DSE Biology Diagnostic: Plant Biology

Unit Test 1: Light-Dependent vs Light-Independent Reactions

Question

A plant is illuminated with light of wavelength 680 nm. Chloroplasts are isolated from this plant and treated with the herbicide DCMU, which blocks electron transfer between photosystem II and the plastoquinone pool.

(a) Explain the role of photosystem II in the light-dependent reaction. In your answer, describe what happens to water molecules and how ATP and NADPH are affected. [4 marks]

(b) With DCMU present, predict what happens to (i) oxygen production, (ii) ATP synthesis, and (iii) NADP$^{+}$ reduction. Explain each prediction. [3 marks]

(c) Despite the DCMU treatment, if an artificial electron donor (such as DCPIP in its reduced form) is supplied to the electron transport chain after the blockage point, NADP$^{+}$ reduction can resume. Explain why. [2 marks]

---

Worked Solution

(a) Photosystem II (PSII) contains a primary pigment (P680) that absorbs light energy at 680 nm. This energy excites electrons in P680 to a higher energy level. The excited electrons are passed along the electron transport chain (via plastoquinone, cytochrome complex, and plastocyanin) to photosystem I.

Meanwhile, photolysis of water occurs at PSII: $2H_{2}O \to 4H^{+} + 4e^{-} + O_{2}$. The electrons from water replace those lost from P680.

As electrons pass through the electron transport chain, they release energy that is used to pump $H^{+}$ ions from the stroma into the thylakoid space, creating a proton gradient. $H^{+}$ ions flow back through ATP synthase, driving the synthesis of ATP (photophosphorylation).

At photosystem I, electrons are re-excited and transferred via ferredoxin to NADP$^{+}$, reducing it to NADPH (with the addition of $H^{+}$ from the stroma).

(b) With DCMU blocking electron transfer from PSII to the plastoquinone pool:

• (i) Oxygen production stops: DCMU prevents the acceptance of electrons from PSII, so the electron transport chain cannot operate. The water-splitting mechanism at PSII halts because there is no way to replace the excited electrons from P680.
• (ii) ATP synthesis stops or is severely reduced: Without electron flow through the chain, no proton gradient is established across the thylakoid membrane, so photophosphorylation cannot occur.
• (iii) NADP$^{+}$ reduction stops: Without electrons reaching PSI (since the electron supply from PSII is blocked), there are no electrons available to reduce NADP$^{+}$ to NADPH.

(c) An artificial electron donor supplies electrons directly to the electron transport chain after the DCMU blockage point. These electrons can then continue through the chain to PSI, where they can be excited and ultimately used to reduce NADP$^{+}$ to NADPH. The bypass restores electron flow downstream of the blockage.

---

Unit Test 2: Transpiration Mechanism and Factors

Question

A potted plant was placed under different environmental conditions, and the rate of water uptake was measured using a potometer.

Condition	Relative rate of water uptake
Still air, $20^{\circ}C$, normal humidity	1.0
Moving air (fan), $20^{\circ}C$, normal humidity	1.8
Still air, $30^{\circ}C$, normal humidity	1.5
Still air, $20^{\circ}C$, high humidity	0.5

(a) Explain the mechanism of water movement through the xylem from the root to the leaf, referring to the transpiration pull and the cohesion-tension theory. [4 marks]

(b) Using the data in the table, explain why moving air increases the rate of water uptake. [2 marks]

(c) Explain why high humidity decreases the rate of water uptake, using the concept of water potential. [2 marks]

---

Worked Solution

(a) The transpiration pull mechanism:

1. Water evaporates from the spongy mesophyll cell walls inside the leaf and diffuses out through the stomata (transpiration).
2. This evaporation creates a water potential gradient: the water potential inside the leaf is lower (more negative) than in the xylem.
3. Water molecules are drawn up the xylem vessels from the roots to replace the water lost.
4. The cohesion-tension theory explains how this pull is transmitted: water molecules form hydrogen bonds with each other (cohesion), creating a continuous column of water. As water is pulled upwards at the top, this tension is transmitted down the entire water column due to the strong cohesive forces between water molecules and the adhesive forces between water and the xylem wall.

The result is a negative pressure (tension) in the xylem that pulls water upwards from the roots.

(b) Moving air (wind) carries away water vapour that has just diffused out of the stomata, maintaining a steep water vapour concentration gradient between the moist air spaces inside the leaf and the drier air outside. A steeper gradient increases the rate of diffusion of water vapour out of the leaf, increasing transpiration and hence water uptake.

(c) At high humidity, the water vapour concentration gradient between the inside of the leaf and the external air is reduced (because the air outside already contains more water vapour). The water potential difference between the leaf air spaces and the atmosphere is smaller, so water vapour diffuses out more slowly. This reduces the transpiration rate and therefore the rate of water uptake.

---

Unit Test 3: Xylem vs Phloem Structure and Translocation

Question

(a) Compare the structure of xylem vessels and phloem sieve tube elements. Your answer should include four structural differences. [4 marks]

(b) Aphids feed by inserting their stylets into phloem sieve tubes. When an aphid's stylet is severed, phloem sap continues to exude from the cut end. Explain this observation in terms of the mass flow hypothesis (pressure flow hypothesis). [4 marks]

(c) A ring of bark (including phloem) is removed from around the entire circumference of a tree trunk (ring-barking). Explain why the tree eventually dies above the ring-barked region. [3 marks]

---

Worked Solution

(a) Four structural differences:

Feature	Xylem Vessels	Phloem Sieve Tube Elements
Cell contents at maturity	Dead (no cytoplasm, no nucleus)	Living (thin layer of cytoplasm, no nucleus)
Cell walls	Thickened with lignin, waterproof	Thin, cellulose cell walls (no lignin)
End walls	End walls broken down, forming continuous tubes	End walls remain as sieve plates with pores
Transport direction	One-way (upward only)	Two-way (upward and downward)

(b) According to the mass flow hypothesis, sugars (mainly sucrose) are actively loaded into phloem sieve tubes at a source (e.g. leaf). This lowers the water potential inside the sieve tube, causing water to enter by osmosis from the xylem and surrounding cells. This creates a high hydrostatic pressure at the source end.

At the sink (e.g. root), sugars are actively unloaded, raising the water potential inside the sieve tube. Water leaves by osmosis, reducing the hydrostatic pressure at the sink end.

This pressure gradient drives the bulk flow of phloem sap from source to sink. When the aphid's stylet is severed, the high hydrostatic pressure inside the sieve tube forces phloem sap to exude from the cut.

(c) Ring-barking removes the phloem, which is responsible for transporting organic nutrients (mainly sucrose) from the leaves (source) to the roots and other parts below the ring (sinks). Without phloem, sugars produced by photosynthesis cannot reach the roots. The roots are unable to respire (no energy supply), cannot absorb water and minerals, and eventually die. Without a functioning root system, the tree above the ring cannot obtain water and minerals, leading to its death.

---

Integration Test 1: Photosynthesis Limiting Factors and Experimental Design

Question

An investigation was carried out to determine the effect of light intensity and carbon dioxide concentration on the rate of photosynthesis in an aquatic plant (Elodea). The number of oxygen bubbles produced per minute was recorded.

Light intensity (arbitrary units)	Rate at 0.04% CO$_{2}$	Rate at 0.10% CO$_{2}$
5	5	12
10	9	22
20	14	30
40	16	33
60	16	34

(a) Explain why the rate of photosynthesis plateaus at high light intensity in both CO$_{2}$ concentrations. [2 marks]

(b) At a light intensity of 40 units, the rate at 0.10% CO$_{2}$ is approximately double the rate at 0.04% CO$_{2}$. Explain this observation. [2 marks]

(c) This experiment uses oxygen bubble count as a measure of photosynthesis rate. Evaluate the reliability of this method and suggest one improvement. [3 marks]

(d) The plant was then transferred to conditions of high light intensity and 0.10% CO$_{2}$, and the temperature was gradually increased from $20^{\circ}C$ to $45^{\circ}C$. Describe and explain the expected change in photosynthesis rate. [4 marks]

---

Worked Solution

(a) At high light intensity, light is no longer the limiting factor. Even though more light energy is available, the rate cannot increase because another factor (in this case, CO$_{2}$ concentration) becomes limiting. The light-dependent reactions are producing ATP and NADPH faster than the light-independent reactions can use them, because there is insufficient CO$_{2}$ for carbon fixation by RuBisCO.

(b) At 0.04% CO$_{2}$, CO$_{2}$ is the limiting factor at this light intensity. At 0.10% CO$_{2}$, more CO$_{2}$ is available for the light-independent reactions (Calvin cycle), so RuBisCO can catalyse more carboxylation reactions, producing more GP and subsequently more TP, leading to a higher overall rate of photosynthesis. The higher CO$_{2}$ concentration raises the point at which CO$_{2}$ becomes limiting.

(c) Limitations of bubble counting:

• Bubble size is not constant, so the number of bubbles does not directly correspond to the volume of oxygen produced.
• Small bubbles may be missed or miscounted.
• The method is subjective and prone to human error.

Improvement: Collect the oxygen gas over water in a graduated cylinder or gas syringe to measure the volume of oxygen produced per unit time, which is a more quantitative and reliable measure.

(d) As temperature increases from $20^{\circ}C$ to approximately $25$--$30^{\circ}C$, the rate of photosynthesis increases because the kinetic energy of molecules increases, leading to more frequent enzyme-substrate collisions (particularly RuBisCO catalysing CO$_{2}$ fixation).

Above approximately $30^{\circ}C$ (the optimum), the rate decreases because the high temperature causes enzymes (especially RuBisCO and other Calvin cycle enzymes) to denature. The tertiary structure of these enzymes unfolds, changing the shape of the active site and reducing their catalytic activity. At $45^{\circ}C$, significant denaturation has occurred, and the photosynthesis rate may be very low.

---

Integration Test 2: Plant Hormones and Phototropism

Question

A young seedling is exposed to unilateral (one-sided) light. The shoot bends towards the light (positive phototropism).

(a) Describe the role of auxin (IAA) in causing the shoot to bend towards the light. [4 marks]

(b) A student proposes that auxin is destroyed on the illuminated side of the shoot. Evaluate this proposal with reference to the Cholodny-Went hypothesis. [3 marks]

(c) If the same experiment is performed on a root instead of a shoot, the root bends away from the light. Explain this difference in response despite both organs receiving the same distribution of auxin. [4 marks]

---

Worked Solution

(a) Auxin (indoleacetic acid, IAA) is produced in the shoot tip and is transported laterally (from the illuminated side to the shaded side) when the shoot receives unilateral light.

The higher concentration of auxin on the shaded side stimulates cell elongation in the zone of elongation. Auxin does this by activating proton pumps in the cell membrane, which lowers the pH of the cell wall. This activates expansin enzymes that break cross-links between cellulose microfibrils, loosening the cell wall. The cells on the shaded side elongate more than those on the illuminated side, causing the shoot to bend towards the light.

(b) The Cholodny-Went hypothesis states that auxin is redistributed (transported laterally) from the illuminated side to the shaded side, rather than being destroyed on the illuminated side.

Evidence against the destruction hypothesis: if auxin were destroyed, the total amount of auxin in the shoot would decrease. However, experiments using agar blocks to collect auxin from illuminated and shaded sides show that the total auxin remains approximately the same; it is simply redistributed so that the shaded side has a higher concentration. The student's proposal is incorrect.

(c) In roots, a higher auxin concentration inhibits cell elongation (the opposite of its effect in shoots). Therefore, the shaded side of the root (which has a higher auxin concentration) experiences reduced elongation, while the illuminated side (lower auxin) elongates more. This differential elongation causes the root to bend away from the light (negative phototropism).

This difference arises because root and shoot cells have different receptor sensitivities and different responses to the same auxin concentration: shoot cells are stimulated to elongate by higher auxin, while root cells are inhibited.

---

Integration Test 3: Transport in Plants -- Multi-Concept Synthesis

Question

A plant is grown in a solution containing radioactive carbon-14 ($^{14}C$) in the form of $^{14}CO_{2}$. The plant is exposed to light and allowed to photosynthesise for several hours.

(a) Describe the pathway by which the radioactive carbon becomes incorporated into sucrose, starting from CO$_{2}$ entering the leaf. Name the enzyme that catalyses the fixation of CO$_{2}$. [5 marks]

(b) The radioactive sucrose is then transported in the phloem from the leaf to a developing fruit. Describe the mechanism by which sucrose is loaded into the phloem at the source. [3 marks]

(c) A researcher traces the radioactive carbon and finds that some of it appears in the xylem sap as amino acids. Explain how the radioactive carbon could have been transferred from the phloem to the xylem. [3 marks]

---

Worked Solution

(a) Pathway of carbon from CO$_{2}$ to sucrose:

1. CO$_{2}$ diffuses through the stomata into the air spaces of the leaf, then through the cell walls and cell surface membrane of mesophyll cells.
2. In the stroma of the chloroplast, CO$_{2}$ combines with ribulose bisphosphate (RuBP), a 5-carbon compound. This reaction is catalysed by the enzyme RuBisCO (ribulose bisphosphate carboxylase/oxygenase).
3. The unstable 6-carbon intermediate immediately splits into two molecules of glycerate-3-phosphate (GP), a 3-carbon compound.
4. GP is reduced to triose phosphate (TP) using ATP and NADPH from the light-dependent reactions.
5. Most TP molecules are used to regenerate RuBP (using ATP), but some TP molecules leave the Calvin cycle and are converted via various intermediates into sucrose (a disaccharide of glucose and fructose). The radioactive $^{14}C$ is now incorporated into the sucrose molecules.

(b) Sucrose loading into the phloem at the source:

1. Sucrose is actively transported from mesophyll cells into companion cells, then into sieve tube elements.
2. This accumulation of sucrose lowers the water potential inside the sieve tubes.
3. Water enters the sieve tubes by osmosis from the xylem and surrounding cells, creating a high hydrostatic pressure at the source end, initiating mass flow.

(c) Transfer of radioactive carbon from phloem to xylem:

The radioactive carbon (originally in sucrose) can be transferred from the phloem to the xylem by a process called phloem-xylem exchange (or lateral transfer). Along the transport pathway, sucrose may be unloaded from the phloem into surrounding parenchyma cells. Some of this sucrose can be converted into amino acids (via transamination, using nitrogen from the soil absorbed by roots and transported in the xylem). These amino acids can then be loaded into the xylem for transport to other parts of the plant. Thus, the radioactive carbon from the original $^{14}CO_{2}$ appears in xylem sap as amino acids.