2025 AQA IAL Biology (9610) 模擬試題連答案詳解

Eukaryotic DNA is linear and wrapped around histone proteins, whereas prokaryotic DNA is circular and is not associated with histones (naked).

1 mark for stating eukaryotic DNA is linear and prokaryotic DNA is circular. 1 mark for stating eukaryotic DNA is associated with histones/proteins and prokaryotic DNA is not/naked. (Accept: Eukaryotic DNA has non-coding introns whereas prokaryotic DNA does not)

A phosphodiester bond is formed via a condensation reaction. The reaction occurs between the 3-carbon of the deoxyribose pentose sugar of one nucleotide and the phosphate group on the 5-carbon of the adjacent nucleotide, with the elimination of a water molecule.

1 mark for stating it is a condensation reaction / releases water. 1 mark for stating the bond forms between the deoxyribose (sugar) of one nucleotide and the phosphate group of the other.

Enzymes lower the activation energy of a reaction. They do this by forming an enzyme-substrate complex, which brings substrates closer together or puts physical strain on their chemical bonds, making it easier for bonds to break or form.

1 mark for stating that enzymes lower the activation energy. 1 mark for describing how the formation of the enzyme-substrate complex facilitates this (e.g. by aligning substrates/putting strain on bonds/bending bonds).

A non-competitive inhibitor binds to the enzyme at an allosteric site (a region other than the active site). This binding causes a change in the tertiary structure of the enzyme, which alters the specific shape of the active site. Consequently, the substrate is no longer complementary to the active site, preventing the formation of enzyme-substrate complexes.

1 mark for stating the inhibitor binds to the allosteric site / a site other than the active site. 1 mark for stating this changes the tertiary structure/shape of the active site so the substrate can no longer bind / form enzyme-substrate complexes.

In response to unilateral light, IAA moves from the light side to the shaded side of the shoot tip. The higher concentration of IAA on the shaded side stimulates greater cell elongation there compared to the light side, resulting in the shoot bending towards the light source.

1 mark for stating that IAA moves/accumulates on the shaded side of the shoot. 1 mark for stating that IAA stimulates cell elongation on the shaded side (causing uneven growth/bending).

According to the acid growth hypothesis, hydrogen ions (\(\text{H}^+\)) are pumped into the cell wall. This lowers the pH, which: 1. Activates specific enzymes called expansins that break the cross-links (hydrogen bonds) between cellulose microfibrils. 2. Increases the elasticity/flexibility of the cell wall, allowing turgor pressure to stretch the cell.

1 mark for mentioning the activation of expansins / enzymes that break bonds/cross-links between cellulose microfibrils. 1 mark for stating that the cell wall becomes more flexible/extensible/loosened, allowing water entry/turgor pressure to expand the cell.

During glycolysis, glucose is phosphorylated by two phosphate groups derived from the hydrolysis of two molecules of ATP, producing hexose bisphosphate. This step is necessary to make the glucose molecule highly reactive (lowering the activation energy for subsequent reactions) and polar, which traps the sugar inside the cell as it cannot cross the cell membrane.

1 mark for stating that glucose is phosphorylated using phosphate from ATP (forming hexose bisphosphate). 1 mark for stating this activates glucose / makes it more reactive / prevents it from diffusing out of the cell.

During the link reaction, pyruvate undergoes decarboxylation and oxidation. The hydrogen removed is used to reduce \(\text{NAD}^+\) to reduced NAD (\(\text{NADH}\)), and the resulting acetate group combines with coenzyme A to form acetyl coenzyme A. Reduced NAD is used in oxidative phosphorylation, and acetyl coenzyme A enters the Krebs cycle.

1 mark for naming reduced NAD / NADH (used in oxidative phosphorylation). 1 mark for naming acetyl coenzyme A / acetyl CoA (used in the Krebs cycle). (Do not accept carbon dioxide as it is a waste product and not used in subsequent stages of aerobic respiration).

Eukaryotic DNA consists of linear chromosomes located in the nucleus, and is tightly wrapped around histone proteins. In contrast, prokaryotic DNA is circular, floats freely in the cytoplasm, and is not associated with histones.

1 mark: Eukaryotic DNA is linear whereas prokaryotic DNA is circular (or eukaryotic DNA has telomeres/ends, prokaryotic does not). 1 mark: Eukaryotic DNA is associated with histone proteins, whereas prokaryotic DNA is 'naked' or not associated with proteins. (Reject: References to presence/absence of nucleus, plasmids, or size, as these are not differences between the DNA molecules themselves.)

A non-competitive inhibitor binds to the enzyme at an allosteric site, which is distinct from the active site. This binding alters the overall tertiary structure of the enzyme, causing the shape of the active site to change. As a result, the substrate is no longer complementary to the active site, meaning substrate molecules can no longer bind and enzyme-substrate complexes cannot form, reducing the rate of reaction.

1 mark: Inhibitor binds to the enzyme at a site other than the active site / allosteric site. 1 mark: This changes the shape / tertiary structure of the active site, so the substrate is no longer complementary / cannot bind / enzyme-substrate complexes (ESCs) cannot form.

When a shoot is exposed to unilateral light, IAA is transported from the illuminated side to the shaded side of the shoot tip. The resulting high concentration of IAA on the shaded side stimulates cells on this side to elongate more rapidly than those on the light side. This uneven rate of elongation causes the shoot to bend towards the light source.

1 mark: IAA moves to / accumulates on the shaded side of the shoot tip. 1 mark: High concentration of IAA on the shaded side stimulates cell elongation (leading to uneven growth/bending). (Do not accept: references to cell division/mitosis instead of cell elongation.)

In the link reaction, pyruvate is decarboxylated and oxidized to produce an acetate group. Coenzyme A binds with this acetate group to form acetyl coenzyme A. The function of coenzyme A is to transport this two-carbon acetyl group into the mitochondrial matrix and deliver it to the Krebs cycle, where it combines with the four-carbon oxaloacetate to form the six-carbon citrate.

1 mark: Combines with acetate / acetyl group (produced in the link reaction) to form acetyl coenzyme A. 1 mark: Transports / delivers the acetyl / acetate group to the Krebs cycle / matrix (to react with oxaloacetate).

A codon consists of a three-nucleotide sequence located on messenger RNA (mRNA) that codes for a specific amino acid. An anticodon is a three-nucleotide sequence on transfer RNA (tRNA) that is complementary to a codon on mRNA. Therefore, they differ in their location (mRNA vs tRNA) and their complementary binding partners.

1 mark: Codon is found on mRNA, whereas anticodon is found on tRNA. 1 mark: Codon determines the amino acid sequence / is complementary to a DNA triplet, whereas an anticodon binds/is complementary to a codon on mRNA (or codon is used in transcription/translation, anticodon is used in translation only).

In anaerobic respiration in yeast (ethanol fermentation), pyruvate is first decarboxylated by pyruvate decarboxylase to produce ethanal (releasing carbon dioxide). Ethanal then acts as the final hydrogen/electron acceptor, receiving hydrogen from reduced NAD (NADH). This oxidizes reduced NAD back to NAD, allowing glycolysis to continue, and converts ethanal into ethanol.

1 mark: Ethanal (accept acetaldehyde). 1 mark: Reduced NAD / NADH (accept NADH + H+). (Do not accept: NAD+ or NAD on its own for the coenzyme oxidized, it must be the reduced form being oxidized.)

When a root is placed horizontally, gravity causes IAA to pool/accumulate on the lower side of the root tip. In root cells, a high concentration of IAA has an inhibitory effect on cell elongation. Consequently, the cells on the upper side of the root (where IAA concentration is low) elongate more rapidly than those on the lower side, causing the root to bend downwards in the direction of gravity.

1 mark: IAA accumulates on the lower side / bottom of the root (due to gravity). 1 mark: High concentration of IAA inhibits cell elongation in roots, causing the upper side to elongate faster / root to bend downwards. (Accept: converse argument for upper side.)

Eukaryotic nuclear DNA and prokaryotic DNA share the same basic double-helix structure consisting of nucleotides joined by phosphodiester bonds. However, they differ significantly in their organization and location: 1. Eukaryotic nuclear DNA is linear, whereas prokaryotic DNA is circular. 2. Eukaryotic DNA is associated with histones (proteins), whereas prokaryotic DNA is not associated with histones (naked). 3. Eukaryotic DNA contains non-coding introns, whereas prokaryotic DNA does not contain introns. 4. Eukaryotic nuclear DNA is enclosed within a double-membrane nucleus, whereas prokaryotic DNA is free in the cytoplasm (nucleoid region). 5. Eukaryotic cells possess a much larger quantity of DNA (longer chromosomes) compared to the smaller, shorter circular DNA molecule of prokaryotes.

1 mark for each correct comparison up to a maximum of 5 marks: Eukaryotic DNA is linear, whereas prokaryotic DNA is circular; Eukaryotic DNA is associated with histones / proteins, whereas prokaryotic DNA is not / is naked; Eukaryotic DNA contains introns / non-coding sequences, whereas prokaryotic DNA does not contain introns; Eukaryotic DNA is enclosed in a nucleus, whereas prokaryotic DNA is free in the cytoplasm / nucleoid; Eukaryotic DNA is longer / has a larger mass / more base pairs than prokaryotic DNA; Accept: Eukaryotic cells do not contain plasmids in their nuclei, whereas prokaryotes often contain plasmids.

To determine the type of inhibition, the researcher should set up reaction mixtures with a constant concentration of the enzyme and the inhibitor. They should then measure the initial rate of reaction across a wide range of increasing substrate concentrations. In the presence of a competitive inhibitor, increasing the substrate concentration increases the rate of reaction until it eventually reaches the same maximum rate (Vmax) as the uninhibited reaction. This occurs because the substrate outcompetes the inhibitor for the active site. In contrast, in the presence of a non-competitive inhibitor, increasing the substrate concentration will not restore the maximum rate of reaction; the Vmax remains significantly lower. This is because the non-competitive inhibitor binds to an allosteric site, permanently altering the active site shape, meaning substrate binding or catalysis is prevented regardless of substrate abundance.

1. Measure the initial rate of reaction at constant enzyme and inhibitor concentrations (1 mark); 2. Vary / increase the concentration of the substrate (1 mark); 3. For competitive inhibition, the rate of reaction will eventually reach the maximum rate (Vmax) / same rate as without inhibitor at high substrate concentrations (1 mark); 4. This is because the substrate outcompetes the inhibitor for the active site (1 mark); 5. For non-competitive inhibition, the maximum rate (Vmax) is never reached / remains lower regardless of substrate concentration (1 mark); 6. This is because the inhibitor binds to an allosteric site, changing the conformation of the active site so that substrates cannot successfully react (1 mark). [Maximum 5 marks]

Unilateral light causes the plant hormone indoleacetic acid (IAA) to migrate from the illuminated side of the shoot tip to the shaded side. This uneven distribution results in a higher concentration of IAA on the shaded side. In shoots, IAA stimulates cell elongation. It does this by stimulating hydrogen ion (\(\text{H}^+\)) pumps in the cell membrane to actively transport protons into the cell wall, lowering the pH. This acidic environment activates enzymes called expansins, which break the chemical bonds/cross-links between cellulose microfibrils in the cell wall, making the wall more flexible and plastic. Due to the high turgor pressure inside the cells, the loosened cell walls stretch, causing the cells on the shaded side to elongate. Because the cells on the shaded side grow longer than those on the light side, the shoot bends towards the light source (positive phototropism).

1. Unilateral light causes IAA to migrate from the light side to the shaded side of the shoot tip (1 mark); 2. This creates a higher concentration of IAA on the shaded side (1 mark); 3. IAA stimulates active transport of hydrogen ions (\(\text{H}^+\)) into the cell wall, lowering the pH (1 mark); 4. The lower pH activates expansins / enzymes that loosen the cellulose microfibrils / cell wall (1 mark); 5. Turgor pressure causes these cells on the shaded side to elongate / stretch more than cells on the light side (1 mark); 6. The uneven elongation causes the shoot to bend towards the light / exhibit positive phototropism (1 mark). [Maximum 5 marks]

In normal respiration, the electron transport chain pumps protons (\(\text{H}^+\) ions) from the matrix into the intermembrane space, establishing an electrochemical proton gradient. Protons normally flow back into the matrix down this gradient exclusively through ATP synthase, driving the phosphorylation of ADP to ATP. When an uncoupler is present, it provides an alternative pathway for protons to leak back across the inner membrane into the matrix, bypassing ATP synthase. This collapses or severely reduces the proton gradient. As a result, little to no ATP is synthesized via oxidative phosphorylation. Because the chemical energy stored in the proton gradient is not captured in the chemical bonds of ATP, it is instead dissipated directly as thermal energy (heat). To compensate for the lack of ATP, the cell increases the rate of glycolysis and the Krebs cycle to attempt to restore ATP levels, which further increases the oxidation of substrates and generates even more heat.

1. Protons flow / leak across the inner mitochondrial membrane into the matrix, bypassing ATP synthase (1 mark); 2. This collapses / reduces the proton gradient / electrochemical gradient across the membrane (1 mark); 3. Less / no ADP is phosphorylated to ATP (via oxidative phosphorylation) (1 mark); 4. The potential energy of the proton gradient is released / dissipated as heat instead of being stored as ATP (1 mark); 5. The rate of substrate oxidation / respiration / oxygen consumption increases in an attempt to compensate for low ATP (1 mark); 6. This hyper-metabolic state generates a large amount of excess body heat (1 mark). [Maximum 5 marks]

In the lock-and-key model, the active site of the enzyme is a rigid structure that is exactly complementary in shape to the substrate before binding. In contrast, the induced-fit model proposes that the active site is flexible and is not fully complementary initially. When the substrate collides with the active site, it induces a conformational change (change in shape) in the enzyme. The active site molds itself around the substrate to form a highly complementary fit, establishing an enzyme-substrate complex. As the active site changes shape, it puts physical strain and tension on specific chemical bonds within the substrate molecule. This destabilizes the bonds, making them easier to break. Alternatively, it aligns the substrate molecules in the precise orientation needed for them to react. Both of these mechanisms significantly lower the activation energy required for the chemical reaction to proceed.

1. In lock-and-key, the active site is rigid/fully complementary before binding; in induced-fit, the active site is flexible / changes shape upon substrate binding (1 mark); 2. Binding of the substrate induces a conformational change of the enzyme's active site (1 mark); 3. This forms a complementary fit around the substrate / forms the enzyme-substrate complex (1 mark); 4. The change in shape of the active site puts strain / stress on the chemical bonds of the substrate (1 mark); 5. This distorts / weakens the bonds, making them easier to break (1 mark); 6. The induced fit aligns the substrates in the correct orientation, facilitating bond formation, lowering the activation energy of the reaction (1 mark). [Maximum 5 marks]

When a root is placed horizontally, specialized organelles called amyloplasts (or statoliths), which contain heavy starch grains, sink to the bottom of the cells in the root tip cap under the influence of gravity. The settling of these amyloplasts on the lower cell membrane triggers the active transport of the plant hormone indoleacetic acid (IAA) to the lower side of the root tip. This leads to an asymmetrical distribution of IAA, with a much higher concentration on the lower side than the upper side. Unlike in shoots where IAA stimulates growth, in roots, a high concentration of IAA inhibits cell elongation. Consequently, cells on the lower side of the root elongate much more slowly than the cells on the upper side. The cells on the upper side continue to elongate at a normal rate, which causes the root to curve and bend downwards towards gravity (positive gravitropism).

1. Amyloplasts / statoliths sink to the bottom of the cells in the root tip due to gravity (1 mark); 2. This triggers the active transport / redistribution of IAA to the lower side of the root (1 mark); 3. This creates a high concentration of IAA on the lower side and a low concentration on the upper side (1 mark); 4. In roots, high concentrations of IAA inhibit cell elongation (1 mark); 5. Cells on the upper side elongate faster / normally compared to cells on the lower side (1 mark); 6. The uneven / differential elongation causes the root to bend downwards / exhibit positive gravitropism (1 mark). [Maximum 5 marks]

The temperature coefficient (\(Q_{10}\)) represents the factor by which the rate of a reaction increases with a \(10\,^\circ\text{C}\) rise in temperature. It is calculated using the formula: \(Q_{10} = \frac{R_2}{R_1}\), where \(R_2\) is the rate at the higher temperature (\(25\,^\circ\text{C}\)) and \(R_1\) is the rate at the lower temperature (\(15\,^\circ\text{C}\)). Substituting the given values: \(Q_{10} = \frac{5.28}{2.4} = 2.2\).

Mark 1: Shows correct formula or substitution of values into the fraction: \(\frac{5.28}{2.4}\).
Mark 2: Correct calculation of intermediate step or shows division setup.
Mark 3: Correct final answer of \(2.2\) (accept with or without 'no units').

First, calculate the total energy captured in the synthesized ATP: \(32\text{ moles of ATP} \times 30.5\text{ kJ mol}^{-1} = 976\text{ kJ}\). Next, calculate the percentage efficiency by dividing the energy stored in ATP by the total energy released from glucose, then multiplying by 100: \(\frac{976}{2870} \times 100 = 34.0069...\%\). Rounding to three significant figures gives \(34.0\%\).

Mark 1: Correct calculation of total energy stored in ATP as \(976\text{ kJ}\) (or shows expression \(32 \times 30.5\)).
Mark 2: Correct calculation setup for percentage efficiency: \(\frac{976}{2870} \times 100\) (or using student's incorrect value of energy from Mark 1 divided by \(2870\)).
Mark 3: Correct final answer of \(34.0\%\) (accept \(34\%\) or \(34.0\)).

First, calculate the total length of the DNA in nanometres (\(\text{nm}\)) by multiplying the number of base pairs by the distance between them: \(4.5 \times 10^3 \times 0.34\text{ nm} = 1530\text{ nm}\). To convert nanometres to micrometres, divide by 1000 (since \(1\,\mu\text{m} = 1000\text{ nm}\)): \(\frac{1530}{1000} = 1.53\,\mu\text{m}\).

Mark 1: Calculates the total length in nanometres as \(1530\text{ nm}\) (or shows expression \(4500 \times 0.34\)).
Mark 2: Shows correct conversion factor from nanometres to micrometres (dividing by 1000 or multiplying by \(10^{-3}\)).
Mark 3: Correct final answer of \(1.53\) (\(\mu\text{m}\)).

1. Find the radius of the capillary tube: \(r = 1.0\text{ mm} \div 2 = 0.5\text{ mm}\). 2. Calculate the volume of oxygen consumed in 15 minutes: \(V = 3.14 \times (0.5)^2 \times 48 = 37.68\text{ mm}^3\). 3. Calculate the rate per hour: \(37.68 \times (60 \div 15) = 150.72\text{ mm}^3\text{ hour}^{-1}\). 4. Calculate the rate per gram per hour: \(150.72 \div 5.0\text{ g} = 30.144\text{ mm}^3\text{ g}^{-1}\text{ hour}^{-1}\). 5. Round to one decimal place: \(30.1\).

1 mark for calculating correct volume of 37.68 \(\text{mm}^3\) (or showing correct substitution into formula). 1 mark for dividing by mass and adjusting for time (hourly rate of 150.72 divided by 5.0, or 30.144). 1 mark for the correct final answer of 30.1 (accept 30.14 if a more precise value for pi was used).

1. Determine the ranges using standard deviations: the IAA group range is \(13.6\text{ to }16.0\text{ mm}\) and the control group range is \(10.8\text{ to }11.6\text{ mm}\). 2. Compare the ranges: the lowest value for the IAA group (13.6 mm) is greater than the highest value for the control group (11.6 mm), so there is no overlap. 3. Draw statistical conclusions: since there is no overlap, the difference is likely significant. A Student's t-test should be used to confirm if \(P \le 0.05\).

1 mark for stating that the standard deviation ranges do not overlap (or showing calculations of ranges: 13.6-16.0 mm vs 10.8-11.6 mm). 1 mark for concluding that the lack of overlap means the difference is likely to be significant (not due to chance). 1 mark for stating that a Student's t-test should be performed to determine if the probability of the difference being due to chance is less than 5% (\(P \le 0.05\)). Reject: chi-squared test or correlation coefficient.