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Thinka Jan 2026 (V2) Cambridge International A Level-Style Mock — Biology (YBI11)

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An original Thinka practice paper modelled on the structure and difficulty of the Jan 2026 (V2) Cambridge International A Level Biology (YBI11) paper. Not affiliated with or reproduced from Cambridge.

PastPaper.section WBI11/01A - Unit 1: Molecules, Diet, Transport and Health

Answer ALL questions. Write your answers in the spaces provided in the Answer Book.
9 PastPaper.question · 80.04 PastPaper.marks
PastPaper.question 1 · multiple choice
8.88 PastPaper.marks
Which of the following statements correctly describes the structural differences between glycogen and amylopectin?
  1. A.Glycogen contains only \(\alpha\)-1,4-glycosidic bonds, whereas amylopectin contains both \(\alpha\)-1,4- and \(\alpha\)-1,6-glycosidic bonds.
  2. B.Glycogen has a higher proportion of \(\alpha\)-1,6-glycosidic bonds than amylopectin, making it more branched and more rapidly hydrolyzed.
  3. C.Amylopectin contains \(\beta\)-glucose monomers joined by \(\beta\)-glycosidic bonds, whereas glycogen contains \(\alpha\)-glucose monomers.
  4. D.Glycogen has a highly unbranched, linear structure that allows for dense storage, whereas amylopectin is highly branched.
PastPaper.showAnswers

PastPaper.workedSolution

Glycogen is highly branched compared to amylopectin because it contains a greater frequency of \(\alpha\)-1,6-glycosidic bonds. This highly branched structure provides more terminal ends, allowing for rapid enzyme hydrolysis to release glucose monomers when energy is needed.

PastPaper.markingScheme

Correct option is B. 8.88 marks awarded for selecting the correct answer.
PastPaper.question 2 · structured
8.88 PastPaper.marks
A clinical study investigated the effect of substituting saturated fatty acids (SFAs) with polyunsaturated fatty acids (PUFAs) on the risk of developing coronary heart disease (CHD). The relative risk (RR) of CHD for individuals on a high-SFA diet was 1.24. This decreased to 0.88 when SFAs were replaced with PUFAs. (a) Calculate the percentage decrease in relative risk when SFAs are replaced with PUFAs. Give your answer to two decimal places. (b) Explain the relationship between dietary SFAs, blood cholesterol levels, and the development of atherosclerosis.
PastPaper.showAnswers

PastPaper.workedSolution

(a) Decrease in Relative Risk = \(1.24 - 0.88 = 0.36\). Percentage decrease = \(\frac{0.36}{1.24} \times 100 = 29.032\%\). Expressed to two decimal places, this is \(29.03\%\). (b) Dietary SFAs increase blood concentrations of Low-Density Lipoproteins (LDLs). LDLs transport cholesterol from the liver to the body tissues. High blood LDL levels lead to excess cholesterol accumulating in the arterial wall at sites of endothelial damage. This initiates an inflammatory response, leading to the formation of plaque (atheroma), which narrows the lumen and hardens the artery wall (atherosclerosis).

PastPaper.markingScheme

Part (a): 3.00 marks total. Method: correct subtraction to find decrease of 0.36 (1 mark). Calculation of percentage: \(\frac{0.36}{1.24} \times 100\) (1 mark). Correct rounding to two decimal places: 29.03% (1 mark). Part (b): 5.88 marks total. Saturated fatty acids increase blood LDL cholesterol levels (1 mark). LDLs transport cholesterol in blood to arteries (1 mark). High LDL levels can lead to cholesterol deposition under damaged arterial endothelium (1 mark). Monocytes engulf cholesterol to form foam cells, leading to plaque/atheroma development (1 mark). This narrows the arterial lumen and restricts blood flow, causing atherosclerosis (1.88 marks).
PastPaper.question 3 · multiple choice
8.88 PastPaper.marks
During DNA replication, a template strand with the sequence 5'-ATGCGATC-3' is replicated. Which of the following represents the correct sequence of the complementary DNA strand synthesized during semi-conservative replication, written in the 5' to 3' direction?
  1. A.5'-TACGCTAG-3'
  2. B.5'-GATCGCAT-3'
  3. C.5'-UTCGCTUG-3'
  4. D.5'-CGATGCAT-3'
PastPaper.showAnswers

PastPaper.workedSolution

The template strand is 5'-ATGCGATC-3'. The complementary strand must run antiparallel, so its sequence in the 3' to 5' direction is 3'-TACGCTAG-5'. Rewriting this complementary strand in the 5' to 3' direction yields 5'-GATCGCAT-3'.

PastPaper.markingScheme

Correct option is B. 8.88 marks awarded for selecting the correct answer.
PastPaper.question 4 · structured
8.88 PastPaper.marks
Explain how the structure of the cell surface membrane restricts the movement of polar molecules, and describe how these molecules are transported across the membrane by facilitated diffusion and active transport.
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PastPaper.workedSolution

The cell surface membrane consists of a phospholipid bilayer where hydrophilic phosphate heads face outwards and hydrophobic fatty acid tails point inwards, forming a non-polar hydrophobic core. This hydrophobic core prevents polar, charged, or hydrophilic substances from dissolving in the lipid bilayer and crossing the membrane. Therefore, polar molecules require specialized transport proteins. Facilitated diffusion is a passive process that transport molecules down their concentration gradient using channel proteins or carrier proteins without using metabolic energy. Active transport uses carrier proteins to move polar molecules against their concentration gradient, which requires the hydrolysis of ATP to provide energy.

PastPaper.markingScheme

Hydrophobic core: Phospholipid bilayer has hydrophobic fatty acid tails facing inwards (1 mark). Polar barrier: Polar molecules are hydrophilic and cannot pass through the non-polar core (1 mark). Facilitated diffusion: Uses channel or carrier proteins (1 mark) to move molecules down their concentration gradient (1 mark). Passive nature: Facilitated diffusion does not require energy from ATP (1 mark). Active transport: Uses specific carrier proteins (1 mark) to move molecules against their concentration gradient (1 mark). ATP requirement: Active transport requires energy from the hydrolysis of ATP (1.88 marks).
PastPaper.question 5 · multiple choice
8.88 PastPaper.marks
An enzyme-catalysed reaction was studied at different temperatures. Which of the following statements correctly explains the effect of increasing temperature from \(20^\circ\text{C}\) to \(35^\circ\text{C}\) on the rate of reaction?
  1. A.The activation energy of the reaction is reduced, causing a greater proportion of substrate molecules to react.
  2. B.The kinetic energy of the molecules increases, leading to more frequent successful collisions between active sites and substrate molecules, forming more enzyme-substrate complexes.
  3. C.The tertiary structure of the enzyme active site changes shape permanently, increasing its complementary fit to the substrate.
  4. D.Hydrogen bonds and ionic bonds within the enzyme break, altering the shape of the active site so that the substrate can no longer bind.
PastPaper.showAnswers

PastPaper.workedSolution

Increasing temperature increases the kinetic energy of both the enzyme and substrate molecules. This increases their speed of movement, leading to a higher frequency of successful collisions. As a result, more enzyme-substrate complexes are formed per unit time, increasing the rate of reaction.

PastPaper.markingScheme

Correct option is B. 8.88 marks awarded for selecting the correct answer.
PastPaper.question 6 · structured
8.88 PastPaper.marks
Two organisms, A (a single-celled amoeba, spherical with radius \(r = 50\text{ }\mu\text{m}\)) and B (a spherical multicellular organism with radius \(r = 250\text{ }\mu\text{m}\)), rely on diffusion for gas exchange. (a) Calculate the surface area to volume ratio (SA:V ratio) for both organisms. Use the formulas: Surface Area of a sphere = \(4\pi r^2\) and Volume of a sphere = \(\frac{4}{3}\pi r^3\). (b) Explain, using Fick's Law of Diffusion, why organism B requires a specialized gas exchange surface and a circulatory system, whereas organism A does not.
PastPaper.showAnswers

PastPaper.workedSolution

(a) For a sphere, the SA:V ratio simplifies to \(\frac{3}{r}\). Organism A: SA:V ratio = \(\frac{3}{50} = 0.06\text{ }\mu\text{m}^{-1}\). Organism B: SA:V ratio = \(\frac{3}{250} = 0.012\text{ }\mu\text{m}^{-1}\). (b) According to Fick's Law, the rate of diffusion is directly proportional to (Surface Area \(\times\) Concentration Gradient) / Diffusion Distance. Organism A has a high SA:V ratio and a short diffusion distance, so simple diffusion is fast enough to meet its metabolic demands. Organism B has a much smaller SA:V ratio, meaning it has less relative surface area to exchange gases. Moreover, the diffusion distance to its inner cells is large, making simple diffusion too slow to sustain life. Therefore, organism B needs a specialized gas exchange surface (to maximize surface area) and a circulatory system (to transport oxygen rapidly and maintain a steep concentration gradient over short distances).

PastPaper.markingScheme

Part (a): 4.00 marks. Correct calculation of SA:V for Organism A as 0.06 (2 marks). Correct calculation of SA:V for Organism B as 0.012 (2 marks). Part (b): 4.88 marks. Reference to Fick's Law showing variables: area, concentration gradient, and diffusion distance (1 mark). Identification that Organism B has a lower SA:V ratio than A (1 mark). Identification that Organism B has a larger diffusion distance to internal tissues (1 mark). Explanation that a specialized respiratory surface increases surface area and a circulatory system maintains concentration gradients to keep diffusion rapid (1.88 marks).
PastPaper.question 7 · multiple choice
8.88 PastPaper.marks
Which of the following describes the correct sequence of events during the blood clotting cascade after an injury to a blood vessel?
  1. A.Thrombin converts prothrombin to thromboplastin, which catalyzes the breakdown of fibrin into fibrinogen.
  2. B.Thromboplastin released from damaged tissue and platelets catalyzes the conversion of prothrombin into active thrombin in the presence of calcium ions, which then converts soluble fibrinogen into insoluble fibrin.
  3. C.Platelets release fibrin, which acts as an enzyme to convert prothrombin to thrombin, leading to the polymerization of thromboplastin.
  4. D.Soluble fibrinogen is converted into insoluble fibrin by thromboplastin, which then activates platelets to release prothrombin.
PastPaper.showAnswers

PastPaper.workedSolution

When a blood vessel is damaged, platelets and damaged tissues release an enzyme called thromboplastin. In the presence of calcium ions, thromboplastin catalyzes the conversion of the inactive plasma protein prothrombin into the active enzyme thrombin. Thrombin then catalyzes the conversion of soluble fibrinogen into insoluble fibrin, which forms a mesh of fibers to trap blood cells and form a clot.

PastPaper.markingScheme

Correct option is B. 8.88 marks awarded for selecting the correct answer.
PastPaper.question 8 · structured
8.88 PastPaper.marks
Cystic fibrosis is caused by mutations in the CFTR gene, which codes for a chloride ion channel protein. (a) Explain how a deletion mutation of three nucleotides in the CFTR gene can result in a non-functional CFTR protein. (b) Describe how a non-functional CFTR protein affects the viscosity of mucus in the respiratory system and explain the consequences this has on gas exchange.
PastPaper.showAnswers

PastPaper.workedSolution

(a) A deletion of three nucleotides removes exactly one codon from the mRNA, which results in the loss of one specific amino acid from the primary structure of the CFTR protein. This change in primary structure alters the folding of the polypeptide chain, changing its final three-dimensional tertiary structure. The altered shape prevents the channel from opening or causes the protein to be misfolded and degraded. (b) A non-functional CFTR channel cannot pump chloride ions out of the epithelial cells into the mucus. As a result, water does not move into the mucus by osmosis to keep it hydrated; instead, water is continuously absorbed into the cells. This leaves the mucus dry, sticky, and highly viscous. This thick mucus cannot be easily swept away by the beating cilia, blocking the bronchioles and airways. This blocks airflow, reducing the effective surface area of alveoli available for gas exchange and increasing the diffusion distance, leading to reduced oxygen absorption.

PastPaper.markingScheme

Part (a): 3.88 marks. Deletion of 3 nucleotides removes one codon/amino acid (1 mark). Primary structure of the protein is altered (1 mark). Misfolding occurs, changing the tertiary structure of the protein (1.88 marks). Part (b): 5.00 marks. Chloride ions cannot be transported out of cells into mucus (1 mark). Water does not move into mucus by osmosis (1 mark). Mucus becomes thick and viscous (1 mark). Cilia cannot clear the sticky mucus, leading to blocked airways (1 mark). Blocked airways reduce surface area for gas exchange and increase diffusion distance (1 mark).
PastPaper.question 9 · Calculations
9 PastPaper.marks
An investigation was conducted to study the effect of a new therapeutic drug, PulmoClear, on the rate of oxygen diffusion across the alveolar-capillary membrane in patients with a chronic inflammatory lung condition.

The drug aims to restore lung function by improving gas exchange parameters. Clinical measurements of a patient were taken before and after a six-week course of PulmoClear.

Before treatment:
- The functional alveolar surface area was \(75.0\text{ m}^2\).
- The concentration gradient of oxygen across the alveolar-capillary membrane was \(5.30\text{ kPa}\).
- The mean thickness of the alveolar-capillary membrane was \(0.60\text{ }\mu\text{m}\).

After treatment:
- The functional alveolar surface area increased by \(15.0\%\).
- The mean thickness of the alveolar-capillary membrane was reduced by \(25.0\%\).
- The concentration gradient of oxygen across the membrane decreased by \(8.00\%\) due to improved systemic circulation and oxygenation.

Using Fick's Law of Diffusion:

\[\text{Rate of diffusion} \propto \frac{\text{Surface Area} \times \text{Difference in Concentration}}{\text{Thickness of Gas Exchange Membrane}}\]

Calculate the percentage change in the rate of diffusion of oxygen across the alveolar-capillary membrane after treatment compared to before treatment. Choose the option that represents the correct percentage change and physiological effect.
  1. A.5.8% increase
  2. B.41.1% increase
  3. C.29.2% increase
  4. D.76.3% increase
PastPaper.showAnswers

PastPaper.workedSolution

To calculate the percentage change in the rate of diffusion, we can use two different valid methods:

**Method 1: Using absolute values**
1. Calculate the initial rate of diffusion (\(R_1\)):
\[R_1 = \frac{75.0 \times 5.30}{0.60} = 662.5\text{ arbitrary units}\]

2. Calculate the post-treatment parameters:
- New surface area (\(A_2\)) = \(75.0 \times 1.15 = 86.25\text{ m}^2\)
- New concentration gradient (\(\Delta C_2\)) = \(5.30 \times (1 - 0.08) = 4.876\text{ kPa}\)
- New membrane thickness (\(T_2\)) = \(0.60 \times (1 - 0.25) = 0.45\text{ }\mu\text{m}\)

3. Calculate the post-treatment rate of diffusion (\(R_2\)):
\[R_2 = \frac{86.25 \times 4.876}{0.45} = \frac{420.555}{0.45} = 934.57\text{ arbitrary units}\]

4. Calculate the percentage change:
\[\text{Percentage Change} = \frac{R_2 - R_1}{R_1} \times 100\%\]
\[\text{Percentage Change} = \frac{934.57 - 662.5}{662.5} \times 100\% = \frac{272.07}{662.5} \times 100\% = 41.07\% \approx 41.1\%\text{ increase}\]

**Method 2: Using proportional factors (Ratios)**
1. Let the initial rate be represented by:
\[R_1 \propto \frac{A \times \Delta C}{T}\]

2. Express the post-treatment rate (\(R_2\)) as a proportion of the initial factors:
- Surface area increases by \(15.0\%\) \(\rightarrow 1.15 A\)
- Concentration gradient decreases by \(8.00\%\) \(\rightarrow 0.92 \Delta C\)
- Membrane thickness decreases by \(25.0\%\) \(\rightarrow 0.75 T\)

3. Calculate the new relative rate:
\[R_2 \propto \frac{(1.15 A) \times (0.92 \Delta C)}{0.75 T} = \frac{1.15 \times 0.92}{0.75} \times \left(\frac{A \times \Delta C}{T}\right) = 1.4107 \times R_1\]

4. Calculate the percentage change:
\[\text{Percentage Change} = (1.4107 - 1) \times 100\% = 41.1\%\text{ increase}\]

Therefore, the correct option is B.

PastPaper.markingScheme

- **Mark 1 (Method):** Correct recall/application of Fick's Law formula by setting up the initial rate calculation or the proportional relationship.
- **Mark 2 (Accuracy):** Calculating the correct value for initial rate \(R_1 = 662.5\) (or equivalent factor in a combined expression).
- **Mark 3 (Method/Accuracy):** Calculating the correct new surface area of \(86.25\text{ m}^2\) (or using multiplier of \(1.15\)).
- **Mark 4 (Method/Accuracy):** Calculating the correct new membrane thickness of \(0.45\text{ }\mu\text{m}\) (or using multiplier of \(0.75\)).
- **Mark 5 (Method/Accuracy):** Calculating the correct new concentration gradient of \(4.876\text{ kPa}\) (or using multiplier of \(0.92\)).
- **Mark 6 (Accuracy):** Calculating the correct value for the post-treatment rate \(R_2 = 934.57\) (or equivalent ratio component of \(1.4107\)).
- **Mark 7 (Method):** Using a valid percentage change formula: \(\frac{\text{final} - \text{initial}}{\text{initial}} \times 100\%\).
- **Mark 8 (Accuracy):** Obtaining the value of \(41.1\%\) (accept \(41\%\) or \(41.07\%\)).
- **Mark 9 (Conclusion):** Selecting Option B (stating it is a \(41.1\%\) increase).

PastPaper.section WBI12/01A - Unit 2: Cells, Development, Biodiversity and Conservation

Answer ALL questions. Write your answers in the spaces provided in the Answer Book.
8 PastPaper.question · 80 PastPaper.marks
PastPaper.question 1 · Structured
10 PastPaper.marks

Part (a) Explain how the arrangement of cellulose microfibrils in plant cell walls contributes to their strength. (3 marks)

Part (b) Compare and contrast the structure of xylem vessels with that of sclerenchyma fibres. (4 marks)

Part (c) A student investigated the tensile strength of hemp fibres. Explain how the student could ensure the investigation was valid and reliable. (3 marks)

PastPaper.showAnswers

PastPaper.workedSolution

Part (a) Cellulose molecules are long, unbranched chains of beta-glucose joined by 1,4-glycosidic bonds. Hydrogen bonds form between adjacent chains to form microfibrils. These microfibrils are laid down at different angles (criss-cross pattern) in a matrix of hemicellulose and pectins, providing high tensile strength and preventing stretching in all directions.

Part (b) Similarities: Both are made of dead cells, both are hollow with no cytoplasm, and both have cell walls thickened with lignin. Differences: Xylem vessels have open ends (vessel elements joined end-to-end to form a continuous column), whereas sclerenchyma fibres have tapered closed ends. Xylem vessels are specialized for transport of water and mineral ions, whereas sclerenchyma is solely for support.

Part (c) To ensure validity, control variables must be kept constant: the length/diameter of the fibres, temperature, and humidity. To ensure reliability, repeat the measurements for each fiber thickness/type multiple times (at least 5 replicates) and calculate a mean, identifying and excluding anomalies.

PastPaper.markingScheme

Part (a) [Max 3 marks]:
1. Cellulose consists of beta-glucose chains joined by 1,4-glycosidic bonds (1)
2. Hydrogen bonds form cross-links between adjacent cellulose chains to form microfibrils (1)
3. Microfibrils are arranged in a criss-cross pattern / net-like structure (embedded in pectin/hemicellulose) to resist tensile forces in multiple directions (1)

Part (b) [Max 4 marks]:
Similarities (Max 2):
1. Both contain cells that are dead at maturity / hollow / lack cytoplasm (1)
2. Both have cell walls thickened with lignin (1)
Differences (Max 2):
3. Xylem has open ends / forms continuous tubes, whereas sclerenchyma has closed/tapered ends (1)
4. Xylem has pits/perforations for lateral water movement, which are absent or different in sclerenchyma (1)
5. Xylem is involved in transport and support, whereas sclerenchyma is only for support (1)

Part (c) [Max 3 marks]:
1. Control variable identified: use fibres of the same length / same diameter / same species / same age (1)
2. Control variable identified: maintain constant environmental conditions such as temperature / humidity (1)
3. Reliability: Repeat testing at least 5 times per category and calculate a mean (excluding anomalies) (1)

PastPaper.question 2 · Structured
10 PastPaper.marks

Part (a) Describe the behavior of chromosomes during metaphase and anaphase of mitosis. (4 marks)

Part (b) A student prepared a root tip squash of Allium cepa to determine the mitotic index.

(i) Describe how to prepare a temporary slide of a root tip to show stages of mitosis. (4 marks)

(ii) In a sample of 250 cells, 35 were found in prophase, 12 in metaphase, 8 in anaphase, and 5 in telophase. Calculate the mitotic index of this tissue. Show your working. (2 marks)

PastPaper.showAnswers

PastPaper.workedSolution

Part (a) During metaphase, chromosomes align individually along the equator/metaphase plate of the cell. Spindle fibres attach to the centromeres. During anaphase, the centromeres split, separating the sister chromatids. The spindle fibres shorten, pulling the chromatids (now chromosomes) to opposite poles of the cell.

Part (b)(i) Cut the terminal 1-2 mm of the root tip. Place in hydrochloric acid to macerate/break down the middle lamella. Rinse in water, then place on a microscope slide. Add a stain (e.g., acetic orcein or Toluidine blue) to stain chromosomes. Gently press down on the coverslip with a thumb (squash) to spread the cells into a single layer, avoiding lateral movement.

Part (b)(ii) Mitotic index = (Number of cells in mitosis / Total number of cells) * 100. Total mitotic cells = 35 + 12 + 8 + 5 = 60. Mitotic index = (60 / 250) * 100 = 24.0% (or 0.24).

PastPaper.markingScheme

Part (a) [Max 4 marks]:
1. (In metaphase) chromosomes condense and line up individually along the cell equator (1)
2. Spindle fibres attach to centromeres (1)
3. (In anaphase) centromeres divide / split (1)
4. Chromatids are pulled to opposite poles (centromere first) as spindle fibres contract/shorten (1)

Part (b)(i) [Max 4 marks]:
1. Cut 1-2 mm from the tip of the root / use root meristem (1)
2. Use of hydrochloric acid to separate cells / break down middle lamella / pectin (1)
3. Add acetic orcein / Toluidine blue / Schiff's reagent (to stain chromosomes) (1)
4. Gently press / squash under a coverslip to obtain a single layer of cells (without twisting/sliding coverslip) (1)

Part (b)(ii) [Max 2 marks]:
1. Correct sum of mitotic cells: 35 + 12 + 8 + 5 = 60 (1)
2. Mitotic index = 24% or 0.24 (accept 24, do not accept 60/250 without calculation) (1)

PastPaper.question 3 · Structured
10 PastPaper.marks

Part (a) Distinguish between totipotent, pluripotent, and multipotent stem cells. (3 marks)

Part (b) Describe how chemical signals can cause a pluripotent stem cell to differentiate into a specialized cell type, such as a cardiomyocyte. (4 marks)

Part (c) Explain why the use of embryonic stem cells is considered controversial by some groups, and suggest how the development of induced pluripotent stem cells (iPSCs) might resolve these ethical issues. (3 marks)

PastPaper.showAnswers

PastPaper.workedSolution

Part (a) Totipotent stem cells can differentiate into any cell type, including extra-embryonic tissues (like placenta). Pluripotent stem cells can differentiate into any embryonic body cell type but not extra-embryonic tissues. Multipotent stem cells can only differentiate into a limited range of specialized cell types within a specific tissue (e.g., hematopoietic stem cells).

Part (b) Chemical signals (transcription factors/morphogens) bind to specific receptors or enter the cell. This activates specific genes while keeping others inactive/silenced. Active genes are transcribed to produce mRNA, which is translated into specific proteins. These proteins permanently alter the cell's structure and function, leading to differentiation into a specialized cell like a cardiomyocyte.

Part (c) Embryonic stem cell extraction requires the destruction of a human blastocyst/embryo, which some believe has the moral status of a human being (right to life). iPSCs are created by reprogramming adult somatic cells (like skin cells) back to a pluripotent state using transcription factors. This avoids the use/destruction of human embryos entirely, bypassing the primary ethical concern.

PastPaper.markingScheme

Part (a) [Max 3 marks]:
1. Totipotent: can differentiate into all cell types, including extra-embryonic cells/placenta (1)
2. Pluripotent: can differentiate into all embryonic/body cell types (but not placenta) (1)
3. Multipotent: can differentiate into only a limited/restricted range of cells (associated with a specific tissue) (1)

Part (b) [Max 4 marks]:
1. Chemical signals/transcription factors activate/turn on specific genes (and switch off others) (1)
2. Active genes undergo transcription to produce mRNA (1)
3. mRNA is translated to synthesize specific proteins (1)
4. These proteins structural/functional proteins permanently modify the cell to become specialized (1)

Part (c) [Max 3 marks]:
1. Controversy: Embryonic stem cell extraction involves destruction of a human embryo / loss of potential life (1)
2. iPSCs are made from adult somatic cells (e.g. skin cells) reprogrammed using transcription factors (1)
3. Resolves ethics because no embryos are destroyed/harmed in the process (1)

PastPaper.question 4 · Structured
10 PastPaper.marks

Part (a) State the role of the acrosome reaction in fertilization and describe how it occurs. (4 marks)

Part (b) Explain how the cortical reaction ensures that only one sperm fertilizes the egg. (3 marks)

Part (c) Compare the structure of a mammalian sperm cell with that of a mature human secondary oocyte. (3 marks)

PastPaper.showAnswers

PastPaper.workedSolution

Part (a) The role is to digest the protective jelly layer (zona pellucida) surrounding the egg so the sperm can reach and fuse with the egg cell membrane. When the sperm contacts the jelly layer, the acrosome membrane fuses with the sperm cell surface membrane, releasing hydrolytic enzymes (such as acrosin) by exocytosis. These enzymes digest a pathway through the zona pellucida.

Part (b) Upon fusion of the sperm and egg cell membranes, calcium ions are released inside the egg. This triggers cortical granules in the egg cytoplasm to fuse with the egg cell membrane and release their contents by exocytosis into the space between the membrane and zona pellucida. The chemicals cause the zona pellucida to thicken and harden, forming a fertilization membrane that prevents any further sperm from entering (polyspermy).

Part (c) Similarities: Both are haploid gametes. Differences: Sperm is small, motile (has a flagellum), has an acrosome, and very little cytoplasm with many mitochondria in the midpiece. The oocyte is large, non-motile, contains a large amount of cytoplasm (food store), is surrounded by a zona pellucida and follicle cells (corona radiata), and lacks a flagellum.

PastPaper.markingScheme

Part (a) [Max 4 marks]:
1. Role: digests the zona pellucida / jelly layer to allow sperm to reach egg membrane (1)
2. Triggered by contact with receptors on the zona pellucida / follicular cells (1)
3. Acrosome membrane fuses with the sperm cell surface membrane (1)
4. Digestive enzymes / acrosin released by exocytosis (1)

Part (b) [Max 3 marks]:
1. Fusion of sperm and egg membranes triggers release of calcium ions (1)
2. Cortical granules fuse with egg cell membrane and release contents by exocytosis (1)
3. Zona pellucida thickens / hardens (forming fertilization membrane) to prevent polyspermy (1)

Part (c) [Max 3 marks]:
Comparison points (Max 3, must include at least one similarity or difference):
1. Both are haploid / contain \( n \) chromosomes (1)
2. Sperm has flagellum / is motile, whereas oocyte is non-motile (1)
3. Sperm contains an acrosome and midpiece with mitochondria, whereas oocyte contains cortical granules / lipid droplets / corona radiata (1)
4. Sperm has very little cytoplasm, whereas oocyte has a large volume of cytoplasm (1)

PastPaper.question 5 · Structured
10 PastPaper.marks

Part (a) Define the terms 'endemism' and 'biodiversity'. (2 marks)

Part (b) A population of an endangered plant species, Silene tomentosa, was analyzed for genetic diversity. In a sample of 80 plants, 24 were found to be heterozygous at a specific gene locus.

(i) Calculate the heterozygosity index (\(H\)) for this locus. (1 mark)

(ii) Explain the significance of a low heterozygosity index for the survival of a population if the environment changes. (3 marks)

(iii) Describe the role of seed banks in the conservation of endangered plant species and explain how seeds are selected and stored to ensure their long-term viability. (4 marks)

PastPaper.showAnswers

PastPaper.workedSolution

Part (a) Endemism refers to a species being found in only one specific, defined geographical area and nowhere else. Biodiversity is the variety of life, including species richness (number of different species) and genetic diversity within species in a habitat.

Part (b)(i) Heterozygosity index \( H = \frac{\text{Number of heterozygotes}}{\text{Total number of individuals in population}} = \frac{24}{80} = 0.30 \).

Part (b)(ii) A low index indicates low genetic diversity / a small gene pool. This means there are fewer different alleles present. If the environment changes (e.g. climate change, new disease), there is a lower probability that some individuals will possess an allele that confers resistance or adaptation, increasing the risk of extinction.

Part (b)(iii) Seed banks store seeds from a wide range of endangered plants to conserve genetic diversity and prevent extinction. Seeds are collected from multiple wild plants (to maximize genetic variety). They are cleaned, X-rayed to ensure they contain living embryos, dried to reduce moisture content (which slows metabolic rate and prevents decay/fungal growth), and stored at sub-zero temperatures (\(-20^{\circ}\text{C}\)). Periodically, samples are germinated to test viability; if germination falls below a threshold, new plants are grown to harvest fresh seeds.

PastPaper.markingScheme

Part (a) [Max 2 marks]:
1. Endemism: a species unique to a single defined geographic location / not found anywhere else (1)
2. Biodiversity: variety of living organisms/species in an area AND their genetic diversity (1)

Part (b)(i) [Max 1 mark]:
1. 0.3 or 0.30 (or 3/10) (1)

Part (b)(ii) [Max 3 marks]:
1. Low index means low genetic diversity / small gene pool / highly homozygous (1)
2. Fewer alleles available in the population (1)
3. Lower chance that some individuals have alleles allowing survival during environmental change / disease outbreak (increasing extinction risk) (1)

Part (b)(iii) [Max 4 marks]:
1. Seeds collected from different plants/locations to maintain genetic diversity (1)
2. Seeds dried to low moisture content (approx 5%) to prevent fungal growth / slow down decay/metabolism (1)
3. Kept at very low temperatures (e.g., \(-20^{\circ}\text{C}\)) to extend survival/viability time (1)
4. Regularly tested for viability by germinating samples (and growing new crops to replenish seeds if viability drops) (1)

PastPaper.question 6 · Structured
10 PastPaper.marks

Part (a) State three differences between the ultrastructure of a typical prokaryotic cell and a typical eukaryotic animal cell. (3 marks)

Part (b) Explain the role of the rough endoplasmic reticulum (RER) and the Golgi apparatus in the production and secretion of a glycoprotein enzyme. (5 marks)

Part (c) Explain why electron microscopes have a higher resolution than light microscopes. (2 marks)

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Part (a) 1. Prokaryotes lack membrane-bound organelles (like nucleus/mitochondria), while eukaryotes have them. 2. Prokaryotes have 70S ribosomes, while eukaryotes have 80S ribosomes. 3. Prokaryotes have circular DNA not associated with histones (naked), while eukaryotes have linear DNA associated with histones inside a nucleus.

Part (b) Ribosomes on the RER translate the mRNA to synthesize the polypeptide chain. The polypeptide enters the lumen of the RER where it folds into its tertiary structure. The RER packages the protein into transport vesicles which bud off and travel to the Golgi apparatus. The vesicles fuse with the Golgi apparatus membrane. Inside the Golgi, the protein is chemically modified by adding carbohydrate chains (glycosylation) to form a glycoprotein. The Golgi packages the final glycoprotein into secretory vesicles, which move to and fuse with the cell surface membrane to release the enzyme via exocytosis.

Part (c) Electron microscopes use a beam of electrons which has a much shorter wavelength than visible light. This allows the electron beam to resolve objects that are much closer together, providing a higher resolution.

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Part (a) [Max 3 marks]:
1. Prokaryotic DNA is circular / free in cytoplasm / not associated with histones, whereas eukaryotic DNA is linear / enclosed in a nucleus / associated with histones (1)
2. Prokaryotes contain 70S ribosomes, whereas eukaryotes contain larger 80S ribosomes (1)
3. Prokaryotes lack membrane-bound organelles (e.g. mitochondria, chloroplasts, RER), which are present in eukaryotes (1)
4. Prokaryotes have a cell wall made of peptidoglycan/murein, whereas animal cells have no cell wall (1)

Part (b) [Max 5 marks]:
1. Ribosomes on RER synthesize the polypeptide chain (1)
2. Polypeptide folds into its 3D structure inside the RER lumen (1)
3. RER packages the protein into transport vesicles that bud off and fuse with the Golgi apparatus (1)
4. Golgi apparatus modifies the protein by adding carbohydrate chains / glycosylation (1)
5. Golgi packages the glycoprotein into secretory vesicles (1)
6. Secretory vesicles fuse with the cell surface membrane to release the protein by exocytosis (1)

Part (c) [Max 2 marks]:
1. Electrons have a much shorter wavelength than light (1)
2. Therefore, electron microscopes have a higher limit of resolution / can distinguish objects that are closer together (up to 0.1-0.5 nm vs 200 nm) (1)

PastPaper.question 7 · Structured
10 PastPaper.marks

Part (a) Plant growth requires inorganic ions. Describe the specific roles of nitrate ions, calcium ions, and magnesium ions in plant development. (3 marks)

Part (b) Cellulose and starch are both carbohydrates found in plants.

(i) Compare and contrast the structures of cellulose and starch (amylose and amylopectin). (4 marks)

(ii) Explain how the structure of starch makes it suitable for its function as a storage molecule. (3 marks)

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Part (a) Nitrate ions are required to synthesize amino acids (and thus proteins), nucleic acids (DNA/RNA), and chlorophyll. Calcium ions are required to form calcium pectate, which cements cell walls together in the middle lamella. Magnesium ions are an essential constituent of chlorophyll, which absorbs light for photosynthesis.

Part (b)(i) Similarities: Both are polysaccharides composed of hexose sugars joined by glycosidic bonds. Differences: Cellulose is made of beta-glucose, whereas starch is made of alpha-glucose. Cellulose is an unbranched, straight chain with only 1,4-glycosidic bonds, whereas starch consists of amylose (unbranched spiral, 1,4-bonds) and amylopectin (branched, 1,4- and 1,6-glycosidic bonds). Cellulose chains lie parallel to each other held by hydrogen bonds to form microfibrils; starch molecules do not form hydrogen-bonded microfibrils.

Part (b)(ii) Starch is insoluble, so it does not affect the water potential of the cell and will not cause osmotic movement of water into the cell. It is compact (coiled structure of amylose), meaning a large amount of energy can be stored in a small space. Its branched structure (amylopectin) means it has many terminal ends, allowing rapid hydrolysis by enzymes to release glucose when needed for respiration.

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Part (a) [Max 3 marks]:
1. Nitrate ions: used to make amino acids / proteins / nucleic acids / chlorophyll (1)
2. Calcium ions: used to make calcium pectate in the middle lamella (holding plant cells together) (1)
3. Magnesium ions: required for the production of chlorophyll (to absorb light) (1)

Part (b)(i) [Max 4 marks]:
Similarities (Max 2):
1. Both are polymers of glucose / contain glycosidic bonds (1)
2. Both contain 1,4-glycosidic bonds (1)
Differences (Max 2):
3. Cellulose is composed of beta-glucose, while starch is composed of alpha-glucose (1)
4. Cellulose is straight/unbranched, whereas starch is coiled/helical (amylose) and branched (amylopectin) (1)
5. Starch has 1,6-glycosidic bonds (in amylopectin), which are absent in cellulose (1)
6. Cellulose molecules form hydrogen bonds between chains (microfibrils), whereas starch does not (1)

Part (b)(ii) [Max 3 marks]:
1. Insoluble: does not affect the water potential / osmotic balance of the plant cell (1)
2. Compact (coiled helical structure): stores a large amount of glucose/energy in a small volume (1)
3. Branched (amylopectin): provides many terminal ends for rapid enzymatic hydrolysis to release glucose (1)

PastPaper.question 8 · Structured
10 PastPaper.marks

Part (a) Distinguish between continuous and discontinuous variation, giving one example of each in humans. (3 marks)

Part (b) Explain how phenotype is determined by both genotype and environmental factors, using a named example. (3 marks)

Part (c) Epigenetic modifications can alter gene expression without changing the DNA base sequence. Explain how DNA methylation and histone modification (acetylation) regulate gene expression. (4 marks)

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Part (a) Continuous variation is quantitative variation where there are no distinct categories but a range of values (e.g., human height). It is usually polygenic (controlled by many genes) and heavily influenced by the environment. Discontinuous variation has distinct, non-overlapping categories with no intermediates (e.g., ABO blood group). It is controlled by one or a few genes (monogenic) and has little or no environmental influence.

Part (b) Taking human height as an example: Genotype determines the potential maximum height an individual can reach (their genetic potential, determined by multiple genes). Environmental factors such as diet/nutrition (e.g., calcium and protein intake) and presence of childhood diseases determine whether this genetic potential is actually reached. If nutrition is poor, the individual will be shorter than their genetic potential.

Part (c) DNA methylation involves adding methyl groups to CpG islands on DNA. This prevents transcription factors from binding to the promoter region, keeping the gene switched off (silenced). Histone acetylation involves adding acetyl groups to histone proteins, which reduces their positive charge. This weakens the attraction between histones and negatively-charged DNA, making the chromatin structure more relaxed/open (euchromatin). This allows RNA polymerase and transcription factors to access the gene easily, increasing transcription/gene expression. (Deacetylation does the opposite, leading to heterochromatin and silencing).

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Part (a) [Max 3 marks]:
1. Continuous variation shows a range of values / no distinct categories (e.g. height, mass) AND is polygenic / affected by environment (1)
2. Discontinuous variation shows distinct categories / no intermediates (e.g. blood group, ability to roll tongue) AND is monogenic / not affected by environment (1)
3. Correct human examples provided for both (1)

Part (b) [Max 3 marks]:
1. Genotype sets the maximum potential limit/boundary of the phenotype (1)
2. Environment determines where within that range the actual phenotype falls (1)
3. Example linked: e.g. height: genes determine potential height, but nutrition / diet determines if that potential is met OR skin color: genes determine baseline melanin, but UV exposure increases melanin production (1)

Part (c) [Max 4 marks]:
1. DNA methylation: addition of methyl groups to DNA/cytosine bases (1)
2. This prevents transcription factors from binding / silences the gene (1)
3. Histone acetylation: addition of acetyl groups to histone proteins (1)
4. This relaxes chromatin structure / makes DNA accessible so RNA polymerase can bind and transcribe the gene (1)

PastPaper.section WBI13/01A - Unit 3: Practical Skills in Biology I

Answer ALL questions. Write your answers in the spaces provided in the Answer Book.
3 PastPaper.question · 49.980000000000004 PastPaper.marks
PastPaper.question 1 · practical
16.66 PastPaper.marks
A student investigated the effect of temperature on the permeability of beetroot cell membranes. The student used six test tubes containing 10 cm3 of distilled water and placed them in water baths at temperatures of 20, 30, 40, 50, 60, and 70 degrees Celsius. After 10 minutes, five beetroot cylinders of equal length and diameter were added to each tube. The tubes were left for 20 minutes. The beetroot cylinders were then removed, and the absorbance of the remaining liquid was measured using a colorimeter with a green filter. Three replicates were completed for each temperature. The raw absorbance values (arbitrary units, a.u.) obtained were: 20 degrees Celsius: 0.04, 0.05, 0.03; 30 degrees Celsius: 0.08, 0.07, 0.09; 40 degrees Celsius: 0.15, 0.18, 0.15; 50 degrees Celsius: 0.38, 0.42, 0.40; 60 degrees Celsius: 0.72, 0.75, 0.78; 70 degrees Celsius: 0.92, 0.95, 0.94. (a) Design a suitable table to present these raw and calculated mean absorbance data. (b) Explain the effect of temperature on the permeability of the beetroot cell membrane, referencing the data and your biological knowledge. (c) Describe how the student could ensure the validity of this investigation by controlling two named variables.
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(a) A properly constructed table should have clear headings with units: 'Temperature / degrees Celsius', 'Raw Absorbance (Replicates 1, 2, and 3) / a.u.', and 'Mean Absorbance / a.u.'. The calculated mean values are: 20 degrees C = 0.04 a.u.; 30 degrees C = 0.08 a.u.; 40 degrees C = 0.16 a.u.; 50 degrees C = 0.40 a.u.; 60 degrees C = 0.75 a.u.; 70 degrees C = 0.94 a.u. (b) As temperature increases, membrane permeability increases, causing more betalain pigment to leak out and increasing light absorbance. Between 20 and 40 degrees C, membrane permeability is low because the membrane remains intact. Between 40 and 60 degrees C, absorbance increases sharply because proteins in the cell membrane denature and phospholipid kinetic energy increases, causing the bilayer to become more fluid and lose its structural integrity. Above 60 degrees C, the rate of increase slows down as concentration equilibrium is reached. (c) Control variables include: 1. Size of beetroot cylinders: Control by using a cork borer of a specific diameter and cutting them to a uniform length (e.g., 2 cm) using a ruler and razor. 2. Volume of distilled water: Use a graduated pipette to measure exactly 10 cm3 of water in each tube.

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Marking Scheme (Total: 16.66 marks, equivalent to 17 marks in standard distribution): (a) Table construction [5 marks]: [1] Table drawn with clear boundaries and columns. [1] Correct column headers with appropriate units (e.g., Temperature / degrees C, Absorbance / a.u.). [1] Raw data correctly transferred to table. [1] All mean values calculated correctly to a consistent number of decimal places (2 d.p.). [1] Standard format where the independent variable is in the left column. (b) Explanation of the trend [8 marks]: [1] General statement that absorbance increases with temperature. [1] Refers to specific data points (e.g., from 0.04 a.u. at 20 degrees C to 0.94 a.u. at 70 degrees C). [1] Identifies the sharpest increase occurring between 40 and 60 degrees C. [1] Explains that increase in temperature increases kinetic energy of pigment and phospholipid molecules. [1] Explains that increased fluidity of the phospholipid bilayer creates gaps. [1] Explains that membrane proteins denature. [1] Disruption of the tonoplast and cell membrane allows betalain pigment to escape. [1] Explains that above 60 degrees C, leakage slows down as equilibrium of concentration is approached. (c) Controlling variables [4 marks]: [1] Identification of beetroot cylinder dimensions as a variable. [1] Description of using a cork borer and cutting with a ruler/scalpel to ensure equal surface area/volume. [1] Identification of volume of water as a variable. [1] Description of using a graduated pipette/measuring cylinder to ensure equal volume of water.
PastPaper.question 2 · practical
16.66 PastPaper.marks
A student carried out an investigation to determine the Vitamin C (ascorbic acid) content of fresh guava juice compared to fresh dragon fruit juice. First, the student prepared standard solutions of ascorbic acid at concentrations of 0.2, 0.4, 0.6, 0.8, and 1.0 mg cm-3. They titrated 1.0 cm3 of 1% DCPIP solution with each standard concentration, recording the volume of ascorbic acid needed to decolourise the DCPIP. The results were: 0.2 mg cm-3: Mean volume = 4.5 cm3; 0.4 mg cm-3: Mean volume = 2.25 cm3; 0.6 mg cm-3: Mean volume = 1.50 cm3; 0.8 mg cm-3: Mean volume = 1.13 cm3; 1.0 mg cm-3: Mean volume = 0.90 cm3. (a) Describe how to plot a suitable calibration graph using these results and how to use it to find the concentration of Vitamin C in the fruit juices. (b) Distinguish between the independent, dependent, and one crucial control variable in this titration phase of the investigation. (c) State two potential sources of error in this method and suggest how the student could minimise their effect to improve the precision of the results.
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(a) Plot a line graph with the concentration of ascorbic acid (mg cm-3) on the x-axis and the mean volume of ascorbic acid solution (cm3) required to decolourise DCPIP on the y-axis. Scale both axes appropriately and plot the five coordinate points. Draw a smooth curve or join points with straight ruled lines. To determine the Vitamin C concentration in a fruit juice, perform the titration using the juice, find the mean volume required, locate this volume on the y-axis, read across to the curve, and read down to find the corresponding concentration on the x-axis. (b) Independent Variable: Concentration of the standard ascorbic acid solutions / type of fruit juice. Dependent Variable: The volume of ascorbic acid or fruit juice needed to decolourise the DCPIP solution. Control Variable: The volume (1.0 cm3) or concentration (1%) of the DCPIP solution used. (c) Error 1: Difficulty in visually determining the exact end-point (colour change from blue to colourless). Minimise by placing the tube on a white tile to observe the colour clearly, and using a standard reference tube of decolourised DCPIP to compare. Error 2: Loss of Vitamin C in fruit juices due to oxidation from exposure to air or high temperature. Minimise by preparing and testing the juices immediately after extraction, and keeping them on ice or covered.

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Marking Scheme (Total: 16.66 marks, equivalent to 17 marks in standard distribution): (a) Graph plotting and usage [7 marks]: [1] X-axis labeled as concentration of ascorbic acid with units (mg cm-3). [1] Y-axis labeled as volume of ascorbic acid with units (cm3). [1] Points plotted accurately with appropriate linear scales. [1] Curve of best fit or ruled straight lines connecting points drawn. [1] Description of titrating the fruit juice to find the average volume required to decolourise the same volume of DCPIP. [1] Use the y-axis to locate the volume of juice used. [1] Read across to the calibration curve and down to the x-axis to determine the equivalent Vitamin C concentration. (b) Variable identification [4 marks]: [1] Correctly identifies Independent Variable as concentration of ascorbic acid / type of fruit juice. [1] Correctly identifies Dependent Variable as volume of solution required to decolourise DCPIP. [1] Correctly identifies Control Variable as volume of DCPIP (or concentration of DCPIP). [1] Explains why keeping DCPIP volume constant ensures validity (same amount of dye to reduce). (c) Sources of error and improvements [6 marks]: [1] Identification of end-point determination subjectivity. [1] Improvement: use of a white background/tile or comparing to a pre-titrated reference tube. [1] Improvement: use a colorimeter to measure transmittance/absorbance changes quantitatively. [1] Identification of oxidation/decay of Vitamin C over time or at high temperature. [1] Improvement: prepare fresh juices immediately before use. [1] Improvement: store juices in a cold container / keep on ice / shield from direct light.
PastPaper.question 3 · practical
16.66 PastPaper.marks
A researcher wanted to compare the tensile strength of vascular bundle fibers extracted from stinging nettle (Urtica dioica) stems with those from flax (Linum usitatissimum) stems. (a) Write a detailed plan describing how the researcher could safely extract the fibers and determine their tensile strength experimentally. Your plan should allow for a valid comparison between the two species. (b) In an experiment, a fiber of flax with a diameter of 0.25 mm broke when a load of 14.7 N was applied. Calculate the tensile strength of this fiber in N mm-2. Use the formula: Tensile Strength = Force / Cross-sectional Area, where Area = pi * r^2. Show your working and give your answer to 3 significant figures. (c) Explain why calculating the tensile strength is a more valid way to compare fibers than simply measuring the maximum breaking force.
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(a) Plan: 1. Soak the stems of stinging nettle and flax in water for several days (a process called retting) to allow bacterial action to break down the surrounding soft tissue, leaving only the vascular bundle fibers. 2. Carefully peel/extract the long vascular fibers from the stems and rinse them with water. 3. Cut the fibers from both species to equal lengths (e.g., 10 cm) to ensure a standardized test length. 4. Clamp one end of a single fiber securely to a clamp stand. Clamp the other end to a weight holder. 5. Gradually add standard weights (e.g., 10 g or 50 g masses) to the holder until the fiber breaks. Record the mass at which the fiber snaps. 6. Repeat this process with at least 5 replicates of fibers from both species to calculate a mean breaking force. 7. Safety measures: Wear protective gloves when handling fresh nettle stems to prevent stings, wear safety goggles to protect eyes if fibers snap suddenly, and place a box filled with bubble wrap or foam under the weights to catch them when they fall. (b) Calculation: Radius (r) = diameter / 2 = 0.25 mm / 2 = 0.125 mm. Cross-sectional Area (A) = pi * (0.125)^2 = 3.14159 * 0.015625 = 0.049087 mm2. Force (F) = 14.7 N. Tensile Strength = 14.7 N / 0.049087 mm2 = 299.47 N mm-2. Rounding to 3 significant figures gives 299 N mm-2. (c) Fibers extracted from different stems will have varying thicknesses (cross-sectional areas). A thicker fiber has more cross-sectional material and will naturally require a greater force to break, regardless of its material properties. Calculating tensile strength factors in the cross-sectional area (force per unit area), allowing for a fair and valid comparison of the intrinsic material strength of the fibers.

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Marking Scheme (Total: 16.66 marks, equivalent to 17 marks in standard distribution): (a) Plan for fiber extraction and tensile strength testing [9 marks]: [1] Mention retting (soaking stems in water) to decompose soft plant tissue. [1] Extract/peel the vascular bundles (fibers) carefully. [1] Standardize the length of the fibers. [1] Securely clamp the fiber at both ends using a clamp stand and weight holder. [1] Add masses sequentially (in small increments) until the fiber snaps. [1] Record the final mass/load that caused breakage. [1] Repeat the experiment with multiple fibers from each plant species (at least 5 replicates) to calculate a mean. [1] Detail a valid safety precaution (e.g., box with padding under weights). [1] Detail another safety precaution (e.g., gloves for stinging nettles / safety glasses). (b) Calculation [5 marks]: [1] Correctly identifies radius = 0.125 mm. [1] Correct calculation of cross-sectional area = 0.0491 mm2 (or 0.049087 mm2). [1] Correct setup of formula: 14.7 / Area. [1] Calculated value of 299.47 (or similar unrounded value). [1] Final answer rounded correctly to 3 significant figures: 299 N mm-2 (must have correct units). (c) Explanation of validity [3 marks]: [1] Fibers have different diameters / thicknesses / cross-sectional areas. [1] Thicker fibers require a larger force to break simply because there is more material, not because the material is stronger. [1] Tensile strength normalises the breaking force against cross-sectional area, making it a fair comparison.

PastPaper.section WBI14/01A - Unit 4: Energy, Environment, Microbiology and Immunity

Answer ALL questions. Write your answers in the spaces provided in the Answer Book.
8 PastPaper.question · 90 PastPaper.marks
PastPaper.question 1 · Analysis
11.25 PastPaper.marks
Temperate deciduous forests are highly productive terrestrial ecosystems. (a) State the relationship between Net Primary Productivity (NPP), Gross Primary Productivity (GPP), and Respiration (R). (b) In a forest ecosystem, the GPP was measured as \( 24500 \text{ kJ m}^{-2} \text{ yr}^{-1} \) and the Respiration loss was \( 14700 \text{ kJ m}^{-2} \text{ yr}^{-1} \). Calculate the percentage of GPP that is converted into NPP. Show your working. (c) Climate change is leading to an increase in mean annual temperature. Explain the biological consequences of an increase in temperature on the NPP of a temperate deciduous forest, assuming water availability is not limiting.
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(a) The relationship is \( NPP = GPP - R \). (b) To calculate the percentage of GPP converted to NPP: \( NPP = GPP - R = 24500 - 14700 = 9800 \text{ kJ m}^{-2} \text{ yr}^{-1} \). Percentage = \( \frac{9800}{24500} \times 100 = 40\% \). (c) An increase in temperature increases the kinetic energy of molecules, leading to more frequent collisions and a higher rate of enzyme-controlled reactions. Up to an optimum temperature, both photosynthesis and respiration rates will increase. However, respiration involves many enzymes throughout the plant and tends to increase exponentially with temperature, whereas photosynthesis may become limited by other factors (such as enzyme denaturation of Rubisco or stomatal closure to limit water loss). If respiration increases more than photosynthesis, GPP - R decreases, resulting in a lower NPP.

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(a) 1 mark: \( NPP = GPP - R \). (b) 1 mark for calculating NPP = 9800. 1 mark for dividing NPP by GPP. 1 mark for final answer of 40%. (c) Max 5.25 marks: Up to 2 marks for effect on enzymes (kinetic energy, enzyme-substrate complexes). Up to 2 marks for explaining that respiration rate increases more rapidly with temperature than photosynthesis rate. 1.25 marks for concluding that because R increases more than GPP, NPP decreases.
PastPaper.question 2 · Structured
11.25 PastPaper.marks
A novel plant pathogen, the Potato Yellowing Virus (PYV), is a single-stranded RNA virus. (a) Describe the roles of reverse transcriptase and DNA polymerase in a reverse transcription polymerase chain reaction (RT-PCR) assay used to detect this virus. (b) Explain how gel electrophoresis is used to separate the resulting DNA fragments after PCR. (c) A researcher ran the PCR products on an agarose gel. Lane 1 contained a DNA ladder, Lane 2 was a negative control (healthy plant), Lane 3 was a positive control (purified PYV RNA), and Lanes 4 and 5 contained samples from symptomatic potato plants. Explain why a band appeared in Lane 5 at the same size as the positive control, but no band appeared in Lane 4.
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(a) Reverse transcriptase uses the viral single-stranded RNA as a template to synthesise a complementary strand of DNA (cDNA). DNA polymerase then uses this cDNA as a template to amplify the double-stranded DNA target sequence through PCR cycles of denaturation, annealing, and extension. (b) DNA fragments are loaded into wells in an agarose gel. An electrical current is applied; DNA has a net negative charge (due to phosphate groups) and migrates towards the positive electrode (anode). The agarose gel matrix acts as a sieve, meaning smaller fragments move faster and further through the pores than larger fragments, separating them by size. (c) The positive control (Lane 3) shows the expected fragment size of amplified PYV DNA. Lane 5 shows a band at the same position, indicating the plant is infected with PYV. Lane 4 has no band, showing the plant does not contain viral RNA (not infected, or viral load is below detection limit). The negative control (Lane 2) ensures there is no contamination during the setup.

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(a) Max 4 marks: 1 mark for reverse transcriptase synthesising cDNA. 1 mark for using viral RNA as a template. 1 mark for DNA polymerase amplifying the DNA. 1 mark for role of primers/free nucleotides. (b) Max 3.25 marks: 1 mark for applying an electrical current. 1 mark for DNA migrating to the anode. 1.25 marks for gel acting as a matrix where smaller fragments move faster. (c) Max 4 marks: 1 mark for identifying Lane 5 is infected. 1 mark for identifying Lane 4 is not infected. 1 mark for role of positive control. 1 mark for role of negative control to prove absence of contamination.
PastPaper.question 3 · Level of Response
11.25 PastPaper.marks
A volcanic eruption formed a new island of bare rock. Scientists studied primary succession on this island over 100 years. (a) Discuss the process of primary succession that leads to the establishment of a climax community on this volcanic island. (b) During the study, scientists measured soil nitrogen content (g per kg of soil) and plant species richness. After 10 years, soil nitrogen was 0.02 g/kg and species richness was 3. After 80 years, soil nitrogen was 1.45 g/kg and species richness was 34. Explain the relationship between soil nitrogen content and plant species richness during succession.
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(a) Primary succession starts with bare rock. Pioneer species (like lichens and mosses) are adapted to survive harsh, soil-free conditions. They erode the rock face and, upon death and decomposition, add organic matter (humus) to form basic soil. This changes the abiotic conditions, making them less hostile. Deeper soil can retain more water and nutrients, enabling herbaceous plants and grasses to establish. These further enrich the soil, allowing shrubs and eventually deep-rooted trees to grow. Over time, dominant tree species establish, forming a stable, self-sustaining climax community. (b) Soil nitrogen increases significantly (from 0.02 to 1.45 g/kg) because early pioneer species (such as nitrogen-fixing plants or cyanobacteria) fix atmospheric nitrogen. As these organisms die and decompose, decomposers release ammonium compounds, which nitrifying bacteria convert into nitrates. Higher nitrate availability allows a wider variety of plant species to grow, as nitrogen is essential for synthesis of amino acids, proteins, and nucleic acids needed for plant growth. This increases species richness from 3 to 34.

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(a) Max 6.25 marks (Level of Response): Level 1 (1-2 marks): Simple description of succession stages or pioneer species. Level 2 (3-4 marks): Explains how pioneer species modify the environment (soil formation, humus) making it less hostile. Level 3 (5-6.25 marks): Detailed explanation of succession up to climax community, referencing changes in abiotic factors, competition, and stability. (b) Max 5 marks: 1 mark for noting positive correlation. 1 mark for explaining the role of nitrogen-fixing bacteria. 1 mark for action of decomposers/nitrifying bacteria. 1 mark for nitrogen being a limiting factor for plant growth. 1 mark for more species being able to compete, increasing richness.
PastPaper.question 4 · Calculations
11.25 PastPaper.marks
The growth of Streptococcus pneumoniae was investigated in the presence of the bacteriostatic antibiotic erythromycin. (a) State the difference between a bacteriostatic antibiotic and a bactericidal antibiotic. (b) In a control culture (no antibiotic), S. pneumoniae cells increased from \( 1.2 \times 10^3 \) cells per mL to \( 4.8 \times 10^6 \) cells per mL in 6 hours of exponential growth. Calculate the exponential growth rate constant (\( k \)) using the formula: \( k = \frac{\ln(N_t) - \ln(N_0)}{t} \), where \( N_t \) is the final cell concentration, \( N_0 \) is the initial cell concentration, and \( t \) is time in hours. Give your answer to 3 significant figures and include units. (c) Explain how conjugation can lead to the rapid spread of erythromycin resistance between different bacterial populations.
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(a) A bacteriostatic antibiotic inhibits the growth and reproduction of bacterial cells, allowing the host's immune system to clear the infection. A bactericidal antibiotic directly kills the bacterial cells. (b) Using the formula: \( N_t = 4.8 \times 10^6 \), \( N_0 = 1.2 \times 10^3 \), \( t = 6 \). \( \ln(N_t) = \ln(4.8 \times 10^6) \approx 15.3837 \). \( \ln(N_0) = \ln(1.2 \times 10^3) \approx 7.0901 \). \( k = \frac{15.3837 - 7.0901}{6} = \frac{8.2936}{6} \approx 1.3822 \). Rounding to 3 significant figures gives 1.38. Units are \( \text{h}^{-1} \) (or \( \text{hour}^{-1} \)). (c) Conjugation is the direct transfer of genetic material between bacterial cells. A donor cell containing a plasmid with the erythromycin resistance gene forms a sex pilus to connect with a recipient cell. The plasmid DNA is replicated and a single strand is transferred across the conjugation tube/pilus to the recipient cell. The recipient synthesises the complementary strand, becoming resistant. Since plasmids can replicate independently and be shared between species, resistance spreads rapidly through horizontal gene transfer.

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(a) 3 marks: 1 mark for bacteriostatic inhibiting growth. 1 mark for bactericidal killing bacteria. 1 mark for bacteriostatic requiring host immunity. (b) Max 4.25 marks: 1 mark for correct substitution. 1 mark for correct calculation of numerator. 1 mark for dividing by 6 to get 1.38. 1.25 marks for correct units (h^-1 or hour^-1) and 3 sig figs. (c) Max 4 marks: 1 mark for plasmid transfer. 1 mark for formation of a sex pilus. 1 mark for replication and transfer of single-stranded plasmid DNA. 1 mark for horizontal gene transfer occurring between different species.
PastPaper.question 5 · Structured
11.25 PastPaper.marks
Mycobacterium tuberculosis is the causative agent of tuberculosis (TB) in humans. (a) Describe how M. tuberculosis survives inside macrophages after phagocytosis. (b) Explain the role of T helper cells in the immune response to a bacterial infection such as TB. (c) Explain why individuals with HIV have a significantly higher risk of developing active tuberculosis than individuals without HIV.
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(a) M. tuberculosis has a thick, waxy cell wall containing mycolic acids, which makes it highly resistant to lysosomal enzymes. Once inside a macrophage via phagocytosis, the bacterium prevents the fusion of the phagosome with lysosomes. This blocks the delivery of acidic hydrolases and reactive oxygen species, allowing the bacterium to survive and replicate inside the phagosome. (b) T helper cells recognize bacterial antigens presented on Antigen-Presenting Cells (APCs) such as macrophages via MHC class II molecules. This activates the T helper cells, causing them to release cytokines (such as interleukins). Cytokines stimulate B cells to differentiate into plasma cells (which produce antibodies) and activate T killer cells to destroy infected host cells, as well as enhancing macrophage activity. (c) HIV targets and infects T helper cells, replicating inside them and eventually causing their destruction. A low count of T helper cells means there is a severe impairment of both humoral and cellular immunity. Without sufficient T helper cells, cytokines are not released, macrophages are not activated, and antibodies/T killer cells are not produced. As a result, the immune system cannot contain the latent bacteria within tubercles, allowing the infection to reactivate and progress to active, life-threatening tuberculosis.

PastPaper.markingScheme

(a) Max 4 marks: 1 mark for referencing waxy cell wall. 1 mark for preventing the fusion of phagosome and lysosome. 1 mark for avoiding destruction by lysosomal enzymes. 1 mark for surviving/replicating inside the macrophage. (b) Max 3.25 marks: 1 mark for binding to antigen presented on MHC II. 1 mark for releasing cytokines. 1.25 marks for cytokines activating B cells to produce antibodies OR activating T killer cells / macrophages. (c) Max 4 marks: 1 mark for HIV infecting and destroying T helper cells. 1 mark for reduced cytokine production. 1 mark for inability to maintain the tubercle/granuloma structure. 1 mark for latent TB reactivating into active TB because the pathogen replicates unchecked.
PastPaper.question 6 · Level of Response
11.25 PastPaper.marks
The boreal spruce bark beetle (Dendroctonus rufipennis) is a pest that infests and kills spruce trees. Warm summer temperatures speed up its life cycle, allowing it to complete its cycle in one year instead of two. (a) Discuss the potential consequences of global warming on the distribution of the spruce bark beetle and the feedback effect this has on global carbon dioxide levels. (b) Explain how scientists can use dendrochronology to reconstruct past climate patterns and support theories of global warming.
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(a) Global warming increases global average temperatures, leading to milder winters and warmer, longer summers. This allows the spruce bark beetle to expand its geographic range further north and to higher altitudes where temperatures were previously too low. The accelerated life cycle results in rapid population growth and massive outbreaks. This leads to widespread mortality of spruce trees. Since living trees act as a major carbon sink (removing CO2 via photosynthesis), their death reduces carbon sequestration. Furthermore, the dead trees undergo decomposition by saprobionts, which release large amounts of CO2 via respiration. If forests catch fire easily due to dead wood, combustion releases even more CO2. This increases atmospheric greenhouse gas concentrations, trapping more infrared radiation, causing further global warming (a positive feedback loop). (b) Dendrochronology is the study of tree ring growth. Every year, a tree produces a new layer of xylem tissue under the bark, forming a visible ring. The width of each annual ring depends on the environmental conditions during that growing season; warmer temperatures and higher water availability stimulate faster cell division and growth, resulting in a wider ring. By taking core samples of living trees and preserved old wood, scientists can match overlapping patterns of ring widths to construct a continuous timeline back thousands of years. Unusually wide rings in recent decades provide strong evidence that temperatures have risen rapidly compared to historical baselines.

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(a) Max 6.25 marks (Level of Response): Level 1 (1-2 marks): Simple description of beetles spreading or killing trees due to warming. Level 2 (3-4 marks): Explains how tree death leads to reduced photosynthesis and increased decomposition, releasing CO2. Level 3 (5-6.25 marks): Coherent discussion of range expansion, life cycle acceleration, and detailed positive feedback loop mechanisms. (b) Max 5 marks: 1 mark for defining dendrochronology as the study of annual tree growth rings. 1 mark for explaining that xylem cell division rate is determined by environmental conditions. 1 mark for linking wider rings to warmer/favorable growing seasons. 1 mark for using overlapping ring patterns of living and dead wood to construct a historical timeline. 1 mark for concluding that rapid increases in ring width in recent years indicate a warming trend.
PastPaper.question 7 · Calculations
11.25 PastPaper.marks
The Hill reaction can be used to investigate the light-dependent reactions of photosynthesis. In an experiment, isolated chloroplasts were mixed with DCPIP, a blue dye that acts as an electron acceptor. (a) State the role of water in the light-dependent reactions of photosynthesis. (b) A student measured the light absorbance of the mixture at 600 nm at 2-minute intervals. The results are shown below: At 0 min: 0.82 absorbance units; At 2 min: 0.65; At 4 min: 0.48; At 6 min: 0.31; At 8 min: 0.14. Calculate the rate of change in absorbance per minute over the 8-minute period. Show your working and include appropriate units. (c) Explain why the absorbance of the mixture decreases over time and how this relates to the light-dependent reactions.
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(a) Water undergoes photolysis (splitting by light energy) in Photosystem II: \( 2\text{H}_2\text{O} \rightarrow 4\text{H}^+ + 4\text{e}^- + \text{O}_2 \). The electrons replace those lost by the chlorophyll photoactivation in PSII. The protons contribute to the proton gradient across the thylakoid membrane and are used to reduce NADP. Oxygen is released as a waste product. (b) Total change in absorbance over 8 minutes = \( 0.14 - 0.82 = -0.68 \) absorbance units. Rate of change = \( \frac{-0.68}{8} = -0.085 \text{ au min}^{-1} \) (or \( \text{min}^{-1} \)). (Accept 0.085 if stated as 'rate of decrease'). (c) DCPIP is a blue dye when oxidized, absorbing light strongly at 600 nm. When chloroplasts are exposed to light, photoactivation of chlorophyll occurs, exciting electrons which travel down an electron transport chain. DCPIP acts as an artificial electron acceptor, intercepting these electrons (and protons) before they reduce NADP. Upon gaining electrons, DCPIP is reduced to a colorless compound, which does not absorb light at 600 nm. Therefore, the absorbance decreases as more DCPIP is reduced.

PastPaper.markingScheme

(a) Max 3 marks: 1 mark for photolysis of water. 1 mark for electrons replacing those lost by chlorophyll/PSII. 1 mark for protons used to reduce NADP. (b) Max 4.25 marks: 1 mark for calculating total change in absorbance (-0.68). 1 mark for dividing by time (8 minutes). 1 mark for correct value of -0.085 or 0.085. 1.25 marks for correct units (absorbance units per minute or \( \text{au min}^{-1} \)). (c) Max 4 marks: 1 mark for stating oxidized DCPIP is blue and reduced DCPIP is colorless. 1 mark for explaining that light excites electrons in chlorophyll. 1 mark for stating DCPIP acts as an electron acceptor. 1 mark for linking the reduction of DCPIP to the receipt of these excited electrons.
PastPaper.question 8 · Structured
11.25 PastPaper.marks
Forensic scientists use entomological and physical evidence to estimate the time of death of a body. (a) Explain how ambient environmental temperature affects the rate of decomposition of a mammalian body. (b) A body was discovered in a woodland area. Forensic scientists collected blowfly (Calliphora vicina) larvae from the body. The average length of the larvae was 14 mm, indicating they were at the third instar stage. At an average ambient temperature of 15°C, the developmental times for C. vicina are: Egg to 1st instar: 24 hours; 1st instar to 2nd instar: 20 hours; 2nd instar to 3rd instar: 32 hours; 3rd instar to wandering stage: 48 hours. Calculate the minimum and maximum estimated times (in hours) since the eggs were laid, assuming they had just entered or were about to leave the 3rd instar stage. (c) Describe how the succession of insect species on a decomposing body can be used to estimate the post-mortem interval (PMI).
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(a) Decomposition is carried out by decomposers (bacteria and fungi) and autolytic enzymes. An increase in ambient temperature increases the kinetic energy of enzymes and substrates, resulting in more frequent successful collisions and a faster rate of enzyme-controlled decomposition reactions. High temperatures also increase the metabolic rate, growth, and reproduction of bacteria, fungi, and detritivorous insects. Extremely high temperatures can denature enzymes, stopping decomposition, while freezing temperatures slow it to a near halt. (b) To reach the start of the 3rd instar, the minimum time elapsed is: \( 24 \text{ h (egg)} + 20 \text{ h (1st)} + 32 \text{ h (2nd)} = 76 \text{ hours} \). To reach the end of the 3rd instar, the additional time is the duration of the 3rd instar itself, which is 48 hours. So, the maximum time is \( 76 + 48 = 124 \text{ hours} \). Thus, the estimated time since the eggs were laid is between 76 hours and 124 hours. (c) Succession on a corpse is a predictable sequence of colonisation by different insect groups over time as the physical and chemical state of the body changes. First colonisers are typically blowflies, attracted by volatile organic compounds released immediately after death. As the body decays, it attracts flesh flies and houseflies. Later stages of dry decay (skeletal/skin remains) attract beetles (such as Dermestid beetles) and moths that can digest keratin and collagen. By identifying the specific community of insects present on the corpse and matching this to known databases for that geographic area and temperature, investigators can estimate the PMI.

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(a) Max 4 marks: 1 mark for role of decomposers or autolytic enzymes. 1 mark for increased temperature increasing kinetic energy and enzyme-substrate complexes. 1 mark for increased growth/reproduction rate of microorganisms. 1 mark for noting extreme temperatures (denaturation or freezing). (b) Max 3.25 marks: 1 mark for calculating the minimum time to reach 3rd instar = 76 hours. 1 mark for calculating the maximum time before wandering = 124 hours. 1.25 marks for stating both values clearly as 76 hours (minimum) and 124 hours (maximum). (c) Max 4 marks: 1 mark for defining succession on a body as a predictable sequence of insect species. 1 mark for early colonisers being blowflies. 1 mark for later colonisers being beetles/moths. 1 mark for matching the species found on the body to a known timeline/database under similar conditions.

PastPaper.section WBI15/01A - Unit 5: Respiration, Internal Environment, Coordination and Gene Technology

Answer ALL questions. Write your answers in the spaces provided in the Answer Book.
10 PastPaper.question · 100 PastPaper.marks
PastPaper.question 1 · Structured
10 PastPaper.marks
Scientists investigated the respiration of a novel marine bacterium using a lipid substrate. The chemical formula of the substrate is tripalmitin (C51H98O6). The balanced equation for its complete aerobic oxidation is:

2 C51H98O6 + 145 O2 -> 102 CO2 + 98 H2O

(a) Calculate the Respiratory Quotient (RQ) for tripalmitin. Show your working.
(b) Explain why the RQ of lipids is different from the RQ of carbohydrates.
(c) An uncoupling agent called 2,4-dinitrophenol (DNP) makes the inner mitochondrial membrane permeable to protons (H+). Explain the effect of DNP on:
(i) ATP synthesis,
(ii) oxygen uptake.
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(a) RQ is calculated using the formula: RQ = CO2 produced / O2 consumed. From the equation, 102 moles of CO2 are produced and 145 moles of O2 are consumed. Therefore, RQ = 102 / 145 = 0.703 (or 0.70).

(b) Lipids contain a lower ratio of oxygen to carbon and hydrogen atoms compared to carbohydrates. This means that more external oxygen is required to oxidise the hydrogen atoms in lipids to water, resulting in a lower RQ value (around 0.7) compared to carbohydrates (which have an RQ of 1.0).

(c) (i) DNP allows protons to leak back across the inner mitochondrial membrane into the matrix, bypassing the ATP synthase channels. This dissipates the proton gradient (proton motive force), meaning ADP cannot be phosphorylated, so ATP synthesis by oxidative phosphorylation ceases. (ii) The electron transport chain (ETC) continues to function as electrons are still transferred along carriers. Oxygen acts as the terminal electron acceptor, combining with electrons and protons to form water, so oxygen uptake continues or even increases as the cell attempts to restore the gradient.

PastPaper.markingScheme

(a) [2 marks total]
- 1 mark for correct working: 102 / 145
- 1 mark for correct answer: 0.70 / 0.703

(b) [3 marks total]
- 1 mark for stating that lipids have a lower ratio of oxygen to carbon/hydrogen than carbohydrates.
- 1 mark for stating that more oxygen is required to oxidise the hydrogen in lipids to water.
- 1 mark for stating that less carbon dioxide is released relative to the oxygen consumed.

(c) [5 marks total]
- 1 mark for stating that protons leak across the inner membrane / bypass ATP synthase.
- 1 mark for stating that the proton gradient / proton motive force is dissipated.
- 1 mark for concluding that ATP synthesis via oxidative phosphorylation is reduced or stopped.
- 1 mark for stating that the electron transport chain continues to function.
- 1 mark for explaining that oxygen uptake continues because oxygen is still needed as the terminal electron acceptor to form water.
PastPaper.question 2 · Structured
10 PastPaper.marks
Muscles contract through the interaction of actin and myosin filaments within sarcomeres.

(a) Describe the roles of calcium ions (Ca2+) and ATP in the contraction of a myofibril.
(b) Compare and contrast the structural and functional properties of slow-twitch and fast-twitch muscle fibres.
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(a) Calcium ions are released from the sarcoplasmic reticulum upon depolarisation. They bind to troponin, causing a conformational change that pulls tropomyosin away from the myosin-binding sites on the actin filament. This allows myosin heads to bind to actin, forming cross-bridges. ATP binds to the myosin head, causing it to detach from the actin filament. ATP is then hydrolysed by ATPase on the myosin head to provide the energy to reset the myosin head to its high-energy state ('cocked' position), ready for another cycle. The release of ADP and Pi during the cycle drives the power stroke.

(b) Both slow-twitch and fast-twitch fibres contain actin and myosin filaments and contract using the sliding filament mechanism. However, slow-twitch fibres are adapted for aerobic respiration; they have more mitochondria, a richer capillary network, and higher myoglobin content, making them fatigue-resistant. Fast-twitch fibres are adapted for anaerobic respiration; they have high glycogen stores, more sarcoplasmic reticulum, and higher creatine phosphate levels, allowing rapid contraction but making them quick to fatigue.

PastPaper.markingScheme

(a) [6 marks total]
- 1 mark for Ca2+ released from sarcoplasmic reticulum.
- 1 mark for Ca2+ binding to troponin, causing a conformational change.
- 1 mark for tropomyosin moving to expose myosin-binding sites on actin.
- 1 mark for myosin heads binding to actin to form cross-bridges.
- 1 mark for ATP binding causing detachment of the myosin head from actin.
- 1 mark for ATP hydrolysis providing energy to reset/cock the myosin head.

(b) [4 marks total]
- 1 mark for similarity: both contain actin/myosin OR both use sliding filament mechanism.
- 1 mark for slow-twitch having more mitochondria / myoglobin / capillaries (aerobic adaptation).
- 1 mark for fast-twitch having higher glycogen / creatine phosphate / sarcoplasmic reticulum (anaerobic adaptation).
- 1 mark for functional difference: slow-twitch are fatigue-resistant/slow-contracting, while fast-twitch contract rapidly/fatigue quickly.
PastPaper.question 3 · Extended Writing
10 PastPaper.marks
During exercise, the mammalian body must adjust cardiac output to meet the increased oxygen demands of active skeletal muscles.

(a) Explain how changes in blood pH and pressure during exercise are detected and lead to an increase in heart rate.
(b) Adrenaline is released during exercise. Describe how adrenaline affects heart rate and explain its physiological significance.
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PastPaper.workedSolution

(a) Increased respiration in muscles produces more carbon dioxide, which dissolves in blood to form carbonic acid, lowering blood pH. This decrease in pH is detected by chemoreceptors in the carotid arteries, aortic arch, and medulla oblongata. Simultaneously, changes in blood pressure are detected by baroreceptors in the carotid sinus and aortic arch. Nerve impulses are sent from these receptors to the cardiovascular control centre in the medulla oblongata. The medulla sends more frequent impulses along the sympathetic nerve to the sinoatrial node (SAN). This stimulates the SAN to depolarise more frequently, increasing the heart rate.

(b) Adrenaline is secreted by the adrenal glands into the blood. It binds to specific beta-receptors on the SAN, increasing the rate of depolarization and cardiac output. This ensures rapid delivery of oxygen and glucose to active muscles for aerobic respiration, and fast removal of metabolic wastes like carbon dioxide and lactate.

PastPaper.markingScheme

(a) [6 marks total]
- 1 mark for stating exercise increases carbon dioxide, which lowers blood pH.
- 1 mark for chemoreceptors detecting low pH/high CO2 in carotid artery/aorta/medulla.
- 1 mark for baroreceptors detecting changes in blood pressure.
- 1 mark for impulses sent to the cardiovascular control centre in the medulla oblongata.
- 1 mark for increased sympathetic nerve impulses sent to the sinoatrial node (SAN).
- 1 mark for noradrenaline release at the SAN increasing the rate of excitation/depolarisation.

(b) [4 marks total]
- 1 mark for adrenaline secreted by the adrenal medulla/glands into blood.
- 1 mark for adrenaline binding to receptors on the SAN.
- 1 mark for causing an increase in the frequency of excitation / heart rate.
- 1 mark for explaining significance: increases blood flow to supply more oxygen/glucose to muscles OR speeds up removal of lactate/CO2.
PastPaper.question 4 · Structured
10 PastPaper.marks
Human rod cells act as photoreceptors in low light conditions.

(a) Describe the molecular and cellular changes that occur in a rod cell when it is exposed to light.
(b) Explain how these changes lead to the generation of action potentials in the optic nerve.
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PastPaper.workedSolution

(a) When light falls on a rod cell, it is absorbed by the pigment rhodopsin. This causes rhodopsin to split (photobleach) into retinal and opsin. Opsin activates a G-protein called transducin, which in turn activates the enzyme phosphodiesterase (PDE). PDE hydrolyses cyclic GMP (cGMP) into GMP. The reduction in cGMP concentration causes sodium channels in the outer segment membrane to close. Since sodium ions are continuously pumped out of the inner segment but can no longer re-enter, the rod cell membrane becomes hyperpolarised.

(b) In the dark, the depolarised rod cell continuously releases the inhibitory neurotransmitter glutamate, which prevents the bipolar cell from depolarising. When light causes hyperpolarisation of the rod cell, the release of glutamate stops. This lack of inhibition depolarises the bipolar cell, which then releases neurotransmitters that depolarise the ganglion cell. This generates an action potential in the ganglion cell, which travels along its axon in the optic nerve to the brain.

PastPaper.markingScheme

(a) [6 marks total]
- 1 mark for light absorption by rhodopsin causing it to split into retinal and opsin.
- 1 mark for opsin activating transducin / G-protein.
- 1 mark for activation of phosphodiesterase (PDE).
- 1 mark for PDE hydrolysing cGMP to GMP.
- 1 mark for closing of sodium channels in the outer segment.
- 1 mark for continuous pumping out of Na+ causing hyperpolarisation.

(b) [4 marks total]
- 1 mark for stating that in the dark, rod cells continuously release the inhibitory neurotransmitter glutamate.
- 1 mark for stating that hyperpolarisation stops/reduces glutamate release.
- 1 mark for stating this depolarises the bipolar cell.
- 1 mark for stating that the bipolar cell excites the ganglion cell, generating action potentials in the optic nerve.
PastPaper.question 5 · Structured
10 PastPaper.marks
Cholinergic synapses are crucial for signal transmission in the nervous system. Organophosphate compounds found in some insecticides act as potent acetylcholinesterase inhibitors.

(a) Describe the sequence of events that occurs at a cholinergic synapse from the arrival of an action potential at the presynaptic membrane to the generation of a post-synaptic action potential.
(b) Explain the physiological effects of an acetylcholinesterase inhibitor on synaptic transmission and muscle contraction.
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PastPaper.workedSolution

(a) The arrival of an action potential at the presynaptic membrane depolarises it, opening voltage-gated calcium channels. Calcium ions (Ca2+) diffuse into the presynaptic neurone down their concentration gradient. This causes synaptic vesicles containing acetylcholine (ACh) to fuse with the presynaptic membrane, releasing ACh into the synaptic cleft by exocytosis. ACh diffuses across the cleft and binds to ligand-gated sodium channels on the postsynaptic membrane, opening them. Na+ ions enter the postsynaptic neurone, depolarising the membrane to threshold, which triggers a new action potential.

(b) Acetylcholinesterase normally hydrolyses ACh into choline and ethanoic acid, terminating the signal. An inhibitor prevents this hydrolysis, causing ACh to remain bound to receptors on the postsynaptic membrane. This results in continuous opening of sodium channels and persistent depolarisation, leading to repeated firing of action potentials. Physically, this causes continuous, uncontrolled muscle spasms, tetanus, and eventually paralysis, which can be fatal if respiratory muscles are affected.

PastPaper.markingScheme

(a) [6 marks total]
- 1 mark for presynaptic depolarisation opening voltage-gated calcium channels.
- 1 mark for calcium ions entering the presynaptic bulb.
- 1 mark for calcium causing vesicles to fuse with the presynaptic membrane and release ACh by exocytosis.
- 1 mark for ACh diffusing across the synaptic cleft.
- 1 mark for ACh binding to ligand-gated sodium channels on the postsynaptic membrane.
- 1 mark for sodium influx causing depolarisation (EPSP) to reach threshold and trigger an action potential.

(b) [4 marks total]
- 1 mark for stating acetylcholinesterase normally breaks down ACh to stop the signal.
- 1 mark for stating the inhibitor prevents ACh breakdown, so ACh remains in the synaptic cleft.
- 1 mark for stating this causes continuous activation of postsynaptic receptors / continuous depolarisation.
- 1 mark for linking this to constant muscle contraction / spasms / paralysis.
PastPaper.question 6 · Calculation
10 PastPaper.marks
Forensic scientists use the Polymerase Chain Reaction (PCR) and gel electrophoresis to analyse DNA samples from crime scenes.

(a) Explain the purpose and temperature requirements of the three main stages of a PCR cycle.
(b) Explain how gel electrophoresis separates the PCR products and how the resulting DNA profile can be used to identify a suspect.
PastPaper.showAnswers

PastPaper.workedSolution

(a) The three stages are:
1. Denaturation: The mixture is heated to 94-98 degrees C to break the hydrogen bonds between complementary base pairs, separating the double-stranded DNA into single template strands.
2. Annealing: The temperature is lowered to 50-65 degrees C to allow primers to bind (anneal) via hydrogen bonds to complementary sequences at the ends of the target DNA region.
3. Extension: The temperature is raised to 72-75 degrees C, which is the optimum temperature for Taq DNA polymerase (thermostable DNA polymerase) to synthesise complementary strands by adding free nucleotides.

(b) DNA fragments are loaded into wells in an agarose gel and an electric current is applied. Because DNA is negatively charged (due to phosphate groups), it migrates towards the positive electrode (anode). Smaller fragments experience less resistance and move faster/further through the gel matrix than larger fragments, separating them by size. The resulting band pattern is the DNA profile. If a suspect's DNA profile matches the crime scene sample perfectly across multiple short tandem repeat (STR) loci, it indicates they were present at the crime scene.

PastPaper.markingScheme

(a) [6 marks total]
- 1 mark for Denaturation stage temperature (94-98 °C) AND 1 mark for explaining it breaks hydrogen bonds to separate DNA strands.
- 1 mark for Annealing stage temperature (50-65 °C) AND 1 mark for explaining it allows primers to bind to complementary DNA sequences.
- 1 mark for Extension stage temperature (72-75 °C) AND 1 mark for explaining it is the optimum temperature for Taq polymerase to synthesise the new strand.

(b) [4 marks total]
- 1 mark for stating DNA is negatively charged and moves towards the positive electrode (anode).
- 1 mark for explaining that smaller DNA fragments move faster/further through the gel.
- 1 mark for stating that the bands represent different sizes of STRs / DNA fragments.
- 1 mark for explaining that a perfect match of all band positions between suspect and sample indicates identity.
PastPaper.question 7 · Extended Writing
10 PastPaper.marks
Golden Rice was genetically engineered to produce beta-carotene in its grains to combat Vitamin A deficiency.

(a) Describe how recombinant DNA technology is used to produce a transgenic plant containing genes from another organism. Include the roles of restriction enzymes, vectors, and promoters.
(b) Discuss the environmental and ethical arguments for and against the cultivation of genetically modified (GM) crops like Golden Rice.
PastPaper.showAnswers

PastPaper.workedSolution

(a) Restriction endonucleases are used to cut the target gene (for beta-carotene) and the vector plasmid at specific recognition sites, leaving matching single-stranded sticky ends. DNA ligase is used to join the target gene to the plasmid vector, forming phosphodiester bonds to create a recombinant plasmid. A tissue-specific promoter (e.g., endosperm promoter) is linked to the gene to ensure it is only transcribed in the grain. The recombinant plasmid is introduced into a vector like Agrobacterium tumefaciens, which infects plant cells and inserts the genes into the plant genome. Transformed plant cells are selected using marker genes and grown into mature plants via tissue culture.

(b) Arguments 'for' include: saving millions of lives from blindness and death caused by Vitamin A deficiency, and potentially lowering farming costs. Arguments 'against' include: the risk of gene flow via cross-pollination to wild relatives, creating 'superweeds'; reduction in biodiversity; and the dependency of poor farmers on seed patents held by multinational corporations.

PastPaper.markingScheme

(a) [6 marks total]
- 1 mark for restriction enzymes cutting the target gene and vector to produce complementary sticky ends.
- 1 mark for DNA ligase joining the gene and plasmid together to form recombinant DNA.
- 1 mark for explaining the role of a promoter (to initiate transcription in specific tissues/endosperm).
- 1 mark for using Agrobacterium tumefaciens (or gene gun) as a vector to deliver the gene into plant cells.
- 1 mark for use of marker genes (e.g., antibiotic resistance) to identify successfully transformed cells.
- 1 mark for using plant tissue culture / micropropagation to grow the GM cells into whole plants.

(b) [4 marks total]
- 1 mark for a valid ethical/humanitarian benefit: reducing blindness/deaths from Vitamin A deficiency.
- 1 mark for a valid environmental benefit: potential for reduced pesticide use (if stacked with other traits).
- 1 mark for a valid environmental concern: gene flow / cross-pollination with wild species or loss of biodiversity.
- 1 mark for a valid ethical/socioeconomic concern: patent ownership of seeds by multinational corporations OR unknown long-term health risks.
PastPaper.question 8 · Structured
10 PastPaper.marks
In mammals, core body temperature is tightly regulated around 37 °C through homeostatic mechanisms.

(a) Explain the negative feedback loop that occurs when a mammal's core body temperature rises above normal.
(b) Contrast the physiological responses of the skin to a decrease in core body temperature with its responses to an increase in core body temperature.
PastPaper.showAnswers

PastPaper.workedSolution

(a) An increase in core body temperature is detected by thermoreceptors in the hypothalamus (and peripheral receptors in the skin). The hypothalamus acts as the coordinator and sends nerve impulses via the autonomic nervous system to various effectors. The sweat glands are stimulated to secrete sweat, which cools the skin surface through evaporation (using latent heat of vaporisation). Arterioles supplying skin surface capillaries dilate (vasodilation) while shunt vessels constrict, allowing more blood to flow close to the skin surface to lose heat by radiation. Erector pili muscles relax, so hairs lie flat, preventing an insulating layer of air from being trapped. Metabolic rate is reduced, minimizing metabolic heat production. This return to normal temperature inhibits the corrective mechanisms, completing the negative feedback loop.

(b) In a cold environment (decrease in temperature), skin arterioles undergo vasoconstriction to reduce blood flow to superficial capillaries, minimizing heat loss; erector pili muscles contract to raise hairs, trapping a warm layer of insulating air next to the skin. In a hot environment (increase in temperature), skin arterioles undergo vasodilation to increase blood flow to surface capillaries to promote heat loss; erector pili muscles relax, causing hairs to lie flat; and sweat glands actively secrete sweat to enable evaporative cooling, whereas they are inactive in the cold.

PastPaper.markingScheme

(a) [6 marks total]
- 1 mark for detection of temperature increase by thermoreceptors in the hypothalamus / skin.
- 1 mark for hypothalamus sending nerve impulses via autonomic nervous system to effectors.
- 1 mark for vasodilation: arterioles dilating and shunt vessels constricting to increase blood flow to skin capillaries.
- 1 mark for increased heat loss by radiation.
- 1 mark for sweat glands secreting sweat for evaporative cooling.
- 1 mark for erector pili muscles relaxing so hairs lie flat OR reduction in metabolic rate.

(b) [4 marks total]
- 1 mark for contrasting arteriole response: vasoconstriction in the cold vs. vasodilation in the heat.
- 1 mark for contrasting hair response: erector pili muscles contracting (hairs raised) in the cold vs. relaxing (hairs flat) in the heat.
- 1 mark for contrasting sweat response: sweat glands inactive in the cold vs. active sweating in the heat.
- 1 mark for describing the functional purpose of at least one of these contrasts (e.g., trapping insulating air in the cold vs. evaporative cooling in the heat).
PastPaper.question 9 · Structured
10 PastPaper.marks
Ventoxin is a respiratory toxin produced by some marine bacteria found near hydrothermal vents. It is a competitive inhibitor of the enzyme succinate dehydrogenase, which catalyses the conversion of succinate to fumarate in the Krebs cycle.

(a) Explain how the inhibition of succinate dehydrogenase by ventoxin affects the production of ATP by oxidative phosphorylation. (4 marks)

(b) Describe and explain the effect of ventoxin on the rate of oxygen consumption in isolated mitochondria. (3 marks)

(c) In an experiment, the rate of oxygen consumption of isolated mitochondria before adding ventoxin was \( 5.4 \text{ nmol mg}^{-1} \text{ min}^{-1} \). After adding ventoxin, the rate of oxygen consumption fell to \( 1.2 \text{ nmol mg}^{-1} \text{ min}^{-1} \).

Calculate the percentage decrease in the rate of oxygen consumption. Give your answer to 3 significant figures. Show your working. (3 marks)
PastPaper.showAnswers

PastPaper.workedSolution

(a) Succinate dehydrogenase catalyses the oxidation of succinate to fumarate, during which FAD is reduced to \( \text{FADH}_2 \). Inhibiting this enzyme reduces the production of \( \text{FADH}_2 \). Consequently, fewer electrons are donated by \( \text{FADH}_2 \) to the electron transport chain (ETC). This decreases the energy released along the ETC, reducing the active transport of protons (\( \text{H}^+ \) ions) across the inner mitochondrial membrane into the intermembrane space. The resulting weaker electrochemical proton gradient decreases the rate of chemiosmosis through ATP synthase, reducing ATP synthesis.

(b) The rate of oxygen consumption will decrease. Oxygen acts as the final electron acceptor at the end of the ETC, combining with electrons and protons to form water. Because the inhibition of the Krebs cycle reduces the supply of reduced coenzymes (\( \text{FADH}_2 \) and subsequently \( \text{NADH} \)), electron flow along the ETC slows down or stops. Therefore, less oxygen is consumed because there are fewer electrons and protons to accept.

(c)
Decrease in rate = \( 5.4 - 1.2 = 4.2 \text{ nmol mg}^{-1} \text{ min}^{-1} \)
Percentage decrease = \( \frac{4.2}{5.4} \times 100 = 77.777...\% \)
To 3 significant figures, this is \( 77.8\% \).

PastPaper.markingScheme

Part (a) [4 marks maximum]:
1. Succinate dehydrogenase converts succinate to fumarate AND reduces FAD / produces reduced FAD / \( \text{FADH}_2 \); (1)
2. (With inhibition) less reduced FAD / \( \text{FADH}_2 \) to deliver protons/electrons to the electron transport chain (ETC); (1)
3. Less energy released from the ETC to pump hydrogen ions / protons into the intermembrane space; (1)
4. Decreased proton / electrochemical gradient OR less chemiosmosis / less movement of protons through ATP synthase; (1)
5. Less phosphorylation of ADP to ATP; (1)

Part (b) [3 marks maximum]:
1. Oxygen consumption decreases / stops; (1)
2. Oxygen is the terminal / final electron acceptor; (1)
3. (With inhibition) fewer electrons / protons pass along the ETC, so less oxygen is reduced / less water is formed; (1)

Part (c) [3 marks maximum]:
1. Correct calculation of the difference in rate: \( 5.4 - 1.2 = 4.2 \); (1)
2. Correct percentage calculation setup: \( \frac{4.2}{5.4} \times 100 \); (1)
3. Correct final answer to 3 significant figures: \( 77.8\% \); (1)

[Note: Award full 3 marks for a correct final answer of 77.8% with or without working shown. Award 2 marks for 77.77% or 78% due to incorrect significant figures.]
PastPaper.question 10 · Structured
10 PastPaper.marks
Ventoxin is a respiratory toxin produced by some marine bacteria found near hydrothermal vents. It is a competitive inhibitor of the enzyme succinate dehydrogenase, which catalyses the conversion of succinate to fumarate in the Krebs cycle.

(a) Explain how the inhibition of succinate dehydrogenase by ventoxin affects the production of ATP by oxidative phosphorylation. (4 marks)

(b) Describe and explain the effect of ventoxin on the rate of oxygen consumption in isolated mitochondria. (3 marks)

(c) In an experiment, the rate of oxygen consumption of isolated mitochondria before adding ventoxin was \( 5.4 \text{ nmol mg}^{-1} \text{ min}^{-1} \). After adding ventoxin, the rate of oxygen consumption fell to \( 1.2 \text{ nmol mg}^{-1} \text{ min}^{-1} \).

Calculate the percentage decrease in the rate of oxygen consumption. Give your answer to 3 significant figures. Show your working. (3 marks)
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PastPaper.workedSolution

(a) Succinate dehydrogenase catalyses the oxidation of succinate to fumarate, during which FAD is reduced to \( \text{FADH}_2 \). Inhibiting this enzyme reduces the production of \( \text{FADH}_2 \). Consequently, fewer electrons are donated by \( \text{FADH}_2 \) to the electron transport chain (ETC). This decreases the energy released along the ETC, reducing the active transport of protons (\( \text{H}^+ \) ions) across the inner mitochondrial membrane into the intermembrane space. The resulting weaker electrochemical proton gradient decreases the rate of chemiosmosis through ATP synthase, reducing ATP synthesis.

(b) The rate of oxygen consumption will decrease. Oxygen acts as the final electron acceptor at the end of the ETC, combining with electrons and protons to form water. Because the inhibition of the Krebs cycle reduces the supply of reduced coenzymes (\( \text{FADH}_2 \) and subsequently \( \text{NADH} \)), electron flow along the ETC slows down or stops. Therefore, less oxygen is consumed because there are fewer electrons and protons to accept.

(c)
Decrease in rate = \( 5.4 - 1.2 = 4.2 \text{ nmol mg}^{-1} \text{ min}^{-1} \)
Percentage decrease = \( \frac{4.2}{5.4} \times 100 = 77.777...\% \)
To 3 significant figures, this is \( 77.8\% \).

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Part (a) [4 marks maximum]:
1. Succinate dehydrogenase converts succinate to fumarate AND reduces FAD / produces reduced FAD / \( \text{FADH}_2 \); (1)
2. (With inhibition) less reduced FAD / \( \text{FADH}_2 \) to deliver protons/electrons to the electron transport chain (ETC); (1)
3. Less energy released from the ETC to pump hydrogen ions / protons into the intermembrane space; (1)
4. Decreased proton / electrochemical gradient OR less chemiosmosis / less movement of protons through ATP synthase; (1)
5. Less phosphorylation of ADP to ATP; (1)

Part (b) [3 marks maximum]:
1. Oxygen consumption decreases / stops; (1)
2. Oxygen is the terminal / final electron acceptor; (1)
3. (With inhibition) fewer electrons / protons pass along the ETC, so less oxygen is reduced / less water is formed; (1)

Part (c) [3 marks maximum]:
1. Correct calculation of the difference in rate: \( 5.4 - 1.2 = 4.2 \); (1)
2. Correct percentage calculation setup: \( \frac{4.2}{5.4} \times 100 \); (1)
3. Correct final answer to 3 significant figures: \( 77.8\% \); (1)

[Note: Award full 3 marks for a correct final answer of 77.8% with or without working shown. Award 2 marks for 77.77% or 78% due to incorrect significant figures.]

PastPaper.section WBI16/01A - Unit 6: Practical Skills in Biology II

Answer ALL questions. Write your answers in the spaces provided in the Answer Book.
4 PastPaper.question · 50 PastPaper.marks
PastPaper.question 1 · Practical Investigation Planning
12.5 PastPaper.marks
Mung bean (Vigna radiata) is a vital legume crop, but its growth and germination are highly sensitive to soil salinity. Design a laboratory investigation to determine the effect of different concentrations of sodium chloride (NaCl) solution on the germination percentage of mung bean seeds. Your answer should include: a clear statement of the hypothesis, an experimental design that includes the control of independent, dependent, and key confounding variables, and a description of how the data will be collected, recorded, and analysed to test your hypothesis.
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PastPaper.workedSolution

Hypothesis: As the concentration of sodium chloride (NaCl) solution increases, the germination percentage of mung bean seeds decreases. Independent Variable: At least five different concentrations of NaCl (e.g., 0.0, 0.1, 0.2, 0.3, 0.4 mol dm^-3) prepared using serial dilution from a stock solution of 1.0 mol dm^-3. Dependent Variable: Percentage germination of mung bean seeds, calculated as (number of germinated seeds / total number of seeds) * 100. Germination is defined as the visible emergence of the radicle (at least 2 mm) from the seed coat within 5 days. Confounding Variables and Control: 1. Temperature: Maintained at a constant 25 degrees Celsius using a thermostatically controlled incubator, as temperature affects enzyme activity during germination. 2. Light conditions: Keep all petri dishes in complete darkness inside the incubator, to eliminate light as a factor triggering or inhibiting germination. 3. Volume of solution: Apply exactly 5.0 cm^3 of the respective NaCl solution to each petri dish containing a double layer of filter paper, to ensure seeds are equally moistened but not submerged (which would cause anaerobic conditions). 4. Seed selection: Use mung bean seeds of the same age, source, and batch, selecting those of similar size and free from visible damage. Method: 1. Surface-sterilise the mung bean seeds by soaking them in a 1% sodium hypochlorite (bleach) solution for 2 minutes, then rinse thoroughly with distilled water to prevent fungal contamination. 2. Label five petri dishes for each concentration (0.0 to 0.4 mol dm^-3) to provide five replicates per concentration (total of 25 dishes). 3. Place 20 sterilised seeds evenly spaced on the filter paper in each petri dish. 4. Add 5.0 cm^3 of the appropriate NaCl solution to each dish and cover with the lid. 5. Place all dishes in the incubator at 25 degrees Celsius. 6. Inspect the seeds daily for 5 days. Record the number of germinated seeds in each dish. Data Analysis: Calculate the mean germination percentage and standard deviation for each NaCl concentration. Plot a line graph of mean germination percentage against NaCl concentration. Perform a Spearman's rank correlation test or calculate Pearson's correlation coefficient to determine if there is a statistically significant negative correlation between salt concentration and germination percentage. Safety: Wear safety goggles and gloves when preparing dilutions and handling the bleach solution. Wash hands thoroughly after handling seeds.

PastPaper.markingScheme

1. Hypothesis [1 mark]: Clear, testable prediction linking increasing NaCl concentration to decreased germination percentage. 2. Independent Variable [2 marks]: Specifies at least 5 distinct NaCl concentrations [1 mark] and explains how they are prepared (e.g., serial or proportional dilution from a stock) [1 mark]. 3. Dependent Variable [2 marks]: Specifies calculating percentage germination [1 mark] and provides a clear, objective criterion for germination (e.g., radicle length > 2 mm) after a fixed time period [1 mark]. 4. Control of Confounding Variables [3 marks]: Identifies three key variables and explains how they are controlled: Temperature using an incubator [1 mark]; light conditions (darkness/constant light) [1 mark]; volume/moisture level of the solution [1 mark]. 5. Reliability and Replicates [1.5 marks]: Mentions using at least 3 to 5 replicates per concentration [1 mark] and calculating mean and standard deviation [0.5 marks]. 6. Experimental Procedure [2 marks]: Details of seed surface-sterilisation to prevent fungal growth [1 mark], setting a fixed observation period (e.g., 5 days) [1 mark]. 7. Analysis & Safety [1 mark]: Proposes an appropriate statistical test (e.g., Spearman's rank) to assess the correlation [0.5 marks] and states standard lab safety precautions (e.g., goggles, washing hands) [0.5 marks].
PastPaper.question 2 · statistical t-test
12.5 PastPaper.marks
An investigation was conducted to determine if there is a significant difference in the mean stomatal density between the upper (adaxial) and lower (abaxial) epidermis of the leaves of a spider plant (Chlorophytum comosum). The student collected 10 leaves, prepared epidermal nail varnish impressions, and counted the number of stomata per mm^2. The summary statistics calculated from the data are as follows: Upper epidermis (Group 1): Mean = 24.6 stomata per mm^2, Variance = 12.8, Sample size = 10. Lower epidermis (Group 2): Mean = 48.2 stomata per mm^2, Variance = 18.4, Sample size = 10. (a) State a null hypothesis for this investigation. (1.5) (b) Use the Student's t-test formula provided below to calculate the t-value for this data. Show your working. Formula: t = (|mean1 - mean2|) / sqrt((var1 / n1) + (var2 / n2)). (4) (c) Determine the degrees of freedom (df) and find the critical value at the 5% (p = 0.05) significance level using the following values: df = 16 (2.120), df = 17 (2.110), df = 18 (2.101), df = 19 (2.093), df = 20 (2.086). (3) (d) State and explain your statistical conclusion based on your calculated t-value and the critical value. (4)
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PastPaper.workedSolution

(a) Null Hypothesis: There is no significant difference in the mean stomatal density between the upper (adaxial) and lower (abaxial) epidermis of the leaves of Chlorophytum comosum. (b) Calculation: Mean 1 = 24.6, Mean 2 = 48.2. Absolute difference = |24.6 - 48.2| = 23.6. Variance 1 / n1 = 12.8 / 10 = 1.28. Variance 2 / n2 = 18.4 / 10 = 1.84. Sum of variances over sample sizes = 1.28 + 1.84 = 3.12. Square root of sum = sqrt(3.12) = 1.7664. t-value = 23.6 / 1.7664 = 13.36 (or 13.4). (c) Degrees of Freedom: df = (n1 + n2) - 2 = (10 + 10) - 2 = 18. Critical value at p = 0.05 for df = 18 is 2.101. (d) Conclusion: The calculated t-value (13.36) is much greater than the critical value (2.101) at the 5% significance level. Therefore, we reject the null hypothesis. There is a statistically highly significant difference in the mean stomatal density between the upper and lower epidermis of Chlorophytum comosum leaves, with the lower epidermis having a significantly higher density.

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(a) Null hypothesis [1.5 marks]: States that there is no significant difference [1 mark] between the mean stomatal densities of the upper and lower epidermis of the leaves [0.5 marks]. (b) Calculation [4 marks]: Correct absolute difference of means (23.6) [1 mark]; correct sum of variances divided by n (1.28 + 1.84 = 3.12) [1 mark]; correct square root calculation (1.766) [1 mark]; correct final t-value of 13.36 or 13.4 (accept 13.3 to 13.4) [1 mark]. (c) Degrees of freedom & Critical value [3 marks]: Correct df formula and value of 18 [1.5 marks]; correct critical value of 2.101 identified from the list [1.5 marks]. (d) Conclusion [4 marks]: Identifies that the calculated t-value is greater than the critical value [1 mark]; rejects the null hypothesis [1 mark]; states that the difference is statistically significant [1 mark]; relates back to the biological context (lower epidermis has higher stomatal density than upper epidermis) [1 mark].
PastPaper.question 3 · data presentation
12.5 PastPaper.marks
An investigation was carried out to study the effect of temperature on the rate of oxygen production by the freshwater pondweed Elodea canadensis. The volume of oxygen gas produced in 10 minutes at five different temperatures was recorded. The data are summarised below: At 15 degrees C: Mean = 13.0 mm^3, SD = 1.0 mm^3. At 25 degrees C: Mean = 30.0 mm^3, SD = 2.0 mm^3. At 35 degrees C: Mean = 45.0 mm^3, SD = 3.0 mm^3. At 45 degrees C: Mean = 18.0 mm^3, SD = 3.0 mm^3. At 55 degrees C: Mean = 2.0 mm^3, SD = 2.0 mm^3. (a) Describe how these data should be presented in a line graph to show the relationship between temperature and rate of oxygen production, including the plotting of error bars. (4) (b) Explain the physiological mechanisms that account for the trend in mean values observed between 15 degrees C and 35 degrees C, and between 35 degrees C and 55 degrees C. (5.5) (c) Discuss what the standard deviation error bars indicate about the reliability and significance of the difference between the rates of oxygen production at 25 degrees C and 35 degrees C. (3)
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PastPaper.workedSolution

(a) Graph Presentation: 1. Plot temperature (degrees C) on the horizontal x-axis as the independent variable, and the mean volume of oxygen produced in 10 minutes (mm^3) on the vertical y-axis as the dependent variable. 2. Use a linear scale starting at 0 for both axes, ensuring the plotted data occupies more than half of the grid area. 3. Plot the mean values as points and connect them with straight lines. 4. Plot error bars by drawing a vertical line through each mean data point, extending 1 SD above the mean and 1 SD below the mean, with horizontal caps at the ends (e.g., at 35 degrees C, the bar extends from 42 to 48 mm^3). (b) Physiological Explanations: 1. From 15 to 35 degrees C: As temperature increases, the kinetic energy of enzymes (such as Rubisco and the water-splitting complex) and substrate molecules increases. This leads to more frequent, high-energy, successful collisions, forming more enzyme-substrate complexes. Consequently, the rate of light-dependent and light-independent reactions increases, leading to a higher rate of photolysis of water and thus greater oxygen gas production. 2. From 35 to 55 degrees C: As temperature rises above the optimum (approx. 35 degrees C), the excessive kinetic energy causes weak bonds (such as hydrogen and ionic bonds) maintaining the tertiary structure of photosynthetic enzymes to break. This denatures the enzymes, changing the shape of their active sites so they are no longer complementary to their substrates. Photosynthesis decreases rapidly, resulting in a sharp decline in oxygen production. (c) Discussion of SD and Significance: 1. The standard deviations at all temperatures are relatively small (1.0 to 3.0 mm^3) compared to their respective means, indicating high precision and low variability between trials. 2. At 25 degrees C, the range is 28.0 to 32.0 mm^3, while at 35 degrees C, the range is 42.0 to 48.0 mm^3. There is no overlap between the error bars of these two temperatures. 3. The lack of overlap strongly suggests that the difference in mean oxygen production between 25 and 35 degrees C is statistically significant and unlikely to be due to random chance.

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(a) Graph description [4 marks]: Identifies temperature on the x-axis and mean volume of oxygen on the y-axis, with correct units [1 mark]; describes appropriate scale occupying >50% of grid [1 mark]; describes plotting points and connecting with straight lines [1 mark]; explains that error bars should be drawn extending 1 SD above and 1 SD below each mean point [1 mark]. (b) Physiological explanation [5.5 marks]: For 15 to 35 degrees C: mentions increased kinetic energy of enzymes/substrates [1 mark], more successful collisions/enzyme-substrate complexes [1 mark], and higher rate of photolysis/photosynthesis [1 mark]. For 35 to 55 degrees C: mentions temperature exceeding optimum denatures enzymes [1 mark], breaking of hydrogen/ionic bonds in tertiary structure [1 mark], active site shape changes so substrate cannot bind [0.5 marks]. (c) Discussion [3 marks]: Small SDs indicate high precision/reliability of data [1 mark]; states the ranges for 25 degrees C and 35 degrees C showing no overlap [1 mark]; concludes that the difference in mean rate is likely statistically significant [1 mark].
PastPaper.question 4 · Practical Investigation Planning
12.5 PastPaper.marks
The rate of photosynthesis in aquatic plants is affected by the wavelength of light. Design a laboratory investigation to determine the effect of different light wavelengths (colours of light) on the rate of oxygen production in the pondweed Cabomba. Your answer should include: a clear statement of the hypothesis, an experimental design that includes the control of independent, dependent, and key confounding variables, and a description of how the data will be collected, recorded, and analysed to test your hypothesis.
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PastPaper.workedSolution

Hypothesis: Red and blue light wavelengths will result in a higher rate of oxygen production (photosynthesis) in Cabomba compared to green light, as chlorophyll absorbs red and blue light but reflects green light. Independent Variable: Five different light wavelengths (colours: red, blue, green, yellow, and white as a control) achieved by placing high-quality coloured filters around a beaker containing the pondweed, or between the light source and the pondweed. Dependent Variable: Rate of photosynthesis measured as the volume of gas (oxygen) produced per minute (mm^3 min^-1). This is measured by collecting the gas in a capillary tube of a photosynthemometer (Audus micro-apparatus) and measuring the length of the gas bubble with a millimetre ruler, then converting to volume using the formula V = pi * r^2 * h. Confounding Variables and Control: 1. Light intensity: Controlled by using a light source (such as an LED lamp to avoid generating heat) kept at a fixed, measured distance (e.g., 20 cm) from the plant. 2. Temperature: Controlled by placing the test tube containing Cabomba in a large water bath of constant temperature (e.g., 20 degrees Celsius), which absorbs any excess thermal energy. 3. Carbon dioxide concentration: Maintained by submerging the Cabomba in a 1.0% sodium hydrogencarbonate (NaHCO3) solution, which provides an abundant and constant supply of dissolved CO2. 4. Plant material: Use the same freshly cut, 10 cm piece of Cabomba stem for all trials, cut at an angle to ensure a clear path for gas bubbles to escape. Method: 1. Set up the photosynthemometer with a freshly cut stem of Cabomba submerged in 1.0% NaCO3 solution inside a boiling tube. 2. Place the boiling tube in a water bath at 20 degrees Celsius. 3. Position the LED lamp 20 cm away. 4. Place the first filter (e.g., red) between the light source and the tube. 5. Allow the plant to acclimate for 10 minutes to reach a steady rate of photosynthesis. 6. Draw the gas bubble into the capillary tube using the syringe and measure its length after exactly 10 minutes. 7. Repeat the measurement three times for each colour filter by resetting the bubble, allowing for acclimation each time. 8. Calculate the mean volume of gas produced per minute for each colour. Data Analysis: Plot a bar chart of the mean rate of photosynthesis (y-axis) against the colour of light (x-axis), with standard deviation error bars. Perform a one-way ANOVA or multiple t-tests to determine if differences in rate between the wavelengths are statistically significant. Safety: Ensure electrical cords and light sources are kept dry and away from the water bath. Mop up any spills immediately to avoid slipping.

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1. Hypothesis [1 mark]: Clearly links light wavelength/colour to rate of photosynthesis (e.g., higher in red/blue than green). 2. Independent Variable [2 marks]: At least 4-5 different wavelengths/colours specified [1 mark], and describes using specific colour filters [1 mark]. 3. Dependent Variable [2 marks]: Describes measuring the volume of oxygen gas using a photosynthemometer / capillary tube and ruler over a fixed time [1 mark], and provides the rate calculation (volume per unit time) [1 mark]. 4. Confounding Variables [3 marks]: Identifies three key variables and explains their control: Light intensity (fixed distance of light source) [1 mark]; temperature (using a water bath) [1 mark]; carbon dioxide concentration (using sodium hydrogencarbonate solution) [1 mark]. 5. Reliability & Replicates [2 marks]: Specifies an acclimation period of at least 5-10 minutes under each light condition [1 mark]; mentions at least 3 replicates per wavelength to calculate mean and SD [1 mark]. 6. Experimental Procedure [1.5 marks]: Describes specific apparatus set up (photosynthemometer/capillary tube with syringe) [1 mark], cutting the stem at an angle to release bubbles [0.5 marks]. 7. Analysis & Safety [1 mark]: Recommends a bar chart with SD error bars and an ANOVA or t-test to assess statistical significance [0.5 marks]; identifies electrical safety with water as a precaution [0.5 marks].

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