plant cells to produce their own energy through photosynthesis. Maintaining a suitable temperature is critical because temperature affects enzyme activity; most plant enzymes function optimally around 25-30°C. Temperatures that are too high can denature enzymes, while temperatures that are too low can slow down metabolic processes to inadequate levels.

Question

Catalase is an enzyme found in many cells. This enzyme controls the breakdown of hydrogen peroxide into water and oxygen.

\( 2H_2O_2 \rightarrow 2H_2O + O_2 \)
hydrogen peroxide water oxygen

A teacher uses this method to investigate the effect of pH on catalase.

Step 1: Cut a cylinder of potato tissue into six equal sized discs.
Step 2: Measure 10 cm³ of hydrogen peroxide solution and place into a boiling tube.
Step 3: Add a pH buffer solution to the tube to keep the pH at 7.
Step 4: Add the six potato discs to the boiling tube.
Step 5: Collect the gas given off by the reaction in an inverted 20 cm³ measuring cylinder.
Step 6: Measure the total volume of gas collected after five minutes.

Repeat steps 1 to 6 using four different pH buffers (pH 4, pH 5, pH 6 and pH 8).

The diagram shows the teacher’s apparatus.

(a) State what is meant by the term enzyme.

(b) Suggest why each potato cylinder was cut into six discs rather than left as one cylinder.

(c)(i) Give the dependent variable in this experiment.
(ii) The teacher controls the time for gas collection and also the volume of hydrogen peroxide used. State two other abiotic variables the teacher should control in this experiment.

(d) The table shows the teacher’s results.

pH of solution	Volume of oxygen after 5 minutes in cm³	Mean rate of reaction in cm³ per minute
4	4	0.8
5	6	1.2
6	7	1.4
7	12	2.4
8	3	0.6

(i) Calculate the percentage change in the mean rate of reaction as the pH is changed from pH 4 to pH 7.
(ii) The teacher calculated the mean rate of reaction by collecting the oxygen released for the first five minutes of the reaction. Explain why the mean rate of reaction changes beyond the first five minutes.(iii) Explain the effect of changing pH on an enzyme-controlled reaction.

Most-appropriate topic codes (Edexcel IGCSE Biology):

• 2(c): Biological molecules — Enzymes — parts (a), (d)(iii)
• 2.10: Role of enzymes as biological catalysts — part (a)
• 2.11, 2.13: Effect of temperature and pH on enzyme function — parts (d)(ii), (d)(iii)
• 2.12, 2.14B: Practical: investigate how enzyme activity is affected by temperature/pH — parts (b), (c)(i), (c)(ii), (d)(i)
• Appendix 3: Mathematical skills — part (d)(i)

▶️ Answer/Explanation

Solution

(a) A biological catalyst / a protein that speeds up (chemical) reactions without being used up.

Explanation: Enzymes are proteins that lower the activation energy for metabolic reactions, increasing the rate of reaction.

(b) To increase the surface area of the potato tissue / so more catalase enzyme is exposed to the hydrogen peroxide.

Explanation: Cutting the cylinder into discs increases the surface area-to-volume ratio. This allows more catalase molecules inside the potato cells to come into contact with the substrate \( (H_2O_2) \), speeding up the reaction and making results easier to measure.

(c)(i) Volume of oxygen gas collected (after 5 minutes) / mean rate of reaction.

Explanation: The dependent variable is what is measured as the outcome of changing the independent variable (pH).

(c)(ii) Two from:
1. Temperature (of the solution / room)
2. Concentration of hydrogen peroxide solution
3. Volume / concentration of the pH buffer added
4. Size / mass / surface area of potato discs

Explanation: These are abiotic (non-living) factors that, if not controlled, could affect the rate of the enzyme reaction and make the results invalid.

(d)(i) \(200\%\)

Working:
Change in rate = \(2.4 – 0.8 = 1.6 \text{ cm}^3/\text{min}\)
Percentage change = \(\frac{1.6}{0.8} \times 100 = 200\%\)
The rate increases by 200% from pH 4 to pH 7.

(d)(ii) An explanation that makes reference to two of:
• The substrate \( (H_2O_2) \) is being used up / its concentration decreases.
• Therefore, there are fewer collisions between enzyme and substrate molecules / fewer enzyme-substrate complexes form.
• The water produced dilutes the hydrogen peroxide.

Explanation: The reaction rate is highest at the start when substrate concentration is highest. As the substrate is converted to products, the rate slows down because there are fewer substrate molecules available to bind with the enzyme’s active sites.

(d)(iii) An explanation that makes reference to:
1. Enzymes have an optimum pH (e.g., pH 7 for catalase). (1)
2. Changing pH away from the optimum denatures the enzyme / alters the bonds in the enzyme’s structure. (1)
3. This changes the shape of the active site, so the substrate no longer fits / enzyme-substrate complexes cannot form. (1)

Explanation: Enzymes are sensitive to pH because hydrogen ions affect the ionic bonds that hold the enzyme’s tertiary structure. At the wrong pH, the active site’s shape is altered (denaturation), reducing or stopping catalytic activity.

Question

Food items can often be spoiled if saprotrophic microorganisms such as mould fungi grow on them.

(a) Describe how a saprotrophic fungus such as mould obtains its food.

(b) A student uses this method to investigate ways of preventing peas from being spoiled.

place three peas in each of three test tubes as shown in the diagram
cover the peas in test tube A with water and keep at 37°C
cover the peas in test tube B with water and keep at 4°C
cover the peas in test tube C with vinegar, which is a weak acid, and keep at 37°C
leave the peas for 24 hours

The student observes the level of cloudiness of the solution to determine how spoiled the peas have become. The level of cloudiness can be used as a measure of fungal growth.

The table shows the student’s results.

(i) Suggest a problem with using the level of cloudiness of the solution to determine how spoiled the peas have become.

(ii) Explain the appearance of the peas in water at 4°C.

(iii) Explain the appearance of the peas in vinegar at 37°C.

Most-appropriate topic codes (Edexcel IGCSE Biology):

• 1(b): Variety of living organisms — part (a)
• 2(c): Biological molecules and enzymes — parts (b)(ii), (b)(iii)
• 2.11–2.13: Enzyme function and factors affecting it — parts (b)(ii), (b)(iii)
• Appendix 6: Suggested practical investigations — part (b) experimental context

▶️ Answer/Explanation

Solution

(a) Saprotrophic fungi like mould obtain their food through extracellular digestion. They secrete enzymes onto the dead or decaying organic matter. These enzymes break down complex molecules like carbohydrates and proteins into simpler, soluble substances. The fungus then absorbs these digested nutrients through its hyphae.

(b)(i) The main problem is that judging cloudiness is subjective and qualitative rather than quantitative. Different people might interpret “very cloudy,” “slightly cloudy,” and “no cloudiness” differently. There’s no precise, numerical measurement, which makes it difficult to compare results accurately or for other scientists to repeat the experiment exactly.

(b)(ii) The peas in water at 4°C showed only slight cloudiness because the low temperature significantly slowed down fungal growth. At 4°C, the enzymes in the saprotrophic fungi work much more slowly due to reduced kinetic energy, leading to fewer enzyme-substrate collisions. This results in less digestion of the peas, less fungal respiration, and therefore less spoilage and cloudiness.

(b)(iii) The peas in vinegar at 37°C showed no cloudiness because the acidic conditions denatured the enzymes of the saprotrophic fungi. Vinegar, being a weak acid, lowers the pH. This change in pH alters the shape of the enzymes’ active sites, preventing them from binding to their substrates and catalyzing the digestive reactions. Consequently, fungal growth is inhibited or the fungi are killed, preventing spoilage and resulting in clear solution.

Question

Pineapple juice contains a protease called bromelain.

A student uses this method to investigate the digestion of solid gelatine protein by bromelain.

place solid gelatine protein into a test tube up to a height of 5 cm
mix 5 cm³ pineapple juice with 1 cm³ of pH 4 buffer
place 1 cm³ of the pineapple juice and buffer solution on top of the gelatine
leave for one hour in a water bath set to 37°C
measure the height of the solid gelatine and use it to calculate the volume of gelatine that has been digested

Repeat the method three more times.

The diagram shows part of the student’s method.

(a) The table shows the student’s results for the volumes of gelatine digested at pH 4.

(i) Calculate the mean volume of gelatine digested in cm³.

Give your answer to two decimal places.

(ii) State what substances are produced when the gelatine protein is digested.

(b) The student repeats the investigation with different pH buffers.

The table shows their results.

(i) Give two variables the student should control.

(ii) Explain the effect of changing the pH on the mean volume of gelatine digested.

(c) Describe how to test for the presence of protein.

Most-appropriate topic codes (Edexcel IGCSE Biology):

• 2(c): Biological molecules — part (a)(ii), (c)
• 2(e): Nutrition — part (a)(ii)
• Appendix 4: Mathematical skills — part (a)(i)
• 2.10, 2.11, 2.13: Enzymes as biological catalysts; effect of temperature and pH — part (b)(ii)
• 2.14B: Practical: investigate how enzyme activity can be affected by changes in pH — part (b)

▶️ Answer/Explanation

Solution

(a)(i) Mean volume = 0.57 cm³

Explanation: To calculate the mean volume of gelatine digested, we first need to identify and exclude any anomalous results. Looking at the data, tube 2 shows a value of 1.89 cm³, which is significantly higher than the other three values (0.55, 0.54, 0.61). This suggests it might be an anomaly, possibly due to experimental error. Therefore, we calculate the mean using the three consistent results:

Sum of volumes = 0.55 + 0.54 + 0.61 = 1.70 cm³

Number of tubes = 3

Mean volume = 1.70 ÷ 3 = 0.5666… cm³

Rounded to two decimal places, this gives us 0.57 cm³.

(a)(ii) Amino acids and peptides

Explanation: Gelatine is a protein. During digestion, proteases like bromelain break down proteins through hydrolysis. The initial breakdown produces smaller polypeptide chains called peptides. Further digestion breaks these peptides down into their monomer units, which are amino acids. Therefore, the end products of protein digestion are primarily amino acids, though some intermediate peptides may also be present.

(b)(i) Two controlled variables:

1. Temperature – The investigation should be conducted at a constant temperature, ideally 37°C as used initially, because enzyme activity is highly dependent on temperature.

2. Volume/Concentration of enzyme (bromelain/pineapple juice) – The same volume and concentration of pineapple juice should be used in each test to ensure the same amount of enzyme is present, making the comparison of pH effects valid.

Other acceptable answers include: volume/mass/concentration of gelatine, volume of buffer, time left in the water bath, or surface area of gelatine.

(b)(ii) Effect of pH:

Explanation: The data shows that the mean volume of gelatine digested increases as the pH moves from 3 to 5, reaching a maximum at pH 5 (0.98 cm³). Beyond pH 5, the volume digested decreases sharply, with very little digestion occurring at pH 11 (0.01 cm³). This pattern is characteristic of enzyme activity. Enzymes have an optimal pH at which they function most efficiently. For bromelain, this appears to be around pH 5. At pH values significantly above or below this optimum, the enzyme’s active site becomes denatured. Denaturation means the shape of the active site changes, so the substrate (gelatine protein) can no longer bind effectively. Consequently, the rate of reaction (digestion) decreases, leading to a smaller volume of gelatine being digested.

(c) Testing for Protein:

Explanation: The standard test for proteins is the Biuret test. Here is a step-by-step description of how to perform it:

Place the sample to be tested in a test tube. If the sample is solid, it should first be crushed and mixed with water to create a liquid suspension.
Add an equal volume of sodium hydroxide solution (Biuret reagent A) to the test tube. Shake gently to mix.
Then, add a few drops of very dilute copper(II) sulfate solution (Biuret reagent B). Do not shake the tube vigorously.
Observe the colour change. A positive result for protein is indicated by a colour change from blue to lilac, purple, or violet. If the solution remains blue, no protein is present.

The principle behind this test is that the copper ions in the reagent form a complex with the peptide bonds in the proteins, producing the characteristic purple colour.

Question

Catalase is an enzyme found in many cells.

It speeds up the breakdown of hydrogen peroxide into water and oxygen.

The equation for the reaction is:

\[ 2H_2O_2 \rightarrow 2H_2O + O_2 \]

A teacher demonstrates the effect of increasing catalase concentration on the initial rate of the reaction.

This is the teacher’s method:

cut five equal size discs from a potato, each 0.2 mm thick
place the discs in a boiling tube with 5 cm³ of buffer solution
add 5 cm³ of hydrogen peroxide solution to the boiling tube
place a bung and delivery tube firmly into the boiling tube
position the other end of the delivery tube under an inverted measuring cylinder
start a timer as soon as the first bubble of oxygen enters the measuring cylinder
measure the volume of oxygen produced in one minute

Repeat this method three times.

The teacher uses this method with different numbers of potato discs, making sure that other conditions are unchanged.

(a) Give the expected relationship between the named independent variable and the named dependent variable in this demonstration.

(b) Give two variables that the teacher controls in this demonstration.

(c) The table shows the teacher’s results.

(i) Calculate the mean volume of oxygen produced in one minute using 5 potato discs.

(ii) Calculate the percentage increase in mean volume of oxygen produced in one minute as the concentration of enzyme changes from 15 to 20 discs.

(iii) Explain the relationship between the concentration of enzyme and mean volume of oxygen produced in one minute.

(d) Suggest why the teacher measures the volume of oxygen after the first minute of the reaction rather than after 10 minutes.

Most-appropriate topic codes (Edexcel IGCSE Biology):

• 2(c): Biological molecules — enzymes as biological catalysts
• 2.12 practical: Investigate how enzyme activity can be affected by changes in temperature — part (a), (b), (c), (d)
• 2.14B practical: Investigate how enzyme activity can be affected by changes in pH — context of controlled variables
• Appendix 4: Mathematical skills — calculations of mean and percentage increase part (c)(i), (c)(ii)
• Appendix 5: Command word taxonomy — “explain”, “suggest” part (c)(iii), (d)

▶️ Answer/Explanation

Solution

(a) As the enzyme concentration (number of potato discs) increases, the initial rate of reaction (volume of oxygen produced in one minute) increases.

Explanation: The independent variable is the enzyme concentration, which is represented by the number of potato discs used. The dependent variable is the rate of reaction, measured by the volume of oxygen gas produced in the first minute. According to enzyme kinetics, increasing the enzyme concentration provides more active sites for the substrate molecules to bind to, which leads to more enzyme-substrate complexes forming per unit time. This results in a higher rate of reaction and therefore more oxygen produced in the same time period.

(b) Any two from: pH, volume of hydrogen peroxide, temperature, time, size/mass of potato discs.

Explanation: In a scientific investigation, it’s crucial to control variables that could affect the results. The teacher keeps these factors constant to ensure that any changes in the volume of oxygen produced are due only to changes in the enzyme concentration (number of potato discs) and not other factors. For example:

pH: Maintained using a buffer solution to keep the enzyme at its optimal pH.
Volume of hydrogen peroxide: Kept constant at 5 cm³ to ensure the same amount of substrate is available.
Temperature: Controlled using a water bath as enzyme activity is temperature-sensitive.
Size of potato discs: Kept equal to ensure the same surface area and enzyme quantity per disc.

(c)(i) Mean = 1.4 cm³

Explanation: To calculate the mean volume for 5 discs, we add the four readings: 1.2 + 1.5 + 0.0 + 1.5 = 4.2 cm³. However, the reading of 0.0 cm³ is likely an outlier (possibly an experimental error). Following the mark scheme guidance, we should use only the three valid readings: 1.2, 1.5, and 1.5. Adding these gives 4.2 cm³, and dividing by 3 gives a mean of 1.4 cm³.

(c)(ii) Percentage increase = 13.7%

Explanation: The percentage increase is calculated using the formula: \[ \frac{\text{New Value} – \text{Original Value}}{\text{Original Value}} \times 100\% \] For enzyme concentration increasing from 15 to 20 discs: \[ \frac{8.3 – 7.3}{7.3} \times 100\% = \frac{1.0}{7.3} \times 100\% = 13.7\% \] This shows that increasing the enzyme concentration by 5 discs resulted in a 13.7% increase in oxygen production.

(c)(iii) As enzyme concentration increases, the volume of oxygen produced increases because there are more active sites available, leading to more frequent collisions and more enzyme-substrate complexes forming.

Explanation: The relationship is directly proportional up to a point. With more potato discs, there’s more catalase enzyme present. Each enzyme molecule has active sites where hydrogen peroxide molecules can bind. When there are more enzymes, there are more active sites available for substrates to bind to, which increases the frequency of successful collisions. This results in more enzyme-substrate complexes forming per unit time, leading to a higher rate of reaction and therefore more oxygen gas being produced in the first minute. However, this relationship may eventually plateau when substrate concentration becomes the limiting factor.

(d) The teacher measures oxygen after one minute to determine the initial rate of reaction when the reaction is fastest, before the substrate concentration decreases significantly.

Explanation: There are two main reasons for measuring oxygen production in the first minute rather than after 10 minutes:

The initial rate is the fastest and most reliable for comparison because at the start of the reaction, the substrate concentration is at its highest. As the reaction proceeds, hydrogen peroxide gets used up, becoming the limiting factor, which slows down the reaction rate.
Measuring after 10 minutes would give the total oxygen produced but not the rate. The reaction might have completed or slowed significantly by 10 minutes, making comparisons between different enzyme concentrations less meaningful.

By measuring in the first minute, the teacher ensures they’re comparing the maximum reaction rates under different enzyme concentrations, which provides valid data for analyzing the effect of enzyme concentration on reaction rate.

Question

Micropropagation is used to produce plant clones.

The process involves growing explants in vitro.

(a) State what is meant by the term in vitro.

(b) The explants grow new roots and shoots.

A student investigates the effect of pH on the growth of new shoots.

The table shows the student’s results.

(i) Explain the relationship between pH and the mean number of shoots per explant.

(ii) Describe a procedure the student could use to obtain explants and produce these results.

(c) Give two benefits of using micropropagation to produce new plants rather than using seeds.

Most-appropriate topic codes (Edexcel IGCSE Biology):

• 5(d): Cloning — parts (a), (c)
• 2(b): Cell structure — part (a) in vitro context
• 2(c): Biological molecules (enzymes) — part (b)(i) enzyme activity & pH
• Appendix 6: Suggested practical investigations — part (b)(ii) experimental design

▶️ Answer/Explanation

Solution

(a) in glass / in test tube / in (Petri) dish

Explanation: The term “in vitro” is a Latin phrase that literally means “in glass.” In biology, it refers to experiments or processes that are conducted outside of a living organism, in an artificial environment controlled by a scientist, such as a test tube, Petri dish, or flask. This is the opposite of “in vivo,” which means experiments conducted within a living organism.

(b)(i)

Answer: The mean number of shoots increases as the pH increases from 4.5 to 6.0, and then it decreases at pH 6.5. This pattern occurs due to the effect of pH on enzyme activity, with an optimum pH around 6.0 for the enzymes involved in shoot growth.

Explanation: When we look at the data, we can see a clear trend. At a very acidic pH of 4.5, the mean number of shoots is low (3.0). As the pH becomes less acidic and moves towards neutral, the number of shoots increases, reaching a maximum of 6.4 at pH 6.0. However, when the pH is increased further to 6.5, the number of shoots drops to 4.3. This pattern is classic for enzyme-controlled processes. The enzymes responsible for cell division and growth in the plant tissues have an optimal pH at which they work most efficiently, which appears to be pH 6.0 in this case. At pH values above and below this optimum, the enzymes become less active (they may denature at extremes), leading to reduced shoot formation.

(b)(ii)

Answer: The student should use sterile techniques to obtain and prepare the explants. Small pieces of plant tissue (explants) are cut from a parent plant using a sterilized scalpel. These explants are then surface-sterilized by washing them in a disinfectant like bleach or ethanol to kill any microorganisms. The sterile explants are placed on a nutrient agar gel in Petri dishes or test tubes. The agar contains essential nutrients, minerals, and plant hormones to support growth. The student would prepare several identical agar plates, each buffered to a specific pH (4.5, 5.0, 5.5, 6.0, 6.5). Multiple explants are placed in each pH condition to ensure the results are reliable. All plates are kept in a controlled environment (e.g., constant temperature and light) for a set period. After this time, the number of new shoots on each explant is counted, and a mean is calculated for each pH level.

Explanation: To get valid and reliable results, the procedure must be very careful and controlled. The key steps involve:

Obtaining Explants: Using a sterile tool like a scalpel or forceps to take small tissue samples from the same part of the same plant species to keep the starting material consistent.
Sterilization: Washing the explants in a disinfectant is crucial. Any bacteria or fungi present would contaminate the nutrient agar and compete with the plant tissue, ruining the experiment.
Growing Medium: The explants are placed on a solid agar medium. This agar is not just a solidifier; it’s a “growth cocktail” containing sugars for energy, mineral ions, vitamins, and plant growth regulators (hormones) like auxins and cytokinins that stimulate shoot and root development.
Controlling Variables: The pH is the independent variable, so it is deliberately changed for each set of plates. All other factors that could affect growth, such as temperature, light intensity, and the composition of the agar (except for pH), must be kept the same for all explants. Using multiple explants (repeats) at each pH allows the student to calculate a mean, making the results more trustworthy and accounting for natural variation between individual explants.
Data Collection: After a predetermined growth period, the student counts the number of shoots that have developed from each explant and calculates the average for each pH group.

(c)

Answer:

It produces genetically identical plants (clones), ensuring desirable characteristics from the parent plant are preserved.
It allows for the rapid production of a large number of plants in a relatively short time, independent of seasonal constraints.

Explanation:

Benefit 1: Genetic Uniformity When you grow plants from seeds, the offspring show genetic variation due to sexual reproduction (cross-pollination). This means the new plants might not have the exact same desirable traits as the parent plant, such as specific flower color, fruit taste, or disease resistance. Micropropagation, however, is an asexual process. All the new plants are clones, meaning they are genetically identical to the original parent plant. This is extremely valuable for farmers and horticulturists who want to propagate a specific cultivar with known, superior qualities reliably.

Benefit 2: Speed and Season Independence Micropropagation can produce a vast number of plants from a single piece of tissue much faster than waiting for seeds to germinate and grow. A single explant can be induced to produce multiple shoots, and each of those shoots can then be divided and cultured again, leading to an exponential increase in plant numbers. Furthermore, this process is carried out in a lab under controlled conditions, meaning it is not dependent on seasons or weather. Plants can be produced all year round, which is not always possible with seeds that may have specific germination requirements.

Other potential benefits include: producing plants that are difficult to grow from seed, and conserving rare or endangered plant species.

Question

Decomposer bacteria are involved in the nitrogen cycle.

The bacteria release an enzyme called urease.

(a) The diagram shows part of one strand of DNA used to make urease.

Complete the diagram by giving the missing bases on the other strand of DNA.

(b) Urease acts on urine to produce ammonia.

The graph shows how pH affects the activity of urease.

(i) Which of these is the optimum pH for urease?

A. 2.5
B. 4.5
C. 7.5
D. 8.5

(ii) Explain the activity of urease at pH 8.5

(c) Describe the role of the other bacteria involved in the nitrogen cycle.

Most-appropriate topic codes (Edexcel IGCSE Biology):

• 3(b): Inheritance — DNA structure and base pairing (part a)
• 2(c): Biological molecules — Enzymes and factors affecting enzyme activity (parts b(i), b(ii))
• 4(c): Cycles within ecosystems — The nitrogen cycle and roles of bacteria (part c)

▶️ Answer/Explanation

Solution

(a)

Explanation: DNA strands are complementary. Adenine (A) always pairs with Thymine (T), and Guanine (G) always pairs with Cytosine (C). Therefore, to complete the complementary strand, we match A with T, T with A, G with C, G with C, C with G, and T with A.

(b)(i) C 7.5

Explanation: The optimum pH is the point at which the enzyme shows its highest activity. From the graph (which would show this visually), urease activity peaks at around pH 7.5, making it the optimum pH for this enzyme’s function.

(b)(ii) At pH 8.5, the activity of urease is lower than its optimum. This happens because the higher pH (alkaline conditions) can cause the enzyme to denature. Denaturation means the enzyme’s active site changes shape, so it can no longer bind effectively to its substrate (urea in this case), resulting in fewer enzyme-substrate complexes and reduced activity.

Explanation: Enzymes are proteins that are sensitive to pH changes. Each enzyme has an optimum pH where its structure is ideal for catalysis. When the pH moves away from this optimum, particularly to extremes like pH 8.5 for urease, the ionic bonds and hydrogen bonds that maintain the enzyme’s specific three-dimensional shape can be disrupted. This alteration in shape, especially of the active site, prevents the substrate from fitting properly, drastically reducing the rate of reaction.

(c) Other bacteria in the nitrogen cycle have crucial roles:

Nitrogen-fixing bacteria: These convert atmospheric nitrogen gas (\(N_2\)) into ammonia (\(NH_3\)), making nitrogen available to plants in a usable form.
Nitrifying bacteria: This is a two-step process. First, some bacteria convert ammonia (\(NH_3\)) into nitrites (\(NO_2^-\)). Then, other bacteria convert these nitrites (\(NO_2^-\)) into nitrates (\(NO_3^-\)), which is the form most easily absorbed by plant roots.
Denitrifying bacteria: These bacteria perform denitrification, which is the conversion of nitrates (\(NO_3^-\)) back into nitrogen gas (\(N_2\)). This process releases nitrogen back into the atmosphere, completing the cycle.

Explanation: The nitrogen cycle is essential for life, as nitrogen is a key component of amino acids and nucleic acids. Different groups of bacteria drive this cycle. Nitrogen-fixing bacteria, often found in root nodules of legumes, “fix” inert atmospheric nitrogen into reactive ammonia. Nitrifying bacteria in the soil then oxidize ammonia to nitrites and then to nitrates, which are soluble and can be taken up by plants. Finally, denitrifying bacteria, typically active in waterlogged, anaerobic soils, reduce nitrates back to nitrogen gas, returning it to the atmosphere and balancing the cycle. Without these bacterial processes, the vast reservoir of nitrogen in the air would be largely inaccessible to living organisms.

Question

Plant root hair cells absorb water from the soil by osmosis.

(a) (i) Explain how the structure of a root hair cell is adapted to absorb water.

(ii) Give one difference between osmosis and diffusion.

(b) A student investigates the effect of light on the volume of water taken up and lost by a plant shoot in one hour.
The table shows the student’s results.

(i) Explain these results.

(ii) Give two abiotic variables the student should control.

(c) Another student uses this apparatus and a stop clock to find the mean (average) rate of water taken up by a plant shoot.

(i) Name the apparatus used by the student.

(ii) Describe how the student could use this apparatus to find the mean rate of water taken up by the plant.

Most-appropriate topic codes (Edexcel IGCSE Biology):

• 2(d): Movement of substances into and out of cells — parts (a)(i), (a)(ii)
• 2(h): Transport in flowering plants — parts (a)(i), (b)(i), (c)(i), (c)(ii)
• 4(a): The organism in the environment — part (b)(ii)
• 2(c): Biological molecules — part (b)(i) [water transport linked to plant metabolism]

▶️ Answer/Explanation

Solution

(a)(i) The root hair cell has a long, thin extension (root hair) that increases its surface area, allowing for more efficient absorption of water from the soil.

Explanation: Root hair cells are specialized for absorption. Their elongated, hair-like projections significantly increase the surface area in contact with the soil water. This larger surface area maximizes the rate at which water can be absorbed via osmosis.

(a)(ii) Osmosis specifically involves the movement of water molecules, while diffusion can involve the movement of any type of molecule or ion.

Explanation: The key distinction is the substance being moved. Osmosis is a special case of diffusion that is exclusively concerned with the passive movement of water molecules across a partially permeable membrane from a region of higher water potential to a region of lower water potential. Diffusion, on the other hand, refers to the net movement of any particles (like oxygen, carbon dioxide, or ions) from a region of higher concentration to a region of lower concentration, and it may or may not involve a membrane.

(b)(i) In the light, both water uptake and water loss are much greater than in the dark. More water is taken up than is lost in both conditions.

Explanation: In the light, the plant’s stomata are open to allow gas exchange for photosynthesis. This opening also increases the rate of transpiration (water loss) from the leaves. The loss of water by transpiration creates a transpiration pull, which draws more water up through the xylem from the roots, leading to the higher uptake. The small difference between uptake and loss (e.g., 10.2 – 9.1 = 1.1 cm³ in light) represents water used for processes like photosynthesis and maintaining cell turgor. In the dark, stomata are mostly closed, drastically reducing both transpiration and, consequently, water uptake.

(b)(ii) 1. Temperature
2. Humidity

Explanation: To ensure a fair test and that the results are solely due to the change in light intensity, other abiotic (non-living) factors that affect transpiration and water uptake must be kept constant. Temperature influences the rate of evaporation. Humidity affects the concentration gradient for water vapor loss; lower humidity increases transpiration. Other valid answers include air movement (wind) and the time allowed for the experiment.

(c)(i) Potometer

Explanation: The apparatus shown in the diagram, designed to measure the rate of water uptake by a plant shoot, is called a potometer. It is important to note that it actually measures the rate of water uptake, which is assumed to be closely related to the rate of transpiration.

(c)(ii) The student should introduce an air bubble into the capillary tube and use the stop clock to measure the time taken for the bubble to move a certain distance along the scale. The distance moved is converted to a volume using the known cross-sectional area of the tube. The rate is calculated as volume divided by time, and the experiment is repeated to find a mean rate.

Detailed Description:
1. Set up the potometer with the plant shoot underwater to ensure no air enters the system.
2. Introduce a single air bubble into the capillary tube.
3. Start the stop clock as the bubble passes a starting point on the scale.
4. Stop the stop clock when the bubble passes a finishing point, and record the time taken.
5. Measure the distance the bubble traveled along the scale.
6. Since the capillary tube has a uniform diameter, the volume of water taken up is equal to the distance moved multiplied by the cross-sectional area of the tube (volume = πr² × distance, where r is the radius).
7. Calculate the rate of water uptake for that single run using: Rate = Volume / Time.
8. Reset the bubble using the reservoir (if available) and repeat the process several times.
9. Calculate the mean (average) rate of water uptake from all the repeat readings to improve reliability.

Question

Blood clotting is an important process in humans. The process is controlled by enzymes.

(a) (i) Give two reasons why blood clotting is important.

(ii) The optimum temperature for the enzymes involved in blood clotting is \(37^\circ C\). Sketch a graph to show how temperature affects the time taken for blood to clot.

(b) Some people cannot make the proteins needed for blood clotting. Cloning is used to produce large numbers of transgenic mammals. These transgenic mammals can make the human blood-clotting proteins. The human blood-clotting proteins can then be removed from the mammals’ milk and injected into people who cannot make the proteins.

(i) Explain why these mammals are described as transgenic.
(ii) Describe how a mammal is cloned.

Most-appropriate topic codes (Edexcel IGCSE Biology):

• 2(h): Transport — Blood clotting (part a(i), a(ii))
• 2(c): Biological molecules — Enzymes (part a(ii))
• 5(c): Genetic modification (genetic engineering) — Transgenic organisms (part b(i))
• 5(d): Cloning — Cloning mammals (part b(ii))

▶️ Answer/Explanation

Solution

(a)(i) Two reasons from:

Prevent loss of blood / stops bleeding (1)
Prevent entry of pathogens / microbes / bacteria / viruses / fungi / prevent infections (1)

(a)(ii) Sketch graph showing a clear minimum (fastest clotting time) at \(37^\circ C\). The line should drop to a low point at \(37^\circ C\) and then rise on both sides.

(b)(i) An explanation that makes reference to:

(They have been given) genetic material / gene / allele / DNA / are genetically altered (1)
From human / a different species (1)

(b)(ii) An answer that makes reference to six of the following points:

Use enucleated egg / empty egg / remove nucleus from egg / eq (1)
Nucleus from body cell / diploid nucleus (placed into empty egg) / fuse adult cell with empty egg (1) (Ignore DNA)
Use of electricity / shock (to fuse cells) (1)
Cell division / mitosis (stimulated) (1)
Embryo forms / develops (1)
Embryo placed into uterus / womb (1)
Surrogate mother (carries embryo to term) (1)

Award marks for any correct and relevant points describing the process of somatic cell nuclear transfer as used in cloning mammals (e.g., Dolly the sheep).

Question

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?

A) asexual reproduction
B) insect pollination
C) mutation
D) wind pollination

(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:

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

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

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

Question

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:

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

Edexcel iGCSE Biology 4BI1 - Paper 2B -Biological molecules- Exam Style Questions- New Syllabus

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