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Question 1

Topic: 6.2

Fig. 1.1 is a diagram of part of a cross-section of a leaf.

 

(a) Explain why a leaf is considered to be an organ.

(b) Explain how layer B and layer C in Fig. 1.1 are adapted for their functions.

(c) A student placed a plant in a very hot room for 12 hours. There was a bright light in the room and the plant was not given any water during the 12-hour period.

Fig. 1.2 shows a series of sketches that the student made of the stomata during the investigation.

 

(i) Identify the cell labelled D on Fig. 1.2.

(ii) State the main function of the stomata.

(iii) State the advantage to the plant of the change to the stomata shown in Fig. 1.2.

(d) The student increased the humidity in the room and repeated the investigation.

Predict and explain the effect of high humidity on the stomata.

▶️ Answer/Explanation
Solution

(a) A leaf is considered an organ because it is made of different tissues (such as palisade mesophyll, spongy mesophyll, and vascular tissues) that work together to perform specific functions like photosynthesis and gas exchange.

Explanation: In biology, an organ is defined as a structure composed of multiple tissues working together to perform specific functions. The leaf contains various tissues like the epidermis for protection, palisade cells for photosynthesis, and vascular tissues for transport, all coordinated to enable the leaf’s functions.

(b) Layer B (palisade layer) is adapted for photosynthesis with tightly packed, columnar cells containing many chloroplasts positioned just below the upper epidermis to maximize light absorption. Layer C (spongy mesophyll) has loosely packed cells with air spaces to facilitate gas exchange and contains vascular bundles (xylem and phloem) for transport.

Explanation: The palisade layer’s vertical cell arrangement and high chloroplast concentration optimize light capture for photosynthesis. The spongy layer’s air spaces allow efficient diffusion of CO₂ in and O₂ out. Xylem vessels have thick walls for water transport, while phloem transports sugars produced during photosynthesis.

(c)(i) Guard cell.

Explanation: The cell labelled D is a guard cell, which is specialized to form and control the stomatal pore.

(c)(ii) The main function of stomata is to regulate gas exchange (CO₂ in for photosynthesis, O₂ out) and control transpiration (water vapor loss).

Explanation: Stomata are crucial pores that balance the plant’s need for CO₂ uptake with the need to conserve water, opening during favorable conditions and closing during stress.

(c)(iii) The advantage is reduced water loss through transpiration, preventing wilting in the hot, dry conditions.

Explanation: By closing the stomata, the plant conserves water that would otherwise be lost through evaporation, maintaining turgor pressure and preventing dehydration damage.

(d) In high humidity, stomata would take longer to close or might not close as completely because the reduced water vapor gradient between the leaf and air means less water is lost, so guard cells remain turgid.

Explanation: High humidity decreases the driving force for water loss (the concentration gradient), so the plant doesn’t need to close stomata as tightly to conserve water. The guard cells stay swollen with water, keeping the stomatal pore more open compared to dry conditions.

Question 2

Topic: 17.1

Fig. 2.1 is a diagram of part of a DNA molecule.

 

(a) Draw a circle on Fig. 2.1 to identify one pair of bases.

(b) The percentage of T bases in the DNA of a species is 29%.

Calculate the percentage of bases that would be base G in the DNA of this species.

Space for working.

(c) (i) State the name given to a length of DNA that codes for a protein.

(ii) Explain how proteins are made by a cell.

(iii) DNA controls cell function by controlling the production of proteins.

State two types of cell membrane proteins.

▶️ Answer/Explanation
Solution

(a)

The student should circle one pair of the inward-pointing horizontal sections opposite each other on the DNA diagram. These represent the complementary base pairs (A-T or C-G).

(b) 21%

Explanation:

According to Chargaff’s rules in DNA:

  • Adenine (A) pairs with Thymine (T), so if T is 29%, then A must also be 29%.
  • This accounts for 58% of the bases (29% A + 29% T).
  • The remaining 42% must be equally divided between Cytosine (C) and Guanine (G) since they pair together.
  • Therefore, G = 42% ÷ 2 = 21%.

(c)(i) gene

A gene is a segment of DNA that contains the instructions for making a specific protein.

(c)(ii)

Explanation of protein synthesis:

  1. Transcription: In the nucleus, the DNA code for a specific gene is copied into messenger RNA (mRNA). This process is called transcription.
  2. mRNA processing: The mRNA molecule is modified and then leaves the nucleus through nuclear pores.
  3. Translation: The mRNA attaches to a ribosome in the cytoplasm. Transfer RNA (tRNA) molecules bring specific amino acids to the ribosome based on the mRNA code.
  4. Peptide bond formation: The ribosome joins the amino acids together with peptide bonds, forming a polypeptide chain.
  5. Protein folding: The polypeptide chain folds into its specific three-dimensional shape to become a functional protein.
  6. Energy requirement: The entire process requires energy in the form of ATP.

(c)(iii) Two types of cell membrane proteins:

  1. Carrier proteins: These help transport molecules across the cell membrane, often against concentration gradients (active transport).
  2. Receptor proteins: These detect and respond to chemical signals from outside the cell, triggering specific cellular responses.
  3. Other possible answers: Channel proteins, enzyme proteins, cell recognition proteins, or structural proteins.
Question 3

Topic: 10.1

(a) Antibiotic resistance is an increasing problem worldwide.

Erythromycin is an antibiotic. Fig. 3.1 shows the daily doses of erythromycin per 1000 people over a 13-year period. The number of bacterial infections resistant to erythromycin per 1000 people is also shown.

 

(i) Calculate the percentage change in the number of bacterial infections resistant to erythromycin per 1000 people between 1993 and 1995. Give your answer to two significant figures.

(ii) Describe the data shown in Fig. 3.1.

(iii) Suggest reasons for the change in the number of bacterial infections resistant to erythromycin from 1993 to 1995 shown in Fig. 3.1.

(iv) Explain how bacteria become resistant to antibiotics.

(b) (i) Bacteria are prokaryotes. State two features of all prokaryotes.

(ii) Some bacteria have a flagellum. State the function of a flagellum.

▶️ Answer/Explanation
Solution

(a)(i) -44%

Explanation: From the graph, in 1993 there were 180 resistant infections per 1000 people, and in 1995 there were 100. The percentage change is calculated as: ((100-180)/180) × 100 = -44.44%, which rounds to -44% to two significant figures. The negative sign indicates a decrease in resistant infections.

(a)(ii)

Explanation: The data shows several key patterns: First, erythromycin usage peaked in 1989 at 2.8 doses per 1000 people. Second, resistant infections weren’t recorded until 1991. Third, resistant infections peaked in 1993 at 180 per 1000 people, four years after the peak in antibiotic usage. Finally, both antibiotic usage and resistant infections showed a declining trend after their respective peaks.

(a)(iii)

Explanation: The decrease in resistant infections from 1993-1995 could be due to: 1) Reduced erythromycin usage leading to less selection pressure for resistant strains, 2) Development of new antibiotics or alternative treatments, 3) Increased public awareness about proper antibiotic use, 4) Improved hygiene and infection control measures in healthcare settings, or 5) Implementation of policies restricting unnecessary antibiotic prescriptions.

(a)(iv)

Explanation: Bacteria develop resistance through several mechanisms: First, random mutations occur in bacterial DNA that may confer resistance. Second, bacteria can acquire resistance genes through horizontal gene transfer via plasmids. Third, when exposed to antibiotics, resistant bacteria survive while non-resistant ones die. Fourth, the resistant bacteria reproduce, passing on their resistance genes. Finally, over time, the resistant strains become dominant in the population through natural selection.

(b)(i)

Explanation: Two key features of all prokaryotes are: 1) They lack a membrane-bound nucleus – their DNA is free in the cytoplasm. 2) They have a cell wall made of peptidoglycan that maintains cell shape and provides protection.

(b)(ii) Movement.

Explanation: The flagellum is a whip-like appendage that rotates to propel the bacterium through liquid environments. This movement helps bacteria find nutrients, avoid harmful substances, or reach favorable environments. The rotation can be clockwise or counterclockwise, allowing for directional movement.

Question 4

Topic: 19.2

(a) A student recorded the information about an aquatic habitat.

Fig. 4.1 shows the student’s notes.

(i) Construct a food web to show the feeding relationships described in Fig. 4.1.

Do not draw the organisms.

(ii) Complete Table 4.1 using the information in Fig. 4.1 by identifying the names of the missing trophic levels and one organism at each different trophic level.

(iii) Outline how the energy in the primary consumers in this aquatic food web is used to produce biomass in the secondary consumers.

(iv) Humans also eat salmon. Predict the impact on the feeding relationships shown in Fig. 4.1 of overharvesting of salmon.

(b) Describe what is meant by the term decomposer.

(c) Animals such as salmon can be farmed for meat. Explain why it is more energy efficient for humans to eat crop plants than to eat livestock that have been fed on crop plants.

▶️ Answer/Explanation
Solution

(a)(i)

The food web should show:

  • aquatic plants → midges → stoneflies → salmon → herons
  • aquatic plants → mayflies → salmon → herons
  • aquatic plants → freshwater shrimps → salmon → herons

Explanation: A food web illustrates all the feeding relationships in an ecosystem. The arrows should point from the food source to the consumer. In this case, aquatic plants are the producers at the base, with multiple pathways leading up to the top predator (herons).

(a)(ii)

Name of the trophic levelOrganism in Fig. 4.1
produceraquatic plants
primary consumermayflies / midges / freshwater shrimps
secondary consumersalmon / stoneflies
tertiary consumersalmon / herons
quaternary consumerherons

Explanation: Trophic levels represent the position an organism occupies in a food chain. Producers (aquatic plants) are at the base, followed by primary consumers (herbivores), then secondary consumers (carnivores that eat herbivores), and so on up to the top predators.

(a)(iii)

Energy transfer occurs through the following process:

  • Primary consumers (like mayflies) obtain chemical energy from eating aquatic plants
  • When secondary consumers (like salmon) eat primary consumers, they gain some of this energy
  • Energy is used for respiration, growth, reproduction, and other life processes
  • Only about 10% of the energy is transferred between trophic levels
  • The remaining energy is lost as heat, in undigested material, or used for metabolism

Explanation: Energy flow through ecosystems is inefficient. Most energy is lost at each transfer, which is why food chains rarely have more than 4-5 trophic levels. The energy that isn’t lost is incorporated into the biomass of the secondary consumers.

(a)(iv)

Overharvesting salmon would likely cause:

  • Decrease in heron population (less food available)
  • Increase in mayflies, stoneflies, and freshwater shrimps (reduced predation)
  • Possible decrease in midges (if stoneflies increase)
  • Variable impact on aquatic plants (may increase or stay stable)

Explanation: Removing a key predator like salmon disrupts the entire food web. Prey species may increase dramatically, which can then affect other species they interact with. This demonstrates the interconnectedness of ecosystems.

(b) A decomposer is an organism that breaks down dead organic matter and waste products, returning nutrients to the ecosystem.

Explanation: Decomposers like bacteria and fungi play a crucial role in nutrient cycling. They break down complex organic compounds into simpler forms that plants can reuse, completing the nutrient cycle.

(c) It’s more energy efficient because:

  • Energy transfer between trophic levels is only about 10% efficient
  • Eating plants directly means accessing more of the original solar energy
  • Less energy is lost to respiration, movement, and heat in plant-based diets
  • Livestock production requires more land, water, and resources

Explanation: The “10% rule” of energy transfer explains why shorter food chains are more efficient. Each time energy moves up a trophic level, 90% is lost. Therefore, eating producers (plants) gives humans access to more of the original energy captured through photosynthesis.

Question 5

Topic: 2.1

Fig. 5.1 is a photomicrograph of a structure found in animal and plant cells.

 

(a) State the name and function of the cell structure labelled A in Fig. 5.1.

(b) State the two pieces of information needed to calculate the actual length of cell structure A in Fig. 5.1.

(c) The actual length of cell structure A is 0.00075 mm.

Convert this value to micrometres (µm).

▶️ Answer/Explanation
Solution

(a) Name: mitochondrion; Function: aerobic respiration

Explanation: The structure labelled A is identified as a mitochondrion based on its characteristic appearance in the photomicrograph. Mitochondria are often described as the “powerhouses” of the cell because their primary function is to carry out aerobic respiration. This process involves breaking down glucose molecules in the presence of oxygen to produce ATP (adenosine triphosphate), which serves as the main energy currency for cellular activities. The inner membrane of mitochondria is highly folded into cristae, which increases the surface area for the enzymes involved in respiration.

(b) 1. Length of the cell structure in the image/picture/photomicrograph/diagram; 2. Magnification

Explanation: To calculate the actual size of a structure from a micrograph, you need two key pieces of information. First, you must measure the length of the structure as it appears in the image (usually in millimeters). Second, you need to know the magnification factor of the image, which tells you how many times larger the image is compared to the actual specimen. The actual size can then be calculated using the formula: Actual size = Image size ÷ Magnification. Without either of these values, the calculation cannot be performed.

(c) 0.75 µm

Explanation: To convert millimeters to micrometers, we use the conversion factor that 1 millimeter equals 1000 micrometers. Therefore, we multiply the given value by 1000: 0.00075 mm × 1000 = 0.75 µm. This conversion is important in microscopy as most cellular structures are measured in micrometers, which are more appropriate for their scale. The mitochondrion’s size of 0.75 µm is typical for these organelles, which usually range from 0.5 to 1.0 µm in diameter and up to 10 µm in length.

Question 6

Topic: 14.5

(a) A student investigated plant growth responses in roots and shoots. They used this method:

  • Damp cotton wool was placed in two Petri dishes.
  • Three bean seedlings were attached to the cotton wool in each Petri dish.
  • Each seedling was orientated so that the roots pointed in a different direction in each Petri dish.
  • Petri dish 1 was kept on its side in a fixed position.
  • Petri dish 2 was kept on its side and rotated constantly.
  • Both Petri dishes were kept in the dark.
  • Both Petri dishes were kept in these conditions for two days.
  • After two days the seedlings were observed.

Fig. 6.1 is a diagram of the apparatus.

Fig. 6.2 shows the seedlings after two days.

 

(i) Describe the pattern of growth of the bean roots and shoots in Petri dish 1 shown in Fig. 6.2.

(ii) State the name of the growth responses observed in the bean roots and shoots.

(iii) Explain how auxin causes the difference in the pattern of growth shown by the shoots of seedlings B and E shown in Fig. 6.2.

(b) Seeds require oxygen and water to germinate.

(i) State one other environmental condition that affects germination.

(ii) Suggest why oxygen and water are required for germination.

▶️ Answer/Explanation
Solution

(a)(i) In Petri dish 1, all roots grew downward (regardless of their initial orientation) showing positive gravitropism, while all shoots grew upward showing negative gravitropism.

Explanation: The fixed position allowed gravity to consistently influence growth. Roots naturally grow toward gravity (positive gravitropism) to anchor the plant and access water/minerals, while shoots grow away from gravity (negative gravitropism) to reach light for photosynthesis.

(a)(ii) Gravitropism (or geotropism).

Explanation: Gravitropism is the growth response of plant parts to gravity. The term “geotropism” is also acceptable though less commonly used in modern biology. This directional growth is crucial for plant survival and orientation.

(a)(iii) Auxin is produced in the shoot tip and moves by diffusion. In seedling B (fixed position), auxin accumulates on the lower side due to gravity, causing more cell elongation on that side and making the shoot curve upward. In seedling E (rotating dish), auxin distribution remains equal on all sides as the rotation counters gravity’s effect, resulting in straight growth.

Detailed explanation:
1. Auxin is synthesized in the shoot apical meristem and moves downward through the plant.
2. In a stationary plant (B), gravity causes auxin to accumulate on the lower side of the shoot.
3. Higher auxin concentration promotes cell elongation on the lower side, causing upward curvature.
4. In the rotating dish (E), constant movement prevents sustained auxin accumulation on any one side.
5. Equal auxin distribution leads to uniform cell elongation and straight growth.
6. This demonstrates how plants use auxin distribution to respond to environmental stimuli like gravity.

(b)(i) Temperature.

Explanation: Temperature affects enzyme activity during germination. Most seeds have an optimal temperature range (typically 20-30°C) where germination occurs most efficiently. Too low temperatures slow metabolic processes, while too high can damage cellular components.

(b)(ii) Oxygen is needed for aerobic respiration to provide energy for growth, while water is required to activate enzymes, soften the seed coat, and serve as a medium for metabolic reactions.

Detailed explanation:
1. Oxygen: Essential for aerobic respiration which produces ATP needed for cell division and growth during germination. The emerging seedling requires energy before it can photosynthesize.
2. Water:
– Rehydrates dry seed tissues, activating metabolic processes
– Softens the seed coat to allow radicle emergence
– Acts as solvent for enzyme reactions and transport of nutrients
– Provides turgor pressure for cell expansion
– Helps break down stored food reserves into usable molecules

Question 7

Topic: 5.1

Complete the sentences about enzymes by writing a suitable word or phrase in each of the spaces provided.

Enzymes are involved in chemical digestion which produces small …… molecules that can be absorbed into the blood.

Two examples of protease enzymes are pepsin and trypsin. Pepsin is produced by the …… and requires acidic conditions. These conditions are created by the release of …… , which provides the optimum pH for pepsin activity and also kills harmful ……

The …… produces trypsin which breaks down protein in …… pH conditions. These conditions are created by a substance called …… , which neutralises the gastric juices and also has an important role in the …… of fats and oils.

▶️ Answer/Explanation
Solution

Completed sentences with explanations:

1. soluble – Enzymes break down large food molecules into smaller, soluble molecules that can be absorbed through the intestinal wall into the bloodstream. This solubility is crucial for nutrient transport throughout the body.

2. stomach – Pepsin is produced by the stomach’s gastric glands as an inactive precursor called pepsinogen, which is then activated by the acidic environment in the stomach.

3. hydrochloric acid – The stomach lining secretes hydrochloric acid (HCl) which creates the strongly acidic environment (pH 1.5-2) needed for pepsin activation and optimal activity.

4. microorganisms / pathogens – The hydrochloric acid serves a dual purpose: it provides the acidic conditions pepsin needs while also killing most harmful bacteria and other pathogens that might be present in ingested food.

5. pancreas / small intestine – Trypsin is produced by the pancreas (as trypsinogen) and released into the small intestine where it’s activated and works on protein digestion.

6. alkaline – Unlike pepsin, trypsin works best in alkaline conditions (pH 7.5-8.5), which is why it’s active in the small intestine rather than the stomach.

7. bile / bile salts – Bile, produced by the liver and stored in the gallbladder, neutralizes the acidic chyme from the stomach, creating the alkaline environment trypsin needs.

8. emulsification – Bile salts emulsify fats, breaking large fat globules into smaller droplets, increasing the surface area for lipase enzymes to work on during fat digestion.

Key concepts: This question covers the complete digestive process involving enzymes, highlighting how different parts of the digestive system create specific pH conditions optimal for different enzymes. It shows the progression from stomach (acidic) to small intestine (alkaline) environments and how the body produces complementary substances (like bile) to facilitate this transition and enhance digestion.

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