2023 AQA AS-Level Biology 7401 模擬試題連答案詳解

Magnification is calculated using the formula: Magnification = Image size / Actual size. First, convert the image size from millimetres to micrometres: \( 36\text{ mm} \times 1000 = 36000\text{ }\mu\text{m} \). Then, divide the image size in micrometres by the actual size in micrometres: \( 36000 / 4.8 = 7500 \).

1 mark for showing correct conversion of units (e.g., 36,000 micrometres or 0.0048 mm) or correct division expression (e.g., 36,000 / 4.8 or 36 / 0.0048).
1 mark for the correct final answer of 7500 (accept \( \times 7500 \)).

Antigenic drift involves mutations that lead to a change in the tertiary structure and shape of the viral antigens. As a result, memory cells or antibodies produced from the vaccine will no longer be complementary in shape to the new antigens, preventing an effective secondary immune response.

1 mark for stating that antigenic drift changes the tertiary structure or shape of the viral antigen.
1 mark for explaining that memory cells or antibodies (produced by vaccination) are no longer complementary in shape, or will no longer recognize/bind to the new antigen.

A peptide bond is formed by a condensation reaction, which releases a molecule of water. This bond is established between the amine (\( -\text{NH}_2 \)) group of one amino acid and the carboxyl (\( -\text{COOH} \)) group of the adjacent amino acid.

1 mark for identifying it as a condensation reaction, or noting that a water molecule is released.
1 mark for stating that the bond forms between the amine group of one amino acid and the carboxyl group of another.

A triglyceride consists of one glycerol molecule chemically bonded to three fatty acids. Conversely, a phospholipid consists of one glycerol molecule bonded to only two fatty acids and has one phosphate group in place of the third fatty acid.

1 mark for stating that triglyceride has three fatty acids whereas phospholipid has two.
1 mark for stating that phospholipid has a phosphate group whereas triglyceride does not.

During vigorous exercise, increased respiration in muscles produces more carbon dioxide. High carbon dioxide concentration lowers the affinity of haemoglobin for oxygen (shifting the oxygen dissociation curve to the right), which means haemoglobin unloads oxygen more readily to the actively respiring tissues.

1 mark for noting that increased carbon dioxide concentration reduces the affinity of haemoglobin for oxygen.
1 mark for explaining that this leads to increased unloading or dissociation of oxygen to the actively respiring muscle tissues.

The mitotic index as a percentage is calculated using the formula: \( \text{Mitotic Index} = (\text{Number of cells in mitosis} / \text{Total number of cells}) \times 100 \). Substituting the values: \( (21 / 140) \times 100 = 15\% \).

1 mark for showing correct working (e.g., \( 21 / 140 \) or \( 0.15 \)).
1 mark for the correct final answer of 15 (accept 15%).

Species richness only counts the number of different species present in a community, but does not take into account the relative abundance or population size of each species. The index of diversity is more useful because it includes both species richness and species evenness (the size of each population), ensuring that a community dominated heavily by a single species is not incorrectly classified as highly diverse.

1 mark for stating that the index of diversity takes into account the population size, abundance, or evenness of each species.
1 mark for stating that species richness only measures the number of species present, ignoring dominance or uneven distribution of populations.

Micelles are formed from bile salts, fatty acids, and monoglycerides. They render these non-polar lipid breakdown products soluble in the aqueous environment of the ileum, transport them to the surface membrane of the epithelial cells, and release them so they can readily diffuse across the phospholipid bilayer.

1 mark for explaining that micelles make fatty acids or monoglycerides soluble in water, or transport them to the epithelial cell surface membrane.
1 mark for explaining that they release the lipid components to allow direct diffusion across the cell membrane.

To calculate magnification, use the formula: Magnification = Image size / Actual size. First, convert both measurements to the same units. Converting 90 mm to micrometres: 90 multiplied by 1000 equals 90,000 micrometres. Now, divide the image size by the actual size: 90,000 micrometres / 4.5 micrometres = 20,000. Therefore, the magnification is x20,000.

1 mark for correct conversion of units (e.g. 90 mm to 90,000 micrometres). 1 mark for the correct final answer of 20,000 or x20,000.

First, find the total number of cells in the three specified stages of mitosis: 28 + 14 + 7 = 49 cells. Next, calculate this total as a percentage of the total number of cells observed: (49 / 350) multiplied by 100 = 14%.

1 mark for showing correct division formula (49 / 350) or establishing that 49 cells are in these stages. 1 mark for the correct final answer of 14%.

In double-stranded DNA, the percentage of adenine equals the percentage of thymine, so thymine is 22%. The total percentage of adenine and thymine combined is 22% + 22% = 44%. The remaining percentage representing cytosine and guanine is 100% - 44% = 56%. Since the percentage of cytosine equals the percentage of guanine, the percentage of cytosine is 56% / 2 = 28%.

1 mark for showing that adenine + thymine = 44% (or that guanine + cytosine = 56%). 1 mark for the correct final answer of 28%.

Active transport requires metabolic energy in the form of ATP to move substances, whereas facilitated diffusion is a passive process that relies only on the kinetic energy of the molecules. Additionally, active transport moves solutes against their concentration gradient (from low to high concentration), whereas facilitated diffusion moves solutes down their concentration gradient (from high to low concentration).

1 mark for stating that active transport requires ATP / energy while facilitated diffusion is passive (or does not require ATP). 1 mark for stating that active transport moves substances against a concentration gradient while facilitated diffusion moves substances down a concentration gradient.

A peptide bond is formed via a condensation reaction. This reaction takes place between the amine (-NH2) group of one amino acid and the carboxyl (-COOH) group of another amino acid. During this process, a covalent peptide bond is formed and a molecule of water (H2O) is released.

1 mark for identifying that it is a condensation reaction that releases a molecule of water. 1 mark for stating that the reaction occurs between the amine group of one amino acid and the carboxyl group of another.

The primary chemical difference lies in the bonds between carbon atoms in the hydrocarbon chain. A saturated fatty acid has only single covalent bonds between carbon atoms (C-C), meaning the carbon chain is fully saturated with hydrogen atoms. An unsaturated fatty acid contains one or more double covalent bonds between carbon atoms (C=C), which creates kinks in the chain and means it holds fewer hydrogen atoms.

1 mark for stating that saturated fatty acids have only single bonds between carbon atoms (or no carbon-carbon double bonds). 1 mark for stating that unsaturated fatty acids have one or more double bonds between carbon atoms.

To estimate the duration of a specific stage of the cell cycle: 1. Identify and count the cells that are in metaphase, as well as the total number of cells in the observed area. 2. Calculate the proportion of cells in metaphase by dividing the metaphase cell count by the total cell count. 3. Multiply this proportion by the total duration of the cell cycle (18 hours or 1080 minutes) to find the estimated time spent in metaphase, assuming the proportion of cells in a stage reflects the time spent in it.

1. Count the number of cells in metaphase and the total number of cells in the sample / field of view. (1 mark) 2. Divide the number of cells in metaphase by the total number of cells (to calculate the mitotic index / proportion). (1 mark) 3. Multiply this proportion/fraction by 18 hours (or 1080 minutes) to find the duration. (1 mark)

A curve shifted to the right indicates that the haemoglobin has a lower affinity for oxygen at any given partial pressure of oxygen. Consequently, oxygen is released or unloaded much more easily and rapidly to respiring tissues. Because small mammals have a high surface area to volume ratio, they lose heat quickly and have a very high metabolic rate to maintain body temperature. Rapid oxygen unloading is crucial to sustain this high rate of aerobic respiration.

1. (Shrew haemoglobin has) lower affinity for oxygen (at any given partial pressure of oxygen). (1 mark) 2. Oxygen is unloaded/dissociates more readily/easily to respiring tissues. (1 mark) 3. To meet high metabolic rate / high rate of respiration (due to high surface area to volume ratio). (1 mark)

Cardiac muscle cells contract continuously and rhythmically, which requires a constant and high yield of ATP. Mitochondria are the sites of aerobic respiration. The highly folded inner membranes (cristae) provide an extremely large surface area for electron transport chain proteins and ATP synthase enzymes. This high concentration of respiratory machinery enables a high rate of oxidative phosphorylation, producing the massive amounts of ATP required to sustain cardiac contraction.

1. Cardiac muscle requires a continuous/constant supply of ATP / energy for contraction. (1 mark) 2. Highly folded inner membrane (cristae) provides a large surface area for enzymes / electron transport chain / ATP synthase. (1 mark) 3. Enabling a high rate of aerobic respiration / oxidative phosphorylation (to produce ATP). (1 mark)

Monoclonal antibodies possess highly specific variable regions with a unique tertiary structure that is complementary to only one specific antigen (the tumor-associated antigen on cancer cells). Because of this specificity, the conjugated drug is delivered selectively to cancer cells where binding occurs. Since healthy, non-cancerous cells do not express this specific antigen, the antibody-drug conjugate does not bind to them, preventing healthy tissue from being exposed to and damaged by the cytotoxic drug, thereby reducing side effects.

1. Monoclonal antibodies have a specific variable region / binding site that is complementary to a specific (tumor-associated) antigen (only on cancer cells). (1 mark) 2. Antibody-drug conjugate only binds to/targets cancer cells. (1 mark) 3. Healthy/normal cells do not have this antigen, so they are not bound / not damaged by the therapeutic drug. (1 mark)

A triglyceride consists of one glycerol molecule bound to three fatty acid chains via ester bonds. In a phospholipid, one of these fatty acid chains is replaced by a hydrophilic phosphate group. This makes the phospholipid molecule amphipathic: it has a polar, hydrophilic (water-attracting) head and two non-polar, hydrophobic (water-repelling) fatty acid tails. In an aqueous environment, phospholipids spontaneously align themselves so that the hydrophilic heads face outwards to interact with the aqueous cytoplasm and extracellular fluid, while the hydrophobic tails face inwards, shielded from water, forming a stable bilayer structure.

1. Phospholipids have two fatty acids and a phosphate group (attached to glycerol), whereas triglycerides have three fatty acids (and no phosphate). (1 mark) 2. The phosphate group/head is hydrophilic (polar) and the fatty acid tails are hydrophobic (non-polar). (1 mark) 3. Hydrophilic heads face out/towards water (aqueous environments) and hydrophobic tails face inwards/away from water, forming a stable bilayer. (1 mark)

During flight, the metabolic rate of an insect increases dramatically. To meet this demand, insects use abdominal pumping: muscular contractions compress the trachea and air sacs, actively forcing air in and out to maintain a steep concentration gradient (mass transport). Additionally, intense muscle activity can lead to anaerobic respiration, producing lactate which lowers the water potential of the muscle cells. Consequently, water moves out of the ends of the tracheoles and into the muscle cells by osmosis. This decreases the volume of liquid in the tracheoles, allowing oxygen gas to diffuse more rapidly directly to the muscle cells since diffusion is much faster through air than through water.

1. Rhythmic contraction of muscles (abdominal pumping) compresses tracheae/air sacs, causing mass flow/movement of air. (1 mark) 2. Anaerobic respiration in flight muscles produces lactate, lowering the water potential of the muscles. (1 mark) 3. Water moves out of the ends of the tracheoles into the muscle cells by osmosis, decreasing diffusion distance / allowing faster diffusion of oxygen through gas rather than liquid. (1 mark)

Sodium ions are actively transported out of the ileum epithelial cells into the blood by the sodium-potassium pump. This continuous removal maintains a much lower concentration of sodium ions inside the epithelial cell compared to the lumen of the ileum. As a result, a concentration gradient is established. Sodium ions then diffuse down this gradient from the lumen into the cell through a specific co-transport protein. This process is coupled with the transport of glucose molecules, which are carried into the cell against their own concentration gradient. This allows glucose to be continuously absorbed even when its concentration in the lumen is low.

1. Active transport of sodium ions out of the cell (via sodium-potassium pump) maintains a low concentration of sodium ions inside the epithelial cell. (1 mark) 2. This creates/maintains a concentration gradient of sodium ions between the lumen and the cell. (1 mark) 3. Sodium ions enter the cell by facilitated diffusion/co-transport down their gradient, carrying glucose into the cell against its concentration gradient (via a co-transport protein). (1 mark)

According to the induced-fit model, the enzyme's active site is not perfectly complementary to the substrate before binding. Instead, as the substrate collides and begins to interact with the active site, the flexible tertiary structure of the enzyme changes shape slightly to fit the substrate more closely. This conformational change puts physical stress or strain on the chemical bonds of the substrate, destabilizing them and thus lowering the activation energy needed to break them. This differs fundamentally from the lock-and-key model, which suggests that the active site is completely rigid and fits the substrate perfectly and statically prior to binding, with no conformational changes occurring.

1. Active site is not (fully) complementary initially, but changes shape / moulds around the substrate as it binds. (1 mark) 2. This conformational change/shaping puts stress/strain on the bonds of the substrate, lowering the activation energy (to break bonds). (1 mark) 3. In the lock-and-key model, the active site is rigid/fixed and perfectly complementary before the substrate binds (no change of shape occurs). (1 mark)

A gene mutation changes the sequence of DNA bases, which alters the primary structure of the polypeptide. This changes where hydrogen, ionic, or disulfide bonds form, altering the final 3D tertiary structure of the protein. If the shape of the active site is altered, the substrate will no longer be complementary, meaning enzyme-substrate complexes cannot form and the enzyme is non-functional.

1. Change in primary structure / sequence of amino acids (Accept: different codon/triplet code). 2. Alters position of hydrogen / ionic / disulfide bonds, changing the tertiary structure / active site shape. 3. Substrate is no longer complementary (to active site) so no enzyme-substrate complexes (ESCs) form. (Reject: 'active site is killed/destroyed')

When sucrose is actively transported into the phloem sieve tube element, it decreases the solute potential and thus lowers the water potential inside the phloem. Water moves from the xylem (which has a higher water potential) into the phloem by osmosis. This accumulation of water increases the hydrostatic pressure at the source, creating a hydrostatic pressure gradient that drives the mass flow of the solution towards the sink where the pressure is lower.

1. Sucrose entry lowers water potential (in phloem / sieve tube element). 2. Water enters by osmosis, increasing hydrostatic pressure (at the source). 3. Creates a hydrostatic pressure gradient, causing mass flow / movement of water and solutes towards the sink (lower hydrostatic pressure).

Helper T cells play a central role in coordinating both the humoral and cellular immune responses. They detect foreign antigens presented on antigen-presenting cells (such as macrophages or B cells) using complementary T-cell receptors. Upon activation, they divide by mitosis and release cytokines. These chemical messengers stimulate B cells to undergo clonal selection and differentiate into plasma cells to produce antibodies, and they also activate cytotoxic T cells and phagocytes to destroy pathogens.

1. Bind to antigen-presenting cells / APCs (using complementary T-cell receptors). 2. Release cytokines / chemical signals. 3. Stimulate B cells (to divide / secrete antibodies) OR stimulate cytotoxic T cells / phagocytes (to engulf/kill pathogens).

The root tip (first 1-2 mm) contains the meristematic region, which is the site of active cell division (mitosis); other parts of the root only contain cells that are elongating or fully differentiated. Pressing down firmly on the coverslip is necessary to squash the tissue into a single, thin layer of cells, allowing light from the microscope to pass through so individual cells and chromosomes are visible. It is crucial not to slide the coverslip sideways, as this can damage the chromosomes or cause cells to overlap and roll, ruining the preparation.

1. (Only using first 1-2 mm because) this is the site of cell division / mitosis / contains the meristem. (Reject: 'growth' on its own). 2. (Pressing down firmly) spreads the cells into a single / thin layer (so light can pass through). 3. (No sliding because) it prevents damage to chromosomes / chromatids OR prevents cells rolling / overlapping.

First, convert the image length from millimeters to micrometers: \(27\text{ mm} \times 1000 = 27,000\ \mu\text{m}\). Next, use the magnification formula: \(\text{Magnification} = \frac{\text{Image size}}{\text{Actual size}}\). Substitute the values into the formula: \(\text{Magnification} = \frac{27,000\ \mu\text{m}}{4.5\ \mu\text{m}} = 6000\).

1 mark for converting \(27\text{ mm}\) to \(27,000\ \mu\text{m}\) OR for showing a correct rearrangement of the formula (e.g., \(\frac{27}{0.0045}\)).
1 mark for the correct answer of 6000 (accept \(\times 6000\)).

First, calculate the resting stroke volume: \(\text{Stroke Volume (rest)} = \frac{\text{Cardiac Output}}{\text{Heart Rate}} = \frac{4.8\text{ dm}^3\text{ min}^{-1}}{64\text{ bpm}} = 0.075\text{ dm}^3\). Next, calculate the exercise stroke volume after a \(25\%\) increase: \(\text{Stroke Volume (exercise)} = 0.075\text{ dm}^3 \times 1.25 = 0.09375\text{ dm}^3\). Finally, calculate the exercise cardiac output: \(\text{Cardiac Output (exercise)} = \text{Stroke Volume (exercise)} \times \text{Heart Rate (exercise)} = 0.09375\text{ dm}^3 \times 120\text{ bpm} = 11.25\text{ dm}^3\text{ min}^{-1}\).

1 mark for calculating the resting stroke volume as \(0.075\text{ dm}^3\) (or \(75\text{ cm}^3\)) OR the exercise stroke volume as \(0.09375\text{ dm}^3\) (or \(93.75\text{ cm}^3\)).
1 mark for the correct final answer of 11.25.

First, calculate the total number of bases in the double-stranded DNA molecule: \(3600\text{ base pairs} \times 2 = 7200\text{ bases}\). Since adenine (A) makes up \(22\%\) of the bases, thymine (T) must also make up \(22\%\) of the bases due to complementary base pairing. Together, A and T account for \(44\%\) of the bases (\(22\% + 22\%\)). The remaining bases (guanine and cytosine) make up \(56\%\) of the total (\(100\% - 44\% = 56\%\)). Since G pairs with C, cytosine accounts for half of this percentage: \(\frac{56\%}{2} = 28\%\). Finally, calculate the total number of cytosine bases: \(0.28 \times 7200 = 2016\).

1 mark for showing a correct method to determine that cytosine makes up \(28\%\) of the total bases OR for calculating the total number of bases as \(7200\).
1 mark for the correct answer of 2016.

First, calculate the total number of cells in metaphase, anaphase, and telophase: \(36 + 18 + 9 = 63\text{ cells}\). Next, determine the proportion of the cell cycle spent in these stages: \(\frac{63}{450} = 0.14\). Convert the total cell cycle duration into minutes: \(20\text{ hours} \times 60\text{ minutes/hour} = 1200\text{ minutes}\). Finally, multiply the proportion of the cell cycle by the total duration in minutes: \(0.14 \times 1200\text{ minutes} = 168\text{ minutes}\).

1 mark for finding the correct proportion (\(0.14\) or \(14\%\)) of cells in these stages OR for converting the total cell cycle duration to \(1200\text{ minutes}\).
1 mark for the correct final answer of 168 (minutes).

Triglycerides are formed by the condensation of one molecule of glycerol and three molecules of fatty acids, creating three ester bonds. Since there are no carbon-to-carbon double bonds (\(\text{C}=\text{C}\)) in its fatty acid chains, the lipid is saturated. Saturated triglycerides have straight hydrocarbon chains that pack closely together, making them solid at room temperature.

C is correct (1 mark). Option A is incorrect because phospholipids have two ester bonds linking fatty acids to glycerol. Option B is incorrect because 'no carbon-carbon double bonds' means it is saturated, not unsaturated. Option D is incorrect because a triglyceride has one glycerol molecule and three fatty acid chains, not three glycerols.

Convert the image size from millimetres to micrometers: \(45\text{ mm} \times 1000 = 45,000\ \mu\text{m}\). Then use the formula for magnification: \(\text{Magnification} = \frac{\text{Image size}}{\text{Actual size}}\). Therefore, \(\text{Magnification} = \frac{45,000\ \mu\text{m}}{1.5\ \mu\text{m}} = 30,000\).

1 mark for correct conversion of units (giving \(45,000\ \mu\text{m}\) or \(0.0015\text{ mm}\)) or showing a correct division step. 1 mark for the correct magnification of \(30,000\) or \(\times 30,000\).

The wall of an arteriole contains a thick layer of smooth muscle. When these muscle fibres contract, they narrow the lumen of the arteriole (vasoconstriction), which restricts the volume of blood flowing into the capillaries.

1 mark for stating that the arteriole wall has a thick layer of muscle tissue / muscle fibres. 1 mark for stating that contraction of this muscle narrows the lumen / restricts blood flow.

First, calculate the surface area: \(\text{SA} = 4 \times \pi \times 3^2 = 36\pi\). Next, calculate the volume: \(\text{Volume} = \frac{4}{3} \times \pi \times 3^3 = 36\pi\). The ratio of surface area to volume is \(\frac{36\pi}{36\pi} = 1\), which is written as \(1:1\).

1 mark for calculating correct surface area (\(36\pi\) or approx. \(113.1\)) and volume (\(36\pi\) or approx. \(113.1\)), or for showing that the ratio simplifying formula is \(\frac{3}{r}\). 1 mark for the correct ratio of \(1:1\) (or \(1\)).

A peptide bond forms through a condensation reaction, where a molecule of water is released. This reaction occurs between the amine group (\(-\text{NH}_2\)) of one amino acid and the carboxyl group (\(-\text{COOH}\)) of the other amino acid.

1 mark for stating it is a condensation reaction / involves the elimination of a water molecule. 1 mark for specifying that the bond is formed between the amine group of one amino acid and the carboxyl group of another.

During prophase, the chromatin condenses into distinct, visible chromosomes, and the nuclear envelope disintegrates. Centrioles also migrate to opposite poles of the cell to begin forming the spindle fibres.

Accept any two of the following for 1 mark each (max 2 marks): 1. Chromosomes condense / shorten and thicken / become visible. 2. Nuclear envelope / membrane breaks down or disintegrates. 3. Centrioles move to opposite poles of the cell. 4. Spindle fibres begin to form.

A triglyceride consists of three fatty acids esterified to one glycerol molecule. A phospholipid, however, contains only two fatty acids and has one of the fatty acids replaced by a phosphate group attached to the glycerol.

1 mark for stating that a triglyceride has three fatty acids while a phospholipid has two fatty acids. 1 mark for stating that a triglyceride has no phosphate group while a phospholipid does contain a phosphate group.

Contraction of abdominal muscles ventilates the tracheal system by pumping air in and out. This movement regularly replaces oxygen-depleted air with oxygen-rich fresh air from the environment and removes carbon dioxide, thereby maintaining a steep concentration gradient between the trachea and the respiring tissues.

1 mark for stating that abdominal movements ventilate the tracheal system / pump air in and out. 1 mark for explaining that this introduces oxygen-rich air / removes carbon-dioxide-rich air to maintain a concentration gradient.

Disulfide bridges are strong covalent bonds that connect the heavy and light polypeptide chains of an antibody together. This stabilizes and maintains the overall tertiary and quaternary structure of the antibody, ensuring the antigen-binding sites retain their specific shape.

1 mark for stating they form covalent bonds between the polypeptide chains (heavy and light chains). 1 mark for explaining that this maintains the specific tertiary / quaternary structure of the antibody / keeps the shape of the antigen-binding site stable.

Active immunity is produced by the host's own immune system in response to exposure to an antigen, leading to the creation of memory cells and long-term protection. Passive immunity is achieved by introducing pre-made antibodies into the individual from an external source, which offers immediate but temporary protection because no memory cells are formed and the antibodies are eventually broken down.

1 mark: Correctly identifies that active immunity produces memory cells/provides long-term immunity while passive immunity does not/is short-term. 1 mark: Correctly identifies that active immunity involves the individual's plasma cells producing antibodies/takes time to develop, whereas passive immunity involves receiving antibodies from an external source/is immediate.

At high altitudes, the partial pressure of oxygen in the air is much lower than at sea level. Having haemoglobin with a high oxygen affinity enables the llama's blood to load oxygen efficiently and achieve high saturation at the gas exchange surface, ensuring sufficient oxygen delivery to tissues to sustain aerobic respiration.

1 mark: Stating that haemoglobin loads oxygen more readily/becomes saturated at lower partial pressures of oxygen in the lungs. 1 mark: Explaining that this ensures enough oxygen is transported/delivered to respiring tissues.

The mitotic index is calculated by dividing the number of cells undergoing mitosis by the total number of cells, then multiplying by 100. Mitotic index = \(\frac{42}{350} \times 100 = 12\%\).

1 mark: Correct working showing \(\frac{42}{350} \times 100\) (or an equivalent ratio like 0.12). 1 mark: Correct final answer of 12%.

If the solution were not isotonic, osmosis would occur. A hypotonic solution would cause water to enter the organelles, leading them to swell and burst (lyse). A hypertonic solution would cause water to leave the organelles, causing them to shrink. Keeping it isotonic prevents water potential differences, preserving organelle structure.

1 mark: To prevent the net movement of water by osmosis (or to maintain the same water potential). 1 mark: To prevent the organelles from swelling and bursting, or shrinking and being damaged.

In a phospholipid, glycerol is bonded to two fatty acids and one phosphate group, whereas in a triglyceride, glycerol is bonded to three fatty acids. The phosphate group is highly polar (hydrophilic), while the fatty acid tails are non-polar (hydrophobic). In water, this causes phospholipids to assemble into bilayers, with their hydrophilic heads facing outward and hydrophobic tails shielded inside, unlike triglycerides which form insoluble droplets.

1 mark: Stating that a phospholipid has two fatty acids and a phosphate group, whereas a triglyceride has three fatty acids and no phosphate group. 1 mark: Explaining that the phosphate group makes the phospholipid polar/hydrophilic at one end, allowing it to form bilayers in water, whereas triglycerides are completely hydrophobic.

An amoeba has a high surface area to volume ratio and its entire cytoplasm is very close to the external environment. Diffusion across its outer membrane is fast enough to supply oxygen and remove carbon dioxide. Large organisms have a much lower surface area to volume ratio and cells located deep inside, creating a long diffusion pathway that makes simple diffusion across the body surface insufficient.

1 mark: Identifies that single-celled organisms have a large surface area to volume ratio and a short diffusion pathway. 1 mark: Explains that simple diffusion across the cell surface membrane is fast enough to meet their metabolic needs (or that large multicellular organisms have a small SA:V ratio and longer diffusion distance, making diffusion too slow).

Sodium ions are actively transported out of the epithelial cells into the blood by the sodium-potassium pump. This lowers the intracellular sodium ion concentration, establishing a concentration gradient between the lumen of the ileum and the inside of the cell. Sodium ions then diffuse down their concentration gradient into the epithelial cell through a co-transporter protein, bringing glucose into the cell against its concentration gradient.

1 mark: Explains that active transport of sodium ions maintains a low concentration of sodium ions inside the epithelial cell (establishing a concentration gradient). 1 mark: Explains that sodium ions diffuse into the cell down their concentration gradient via a co-transporter protein, carrying glucose into the cell with them (against its concentration gradient).

According to the lock-and-key model, the substrate fits into a rigid active site that does not change shape. In contrast, the induced-fit model proposes that the active site is flexible and molds around the substrate as the enzyme-substrate complex forms. This conformational change puts strain on the bonds of the substrate, lowering the activation energy required for the reaction.

1 mark: Stating that in the induced-fit model, the active site changes shape / molds around the substrate as it binds (whereas in lock-and-key, the active site is rigid/fixed). 1 mark: Stating that in the lock-and-key model, the active site is fully complementary before the substrate binds (whereas in induced-fit, it only becomes complementary upon binding).

Plasma proteins normally lower the water potential of blood inside the capillaries, establishing a water potential gradient that draws water back into the circulatory system at the venule end by osmosis. When plasma protein levels are low, the water potential of the blood is higher (less negative) than normal. This reduces the water potential gradient between the tissue fluid and the blood, meaning less water is reabsorbed back into the capillary. Consequently, excess tissue fluid accumulates in the surrounding tissues.

1. Water potential of blood (in capillary) is higher / less negative OR there is a reduced water potential gradient (between the blood and tissue fluid); [1 mark]
2. Less water is reabsorbed / enters the capillary by osmosis; [1 mark]

Reject: 'No water is reabsorbed' (must imply less/reduced reabsorption).

A low concentration of plasma proteins means the water potential of the blood is higher (less negative) than normal. This reduces the water potential gradient between the blood and the tissue fluid, so less water is reabsorbed back into the capillary at the venule end by osmosis, causing tissue fluid to accumulate.

1. Water potential of blood is higher / less negative OR reduced water potential gradient (between blood and tissue fluid); [1 mark]
2. Less water is reabsorbed into the capillary / blood by osmosis; [1 mark]

Reject: 'No water is reabsorbed' / 'water moves out of blood by osmosis'.

Spindle fibers fail to form or attach properly to the centromeres. Because of this, sister chromatids cannot be separated during anaphase to move to opposite poles of the spindle. The resulting daughter cells do not receive a complete or correct set of chromosomes (aneuploidy), or the cell cycle arrests, preventing division.

1. Mark for stating that spindle fibers cannot form, contract, or attach to centromeres.
2. Mark for explaining that sister chromatids cannot separate / be pulled to opposite poles of the cell (preventing anaphase).
3. Mark for stating that this leads to genetically unequal daughter cells / cells with incorrect numbers of chromosomes / aneuploidy (or cell division is halted entirely).

An increased rate of respiration increases the partial pressure of carbon dioxide in the blood, which reacts to form carbonic acid, lowering the pH. This decrease in pH causes a conformational change in the tertiary structure of hemoglobin, which reduces its affinity for oxygen. Consequently, hemoglobin releases / unloads its oxygen more readily to the tissues that require it.

1. Mark for identifying that increased carbon dioxide concentration lowers the pH / increases acidity.
2. Mark for explaining that the lower pH changes the shape / tertiary structure of hemoglobin, reducing its affinity for oxygen.
3. Mark for stating that hemoglobin unloads / dissociates oxygen more readily / easily to the respiring tissues (at the same partial pressure of oxygen / \(pO_2\)).

A light microscope has a limited resolution (about 200 nm) due to the long wavelength of light, which cannot resolve the detailed internal membranes of mitochondria. An electron microscope has a much shorter wavelength and higher resolution. A TEM must be used instead of an SEM because a TEM passes an electron beam through a thin section of the specimen to show the internal ultrastructure (like cristae), whereas an SEM only bounces electrons off the surface to produce a 3-D surface image.

1. Mark for stating that the TEM has a much higher resolution / resolving power than a light microscope (due to the shorter wavelength of the electron beam).
2. Mark for explaining that the TEM passes electrons through a thin specimen to reveal internal structures / organelles (like cristae / matrix).
3. Mark for contrasting with SEM, which only scans the surface of the specimen to produce a 3-D surface image (cannot show internal details).

The first vaccine dose triggers a primary immune response, generating specific memory B and T cells. The booster dose acts as a secondary exposure, stimulating these existing memory B cells to rapidly undergo clonal selection, mitosis, and differentiation into plasma cells. This leads to the rapid production of a much higher concentration of antibodies, which remain in circulation to neutralize the pathogen immediately upon real infection.

1. Mark for identifying that the first dose produced memory cells (B or T lymphocytes).
2. Mark for explaining that the booster stimulates these memory cells to undergo rapid clonal selection / mitosis / division into plasma cells.
3. Mark for explaining that this leads to faster production of a much higher concentration / titer of antibodies (which remain in the blood to destroy the pathogen quickly).

A phospholipid has a polar, hydrophilic phosphate head which is attracted to water, and two non-polar, hydrophobic fatty acid tails which repel water. When surrounded by aqueous extracellular fluid and cytoplasm, they self-assemble into a double layer (bilayer). The hydrophilic heads face outwards to interact with water, while the hydrophobic fatty acid tails face inwards, forming a central hydrophobic core that acts as a barrier to water-soluble molecules.

1. Mark for stating that a phospholipid consists of a hydrophilic / polar phosphate head and two hydrophobic / non-polar fatty acid tails.
2. Mark for explaining that in an aqueous environment, they arrange into a bilayer with heads pointing outwards (towards the water/cytoplasm) and tails pointing inwards (away from water).
3. Mark for explaining that this creates a hydrophobic core / barrier that prevents the passage of polar / water-soluble substances.

Small mammals have a much larger surface area to volume ratio (\(SA:V\)) than larger mammals. This means they lose metabolic heat to their colder surroundings much more rapidly per unit of body mass. To maintain a constant core body temperature, they must increase their rate of aerobic respiration to generate heat, which directly leads to an increased rate of oxygen consumption per gram of tissue.

1. Mark for stating that small mammals have a larger surface area to volume (\(SA:V\)) ratio.
2. Mark for explaining that they lose heat more rapidly to the environment (per unit body mass).
3. Mark for explaining that they must maintain a higher rate of respiration / metabolism to generate enough heat to maintain a constant body temperature (thus consuming more oxygen).

Sodium ions (\(Na^+\)) are actively pumped out of epithelial cells into the blood by the sodium-potassium pump. This reduces the concentration of sodium ions inside the cell, establishing a concentration gradient between the lumen of the ileum and the cytoplasm of the epithelial cell. Sodium ions then diffuse down this gradient back into the cell via a co-transporter protein, which co-transports glucose molecules into the cell against the glucose concentration gradient.

1. Mark for explaining that the active transport of sodium ions (\(Na^+\)) out of the cell into the blood (via the sodium-potassium pump) maintains a low concentration of \(Na^+\) inside the cell.
2. Mark for stating that this creates a concentration gradient for \(Na^+\) between the lumen of the ileum and the inside of the cell.
3. Mark for explaining that \(Na^+\) enters the cell down its gradient via a co-transporter protein, bringing glucose into the cell with it (against the glucose concentration gradient).

As temperature rises above the optimum, the rate of reaction drops rapidly. Increased thermal energy increases the kinetic energy of the enzyme, causing vibration that breaks weak hydrogen and ionic bonds that maintain the protein's tertiary structure. This causes the enzyme to denature, changing the precise shape of its active site so that it is no longer complementary to the substrate, preventing the formation of enzyme-substrate complexes.

1. Mark for stating that the rate of reaction decreases / drops to zero.
2. Mark for explaining that high temperatures break hydrogen / ionic bonds (and disulfide bridges) in the enzyme.
3. Mark for explaining that this changes the tertiary structure and the shape of the active site, meaning the substrate is no longer complementary / cannot bind, so no enzyme-substrate complexes can form (the enzyme is denatured).

At lower temperatures, cell membranes tend to lose fluidity and become more rigid as the kinetic energy of the molecules decreases. Unsaturated fatty acids contain one or more carbon-to-carbon double bonds (C=C), which introduces a kink or bend in the hydrocarbon tail. These kinks prevent the phospholipid molecules from packing closely together in the bilayer. This lower packing density maintains membrane fluidity, ensuring that membrane proteins can function and substances can still be transported across the membrane effectively at 15 °C.

1. Unsaturated fatty acids have carbon-carbon double bonds (C=C) which produce kinks/bends in the hydrocarbon chains; 2. (Kinks) prevent the phospholipids packing closely together (at low temperatures); 3. This maintains membrane fluidity / prevents the membrane solidifying (so transport/membrane proteins can function). [Accept: 'rigid' instead of 'solidified' for mark point 3. Reject: 'creates space for molecules to pass' unless linked to maintaining fluidity.]

A leftward shift of the oxygen dissociation curve means that llama haemoglobin has a higher affinity for oxygen at any given partial pressure compared to human haemoglobin. At high altitudes, the partial pressure of oxygen in the lungs is lower. Because of the higher affinity, llama haemoglobin can load oxygen more readily and become fully saturated even under these low oxygen conditions. This ensures that an adequate volume of oxygen is loaded in the lungs and subsequently unloaded/delivered to respiring tissues to sustain aerobic respiration.

1. Llama haemoglobin has a higher affinity for oxygen (at any given partial pressure of oxygen); 2. This allows haemoglobin to load oxygen more readily / become saturated at the lower partial pressure of oxygen in the lungs; 3. This ensures sufficient oxygen is transported/delivered to respiring tissues (for aerobic respiration). [Accept: 'association' / 'binding' for 'loading'. Reject: 'holds onto oxygen tighter' if it implies it cannot unload it at all at tissues.]

First, calculate the surface area (SA) of the bacterium: \(\text{SA} = 2 \times \pi \times 0.6 \times 4.5 + 2 \times \pi \times 0.6^2 = 5.4\pi + 0.72\pi = 6.12\pi \approx 19.23\ \mu\text{m}^2\). Next, calculate the volume (V) of the bacterium: \(\text{V} = \pi \times 0.6^2 \times 4.5 = 1.62\pi \approx 5.09\ \mu\text{m}^3\). Finally, divide surface area by volume to find the ratio: \(19.23 / 5.09 \approx 3.78\). The surface area to volume ratio is therefore 3.78 : 1.

1 mark for calculating surface area (\(19.23\ \mu\text{m}^2\) or \(6.12\pi\)) and volume (\(5.09\ \mu\text{m}^3\) or \(1.62\pi\)). 1 mark for the correct final ratio of 3.78 : 1 (accept 3.78).

First, convert the image length into the same units as the actual length. Since \(1\text{ mm} = 1000\ \mu\text{m}\), the image length is \(48 \times 1000 = 48,000\ \mu\text{m}\). Use the formula: \(\text{Magnification} = \text{Image size} / \text{Actual size}\). Therefore, \(\text{Magnification} = 48,000 / 6 = 8000\).

1 mark for converting the image length to micrometres (\(48,000\ \mu\text{m}\)) or actual length to millimetres (\(0.006\text{ mm}\)). 1 mark for the correct answer of 8000 (accept \(\times 8000\) or \(8,000\)).

Use the formula: \(\text{Cardiac output} = \text{Heart rate} \times \text{Stroke volume}\). First, calculate cardiac output in \(\text{cm}^3\text{ min}^{-1}\): \(\text{Cardiac output} = 145 \times 95 = 13,775\text{ cm}^3\text{ min}^{-1}\). To convert this into \(\text{dm}^3\text{ min}^{-1}\), divide by 1000: \(13,775 / 1000 = 13.775\text{ dm}^3\text{ min}^{-1}\). Rounding to 3 significant figures gives \(13.8\text{ dm}^3\text{ min}^{-1}\).

1 mark for calculating cardiac output in \(\text{cm}^3\text{ min}^{-1}\) (\(13,775\)) or for dividing by 1000 to convert units. 1 mark for the correct answer of 13.8.

First, find \(N\): \(N = 12 + 8 + 5 + 3 = 28\). Thus, \(N(N-1) = 28 \times 27 = 756\). Next, calculate \(\sum n(n-1)\) for each species: Oak = \(12 \times 11 = 132\); Birch = \(8 \times 7 = 56\); Beech = \(5 \times 4 = 20\); Rowan = \(3 \times 2 = 6\). Sum of \(n(n-1) = 132 + 56 + 20 + 6 = 214\). Finally, calculate \(d = 756 / 214 \approx 3.5327\). Rounded to 2 decimal places, \(d = 3.53\).

1 mark for calculating the correct values for \(N(N-1) = 756\) and \(\sum n(n-1) = 214\). 1 mark for the correct index of diversity of 3.53.

First, calculate the increase in population density: \(22.3 - 14.2 = 8.1\text{ plants per m}^2\). Next, calculate this increase as a percentage of the original 2012 value: \(\text{Percentage increase} = (8.1 / 14.2) \times 100\% \approx 57.04225\%\). Rounded to 1 decimal place, the percentage increase is 57.0%.

1 mark for showing the correct increase divided by the original value: \((8.1 / 14.2) \times 100\) or an unrounded percentage of 57.04% (or 57%). 1 mark for the correct answer of 57.0 (accept 57.0%).

First, calculate the proportion of cells that are in metaphase: \(\frac{14}{350} = 0.04\). Next, convert the total duration of the cell cycle from hours into minutes: \(24 \text{ hours} \times 60 \text{ minutes/hour} = 1440 \text{ minutes}\). Finally, multiply the proportion of cells in metaphase by the total duration in minutes to find the time spent in metaphase: \(0.04 \times 1440 \text{ minutes} = 57.6 \text{ minutes}\).

1 mark for the correct option A (57.6 minutes).

An increase in carbon dioxide concentration reduces the pH of the blood because carbon dioxide reacts with water to form carbonic acid, which dissociates to release hydrogen ions. This lower pH alters the tertiary structure of haemoglobin, reducing its affinity for oxygen. Consequently, the oxygen dissociation curve shifts to the right (the Bohr effect), allowing oxygen to be unloaded more readily to respiring tissues.

1 mark for the correct option B (An increase in blood carbon dioxide concentration decreases pH, causing the oxygen dissociation curve to shift to the right, which decreases the affinity of haemoglobin for oxygen).

Phagocytes are attracted to pathogens by chemicals released by the pathogen or damaged cells (chemotaxis). Receptor proteins on the phagocyte membrane bind to foreign antigens on the pathogen. The phagocyte's membrane folds around the pathogen, engulfing it in a vesicle called a phagosome. Lysosomes within the phagocyte fuse with the phagosome to form a phagolysosome. Hydrolytic enzymes, such as lysozymes, are released into the vesicle, digesting and destroying the pathogen. The phagocyte then processes the pathogen's antigens and presents them on its own cell surface membrane, becoming an antigen-presenting cell (APC). Specific helper T cells (T lymphocytes) with complementary receptor proteins bind to these presented antigens. This binding, along with chemical signals, activates the T cell to undergo clonal expansion by mitosis.

1. Phagocyte moves towards pathogen due to chemotaxis / chemical attractants AND binds to pathogen antigens. 2. Phagocyte engulfs pathogen to form a phagosome / vesicle. 3. Lysosomes fuse with the phagosome and release lysozymes / hydrolytic enzymes to digest / hydrolyse the pathogen. 4. Phagocyte presents pathogen antigens on its cell surface membrane (to become an antigen-presenting cell). 5. T helper cell with complementary receptor binds to the presented antigen, activating the T cell (to divide by mitosis / undergo clonal selection). Accept: T lymphocyte for T helper cell. Reject: 'kills' or 'eats' the pathogen without reference to hydrolysis/digestion. (Max 5 marks)

At the arteriole end of a capillary bed, the hydrostatic pressure of the blood is higher than the hydrostatic pressure of the tissue fluid outside. This pressure gradient forces water and small dissolved solutes (like glucose, amino acids, and ions) out of the capillary through the small gaps in the capillary wall, forming tissue fluid. Large molecules, such as plasma proteins and red blood cells, are too large to pass through and remain in the blood. This retention of plasma proteins lowers the water potential of the blood inside the capillary. At the venule end of the capillary, the hydrostatic pressure is significantly lower due to friction and distance from the heart. The osmotic pressure (pulling water back in due to the lower water potential of the blood) exceeds the outward hydrostatic pressure. Consequently, water re-enters the capillary by osmosis down a water potential gradient. Any excess tissue fluid that does not re-enter the capillary directly is drained into lymphatic capillaries, forming lymph, which is eventually returned to the circulatory system via the thoracic duct.

1. High hydrostatic pressure at the arteriole end of the capillary forces water and small solutes out. 2. Large proteins / cells remain in the capillary because they are too large to pass through capillary walls. 3. This lowers the water potential of the blood at the venule end (establishing a water potential gradient). 4. Water re-enters the capillary at the venule end by osmosis down the water potential gradient. 5. Excess tissue fluid is drained into the lymphatic system / lymph vessels (which returns it to the bloodstream). Accept: ultrafiltration for forcing out of fluid. Reject: 'active transport' of water. (Max 5 marks)