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For part (a), two other environmental factors that affect transpiration must be controlled, such as light intensity and temperature. For part (b), the potometer measures water uptake, but some water is used by the plant for photosynthesis or to maintain cell turgidity, so water uptake is slightly higher than transpiration. For part (c), the radius (r) of the capillary tube is half the diameter, so r = 0.4 mm. The cross-sectional area of the tube is \( \pi r^2 = \pi \times 0.4^2 = 0.5027 \text{ mm}^2 \). The volume of water taken up in 10 minutes is \( \text{area} \times \text{distance} = 0.5027 \times 42 = 21.11 \text{ mm}^3 \). Since there are six 10-minute intervals in one hour, the volume taken up per hour is \( 21.11 \times 6 = 126.67 \text{ mm}^3 \). Rounded to 3 significant figures, this is 127 mm³. For part (d), increased wind speed blows away water vapour from around the leaf surface, maintaining a steep concentration gradient of water vapour between the inside and the outside of the leaf, increasing the rate of diffusion out through the stomata.

Part (a): 1 mark for each correct factor named (e.g. temperature, light intensity, relative humidity). Max 2 marks. Part (b): 1 mark for stating that water is used in photosynthesis; 1 mark for stating that water is used to keep cells turgid. Max 2 marks. Part (c): 1 mark for calculating the correct radius (0.4 mm); 1 mark for calculating the cross-sectional area (approx 0.503 mm²); 1 mark for calculating the volume in 10 minutes (21.1 mm³); 1 mark for multiplying by 6 to find the hourly rate; 1 mark for correct rounding to 3 sig figures (127 mm³). Max 4.2 marks. Part (d): 1 mark for stating wind moves water vapour away from leaf surface; 1 mark for explaining this maintains a steep diffusion/concentration gradient; 1 mark for explaining that diffusion of water vapour out of stomata occurs faster. Max 4 marks.

For part (a), heterozygous means having two different alleles for a particular gene (e.g., Ww). For part (b), the heterozygous man has genotype Ww and the normal hair woman has genotype ww. The gametes are W and w from the father, and w and w from the mother. The offspring genotypes are 50% Ww (woolly hair) and 50% ww (normal hair). The probability of woolly hair is 0.5 (or 50%). For part (c), cystic fibrosis is recessive (f), so the parents are both Ff. The offspring genotypes from an Ff x Ff cross are FF (normal, non-carrier), Ff (carrier), and ff (has cystic fibrosis). The carriers are Ff, which make up 2 out of 4 possible outcomes, giving a probability of 0.5 (or 50%).

Part (a): 1 mark for mentioning having two different alleles; 1 mark for mentioning for a specific gene. Max 2 marks. Part (b): 1 mark for correct parental genotypes (Ww and ww); 1 mark for correct gametes (W, w and w); 1 mark for correct offspring genotypes (Ww and ww); 1 mark for correct offspring phenotypes linked to genotypes; 2 marks for stating the correct probability of 0.5, 50%, or 1 in 2. Max 6 marks. Part (c): 1 mark for identifying parental genotypes as Ff; 1 mark for identifying offspring genotypes as FF, Ff, and ff; 1 mark for identifying that Ff are carriers; 1.2 marks for stating the correct probability is 0.5 or 50% (or 2 in 4). Max 4.2 marks.

For part (a), the experiment involves measuring the breathing rate of a person at rest by counting the number of breaths in one minute. The person then performs a specific exercise (e.g., stepping on a bench) at a controlled intensity for a set time (e.g., 3 minutes). Immediately after exercise, the breathing rate is counted again. The experiment is repeated with different intensities of exercise, and with multiple subjects to calculate averages. For part (b), exercise increases muscle contraction, which requires more energy from aerobic respiration. This increases the demand for oxygen and produces more carbon dioxide. The breathing rate increases to obtain more oxygen and excrete the excess carbon dioxide. For part (c), during inhalation, the external intercostal muscles contract, pulling the ribcage upwards and outwards. At the same time, the diaphragm muscles contract, causing the diaphragm to flatten. This increases the volume of the thorax and decreases the pressure, forcing air into the lungs.

Part (a): 1 mark for measuring resting breathing rate; 1 mark for defining a standardized exercise (duration/type); 1 mark for measuring breathing rate immediately after exercise; 1 mark for repeating the test at different intensities; 1 mark for repeating with other people/calculating average. Max 5 marks. Part (b): 1 mark for stating muscles contract more; 1 mark for stating increased respiration is required; 1 mark for stating more oxygen is needed; 1.2 marks for stating more carbon dioxide must be removed. Max 4.2 marks. Part (c): 1 mark for stating external intercostal muscles contract; 1 mark for stating ribcage moves up and out; 1 mark for stating diaphragm contracts and flattens. Max 3 marks.

For part (a), using CORMS: Change (C) the concentration of auxin (e.g., 0, 10, 50, 100 ppm). Organism (O) should be germinating seeds of the same species and age (e.g., mung beans). Repeat (R) the experiment with at least 10 seeds per concentration to calculate a reliable average. Measure (M) the change in root length in millimetres using a ruler after a set time, such as 5 days. Same (S) environmental conditions must be controlled, such as keeping the temperature constant (using an incubator) and keeping them in complete darkness to avoid phototropic responses. For part (b), auxin is produced at the shoot tip. When light shines on one side of the shoot, auxin diffuses to the shaded side. Higher auxin concentration on the shaded side stimulates cell elongation, causing the shaded side to grow faster, bending the shoot towards the light. For part (c), nervous communication uses electrical impulses along neurones, which is very rapid and has localized effects. Hormonal communication uses chemical hormones transported in the blood, which is slower and often has widespread effects.

Part (a): 1 mark for C: at least three different auxin concentrations; 1 mark for O: same species/age/source of seeds; 1 mark for R: repeat for each concentration and calculate mean; 1 mark for M1: measure root length in mm; 1 mark for M2: after a specific time interval (e.g., 5 days); 1 mark for S: control temperature or light/darkness. Max 6 marks. Part (b): 1 mark for stating auxin is produced in the tip; 1 mark for stating auxin moves to the shaded side; 1 mark for stating auxin stimulates cell elongation in shoots; 1.2 marks for explaining this causes bending towards the light source. Max 4.2 marks. Part (c): 1 mark for each valid difference (e.g., speed of transmission, electrical vs chemical, duration of effect, localized vs widespread). Max 2 marks.

For part (a), the graph should have pH on the independent x-axis and the rate of reaction on the dependent y-axis. The axes must be labeled with appropriate units, use a linear scale covering more than half the grid, and points should be plotted accurately and joined with straight lines or a smooth curve. For part (b), the optimum pH for amylase is pH 7.0, where the rate of reaction is highest (95 au) because the shape of the active site is complementary to the starch substrate. As pH moves away from optimum (either more acidic or alkaline), the hydrogen and ionic bonds holding the protein's tertiary structure are disrupted. This changes the shape of the active site, meaning starch can no longer bind (the enzyme is denatured), reducing the rate of reaction. For part (c), first, add iodine solution to a sample; if it remains orange-brown (does not turn blue-black), all starch has been broken down. Second, perform the Benedict's test by adding Benedict's solution and heating in a water bath at 80 degrees Celsius; a change from blue to red/orange indicates the presence of glucose.

Part (a): 1 mark for correct axes labels (pH on x-axis, rate of reaction on y-axis); 1 mark for linear and appropriate scales; 1 mark for plotting points accurately and joining with a line. Max 3 marks. Part (b): 1 mark for identifying optimum pH is 7; 1 mark for explaining that active site shape is complementary to starch; 1 mark for stating extreme pH changes the active site shape; 1 mark for stating the enzyme denatures; 1 mark for explaining substrate can no longer fit/bind; 1 mark for referencing the data (e.g., rate drops to 5 at pH 9). Max 6 marks. Part (c): 1 mark for iodine test remaining orange-brown to confirm absence of starch; 1 mark for adding Benedict's reagent and heating; 1.2 marks for stating a color change to red/orange confirms glucose is present. Max 3.2 marks.

For part (a), the word equation is glucose -> ethanol + carbon dioxide. For part (b), the layer of liquid paraffin acts as a physical barrier to prevent oxygen from dissolving into the mixture from the air, ensuring anaerobic conditions. For part (c), you can connect the flask containing yeast to a delivery tube leading into a gas syringe to measure the volume of gas collected in a set time, or bubble the gas into water and count the number of bubbles produced per minute. For part (d), above 45 °C, the high temperature breaks the bonds holding the yeast's respiratory enzymes in shape, denaturing the active sites so substrates can no longer bind, slowing down and eventually stopping respiration. For part (e), glucose is a monosaccharide (simple sugar) that can be directly absorbed and used in glycolysis, whereas starch is a large, insoluble polysaccharide that must first be broken down by amylase, which takes time.

Part (a): 2 marks for 'glucose -> ethanol + carbon dioxide' (1 mark if ethanol or carbon dioxide is missing). Max 2 marks. Part (b): 2 marks for explaining it prevents oxygen entering the mixture to maintain anaerobic conditions. Max 2 marks. Part (c): 1 mark for counting bubbles or using a gas syringe; 1 mark for specifying a time frame (e.g., per minute); 1 mark for mentioning a delivery tube setup. Max 3 marks. Part (d): 1 mark for mentioning enzymes; 1 mark for stating that high temperature causes denaturing; 1.2 marks for stating active site changes shape so glucose can no longer bind. Max 3.2 marks. Part (e): 1 mark for stating glucose is a small/soluble molecule; 1 mark for stating starch requires digestion/hydrolysis before use. Max 2 marks.

For part (a), restriction enzymes are used to cut open the plasmid vector and isolate the human insulin gene from DNA, leaving complementary 'sticky ends'. Ligase enzymes are used to join the sticky ends of the human gene and the plasmid together, forming recombinant DNA. For part (b), the insulin gene is cut out using restriction enzymes. Plasmids from E. coli are cut with the same restriction enzymes. The gene and plasmid are joined using ligase to form a recombinant plasmid, which is inserted back into the bacteria. The bacteria are grown in an industrial fermenter where temperature, pH, and oxygen are controlled to maximize growth. The bacteria express the gene, and the insulin is extracted and purified. For part (c), advantages include: the insulin produced is identical to human insulin, so there is no allergic reaction; it avoids ethical/religious objections associated with slaughtering animals; and it can be produced in unlimited quantities rapidly.

Part (a): 1 mark for restriction enzyme cutting DNA; 1 mark for leaving sticky ends; 1 mark for ligase joining DNA fragments; 1 mark for forming recombinant DNA. Max 4 marks. Part (b): 1 mark for isolating human insulin gene; 1 mark for cutting plasmid vector; 1 mark for inserting plasmid into bacteria; 1 mark for multiplying bacteria in a fermenter; 1.2 marks for controlling conditions in fermenter (e.g. nutrients, pH, O2) to optimize yield. Max 5.2 marks. Part (c): 1 mark for each valid advantage explained (e.g., identical to human insulin/fewer side effects, ethical/religious acceptability, cheaper/larger quantities). Max 3 marks.

For part (a), potato cylinders must be dried to remove any excess water on their surface, which would add to the measured mass but is not part of the internal potato cell mass. For part (b), percentage change accounts for differences in the starting (initial) masses of the potato cylinders, allowing for a fair comparison between them. For part (c), the change in mass is \( 2.15 - 2.45 = -0.30 \text{ g} \). The percentage change is \( \frac{-0.30}{2.45} \times 100 = -12.24 \% \). To 3 significant figures, this is -12.2%. For part (d), the 0.8 mol/dm³ sucrose solution has a lower water potential (is hypertonic) compared to the cytoplasm of the potato cells. Water moves out of the potato cells, down the water potential gradient, by osmosis across the selectively permeable cell membranes, reducing the mass of the tissue.

Part (a): 2 marks for explaining that removing surface liquid ensures that only the actual mass of the potato tissue is measured. Max 2 marks. Part (b): 1 mark for stating initial masses are different; 2 marks for explaining that percentage change normalizes the results for fair comparison. Max 3 marks. Part (c): 1 mark for calculating correct change in mass (-0.30 g); 1 mark for dividing by initial mass (2.45 g) and multiplying by 100; 1.2 marks for correct calculation and rounding to -12.2%. Max 3.2 marks. Part (d): 1 mark for stating sucrose solution has lower water potential than the potato cells; 1 mark for stating water moves out of the cells; 1 mark for stating this occurs by osmosis; 1 mark for mentioning movement across a selectively permeable membrane. Max 4 marks.

(a) Setup and use:
1. Cut the leafy shoot underwater to prevent air bubbles from entering xylem vessels.
2. Assemble the apparatus underwater and seal all joints with petroleum jelly (Vaseline) to make it airtight.
3. Introduce a single air bubble into the capillary tube.
4. Record the starting position of the bubble and measure the distance it travels along the scale in a set time (e.g., 10 minutes) using a stopwatch.

(b) CORMS design:
- C (Change): Vary wind speed using an electric fan set at different distances (e.g., 0.5 m, 1.0 m, 1.5 m) or different speed settings (low, medium, high).
- O (Organism): Use the same leafy shoot of the same plant species.
- R (Repeat): Repeat the measurement three times at each wind speed to calculate a mean and identify anomalies.
- M1 (Measure 1): Measure the distance moved by the bubble in the capillary tube.
- M2 (Measure 2): Record this over a set period of time (e.g., 10 minutes).
- S (Same): Keep light intensity (constant lamp distance) and room temperature constant.

(c) Explanation:
Wind blows away water vapour that accumulates outside the stomata. This maintains a steep concentration gradient of water vapour between the inside of the leaf (high concentration) and the air outside the leaf (low concentration), which increases the rate of diffusion of water vapour out of the leaf.

(a) (Maximum 4 marks)
- MP1: Cut shoot underwater; [1]
- MP2: Seal joints with Vaseline / make airtight; [1]
- MP3: Introduce an air bubble into the capillary tube; [1]
- MP4: Measure distance bubble travels; [1]
- MP5: Record time using a stopwatch / timer; [1]

(b) (Maximum 6 marks)
- C: change wind speed using a fan at different distances / settings; [1]
- O: same species / age / leaf area of shoot; [1]
- R: repeat at each wind speed to calculate mean / identify anomalies; [1]
- M1: measure distance moved by bubble; [1]
- M2: in a specified time period (e.g., minutes); [1]
- S: keep temperature / light intensity / humidity constant; [allow up to 2 marks - 1 mark for each named control variable]

(c) (Maximum 2 marks)
- MP1: Wind removes water vapour near leaf surface / stomata; [1]
- MP2: Maintained / increased concentration gradient of water vapour; [1]
- MP3: Faster diffusion of water vapour out of the leaf; [1]

Part (a): 1. Convert water loss from grams to milligrams: \(5.4\text{ g} = 5400\text{ mg}\). 2. Calculate water loss per unit area: \(5400\text{ mg} / 120\text{ cm}^2 = 45\text{ mg cm}^{-2}\). 3. Calculate the rate per hour: \(45\text{ mg cm}^{-2} / 2.5\text{ hours} = 18\text{ mg cm}^{-2}\text{ hour}^{-1}\). Part (b): 1. Increasing wind speed moves humid air (water vapour) away from the leaf surface and stomatal openings. 2. This maintains a steep concentration gradient of water vapour between the intercellular air spaces inside the leaf and the outside air, increasing the rate of transpiration via diffusion. 3. At extremely high wind speeds, the rate levels off because the stomata close to prevent excessive water loss, or another factor (such as stomatal aperture or water availability) becomes the limiting factor.

Part (a) [4.6 marks]: - 1.5 marks for converting grams to milligrams (5.4 g = 5400 mg). - 1.5 marks for dividing mass by area (45 mg cm^-2). - 1.6 marks for dividing by time to get the correct final answer of 18 mg cm^-2 hour^-1. Part (b) [7.0 marks]: - 2.0 marks for stating that wind removes water vapour/boundary layer of air around the leaf. - 2.0 marks for explaining that this maintains a steep diffusion/concentration gradient. - 2.0 marks for explaining that stomata close at very high wind speeds to reduce water loss / transpiration. - 1.0 mark for mentioning that stomatal resistance or water availability becomes the limiting factor.

Part (a): A represents the stimulation of body cells to take up glucose from the blood and the activation of enzymes in the liver and muscle cells to convert glucose into stored glycogen (glycogenesis). B is the alpha-cells of the islets of Langerhans in the pancreas. C is 'Increases'. D is the rapid breakdown of glycogen to glucose (glycogenolysis) and secretion of glucose into the blood, as well as increasing heart rate and blood flow. Part (b): Insulin and glucagon act antagonistically: when blood glucose rises (e.g., after a meal), insulin is released to lower it; when blood glucose falls (e.g., during exercise), glucagon is released to raise it. This is negative feedback because the deviation from the normal set point triggers a corrective response that reverses the change, restoring blood glucose to its optimum level. During exercise, muscle cells consume glucose rapidly for aerobic respiration; glucagon ensures glucose is released from liver stores to maintain adequate blood glucose levels for the brain and muscles.

Part (a) [4.6 marks]: - 1.15 marks for each correct identification (A, B, C, D). Accept equivalent wording (e.g., glycogenesis for A, alpha-cells for B, raises for C, glycogenolysis/flight-or-fight for D). Part (b) [7.0 marks]: - 2.0 marks for defining antagonistic action (insulin decreases, glucagon increases blood glucose). - 2.0 marks for explaining negative feedback (deviation from norm triggers a response that opposes/reverses the change). - 2.0 marks for linking exercise to increased respiration and glucose demand. - 1.0 mark for stating glucagon maintains glucose levels to prevent hypoglycemia / fuel vital organs.

Part (a): 1. Maximum phosphate concentration is 95 micrograms/L in Month 2. 2. Since \(1\text{ dm}^3 = 1\text{ L}\), the concentration is 95 micrograms/dm^3. 3. Convert micrograms to milligrams by dividing by 1000: \(95 / 1000 = 0.095\text{ mg/dm}^3\). Part (b): 1. High phosphate in Month 2 acts as a limiting nutrient, causing rapid growth of algae (algal bloom) at the surface. 2. The algae block sunlight, causing submerged plants and algae in deeper layers to die because they cannot photosynthesise. 3. In Month 3 and 4, the population of decomposers (aerobic bacteria) increases dramatically (from 400 to 8000 cells/mL) as they feed on the dead plant matter. 4. These aerobic bacteria respire rapidly, using up dissolved oxygen. This leads to the severe depletion of dissolved oxygen (to 1.5 mg/L in Month 4), causing fish and other aquatic organisms to suffocate and die.

Part (a) [4.6 marks]: - 1.5 marks for identifying the maximum concentration as 95 micrograms/L. - 1.5 marks for using the conversion factor (divide by 1000). - 1.6 marks for the correct final value with units (0.095 mg/dm^3). Part (b) [7.0 marks]: - 2.0 marks for describing the algal bloom caused by high phosphate and blocking of sunlight. - 2.0 marks for linking death of plants to increased food supply for decomposers/bacteria. - 2.0 marks for explaining that aerobic bacteria multiply and consume oxygen during respiration. - 1.0 mark for using the data (e.g., bacteria increasing to 8000 cells/mL coinciding with oxygen dropping to 1.5 mg/L).

Part (a): The curly wool allele is recessive because it disappears in the F1 generation and reappears in the F2 generation in a 3:1 ratio. Parents of F2 are both heterozygous (Cc). Punnett square: Gametes C and c from both parents yield offspring genotypes: CC (Normal), Cc (Normal), Cc (Normal), and cc (Curly). This gives the 3:1 phenotype ratio of normal to curly. Part (b): 1. In the F2 generation, the expected genotypes are 25% CC, 50% Cc, and 25% cc. Therefore, 50% of the F2 generation is expected to be heterozygous. 2. The normal-wooled F2 sheep can only have the genotypes CC or Cc (total of 3 parts out of the 4). Of these normal-wooled sheep, the carriers (heterozygotes, Cc) represent 2 parts. Thus, the probability that a normal-wooled sheep is a carrier is 2/3 or approximately 0.67 (66.7%).

Part (a) [6.0 marks]: - 1.0 mark for identifying curly wool as recessive and normal as dominant. - 2.0 marks for drawing a correct Punnett square with correct parental gametes (C and c). - 2.0 marks for showing correct offspring genotypes (CC, Cc, cc) and linking them to phenotypes. - 1.0 mark for explaining that F1 must be heterozygous (Cc). Part (b) [5.6 marks]: - 2.0 marks for stating that 50% of the F2 offspring are heterozygous. - 3.6 marks for calculating the probability of a normal sheep being a carrier as 2/3 or 0.67 / 66.7% (method: dividing 2 heterozygous parts by the 3 total normal parts; do not accept 2/4 or 0.50, which is the overall probability of any offspring being a carrier, not specifically a normal-wooled one).

Part (a): 1. Find lactic acid concentration at hour 2: 3.5 g/L. At hour 4: 12.0 g/L. 2. Calculate the increase in lactic acid: 12.0 - 3.5 = 8.5 g/L. 3. Convert g/L to micrograms/mL: 8.5 g/L = 8.5 mg/mL = 8500 micrograms/mL. 4. Divide by the time interval (4 - 2 = 2 hours): 8500 micrograms/mL / 2 hours = 4250 micrograms mL^-1 hour^-1. Part (b): 1. Lactobacillus bulgaricus ferments lactose (the milk sugar) anaerobically to produce lactic acid. 2. The production of lactic acid increases the concentration of hydrogen ions, which lowers the pH of the milk. 3. The low pH (acidic conditions) denatures/coagulates milk proteins (casein), causing them to precipitate and clot, giving yoghurt its thick texture. 4. The low pH also inhibits the enzymes and growth of pathogenic or spoilage micro-organisms, acting as a natural preservative to prevent contamination.

Part (a) [4.6 marks]: - 1.5 marks for calculating the absolute change in lactic acid concentration (8.5 g/L). - 1.5 marks for converting g/L to micrograms/mL (8500 micrograms/mL). - 1.6 marks for dividing by the time interval (2 hours) to obtain 4250 micrograms mL^-1 hour^-1. Part (b) [7.0 marks]: - 2.0 marks for stating that Lactobacillus ferments lactose anaerobically to produce lactic acid. - 2.0 marks for explaining that lactic acid lowers pH, causing milk protein (casein) to denature, coagulate, and thicken. - 2.0 marks for explaining that low pH prevents the growth/reproduction of other bacteria/pathogens. - 1.0 mark for mentioning that this serves as a preservation method.

Part (a): 1. Calculate Q10: Q10 = (Rate at 30 degrees C) / (Rate at 20 degrees C) = 0.99 / 0.45 = 2.2. 2. Convert the rate at 30 degrees C (0.99 mg dm^-3 s^-1) into micrograms dm^-3 min^-1: - Convert milligrams to micrograms: 0.99 mg = 990 micrograms. - Convert seconds to minutes by multiplying by 60: 990 micrograms s^-1 * 60 s/min = 59400 micrograms dm^-3 min^-1. Part (b): 1. As temperature increases from 30 degrees C to the optimum (typically around 37-40 degrees C), kinetic energy increases, leading to more frequent collisions and more enzyme-substrate complexes. 2. However, as the temperature continues to rise towards 60 degrees C, the rate of reaction drops rapidly to zero. 3. This is because high thermal energy breaks the hydrogen bonds, ionic bonds, and hydrophobic interactions maintaining the enzyme's tertiary structure. 4. The active site of the amylase enzyme changes shape (denatures) and is no longer complementary to the starch substrate, so no enzyme-substrate complexes can form.

Part (a) [4.6 marks]: - 1.5 marks for calculating Q10 correctly as 2.2. - 1.5 marks for converting milligrams to micrograms (0.99 mg = 990 micrograms). - 1.6 marks for multiplying by 60 to convert per-second to per-minute to get 59400 micrograms dm^-3 min^-1. Part (b) [7.0 marks]: - 2.0 marks for explaining that initially, temperature increases kinetic energy and collision frequency. - 2.0 marks for stating that above the optimum, high temperature breaks bonds holding the enzyme's shape. - 2.0 marks for explaining denaturation (active site changes shape, substrate can no longer bind / complementary fit is lost). - 1.0 mark for stating that enzyme-substrate complexes can no longer form, reducing rate to zero.