2023 Cambridge IAL Biology (9700) Practice Paper with Answers

Inhibitor X increases the \(K_m\) from \(2.0\) to \(5.0\text{ mmol dm}^{-3}\) while \(V_{max}\) remains unchanged at \(100\text{ a.u.}\). This is characteristic of a competitive inhibitor, which binds reversibly to the active site. Thus, statement 1 is correct. Inhibitor Y decreases \(V_{max}\) from \(100\) to \(50\text{ a.u.}\) while \(K_m\) remains unchanged. This is characteristic of a non-competitive inhibitor, which does not change the affinity of the enzyme (hence \(K_m\) is unchanged). Thus, statement 2 is incorrect. Because X is a competitive inhibitor, high substrate concentrations can outcompete the inhibitor, overcoming its effect and reaching the original \(V_{max}\). Because Y is a non-competitive inhibitor, increasing the substrate concentration does not overcome the inhibition. Thus, statement 3 is correct. Therefore, statements 1 and 3 are correct.

[1 mark] C is correct. Award 1 mark for identifying that competitive inhibition (X) is reversible at the active site and can be overcome by high [S], whereas non-competitive inhibition (Y) does not alter Km and cannot be overcome.

The Casparian strip is made of suberin, which is hydrophobic and completely impermeable to water and dissolved solutes, blocking the apoplast pathway. Therefore, all substances must enter the symplast pathway (inside the cytoplasm) to cross the endodermis. This requires mineral ions to be transported across the plasma membrane of endodermal cells (using active transport or facilitated diffusion), making statement 1 correct. Water must also cross the plasma membrane of these cells via osmosis (facilitated by aquaporins), making statement 2 correct. Water and solutes cannot continue through the cell walls (apoplast) at the Casparian strip because it is blocked; they must travel through the protoplasts (symplast). Thus, statement 3 is incorrect.

[1 mark] A is correct. Award 1 mark for identifying that blocking the apoplast pathway requires substances to cross the cell surface membranes of endodermal cells (statements 1 and 2) and that apoplastic transport through endodermal walls is blocked (statement 3 is incorrect).

First, determine the mRNA sequence transcribed from the DNA template strand. Since RNA transcription is complementary and antiparallel:
DNA template: \(3'\text{ - T A C G G T C C G A A A - } 5'\)
mRNA strand: \(5'\text{ - A U G C C A G G C U U U - } 3'\)
Divide the mRNA into codons (from \(5'\) to \(3'\)):
Codon 1: \(5'\text{-AUG-3'}\)
Codon 2: \(5'\text{-CCA-3'}\)
Codon 3: \(5'\text{-GGC-3'}\)
Codon 4: \(5'\text{-UUU-3'}\)
Determine the complementary tRNA anticodons. Remember that codon-anticodon base pairing is also antiparallel:
For Codon 1 \(5'\text{-AUG-3'}\) \(\rightarrow\) Anticodon is \(3'\text{-UAC-5'}\), which is written as \(5'\text{-CAU-3'}\).
For Codon 2 \(5'\text{-CCA-3'}\) \(\rightarrow\) Anticodon is \(3'\text{-GGU-5'}\), which is written as \(5'\text{-UGG-3'}\).
For Codon 3 \(5'\text{-GGC-3'}\) \(\rightarrow\) Anticodon is \(3'\text{-CCG-5'}\), which is written as \(5'\text{-GCC-3'}\).
For Codon 4 \(5'\text{-UUU-3'}\) \(\rightarrow\) Anticodon is \(3'\text{-AAA-5'}\), which is written as \(5'\text{-AAA-3'}\).
Therefore, the correct sequence is A.

[1 mark] A is correct. Award 1 mark for correctly transcribing the DNA template into mRNA codons, and then finding the complementary anticodons written in the 5' to 3' direction.

ADH normally binds to GPCRs on the collecting duct cells, initiating a cascade that leads to the fusion of aquaporin-containing vesicles with the luminal membrane. Since the drug is an antagonist blocking these receptors, vesicle fusion is prevented. Thus, statement 1 is correct. Without aquaporins on the luminal membrane, less water is reabsorbed from the collecting duct into the hypertonic renal medulla. This means more water remains in the filtrate. As a result, the urine becomes more dilute, which means its water potential will become less negative (higher), not more negative. Thus, statement 2 is incorrect. Because less water is reabsorbed, a larger volume of dilute urine is excreted. Thus, statement 3 is correct. Therefore, only statements 1 and 3 are correct.

[1 mark] B is correct. Award 1 mark for identifying that blocking ADH receptors prevents aquaporin insertion (statement 1) and increases urine volume (statement 3), while realizing that urine water potential rises (becomes less negative) due to dilution, making statement 2 incorrect.

This is an example of recessive epistasis where the locus A/a is epistatic to the locus B/b. The parental cross is \(AaBb \times AaBb\). The standard dihybrid ratio of genotypes in the offspring is:
- \(9\ A\_B\_\): Has the precursor (from A) and converts it to purple (from B) \(\rightarrow\) Purple.
- \(3\ A\_bb\): Has the precursor (from A) but cannot convert it to purple (recessive bb), so it becomes red \(\rightarrow\) Red.
- \(3\ aaB\_\): Lacks the precursor (recessive aa), so no pigment is produced \(\rightarrow\) White.
- \(1\ aabb\): Lacks the precursor (recessive aa), so no pigment is produced \(\rightarrow\) White.
Summing up the white phenotypes: \(3 (aaB\_) + 1 (aabb) = 4\) white. The expected phenotypic ratio is therefore 9 purple : 3 red : 4 white.

[1 mark] A is correct. Award 1 mark for applying the principles of recessive epistasis to determine that the standard 9:3:3:1 ratio is modified to a 9:3:4 ratio.

Cyclic photophosphorylation involves only Photosystem I (PSI). Photolysis of water only occurs at Photosystem II (PSII) to replace lost electrons during non-cyclic photophosphorylation. Therefore, photolysis does not occur during cyclic photophosphorylation. The electron flow in cyclic photophosphorylation generates a proton gradient that drives ATP synthesis via chemiosmosis, but does not reduce NADP+ and does not produce oxygen. Thus, the only product is ATP. Therefore, Row A is correct.

[1 mark] A is correct. Award 1 mark for identifying that cyclic photophosphorylation involves PSI only, does not feature photolysis of water, and produces only ATP.

One molecule of glucose yields 2 molecules of pyruvate during glycolysis. In the Link reaction (per molecule of glucose, i.e., for 2 molecules of pyruvate):
\(2\text{ Pyruvate} \rightarrow 2\text{ Acetyl CoA} + 2\text{ CO}_2 + 2\text{ reduced NAD}\)
In the Krebs cycle (per molecule of glucose, i.e., for 2 turns of the cycle using 2 Acetyl CoA):
\(2\text{ Acetyl CoA} \rightarrow 4\text{ CO}_2 + 6\text{ reduced NAD} + 2\text{ reduced FAD} + 2\text{ ATP}\)
Combined totals for Link reaction and Krebs cycle:
- Carbon dioxide: \(2\text{ (Link)} + 4\text{ (Krebs)} = 6\)
- Reduced NAD: \(2\text{ (Link)} + 6\text{ (Krebs)} = 8\)
- Reduced FAD: \(0\text{ (Link)} + 2\text{ (Krebs)} = 2\)
Therefore, Row B is the correct choice.

[1 mark] B is correct. Award 1 mark for correctly accounting for 2 pyruvates per glucose and calculating the combined yield of the Link reaction and the Krebs cycle.

The immunity is 'passive' because the patient receives pre-formed antibodies rather than generating an immune response using their own cells. It is 'artificial' because the antibodies are introduced via a medical injection rather than through a natural pathway (e.g., colostrum or placental transfer). Since the patient's own B-lymphocytes are not activated to undergo clonal selection and expansion, no memory cells are produced. The injected antibodies are foreign proteins and are gradually cleared from the body, providing only temporary, short-term protection. Therefore, Row B is correct.

[1 mark] B is correct. Award 1 mark for identifying the type of immunity as passive artificial, which does not produce memory cells and only provides short-term protection.

Competitive inhibitors bind reversibly to the active site of the enzyme, competing directly with the substrate. This increases the apparent \(K_m\) because a higher substrate concentration is required to achieve half the maximum rate (\(\frac{1}{2} V_{\max}\)). However, at infinitely high substrate concentrations, the inhibitor is fully displaced, so the \(V_{\max}\) remains unchanged. Doubling the enzyme concentration doubles the total number of active sites, which doubles the \(V_{\max}\). However, the affinity of the enzyme for its substrate (\(K_m\)) remains constant, as it is an intrinsic property of the enzyme-substrate system and is independent of enzyme concentration.

1 mark for selecting option A. All other options are incorrect because they fail to properly distinguish how inhibitors and enzyme concentration independently affect \(K_m\) and \(V_{\max}\).

Statement 1 is incorrect because active transport of hydrogen ions (protons) out of the companion cell cytoplasm into the apoplast (cell wall) increases the proton concentration, which decreases the pH of the cell wall. Statement 2 is correct because the inward movement of protons down their electrochemical gradient provides the energy for the co-transporter protein to move sucrose against its concentration gradient into the companion cell. Statement 3 is incorrect because the accumulation of sucrose in the sieve tube element at the source decreases (makes more negative) the water potential, causing water to enter by osmosis. Statement 4 is correct because unloading of sucrose at the sink causes water to leave the sieve tube by osmosis, reducing the volume of liquid and thus lowering the hydrostatic pressure at the sink, establishing a mass flow gradient from high hydrostatic pressure at the source to low hydrostatic pressure at the sink.

1 mark for selecting option D (statements 2 and 4 are correct). Reject options containing statements 1 and/or 3.

First, find the sequence of the transcribed mRNA. The mRNA is synthesized complementary and antiparallel to the DNA template strand (3'-TAC CGA CGT AGT-5'). Thus, the mRNA sequence is 5'-AUG GCU GCA UCA-3'. Next, determine the tRNA anticodons. The anticodons must be complementary and antiparallel to the mRNA codons. For Codon 1 (5'-AUG-3'), the complementary tRNA anticodon is 3'-UAC-5'. For Codon 2 (5'-GCU-3'), the complementary tRNA anticodon is 3'-CGA-5'. For Codon 3 (5'-GCA-3'), the complementary tRNA anticodon is 3'-CGU-5'. For Codon 4 (5'-UCA-3'), the complementary tRNA anticodon is 3'-AGU-5'. Therefore, option A is correct. Option C is incorrect because the 5' and 3' ends are reversed, which would prevent correct antiparallel pairing. Options B and D represent either incorrect base sequences or incorrect orientations.

1 mark for selecting option A. Correct orientation and base-pairing must be demonstrated for both transcription (DNA to mRNA) and translation (mRNA to tRNA).

When ADH binds to its receptor on the basolateral membrane (facing the blood/interstitial fluid), it activates a G-protein which subsequently activates adenylyl cyclase to convert ATP into the second messenger cyclic AMP (cAMP). This initiates a phosphorylation cascade that causes vesicles containing aquaporins to move to and fuse with the luminal (apical) membrane (facing the fluid in the collecting duct). This allows water to be reabsorbed from the filtrate back into the cell by osmosis. As a result, more water is retained by the body, leaving a smaller, more concentrated volume of urine, which means the water potential of the urine produced decreases.

1 mark for selecting option A. Distractors are incorrect because they specify either the wrong second messenger (calcium ions), the wrong membrane for aquaporin insertion (basolateral membrane is already highly permeable to water), or the incorrect trend in urine water potential (increases).

This is an example of recessive epistasis where the homozygous recessive alleles of gene A (\(aa\)) mask the expression of gene B. When a dihybrid heterozygous plant (\(AaBb\)) is self-pollinated, the genotypes of the offspring are produced in a classic 9:3:3:1 ratio:
- \(9/16\) \(A\_B\_\): possess both dominant alleles \(A\) and \(B\), so they produce purple flowers.
- \(3/16\) \(A\_bb\): possess dominant \(A\) but homozygous recessive \(bb\), so they produce pink flowers.
- \(3/16\) \(aaB\_\) and \(1/16\) \(aabb\): both have the homozygous recessive \(aa\) genotype, which epistatically suppresses the expression of the B/b locus, leading to white flowers (totaling \(3/16 + 1/16 = 4/16\)).
Therefore, the expected phenotypic ratio is 9 purple : 3 pink : 4 white.

1 mark for selecting option A. Distractors B, C, and D represent other epistatic ratios (9:6:1, 12:3:1, and 9:7 respectively) which do not match the genetic details provided.

In cyclic photophosphorylation, excited electrons leave Photosystem I (PSI) and are passed along an electron transport chain before returning to PSI. This flow of electrons drives the pumping of protons across the thylakoid membrane, generating ATP via ATP synthase, but does not involve Photosystem II (PSII) or NADP reductase, so no oxygen is produced (as there is no photolysis of water) and no reduced NADP is formed. Thus, the only product is ATP.
In non-cyclic photophosphorylation, both PSI and PSII are involved. Photolysis of water at PSII provides electrons (releasing \(\text{O}_2\) as a waste product and protons), which are passed through an electron transport chain to PSI, generating ATP. The electrons from PSI are then used to reduce NADP to reduced NADP. Thus, the products are ATP, reduced NADP, and \(\text{O}_2\).

1 mark for selecting option A. Other options represent incorrect configurations of the products of cyclic and non-cyclic photophosphorylation.

For each molecule of glucose (\(\text{C}_6\text{H}_{12}\text{O}_6\)), glycolysis in the cytoplasm produces 2 molecules of pyruvate (\(\text{C}_3\)). Both molecules of pyruvate enter the mitochondrial matrix, where they undergo the Link reaction. Each pyruvate is decarboxylated and dehydrogenated, converting it to acetyl-CoA and producing 1 molecule of reduced NAD. Therefore, for one molecule of glucose, the Link reaction produces 2 molecules of reduced NAD. The 2 molecules of acetyl-CoA then enter the Krebs cycle. For each turn of the Krebs cycle (representing 1 acetyl-CoA), 3 molecules of reduced NAD and 1 molecule of reduced FAD are produced. Since 2 acetyl-CoA molecules are processed per glucose molecule, the Krebs cycle yields 6 molecules of reduced NAD and 2 molecules of reduced FAD. Thus, option A is correct.

1 mark for selecting option A. Incorrect choices fail to account for the fact that one glucose molecule yields two pyruvates, or they miscalculate the yields per turn of the Krebs cycle.

During a cellular and humoral immune response:
- B-lymphocytes bind to specific antigens, undergo clonal selection, and differentiate into plasma cells, which secrete large quantities of antigen-specific antibodies.
- T-helper lymphocytes recognize antigens presented on MHC class II molecules and secrete cytokines (such as interleukins) that stimulate B-lymphocytes and T-killer cells to undergo clonal expansion.
- T-killer (cytotoxic T) lymphocytes recognize foreign antigens on the surface of infected host cells and release cytotoxic proteins like perforins, which form pores in the cell membrane and destroy the infected cell.
Therefore, row A is correct.

1 mark for selecting option A. Other options contain incorrect physiological details (such as lymphocytes performing phagocytosis, T-helper cells secreting antibodies, or T-killer cells binding to free circulating antigens).

Inhibitor X increases the \(K_m\) (from 2.5 to 5.0) but does not change the \(V_{max}\) (120), which indicates it is a competitive inhibitor. Competitive inhibitors bind reversibly to the active site of the enzyme and compete with the substrate. Inhibitor Y does not change the \(K_m\) (2.5) but decreases the \(V_{max}\) (from 120 to 60), indicating it is a non-competitive inhibitor. Non-competitive inhibitors bind to an allosteric site (a site other than the active site), altering the tertiary structure of the enzyme and the shape of the active site so that the substrate can no longer bind or be converted to product.

1 mark for identifying the correct mode of action for both inhibitors. Correct option is A.

During active loading, companion cells use ATP-driven proton pumps to actively transport protons (\(\text{H}^+\)) out of their cytoplasm into the cell wall (apoplast). This creates a high proton concentration in the apoplast, establishing a proton electrochemical gradient. Protons then diffuse back into the companion cell cytoplasm down their gradient through a co-transporter protein, which simultaneously transports sucrose molecules against their concentration gradient into the companion cell.

1 mark for the correct description of proton pumping and sucrose co-transport. Correct option is A.

1) Number of amino acids: The mRNA is 360 nucleotides long. Since 3 nucleotides form 1 codon, there are \(360 / 3 = 120\) codons. One of these is the stop codon, which does not code for an amino acid (it terminates translation). Thus, the maximum number of amino acids in the polypeptide chain is \(120 - 1 = 119\).
2) Number of adenine bases: In double-stranded DNA, Cytosine (C) = Guanine (G). Since C = 22%, G = 22%. Together, C + G = 44%. The remaining bases (Adenine and Thymine) make up 56% (\(100\% - 44\% = 56\%\)). Since A = T, Adenine (A) = \(56\% / 2 = 28\%\). The total number of nucleotides in both strands combined is 1200, so the number of Adenine bases is \(28\% \text{ of } 1200 = 0.28 \times 1200 = 336\).

1 mark for the correct calculation of both values. Correct option is A.

Glucagon binds to its cell-surface receptor, activating a G-protein. The active G-protein activates the membrane-bound enzyme adenylyl cyclase (enzyme X). Adenylyl cyclase converts ATP into cyclic AMP (cAMP, substance Y), which acts as a second messenger. cAMP binds to and activates protein kinase A (enzyme Z), which initiates an enzyme cascade that eventually activates glycogen phosphorylase, leading to glycogenolysis (glycogen breakdown).

1 mark for correctly identifying all three components in the signaling pathway. Correct option is A.

A cross between two double heterozygotes (\(AaBb \times AaBb\)) yields the standard dihybrid genotypic ratio in the \(F_2\) generation:
- \(9\ A\_B\_\): possess both functional enzyme A and functional enzyme B. Precursor \(\rightarrow\) yellow intermediate \(\rightarrow\) purple pigment. Phenotype = Purple.
- \(3\ A\_bb\_\): possess functional enzyme A but non-functional enzyme B. Precursor \(\rightarrow\) yellow intermediate (which accumulates). Phenotype = Yellow.
- \(3\ aaB\_\): possess non-functional enzyme A. No yellow intermediate is produced. Precursor remains colorless. Phenotype = White.
- \(1\ aabb\): possess non-functional enzyme A and non-functional enzyme B. Precursor remains colorless. Phenotype = White.
Combining the white phenotypes: \(3 + 1 = 4\) white plants. The ratio is therefore 9 purple : 3 yellow : 4 white. This is an example of recessive epistasis.

1 mark for correctly analyzing the gene interaction and determining the phenotypic ratio. Correct option is A.

Let us analyze each statement:
- Statement 1: Photolysis of water occurs only during non-cyclic photophosphorylation to provide electrons to replace those lost by Photosystem II (PSII). Cyclic photophosphorylation does not involve PSII or the photolysis of water.
- Statement 2: Both cyclic and non-cyclic photophosphorylation involve an electron transport chain that pumps protons (\(\text{H}^+\)) from the stroma into the thylakoid lumen, creating a proton gradient used to synthesize ATP via ATP synthase.
- Statement 3: Both processes utilize Photosystem I (PSI) to absorb light energy and emit photoexcited electrons.
- Statement 4: Reduced NADP (\(\text{NADPH}\)) is only produced in non-cyclic photophosphorylation when electrons from PSI are transferred to NADP. In cyclic photophosphorylation, electrons return to the electron transport chain and no reduced NADP is formed.
Therefore, only statements 2 and 3 are common to both processes.

1 mark for identifying that only statements 2 and 3 are common to both. Correct option is A.

For one molecule of glucose:
1) Glycolysis produces two molecules of pyruvate, which enter the mitochondrial matrix.
2) In the link reaction, each pyruvate is converted into acetyl-CoA, producing 1 molecule of reduced NAD (\(\text{NADH}\)) per pyruvate. Thus, the link reaction produces 2 molecules of reduced NAD in total.
3) The two molecules of acetyl-CoA enter the Krebs cycle. Each turn of the Krebs cycle produces 3 molecules of reduced NAD and 1 molecule of reduced FAD (\(\text{FADH}_2\)). For two turns of the cycle (one per acetyl-CoA), the Krebs cycle produces 6 molecules of reduced NAD and 2 molecules of reduced FAD.
Combining both stages: reduced NAD = \(2 + 6 = 8\) molecules; reduced FAD = \(0 + 2 = 2\) molecules.

1 mark for correct calculations of coenzyme yields for both the link reaction and Krebs cycle combined. Correct option is A.

In monoclonal antibody production, B-lymphocytes (specifically, antibody-producing plasma cells) isolated from the spleen of an immunized animal are fused with cancerous plasma cells (myeloma cells). The resulting cell mixture is plated on a selective medium (like HAT medium). Unfused B-lymphocytes die naturally because they cannot divide in vitro, while unfused myeloma cells die because they lack a functional salvage pathway enzyme (such as HGPRT) required to replicate DNA in the presence of the medium's drugs. Only the hybridoma cells, which inherit the cellular machinery from the B-cell and immortality from the myeloma cell, can survive and proliferate.

1 mark for identifying the correct cells fused and the role of the selective medium. Correct option is A.

A competitive inhibitor competes with the substrate for the active site. At very high substrate concentrations, the substrate outcompetes the inhibitor, so the maximum velocity (\(V_{max}\)) remains unchanged. However, more substrate is required to reach half of \(V_{max}\), meaning the Michaelis-Menten constant (\(K_m\)) is increased (reflecting a decreased apparent affinity). Therefore, inhibitor X is a competitive inhibitor that increases the \(K_m\).

1 mark: Correctly identifies that a competitive inhibitor increases the \(K_m\) while keeping the \(V_{max}\) unchanged. Reject options stating it increases affinity or is non-competitive.

The Casparian strip is made of suberin, which is impermeable to water and solutes. This blocks the apoplast pathway (through cell walls) and forces water and solutes into the symplast pathway (through the cytoplasm/protoplast). Solutes must cross the selectively permeable plasma membrane of the endodermal cells, which allows the cell to regulate entry using active transport or specific transport proteins.

1 mark: Correctly identifies that apoplastic water is blocked by suberin and forced into the symplast pathway, and that solutes must cross the plasma membrane selectively.

The DNA segment has 1200 base pairs, meaning the template strand is 1200 nucleotides long. Transcription produces a pre-mRNA of 1200 nucleotides. Splicing removes 25% of this pre-mRNA as introns (0.25 * 1200 = 300 nucleotides). The mature mRNA containing only exons is 1200 - 300 = 900 nucleotides. Each codon consists of 3 nucleotides, so there are 900 / 3 = 300 codons. Since the stop codon does not code for an amino acid, the maximum number of amino acids in the polypeptide is 300 - 1 = 299.

1 mark: Correct calculation of mature mRNA nucleotides (900), dividing by 3 to get 300 codons, and subtracting 1 for the stop codon to yield 299 amino acids.

ADH is a peptide hormone that cannot pass through the phospholipid bilayer. It binds to specific receptors on the basolateral membrane of the collecting duct cells. This activates G-proteins, which in turn activate the enzyme adenylyl cyclase. Adenylyl cyclase converts ATP to cyclic AMP (cAMP), a second messenger. cAMP triggers a cell signaling cascade that causes intracellular vesicles containing aquaporins to move to and fuse with the luminal (apical) membrane of the cells, increasing their water permeability.

1 mark: Correct sequential pathway identifying G-proteins, adenylyl cyclase, cAMP, and vesicle fusion with the luminal membrane.

Recombination frequency is calculated as (number of recombinant offspring / total number of offspring) * 100. The parental phenotypes are those corresponding to the original linkage: Grey body/normal wings (\(BbVv\)) and Black body/vestigial wings (\(bbvv\)). The recombinant phenotypes are Grey body/vestigial wings (92) and Black body/normal wings (88). Total recombinants = 92 + 88 = 180. Total offspring = 412 + 408 + 92 + 88 = 1000. Recombination frequency = (180 / 1000) * 100 = 18.0%.

1 mark: Correctly identifies recombinants (180) and total offspring (1000), calculating the percentage as 18.0%.

To fix 6 molecules of \(CO_2\), 6 molecules of RuBP are combined with \(CO_2\) to produce 12 molecules of GP (glycerate 3-phosphate). 12 molecules of ATP and 12 molecules of reduced NADP are used to reduce these 12 GP molecules to 12 TP (triose phosphate) molecules. Out of these 12 TP molecules, 2 are used to synthesize 1 molecule of glucose, leaving 10 TP molecules. To regenerate the 6 molecules of RuBP from these 10 TP molecules, another 6 molecules of ATP are consumed. Thus, 12 ATP are used in GP reduction, 6 ATP in RuBP regeneration, and 12 reduced NADP are used in total.

1 mark: Correct allocation of ATP (12 for GP reduction, 6 for RuBP regeneration) and reduced NADP (12) per glucose molecule.

In mammalian muscle cells, anaerobic respiration produces lactate from pyruvate without releasing carbon dioxide, and the process is reversible once oxygen becomes available. In yeast, anaerobic respiration (fermentation) produces ethanol and carbon dioxide, and this is an irreversible pathway because yeast cells cannot convert ethanol back into pyruvate.

1 mark: Correct comparison of end products, carbon dioxide release, and reversibility between mammalian and yeast anaerobic respiration.

In G1 phase, the diploid cell has 12 chromosomes, each consisting of a single DNA molecule. After S phase (DNA replication), each chromosome consists of two identical sister chromatids held together by a centromere. During metaphase of mitosis, the centromeres have not yet split, so there are still 12 chromosomes, but there are now 24 chromatids in total. Because of DNA replication during S phase, the DNA mass is double that of the G1 phase.

1 mark: Correctly identifies 12 chromosomes, 24 chromatids, and double DNA content at metaphase compared to G1.

Inhibitor X does not change the maximum rate of reaction (Vmax remains 120) but increases the Michaelis-Menten constant (Km increases from 2.5 to 6.0). An increased Km indicates a lower affinity of the enzyme for its substrate, which is characteristic of a competitive inhibitor as its effects can be overcome by higher substrate concentrations. Inhibitor Y reduces Vmax (from 120 to 60) but leaves Km unchanged (2.5). This indicates non-competitive inhibition, where the inhibitor binds to an allosteric site, reducing the concentration of active enzyme without affecting the substrate affinity of the remaining active enzymes.

[1 mark] B is correct. Competitive inhibitors increase the apparent Km (decreasing substrate affinity) without changing Vmax. Non-competitive inhibitors decrease Vmax without changing Km.

In phloem loading, hydrogen ions (H+) are actively pumped out of the companion cells into the cell wall space (apoplast) using ATP (active transport). This creates a high electrochemical gradient of H+ outside. H+ ions then diffuse back into the companion cells down this gradient through a co-transporter protein (facilitated diffusion). The entry of H+ is coupled with the transport of sucrose into the companion cells against its concentration gradient.

[1 mark] A is correct. H+ is actively pumped out (active transport) and enters back down its gradient (facilitation diffusion) carrying sucrose with it against its concentration gradient.

The mRNA sequence is 5'-AUG-CCA-GCU-AAU-3'. The DNA template strand must be complementary and antiparallel to this mRNA strand. Therefore, the 5' end of the mRNA corresponds to the 3' end of the DNA template strand, and the 3' end of the mRNA corresponds to the 5' end of the DNA template. Matching the complementary bases (A with T, U with A, G with C, C with G) gives: mRNA 5'-AUG-3' complementary to DNA template 3'-TAC-5'; mRNA 5'-CCA-3' complementary to DNA template 3'-GGT-5'; mRNA 5'-GCU-3' complementary to DNA template 3'-CGA-5'; mRNA 5'-AAU-3' complementary to DNA template 3'-TTA-5'. Combining these in order gives the DNA template strand: 3'-TAC-GGT-CGA-TTA-5'.

[1 mark] D is correct. The template strand runs antiparallel (3' to 5') to the mRNA (5' to 3') and contains thymine (T) instead of uracil (U).

Substance 1 is present in high concentrations in blood plasma (80.0 g dm^-3) but completely absent from glomerular filtrate, indicating it is too large to pass through the basement membrane. This corresponds to protein. Substance 2 is filtered into the nephron (1.0 g dm^-3) but is completely reabsorbed by the end of the proximal convoluted tubule, indicating it is glucose. Substance 3 is filtered (0.3 g dm^-3), its concentration doubles by the end of the proximal convoluted tubule due to water reabsorption, and it becomes highly concentrated in urine (18.0 g dm^-3) because it is a waste product that is not actively reabsorbed. This corresponds to urea.

[1 mark] B is correct. Substance 1 is protein (not filtered), Substance 2 is glucose (filtered and fully reabsorbed), and Substance 3 is urea (filtered and concentrated in urine).

Since the parents are unaffected but have an affected daughter, the condition cannot be dominant. It cannot be X-linked recessive because an affected daughter must inherit an affected allele from both parents, meaning the father would have to be affected (which he is not). Therefore, the condition must be autosomal recessive, and both parents are heterozygous carriers (Aa). The probability of their next child being affected (aa) is 0.25. The probability of the child being male is 0.5. Since these are independent events, the probability of an affected male is 0.25 * 0.5 = 0.125.

[1 mark] A is correct. Autosomal recessive inheritance is deduced because unaffected parents have an affected daughter. The combined probability is calculated as 0.25 (affected) * 0.5 (male) = 0.125.

In cyclic photophosphorylation, only Photosystem I (PSI) is involved, electrons are recycled, and ATP is the sole product. In non-cyclic photophosphorylation, both Photosystem I and Photosystem II (PSII) are involved, water is photolysed to provide electrons (releasing oxygen), and both ATP and reduced NADP are produced.

[1 mark] C is correct. Cyclic photophosphorylation uses only PSI and produces only ATP, whereas non-cyclic photophosphorylation uses both PSI and PSII, producing ATP, reduced NADP, and oxygen.

One molecule of glucose yields two molecules of acetyl-CoA through glycolysis and the link reaction. Each acetyl-CoA enters the Krebs cycle, driving it for one turn. One turn of the Krebs cycle produces 3 reduced NAD, 1 reduced FAD, and 2 carbon dioxide molecules. Therefore, for one molecule of glucose (two turns of the cycle), the yields are 6 reduced NAD, 2 reduced FAD, and 4 carbon dioxide molecules.

[1 mark] B is correct. Yields per glucose molecule in the Krebs cycle are double the yields of a single turn (6 reduced NAD, 2 reduced FAD, 4 CO2).

In a secondary immune response, memory B-lymphocytes bind to the specific antigen (clonal selection) and are stimulated to divide rapidly by mitosis (clonal expansion). They differentiate into plasma cells, which are the specialized cells that secrete large quantities of antibodies. This response is much faster and produces a higher concentration of antibodies than the primary response.

[1 mark] B is correct. Memory B-cells undergo selection, mitosis, and differentiation into plasma cells, which secrete antibodies.

(a) State the definition of \(K_m\) as the substrate concentration at half \(V_{\max}\). Connect \(K_m\) inversely with affinity. Explain that \(K_m\) allows comparison of enzyme affinities.

(b) (i) Identify as competitive inhibition because \(V_{\max}\) is unchanged but \(K_m\) is increased. Use the provided numerical values from the data to support the explanation.
(ii) Suggest increasing the substrate concentration so that substrate molecules outcompete the inhibitor molecules for active sites.

(c) Explain denaturation in terms of: kinetic energy and vibrations; breaking of hydrogen, ionic, and hydrophobic bonds; alteration of tertiary structure; and the loss of complementary fit between the active site and substrate, preventing enzyme-substrate (ES) complex formation.

(a) [Max 3 marks]
- \(K_m\) is the substrate concentration at which the reaction rate is half of \(V_{\max}\) / half the maximum rate (1 mark)
- Inverse relationship with affinity: a low \(K_m\) indicates a high affinity / a high \(K_m\) indicates a low affinity of the enzyme for its substrate (1 mark)
- Allows comparison of the affinity of different enzymes for the same substrate / of the same enzyme under different conditions (1 mark)

(b) (i) [Max 3 marks]
- Type of inhibition: Competitive (1 mark)
- Explanation: \(V_{\max}\) remains unchanged / is \(4.2 \text{ mmol dm}^{-3} \text{ min}^{-1}\) in both (1 mark)
- Explanation: \(K_m\) increases (from \(1.2\) to \(3.6 \text{ mmol dm}^{-3}\)), showing that more substrate is needed to reach half \(V_{\max}\) (1 mark)

(b) (ii) [Max 1 mark]
- Increase the substrate concentration (1 mark)

(c) [Max 3 marks]
- Increased kinetic energy results in increased molecular vibrations (1 mark)
- Breaks hydrogen bonds / ionic bonds / hydrophobic interactions (1 mark) [Reject: peptide bonds / covalent bonds]
- Loss of tertiary structure / permanent alteration of active site shape, so substrate is no longer complementary / enzyme-substrate complexes can no longer form (1 mark)

(a) Correlate specific structural features of sieve tube elements (lack of organelles, sieve plates with pores, elongated arrangement, plasmodesmata, rigid cellulose walls) to their respective functional benefits (reducing resistance, facilitating flow, enabling companion cell support, withstanding high hydrostatic pressure).

(b) Detail the step-by-step mechanism of active loading. Mention the active pumping of \(\text{H}^+\) using ATP, the establishment of the proton gradient, the facilitated diffusion of \(\text{H}^+\) back through co-transporter proteins, the inward movement of sucrose against its gradient, and the subsequent diffusion into sieve tubes via plasmodesmata.

(c) Explain that although the mass flow within the sieve tube is driven by hydrostatic pressure gradients (a physical, passive process), this gradient is established and maintained solely by the active (ATP-dependent) loading of sucrose at the source.

(a) [Max 4 marks]
- Minimal cytoplasm / no nucleus / no vacuole / few organelles to reduce resistance to the flow of phloem sap (1 mark)
- Presence of sieve plates with pores to allow easy flow of organic solutes between adjacent elements (1 mark)
- Elongated cells joined end-to-end to form a continuous conduit (1 mark)
- Plasmodesmata connections to companion cells to receive ATP / metabolic support (1 mark)
- Cellulose cell wall to withstand high hydrostatic pressure (1 mark)

(b) [Max 4 marks]
- Hydrogen ions / protons / \(\text{H}^+\) are actively pumped out of companion cells into the cell wall / apoplast (1 mark)
- Driven by proton pumps using ATP (1 mark)
- Creates a proton / electrochemical concentration gradient (higher outside than inside) (1 mark)
- Protons diffuse back into the companion cell through co-transporter proteins (1 mark)
- Sucrose is transported into companion cells against its concentration gradient alongside the protons (1 mark)
- Sucrose diffuses into the sieve tube element via plasmodesmata (1 mark)

(c) [Max 2 marks]
- Loading of sucrose at the source is an active process that requires ATP (1 mark)
- Active loading lowers the water potential, causing water to enter and generate the high hydrostatic pressure that drives mass flow (1 mark)

(a) Systematically contrast DNA replication and transcription. Focus on four main contrast points: enzymes (DNA polymerase vs RNA polymerase), templates (both strands vs template strand), products (double-stranded DNA vs single-stranded mRNA), and nucleotides (deoxyribonucleotides vs ribonucleotides).

(b) (i) Convert the original `3'-CTC-5'` DNA template to complementary mRNA: C matches G, T matches A, C matches G, giving `5'-GAG-3'`. Convert mutated `3'-CAC-5'` to complementary mRNA: C matches G, A matches U, C matches G, giving `5'-GUG-3'`.
(ii) Use the table to find the amino acid change: GAG is Glutamic Acid, GUG is Valine. Note the substitution.

(c) Explain the concept of the universality of the genetic code. Then explain why post-translational modifications (like disulfide bonds) are absent or incorrect in prokaryotes by referencing the lack of Golgi apparatus, rough endoplasmic reticulum, and associated eukaryotic processing enzymes.

(a) [Max 4 marks - 1 mark for each direct contrast]:
- Replication uses DNA polymerase, whereas transcription uses RNA polymerase (1 mark)
- Replication uses both DNA strands as templates, whereas transcription uses only one strand / the antisense / the template strand (1 mark)
- Replication produces a double-stranded DNA molecule, whereas transcription produces single-stranded mRNA (1 mark)
- Replication uses deoxyribose-containing nucleotides / DNA nucleotides / thymine, whereas transcription uses ribose-containing nucleotides / RNA nucleotides / uracil (1 mark)

(b) (i) [2 marks]:
- Original mRNA codon: `5'-GAG-3'` / GAG (1 mark)
- Mutated mRNA codon: `5'-GUG-3'` / GUG (1 mark)

(b) (ii) [1 mark]:
- Glutamic acid is replaced by valine (1 mark)

(c) [Max 3 marks]:
- The genetic code is universal / same codons code for the same amino acids in prokaryotes and eukaryotes (1 mark)
- Bacteria are prokaryotes and lack membrane-bound organelles / rough endoplasmic reticulum (RER) / Golgi apparatus (1 mark)
- Bacteria do not have the necessary enzymes/machinery to carry out post-translational modifications / disulfide bond formation / folding / glycosylation (1 mark)

(a) (i) Select structures that are below the resolution limit of a light microscope (approx. \(200\text{ nm}\)), such as ribosomes (\(20-30\text{ nm}\)), nuclear pores, or lysosomes.
(ii) Trace the movement: translation on RER ribosomes -> RER lumen -> transport vesicles -> Golgi apparatus (modification) -> secretory vesicles -> exocytosis at the plasma membrane.

(b) Apply the formula \(A = \frac{I}{M}\). First, convert the image measurement from millimeters to micrometers by multiplying by \(1000\) (\(18\text{ mm} = 18\,000\text{ }\mu\text{m}\)). Then divide by \(15\,000\) to obtain \(1.2\text{ }\mu\text{m}\).

(c) Describe the beta-glucose monomer structure, the \(180^\circ\) rotation of alternate units, the resulting straight/unbranched nature of the chains, and the extensive hydrogen bonding between adjacent chains to form microfibrils.

(a) (i) [2 marks]:
- Ribosomes / lysosomes / microtubules / centrioles / nuclear pores / membrane bilayers (Accept any two, 1 mark each)

(a) (ii) [Max 4 marks]:
- Synthesized on ribosomes bound to RER and enters the RER lumen (1 mark)
- Packaged into transport vesicles which bud off from the RER (1 mark)
- Vesicles fuse with the Golgi apparatus, where the protein is modified / glycosylated (1 mark)
- Packaged into secretory vesicles which bud off from the Golgi (1 mark)
- Vesicles move along microtubules / cytoskeleton to the plasma membrane (1 mark)
- Vesicle membrane fuses with the plasma membrane, releasing the protein by exocytosis (1 mark)

(b) [2 marks]:
- Correct conversion of mm to \(\mu\text{m}\) (\(18\text{ mm} = 18\,000\text{ }\mu\text{m}\)) (1 mark)
- Correct calculation: \(1.2\text{ }\mu\text{m}\) (1 mark)
- (Award 2 marks for correct answer with units without working shown)

(c) [Max 2 marks]:
- Composed of \(\beta\)-glucose with alternate monomers rotated \(180^\circ\), resulting in straight / unbranched chains (1 mark)
- Many parallel chains run alongside each other and are linked by many hydrogen bonds to form strong microfibrils (1 mark)

(a) Complete the table based on standard anatomical knowledge of the respiratory tract tissues. Cartilage is present only in trachea and bronchi (absent in bronchioles and alveoli). Ciliated epithelium is found in trachea and bronchioles (absent in alveoli). Smooth muscle is present in trachea and bronchioles (absent in alveoli).

(b) Connect tar and nicotine to bronchitis and emphysema (the components of COPD). Discuss goblet cell hypertrophy, cilia paralysis, mucus buildup, secondary bacterial infections, neutrophil infiltration, elastase secretion, and elastic fibre degradation in emphysema.

(c) Describe the elastic properties: stretching during inhalation (to increase surface area and prevent damage) and passive recoil during exhalation (to expel air).

(a) [3 marks - 1 mark for each fully correct row]:
- Row 1 (Trachea): \(\checkmark\), \(\checkmark\), \(\checkmark\) (1 mark)
- Row 2 (Bronchiole): \(\times\), \(\checkmark\), \(\checkmark\) (1 mark)
- Row 3 (Alveolus): \(\times\), \(\times\), \(\times\) (1 mark)

(b) [Max 4 marks]:
- Tar stimulates goblet cells / mucus glands to secrete excess mucus (1 mark)
- Tar paralyses / destroys cilia, preventing the clearance of mucus (1 mark)
- Accumulated mucus obstructs airways and acts as a breeding ground for pathogens / bacteria (leading to chronic bronchitis) (1 mark)
- In response to infection, phagocytes / neutrophils release elastase (1 mark)
- Elastase breaks down the elastic fibres in alveolar walls, leading to the permanent enlargement of air spaces / emphysema (1 mark)
- Nicotine causes bronchoconstriction / narrowing of airways (1 mark)

(c) [Max 3 marks]:
- Elastic fibres stretch during inhalation / inspiration to allow alveoli to expand (1 mark)
- This expansion increases the surface area for gas exchange / prevents alveoli from bursting (1 mark)
- Elastic fibres recoil passively during exhalation / expiration to help force air out of the lungs (1 mark)

(a) (i) Explain the secondary immune response. Emphasize the formation of memory B-cells during the primary response, and how they quickly divide (mitosis) and differentiate into plasma cells upon re-exposure.
(ii) Contrast active and passive immunity across three parameters: source of antibodies (self-produced vs external), presence of memory cells (present vs absent), and duration of protection (long-term vs temporary).

(b) Structure-function relationships of immunoglobulin: Describe the variable regions (complementary fit), constant region (phagocyte binding), hinge region (flexibility for agglutination), and disulfide bonds (quaternary structure stability).

(a) (i) [Max 3 marks]:
- First exposure / primary response produces memory (B and T) cells that persist in the blood / lymph nodes (1 mark)
- On second exposure, memory B-cells recognize the antigen immediately (1 mark)
- Memory cells undergo rapid clonal expansion / mitosis to produce large numbers of plasma cells (1 mark)
- Plasma cells secrete specific antibodies at a faster rate and in much larger quantities (1 mark)

(a) (ii) [Max 3 marks - 1 mark for each direct contrast]:
- Active involves the host's own body / immune system producing antibodies, whereas passive involves receiving antibodies from another organism (1 mark)
- Active produces memory cells, whereas passive does not produce memory cells (1 mark)
- Active provides long-term immunity, whereas passive provides immediate but short-term / temporary immunity (1 mark)

(b) [Max 4 marks]:
- Variable regions have a specific tertiary structure complementary to a specific antigen, allowing binding (1 mark)
- Two antigen-binding sites allow agglutination / clumping of pathogens (1 mark)
- Constant region binds to receptors on phagocytes to act as an opsonin / facilitate phagocytosis (1 mark)
- Hinge region provides flexibility, allowing the antibody to bind to antigens at different distances / angles (1 mark)
- Disulfide bonds hold heavy and light polypeptide chains together / stabilize the quaternary structure (1 mark)

Part (a) focuses on the physiological detection. A lower blood water potential creates an osmotic gradient, drawing water out of osmoreceptor cells, causing them to shrink. This cellular physical change generates nerve impulses sent to the posterior pituitary to stimulate ADH secretion.

Part (b) covers the molecular signaling pathway. ADH acts as the first messenger, binding to receptors, activating G-proteins, upregulating cAMP, and ultimately driving the translocation of aquaporin-containing vesicles to the luminal membrane.

Part (c) integrates the countercurrent multiplier system. The ascending limb actively pumps out ions to establish a hypertonic medulla. Water moves out of the collecting duct into this hypertonic space by osmosis, driven by the osmotic gradient created by the loop of Henle.

(a) Maximum of 3 marks:
1. Blood water potential decreases, creating a water potential gradient between osmoreceptors and blood [1];
2. Water leaves osmoreceptor cells by osmosis, causing them to shrink [1];
3. This stimulates nerve impulses to the hypothalamus/posterior pituitary [1];
4. ADH is released from the posterior pituitary gland into capillaries [1].

(b) Maximum of 4 marks:
1. ADH binds to complementary receptors on the basolateral membrane of collecting duct epithelial cells [1];
2. Activation of G-protein and adenyl cyclase / production of cyclic AMP (cAMP) as a secondary messenger [1];
3. Kinase cascade leads to movement of intracellular vesicles containing aquaporins [1];
4. Vesicles fuse with the luminal (apical) cell surface membrane [1];
5. This increases the number of water channels, increasing permeability [1].

(c) Maximum of 3 marks:
1. Active transport of \(Na^+\) and \(Cl^-\)\ ions out of the thick ascending limb [1];
2. Ascending limb is impermeable to water, establishing a high solute concentration / very low water potential in the medulla tissue [1];
3. Water moves out of the collecting duct down the water potential gradient by osmosis into the medulla [1];
4. Water is reabsorbed into blood capillaries, resulting in concentrated urine [1].

Part (a) requires a clear contrast: cyclic involves only PSI and produces only ATP, while non-cyclic involves both photosystems and yields ATP, reduced NADP, and oxygen.

Part (b) describes the reducing steps in the stroma (Calvin cycle). GP is reduced to TP using the ATP (energy/phosphate source) and reduced NADP (reducing power) produced in the thylakoid membrane.

Part (c) relates the block in light-dependent electron flow to the light-independent stage. Without reduced NADP, GP cannot form TP. This starves the regeneration phase of RuBP, which is required by Rubisco to fix incoming carbon dioxide.

(a) Maximum of 3 marks:
1. Cyclic involves only Photosystem I (PSI) / P700 whereas non-cyclic involves both PSI and Photosystem II (PSII) / P680 [1];
2. Cyclic produces only ATP (no photolysis of water) [1];
3. Non-cyclic produces ATP, reduced NADP (NADPH), and oxygen [1].

(b) Maximum of 4 marks:
1. Glycerate 3-phosphate (GP) is reduced to triose phosphate (TP) [1];
2. ATP provides energy / phosphate group [1];
3. Reduced NADP provides hydrogen / electrons [1];
4. Five out of six TP molecules are used to regenerate ribulose bisphosphate (RuBP) [1].

(c) Maximum of 3 marks:
1. Reduced NADP is not produced / NADP is not reduced [1];
2. Glycerate 3-phosphate (GP) cannot be converted to triose phosphate (TP) [1];
3. Ribulose bisphosphate (RuBP) cannot be regenerated [1];
4. No RuBP is available to combine with carbon dioxide / carbon dioxide cannot be fixed by Rubisco [1].

Part (a) requires the location (mitochondrial matrix) and the stoichiometry of the link reaction. Pyruvate (3C) is converted to acetyl CoA (2C), releasing CO2 and reducing NAD.

Part (b) covers the details of Krebs cycle enzymes. Dehydrogenation provides the reducing equivalents (NADH and FADH2) for the electron transport chain. Decarboxylation step-wise reduces the carbon structure, returning the cycle from citrate (6C) back to oxaloacetate (4C).

Part (c) details the electron transport chain and chemiosmotic theory. Oxygen is required to accept electrons. Without it, the electron carriers remain reduced, halting the proton pump and collapsing the proton gradient needed to drive ATP synthase.

(a) Maximum of 3 marks:
1. Occurs in the mitochondrial matrix [1];
2. Pyruvate is decarboxylated (CO2 removed) and dehydrogenated (hydrogen removed) [1];
3. Pyruvate combines with coenzyme A to produce acetyl coenzyme A [1];
4. Yields 1 acetyl CoA, 1 reduced NAD, and 1 \(CO_2\) per pyruvate [1].

(b) Maximum of 4 marks:
1. Dehydrogenation: hydrogen atoms are removed from intermediates [1];
2. Accepted by NAD and FAD to form reduced NAD and reduced FAD [1];
3. High-energy electrons from hydrogen are delivered to the electron transport chain [1];
4. Decarboxylation: carbon dioxide is removed from intermediates [1];
5. Reduces carbon chain length from 6C to 5C to 4C [1].

(c) Maximum of 3 marks:
1. Oxygen is the final / terminal electron acceptor [1];
2. Oxygen combines with electrons and protons to form water [1];
3. In its absence, electron transport chain stops and reduced NAD/FAD cannot be reoxidized [1];
4. Proton gradient across the inner membrane cannot be maintained, stopping ATP synthesis via ATP synthase / chemiosmosis [1].

Part (a) defines epistasis as gene-gene interaction where one gene masks another.

Part (b)(i) cross: aaBB x AAbb. The gametes are aB and Ab, giving the F1 genotype AaBb. Since both dominant alleles A and B are present, the precursor is converted to yellow, which is then converted to red. Hence, the F1 phenotype is red.

Part (b)(ii) selfing AaBb x AaBb: Standard dihybrid cross gametes: AB, Ab, aB, ab.
- A_B_ has both functional enzymes: precursor -> yellow -> red (Red, 9/16).
- A_bb has functional enzyme A but non-functional B: precursor -> yellow (Yellow, 3/16).
- aaB_ has non-functional A, so precursor remains colorless regardless of B (Colorless, 3/16).
- aabb has non-functional A and B: precursor remains colorless (Colorless, 1/16).
This results in a 9 Red : 3 Yellow : 4 Colorless ratio (recessive epistasis).

Part (c) explains applying the chi-squared formula, finding degrees of freedom (df = number of classes - 1 = 2), and comparing to the critical value at the 5% level.

(a) Maximum of 2 marks:
1. Interaction between different genes / genes at different loci [1];
2. One gene masks / alters / suppresses the expression of another gene [1].

(b)(i) Maximum of 2 marks:
1. Genotype: AaBb [1];
2. Phenotype: Red [1].

(b)(ii) Maximum of 4 marks:
1. Correct gametes of F1: AB, Ab, aB, ab [1];
2. Correct identification of genotypes corresponding to phenotypes: A_B_ is Red, A_bb is Yellow, and aaB_ and aabb are Colorless [1];
3. Correct grouping of aaB_ and aabb together as Colorless [1];
4. Expected ratio of 9 Red : 3 Yellow : 4 Colorless [1].

(c) Maximum of 2 marks:
1. Use formula \(\chi^2 = \sum \frac{(O-E)^2}{E}\) to calculate \(\chi^2\) value [1];
2. Compare calculated value to critical value from tables at 2 degrees of freedom and \(p = 0.05\) [1].

Part (a) focuses on the mechanism of selection pressures. Biotic and abiotic factors select which individuals survive to reproduce, shifting allele frequencies.

Part (b) compares stabilizing selection (preserves the intermediate phenotype, reduces variance) and directional selection (shifts the phenotypic mean towards an extreme in response to environmental change).

Part (c) details allopatric speciation. Physical barriers block gene flow, allowing separate populations to accumulate genetic differences until reproductive barriers prevent them from merging back.

(a) Maximum of 3 marks:
1. Biotic / abiotic factors (such as predators, competition, or climate) limit population survival [1];
2. Organisms with advantageous phenotypes / alleles have a selective advantage [1];
3. Survival of the fittest: they survive and reproduce [1];
4. Pass on advantageous alleles to offspring, increasing their frequency [1].

(b) Maximum of 4 marks:
1. Stabilizing selection occurs in a stable / unchanging environment [1];
2. It selects against extreme phenotypes / favors intermediate phenotypes [1];
3. Reduces phenotypic variation / keeps the mean phenotype the same [1];
4. Directional selection occurs when environmental conditions change [1];
5. It favors one extreme phenotype [1];
6. Shifts the mean of the population towards that extreme [1].

(c) Maximum of 3 marks:
1. Physical barrier (e.g., river, mountain) prevents gene flow / interbreeding between populations [1];
2. Separate populations experience different environmental selection pressures / experience independent mutations [1];
3. Leads to divergence of gene pools [1];
4. Results in reproductive isolation (inability to interbreed and produce fertile offspring) [1].

Part (a) details resting potential creation: the Na+/K+ pump moves ions against gradients, while differential permeability (K+ leak channels) allows K+ to leave faster than Na+ can enter, causing a negative interior.

Part (b) describes synaptic transmission step-by-step: depolarization -> Ca2+ entry -> exocytosis of acetylcholine -> diffusion -> receptor binding -> opening of ligand-gated Na+ channels -> postsynaptic depolarization.

Part (c) relates to the disruption of this mechanism. Preventing ACh breakdown leads to chronic postsynaptic stimulation, excitotoxicity, and muscle paralysis/failure.

(a) Maximum of 3 marks:
1. Sodium-potassium pump actively transports 3 \(Na^+\)\ out and 2 \(K^+\)\ into the axon [1];
2. Requires energy from ATP hydrolysis [1];
3. Membrane is more permeable to \(K^+\)\ than \(Na^+\)\ due to open \(K^+\)\ leak channels [1];
4. Potassium ions diffuse out of the axon down their concentration gradient, leaving a net negative charge inside [1].

(b) Maximum of 5 marks:
1. Action potential depolarizes the presynaptic membrane, opening voltage-gated \(Ca^{2+}\)\ channels [1];
2. Calcium ions diffuse into the presynaptic neurone down their concentration gradient [1];
3. Calcium ions cause synaptic vesicles containing acetylcholine (ACh) to fuse with the presynaptic membrane [1];
4. ACh is released into the synaptic cleft by exocytosis [1];
5. ACh diffuses across the cleft and binds to complementary receptors on ligand-gated \(Na^+\)\ channels [1];
6. Sodium channels open, sodium ions enter the postsynaptic neurone, causing depolarization [1].

(c) Maximum of 2 marks:
1. Acetylcholine is not hydrolyzed / remains bound to receptors [1];
2. Sodium channels remain open, causing continuous depolarization / continuous firing of action potentials [1];
3. Leads to muscle spasms / tetany / paralysis / receptor desensitization [1].

Part (a) describes the temperature cycles: 95°C breaks hydrogen bonds (denaturation); 55°C allows primers to attach (annealing); 72°C is the optimum for Taq polymerase elongation.

Part (b) explains why thermostability is needed. Human polymerase denatures at high temperatures, whereas Taq remains stable and active, enabling continuous automation.

Part (c) discusses gel electrophoresis mechanics. DNA's negative charge drives movement to the positive anode, while the physical matrix separates them by size (molecular sieving).

Part (d) lists two standard applications of DNA profiling.

(a) Maximum of 3 marks:
1. \(95\,^{\circ}\text{C}\): Denaturation of DNA / breaks hydrogen bonds to separate strands [1];
2. \(55\,^{\circ}\text{C}\): Annealing of primers / primers bind to single-stranded templates [1];
3. \(72\,^{\circ}\text{C}\): Extension / synthesis / optimum temperature for *Taq* polymerase [1].

(b) Maximum of 2 marks:
1. *Taq* polymerase is thermostable / heat-tolerant [1];
2. Does not denature at \(95\,^{\circ}\text{C}\) [1];
3. Human polymerase would denature / lose activity at high temperature, requiring manual replenishment [1].

(c) Maximum of 3 marks:
1. DNA is negatively charged due to phosphate groups [1];
2. DNA fragments move towards the positive electrode / anode [1];
3. Agarose gel act as a sieve [1];
4. Smaller fragments move faster / further than larger fragments [1].

(d) Maximum of 2 marks:
1. Forensic analysis / matching suspect DNA to crime scene samples [1];
2. Paternity / relationship testing [1];
3. Medical diagnostics / identifying genetic diseases [1];
4. Phylogenetics / evolutionary studies [1].

Part (a) requires clear definitions. Richness is a simple count of species; evenness refers to how close in numbers each species is.

Part (b)(i) asks for the calculation of Simpson's Index. First find the sum of squared fractions for each site, then subtract from 1.
For A: \(0.85^2 + 0.05^2 + 0.04^2 + 0.06^2 = 0.7225 + 0.0025 + 0.0016 + 0.0036 = 0.7302\). \(D_A = 1 - 0.7302 = 0.27\).
For B: \(0.3^2 + 0.25^2 + 0.22^2 + 0.23^2 = 0.09 + 0.0625 + 0.0484 + 0.0529 = 0.2538\). \(D_B = 1 - 0.2538 = 0.75\).

Part (b)(ii) compares these values. High evenness in B makes it more stable and diverse than the highly skewed distribution in A.

Part (c) highlights the value of biodiversity, contrasting ecological functions with tangible economic gains.

(a) Maximum of 2 marks:
1. Species richness: number of different species in an area/community [1];
2. Species evenness: relative abundance of individuals within each species [1].

(b)(i) Maximum of 4 marks:
1. Correct calculation of \(\sum(n/N)^2\) for Woodland A (\(0.7302\)) OR Woodland B (\(0.2538\)) [1];
2. Correct \(D\) value for Woodland A as \(0.27\) or \(0.2698\) [1];
3. Correct \(D\) value for Woodland B as \(0.75\) or \(0.7462\) [1];
4. Working shown clearly with division by 100 [1].

(b)(ii) Maximum of 2 marks:
1. Woodland B has higher biodiversity because it has a higher index value (\(0.75\) vs \(0.27\)) [1];
2. Both have the same species richness, but Woodland B has higher species evenness / Woodland A is dominated by Species 1 [1].

(c) Maximum of 2 marks:
1. Ecological: stability of food webs / ecosystem services / nutrient cycling / pollination [1];
2. Economic: source of medicines / tourism / genes for crop breeding / resources [1].

(a) Normal oxidative phosphorylation involves the transfer of hydrogen atoms from reduced NAD and reduced FAD to the electron transport chain in the inner mitochondrial membrane. As electrons pass along the carriers, energy is released and used to pump protons (H+) from the matrix into the intermembrane space. This creates a high concentration of protons in the intermembrane space, forming a proton / electrochemical gradient. Protons diffuse back into the matrix down this gradient through ATP synthase, which catalyzes the phosphorylation of ADP and inorganic phosphate to form ATP.

(b)(i) When UCP1 is active, the rate of ATP synthesis decreases significantly. This is because protons diffuse through UCP1 back into the matrix instead of passing through ATP synthase, which reduces or dissipates the proton gradient/motive force.

(b)(ii) The energy released from electron transport is normally conserved as potential energy in the proton gradient. When protons diffuse through UCP1, this potential energy is not used to synthesize ATP but is instead released directly as thermal energy (heat).

(c) Active UCP1 in BAT allows hibernating mammals to rapidly increase their core body temperature (rewarm) when emerging from torpor. This non-shivering thermogenesis is highly efficient and does not require muscular contraction, which conserves valuable energy and ATP for other vital physiological processes during arousal.

(a) [Max 4 marks]
1. Reduced NAD / reduced FAD release electrons to the electron transport chain (ETC) / electron carriers;
2. Electrons pass along the ETC, releasing energy;
3. Energy is used to pump protons (H+) from the matrix into the intermembrane space;
4. An electrochemical / proton gradient (or proton motive force) is established;
5. Protons diffuse back into the matrix through ATP synthase;
6. This drives the phosphorylation of ADP to ATP;

(b)(i) [Max 2 marks]
1. Rate of ATP synthesis decreases / stops;
2. Protons flow through UCP1 instead of ATP synthase;
3. This dissipates/reduces the proton gradient / proton motive force;

(b)(ii) [Max 2 marks]
1. Energy from electron transport chain / proton gradient is not captured as chemical energy / ATP;
2. Energy is released/dissipated as thermal energy / heat;

(c) [Max 2 marks]
1. Allows rapid rewarming / elevation of body temperature to active physiological levels;
2. Prevents the need for shivering / muscle contraction;
3. Conserves ATP/energy for other metabolic / survival activities;

(a) Rubisco is located in the stroma of the chloroplast. Its role in carbon fixation is to catalyze the carboxylation reaction between carbon dioxide (1C) and ribulose bisphosphate (RuBP, 5C). This produces an unstable 6C intermediate, which immediately splits into two molecules of glycerate 3-phosphate (GP, 3C).

(b)(i) When oxygen reacts with RuBP, the products are one molecule of glycerate 3-phosphate (GP) and one molecule of phosphoglycolate (2C).

(b)(ii) Photorespiration reduces the rate of photosynthesis and crop yield because only one molecule of GP is produced instead of two, meaning less GP is available for conversion to triose phosphate (TP). Furthermore, the cell must spend energy in the form of ATP and reducing power (reduced NADP/ferredoxin) in metabolic pathways to salvage the phosphoglycolate and recycle it, reducing the efficiency of the light-independent stage. With less net TP produced, there is less carbohydrate available for the plant to allocate to growth, storage organs, and crop yield.

(c) C4 plants exhibit Kranz anatomy, characterized by a ring of mesophyll cells surrounding a ring of bundle sheath cells. In mesophyll cells, PEP carboxylase fixes carbon dioxide into a 4C compound (malate or oxaloacetate) with high affinity and no oxygenase activity. This 4C compound is transported to bundle sheath cells, where it is decarboxylated to release CO2 directly around Rubisco. This maintains a high local CO2:O2 ratio, preventing Rubisco from binding oxygen and thus suppressing photorespiration.

(a) [Max 3 marks]
1. (Location) Stroma (of chloroplast);
2. Rubisco catalyzes the reaction between carbon dioxide and ribulose bisphosphate / RuBP;
3. To produce an unstable 6-carbon compound;
4. Which splits into two molecules of glycerate 3-phosphate / GP;

(b)(i) [Max 2 marks]
1. (One molecule of) glycerate 3-phosphate / GP;
2. (One molecule of) phosphoglycolate / glycolate;

(b)(ii) [Max 3 marks]
1. Only one GP is produced per RuBP (instead of two) / less GP is formed;
2. ATP / reduced NADP is consumed / wasted to salvage/recycle phosphoglycolate;
3. Less triose phosphate (TP) is formed (per unit energy/time);
4. Less organic material / sugar / starch is available for growth / storage / yield;

(c) [Max 2 marks]
1. Kranz anatomy / physical separation of mesophyll and bundle sheath cells;
2. Mesophyll cells surround bundle sheath cells (protecting them from air spaces/oxygen);
3. PEP carboxylase fixes carbon dioxide in mesophyll cells to form a 4C compound;
4. 4C compound is transported to bundle sheath cells where CO2 is released, maintaining high CO2 concentration around Rubisco / preventing oxygen from binding to Rubisco;

(a) To prepare the serial dilution:
1. Prepare five beakers labeled 1.0%, 0.50%, 0.25%, 0.125%, and 0.0625%.
2. Place 200 cm\(^3\) of the 1.0% stock solution into the beaker labeled 1.0%.
3. Transfer 100 cm\(^3\) of the 1.0% stock solution into the 0.50% beaker, and add 100 cm\(^3\) of distilled water. Mix thoroughly. This leaves 100 cm\(^3\) of 1.0% solution in the first beaker and creates 200 cm\(^3\) of 0.50% solution in the second beaker.
4. Transfer 100 cm\(^3\) of the 0.50% solution into the 0.25% beaker, and add 100 cm\(^3\) of distilled water. Mix thoroughly.
5. Repeat this process by transferring 100 cm\(^3\) of the 0.25% solution to the 0.125% beaker (adding 100 cm\(^3\) of distilled water), and then 100 cm\(^3\) of the 0.125% solution to the 0.0625% beaker (adding 100 cm\(^3\) of distilled water).
6. Finally, discard 100 cm\(^3\) of the solution from the 0.0625% beaker so that all five beakers contain exactly 100 cm\(^3\) of their respective concentrations.

(b) Method details:
1. Cut a fresh shoot of Cabomba caroliniana of a standard length (e.g., 8 cm) using a sharp scalpel. Ensure the cut is made obliquely under water to prevent air bubbles from blocking xylem vessels.
2. Place the shoot upside down in a boiling tube containing the 1.0% \(\text{NaHCO}_3\) solution.
3. Place the boiling tube in a thermostatically controlled water bath kept at a constant temperature (e.g., 20 \(^\circ\)C).
4. Set up an LED bench lamp (which minimizes heat transfer) at a fixed distance (e.g., 15 cm) from the boiling tube to maintain a constant light intensity.
5. Place a transparent glass shield or a beaker of water between the lamp and the boiling tube as an additional heat shield to absorb infra-red radiation.
6. Allow the plant shoot to acclimate to the conditions for at least 5 minutes until a steady stream of bubbles is observed from the cut stem.
7. Count the number of bubbles released from the cut stem over a 3-minute period, or use a capillary tube connected to a syringe to collect the gas and measure the volume of gas produced (in mm\(^3\)) over 5 minutes. Record the value.
8. Repeat the measurement at least three times at this concentration to calculate a mean rate of photosynthesis.
9. Repeat the entire procedure (steps 2 to 8) using fresh shoots of identical length (or the same shoot after rinsing) with each of the other prepared \(\text{NaHCO}_3\) concentrations (0.50%, 0.25%, 0.125%, and 0.0625%).
10. Safety: Keep all electrical connections and light cables away from water sources. Wear safety goggles when handling solutions.

(c) Statistical analysis:
1. The student should use Spearman's rank correlation coefficient (\(r_s\)).
2. This test is appropriate because both variables (concentration of \(\text{NaHCO}_3\) and rate of photosynthesis) represent continuous quantitative data, and the test determines if there is a significant correlation/association between them without assuming a normal distribution or a strictly linear relationship (it tests for a monotonic trend).
3. The calculated value of \(r_s\) is compared to critical values at a significance level of \(p = 0.05\) to accept or reject the null hypothesis.

Part (a): Serial Dilution [Max 4 marks]
- Description of transferring 100 cm\(^3\) of solution from one beaker to the next and adding 100 cm\(^3\) of distilled water [1]
- Correctly specifies the sequence of concentrations produced: 1.0%, 0.50%, 0.25%, 0.125%, 0.0625% [1]
- Correctly specifies starting with 200 cm\(^3\) of stock and discarding 100 cm\(^3\) of the final dilution to ensure equal 100 cm\(^3\) volumes [1]
- Mentions thorough mixing/stirring at each dilution step [1]

Part (b): Method [Max 8 marks]
- Cut a shoot of Cabomba of a standardized length (e.g., 5-10 cm) obliquely under water [1]
- Place the shoot upside down in a boiling/test tube containing the bicarbonate solution [1]
- Use of an LED lamp / use of a heat shield / water bath to keep temperature constant [1]
- Keep light intensity constant by keeping the light source at a fixed, measured distance [1]
- Allow an equilibration/acclimation period of at least 5 minutes before counting [1]
- Measure rate by counting bubbles per unit time OR measuring gas volume using a capillary tube/syringe over a set time [1]
- Replicate measurements at each concentration at least 3 times to calculate a mean [1]
- Repeat the procedure across all five concentrations of carbon dioxide [1]
- Safety precaution: keeping electrical devices away from water OR wearing eye protection [1]

Part (c): Statistical Test [Max 3 marks]
- Name of test: Spearman's rank correlation coefficient (accept: Pearson's linear correlation coefficient only if normal distribution is assumed) [1]
- Reason: To determine the strength and direction of a correlation/association between two continuous variables [1]
- Interpretation: Compare calculated coefficient value against critical values at \(p = 0.05\) (or 5% significance level) to reject or accept the null hypothesis [1]

(a)
- Independent variable: Substrate concentration (\([\text{S}]\)) (accept: presence/absence of lead ions / inhibitor).
- Dependent variable: Initial rate of reaction (\(V\)) / rate of p-nitrophenol production (accept: rate of increase in absorbance at 405 nm).

(b)
- At the start of the reaction, substrate concentration is known and has not been significantly reduced by enzyme-substrate conversions.
- Measuring the initial rate ensures that the rate is not limited by substrate depletion, accumulation of products that may cause end-product inhibition, or enzyme denaturation over time, ensuring a valid comparison.

(c)(i)
- The student must calculate the reciprocal of each substrate concentration, \(1/[\text{S}]\) (e.g., \(1/0.10 = 10.0\), \(1/0.20 = 5.0\), etc.).
- The student must calculate the reciprocal of each initial rate of reaction, \(1/V\) for both the control and inhibited trials (e.g., for control, \(1/12.0 = 0.083\)).
- Plot \(1/V\) on the y-axis against \(1/[\text{S}]\) on the x-axis, drawing a line of best fit for each of the two sets of data.

(c)(ii)
- The y-intercept of the plotted line is equal to \(1/V_{\text{max}}\). Therefore, \(V_{\text{max}}\) is calculated as the reciprocal of the y-intercept (\(V_{\text{max}} = 1/\text{y-intercept}\)).
- The x-intercept of the plotted line is equal to \(-1/K_{\text{m}}\). Therefore, \(K_{\text{m}}\) is calculated as the negative reciprocal of the x-intercept (\(K_{\text{m}} = -1/\text{x-intercept}\)).

(c)(iii)
- Both lines (control and inhibited) will be straight lines that intersect the negative x-axis at exactly the same point (identical x-intercept of \(-1/K_{\text{m}}\)). This is because non-competitive inhibitors do not affect the affinity of the enzyme for the substrate, so \(K_{\text{m}}\) is unchanged.
- The line for the inhibited reaction (with lead ions) will have a higher intercept on the y-axis than the line for the control. This is because non-competitive inhibitors decrease \(V_{\text{max}}\), which increases the reciprocal value \(1/V_{\text{max}}\).
- The line for the inhibited reaction will have a steeper slope than the control line because the slope is \(K_{\text{m}}/V_{\text{max}}\); as \(V_{\text{max}}\) decreases, the slope increases.

(d)
Three controlled variables and their controls:
1. Temperature: Controlled by placing all reaction tubes in a thermostatically controlled water bath set to a constant temperature (e.g., 37 \(^\circ\)C).
2. pH: Controlled by using a specific buffer solution (such as a carbonate-bicarbonate buffer at pH 9.8) to dissolve both the enzyme and the substrate.
3. Enzyme concentration: Controlled by using the exact same concentration and volume (e.g., 1.0 cm\(^3\) of a 0.1% solution) of alkaline phosphatase in every trial.

Part (a): Variables [2 marks]
- Independent variable: substrate concentration (accept: concentration of pNPP) [1]
- Dependent variable: initial rate of reaction / rate of p-nitrophenol production (accept: rate of change in absorbance per minute) [1]

Part (b): Initial Rate [2 marks]
- Substrate concentration has not significantly decreased / is constant at the start [1]
- Prevents limitations caused by substrate depletion / product accumulation / end-product inhibition / enzyme denaturation [1]

Part (c)(i): Data Processing [2 marks]
- Calculates the reciprocal of both substrate concentration (\(1/[\text{S}]\)) and rate (\(1/V\)) [1]
- Plots \(1/V\) on the y-axis and \(1/[\text{S}]\) on the x-axis, drawing straight lines of best fit [1]

Part (c)(ii): Intercepts [2 marks]
- \(V_{\text{max}} = 1/\text{y-intercept}\) (or y-intercept represents \(1/V_{\text{max}}\)) [1]
- \(K_{\text{m}} = -1/\text{x-intercept}\) (or x-intercept represents \-1/K_{\text{m}}\) [1]

Part (c)(iii): Appearance of Plot [4 marks]
- Both lines are straight/linear [1]
- Both lines intersect at the same point on the negative x-axis (same \(K_{\text{m}}\)) [1]
- The inhibited line has a higher intercept on the y-axis (smaller \(V_{\text{max}}\)) [1]
- The inhibited line is steeper / has a greater slope than the control line [1]

Part (d): Controlled Variables [3 marks]
- Temperature: controlled using a thermostatically controlled water bath [1]
- pH: controlled using a buffer solution [1]
- Enzyme concentration/volume: controlled by using the same concentration and volume of alkaline phosphatase [1]