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A Level H2 Biology Practice Paper 1

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A Level H2 Biology AI Generated Generated by Qwen3.6 Plus Updated 2026-08-17
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TuitionGoWhere Practice Paper - Biology H2 A-Level

Answer Key and Marking Scheme

Subject: Biology H2
Topic Focus: Cells & Biomolecules
Version: 1 of 5

Section A: Structured Questions

1. (a) Identification: Cholesterol [1].
Function: Regulates membrane fluidity / stabilises the membrane / reduces permeability to small water-soluble molecules [1].
(b) Fluid: Phospholipids and proteins can move laterally within the layer [1].
Mosaic: Proteins are embedded in the bilayer in a scattered pattern [1].
(c) Difference 1: Simple diffusion does not require a transport protein; facilitated diffusion requires a channel or carrier protein [1].
Difference 2: Simple diffusion is not saturable (rate increases linearly with concentration); facilitated diffusion is saturable (reaches VmaxV_{max}) [1].
(Accept: Simple diffusion moves down concentration gradient only; facilitated can be gated. Do not accept 'active transport' references.)

2. (a) The minimum amount of energy required for reactants to undergo a chemical reaction [1].
(b) (i) As temperature increases, kinetic energy of molecules increases [1]. This leads to more frequent and successful collisions between enzyme and substrate [1].
(ii) High temperature breaks hydrogen bonds and other interactions maintaining the tertiary structure [1]. The active site changes shape (denaturation) [1], so substrate can no longer bind.
(c) VmaxV_{max}: Decreases [1].
Explanation: The inhibitor binds to an allosteric site, changing the shape of the active site, so fewer enzyme-substrate complexes form even at high substrate concentrations [1].
KmK_m: Remains unchanged (or increases slightly depending on definition, but typically unchanged in pure non-competitive) [1]. Note: In A-Level H2, non-competitive inhibitors are often taught to decrease VmaxV_{max} and leave KmK_m unchanged or effectively increase apparent KmK_m if mixed. Accept: KmK_m unchanged as affinity of remaining active enzymes is same.

3. (a) Water molecules are polar (dipole) [1]. The positive hydrogen ends attract negative ions (anions) and negative oxygen ends attract positive ions (cations), surrounding them and keeping them in solution [1].
(b) Hydrogen bonds form between polar R-groups or between backbone atoms (C=O and N-H) [1]. These bonds stabilise the secondary (alpha-helix/beta-sheet) and tertiary structures [1].
(c) Property: High specific heat capacity [1].
Explanation: Much energy is required to break hydrogen bonds between water molecules, so water temperature changes slowly, buffering organisms against temperature fluctuations [1].
(Accept: High latent heat of vaporisation for cooling via sweating.)

4. (a) Inner membrane folded into cristae to increase surface area for electron transport chain enzymes [1]. Matrix contains enzymes for the Krebs cycle [1]. Double membrane creates compartments for chemiosmosis [1].
(b) Electrons from reduced NAD/FAD pass through carrier proteins [1]. Energy released is used to pump protons (H+H^+) from the matrix to the intermembrane space [1]. This creates an electrochemical gradient [1].

5. (a) 1. DNA is double-stranded; RNA is single-stranded [1].
2. DNA contains deoxyribose sugar; RNA contains ribose sugar [1].
3. DNA contains thymine; RNA contains uracil [1].
(b) DNA helicase unwinds the helix [1]. DNA polymerase adds free nucleotides to the 3' end of the growing strand [1]. It joins nucleotides via phosphodiester bonds in a 5' to 3' direction [1].

6. (a) Proteins are separated based on their size/molecular mass and charge [1]. An electric field is applied, causing charged proteins to migrate through the gel matrix at different rates [1].
(b) Heterozygotes have two different alleles (HbA and HbS) [1]. These alleles code for haemoglobin proteins with different charges/masses, so they migrate to different positions, forming two distinct bands [1].
(c) Heterozygotes (or those with some HbS) are resistant to malaria because the sickle-shaped cells inhibit the growth of the Plasmodium parasite [1]. This provides a survival advantage in malaria-endemic regions [1].

7. (a) Mitosis: Produces 2 daughter cells [1], genetically identical to parent [1].
Meiosis: Produces 4 daughter cells [1], genetically different/haploid [1]. (Award marks for clear comparison.)
(b) Crossing over involves the exchange of genetic material between non-sister chromatids of homologous pairs [1]. This creates new combinations of alleles (recombinants), increasing genetic variation [1].

8. (a) Condensation reaction between the carboxyl group of one amino acid and the amino group of another [1]. A molecule of water is removed, forming a peptide bond [1].
(b) The primary structure determines the folding pattern [1]. A change in amino acid may alter R-group interactions (e.g., ionic, hydrogen, disulphide bonds) [1]. This changes the 3D shape (tertiary structure), potentially altering the active site and preventing substrate binding [1].

9. (a) The movement of molecules/ions against their concentration gradient [1], using energy (ATP) and carrier proteins [1].
(b) 1. Na+ binds to the pump inside the cell [1].
2. ATP phosphorylates the pump, causing a conformational change [1].
3. Na+ is released outside the cell [1].
4. K+ binds from outside, causing dephosphorylation and return to original shape, releasing K+ inside [1].

10. (a) Starch: Helical/compact, insoluble, alpha-glucose, 1,4 and 1,6 glycosidic bonds, storage [1].
Cellulose: Straight chains, H-bonds between chains form microfibrils, beta-glucose, 1,4 glycosidic bonds, structural [1].
(Award up to 4 marks for clear comparative points.)
(b) Glycogen is highly branched [1], allowing for rapid hydrolysis to release glucose when needed [1]. It is compact and insoluble, so it does not affect water potential [1].

11. (a) Graph:

  • Axes labelled correctly (Substrate Conc vs Rate) with units [1].
  • Scale appropriate and uniform [1].
  • Points plotted correctly [1].
  • Smooth curve drawn (not dot-to-dot) showing plateau [1].
    (b) At high substrate concentrations, all active sites are occupied (saturated) [1]. The enzyme is working at VmaxV_{max} [1]. Adding more substrate cannot increase the rate further as there are no free active sites [1].
    (c) The rate would increase at all substrate concentrations below saturation [1]. The curve would rise more steeply initially [1]. VmaxV_{max} would be higher because higher kinetic energy leads to more successful collisions [1].

12. (a) Light-dependent: Thylakoid membrane/grana [1].
Light-independent: Stroma [1].
(b) ATP provides energy for the reduction of GP to TP [1]. Reduced NADP provides hydrogen/electrons for the reduction of GP to TP [1]. Both are regenerated and return to the light-dependent reaction [1].
(c) Thylakoid membranes contain photosystems (chlorophyll) and electron transport chain components [1]. The large surface area allows for maximum light absorption and ATP synthesis [1]. The lumen allows for the accumulation of protons to create a gradient [1].

13. (a) Mix sample with ethanol and shake [1]. Add water; a cloudy white emulsion indicates the presence of lipids [1].
(b) Triglycerides have a high ratio of energy-storing C-H bonds to carbon atoms [1]. They are hydrophobic and store more energy per gram than carbohydrates (which are hydrated) [1].
(c) Phospholipids have hydrophilic phosphate heads and hydrophobic fatty acid tails [1]. In water, heads face outward towards water, and tails face inward away from water [1], forming a stable bilayer barrier [1].

14. (a) At high temperatures, cholesterol restricts phospholipid movement, reducing fluidity [1]. At low temperatures, it prevents phospholipids from packing too closely, maintaining fluidity [1]. It stabilises the membrane [1].
(b) Organisms in cold environments will have more unsaturated fatty acids [1]. The kinks in unsaturated tails prevent tight packing, keeping the membrane fluid at low temperatures [1]. Organisms in hot environments may have more saturated fatty acids to prevent excessive fluidity [1].

15. (a) The tendency of water molecules to move from one region to another [1]. Or: The chemical potential of water.
(b) Water leaves the cell by osmosis [1] because the external solution has a lower (more negative) water potential [1]. The cytoplasm shrinks and the cell membrane pulls away from the cell wall (plasmolysis) [1].
(c) Ψ=Ψs+Ψp\Psi = \Psi_s + \Psi_p
Ψ=−800+200\Psi = -800 + 200
Ψ=−600\Psi = -600 kPa [2].

16. (a) 1. Enzymes can be reused/recovered easily [1].
2. Product is not contaminated with enzyme [1].
(Accept: Process can be continuous.)
(b) Method: Entrapment in alginate beads [1]. Or adsorption onto clay/resin [1]. Or covalent bonding to a surface [1].
(c) Diffusion of substrate to the enzyme may be slower [1]. The immobilisation process may slightly alter the active site shape [1].

17. (a) The sugar-phosphate backbone is on the outside, protecting the nitrogenous bases (genetic code) on the inside from chemical damage [1]. The double strand allows for repair mechanisms using the complementary strand as a template [1].
(b) Each new DNA molecule consists of one original (parental) strand and one newly synthesized strand [1]. This ensures genetic continuity and accuracy across generations [1]. It allows for semi-conservative replication which is efficient [1].

18. (a) Adenine (nitrogenous base) [1], Ribose sugar [1], Three phosphate groups [1].
(b) It releases a small, manageable amount of energy suitable for cellular reactions (not too much to cause damage) [1]. It can be rapidly regenerated from ADP and Pi [1]. It is soluble and can move easily within the cell [1].

19. (a) Signalling molecule (ligand) binds to specific receptor on membrane [1]. This causes a conformational change in the receptor [1]. This triggers a cascade of intracellular events (second messengers) leading to a cellular response [1].
(b) Receptors have specific 3D shapes that only fit specific ligands (lock and key) [1]. Only cells with the correct receptor will respond to the signal [1].

20. (a) Genetic material (DNA or RNA) enclosed in a protein coat (capsid) [1]. Some have an envelope derived from the host membrane [1].
(b) They cannot reproduce independently (require host machinery) [1]. They do not carry out metabolism/respiration [1].
(c) The virus envelope fuses with the host cell membrane [1]. This fusion requires the fluidity of the lipid bilayer to merge the two membranes [1], allowing the viral capsid to enter [1].