A Level H2 Biology Practice Paper 5

TuitionGoWhere Practice Paper - Biology H2 A-Level

Answer Key and Marking Scheme

Subject: Biology H2 (9477) Paper: Practice Paper 5 (Cells & Biomolecules) Total Marks: 75

---

Section A: Structured Questions (30 marks)

---

Question 1

(a) [3 marks]

• P: Mitochondrion / Mitochondria [1]
• Q: Rough endoplasmic reticulum / RER [1]
• R: Golgi apparatus / Golgi body / Golgi complex [1]

(b) [2 marks]

• Semi-autonomous means the organelle can replicate independently of the cell cycle / contains its own genetic material and ribosomes / can synthesise some of its own proteins [1].
• Evidence: Mitochondria contain their own circular DNA (mtDNA) / mitochondria contain 70S ribosomes similar to prokaryotes / mitochondria divide by binary fission [1].

(c) [2 marks]

• Similarity: Both are composed of rRNA and proteins / both have a large and small subunit [1].
• Difference: Organelle Q (RER ribosomes) are 80S ribosomes, while prokaryotic ribosomes are 70S / RER ribosomes are larger than prokaryotic ribosomes [1].

---

Question 2

(a) [1 mark]

• As temperature increases, the distance diffused increases / positive correlation between temperature and distance diffused [1].

(b) [2 marks]

• At higher temperatures, particles have greater kinetic energy [1].
• Particles move faster, so they diffuse more rapidly / collide more frequently, spreading further in the same time period [1].

(c) [2 marks]

• The suggestion is incorrect / unlikely to be correct [1].
• At very high temperatures, the agar jelly would melt / the protein structure of the agar would denature / the medium would break down, so diffusion cannot be measured / at extremely high temperatures, the crystals may decompose [1].

---

Question 3

(a) [3 marks]

• X: Phospholipid / phospholipid bilayer [1]
• Y: Integral protein / intrinsic protein / channel protein / carrier protein [1]
• Z: Cholesterol [1]

(b) [2 marks]

• Phospholipids have a hydrophilic (polar) phosphate head and two hydrophobic (non-polar) fatty acid tails [1].
• In an aqueous environment, the hydrophilic heads face outwards towards the water, while the hydrophobic tails face inwards away from the water, forming a bilayer [1].

(c) [2 marks]

• Any two from:
  • Cell recognition / cell-to-cell recognition [1]
  • Cell adhesion / binding cells together [1]
  • Receptor sites for hormones / neurotransmitters / signalling molecules [1]
  • Antigens / involved in immune response [1]
  • Forming the glycocalyx [1]

---

Question 4

(a) [2 marks]

• The water potential of the potato cells is approximately equal to the water potential of the sucrose solution where there is no net change in mass / where the percentage change is zero [1].
• From the graph/interpolation, this occurs at approximately 0.45–0.50 mol dm⁻³ sucrose / between 0.4 and 0.6 mol dm⁻³ where the line crosses zero [1].

(b) [2 marks]

• The 0.0 mol dm⁻³ sucrose solution (distilled water) has a higher water potential (less negative / zero) than the potato cells [1].
• Water enters the cells by osmosis down the water potential gradient, causing the cells to swell and gain mass [1].

(c) [2 marks]

• The prediction is likely to be correct because the trend shows increasing mass loss with increasing sucrose concentration [1].
• However, the relationship may not be linear at very high concentrations / the cells may become fully plasmolysed and mass loss may plateau / at very high concentrations, the water potential gradient may not increase proportionally [1].

---

Question 5

(a) [1 mark]

• Maltose [1]

(b) [1 mark]

• α-1,4-glycosidic bond / alpha-1,4-glycosidic bond [1]

(c) [2 marks]

• Maltose is formed by a condensation reaction between two α-glucose molecules [1].
• A molecule of water is removed, and a glycosidic bond forms between carbon-1 of one glucose and carbon-4 of the other [1].

(d) [2 marks]

• Maltose has a free anomeric carbon / free carbon-1 on one of the glucose residues that is not involved in the glycosidic bond [1].
• This free anomeric carbon can open to form an aldehyde group, which can reduce Benedict's reagent / act as a reducing agent [1].

---

Section B: Data-Based and Extended Response Questions (30 marks)

---

Question 6

(a) [2 marks]

• Bacillus amylase has an optimum pH of approximately 8.0 (activity = 88 arbitrary units) [1].
• Aspergillus amylase has an optimum pH of approximately 6.0 (activity = 85 arbitrary units) / Aspergillus amylase has a lower pH optimum than Bacillus amylase [1].

(b) [3 marks]

• At pH values above and below the optimum, the charges on the amino acid side chains in the active site are altered [1].
• This disrupts the ionic and hydrogen bonds that maintain the specific three-dimensional shape (tertiary structure) of the enzyme [1].
• The active site is no longer complementary to the substrate, so the enzyme-substrate complex cannot form / the enzyme is denatured at extreme pH [1].

(c) [3 marks]

• The enzyme has a high proportion of disulfide bonds / more ionic bonds / more hydrophobic interactions that stabilise its tertiary structure at high temperatures [1].
• The enzyme's amino acid composition results in an active site with charges that are optimal at pH 8.0 [1].
• The enzyme is adapted to the hot spring environment, so it is not denatured at 70°C and functions most efficiently at pH 8.0, allowing the bacterium to digest starch in its habitat [1].

---

Question 7

(a) [3 marks]

• A: Outer mitochondrial membrane [1]
• B: Crista / cristae [1]
• C: Matrix [1]

(b) [3 marks]

• The inner membrane is highly folded into cristae, which greatly increases the surface area for the electron transport chain and ATP synthase [1].
• The membrane contains the protein complexes of the electron transport chain (NADH dehydrogenase, cytochrome b-c1 complex, cytochrome c oxidase) for electron transfer and proton pumping [1].
• The membrane contains ATP synthase / stalked particles, which use the proton gradient to synthesise ATP via chemiosmosis [1].
• The membrane is impermeable to protons (H⁺), allowing a proton gradient to be established [1].

(c) [2 marks]

• Graph should show:
  • A steady decrease in oxygen concentration before point X (indicating oxygen consumption by respiring mitochondria) [1].
  • After point X, the rate of oxygen consumption decreases significantly / the line becomes almost flat (indicating inhibition of the electron transport chain) [1].

(d) [2 marks]

• Sodium azide inhibits cytochrome c oxidase (Complex IV), the final enzyme in the electron transport chain [1].
• This prevents the transfer of electrons to oxygen (the final electron acceptor), so oxygen is no longer reduced to water, and oxygen consumption stops or decreases dramatically [1].

---

Question 8

(a) [4 marks]

• Primary structure: The repeating amino acid sequence Gly-X-Y, where X is often proline and Y is often hydroxyproline / a specific sequence rich in glycine, proline, and hydroxyproline [1].
• Secondary structure: Each polypeptide chain forms a left-handed helix / a tight, extended helix (not an α-helix) [1].
• Tertiary structure: Three polypeptide chains wind around each other to form a right-handed triple helix / tropocollagen molecule [1].
• The structure is stabilised by hydrogen bonds between the chains and covalent cross-links between lysine and hydroxylysine residues [1].

(b) [3 marks]

• Collagen molecules assemble into fibrils, which are further bundled into fibres, providing high tensile strength [1].
• The triple helix structure and covalent cross-links between molecules make collagen strong and resistant to stretching [1].
• This allows tendons to transmit the force of muscle contraction to bones without breaking / collagen provides structural support in tendons [1].

(c) [3 marks]

• Vitamin C is a cofactor for the enzymes prolyl hydroxylase and lysyl hydroxylase, which catalyse the hydroxylation of proline and lysine residues in collagen [1].
• Without hydroxylation, fewer hydrogen bonds form between the three polypeptide chains, so the triple helix is less stable [1].
• The collagen molecules are weaker and cannot form strong fibres, leading to weakened connective tissues, fragile blood vessels, and poor wound healing [1].

---

Question 9

(a) [3 marks]

• D: Phosphate group [1]
• E: Ribose / pentose sugar [1]
• F: Nitrogenous base / adenine / guanine / cytosine / uracil [1]

(b) [1 mark]

• The sugar is ribose (in RNA) instead of deoxyribose (in DNA) / RNA contains uracil instead of thymine [1].

(c) [2 marks]

• Nucleotides are joined by phosphodiester bonds formed through condensation reactions [1].
• The phosphate group of one nucleotide bonds to the hydroxyl group on carbon-3' of the sugar of the next nucleotide, forming a sugar-phosphate backbone [1].

---

Section C: Essay Question (15 marks)

---

Question 10

Marking scheme for essay question:

The essay should demonstrate a comprehensive understanding of the role of hydrogen bonding in biological molecules. Marks are allocated for content (maximum 12 marks) and for the quality of written communication and organisation (maximum 3 marks).

Content marks (maximum 12):

Candidates should refer to at least three different types of biological molecules. Award marks for accurate, detailed explanations of the role of hydrogen bonding in each.

1. Water (up to 4 marks):

• Water molecules are polar, with δ+ on hydrogen atoms and δ- on oxygen atom [1].
• Hydrogen bonds form between the δ+ hydrogen of one water molecule and the δ- oxygen of another [1].
• Hydrogen bonding gives water its properties: high specific heat capacity (many H-bonds absorb energy before temperature rises), high latent heat of vaporisation (H-bonds must be broken for evaporation), cohesion and surface tension (H-bonds between water molecules), and solvent properties (H-bonds with polar solutes) [up to 2 marks for explaining at least two properties].

2. Proteins (up to 4 marks):

• Hydrogen bonds form between the C=O and N-H groups of amino acids in the polypeptide backbone [1].
• In secondary structure: H-bonds stabilise α-helices (between every fourth amino acid) and β-pleated sheets (between adjacent polypeptide strands) [1].
• In tertiary structure: H-bonds form between R-groups of amino acids (e.g., between serine, threonine, tyrosine residues) [1].
• Hydrogen bonding is essential for maintaining the specific three-dimensional shape of proteins, which determines their function (e.g., enzyme active sites, antibody binding sites) [1].

3. DNA (up to 4 marks):

• Hydrogen bonds form between complementary nitrogenous bases on the two antiparallel polynucleotide strands [1].
• Adenine forms two hydrogen bonds with thymine; guanine forms three hydrogen bonds with cytosine [1].
• Hydrogen bonding holds the two strands together in the double helix, but allows them to separate during replication and transcription [1].
• The specificity of base pairing (complementary) ensures accurate replication and transmission of genetic information [1].

4. Polysaccharides / Carbohydrates (up to 3 marks):

• Hydrogen bonds form between adjacent cellulose chains in plant cell walls [1].
• These H-bonds cross-link cellulose microfibrils, giving cellulose high tensile strength [1].
• This enables plant cell walls to withstand turgor pressure and provide structural support [1].

5. Collagen (up to 3 marks):

• Hydrogen bonds form between the three polypeptide chains in the collagen triple helix [1].
• These H-bonds, along with covalent cross-links, stabilise the tropocollagen structure [1].
• This contributes to the high tensile strength of collagen fibres in connective tissues [1].

Quality of written communication (maximum 3 marks):

• 3 marks: Answer is well-structured, uses appropriate scientific terminology accurately, and presents a logical and coherent argument with clear links between structure and function.
• 2 marks: Answer is mostly well-organised with some use of scientific terminology; ideas are generally clear but may lack full coherence.
• 1 mark: Answer has some relevant content but lacks organisation; limited or inaccurate use of scientific terminology.
• 0 marks: Answer is poorly organised with little or no relevant scientific content.

---

END OF ANSWER KEY

---

© TuitionGoWhere 2025. This is an AI-generated answer key intended for educational use.