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A Level H1 Biology Practice Paper 2

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TuitionGoWhere Practice Paper - Biology H1 A-Level

TuitionGoWhere Practice Paper (AI)

Subject: Biology H1
Level: A-Level
Paper: 2 (Structured & Free Response)
Duration: 2 hours
Total Marks: 80
Version: 2 of 5

Name: _________________________
Class: _________________________
Date: _________________________

Instructions to Candidates

This paper consists of two sections: Section A (Structured Questions) and Section B (Free Response Questions).
Answer all questions in Section A. Section A is worth 50 marks.
Answer one question from Section B. Section B is worth 30 marks.
Write your answers in the spaces provided. Additional paper may be used if necessary.
The use of an approved calculator is permitted.
You are advised to spend about 1 hour 15 minutes on Section A and 45 minutes on Section B.
Marks are indicated in brackets [ ] at the end of each part-question.

Section A: Structured Questions (50 marks)

Answer all questions in this section.

Question 1: Cell Membranes and Transport

Fig. 1.1 shows a diagram of a cell surface membrane.

(A simple labelled diagram of the fluid mosaic model is provided, showing phospholipid bilayer, channel protein, carrier protein, glycoprotein, and cholesterol.)

(a) With reference to Fig. 1.1, describe the arrangement of phospholipids in the membrane and explain how this arrangement contributes to the selective permeability of the membrane. [3]

(b) Name the type of transport that would be used to move glucose into a cell against its concentration gradient, and explain the role of the structure labelled X (a carrier protein) in this process. [3]

(c) Explain why cholesterol is an important component of animal cell membranes. [2]

(d) A student carried out an investigation into the effect of temperature on the permeability of beetroot cell membranes. The student placed identical discs of beetroot in water at different temperatures and measured the absorbance of the surrounding solution after 30 minutes. The results are shown in Table 1.1.

Table 1.1

Temperature / °C	Absorbance (arbitrary units)
10	0.12
20	0.15
30	0.18
40	0.25
50	0.48
60	0.92
70	1.35

(i) Describe the relationship between temperature and absorbance shown in Table 1.1. [2]

(ii) Explain the results at temperatures above 50 °C. [3]

[Total: 13 marks]

Question 2: Biological Molecules

(a) Fig. 2.1 shows the structure of a molecule of a disaccharide.

(A diagram of maltose is shown, with two glucose units linked by an α-1,4 glycosidic bond.)

(i) Name the disaccharide shown in Fig. 2.1. [1]

(ii) Name the type of reaction that forms the bond between the two monosaccharides, and explain why this bond is called a glycosidic bond. [2]

(b) Describe the structure of a triglyceride molecule and explain how its structure is related to its function as an energy store. [4]

(c) Collagen and haemoglobin are both proteins but have very different structures and functions. Compare and contrast the structure of collagen and haemoglobin. [4]

[Total: 11 marks]

Question 3: Enzymes

(a) Define the term enzyme and explain why enzymes are described as biological catalysts. [2]

(b) Fig. 3.1 shows the effect of substrate concentration on the rate of an enzyme-catalysed reaction.

(A graph is shown with rate of reaction on the y-axis and substrate concentration on the x-axis. The curve rises steeply initially and then plateaus.)

(i) Explain why the rate of reaction increases rapidly at low substrate concentrations. [2]

(ii) Explain why the rate of reaction reaches a maximum (Vmax) at high substrate concentrations. [2]

(c) Methanol is a competitive inhibitor of the enzyme alcohol dehydrogenase. Explain how methanol inhibits the enzyme and suggest how the inhibition could be overcome. [3]

[Total: 9 marks]

Question 4: DNA and Protein Synthesis

(a) Fig. 4.1 shows a short section of a DNA molecule.

(A diagram of DNA double helix is shown, with labels indicating deoxyribose, phosphate, and bases.)

(i) Name the type of bond labelled P (between phosphate and deoxyribose) and the type of bond labelled Q (between complementary bases). [2]

(ii) Explain how the structure of DNA allows it to replicate accurately. [3]

(b) A gene has the following base sequence on the template strand:

3' – TAC GGA TTC ACT – 5'

(i) Write the sequence of the mRNA molecule that would be transcribed from this gene. [1]

(ii) Using the genetic code table provided (Table 4.1), determine the amino acid sequence that would be translated from this mRNA.

Table 4.1 (extract)

Codon	Amino Acid
AUG	Methionine
CCU	Proline
AAG	Lysine
UGA	Stop

[2]

(iii) A mutation changes the fourth base in the template strand from G to C. Explain how this mutation could affect the protein produced. [3]

[Total: 11 marks]

Question 5: Cell Cycle and Mitosis

(a) Fig. 5.1 shows the amount of DNA per cell during the cell cycle.

(A graph is shown with DNA amount per cell on the y-axis and time on the x-axis. The graph shows a stepwise pattern: DNA amount is constant at 2 arbitrary units, then doubles to 4 units, remains constant, then halves back to 2 units.)

(i) Identify the phases of the cell cycle labelled A, B, and C on Fig. 5.1. [3]

(ii) Explain the changes in DNA amount that occur during phase B. [2]

(b) Describe the behaviour of chromosomes during anaphase of mitosis and explain how this ensures genetic stability. [3]

[Total: 8 marks]

Section B: Free Response Questions (30 marks)

Answer one question from this section.
Write your answer on separate paper.
Marks will be awarded for the quality of written communication and the logical presentation of ideas.

Question 6

(a) Describe the structure of a chloroplast and explain how its structure is adapted for the light-dependent reactions of photosynthesis. [10]

(b) Discuss the significance of membrane transport processes in the functioning of a plant leaf. In your answer, you should consider the movement of carbon dioxide, water, and mineral ions, and explain how these movements are essential for photosynthesis and transpiration. [20]

[Total: 30 marks]

Question 7

(a) Explain how the structure of DNA is related to its functions in storing genetic information and directing protein synthesis. [10]

(b) Discuss the importance of mitosis in the life of a multicellular organism. In your answer, you should refer to growth, repair, and asexual reproduction, and explain how mitosis maintains genetic stability. [20]

[Total: 30 marks]

End of Paper

Answers

TuitionGoWhere Practice Paper - Biology H1 A-Level

Answer Key and Marking Scheme

Version 2 of 5

Section A: Structured Questions (50 marks)

Question 1: Cell Membranes and Transport (13 marks)

(a) With reference to Fig. 1.1, describe the arrangement of phospholipids in the membrane and explain how this arrangement contributes to the selective permeability of the membrane. [3]

Answer:
Phospholipids form a bilayer with hydrophilic phosphate heads facing outwards towards the aqueous environment on both sides of the membrane, and hydrophobic fatty acid tails facing inwards, away from water. This arrangement creates a hydrophobic core that acts as a barrier to large, polar, or charged molecules, while allowing small, non-polar molecules (e.g., O₂, CO₂) and lipid-soluble substances to diffuse through. The bilayer thus contributes to selective permeability by restricting the passage of most water-soluble substances.

Marking:

1 mark: bilayer with heads out, tails in.
1 mark: hydrophobic core restricts polar/charged molecules.
1 mark: allows non-polar/lipid-soluble molecules to pass.

Answer:
Active transport. The carrier protein (X) binds glucose on the side of the membrane where its concentration is lower. Using energy from ATP hydrolysis, the carrier protein undergoes a conformational change, moving glucose across the membrane and releasing it on the side of higher concentration. The protein then returns to its original shape.

Marking:

1 mark: active transport.
1 mark: binding of glucose and conformational change.
1 mark: energy from ATP used / reference to ATP hydrolysis.

(c) Explain why cholesterol is an important component of animal cell membranes. [2]

Answer:
Cholesterol fits between phospholipid molecules, reducing the fluidity of the membrane at high temperatures (by restraining phospholipid movement) and preventing the membrane from becoming too rigid at low temperatures (by disrupting close packing of fatty acid tails). This helps maintain membrane stability and proper permeability over a range of temperatures.

Marking:

1 mark: regulates fluidity / prevents excessive fluidity at high temp.
1 mark: prevents rigidity at low temp / maintains stability.

(d) (i) Describe the relationship between temperature and absorbance shown in Table 1.1. [2]

Answer:
As temperature increases, absorbance increases. The increase is gradual from 10 °C to 40 °C, but becomes much steeper above 50 °C, with a sharp rise between 50 °C and 70 °C.

Marking:

1 mark: positive correlation / absorbance increases with temperature.
1 mark: reference to gradual increase then steep rise above 50 °C.

(ii) Explain the results at temperatures above 50 °C. [3]

Answer:
At temperatures above 50 °C, the high kinetic energy disrupts the hydrogen bonds and hydrophobic interactions that hold membrane proteins in their tertiary structure, causing them to denature. The phospholipid bilayer also becomes more fluid, and gaps may form. This increases membrane permeability, allowing more betalain pigment to leak out of the vacuole and cell, resulting in higher absorbance of the surrounding solution.

Marking:

1 mark: proteins denature / tertiary structure disrupted.
1 mark: membrane becomes more fluid / gaps form.
1 mark: more pigment leaks out → higher absorbance.

Question 2: Biological Molecules (11 marks)

(a) (i) Name the disaccharide shown in Fig. 2.1. [1]

Answer: Maltose.

(ii) Name the type of reaction that forms the bond between the two monosaccharides, and explain why this bond is called a glycosidic bond. [2]

Answer:
Condensation reaction. The bond is called a glycosidic bond because it forms between the anomeric carbon (carbon 1) of one glucose and the hydroxyl group on carbon 4 of the other glucose, with the elimination of a water molecule. The bond involves a sugar (glycose) molecule.

Marking:

1 mark: condensation reaction.
1 mark: bond between carbon atoms of sugars / involves sugar residues.

(b) Describe the structure of a triglyceride molecule and explain how its structure is related to its function as an energy store. [4]

Answer:
A triglyceride consists of one glycerol molecule esterified to three fatty acid chains. The fatty acids are long hydrocarbon chains, which are hydrophobic and contain many carbon–hydrogen bonds. When oxidised during respiration, these bonds release a large amount of energy (more than carbohydrates per gram). The hydrophobic nature allows triglycerides to be stored in a compact, anhydrous form in adipose tissue, without adding bulk from associated water. This makes them an efficient, lightweight energy reserve.

Marking:

1 mark: glycerol + 3 fatty acids / ester bonds.
1 mark: long hydrocarbon chains with many C–H bonds.
1 mark: high energy yield on oxidation.
1 mark: hydrophobic / stored without water → compact energy store.

(c) Collagen and haemoglobin are both proteins but have very different structures and functions. Compare and contrast the structure of collagen and haemoglobin. [4]

Answer:
Similarities: Both are proteins with quaternary structure (collagen: three polypeptide chains; haemoglobin: four subunits). Both contain hydrogen bonds stabilising their structure.
Differences: Collagen is a fibrous protein with a triple helix structure, while haemoglobin is a globular protein with a roughly spherical shape. Collagen has a repeating Gly-X-Y sequence and is insoluble, providing tensile strength in connective tissues. Haemoglobin has a prosthetic haem group, is soluble, and its quaternary structure allows cooperative oxygen binding for transport.

Marking:

1 mark: both have quaternary structure.
1 mark: collagen fibrous/triple helix vs. haemoglobin globular.
1 mark: collagen insoluble/structural vs. haemoglobin soluble/transport.
1 mark: haemoglobin has haem group / cooperative binding.

Question 3: Enzymes (9 marks)

(a) Define the term enzyme and explain why enzymes are described as biological catalysts. [2]

Answer:
An enzyme is a globular protein that speeds up a specific biochemical reaction by lowering the activation energy, without being consumed in the process. They are biological catalysts because they are produced by living cells and catalyse metabolic reactions.

Marking:

1 mark: protein that speeds up reaction / lowers activation energy.
1 mark: not consumed / biological origin.

(b) (i) Explain why the rate of reaction increases rapidly at low substrate concentrations. [2]

Answer:
At low substrate concentrations, many enzyme active sites are unoccupied. As substrate concentration increases, more enzyme–substrate complexes can form, so the rate of reaction increases proportionally. The increase is rapid because the availability of active sites is not yet limiting.

Marking:

1 mark: more active sites occupied as substrate increases.
1 mark: rate proportional to substrate concentration / not limited by enzyme.

(ii) Explain why the rate of reaction reaches a maximum (Vmax) at high substrate concentrations. [2]

Answer:
At high substrate concentrations, all enzyme active sites are saturated with substrate. The enzyme molecules are working at their maximum turnover rate. Adding more substrate cannot increase the rate because there are no free active sites available; the enzyme concentration becomes the limiting factor.

Marking:

1 mark: all active sites occupied / saturation.
1 mark: enzyme concentration is limiting / maximum turnover rate.

(c) Methanol is a competitive inhibitor of the enzyme alcohol dehydrogenase. Explain how methanol inhibits the enzyme and suggest how the inhibition could be overcome. [3]

Answer:
Methanol has a similar shape to the normal substrate (ethanol) and can bind to the active site of alcohol dehydrogenase, preventing ethanol from binding. This reduces the rate of ethanol breakdown. The inhibition can be overcome by increasing the concentration of ethanol, which outcompetes methanol for the active site, as competitive inhibition is reversible and depends on relative concentrations.

Marking:

1 mark: methanol binds to active site / competes with ethanol.
1 mark: prevents substrate binding / reduces enzyme activity.
1 mark: overcome by increasing ethanol concentration / outcompeting.

Question 4: DNA and Protein Synthesis (11 marks)

(a) (i) Name the type of bond labelled P (between phosphate and deoxyribose) and the type of bond labelled Q (between complementary bases). [2]

Answer:
P: Phosphodiester bond.
Q: Hydrogen bond.

(ii) Explain how the structure of DNA allows it to replicate accurately. [3]

Answer:
DNA is a double helix with two antiparallel strands held together by hydrogen bonds between complementary base pairs (A–T and C–G). During replication, the strands separate, and each acts as a template. Free DNA nucleotides align opposite their complementary bases on the template strand (A with T, C with G) and are joined by DNA polymerase, forming new complementary strands. This semi-conservative mechanism ensures that each new DNA molecule contains one original strand and one new strand, preserving the genetic code.

Marking:

1 mark: complementary base pairing (A–T, C–G).
1 mark: each strand acts as a template.
1 mark: semi-conservative replication / accurate copying.

(b) (i) Write the sequence of the mRNA molecule that would be transcribed from this gene. [1]

Answer:
5' – AUG CCU AAG UGA – 3'
(Note: template strand 3'→5' TAC GGA TTC ACT, so mRNA is complementary: AUG CCU AAG UGA.)

(ii) Using the genetic code table, determine the amino acid sequence. [2]

Answer:
AUG – Methionine
CCU – Proline
AAG – Lysine
UGA – Stop
Sequence: Methionine – Proline – Lysine (stop codon does not code for an amino acid).

Marking:

1 mark: correct amino acids for first three codons.
1 mark: stop codon recognised / no amino acid for UGA.

(iii) A mutation changes the fourth base in the template strand from G to C. Explain how this mutation could affect the protein produced. [3]

Answer:
Original template: TAC GGA TTC ACT → mRNA: AUG CCU AAG UGA.
Mutation: fourth base G → C, so template becomes TAC CCA TTC ACT → mRNA: AUG GGU AAG UGA.
The second codon changes from CCU (Proline) to GGU (Glycine). This is a missense mutation, altering the primary structure of the polypeptide. The substitution of a different amino acid may affect the folding of the protein (secondary/tertiary structure), potentially altering its function. If the new amino acid has different properties (e.g., charge, size), the protein may become non-functional.

Marking:

1 mark: identifies change in codon / amino acid (Pro → Gly).
1 mark: missense mutation / change in primary structure.
1 mark: may affect folding/function of protein.

Question 5: Cell Cycle and Mitosis (8 marks)

(a) (i) Identify the phases of the cell cycle labelled A, B, and C on Fig. 5.1. [3]

Answer:
A: G1 phase (or Gap 1).
B: S phase (DNA synthesis).
C: G2 phase (or Gap 2) / mitosis (if the drop is at division; depending on graph, C is the phase after DNA doubling and before halving, so G2. The halving occurs at mitosis/cytokinesis, which might be labelled separately. Assuming graph shows constant 4 units then drop to 2, C is the period of 4 units before drop, so G2. Accept G2 or G2 + mitosis if the drop is part of C.)

Marking:

1 mark each for correct identification: A = G1, B = S, C = G2 (or G2/M).

(ii) Explain the changes in DNA amount that occur during phase B. [2]

Answer:
During S phase, DNA replication occurs. Each chromosome is duplicated, producing two identical sister chromatids held together at the centromere. The amount of DNA per cell doubles from 2 arbitrary units to 4 arbitrary units as DNA polymerase synthesises new complementary strands using the existing strands as templates.

Marking:

1 mark: DNA replication / semi-conservative replication.
1 mark: DNA amount doubles / sister chromatids formed.

(b) Describe the behaviour of chromosomes during anaphase of mitosis and explain how this ensures genetic stability. [3]

Answer:
During anaphase, the centromeres divide, and the sister chromatids of each chromosome are pulled apart to opposite poles of the cell by the shortening of spindle fibres. This ensures that each daughter cell receives an identical set of chromosomes (same number and genetic content) as the parent cell. Genetic stability is maintained because the genetic material is equally and accurately distributed.

Marking:

1 mark: centromeres divide / sister chromatids separate.
1 mark: chromatids move to opposite poles via spindle fibres.
1 mark: each daughter cell gets identical genetic information.

Section B: Free Response Questions (30 marks)

Question 6

(a) Describe the structure of a chloroplast and explain how its structure is adapted for the light-dependent reactions of photosynthesis. [10]

Answer (indicative content):

Chloroplast is surrounded by a double membrane (envelope).
Inside is the stroma, containing enzymes for the Calvin cycle, but the light-dependent reactions occur on the thylakoid membranes.
Thylakoids are flattened membrane sacs, often stacked into grana, providing a large surface area for the attachment of photosystems, electron carriers, and ATP synthase.
The thylakoid membrane contains chlorophyll and other photosynthetic pigments organised into photosystems I and II, which absorb light energy efficiently.
The arrangement of pigments allows resonance energy transfer to the reaction centre.
The thylakoid space (lumen) is small, allowing a proton gradient to build up quickly during photophosphorylation.
ATP synthase is embedded in the thylakoid membrane, allowing protons to flow back into the stroma, driving ATP synthesis (chemiosmosis).
The membrane also contains electron carriers (plastoquinone, cytochrome complex, plastocyanin) arranged in a chain for electron transport.
The stroma contains NADP⁺, which is reduced to NADPH at the end of the electron transport chain.
The large surface area of thylakoid membranes maximises light absorption and the rate of ATP and NADPH production.

Marking:

1–2 marks: basic description of chloroplast structure (double membrane, stroma, thylakoids, grana).
3–4 marks: description of thylakoid membrane components (photosystems, pigments, electron carriers, ATP synthase).
5–6 marks: explanation of adaptations for light absorption (large surface area, pigment arrangement).
7–8 marks: explanation of adaptations for ATP synthesis (proton gradient, chemiosmosis, ATP synthase location).
9–10 marks: comprehensive, well-structured answer linking all adaptations to light-dependent reactions, with correct terminology and logical flow.

(b) Discuss the significance of membrane transport processes in the functioning of a plant leaf. [20]

Answer (indicative content):

Introduction: Plant leaves are the main sites of photosynthesis and transpiration; membrane transport is crucial for gas exchange, water movement, and nutrient uptake.
Carbon dioxide uptake: CO₂ diffuses into the leaf through stomata, then across the cell wall, cell membrane, and chloroplast envelope into the stroma. The movement is by simple diffusion down a concentration gradient, maintained by CO₂ fixation in the Calvin cycle. The thin, flat shape of leaf cells and the large surface area of mesophyll cells facilitate rapid diffusion.
Water movement: Water enters root hair cells by osmosis and moves through the plant via the xylem to leaf cells. In the leaf, water moves from xylem to mesophyll cells by osmosis (apoplast/symplast pathways). Water evaporates from moist cell walls into air spaces and diffuses out through stomata (transpiration). The cohesion-tension theory explains the upward pull. Membrane transport of water is essential for maintaining turgor pressure, which keeps leaves expanded for light capture and stomatal opening.
Mineral ion uptake: Mineral ions (e.g., Mg²⁺, NO₃⁻, K⁺) are absorbed by root hair cells via active transport and facilitated diffusion, then transported in xylem to leaves. In leaf cells, ions are taken up across cell membranes. Mg²⁺ is a component of chlorophyll; NO₃⁻ is needed for amino acid and protein synthesis; K⁺ regulates stomatal opening by moving into guard cells, lowering water potential, causing water entry and guard cell swelling.
Significance for photosynthesis: CO₂ entry is directly needed for the Calvin cycle. Water is a raw material for photolysis in the light-dependent reactions, providing electrons and protons. Mineral ions are cofactors and structural components. Membrane transport ensures these substances reach the chloroplasts.
Significance for transpiration: Transpiration stream delivers water and dissolved minerals; evaporation cools the leaf. Membrane transport (osmosis) maintains the water potential gradient that drives transpiration.
Integration: The coordinated function of membrane transport processes ensures that photosynthesis can proceed at an optimal rate, supporting plant growth and productivity. Disruption of any transport process (e.g., stomatal closure, membrane damage) reduces photosynthetic efficiency.

Marking:

1–4 marks: basic description of one or two transport processes with limited linkage to leaf function.
5–8 marks: describes CO₂, water, and ion transport with some explanation of significance; may lack detail or integration.
9–12 marks: clear explanation of each transport process, with good linkage to photosynthesis and transpiration; uses appropriate terminology.
13–16 marks: detailed, accurate explanation of all processes, with strong integration and discussion of significance; well-structured.
17–20 marks: comprehensive, fluent essay demonstrating deep understanding; all processes fully explained and linked to leaf function; may include examples of consequences if transport fails; excellent written communication.

Question 7

(a) Explain how the structure of DNA is related to its functions in storing genetic information and directing protein synthesis. [10]

Answer (indicative content):

DNA is a double-stranded polymer of nucleotides, each consisting of deoxyribose, phosphate, and a nitrogenous base (A, T, C, G).
The two strands are antiparallel and held together by hydrogen bonds between complementary base pairs (A–T, C–G).
Storing genetic information: The sequence of bases along the DNA strand forms the genetic code. The code is a triplet code—each sequence of three bases (codon) specifies one amino acid. The order of bases determines the primary structure of all proteins. The double-stranded structure with complementary base pairing allows accurate replication, ensuring genetic information is passed to daughter cells.
Directing protein synthesis: During transcription, one strand of DNA acts as a template for the synthesis of mRNA. RNA polymerase binds to the promoter and synthesises a complementary mRNA strand using base-pairing rules (A–U, T–A, C–G, G–C). The mRNA carries the genetic code from the nucleus to ribosomes. During translation, the sequence of codons on mRNA is read by ribosomes, and tRNA molecules with complementary anticodons bring specific amino acids. The amino acids are joined by peptide bonds to form a polypeptide. The specificity of base pairing ensures the correct amino acid sequence.
The stability of DNA (due to hydrogen bonds and the sugar-phosphate backbone) protects the genetic information from degradation. The double helix allows for compact storage in chromosomes.
Mutations (changes in base sequence) can alter the protein produced, demonstrating the direct relationship between DNA sequence and protein function.

Marking:

1–2 marks: basic description of DNA structure.
3–4 marks: explanation of how base sequence stores information (triplet code).
5–6 marks: description of transcription and translation with reference to DNA’s role.
7–8 marks: clear linkage between structure (complementary base pairing, template strand) and accurate information transfer.
9–10 marks: comprehensive answer with correct terminology, logical flow, and reference to replication, transcription, and translation.

(b) Discuss the importance of mitosis in the life of a multicellular organism. [20]

Answer (indicative content):

Introduction: Mitosis is nuclear division that produces two genetically identical daughter nuclei. It is essential for growth, repair, and asexual reproduction.
Growth: Multicellular organisms grow by increasing cell number through mitosis. After fertilisation, the zygote undergoes repeated mitotic divisions to form an embryo. In plants and animals, mitosis in meristems and growth plates allows increase in size. All cells produced are genetically identical, ensuring tissue uniformity.
Repair and replacement: Damaged or worn-out cells are replaced by mitosis. For example, skin cells, red blood cells, and gut lining cells are constantly renewed. Mitosis ensures that new cells have the same genetic information as the cells they replace, maintaining tissue function.
Asexual reproduction: Many organisms reproduce asexually via mitosis, producing offspring that are genetically identical clones. This is advantageous in stable environments, allowing rapid population increase. Examples: budding in yeast, runners in strawberries, binary fission in bacteria (though prokaryotes, analogous process).
Genetic stability: Before mitosis, DNA replicates during S phase, producing two identical sister chromatids per chromosome. During mitosis, the chromatids separate precisely to opposite poles, ensuring each daughter cell receives an exact copy of the genome. This maintains the chromosome number and genetic consistency across cell generations.
Consequences of errors: If mitosis goes wrong (e.g., non-disjunction), daughter cells may have abnormal chromosome numbers, leading to genetic disorders or cancer. The importance of checkpoints and controlled mitosis is highlighted.
Importance in development: Cell differentiation often follows mitotic divisions; the genetically identical cells can then specialise, forming tissues and organs.
Conclusion: Mitosis is fundamental to life, enabling growth, maintenance, and reproduction while preserving genetic integrity.

Marking:

1–4 marks: superficial description of mitosis or one aspect of its importance.
5–8 marks: describes growth, repair, and asexual reproduction with some explanation; may lack detail on genetic stability.
9–12 marks: clear explanation of each role, with reference to genetic stability; uses examples.
13–16 marks: detailed, accurate discussion integrating all aspects; explains how mitosis maintains genetic stability; well-structured.
17–20 marks: comprehensive essay demonstrating deep understanding; discusses consequences of errors, checkpoints, and differentiation; excellent written communication and logical flow.

End of Answer Key