2025 Cambridge IGCSE Biology (0610) 模擬試題連答案詳解

The rate of transpiration (and thus water uptake in a potometer) is lowest when the concentration gradient of water vapour between the inside of the leaf and the atmosphere is minimized, and when stomatal opening is reduced. Still air allows a boundary layer of humid air to build up around the leaf, reducing diffusion. Low temperature reduces the kinetic energy of water molecules, slowing evaporation. High humidity reduces the water potential gradient between the inside of the leaf and the outside air. Low light intensity causes stomata to close. Therefore, condition A results in the lowest rate of water uptake.

1 mark for selecting option A. Reject all other options.

Flower 1 has large, brightly colored petals to attract insects, sticky stigmas to catch pollen from insects' bodies, and spiky pollen to adhere to insects. These are adaptations for insect-pollination. Flower 2 has dull green petals (as it does not need to attract insects), feathery stigmas with a large surface area hanging outside to catch wind-blown pollen, and light, smooth pollen to be easily carried by the wind. These are adaptations for wind-pollination.

1 mark for selecting option A.

In a double circulatory system, blood passes through the heart twice for each complete circuit of the body. The first circuit (pulmonary) pumps deoxygenated blood at lower pressure to the lungs for gas exchange. The oxygenated blood returns to the heart to be pumped out at high pressure in the second circuit (systemic) to body tissues, allowing rapid and efficient delivery of oxygen.

1 mark for selecting option B.

Eutrophication begins with fertilizer runoff containing nitrates and phosphates, causing rapid algal growth (algal bloom). This blocks light, causing submerged producers to die. Decomposers (bacteria) feed on the dead matter and respire aerobically, consuming the dissolved oxygen. This lack of oxygen leads to the death (suffocation) of fish.

1 mark for selecting option A.

Organisms of the same genus are closely related evolutionarily, meaning they share more similar DNA base sequences than organisms belonging to different genera. They cannot produce fertile offspring (since they are different species), they share more physical characteristics than organisms in different families, and the second word of their binomial name (the species name) is different.

1 mark for selecting option C.

First, convert all measurements to the same unit (e.g., millimeters). 4.0 cm is equal to 40 mm. Using the magnification formula: Magnification = Image size / Actual size = 40 mm / 0.08 mm = 500. Thus, the magnification is \(\times 500\).

1 mark for selecting option C.

During inspiration, the external intercostal muscles contract, pulling the ribcage up and out. Simultaneously, the diaphragm contracts and flattens. This increases the volume of the thorax, which decreases the pressure inside the thorax below atmospheric pressure, causing air to move in.

1 mark for selecting option A.

Denitrifying bacteria are responsible for denitrification, which converts nitrate ions in the soil back into nitrogen gas in the atmosphere, typically under anaerobic conditions such as waterlogged soil.

1 mark for selecting option C.

When light is present (Conditions 1 and 2), stomata are open, allowing transpiration to occur. To investigate the effect of wind speed when stomata are open, you need to compare still air (Condition 1) with moving air (Condition 2) under light. In the dark (Conditions 3 and 4), stomata are closed, so wind speed has little to no effect on transpiration because water vapor cannot leave the leaves easily.

Award 1 mark for identifying that comparing Condition 1 and Condition 2 shows the effect of wind speed under light conditions (when stomata are open).

Wind-pollinated flowers produce pollen grains that are small, smooth, and lightweight, and they must be produced in huge quantities (Type R) to increase the chance of successful pollination because wind direction is random. Their stigmas must be large and feathery (Type W) and hang outside the flower to maximize the surface area for catching passing wind-borne pollen grains.

Award 1 mark for selecting the correct combination of wind-pollination adaptations (Pollen R and Stigma W).

In a fish (single circulation), the blood pressure is highest as it leaves the ventricle and goes directly into the gills. The pressure drops significantly as it passes through the gill capillaries, meaning pressure in the gills is higher than pressure in the body tissues. In a mammal (double circulation), blood is pumped to the lungs at a lower pressure (from the right ventricle) to avoid damaging the delicate lung tissues, and returns to the heart to be pumped to body tissues at a much higher pressure (from the left ventricle). Therefore, in mammals, pressure in the lungs is lower than in the body tissues.

Award 1 mark for the correct comparison of relative blood pressure in fish and mammals (Row C).

When excess fertilizers or sewage enter a water body, they cause an algal bloom. When the algae die, they are decomposed by aerobic bacteria. These bacteria multiply rapidly and respire aerobically, consuming the dissolved oxygen in the water. This depletion of oxygen directly leads to the death of fish and other aquatic organisms.

Award 1 mark for identifying aerobic respiration by decomposers (bacteria) as the direct cause of oxygen depletion.

Organisms that share a more recent common ancestor have DNA base sequences that are more similar. Species W and X share 94% similarity, whereas W and Z share only 71% similarity. Therefore, W and X share a more recent common ancestor than W and Z.

Award 1 mark for identifying that higher sequence similarity indicates a more recent common ancestor.

Using the formula: \(A = \frac{I}{M}\).
First, convert the image length from centimeters to micrometers (\(\mu\text{m}\)):
\(4.8\text{ cm} = 48\text{ mm} = 48,000\ \mu\text{m}\).
Now, calculate the actual length:
\(A = \frac{48,000\ \mu\text{m}}{1200} = 40\ \mu\text{m}\).

Award 1 mark for using the correct units conversion and calculating the actual size to be 40 micrometers.

Nitrogen gas in the air is chemically unreactive and cannot be absorbed or used by the human body during gas exchange. Since it is neither consumed nor produced by metabolic processes, the quantity inhaled is the same as the quantity exhaled, keeping its percentage in the air constant.

Award 1 mark for identifying that nitrogen is not absorbed or used by the body during gas exchange.

- Nitrifying bacteria convert ammonium ions to nitrate ions (Process Y).
- Denitrifying bacteria convert nitrate ions to nitrogen gas under anaerobic conditions (Process X).
- Nitrogen-fixing bacteria convert atmospheric nitrogen gas into ammonium ions/nitrogenous compounds (Process Z).

Award 1 mark for the correct match of nitrifying, denitrifying, and nitrogen-fixing bacteria to their respective processes (Row A).

Water molecules are polar and form hydrogen bonds with each other, resulting in cohesion. They also form bonds with the cellulose in the walls of the xylem vessels, resulting in adhesion. Cohesion and adhesion together maintain a continuous, unbroken column of water, preventing it from breaking under the tension created by transpiration.

1 mark: Identifies that cohesion and adhesion are the forces responsible for maintaining a continuous water column.

Removing the anthers before they mature (emasculation) prevents the flower from producing its own pollen (preventing self-pollination). Covering the flower with a fine mesh bag allows gas exchange but prevents insects from introducing foreign pollen until the breeder performs the controlled cross-pollination.

1 mark: Identifies anther removal and bagging as the correct combination to prevent self-pollination and uncontrolled cross-pollination.

In a single circulatory system (e.g., in fish), blood pressure drops significantly as it passes through the gas exchange organs (gills) before flowing to the rest of the body. In a double circulatory system, blood is returned to the heart to be pumped again after oxygenation, allowing it to be sent to the systemic tissues at a much higher pressure and flow rate, delivering oxygen more rapidly.

1 mark: Identifies that a double circulation allows oxygenated blood to be pumped to body tissues under high pressure to deliver oxygen rapidly.

Fertiliser runoff causes an algal bloom. This blocks sunlight from reaching submerged plants, which can no longer photosynthesise and die. Decomposers (bacteria) feed on the dead plant matter and multiply, respiring aerobically. This uses up the dissolved oxygen in the water, causing aquatic animals like fish to suffocate and die.

1 mark: Identifies the correct chronological sequence of eutrophication events.

A higher percentage similarity in DNA base sequences indicates that two species are more closely related and shared a more recent common ancestor. Species X and Species 1 share 95% similarity, whereas Species X and Species 4 share only 61% similarity, meaning Species X and 1 are more closely related.

1 mark: Correctly identifies that high DNA similarity indicates a more recent common ancestor.

First, convert both values to the same unit. Actual diameter = 8 \(\mu\)m = 0.008 mm. Image diameter = 24 mm. Using the formula Magnification = Image size / Actual size: Magnification = 24 mm / 0.008 mm = 3000. Alternatively, 24 mm = 24000 \(\mu\)m, so Magnification = 24000 / 8 = 3000.

1 mark: Converts units correctly and applies the magnification formula to calculate \(\times\) 3000.

During inspiration, the diaphragm contracts and flattens (moves downwards) to increase thoracic volume. At the same time, the external intercostal muscles contract to pull the ribcage upwards and outwards, further increasing the volume of the thorax.

1 mark: Correctly identifies that both the diaphragm and external intercostal muscles contract during inspiration.

Nitrifying bacteria convert ammonium ions into nitrites and then into nitrates (nitrification). Denitrifying bacteria convert nitrates back to nitrogen gas (denitrification). Nitrogen-fixing bacteria convert atmospheric nitrogen gas into nitrogen-containing compounds like ammonia/ammonium. Decomposers break down proteins and urea into ammonium ions.

1 mark: Correctly matches nitrifying bacteria to the conversion of ammonium ions to nitrate ions.

(a) Transpiration is officially defined as the loss of water vapor from plant leaves by evaporation of water at the surfaces of the mesophyll cells followed by diffusion of water vapor through the stomata.

(b) Other environmental factors that affect transpiration must be kept constant to ensure a fair test. These include light intensity, temperature, and ambient humidity.

(c) Increased wind speed blows away the boundary layer of moist air (water vapor) that accumulates around the surface of the leaf. This maintains a steep diffusion/concentration gradient for water vapor between the moist air spaces inside the leaf and the drier air outside, increasing the rate of diffusion through the stomata.

(d) Water leaves the xylem vessels and moves into the surrounding mesophyll cells of the leaf. It then evaporates from the damp surfaces of the cell walls of the mesophyll cells into the air spaces within the leaf. Finally, the water vapor diffuses down its concentration gradient out of the intercellular air spaces into the atmosphere via the stomata.

(e) Guard cells have cell walls of uneven thickness; the inner wall next to the stoma is thicker and less elastic than the outer wall. When water enters the guard cells by osmosis, they swell and become turgid, causing the outer thin walls to stretch more and curve outward, which pulls the stoma open. When water leaves the guard cells, they become flaccid and the stoma closes.

(a) [Max 2 marks]
- Loss of water vapor from leaves / plant parts [1]
- Evaporation at the surfaces of mesophyll cells [1]
- Diffusion of water vapor through the stomata [1]

(b) [Max 2 marks]
- Light intensity [1]
- Temperature [1]
- Humidity / moisture in air [1]
(Reject: wind speed, water availability in soil)

(c) [Max 3 marks]
- Increases transpiration rate [1]
- Removes / blows away water vapor surrounding the leaf surface / stomata [1]
- Maintains / increases concentration gradient of water vapor (between leaf interior and atmosphere) [1]
- Increases rate of diffusion of water vapor [1]

(d) [Max 4 marks]
- Water moves out of xylem (into mesophyll cells) [1]
- Evaporates from surfaces of mesophyll cell walls [1]
- Into intercellular / air spaces [1]
- Water vapor diffuses [1]
- Out of stomata into the atmosphere [1]

(e) [Max 2 marks]
- Guard cells have thicker inner walls (and thinner outer walls) [1]
- When turgid / water enters, guard cells bend / curve outwards to open stoma [1]
- When flaccid / water leaves, guard cells straighten to close stoma [1]

(a) Pollination is the physical transfer of pollen grains from an anther (male part) to a receptive stigma (female part) of a flower. Fertilization, however, is a chemical/biological process involving the fusion of the haploid male gamete nucleus from the pollen grain with the haploid female gamete nucleus in the ovule.

(b) Insect-pollinated flowers show distinct adaptations:
1. Large, brightly colored petals to visually attract insects from a distance.
2. Production of nectar and scent from nectaries, which act as a food reward and olfactory attractant.
3. Sticky or spiky pollen grains, which allow the pollen to easily adhere to the bodies of visiting insects as they brush against the anthers.

(c) Once a compatible pollen grain lands on the stigma, it absorbs nutrients and germinates. It grows a pollen tube that digests its way down through the style tissue, guided by chemical signals towards the ovary. The pollen tube enters the ovary and penetrates an ovule through a tiny opening called the micropyle. The male gamete nucleus travels down this tube and is released into the ovule, where it fuses with the female egg cell nucleus, completing fertilization to form a diploid zygote.

(a) [Max 2 marks]
- Pollination: transfer of pollen from anther to stigma [1]
- Fertilization: fusion of male and female gamete nuclei [1]

(b) [Max 6 marks - 1 mark for adaptation, 1 mark for corresponding explanation, up to 3 pairs]:
- Brightly colored / large petals [1] to attract insects visually [1]
- Nectar / nectaries / sweet scent [1] to act as food reward / attract insects by smell [1]
- Sticky / spiky pollen grains [1] to cling to insect bodies [1]
- Enclosed anthers / stigma within petals [1] so insects must brush against them to reach nectar [1]
- Lobed / sticky stigma [1] to easily catch pollen from visiting insects [1]

(c) [Max 5 marks]
- Pollen grain germinates (on the stigma) [1]
- Pollen tube grows [1]
- Down the style [1]
- Enters the ovary / ovule [1]
- Passes through the micropyle [1]
- Male gamete / pollen nucleus moves down the tube [1]
- Nuclei fuse / fertilization occurs to produce a diploid zygote [1]

(a) Double circulation refers to a system where blood flows through the heart twice for every one complete circuit around the entire body (the pulmonary circulation to the lungs, and systemic circulation to the body organs). The major advantage of this is that blood pressure can be boosted after passing through the lungs. This allows oxygenated blood to be transported to tissues at high pressure and speed, supporting high metabolic rates.

(b)
(i) The hepatic artery supplies oxygenated blood to the liver.
(ii) The renal vein carries deoxygenated blood away from the kidneys back to the vena cava.
(iii) The hepatic portal vein transports blood containing dissolved products of digestion from the small intestine to the liver.

(c) The septum is a thick muscular wall that completely divides the left and right sides of the heart. It prevents the mixing of oxygenated blood (on the left side) with deoxygenated blood (on the right side). This ensures that only fully oxygenated blood is sent to body tissues, maintaining a high concentration gradient for oxygen diffusion at respiring cells.

(d) Arteries have a thick outer wall, a thick layer of muscle and elastic fibers, and a narrow lumen to withstand and maintain high blood pressure from the heart. Veins have much thinner walls with less muscle and elastic tissue because blood travels at a much lower pressure. They have a wider lumen to reduce resistance to blood flow and contain semilunar valves to prevent the backflow of blood against gravity.

(a) [Max 3 marks]
- Definition: Blood passes through the heart twice for each complete circuit of the body [1]
- Pulmonary and systemic circuits mentioned [1]
- Advantage: Blood pressure is restored / kept high after the lungs [1]
- Speeds up delivery of oxygen / nutrients to tissues [1]

(b) [Max 3 marks]
- (i) Hepatic artery [1]
- (ii) Renal vein [1]
- (iii) Hepatic portal vein [1]

(c) [Max 3 marks]
- Partition / wall that separates right and left sides of the heart [1]
- Prevents mixing of oxygenated and deoxygenated blood [1]
- Ensures maximum oxygen saturation of blood sent to tissues [1]
- Keeps systemic circulation efficient / supports high metabolic rate [1]

(d) [Max 4 marks]
- Artery has thick muscular / elastic walls AND vein has thin walls [1]
- Artery has a narrow lumen AND vein has a wide lumen [1]
- Artery wall withstands / maintains high pressure [1]
- Veins have valves (to prevent backflow) AND arteries do not [1]
- Vein's wide lumen reduces friction / resistance to low-pressure blood flow [1]

(a) When fertilizers or sewage enter a river, they introduce high concentrations of nitrates and phosphates. This leads to an 'algal bloom' where algae multiply rapidly on the water surface. This thick layer of algae blocks sunlight from reaching submerged aquatic plants, preventing them from photosynthesizing and causing them to die. Aerobic bacteria and decomposers feed on the dead plant material, and their populations explode. As these bacteria reproduce, they respire aerobically, consuming vast amounts of dissolved oxygen from the water. The oxygen levels drop drastically, causing fish and other aquatic animals to suffocate and die due to anoxia.

(b) Sources of chemical land pollution include:
1. Untreated industrial waste / heavy metals discharged from factories.
2. Leaching of toxic chemicals / household waste from landfill sites.
3. Improper disposal of electronic waste (e-waste) or batteries.

(c) Non-biodegradable plastics persist in marine environments for hundreds of years. Animals can mistake plastic debris for food (e.g., turtles eating plastic bags thinking they are jellyfish), leading to physical blockages in their digestive systems, starvation, and death. Animals can also become entangled in abandoned plastic fishing nets and packaging rings, causing injury or drowning. Furthermore, plastics break down slowly into microplastics, which absorb toxic chemical pollutants and bioaccumulate up marine food chains, poisoning marine life.

(a) [Max 7 marks]
- Inflow of nitrates / phosphates / nutrients [1]
- Rapid growth of algae / algal bloom [1]
- Algae blocks sunlight [1]
- Submerged aquatic plants cannot photosynthesize and die [1]
- Bacteria / decomposers feed on dead plants [1]
- Bacteria population increases / multiplies rapidly [1]
- Bacteria respire aerobically [1]
- Oxygen concentration in water decreases / depleted [1]
- Fish suffocate / die due to lack of oxygen [1]

(b) [Max 2 marks]
- Industrial effluent / heavy metal waste from factories [1]
- Insecticides / pesticides / herbicides [1]
- Landfill site leachates / household chemical waste [1]
- Acid rain / mining waste [1]

(c) [Max 4 marks]
- Ingestion / swallowing of plastics causes intestinal blockage / starvation [1]
- Entanglement / trapping of animals in nets / plastic rings [1]
- Degradation into microplastics [1]
- Bioaccumulation / biomagnification of toxic chemicals up food chains [1]
- Smothering / damage to coral reefs / benthic habitats [1]

(a) Organisms in the Fungi kingdom are eukaryotic, possess cell walls made of chitin, do not photosynthesize (as they lack chlorophyll), are heterotrophic (usually saprophytic decomposers or parasitic), and reproduce via spores. Their bodies are typically structured as a network of threads called hyphae.

(b) Historically, classification relied on morphological (structural) and anatomical characteristics. DNA base sequence analysis compares the actual genetic code. Organisms with a higher percentage of identical or highly similar base sequences are classified as more closely related. This biochemical comparison helps resolve evolutionary links that morphology might obscure due to convergent evolution (where unrelated species look similar because of similar environments).

(c) Organisms sharing a recent common ancestor diverged from each other relatively recently in evolutionary terms. Consequently, there has been less time for random DNA mutations to occur and accumulate, resulting in fewer differences in their DNA base sequences.

(d) The binomial name of the domestic cat is written as *Felis catus*. The rules are:
1. The genus name is written first and starts with a capital letter.
2. The species name is written second and starts with a lowercase letter.
3. The scientific name must be written in italics (or underlined if handwritten).

(a) [Max 3 marks]
- Have cell walls made of chitin [1]
- Do not perform photosynthesis / lack chlorophyll [1]
- Saprophytic / parasitic / heterotrophic nutrition [1]
- Reproduce by spores [1]
- Body made of hyphae / mycelium [1]
(Accept: eukaryotic / single-celled yeast)

(b) [Max 4 marks]
- DNA base sequence determines genetic code [1]
- More similar base sequences show closer evolutionary relationships [1]
- Quantifies similarity / more accurate than using structure alone [1]
- Helps identify convergent evolution (where unrelated species look alike) [1]
- Allows revision of incorrect traditional groupings [1]

(c) [Max 2 marks]
- Less evolutionary time has passed since divergence [1]
- Fewer mutations have occurred to alter the DNA sequence [1]

(d) [Max 4 marks]
- *Felis catus* (must be italicized or underlined) [1]
- Genus starts with a capital letter [1]
- Species starts with a lowercase letter [1]
- Written in italics / underlined [1]

(a) Alveoli are adapted for gas exchange in several distinct ways:
1. They provide an extremely large total surface area due to millions of tiny spherical sacs.
2. The alveolar wall is only one cell thick, which provides an incredibly short diffusion pathway for gases.
3. The inner surface of the alveolus is lined with a moist film of water, allowing oxygen to dissolve before diffusing across the membrane.
4. Each alveolus is wrapped in an extensive network of capillaries, maintaining a continuous flow of blood to transport gases away and keep concentration gradients steep.

(b) Inside body cells, aerobic respiration continuously consumes oxygen and produces carbon dioxide as a waste product. Consequently, blood arriving at the lungs has a low oxygen concentration and a high carbon dioxide concentration compared to the air inside the alveoli. Oxygen diffuses out of the alveolar air into the blood, reducing its concentration to \( 16\% \) in the expired air. Carbon dioxide diffuses out of the blood into the alveoli, increasing its concentration to \( 4\% \) in the expired air.

(c) Goblet cells and ciliated cells line the trachea, bronchi, and bronchioles. Goblet cells synthesize and secrete sticky mucus, which traps inhaled dust particles, bacteria, and other pathogens. Ciliated cells possess tiny hair-like cytoplasmic extensions called cilia. These cilia beat rhythmically in a coordinated fashion, sweeping the trapped mucus up and away from the lungs towards the back of the throat (pharynx), where it is swallowed into the acidic stomach and destroyed.

(a) [Max 5 marks]
- Large total surface area (due to large numbers of alveoli) [1]
- Very thin walls / one cell thick [1]
- Short diffusion distance / pathway [1]
- Moist lining (to dissolve gases) [1]
- Excellent capillary / blood supply [1]
- Constant ventilation / blood flow maintains steep concentration gradient [1]

(b) [Max 4 marks]
- Cells use oxygen for aerobic respiration [1]
- Oxygen diffuses from alveoli into the blood [1]
- Cells produce carbon dioxide during respiration [1]
- Carbon dioxide diffuses from blood into alveoli [1]
- Reference to numerical changes (e.g. oxygen decrease by \( 5\% \) or CO2 increase by 100-fold) [1]

(c) [Max 4 marks]
- Goblet cells secrete / produce mucus [1]
- Mucus is sticky and traps pathogens / bacteria / dust [1]
- Ciliated cells have cilia / hair-like structures [1]
- Cilia beat / sweep mucus upwards [1]
- Away from the lungs / towards the throat / to be swallowed [1]

Part (a)(i): The independent variable is the wind speed (m/s). The dependent variable is the distance moved by the bubble in 10 minutes (mm). Part (a)(ii): Mean distance = (41 + 48 + 43) / 3 = 132 / 3 = 44 mm. Part (a)(iii): A potential source of error is air leaks in the potometer system. An improvement is to apply petroleum jelly (Vaseline) to all connection joints to ensure an airtight seal. (Alternative error: variation in temperature/light. Alternative improvement: carry out the experiment in a temperature-controlled room with constant artificial light). Part (b): To plot the bar chart: 1. Label the x-axis as 'Wind speed / m/s' and the y-axis as 'Mean distance moved by bubble in 10 minutes / mm'. 2. Choose a linear scale for the y-axis where 10 mm on the grid represents 10 mm of bubble movement, ensuring the data occupies more than half of the grid. 3. Draw three discrete bars of equal width representing 0 m/s, 2 m/s, and 5 m/s with clear gaps between them. 4. Plot the heights of the bars accurately at 12 mm, 25 mm, and 44 mm using a ruler. Part (c): Moving air (wind) sweeps away the boundary layer of moist air / water vapor that accumulates on the leaf surface and outside the stomata. This maintains a steep water vapor concentration gradient between the internal air spaces of the leaf and the drier external atmosphere, which increases the rate of diffusion of water vapor out of the stomata.

Part (a)(i) [2 marks]: - Independent variable identified as wind speed [1] - Dependent variable identified as distance moved by bubble in 10 minutes [1] Part (a)(ii) [2 marks]: - Correct working shown: (41 + 48 + 43) / 3 [1] - Correct calculation of mean: 44 (mm) [1] Part (a)(iii) [2 marks]: - Any valid experimental error (e.g., air leaks in apparatus / variation in temperature or light / bubble not reset properly) [1] - Corresponding valid improvement (e.g., seal joints with petroleum jelly / perform in a temperature-controlled room / use syringe to reset bubble) [1] Part (b) [4 marks]: - Axes correctly labeled with appropriate units: 'Wind speed / m/s' on x-axis AND 'Mean distance moved (by bubble in 10 minutes) / mm' on y-axis [1] - Suitable scale chosen where plotted bars occupy more than half the grid height [1] - Bars plotted at correct heights: 12 mm, 25 mm, and 44 mm [1] - Bars drawn with a ruler, having equal width and distinct gaps between them [1] Part (c) [3 marks]: - Wind removes water vapor / moist air from the surface of the leaf [1] - This maintains / increases the water vapor concentration gradient [1] - Leading to a faster rate of diffusion of water molecules through the stomata [1]

Part (a): To plot the line graph: 1. Label the x-axis as 'Sucrose concentration / %' and the y-axis as 'Percentage of germinated pollen grains / %'. 2. Choose a linear scale for both axes where the data points occupy more than half of the grid area. 3. Plot the five coordinate points accurately: (0, 0), (5, 15), (10, 58), (15, 82), and (20, 40) using small crosses (x) or circled dots. 4. Connect the points point-to-point with straight, clean, ruled lines or draw a smooth, appropriate curve. Part (b): As sucrose concentration increases from 0% to 15%, the percentage of germinated pollen grains increases. The optimum sucrose concentration for germination is 15%, which results in a maximum germination of 82%. As the sucrose concentration increases above 15%, the percentage of germinated pollen grains decreases to 40% at 20% sucrose concentration. Part (c)(i): In pure water (0% sucrose), there is a high water potential outside the pollen grain. Water enters the pollen grain rapidly by osmosis, causing it to swell and undergo osmotic lysis (bursting) before germination can occur. Part (c)(ii): Temperature (e.g., kept constant at 25 degrees Celsius using an incubator) and light intensity (or humidity). Part (d): 1. Use a sharp pencil to draw single, clean, continuous lines with no sketching or overlapping. 2. Do not use shading, stippling, or color of any kind. 3. Ensure the drawing is large and occupies at least half of the available page space with accurate proportions of all structures shown.

Part (a) [4 marks]: - Axes labeled with units: 'Sucrose concentration / %' on x-axis and 'Percentage of germinated pollen grains / %' on y-axis [1] - Scale is linear, appropriate, and uses more than half of the grid area [1] - All 5 points plotted accurately within half a small grid square [1] - Points connected point-to-point with straight, clean, ruled lines or a smooth curve [1] Part (b) [3 marks]: - Germination percentage increases as sucrose concentration increases from 0% to 15% [1] - The maximum / optimum germination percentage is at 15% sucrose concentration [1] - Germination percentage decreases as sucrose concentration increases above 15% / to 20% [1] - Correct use of data values to support description (e.g., peak of 82% germination at 15% sucrose vs 0% at 0% sucrose) [1] (Max 3 marks) Part (c) [3 marks]: - (i) Water enters by osmosis down a water potential gradient causing cells to swell and burst / osmotic lysis [1] - (ii) Temperature [1] - (ii) Time / duration of incubation (2 hours) [1] Part (d) [3 marks]: - Draw with a sharp pencil using single, thin, continuous lines / no sketching [1] - No shading or coloring is present in the drawing [1] - Large size (drawing occupies at least 50% of the space provided) / accurate proportions [1]

Part (a)(i): As the concentration of the organic pollutant increases, the time taken for methylene blue to decolorize decreases. This is because higher pollutant concentrations provide more organic substrate (nutrients like lactose and proteins in milk) for the yeast. This increases the rate of aerobic respiration, leading to faster consumption and depletion of dissolved oxygen, which reduces and decolorizes the methylene blue more rapidly. Part (a)(ii): The control tube shows that the decolorisation of methylene blue does not occur without the organic pollutant (food source), proving that the reaction is driven by the respiration of yeast consuming the organic substrate. Part (b): Time taken at 10% pollutant = 95 seconds. Rate = (1 / 95) * 1000 = 10.5263... When rounded to 2 decimal places, the rate is 10.53 arbitrary units. Part (c): 1. Independent variable: Temperature. Use at least 5 different temperatures (e.g., 10, 20, 30, 40, and 50 degrees Celsius) maintained using thermostatically-controlled water baths. 2. Dependent variable: Time taken for the blue color of methylene blue to completely disappear, measured using a stopwatch. 3. Control variables: Volume and concentration of yeast suspension (e.g., 5 cm³ of 2% yeast), volume and concentration of milk (e.g., 5 cm³ of 5% milk), and volume/concentration of methylene blue indicator (e.g., 3 drops of 0.1% solution). 4. Method: Pre-incubate the yeast suspension and milk-methylene blue solutions in separate test tubes at each target temperature for 5 minutes before mixing to ensure they reach the correct temperature. Mix them, start the stopwatch immediately, and record the time when the blue color completely disappears. 5. Replication: Repeat the procedure three times at each temperature to calculate a mean and identify anomalies. 6. Safety: Wear safety goggles and gloves when handling methylene blue to avoid skin/eye irritation and staining.

Part (a) [4 marks]: - (i) Shorter decolorisation times are observed at higher pollutant concentrations [1] - More organic substrate / food source is available for yeast respiration [1] - Rate of aerobic respiration increases, leading to faster consumption / depletion of dissolved oxygen [1] - (ii) Control confirms that the organic pollutant is necessary for respiration / oxygen depletion to occur / methylene blue does not decolorize spontaneously without substrate [1] Part (b) [3 marks]: - Formula substitution shown: 1 / 95 [1] - Calculation of value: 10.5263... [1] - Correct rounding to 2 decimal places: 10.53 [1] Part (c) [6 marks]: - Independent variable: At least 5 different temperatures specified (e.g., 10, 20, 30, 40, 50 degrees Celsius) [1] - Method for temperature control: Use of thermostatically-controlled water baths [1] - Dependent variable: Measure time taken for methylene blue to decolorize using a stopwatch [1] - Control variables: Volume/concentration of yeast AND volume/concentration of milk AND volume/drops of methylene blue kept constant (at least two named) [1] - Experimental detail: Pre-incubate solutions separately at target temperatures before mixing / repeat each temperature trial 3 times [1] - Safety precaution: Wear goggles or gloves due to methylene blue being an irritant/dye, or handle hot water baths with tongs [1]