A Level H1 Biology Practice Paper 3

TuitionGoWhere Practice Paper – Biology H1 A-Level

Answer Key and Mark Scheme (Version 3)

Section A: Structured Questions

Question 1: Cell Membrane and Transport

(a) (i) A: Phospholipid (or phospholipid bilayer) [1]
(ii) B: Channel protein / integral protein / transport protein [1]
(iii) C: Glycoprotein / glycocalyx [1]

(b) One function of glycoprotein:

• Cell recognition / cell adhesion / acts as receptor for hormones/signals / antigen (any one, 1 mark) [1]

(c)

• Phospholipids form a bilayer (1).
• Hydrophilic phosphate heads face outward toward the aqueous environment on both sides (1).
• Hydrophobic fatty acid tails face inward, away from water (1).
• This arrangement creates a hydrophobic core that is impermeable to most polar/charged substances, maintains membrane integrity, and allows lipid-soluble molecules to diffuse through (any link to function, 1 mark for arrangement + 1 mark for functional importance, up to 3 marks). [3]

(d)

• Active transport (1).
• Calcium ions bind to a specific carrier protein (B) embedded in the membrane (1).
• ATP is hydrolysed, causing a conformational change in the carrier protein (1).
• The protein releases the ion on the opposite side of the membrane, moving the ion against its concentration gradient (from low to high concentration) (1). [3]

(e)

• Water enters the cell by osmosis (down a water potential gradient / from higher water potential outside to lower water potential inside) (1).
• The cell swells (1).
• Because red blood cells have no cell wall, they eventually burst (haemolysis) (1). [3]

(f)

• Fluidity: phospholipids can move laterally / cholesterol modulates fluidity / unsaturated fatty acid tails increase fluidity (any one, 1 mark)
• Mosaic character: proteins (and glycoproteins/glycolipids) are embedded in the phospholipid bilayer, giving a patchwork appearance (1 mark). [2]

Total: 15 marks.

Question 2: Biological Molecules and Enzymes

(a) Graph:

• Axes correctly labelled: x-axis “Temperature (°C)”, y-axis “Rate of oxygen production (cm³ min⁻¹)” (1).
• Suitable title (1).
• Points plotted accurately (1).
• Smooth curve drawn, showing increase to a maximum around 40 °C then sharp decline (1). [4]

(b) Relationship:
As temperature increases from 10 °C to 40 °C, the rate of reaction increases (1). The rate rises most steeply between 20 °C and 40 °C, approximately doubling or more per 10 °C rise (1).
Accept description of positive correlation, exponential-like increase, etc. [2]

(c)

• Increase in kinetic energy of enzyme and substrate molecules (1).
• Increased frequency of successful collisions, leading to more enzyme–substrate complexes formed per unit time (1). [2]

(d)

• Above 50 °C, the enzyme begins to denature (1).
• High temperature breaks hydrogen bonds, ionic bonds, and hydrophobic interactions that stabilise the enzyme’s tertiary structure (1).
• The active site loses its specific complementary shape; the substrate can no longer bind, and rate falls to zero when the enzyme is completely denatured (1). [3]

(e)

• The small additional rise at 40 °C compared with 37 °C could be due to:
  • The experiment not exactly mimicking human body temperature; perhaps the enzyme used was from a different source (e.g., yeast or liver) with a slightly higher optimum temperature (1).
  • Additional kinetic energy overriding any minor denaturation at 40 °C in the short measurement period (1). [2]

(f)

• A competitive inhibitor has a shape similar to the substrate and competes for the active site (1). It temporarily blocks the active site, preventing substrate binding and reducing rate.
• Increasing substrate concentration increases the chance of substrate molecules occupying the active site rather than the inhibitor, overcoming the inhibition (1). [2]

Total: 15 marks.

Question 3: DNA Replication and Protein Synthesis

(a) (i) DNA helicase [1]
(ii) Single-stranded binding proteins bind to the separated DNA strands and prevent them from re-annealing (re-pairing) / stabilise the single-stranded DNA. [1]
[2 marks]

(b)

• During replication, each new DNA molecule consists of one original (parental) polynucleotide strand and one newly synthesised strand (1).
• This was demonstrated by the Meselson–Stahl experiment (or brief description: DNA was labelled with heavy nitrogen; after one round of replication in light nitrogen, all DNA was of intermediate density) (1). [2]

(c)

• In both species, the percentage of adenine equals that of thymine, and the percentage of guanine equals that of cytosine (1).
• This illustrates Chargaff’s rule: in any double-stranded DNA, A = T and G = C. The absolute proportions vary between species (1). [2]

(d)
(i) A mutation is a change in the sequence of nucleotide bases in DNA (or in the structure of a chromosome). [1]
(ii) A substitution mutation replaces one base with another:

• The altered codon may specify a different amino acid (missense) (1).
• The substituted amino acid may have different R-group properties (e.g., hydrophilic → hydrophobic), altering the primary structure (1).
• This can change the folding of the polypeptide, disrupting the tertiary structure and the specific shape of the active site, so the enzyme is non‑functional (1).
  Example (not required): sickle‑cell anaemia, where glutamic acid is replaced by valine. [3]

(e)

• mRNA is synthesised during transcription in the nucleus (1); it carries the genetic code as a sequence of codons (1).
• mRNA moves to the cytoplasm and binds to a ribosome (1).
• tRNA molecules have specific anticodons complementary to mRNA codons, and each carries a specific amino acid (1).
• In the ribosome, the anticodon of the tRNA pairs with the complementary codon on mRNA, ensuring the correct amino acid is added to the growing polypeptide chain (1).
• Peptide bonds form between adjacent amino acids, and the ribosome moves along the mRNA; the process continues until a stop codon is reached, releasing the completed polypeptide (1).
  [5 marks] – Award marks for key role of each and correct sequence.

Total: 15 marks.

Question 4: Inheritance and Genetic Disease

(a)
(i) I-1: X^H Y (normal male) [1]; I-2: X^H X^h (carrier female) [1].
(ii) Because she has an affected son (II‑1) with genotype X^h Y (1). The son inherited his X chromosome from his mother; as his only X carries the recessive allele, the mother must have one recessive allele (she is a carrier). [1]
[3 marks]

(b)
The father passes his Y chromosome to a son, not his X chromosome (1). The haemophilia allele is on the X chromosome, so a father-to-son transmission is impossible. [1]

(c)

• The mutation alters the DNA sequence of the factor VIII gene (1).
• It may cause a change in the amino acid sequence of the protein (1).
• The altered protein is non‑functional / cannot participate in the blood clotting cascade (1).
• Without functional factor VIII, fibrinogen is not converted to fibrin (or clotting pathway is blocked), so bleeding continues for longer (1). [3]

(d)
Genetic diagram (4 marks):

• Parental phenotypes: carrier female × normal male
• Parental genotypes: X^H X^h × X^H Y (1)
• Gametes: female produces X^H and X^h; male produces X^H and Y (1)
• Punnett square or forked-line:
  • X^H X^H (normal female)
  • X^H X^h (carrier female)
  • X^H Y (normal male)
  • X^h Y (haemophiliac male) (1)
• Statement: Probability that a son is haemophiliac = 1/2 (or 50%) (1).
  [Award full marks for correct diagram and probability even if format differs slightly.]

(e)

• Genetic counselling: allows individuals to make informed reproductive choices.
• Prenatal diagnosis: allows early detection and preparation.
• Family planning: may decide to adopt or use IVF with pre-implantation genetic diagnosis.
  (Any one reasonable, 1 mark) [1]

(f)

• Independent assortment of chromosomes during metaphase I of meiosis: the X chromosome carrying the X^h allele assorts independently of other chromosomes, so it can be passed to a gamete along with different combinations of alleles (1).
• Crossing over (though less common for X chromosome) can also shuffle alleles between homologous chromosomes (1). [2]

(g)

• The allele is maintained in the population because female carriers do not suffer severe symptoms (they have a normal allele) and can pass the allele to offspring (1).
• New mutations may also arise.
  (Any one valid point, 1 mark) [1]

Total: 15 marks.

Section B: Essay Questions

Question 5 (20 marks) – Mark each part out of 10, then sum.

(a) Structure of cell membrane and control of movement (10 marks)

Indicative content:

• Fluid mosaic model: phospholipid bilayer, proteins embedded or attached; cholesterol in animal membranes for fluidity.
• Phospholipid arrangement: hydrophilic heads outwards, hydrophobic tails inwards → hydrophobic core impermeable to ions, polar molecules.
• Simple diffusion: small, non‑polar molecules (O₂, CO₂) pass directly.
• Facilitated diffusion: channel proteins (e.g., aquaporins for water) and carrier proteins for glucose; requires specific binding, no energy, down concentration gradient.
• Active transport: carrier proteins use ATP to move substances against gradient (e.g., Na⁺/K⁺-ATPase).
• Osmosis: movement of water through aquaporins or between phospholipids, down water potential gradient.
• Endocytosis/exocytosis: bulk transport via vesicles (e.g., phagocytosis, receptor‑mediated endocytosis).

Mark scheme:

• Clear description of phospholipid bilayer and embedded proteins (2 marks).
• Explanation of selective permeability based on hydrophobic core (1 mark).
• Accurate description of at least two passive transport mechanisms (facilitated diffusion, simple diffusion, osmosis) with examples (3 marks).
• Accurate description of active transport with an example (2 marks).
• Mention of bulk transport if needed (1 mark).
• Logical organisation, use of correct terminology (1 mark).

(A maximum of 10 marks for part (a).)

(b) Membrane transport in nerve impulses (10 marks)

Indicative content:

• Resting potential maintained by Na⁺/K⁺ pump (3 Na⁺ out, 2 K⁺ in; active transport using ATP).
• Membrane more permeable to K⁺ than Na⁺ at rest (via K⁺ leak channels), inside negative.
• Depolarisation: voltage‑gated Na⁺ channels open, Na⁺ influx by facilitated diffusion (down electrochemical gradient).
• Repolarisation: Na⁺ channels close, voltage‑gated K⁺ channels open, K⁺ efflux (facilitated diffusion).
• Hyperpolarisation, then return to resting potential by pump.
• Propagation: local currents cause sequential opening of channels along axon (saltatory conduction in myelinated neurones – ions exchange only at nodes of Ranvier).
• Synaptic transmission: Ca²⁺ influx via voltage‑gated channels triggers exocytosis of neurotransmitter vesicles.

Mark scheme:

• Correct description of resting potential and role of Na⁺/K⁺ pump (active transport) (2 marks).
• Clear explanation of ion movements during action potential (Na⁺ and K⁺ channels, facilitated diffusion) (4 marks).
• Role of membrane transport in propagation (local currents, myelination) (2 marks).
• Synaptic transmission: Ca²⁺-mediated exocytosis (bulk transport) (1 mark).
• Coherent answer with appropriate terminology (1 mark).

(A maximum of 10 marks for part (b).)

Question 6 (20 marks)

(a) DNA replication fidelity and frameshift mutation (10 marks)

Indicative content:

• Semi‑conservative replication: each new DNA has one old and one new strand.
• Enzymes: helicase unwinds, DNA polymerase adds nucleotides (5’→3’), DNA ligase joins Okazaki fragments.
• Fidelity: Proofreading by DNA polymerase (3’→5’ exonuclease activity) removes incorrect bases. Complementary base pairing (A‑T, G‑C) ensures accuracy.
• Mismatch repair mechanisms post‑replication.
• Frameshift mutation: insertion or deletion of a number of nucleotides not divisible by three.
• Effect: shifts the reading frame; all codons downstream are changed → completely different amino acid sequence → non‑functional protein likely.
• Contrast with substitution.

Mark scheme:

• Description of semi‑conservative replication and enzyme roles (3 marks).
• Explanation of proofreading and error correction mechanisms (2 marks).
• Definition of frameshift mutation and how it arises (1 mark).
• Explanation of consequence on polypeptide sequence (2 marks).
• Example, e.g., cystic fibrosis deletion, could be given (1 mark).
• Logical flow and terminology (1 mark).

(A maximum of 10 marks for part (a).)

(b) Meiosis, fertilisation, genetic variation and natural selection (10 marks)

Indicative content:

• Meiosis I: crossing over (prophase I) → exchange of alleles between homologous chromosomes → new combinations of alleles.
• Independent assortment (metaphase I) → random alignment of bivalents, 2ⁿ combinations (for humans, 2²³).
• Meiosis II: separation of chromatids; no further crossing over.
• Fertilisation: random fusion of gametes → enormous number of zygote genotypes.
• Genetic variation provides raw material for natural selection.
• Individuals with advantageous alleles more likely to survive and reproduce → allele frequencies change over generations.
• Examples: antibiotic resistance, beak size in finches.

Mark scheme:

• Clear explanation of crossing over and its genetic consequences (2 marks).
• Clear explanation of independent assortment (2 marks).
• Role of random fertilisation in increasing variation (1 mark).
• Link between genetic variation and natural selection (1 mark).
• Mechanism of natural selection explained (differential survival/reproduction) (2 marks).
• Use of at least one example (1 mark).
• Coherent argument (1 mark).

(A maximum of 10 marks for part (b).)

End of Mark Scheme