A Level H1 Biology Plant Biology Quiz

A-Level Biology H1 Quiz - Plant Biology – Answers and Marking Scheme

Total marks: 60

Section A – Structured Questions

1.

Answer: Chloroplast contains thylakoids/grana, while mitochondrion does not; OR Chloroplast has three membranes (outer, inner, thylakoid) whereas mitochondrion has two; OR Chloroplast has internal membrane structures (thylakoids) whereas mitochondrion does not.
Any one correct structural difference visible under EM – accept thylakoids vs. cristae (though note both have internal membranes, so distinguish properly).
[1] for a clearly stated difference.

2.

Answer: Thylakoid membrane / thylakoid / grana (accept ‘thylakoid space’ as location).
[1]

3.

Answer: Many chloroplasts increase the surface area for absorption of light / contain more chlorophyll / more photosystems. This allows a higher rate of light‑dependent reactions, producing more ATP and NADPH for the Calvin cycle, thus increasing overall photosynthesis.
[2] – 1 mark for surface area/chlorophyll, 1 mark for linking to increased light‑dependent reaction products.

4.

Answer: 6 CO₂ + 6 H₂O → C₆H₁₂O₆ + 6 O₂
[2] – 1 mark for correctly placing carbon dioxide and water on left, 1 mark for glucose and oxygen on right. (Order of water and CO₂ may be reversed; accept balanced equation.)

5.

(a) Ribulose‑1,5‑bisphosphate carboxylase/oxygenase. [1]
(b) RuBisCO has a low affinity for CO₂ / catalyses a slow reaction. A high concentration ensures that the rate of carbon fixation is sufficient to meet the plant’s metabolic needs.
[2] – 1 for low affinity/slow turnover, 1 for linking to carbon fixation rate.

6.

(a) Negative values indicate that respiration is releasing more CO₂ than photosynthesis is consuming / respiration rate exceeds photosynthesis rate. The leaf is a net producer of CO₂ at these low light intensities.
[2] – 1 for stating respiration > photosynthesis, 1 for net CO₂ output.
(b) 40 arbitrary units. [1] – where net CO₂ uptake = 0.
(c) Light is no longer the limiting factor; another factor such as CO₂ concentration, temperature, or chlorophyll content is now limiting the rate.
[1] – any valid limiting factor.

7.

(a) Xylem. [1]
(b) Evidence: (any one)

• When a xylem vessel is cut, the column of water does not recede (xylem sap remains under tension);
• The diameter of tree trunks shrinks during the day when transpiration is high (tension pulls in the walls);
• If the water column is broken by cavitation, a ‘clicking’ sound is produced and the vessel becomes non‑functional;
• The column of water does not break when a plant is cut.
  [1] for one credible piece of evidence.

8.

(a) The phloem sap continues to ooze/flow from the cut style for an extended period (not just a brief drip). [1]
(b) This indicates that the phloem sap is under positive pressure. In the mass‑flow hypothesis, solute loading at the source lowers water potential; water enters by osmosis, creating high hydrostatic pressure that pushes sap towards the sink. The continued oozing suggests that pressure is maintained.
[2] – 1 for pressure, 1 for linking to osmotically generated flow.

9.

(a) Nitrate ions are required for the synthesis of amino acids, proteins, nucleic acids (DNA, RNA), and chlorophyll. (Accept: essential for making proteins/nitrogenous compounds.) [1]
(b) Without nitrate ions, the plant cannot synthesise sufficient amino acids and proteins, which are necessary for cell division, enzyme production, and structural growth. This reduces the rate of growth, resulting in stunted appearance.
[2] – 1 for linking to protein synthesis, 1 for explaining consequence on growth.

10.

(a) B – anther, C – stigma, D – ovary. [3] – 1 per correct label.
(b) The anther (structure B) contains pollen sacs where pollen grains develop. The wall of the anther is thin and the epidermal layer may have a fibrous thickening that aids in dehiscence (splitting) to release pollen. It is positioned at the top of the filament where it can be easily contacted by pollinators or wind.
[2] – 1 for structure (pollen sacs, dehiscence mechanism), 1 for adaptation to pollination.

Section B – Data‑Based Questions

11.

(a) Wavelength / colour of light. [1]
(b) The bacteria are aerobic and move towards regions of higher oxygen concentration. The algae produce more oxygen in red and blue light because these wavelengths are absorbed most strongly by chlorophyll, leading to a higher rate of photosynthesis. Therefore, bacteria cluster where photosynthesis (and O₂ release) is greatest.
[2] – 1 for oxygen production by photosynthesis, 1 for linking red/blue absorption to high O₂.
(c) Use a green filter / illuminate the alga with green light only, and record the distribution of bacteria (they should remain more dispersed). [1]

12.

(a) As humidity increases, transpiration rate decreases. Under still air, it drops from 18 to 12 g m⁻² h⁻¹; under breeze, from 28 to 21 g m⁻² h⁻¹. [1] – must quote data or describe trend.
(b) Water vapour diffuses out of the stomata driven by the water vapour concentration gradient between the leaf’s intercellular spaces and the outside air. In a breeze, water vapour that accumulates near the leaf surface is blown away, maintaining a steeper concentration gradient. This increases the rate of diffusion of water vapour out of the leaf.
[2] – 1 for concentration gradient, 1 for effect of wind removing boundary layer.
(c) Predicted rate: ~12 g m⁻² h⁻¹ (or slightly lower, such as 10–12). At 95% humidity, the external air is almost saturated, so the water vapour concentration gradient is very small, reducing transpiration. The value would be lower than the rate at 80% humidity because the gradient is even smaller.
[2] – 1 for a sensible prediction, 1 for explanation relating to gradient/humidity.

13.

(a) As distance from the lamp increases, the rate of photosynthesis (bubbles per minute) decreases. OR Negative correlation. [1]
(b) As distance increases from 10 cm to 20 cm, the light intensity reaching the plant decreases (inverse square law). The lower light intensity reduces the rate of the light‑dependent reactions, so less ATP and NADPH are produced. Consequently, the Calvin cycle proceeds more slowly, and the overall rate of photosynthesis falls.
[2] – 1 for describing light intensity decrease, 1 for linking to reduced photochemical products.
(c) Use a brighter lamp with a heat‑absorbing screen (e.g., water filter), or move the lamp closer but use a fan to dissipate heat, or measure the rate under natural sunlight with a shade to control intensity.
[1] for a reasonable and safe alternative.

14.

(a) (16 – 3) / 3 × 100 = 433.3% (accept 433% or 430%) [1] – correct calculation and answer.
(b) The endosperm contains stored food reserves, mainly starch. During germination, the embryo secretes gibberellin, which stimulates the aleurone layer to produce amylase. Amylase hydrolyses starch into sugars, which are transported to the growing embryo and used for respiration and growth. This breakdown causes the dry mass of the endosperm to decrease.
[2] – 1 for hydrolysis of starch, 1 for role of gibberellin/amylase and utilisation by embryo.
(c) Respiration uses up some of the stored carbohydrates, releasing carbon dioxide and water, which are lost as gas. Also, some water is used in hydrolysis and later lost.
[1] for respiration as reason for mass loss.

Section C – Extended‑Response Questions

15.

(a) Photolysis is the light‑driven splitting of water molecules into protons (H⁺), electrons, and oxygen gas. The reaction occurs at photosystem II in the thylakoid membrane: 2 H₂O → 4 H⁺ + 4 e⁻ + O₂.
[2] – 1 for stating splitting of water, 1 for identifying products and location.
(b)

• The electrons are passed to the electron transport chain, replacing the electrons lost by chlorophyll at PSII.
• The protons (H⁺) accumulate in the thylakoid space, contributing to the proton gradient used in chemiosmosis for ATP synthesis.
• Oxygen is released as a by‑product into the atmosphere.
  [3] – 1 mark each for the use of electrons, protons, and oxygen.
  (c) Photolysis supplies the electrons that replace those lost from PSII. Without this replacement, the electron transport chain would stop, and non‑cyclic photophosphorylation could not continue. The proton gradient would also not be generated effectively, reducing ATP and NADPH production. Thus photolysis is essential for sustaining the light‑dependent reactions.
  [2] – 1 for electron replacement, 1 for linking to continuation of phosphorylation.

16.

(a) Diagram should show a circular/oval section with:

• Epidermis as a single outer layer, cells with thick outer wall and waxy cuticle.
• Cortex composed of several layers of thin‑walled parenchyma cells, with intercellular spaces.
• Vascular bundles arranged in a ring; each bundle with outer phloem and inner xylem, cambium between (for dicot).
• Pith in the centre, composed of parenchyma cells.
  All four labels correctly placed.
  [4] – 1 for each correctly labelled tissue; deduct if any major anatomical error for a dicot stem.
  (b)
  (i) Xylem vessels: transport water and dissolved mineral ions from the roots to the leaves; also provide mechanical support because of lignified walls.
  (ii) Phloem sieve tubes: transport organic solutes (mainly sucrose) from the source (photosynthetic tissue) to sinks (growing regions, storage organs).
  [2] – 1 for each function; accept correct additional details.

17.

(a) Protein content in wheat is controlled by a single gene with two alleles showing incomplete dominance / codominance (medium‑protein heterozygote).

• Parental phenotypes: high‑protein × low‑protein
• Genotypes: PP × pp
• Gametes: P and p
• F₁: all Pp, medium‑protein
• F₂ cross: Pp × Pp; gametes P, p and P, p; offspring: 1 PP (high), 2 Pp (medium), 1 pp (low).
  [4] – 1 for stating incomplete dominance/codominance, 1 for correct notation, 1 for gamete formation, 1 for correct F2 ratio.
  (b) The breeder could self‑pollinate individual F₂ high‑protein plants (PP or Pp?). However, all high‑protein plants are PP because the trait is completely homozygous for the dominant allele in this scenario – but careful: incomplete dominance means medium is Pp, high is PP. To test for true‑breeding, he could self‑fertilise the high‑protein plants and see if the offspring are all high‑protein (if heterozygous Pp, medium‑protein offspring would appear). Alternatively, he could cross the high‑protein plant with a low‑protein pp plant; if offspring are all medium‑protein, the high‑protein plant is PP; if half medium and half low, it is heterozygous (but in this case, high‑protein is PP only, so all high‑protein plants are true‑breeding). Actually, given the F₂ ratio: 1 PP:2 Pp:1 pp, the high‑protein are PP, so they are already homozygous. So identification is simply growing the seeds from those high‑protein plants and checking that all offspring are high‑protein.
  [2] – 1 for test‑cross or selfing method, 1 for expected outcome indicating homozygosity.
  Accepted answer: grow seeds from the high‑protein plants separately and see if all progeny are high‑protein; a test cross with pp producing all medium would confirm PP.

18.

(a) Stomatal closure reduces transpiration, so the plant loses less water. During drought, this conserves water in the tissues and prevents wilting and desiccation, helping the plant survive until water becomes available again.
[2] – 1 for reducing water loss, 1 for survival advantage.
(b)

• Disadvantage 1: Reduced CO₂ uptake – stomata are the main entry for carbon dioxide needed for the Calvin cycle. Prolonged closure reduces the rate of photosynthesis, limiting carbohydrate production for growth, storage, and energy.
• Disadvantage 2: Reduced transpiration also reduces the transpiration pull that drives water and mineral uptake from the roots. Without the upward movement of water, mineral ions (e.g., nitrates) are not transported effectively, which can lead to nutrient deficiencies and impaired metabolism.
  (Also accept reduced cooling effect, leading to possible overheating; or build‑up of oxygen in the leaf reducing photosynthesis.)
  [4] – 2 marks for each well‑explained disadvantage; 1 for stating impact on CO₂/photosynthesis, 1 for impact on mineral transport/water movement.

19.

Description of chloroplast structure and function: (6 marks)

• The chloroplast is bounded by a double membrane (outer and inner envelope). The inner membrane regulates the passage of substances in and out.
• Inside, the stroma is a fluid‑filled matrix containing enzymes (e.g., RuBisCO) for the Calvin cycle and the chloroplast’s own DNA and ribosomes.
• Embedded in the stroma is an extensive system of membranes called thylakoids. Thylakoids are flattened sacs stacked into grana (singular granum).
• The thylakoid membrane contains chlorophyll, other accessory pigments, and electron carriers. This is where the light‑dependent reactions occur – light absorption, ATP synthesis (via chemiosmosis), and NADP reduction.
• The thylakoid space (lumen) is the enclosed interior, where protons accumulate to create a proton gradient for ATP production.
• Stroma lamellae (intergranal lamellae) connect grana and contain photosystem I and ATP synthase, facilitating non‑cyclic electron flow.
  Marking: 1 mark for each labelled structure with its function, up to 6 marks. Typical answer might mention: outer/inner membrane (regulation), stroma (Calvin cycle), thylakoid membrane (light reactions, photosystems), grana (stacking for large surface area), thylakoid lumen (proton gradient), stroma lamellae (connectivity).

[6] – Award 1 mark for each accurate structure‑function pair; must include at least four distinct structures with linked functions to earn full marks.

20.

(a) Auxin (IAA – indole‑3‑acetic acid). [1]
(b) Auxin stimulates cell division and cell elongation in the stem tissues at the cut end. It promotes the differentiation of some parenchyma cells into root primordia (meristematic cells that develop into adventitious roots). Auxin also activates genes concerned with root development, increasing the success of vegetative propagation.
[2] – 1 for stimulating cell division/differentiation, 1 for specifically leading to root formation.
(c) Commercial use: (any one)

• Promotes fruit set in some plants (e.g., tomatoes) without pollination (parthenocarpy).
• Used as a selective herbicide (e.g., 2,4‑D kills broad‑leaved weeds).
• Prevents premature fruit drop in orchards.
• Used in tissue culture to promote callus and root formation.
  [1] for any valid commercial application.

END OF ANSWERS