2024 IB DP Biology Practice Paper with Answers

The scenario describes end-product inhibition, which is a form of non-competitive inhibition. The end product (D) binds to an allosteric site on the first enzyme (E1), changing the conformation of the active site so that the substrate (reactant A) can no longer bind. Since the inhibitor binds to a site other than the active site, increasing the concentration of substrate A cannot overcome the inhibition.

Award [1] mark for the correct option (A). [1] marks are given for selecting the option that correctly identifies end-product/allosteric non-competitive inhibition and recognizes that increasing the substrate concentration does not overcome this inhibition.

The total water potential of the plant cell is calculated as \(\psi_w = \psi_s + \psi_p = -0.8\text{ MPa} + 0.3\text{ MPa} = -0.5\text{ MPa}\). The water potential of the solution in the open beaker is equal to its solute potential, \(\psi_w = -0.5\text{ MPa}\), since pressure potential is zero in an open container. Because both the plant cell and the solution have the same water potential (\(-0.5\text{ MPa}\)), there is no net movement of water between the cell and the solution.

Award [1] mark for the correct option (C). No marks are awarded for incorrect calculations of water potential or incorrect directions of water movement.

The autotrophs represent the first trophic level (Producers) with \(24,000\text{ kJ m}^{-2}\text{ yr}^{-1}\). The second trophic level (Primary consumers) receives \(10\%\) of this energy: \(2,400\text{ kJ m}^{-2}\text{ yr}^{-1}\). The third trophic level (Secondary consumers) receives \(10\%\) of that: \(240\text{ kJ m}^{-2}\text{ yr}^{-1}\). The fourth trophic level (Tertiary consumers) receives \(10\%\) of the secondary consumer level: \(24\text{ kJ m}^{-2}\text{ yr}^{-1}\).

Award [1] mark for the correct numerical calculation leading to option (C).

Phloem loading occurs via active transport. Proton pumps (H+-ATPases) actively pump hydrogen ions out of the companion cell cytoplasm into the cell wall/apoplast, generating a high proton concentration gradient. Protons then flow back down their concentration gradient into the companion cell through a co-transporter protein (symport), carrying sucrose molecules with them against their concentration gradient.

Award [1] mark for identifying the correct mechanism (B) involving proton gradients and co-transport of sucrose.

In myelinated axons, the myelin sheath acts as an electrical insulator. Depolarisation can only occur at the nodes of Ranvier, where voltage-gated sodium channels are highly concentrated. This causes the action potential to 'jump' from one node to the next (saltatory conduction), which is much faster than the continuous propagation found in unmyelinated axons.

Award [1] mark for identifying the correct structural and functional distinction regarding myelinated vs unmyelinated propagation (C).

A mutation that changes a codon coding for an amino acid into a translation termination (stop) codon is classified as a nonsense mutation. This leads to premature termination of translation, resulting in a truncated and typically non-functional polypeptide.

Award [1] mark for selecting the correct mutation classification and its outcome (C).

Natural selection operates on pre-existing genetic variation. Random mutations occurring during DNA replication produce alleles conferring resistance before exposure to the antibiotic. When the population is exposed to the antibiotic (which acts as the selective pressure), non-resistant bacteria are killed, while resistant bacteria survive and reproduce. Over time, the frequency of the resistant allele increases in the population.

Award [1] mark for correctly identifying natural selection steps, emphasizing that the mutation is pre-existing rather than induced by the environment (B).

The link reaction takes place within the mitochondrial matrix. In this reaction, each 3-carbon pyruvate molecule is decarboxylated and oxidized to a 2-carbon acetyl group, which is then attached to Coenzyme A to form Acetyl-CoA. This process yields \(1\text{ acetyl-CoA}\), releases \(1\text{ CO}_2\), and reduces \(1\text{ NAD}^+\) to \(1\text{ NADH}\).

Award [1] mark for identifying both the correct location (mitochondrial matrix) and correct yields per single pyruvate molecule (B).

Inhibitor X is competitive because its inhibitory effect can be overcome by increasing the substrate concentration (reaching \(V_{\max}\)). Competitive inhibitors bind to the active site of the enzyme. Inhibitor Y is non-competitive because it reduces the \(V_{\max}\) regardless of substrate concentration. Non-competitive inhibitors bind to an allosteric site, altering the shape of the active site so that the substrate can no longer bind or be converted to product.

Award 1 mark for the correct choice A. No partial marks are awarded for multiple choice questions.

The water potential of the plant cell is calculated as \(\psi_w = \psi_s + \psi_p = -0.8\text{ MPa} + 0.2\text{ MPa} = -0.6\text{ MPa}\). The water potential of the solution in the open beaker is equal to its solute potential, which is \(-0.5\text{ MPa}\) (since \(\psi_p = 0\)). Water moves from an area of higher water potential (\(-0.5\text{ MPa}\)) to an area of lower water potential (\(-0.6\text{ MPa}\)), so there is a net movement of water into the cell, making it more turgid. Plant cells do not burst due to the presence of the cell wall.

Award 1 mark for the correct choice A. No partial marks are awarded for multiple choice questions.

Ecological efficiency is calculated as the percentage of energy transferred from one trophic level to the next. From primary producers to primary consumers, the efficiency is \((2,400 / 24,000) \times 100\% = 10\%\).

Award 1 mark for the correct choice B. No partial marks are awarded.

Phloem loading is an active process. Protons (\(\text{H}^+\)) are actively pumped out of the companion cells into the cell wall spaces (apoplast) using ATP. This generates an electrochemical proton gradient. Protons then flow back into the companion cells down their concentration gradient through a co-transporter protein, which co-transports sucrose molecules against their concentration gradient into the companion cells.

Award 1 mark for the correct choice B. No partial marks are awarded.

Depolarization occurs when the membrane potential becomes more positive. This is caused by the opening of voltage-gated sodium channels, allowing sodium ions (\(\text{Na}^+\)) to diffuse rapidly into the axon down their electrochemical gradient. The sodium-potassium pump restores resting concentrations later, and potassium efflux causes repolarization.

Award 1 mark for the correct choice C.

The mutation changes a codon for an amino acid (CAG, Glutamine) into a stop codon (UAG). This is a nonsense mutation, which leads to premature termination of translation and results in a truncated (shortened) and likely non-functional polypeptide.

Award 1 mark for the correct choice B.

Let \(X^B\) be the normal allele and \(X^b\) be the color-blind allele. The woman has normal vision but her father was color-blind (\(X^b Y\)), meaning she must be a carrier with the genotype \(X^B X^b\). The man has normal vision with the genotype \(X^B Y\). The possible offspring are: \(X^B X^B\) (normal female, 25%), \(X^B X^b\) (carrier female, 25%), \(X^B Y\) (normal male, 25%), and \(X^b Y\) (color-blind male, 25%). Therefore, the probability that their first child is a color-blind male is 25% (or 1 in 4).

Award 1 mark for the correct choice B.

Natural selection occurs when environmental selective pressures (such as predators) favor individuals with certain heritable traits. In this case, visual predators (birds) selectively preyed upon the light-colored moths that were no longer camouflaged on the dark, soot-covered tree trunks. The dark moths had higher survival and reproductive success, leading to an increase in the frequency of the dark allele in the population over time.

Award 1 mark for the correct choice C.

Competitive inhibitors bind to the active site and block substrate binding. Since they compete with the substrate, increasing the substrate concentration increases the probability of the substrate binding instead of the inhibitor, thereby overcoming the inhibition and reaching the same maximum rate (\(V_{\max}\)).

Award 1 mark for identifying both the competitive nature of the inhibition (as maximum velocity is unaffected at high substrate concentrations) and that competitive inhibitors bind to the active site.

The initial water potential of the plant cell is calculated using the formula: \(\psi_{\text{cell}} = \psi_s + \psi_p = -0.8\text{ MPa} + 0.3\text{ MPa} = -0.5\text{ MPa}\). The water potential of the surrounding sucrose solution is \(-0.6\text{ MPa}\). Since water moves from a region of higher (less negative) water potential to a region of lower (more negative) water potential, water will move from the cell (\(-0.5\text{ MPa}\)) to the solution (\(-0.6\text{ MPa}\)). The loss of water from the cell will cause the protoplast to shrink and decrease the turgor pressure (pressure potential).

Award 1 mark for the correct calculation of cell water potential (\(-0.5\text{ MPa}\)) and identifying that water moves down the water potential gradient out of the cell, which decreases its turgor pressure.

Reaching the threshold potential of \(-55\text{ mV}\) triggers the rapid opening of voltage-gated \(Na^+\) channels, allowing \(Na^+\) to enter the cell down its concentration and electrical gradients, causing depolarization. Shortly after, these channels close and voltage-gated \(K^+\) channels open, causing \(K^+\) efflux, which repolarizes the membrane.

Award 1 mark for identifying that threshold potential crossing leads first to the opening of voltage-gated \(Na^+\) channels (influx) and subsequently to the opening of voltage-gated \(K^+\) channels (efflux).

In the CRISPR-Cas9 system, the guide RNA (gRNA) is engineered with a specific nucleotide sequence complementary to the target DNA sequence, ensuring precise targeting. The Cas9 protein is an endonuclease that binds to the gRNA and introduces a double-strand break (DSB) at the designated genomic location.

Award 1 mark for identifying that gRNA is responsible for sequence-specific targeting through complementary base pairing, and Cas9 is the endonuclease that cleaves the DNA strands.

Natural selection acts on existing genetic variations. The melanic (dark) form of the moth already existed at low frequencies. When the tree trunks became dark due to pollution, the melanic moths were better camouflaged against predators, resulting in differential survival and reproduction. Thus, they passed on the melanic allele, increasing its frequency over time.

Award 1 mark for identifying that melanic moths had a camouflage advantage on dark tree trunks, leading to higher survival and reproduction rates (natural selection) which increased the frequency of the allele.

Protons are pumped from the matrix into the intermembrane space by the electron transport chain, creating an electrochemical proton gradient. Protons then flow back down their concentration gradient into the matrix via the enzyme ATP synthase. This passive movement (chemiosmosis) releases energy, which ATP synthase uses to phosphorylate ADP to ATP.

Award 1 mark for identifying that active pumping of protons creates a gradient, and their subsequent facilitated diffusion back into the matrix through ATP synthase is used to synthesize ATP.

The rough endoplasmic reticulum (rER) is covered with ribosomes, which specialize in synthesizing proteins destined for secretion, lysosomal targeting, or insertion into membranes. The Golgi apparatus receives these proteins, modifies them (e.g., glycosylation), and packages them into secretory vesicles. Thus, a high density of both organelles indicates a cell highly active in secretory protein synthesis.

Award 1 mark for explaining that the primary roles of rER and Golgi apparatus are protein synthesis and modification/packaging for secretion, respectively, pointing to the secretion of extracellular proteins.

The sodium-potassium pump is an active transporter (antiport) that maintains resting potential in cells. Each cycle of the pump actively moves 3 sodium ions (\(Na^+\)) out of the cell and 2 potassium ions (\(K^+\)) into the cell, against their concentration gradients. This process is driven by the conformational change triggered by the hydrolysis of one molecule of ATP.

Award 1 mark for correctly identifying that 3 \(Na^+\) are pumped out, 2 \(K^+\) are pumped in, and that the pump is powered by the hydrolysis of one ATP molecule.

Competitive inhibitors compete with substrate molecules for the active site. At very high substrate concentrations, the substrate outcompetes the inhibitor, allowing the reaction to reach the same maximum velocity (\(V_{max}\)) as without the inhibitor.

Award 1 mark for correct answer A. Other choices are incorrect because they misidentify the inhibitor type or the binding site.

Water moves from a region of higher (less negative) water potential to a region of lower (more negative) water potential. Since \(-0.4\text{ MPa} > -0.7\text{ MPa}\), water will leave the cell down the water potential gradient, causing the protoplast to shrink away from the cell wall (plasmolysis).

Award 1 mark for correct answer B. Other choices are incorrect because they misidentify the direction of water movement or the cellular state of a plant cell losing water.

When ventricular systole begins, the pressure in the ventricles rises rapidly and exceeds the pressure in the atria. This pressure difference immediately forces the atrioventricular (AV) valves to close, preventing backflow and producing the first heart sound ('lub').

Award 1 mark for correct answer B. Distractors are incorrect because semilunar valves close at the start of diastole, AV valves do not open during systole, and the second heart sound is caused by semilunar valve closure.

At the peak of depolarization, voltage-gated sodium channels rapidly close or inactivate, halting the influx of sodium ions. Simultaneously, voltage-gated potassium channels open, allowing potassium ions to diffuse out of the cell and initiating repolarization.

Award 1 mark for correct answer B. Distractors are incorrect because sodium channels are inactive/closed at peak, and potassium channels are open.

A silent mutation occurs when a single nucleotide change does not alter the amino acid sequence of the polypeptide. This is possible because the genetic code is degenerate, meaning multiple codons can specify the same amino acid.

Award 1 mark for correct answer B. Distractors are incorrect because missense mutations change the amino acid, nonsense mutations introduce a premature stop codon, and frameshift mutations alter the entire reading frame (and are not simple single base substitutions).

Natural selection acts on existing variation. The antibiotic serves as a selective pressure that kills susceptible bacteria while allowing those with pre-existing resistance mutations to survive and reproduce, increasing the frequency of resistance alleles in the population over time.

Award 1 mark for correct answer B. Distractors are incorrect because antibiotics do not direct or induce mutations to occur, individual bacteria cannot adapt their genome during their lifetime, and antibiotics do not intentionally increase the mutation rate to yield beneficial alleles.

(a) The independent variable is the substrate (lactose) concentration.

(b) A competitive inhibitor increases the \(K_m\) (requires a higher substrate concentration to reach half of \(V_{max}\)) but does not change the ultimate maximum rate of reaction (\(V_{max}\)) because high substrate concentrations can outcompete the inhibitor. In contrast, a non-competitive inhibitor decreases the \(V_{max}\) because it reduces the overall number of active enzymes available, but it does not change the affinity of the remaining functional active sites, leaving the \(K_m\) unchanged.

(c) A non-competitive inhibitor binds to an allosterive site (a site other than the active site) on the enzyme. This binding causes a conformational (shape) change in the active site of the enzyme. As a result, the substrate can no longer bind properly, or the enzyme can no longer catalyze the reaction, reducing the overall rate of reaction regardless of the substrate concentration.

(a) [1 mark max]
- Lactose / substrate concentration [1]

(b) [3 marks max]
- Competitive inhibitor increases \(K_m\) whereas non-competitive inhibitor does not change \(K_m\) [1]
- Competitive inhibitor does not change \(V_{max}\) whereas non-competitive inhibitor decreases \(V_{max}\) [1]
- Explanation/clarification that competitive inhibition can be overcome by high substrate concentrations but non-competitive cannot [1]

(c) [3 marks max]
- Inhibitor binds to an allosteric site (on the enzyme) [1]
- Causes a conformational change in the tertiary structure / shape of the active site [1]
- Substrate can no longer bind (or active site is no longer functional / enzyme-substrate complex cannot form) [1]

(a) When a threshold potential is reached, voltage-gated sodium channels open, causing a rapid influx of sodium ions (\(Na^+\)) into the axon, depolarizing the membrane. Subsequently, sodium channels close and voltage-gated potassium channels open, allowing potassium ions (\(K^+\)) to flow out of the axon, which repolarizes the membrane back towards resting potential.

(b) Myelin acts as an electrical insulator, preventing ions from leaking through the axonal membrane. Consequently, depolarization can only occur at the unmyelinated gaps. The nerve impulse 'jumps' from one node of Ranvier to the next in a process called saltatory conduction, which greatly increases the speed of transmission compared to continuous propagation along an unmyelinated axon.

(c) Nodes of Ranvier.

(a) [3 marks max]
- (Threshold potential triggers) opening of voltage-gated sodium channels causing sodium (\(Na^+\)) influx / depolarization [1]
- Opening of voltage-gated potassium channels causes potassium (\(K^+\)) efflux / repolarization [1]
- Sequence/coordination: sodium channels close before potassium channels fully open / restoration of resting potential [1]

(b) [3 marks max]
- Myelin acts as an insulator / prevents ion leakage [1]
- Depolarization/action potential only occurs at the nodes of Ranvier [1]
- Nerve impulses jump from node to node / saltatory conduction, increasing conduction speed [1]

(c) [1 mark max]
- Node(s) of Ranvier [1]

(a) The gaseous waste product is carbon dioxide (\(CO_2\)). During baking, the yeast ferments sugars and releases \(CO_2\) gas, which forms pockets of gas inside the dough, causing the bread to rise and giving it a light, porous texture.

(b) Aerobic cell respiration produces a much higher yield of ATP (approximately 30 to 32 ATP molecules per glucose molecule) because it involves the link reaction, Krebs cycle, and oxidative phosphorylation. Anaerobic respiration yields only 2 ATP molecules per glucose molecule, which are generated solely during glycolysis.

(c) In glycolysis, \(NAD^+\) is reduced to \(NADH\) when glucose is broken down to pyruvate. Because oxygen is absent to act as the terminal electron acceptor in the electron transport chain, \(NADH\) cannot transfer its electrons there. Yeast cells solve this by reducing pyruvate to ethanol and carbon dioxide, which oxidizes \(NADH\) back to \(NAD^+\). This regenerated \(NAD^+\) is crucial because it allows glycolysis to continue and produce ATP anaerobically.

(a) [2 marks max]
- Carbon dioxide / \(CO_2\) [1]
- Causes the dough to rise / creates bubbles or air pockets in bread [1]

(b) [2 marks max]
- Aerobic respiration produces a high yield of ATP / approx 30-32 ATP per glucose [1]
- Anaerobic respiration produces a low yield of ATP / only 2 ATP per glucose [1]

(c) [3 marks max]
- Glycolysis requires oxidized \(NAD^+\) to continue converting glucose into pyruvate [1]
- Under anaerobic conditions, \(NADH\) transfers its electrons to pyruvate/acetaldehyde [1]
- This oxidizes/regenerates \(NAD^+\), preventing glycolysis from halting / maintaining ATP production [1]

(a) Somatic mutations occur in non-reproductive body cells and are not passed on to offspring, meaning they have no direct evolutionary impact on the gene pool of a population. Germline mutations occur in gametes (sperm or egg cells) and can be inherited by future generations, introducing new alleles and thus acting as a major source of genetic variation for natural selection and evolution.

(b) The guide RNA (gRNA) is designed with a specific sequence of nucleotides that is complementary to the target DNA sequence, guiding the Cas9 complex to the exact genomic locus of interest. The Cas9 enzyme acts as molecular scissors, binding to the DNA at this target site guided by the gRNA and introducing a double-strand break in the DNA helix, which can then be repaired or modified.

(c) One major ethical concern is that any edits made to human germline cells are heritable, meaning they will be passed on to future generations without their consent. There is also a risk of 'off-target' mutations (unintended edits in other parts of the genome) that could cause genetic disorders or cancer, which would then persist in the human gene pool.

(a) [2 marks max]
- Somatic mutations are not heritable / have no direct effect on the population's gene pool / evolution [1]
- Germline mutations are inherited by offspring / can alter the population's gene pool / drive evolution [1]

(b) [3 marks max]
- gRNA binds to/recognizes a specific target DNA sequence through complementary base pairing [1]
- Cas9 is an endonuclease enzyme / acts as molecular scissors [1]
- Cas9 cuts both strands of DNA / creates a double-strand break at the site designated by the gRNA [1]

(c) [2 marks max]
- Changes are passed to future generations who cannot give consent [1]
- Potential for unintended off-target mutations/long-term harmful consequences [1]
- Equity concerns / potential for enhancement ('designer babies') leading to societal inequality [1]
(Accept any one well-explained ethical point for 2 marks, or 1 mark for basic point + 1 mark for elaboration)

(a) Water potential is the potential energy of water per unit volume relative to pure water, determining the direction of water movement (water moves from an area of higher water potential to an area of lower water potential).

(b) Potato cylinders are weighed before and after immersion in different concentrations of sucrose. The percentage change in mass is calculated for each concentration. These percentage changes are plotted on a graph against sucrose concentration. The point on the graph where the line of best fit crosses the x-axis (where there is 0% change in mass) represents the point of equilibrium, which corresponds to the osmolarity/solute concentration of the potato tissue.

(c) In a highly hypertonic sucrose solution, the solution has a lower water potential than the cytoplasm of the potato cells. Water will leave the cells by osmosis down the water potential gradient. This will cause the cells to lose turgor pressure and become plasmolysed (flaccid), as the cell membrane pulls away from the cell wall.

(a) [1 mark max]
- Measure of the tendency of water to move from one area to another / potential energy of water per unit volume relative to pure water [1]

(b) [3 marks max]
- Measure initial and final mass (or calculate percentage change in mass) of potato cylinders [1]
- Plot percentage change in mass against concentration of sucrose solution [1]
- The concentration at which there is zero change in mass (where the line crosses the x-axis) is equal to the osmolarity of the potato tissue [1]

(c) [3 marks max]
- Water potential of the solution is lower than that of the potato cells / solution is hypertonic [1]
- Water leaves the cells by osmosis [1]
- Cells become plasmolysed / flaccid / cell membrane detaches from cell wall [1]

a) Activation energy is the minimum amount of energy that must be overcome for a chemical reaction to occur, representing the energy required to reach the unstable transition state and break chemical bonds in the reactants. Enzymes are biological catalysts that speed up metabolic reactions without being consumed in the process. They work by binding to substrate molecules at a specific region called the active site, forming an enzyme-substrate complex. This binding puts physical strain on the bonds of the substrate or aligns the reactants in an optimal orientation, thereby lowering the activation energy barrier. As a result, more reactant molecules have the energy to undergo the reaction at body temperature, increasing the reaction rate while leaving the overall net energy change of the reaction unchanged.

b) The rate of enzyme-catalyzed reactions is heavily influenced by physical factors such as temperature and pH:
- Temperature: As temperature increases, the kinetic energy of both enzyme and substrate molecules increases. This leads to faster molecular movement and more frequent, high-energy collisions, resulting in an increased rate of reaction up to an optimum temperature. However, if the temperature rises beyond this optimum, the high thermal energy causes the enzyme molecule to vibrate violently. This disrupts the hydrogen bonds, ionic bonds, and other intermolecular forces holding the enzyme's specific three-dimensional tertiary structure. The shape of the active site changes permanently, a process called denaturation, meaning the substrate can no longer fit, and the reaction rate rapidly drops to zero.
- pH: Every enzyme has an optimum pH where its catalytic activity is highest. Deviations from this optimum pH alter the concentration of hydrogen ions (\(H^+\)) or hydroxide ions (\(OH^-\)) in the environment. These charged ions interact with the acidic and basic amino acid side chains within the enzyme. This alters the overall charge distribution, especially at the active site, and disrupts ionic bonds and hydrogen bonds. This leads to a conformational change in the enzyme's shape (denaturation), preventing substrate binding and reducing or halting enzymatic activity.

c) Inhibitors are substances that reduce the rate of enzyme-controlled reactions by interfering with the active site or changing the enzyme's structure:
- Competitive Inhibition: Competitive inhibitors have a chemical structure structurally similar to the substrate. They compete directly with substrate molecules for binding to the enzyme's active site. When a competitive inhibitor binds to the active site, it physically blocks the substrate from entering. This inhibition can be overcome by significantly increasing the substrate concentration, as the substrate will outcompete the inhibitor for the active sites, allowing the reaction to eventually reach its maximum rate (\(V_{max}\)).
- Non-Competitive Inhibition: Non-competitive inhibitors do not share structural similarity with the substrate. Instead of binding to the active site, they bind to an alternative site on the enzyme known as an allosteric site. The binding of the inhibitor causes a conformational (shape) change in the enzyme's tertiary structure, which alters the shape of the active site. Consequently, the substrate can no longer bind effectively, or the enzyme can no longer catalyze the reaction. Because they bind to a different site, increasing the substrate concentration does not overcome non-competitive inhibition; the maximum possible rate of the reaction (\(V_{max}\)) is permanently reduced.

Part (a) [Max 4 marks]
- Activation energy is the minimum energy required to start a chemical reaction / reach the transition state. [1]
- Enzymes lower the activation energy barrier. [1]
- Enzymes bind to substrates at the active site to form an enzyme-substrate complex. [1]
- Substrate bonds are stressed, weakened, or aligned properly when bound to the active site. [1]
- The overall net energy change (free energy change) of the reaction remains unaltered. [1]

Part (b) [Max 6 marks]
- Increasing temperature increases the kinetic energy of enzymes and substrates, leading to more frequent collisions. [1]
- The rate increases up to an optimum temperature. [1]
- Temperatures above the optimum disrupt weak bonds (hydrogen/ionic bonds) maintaining the tertiary structure. [1]
- This causes denaturation, altering the shape of the active site so substrates can no longer bind. [1]
- Enzymes also have an optimum pH where their activity is maximized. [1]
- Extreme pH values (too acidic or basic) alter the charges on amino acid side chains. [1]
- This disruption of ionic and hydrogen bonds causes denaturation and loss of function. [1]

Part (c) [Max 5 marks]
- Competitive inhibitors are structurally similar to the substrate. [1]
- They bind directly to the active site, blocking substrate entry. [1]
- Competitive inhibition can be overcome by increasing substrate concentration (reaching normal \(V_{max}\)). [1]
- Non-competitive inhibitors bind to an allosteric site (a site other than the active site). [1]
- Binding causes a conformational change in the active site, preventing substrate binding or catalysis. [1]
- Non-competitive inhibition cannot be overcome by increasing substrate concentration (reduces \(V_{max}\)). [1]

Quality Mark [1]
- 1 mark is awarded if the candidate writes a clear, coherent, and logical response using accurate biological terminology throughout all parts of the question, without unnecessary repetition.

(a) There is a positive correlation / direct relationship between substrate concentration and the volume of oxygen produced (rate of reaction) up to 1.5%.
(b) At concentrations of 2.0% and above, the enzyme's active sites are fully saturated with substrate. The enzyme concentration becomes the limiting factor, meaning further increases in substrate concentration do not increase the rate.
(c) Controlled variables include: catalase/enzyme concentration, volume of enzyme solution, volume of substrate solution, pH of the buffer, and the source/batch of the catalase.

(a) [1 max]
- Award 1 mark for describing the positive correlation / direct relationship between substrate concentration and rate of oxygen production.

(b) [2 max]
- Award 1 mark for stating that enzyme active sites are saturated/occupied.
- Award 1 mark for explaining that enzyme concentration is the limiting factor (or that substrate concentration is no longer limiting).

(c) [2 max]
- Award 1 mark for each valid controlled variable (e.g., pH, enzyme concentration, enzyme volume, total volume of reaction mixture) up to 2 marks. Reject: temperature (stated in question).

(a) The isotonic concentration is 0.4 \(\text{mol dm}^{-3}\) because at this concentration there is 0% change in mass, indicating no net movement of water.
(b) The water potential of the 0.0 \(\text{mol dm}^{-3}\) solution is higher (hypotonic) than that of the cytoplasm of the potato cells. Water moves into the potato cells by osmosis down the water potential gradient, increasing the cell turgor and overall mass.
(c) Potato cylinders may have slightly different initial masses; calculating percentage change allows for a fair comparison by standardizing the starting point.
(d) The cylinders should be gently blotted with a paper towel to remove any excess water adhering to the outer surface before weighing.

(a) [1 max]
- Award 1 mark for 0.4 \(\text{mol dm}^{-3}\) (accept 0.4 M).

(b) [2 max]
- Award 1 mark for identifying that water moves into cells by osmosis (down the water potential gradient / from hypotonic to hypertonic).
- Award 1 mark for linking this water intake to an increase in cell turgor/mass.

(c) [1 max]
- Award 1 mark for stating that it accounts for variations in initial mass of the cylinders.

(d) [1 max]
- Award 1 mark for blotting the surface dry / removing surface water.

(a) Light intensity is inversely proportional to the square of the distance (inverse square law). As distance increases, light intensity decreases.
(b) The light bulb emits heat as well as light. Placing a water shield between the light and the plant absorbs the heat, keeping the temperature constant so that temperature does not become an uncontrolled confounding variable that affects photosynthetic enzyme activity.
(c) The dependent variable is the rate of photosynthesis (measured as bubbles per minute). A more accurate method to measure gas production is using a gas syringe or a capillary tube with a syringe (Audus apparatus) to measure the actual volume of oxygen gas produced, or using a dissolved oxygen probe.

(a) [1 max]
- Award 1 mark for stating that light intensity is inversely proportional to distance (or distance squared) / decreases as distance increases.

(b) [2 max]
- Award 1 mark for stating that the water shield absorbs heat / prevents temperature changes.
- Award 1 mark for explaining that temperature affects enzyme activity / rate of photosynthesis, so it must be controlled.

(c) [2 max]
- Award 1 mark for identifying the dependent variable as rate of oxygen production / rate of photosynthesis.
- Award 1 mark for suggesting a valid alternative method (e.g., using a gas syringe, capillary tube, oxygen probe, or gas sensor).

During exercise, cellular respiration increases, producing more carbon dioxide. CO2 dissolves in blood to form carbonic acid, decreasing blood pH. Chemoreceptors in the medulla, carotid arteries, and aorta detect this change and send signals to the cardiovascular centre of the medulla oblongata. The medulla sends sympathetic nerve signals via the cardiac nerve to the sinoatrial (SA) node, increasing its rate of firing. Additionally, stress/exercise triggers the release of epinephrine from the adrenal medulla, which is carried in the bloodstream to the SA node, further accelerating the heart rate to ensure rapid transport of oxygen and nutrients.

Award [1] for each of the following up to [5 max]: 1. Exercise increases cellular respiration, producing more CO2 / lowering blood pH. 2. Chemoreceptors (in aorta/carotid arteries/medulla) detect the drop in pH / increase in CO2. 3. Nerve impulses are sent to the cardiovascular centre in the medulla oblongata. 4. The medulla sends impulses along the sympathetic (cardiac) nerve to the sinoatrial (SA) node (pacemaker) to increase heart rate. 5. Epinephrine (adrenaline) is released from the adrenal glands into the blood. 6. Epinephrine acts on the SA node to increase heart rate (sustained/rapid response).

Blood glucose levels are regulated by negative feedback. Following a meal, blood glucose levels rise, which is detected by the pancreas. The beta cells in the islets of Langerhans secrete insulin directly into the bloodstream. Insulin targets muscle and liver cells, causing them to increase uptake of glucose by inserting more glucose transporters into their membranes. In liver and muscle cells, insulin activates enzymes that convert glucose into insoluble glycogen for storage. This process removes glucose from the blood, lowering blood glucose levels back to normal, which then reduces further insulin secretion.

Award [1] for each of the following up to [5 max]: 1. Rise in blood glucose is detected by the pancreas (specifically beta cells in islets of Langerhans). 2. Beta cells secrete the hormone insulin into the blood. 3. Insulin targets liver and/or muscle cells. 4. Insulin increases the permeability of cells to glucose / increases glucose uptake from blood. 5. Insulin stimulates the conversion of glucose to glycogen (glycogenesis) for storage. 6. This represents a negative feedback loop as blood glucose levels return to normal, decreasing insulin secretion.

About 70% of carbon dioxide is transported as bicarbonate ions (HCO3-) in the plasma. In red blood cells, the enzyme carbonic anhydrase catalyzes the reaction between CO2 and H2O to form carbonic acid (H2CO3), which dissociates into H+ and HCO3-. Bicarbonate ions diffuse out of the red blood cell into the plasma. The remaining CO2 is transported either dissolved directly in blood plasma or bound to hemoglobin as carbaminohemoglobin. The accumulated H+ ions in red blood cells bind to hemoglobin, causing conformational changes that lower its affinity for oxygen. This is known as the Bohr shift, which shifts the oxygen dissociation curve to the right, causing hemoglobin to release oxygen more readily in actively respiring tissues where CO2 levels are high.

Award [1] for each of the following up to [5 max]: 1. Carbon dioxide is transported in three ways: dissolved in plasma, bound to hemoglobin, and as hydrogen carbonate (bicarbonate) ions. 2. Inside red blood cells, carbon dioxide reacts with water to form carbonic acid, catalyzed by the enzyme carbonic anhydrase. 3. Carbonic acid dissociates into hydrogen ions (H+) and bicarbonate ions (HCO3-). 4. Bicarbonate ions diffuse out into the plasma (while chloride ions enter - chloride shift). 5. Increased carbon dioxide and hydrogen ion concentration lower blood pH. 6. Hydrogen ions bind to hemoglobin, causing a conformational change that decreases its affinity for oxygen (Bohr shift), shifting the oxygen dissociation curve to the right and releasing oxygen to tissues.

The nervous system and endocrine system coordinate physiological processes. 1. Transmission mechanism: The nervous system uses electrical impulses along axons and chemical neurotransmitters across synaptic clefts, whereas the endocrine system uses chemical messengers called hormones transported via the bloodstream. 2. Speed of transmission and response: Nervous communication is extremely rapid (milliseconds), whereas endocrine communication is generally slower (seconds to hours) as hormones must travel through circulation. 3. Target range/specificity: Nervous signals are highly localized and specific to individual cells, muscles, or glands, while endocrine signals are often widespread, affecting any cell in the body that possesses the specific receptor. 4. Duration of effect: Nervous responses are brief and immediate, whereas endocrine responses are typically long-lasting and sustained.

Award [1] for each of the following up to [5 max]: 1. Nervous system uses electrical impulses along neurons and neurotransmitters at synapses, while endocrine system uses hormones carried in the bloodstream. 2. Nervous transmission/response is much faster (takes milliseconds) compared to slower endocrine transmission (seconds to days). 3. Nervous responses are highly localized (specific target cells/muscles), whereas endocrine hormones act on widespread target organs/cells with appropriate receptors. 4. Nervous responses are short-lived/brief, while endocrine responses have a longer duration of effect. 5. Both systems are involved in homeostatic regulation and integration of body activities. 6. Accept comparisons structured in a table or prose format.