igcse-edexcel-biology-4bi1-2b-que-2023-Nov

Question 1

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

Plants to the rescue

Since the early 20th century, the concentration of carbon dioxide in the atmosphere has increased rapidly. This has further increased in recent years due to more cars and the increased demand for electricity for homes and industry.

Scientists have found that plants play a critical part in removing this excess carbon dioxide from the atmosphere. Using computer models, the scientists concluded that photosynthesis has increased by 30 per cent.

The scientists measured carbonyl sulfide found in ice cores and air samples. In addition to carbon dioxide, plants take in carbonyl sulfide gas during their natural carbon cycle, and this is frequently used as a measure of photosynthesis on a global scale. Terrestrial plants are removing about 29 per cent of carbon dioxide emissions that would otherwise contribute to an increase in the atmospheric carbon dioxide concentration.

A carbon sink is an ecosystem, such as a forest, that absorbs more carbon dioxide than it releases. The size of the carbon sink depends on the rate of photosynthesis but also on the levels of deforestation and respiration. The model the scientists used showed that the role of photosynthesis in producing a carbon sink in land plants is larger than estimated in most other models.

Other scientists are less confident about using carbonyl sulfide as a measure of photosynthesis. Plant absorption of carbonyl sulfide can vary depending upon the amount of light the plants receive. Therefore, the measure of global photosynthesis could be overestimated.

Regardless of the rate at which photosynthesis has increased, scientists agree that excess carbon dioxide is boosting the growth of plants. Trees are becoming leafier, and there is more wood. The wood is where most of the carbon is stored in the plant.

In experimental research, scientists exposed plants to double the normal concentration of carbon dioxide found in the atmosphere. Under these increased carbon dioxide conditions, the composition of the leaf tissues changed. This made the leaves tougher for herbivores to eat and made it harder for insect larvae to grow.

Scientists have also observed that when plants are exposed to increasing levels of carbon dioxide, the size of the stomatal pores on a leaf increases.

(a) Explain why more cars would result in an increase in atmospheric carbon dioxide. (Lines 1-3)

(b) Explain how increased carbon dioxide can cause climate change.

(c) Explain why increasing carbon dioxide concentration can increase photosynthesis in plants.

(d) Explain why the carbon sink depends upon respiration and deforestation as well as photosynthesis. (Lines 13 to 15)

(e) Light affects gas exchange in leaves. (Lines 19 and 20)
Describe an experiment that shows how the net gas exchange from a leaf varies with light intensity.

(f) The photograph shows stomata on a small square of the lower surface of a leaf.

The total area of the lower surface of the leaf is \(150 \text{ cm}^2\).
Using the photograph and the total lower surface area of the leaf, estimate the number of stomata on the lower surface of the leaf.

(g) Stomata also have a role in water transport in the plant. Explain the role of stomata in water transport.

Most-appropriate topic codes (Edexcel IGCSE Biology):

• 2(e): Nutrition (Photosynthesis) — parts (c), (e)
• 5(d): Human influences on the environment — parts (a), (b)
• 5(c): Cycles within ecosystems (Carbon cycle) — part (d)
• 2(g): Gas exchange — parts (e), (f), (g)
• 3(h): Transport in plants — part (g)

▶️ Answer/Explanation

Solution

(a)
An explanation that makes reference to:
• Burning / combustion of petrol / diesel / fuel in car engines. (1 mark)
• This combustion reaction releases carbon dioxide (\( \text{C} + \text{O}_2 \rightarrow \text{CO}_2 \)). (1 mark)
More cars mean more fuel burned, directly increasing \( \text{CO}_2 \) emissions.

(b)
An explanation that makes reference to two of the following:
• Carbon dioxide is a greenhouse gas. (1 mark)
• It traps / absorbs infrared (IR) radiation (heat) from the Earth, preventing its escape into space. (1 mark)
• This leads to an enhanced greenhouse effect, causing global warming / climate change. (1 mark)
(Maximum 2 marks)

(c)
An explanation that makes reference to:
• Carbon dioxide is a reactant / raw material needed for photosynthesis. (1 mark)
• At lower concentrations, \( \text{CO}_2 \) can be a limiting factor for photosynthesis; increasing its concentration can increase the rate up to a point. (1 mark)
The photosynthesis equation is: \( 6\text{CO}_2 + 6\text{H}_2\text{O} \xrightarrow{\text{light}} \text{C}_6\text{H}_{12}\text{O}_6 + 6\text{O}_2 \).

(d)
An explanation that makes reference to two of the following:
• A carbon sink absorbs more \( \text{CO}_2 \) than it releases. (1 mark)
• Respiration (by plants, animals, decomposers) releases \( \text{CO}_2 \) back into the atmosphere, reducing net absorption. (1 mark)
• Deforestation (cutting down trees) reduces the number of plants for photosynthesis and often involves burning/decay, releasing stored carbon as \( \text{CO}_2 \). (1 mark)
(Maximum 2 marks)

(e)
A description that makes reference to three of the following:
• Place leaves (or aquatic plants like pondweed) in test tubes containing hydrogen-carbonate indicator. (1 mark)
• Expose one setup to bright light and another to darkness (or vary light intensity using a lamp at different distances). (1 mark)
• In bright light, the indicator turns purple/dark red (due to net \( \text{CO}_2 \) uptake in photosynthesis). In darkness, it turns yellow (due to net \( \text{CO}_2 \) release from respiration). (1 mark)
• Control other variables: use leaves of the same species, size, age; same volume and concentration of indicator; same temperature; same time period. (1 mark)
(Maximum 3 marks)

(f)
Step-by-step calculation:
1. Area in photograph: side = \( 400 \mu\text{m} = 0.4 \text{ mm} = 0.04 \text{ cm} \).
2. Area of square = \( (0.04 \text{ cm})^2 = 0.0016 \text{ cm}^2 \).
3. Number of stomata in this area = 2.
4. Stomatal density = \( \frac{2}{0.0016} = 1250 \) stomata per \( \text{cm}^2 \).
5. Total number on leaf = \( 1250 \times 150 = 187500 \).
Answer: \( \mathbf{187500} \) stomata. (3 marks)

(g)
An explanation that makes reference to:
• Stomata allow water vapour to evaporate / be lost from the leaf in a process called transpiration. (1 mark)
• This transpiration pull creates a tension / negative pressure in the xylem, drawing a continuous column of water up from the roots, through the stem, and into the leaves. (1 mark)
This is known as the transpiration stream, and it relies on stomatal opening for water movement against gravity.

Question 2

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 3

(a) The diagram shows a nephron from a human kidney with some structures labelled.

(a) (i) Which structure is the Bowman’s capsule?

A) P
B) R
C) S
D) T

(ii) Which structure is the loop of Henle?

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

(iii) Which structure is affected by ADH?

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

(b) (i) Blood plasma contains much glucose, but urine normally does not. Explain what happens to glucose in the kidney.
(ii) Describe how a sample of urine could be tested for glucose.

(c) As a person sweats, they may become dehydrated.
Describe the changes in a person’s urine if they become dehydrated.

Most-appropriate topic codes (Edexcel IGCSE Biology):

• 3(i): Excretion (Humans) — parts (a)(i), (a)(ii), (a)(iii), (b)(i), (c)
• 2(b): Cell structure / Biological molecules — part (b)(ii) link to food tests
• 3(j): Co-ordination and response (Homeostasis) — part (a)(iii), (c) link

▶️ Answer/Explanation

Solution

(a)(i) C (S)
Explanation: The Bowman’s capsule (S) is the cup-like sac at the beginning of the nephron that surrounds the glomerulus and receives the filtrate.

(a)(ii) B (Q)
Explanation: The loop of Henle (Q) is a U-shaped tubule that descends into and ascends from the medulla of the kidney. It is crucial for creating a concentration gradient for water reabsorption.

(a)(iii) A (P)
Explanation: ADH (Antidiuretic Hormone) affects the collecting duct (P). ADH increases the permeability of the collecting duct walls to water, allowing more water to be reabsorbed back into the blood, producing more concentrated urine.

(b)(i) An explanation that makes reference to three of the following points:
• Glucose passes from the blood in the glomerulus (R) into the Bowman’s capsule / renal capsule (S) during ultrafiltration. (1)
• (All) glucose is (then) reabsorbed / absorbed back into the blood / eq. (1)
• This reabsorption occurs in the proximal convoluted tubule / PCT (T). (1)
• It is reabsorbed by active transport (which requires energy). (1)
Explanation: During ultrafiltration, small molecules like glucose enter the nephron. The body cannot afford to lose this valuable energy source, so 100% of filtered glucose is normally reclaimed from the filtrate in the proximal convoluted tubule via active transport against its concentration gradient.

(b)(ii) A description that makes reference to two of the following points:
• Add Benedict’s solution to the urine sample (and heat). (1)
• A positive result is indicated by a colour change to green / yellow / orange / brick-red. (1)
Alternative: Use a test strip (e.g., Clinistix) which changes colour (e.g., to brown) in the presence of glucose. (1 each)
Explanation: Benedict’s test is a standard biochemical test for reducing sugars like glucose. Heating with Benedict’s reagent causes a reduction reaction, producing a coloured precipitate of copper(I) oxide.

(c) A description that makes reference to two of the following points:
• Less urine is produced / lower volume. (1)
• The urine becomes more concentrated / contains less water / appears darker in colour. (1)
• (It may contain) a higher concentration of urea / other solutes. (1)
Explanation: Dehydration lowers the water potential of the blood. This is detected by osmoreceptors, leading to increased secretion of ADH. ADH causes more water to be reabsorbed from the collecting duct back into the blood, conserving water. This results in a smaller volume of more concentrated, darker yellow urine.

Question 4

Mutation can cause changes in the enzymes produced during protein synthesis.

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

(b) Explain the different effects that a single base mutation can have on transcription and translation and the production and activity of an enzyme.

(c) Scientists investigate the rate of spontaneous mutations in yeast. They record the number of mutations per cell during mitosis and during meiosis. They count the mutations in three different strains of yeast and repeat each count. The mutation rates are counted per division per cell.

Yeast strain	Mutation rate \( \times 10^9 \) per division per cell
	mitosis	meiosis
5160	6.53	45.4
5160	4.86	29.6
5207	5.90	132.6
5207	5.03	43.2
5209	0.21	2.20
5209	0.28	1.42

(i) Calculate the mean mutation rate for strain 5207 during meiosis.

(ii) Suggest why the mutation rate is calculated per division per cell rather than just per cell.

(iii) Comment on the differences in rates of mutation for mitosis and meiosis in the different yeast strains.

Most-appropriate topic codes (Edexcel IGCSE Biology):

• 4(b): Inheritance — parts (a), (b), (c)(ii), (c)(iii)
• 4(b) B: DNA, protein synthesis, and mutation — parts (a), (b)
• Appendix 3: Mathematical skills — part (c)(i)

▶️ Answer/Explanation

Solution

(a) A mutation is a rare, random change in the DNA / genetic material (1). Mutations can be caused by factors like radiation or chemicals and can be inherited.

(b) An explanation that makes reference to the following points:
• During transcription, mRNA is produced from DNA (1). A single base mutation changes the codon / triplet / base sequence on the mRNA (1).
• During translation, a different tRNA with a different anticodon may bind, leading to a different amino acid being incorporated (1).
• This results in a different sequence of amino acids in the polypeptide chain (1).
• The final protein / enzyme may have a different structure / shape (1).
• If the enzyme’s active site is altered, it may no longer fit the substrate, reducing or stopping enzyme activity (1).
• However, due to the degenerate genetic code, the mutation may code for the same amino acid and have no effect on the protein (1).
Up to 5 marks available.

(c)(i)
• Values for strain 5207 during meiosis: \( 132.6 \) and \( 43.2 \).
• Sum = \( 132.6 + 43.2 = 175.8 \).
• Mean = \( \frac{175.8}{2} = 87.9 \).
• Mean mutation rate = \( \mathbf{87.9 \times 10^9} \) per division per cell (2).
Accept 88 for full marks.

(c)(ii) The rate is calculated per division per cell because meiosis involves two divisions (meiosis I and II), while mitosis involves only one division (1). This allows for a fair comparison between the two processes.

(c)(iii) Comment making reference to the following observations:
• In all strains, the mutation rate is higher during meiosis than during mitosis (1).
• The largest difference between mitosis and meiosis rates is seen in strain 5207 (1).
• The smallest difference is in strain 5209 (1).
• Strain 5209 shows the lowest mutation rates for both mitosis and meiosis (1).
• The highest mitosis rate is in strain 5160 (first reading 6.53) (1).
• The highest meiosis rate is in strain 5207 (first reading 132.6) (1).
• There is variation between the repeated measurements for each strain, indicating natural variability or experimental error (1).
Up to 4 marks available.

Question 5

The diagram shows the nitrogen cycle with some processes labelled.

(a) (i) Explain why plants need nitrate ions.
(ii) Name the processes \( V \), \( X \) and \( Y \).

(b) Scientists measure the dissolved nitrate levels in a river that passes through farmland.

They also measure the nitrogen compounds being used on the farmland either as fertiliser or as manure from livestock.

The graph shows how the nitrate levels in the river and the nitrogen compounds applied to the farmland changed from 1950 to 2005.

(i) Comment on the relationship between nitrate levels in the river and the nitrogen compounds applied to the farmland.

(ii) Describe the changes within the river that would have occurred between 1950 and 2000.

Most-appropriate topic codes (Edexcel IGCSE Biology):

• 5(c): Cycles within ecosystems — part (a)
• 5.11B: Describe the stages in the nitrogen cycle — part (a)(ii)
• 2.22: Understand that plants require mineral ions for growth — part (a)(i)
• 5(d): Human influences on the environment — parts (b)(i), (b)(ii)
• 5.17: Understand the biological consequences of eutrophication — part (b)(ii)
• 5.15: Understand how human activities contribute to greenhouse gases / pollution — part (b)(i)

▶️ Answer/Explanation

Solution

(a)(i)
Explanation: Plants need nitrate ions (\( NO_3^- \)) because:
1. Nitrates are a source of nitrogen, which is a key element needed to synthesize amino acids.
2. Amino acids are the building blocks for proteins, which are essential for plant growth, enzyme function, and cell structure.
Additional detail: Nitrogen is also required for making other important compounds like chlorophyll, DNA, and RNA.

(a)(ii)
Process Names:
• \( V \): Nitrogen fixation (carried out by nitrogen-fixing bacteria).
• \( X \): Nitrification (carried out by nitrifying bacteria, converting ammonium to nitrite then nitrate).
• \( Y \): Denitrification (carried out by denitrifying bacteria, converting nitrates back to nitrogen gas).

(b)(i)
Comment on the relationship:
1. There is a general positive correlation between the two variables from 1950 to 1970; as nitrogen application increased, river nitrate levels also increased.
2. From 1980 onwards, the relationship becomes less clear/more variable.
3. The amount of nitrogen applied to fields fluctuates more (shows greater variability) than the nitrate levels in the river, which remain relatively more stable.
4. This can be explained by factors such as:
   – Leaching of excess nitrate from fields into the river.
   – Variations in fertilizer application rates by farmers.
   – The river receiving nitrates from other sources (e.g., sewage, natural decay).

(b)(ii)
Describe changes in the river (eutrophication process):
1. Increased nitrate levels led to excessive algal growth / algal blooms (eutrophication).
2. The increase in algae and later their death led to a rise in decomposer bacteria that break down the organic matter.
3. These decomposers respire, using up oxygen, leading to reduced oxygen levels (anoxia) in the water.
4. This resulted in increased turbidity (cloudiness) and reduced light penetration, harming aquatic plants.
5. Ultimately, there was a loss of biodiversity, including death of fish and other oxygen-dependent organisms.

Question 6

(a) Describe how stem cells are different from other cell types.

(b) Stem cells from a donor can be used for stem cell therapy. Human blood cells are produced in bone marrow. Explain how using stem cells from the bone marrow enables doctors to treat a number of different blood conditions.

(c) One type of stem cell transplant uses a patient’s own stem cells. They can be used if the patient is producing enough healthy bone marrow cells. These cells can be collected, frozen and stored for later use. Explain the advantage of using the patient’s own stem cells rather than using stem cells from a donor.

Most-appropriate topic codes (Edexcel IGCSE Biology):

• 2(b): Cell structure — part (a)
• 2.5B: Cell differentiation in development of specialised cells — part (a)
• 2.6B: Advantages and disadvantages of using stem cells in medicine — parts (b) and (c)
• 3(h): Transport — part (b) (blood cell production)
• 3.12: Immune system response — part (c) (rejection)

▶️ Answer/Explanation

Solution

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

Undifferentiated / unspecialised / can differentiate / can become specialised cells (1)
And can become different cell types (1)
Can continue to divide / continue to multiply / can divide several times (1)

Note: “Can divide whilst remaining undifferentiated” scores 2 marks.

Explanation: Stem cells are unique because they are not yet specialized for a specific function (like muscle or nerve cells). They retain the ability to divide and produce more stem cells or differentiate into various specialized cell types when needed.

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

Can make different (blood) cells / red cells and white cells / red cells and platelets / white cells and platelets (1) (Accept any two types of blood cell)
Max two from:
- Red blood cells to treat anaemia / ensure oxygen transport around body (1) (Accept to treat sickle cell anaemia / thalassemia)
- White blood cells to kill infections / kill bacteria / remove viruses / improve immunity / increase antibodies (1)
- Platelets to treat blood clotting problems (1)
- Any one blood cell type to treat bone marrow cancer / treat leukaemia / treat lymphoma / treat myeloma / recover from chemotherapy (1)

Explanation: Bone marrow contains hematopoietic stem cells that can differentiate into all types of blood cells. This makes them extremely valuable in medicine. They can be transplanted to replace faulty or destroyed bone marrow, enabling the production of healthy red blood cells (to treat anaemia), white blood cells (to fight infections and restore immunity), and platelets (to aid clotting). This is crucial in treating conditions like leukaemia, lymphoma, or after high-dose chemotherapy.

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

Have same proteins / antigens (1)
Genetically identical / same genes (1)
No rejection / immune response / no need for immunosuppressant drugs / cells are accepted by body / cells are compatible / cells have same blood group (1)
No transfer of other disease / infections / viruses (1)

Explanation: Using a patient’s own stem cells (an autologous transplant) eliminates the risk of immune rejection because the cells are genetically identical to the patient’s own tissues. There is no need for powerful immunosuppressant drugs, which have serious side effects. Additionally, it removes the risk of transmitting infectious diseases from a donor. This makes the transplant safer and increases the likelihood of successful engraftment.

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