A Level H2 Biology Practice Paper 4

Answer Key - Biology H2 Practice Paper (Version 4)

Section A

Question 1 (a)

• Electrons are excited/passed from carriers in the ETC (Complex I-IV). [1]
• Energy released from electron transfer is used to pump H⁺ ions from the matrix into the intermembrane space. [1]
• This creates a proton gradient/electrochemical gradient. [1]
• H⁺ ions flow back into the matrix through ATP synthase (chemiosmosis), providing energy to phosphorylate ADP to ATP. [1]

(b)

• Rate of oxygen consumption will decrease/stop. [1]
• Oxygen is the final electron acceptor at Complex IV. [1]
• If electrons cannot reach Complex IV, oxygen cannot be reduced to water, halting the ETC. [1]

(c)

• ATP synthase requires ADP and Pi as substrates to produce ATP. [1]
• If ADP/Pi are depleted, ATP synthesis stops. [1]
• This leads to a buildup of the proton gradient, which eventually makes it energetically unfavorable for the ETC to pump more H⁺, thus slowing the ETC. [1]

Question 2 (a)

• Misfolded proteins expose hydrophobic R-groups/residues that are normally buried in the core. [1]
• These hydrophobic regions seek to avoid the aqueous environment of the cytoplasm. [1]
• They interact with hydrophobic regions of other misfolded proteins via hydrophobic interactions, leading to aggregation. [1]

(b)

• Primary: Linear sequence of amino acids linked by peptide bonds. [1]
• Tertiary: Three-dimensional folding of a single polypeptide chain. [1]
• A change in one amino acid (primary) can change the R-group properties (e.g., polar to non-polar). [1]
• This disrupts specific bonds (e.g., ionic, disulfide, hydrogen) required for correct folding, altering the active site/binding site and causing loss of function. [1]

(c)

• Gel electrophoresis separates proteins based on charge and size. [1]
• An electric field/potential difference is applied across the gel. [1]
• Proteins migrate toward the opposite charge; a variant protein with a different net charge will migrate at a different speed/distance. [1]
• This results in different banding patterns (positions) on the gel for the normal vs variant protein. [1]

Question 3 (a)

• Inducible: Operon that is usually "off" but can be turned "on" by a specific substrate/inducer. [1]
• Repressible: Operon that is usually "on" but can be turned "off" when the end-product accumulates. [1]

(b)

• Tryptophan is an essential amino acid needed for protein synthesis. [1]
• It is wasteful to synthesize it if it is already available in the environment. [1]
• A repressible operon allows the cell to stop production when levels are sufficient, conserving energy and raw materials. [1]

(c)

• In the absence of lactose, the repressor binds to the operator, blocking RNA polymerase. [1]
• When lactose is present, allolactose (inducer) binds to the repressor protein. [1]
• This causes a conformational change in the repressor, preventing it from binding to the operator. [1]
• RNA polymerase can then bind to the promoter and transcribe the genes for lactose metabolism. [1]

Question 4 (a)

• Both use membrane proteins (carriers/channels). [1]
• Facilitated diffusion: passive, moves down concentration gradient, no ATP. [1]
• Active transport: requires energy (ATP), moves against concentration gradient. [1]
• Facilitated diffusion can use channels; active transport requires specific carrier proteins (pumps). [1]

(b)

• The pump uses ATP to actively transport 3 Na⁺ ions out of the cell and 2 K⁺ ions into the cell. [1]
• This creates a concentration gradient for both ions. [1]
• The net export of positive charges (3 out vs 2 in) contributes to a negative internal charge. [1]
• This maintains the electrochemical gradient necessary for the resting potential. [1]

(c)

• Inhibitor binds to the allosteric site (not the active site). [1]
• This causes a conformational change in the enzyme's active site. [1]
• The substrate can no longer bind to the active site, reducing the rate of reaction regardless of substrate concentration. [1]

Question 5 (a)

• Genetic material: Prokaryotes have circular DNA in a nucleoid (no nucleus); Eukaryotes have linear DNA enclosed in a nuclear envelope. [2]
• Organelles: Prokaryotes lack membrane-bound organelles; Eukaryotes have mitochondria, chloroplasts, ER, Golgi, etc. [2]

(b)

• Lysosomes contain hydrolytic enzymes (acid hydrolases). [1]
• Compartmentalization prevents these enzymes from digesting the rest of the cell's components. [1]
• It allows the maintenance of an acidic pH inside the lysosome, which is optimal for these enzymes. [1]

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Section B

Question 6 (a)

• Different alleles have different restriction enzyme recognition sites. [1]
• A restriction enzyme cuts DNA at specific sequences. [1]
• Individuals A and C are homozygous; both alleles are identical and cut at the same place, producing one fragment size. [1]
• Individuals B and D are heterozygous; they possess two different alleles, each producing a fragment of a different size. [1]

(b)

• Individual A: Homozygous (Normal/Healthy) [1]
• Individual B: Heterozygous (Carrier) [1]
• Individual C: Homozygous (Affected/Disease) [1]
• Individual D: Heterozygous (Carrier) [1] (Note: A and C can be swapped depending on which band is associated with the disease allele)

(c)

• RFLP only detects changes at specific restriction sites. [1]
• A disease might be caused by mutations that do not alter the restriction site, leading to false negatives. [1]
• Complex diseases are often polygenic (influenced by multiple genes); analyzing one locus provides an incomplete picture. [1]
• Environmental factors also play a role in disease onset, which DNA analysis cannot detect. [1]
• Penetrance: carrying the genotype does not always guarantee the phenotype. [1]
• Therefore, RFLP provides a correlation/risk factor but not a definitive diagnosis for complex traits. [1]

Question 7 (a)

• Helicase unwinds the DNA double helix by breaking hydrogen bonds between complementary bases. [1]
• DNA polymerase adds complementary nucleotides to the leading and lagging strands. [1]
• It synthesizes DNA in the 5' to 3' direction. [1]
• On the lagging strand, DNA is synthesized in short Okazaki fragments. [1]
• DNA ligase joins these Okazaki fragments together by forming phosphodiester bonds. [1]
• The process is semi-conservative, resulting in two identical DNA molecules. [1]

(b)

• Complementary base pairing (A-T, C-G) ensures accurate replication/transmission of information. [1]
• The sugar-phosphate backbone is held by strong covalent phosphodiester bonds, providing stability. [1]
• Hydrogen bonds between bases are weak enough to be "unzipped" for replication/transcription. [1]
• The double helix structure protects the nitrogenous bases (the code) inside the helix. [1]
• Anti-parallel strands allow for the specific orientation required by polymerase enzymes. [1]
• The sequence of bases allows for virtually infinite combinations to encode diverse genetic information. [1]