Bioenergetics — Diagnostic Tests

Unit Tests

UT-1: Photosynthesis

Question: (a) Write the balanced symbol equation for photosynthesis. (b) Explain how the structure of a leaf is adapted for photosynthesis. (c) A leaf with surface area 50 \text{ cm^2 absorbs 2.5 \text{ mg of CO$_2$ per hour. Calculate the rate of photosynthesis in mg CO$_2$/cm$^2$/hour. (d) Explain why most plants appear green.

Solution:

(a) 6\text{CO_2 + 6\text{H_2\text{O \xrightarrow{\text{light} \text{C_6\text{H_{12}\text{O_6 + 6\text{O_2

(b) Leaf adaptations: (1) Broad, flat shape — maximises surface area for light absorption. (2) Thin — short diffusion distance for CO$_2$ and O$_2$. (3) Waxy cuticle — reduces water loss by evaporation. (4) Palisade mesophyll — cells at the top of the leaf, packed with chloroplasts, receive the most light. (5) Spongy mesophyll — air spaces allow gas circulation (CO$_2$ and O$_2$ diffusion). (6) Stomata — pores on the lower surface () that allow CO$_2$ to enter and O$_2$ to exit; controlled by guard cells. (7) Chlorophyll — green pigment in chloroplasts that absorbs red and blue light.

(c) Rate $= 2.5 / 50 = 0.05$ mg CO$_2$/cm$^2$/hour.

(d) Chlorophyll absorbs red and blue wavelengths of light but reflects green light. Since green light is not absorbed (it is reflected), this is the colour we perceive. The reflected green light reaches our eyes, and our brain interprets it as the colour green.

UT-2: Aerobic and Anaerobic Respiration

Question: (a) Write the word equations for aerobic respiration and anaerobic respiration in: (i) animal cells, (ii) yeast cells. (b) Calculate the volume of oxygen consumed by a human at rest if they respire aerobically at a rate that produces 360 L of CO$_2$ per hour. (c) Explain the difference between aerobic and anaerobic respiration in terms of: ATP yield, oxygen requirement, and products. (d) Explain why anaerobic respiration is important for yeast in bread-making and brewing.

Solution:

(a) Aerobic respiration: glucose + oxygen $\to$ carbon dioxide + water (+ ATP).

Anaerobic respiration: (i) Animal cells: glucose $\to$ lactic acid (+ ATP). (ii) Yeast cells (fermentation): glucose $\to$ ethanol + carbon dioxide (+ ATP).

(b) In aerobic respiration, the molar ratio of CO$_2$ produced to O$_2$ consumed is 1:1. Therefore, O$_2$ consumed $=$ CO$_2$ produced $= 360$ L/hour.

(c)	Feature	Aerobic	Anaerobic (animal)	Anaerobic (yeast)
ATP yield	High (~38 ATP per glucose)	Low (2 ATP per glucose)	Low (2 ATP per glucose)
Oxygen	Required	Not required	Not required
Products	CO$_2$ + H$_2$O	Lactic acid	Ethanol + CO$_2$
Speed	Slower (complete breakdown)	Faster (quick energy)	Slower than aerobic

(d) In bread-making, yeast undergoes anaerobic respiration (fermentation), producing CO$_2$ gas. The CO$_2$ gets trapped in the dough, causing it to rise (leavening). The ethanol evaporates during baking.

In brewing, yeast ferments sugars in the wort (malted barley liquid), producing ethanol (alcohol) and CO$_2$. The ethanol is the desired product in alcoholic beverages. The CO$_2$ can carbonate the drink or be released.

UT-3: Limiting Factors and Rate of Photosynthesis

Question: (a) Define the term “limiting factor”. (b) Explain how light intensity, CO$_2$ concentration, and temperature affect the rate of photosynthesis. (c) A farmer wants to increase crop yields in a greenhouse. Suggest three methods, explaining the biology behind each. (d) Explain why increasing temperature above $25^\circ$C does not continue to increase the rate of photosynthesis.

Solution:

(a) A limiting factor is the environmental factor that is in shortest supply (or furthest from its optimum), which limits the rate of a process. The rate cannot increase beyond the point where the limiting factor restricts it, regardless of improvements in other factors.

(b) Light intensity: Increasing light intensity increases the rate of photosynthesis (more energy for the light-dependent reactions) until another factor becomes limiting. At very high light, the rate plateaus because CO$_2$ or temperature is now limiting.

CO$_2$ concentration: CO$_2$ is a reactant in the light-independent reactions (Calvin cycle). Increasing CO$_2$ increases the rate until another factor limits. At very high CO$_2$Light or temperature becomes limiting.

Temperature: Increasing temperature increases the rate of enzyme-controlled reactions in photosynthesis (up to the optimum). Above the optimum ($\approx 25$—$30^\circ$C), enzymes begin to denature and the rate decreases.

(c) Three methods: (1) Artificial lighting — extends the hours of photosynthesis beyond daylight, increasing total glucose production. (2) CO$_2$ enrichment — pumping CO$_2$ into the greenhouse increases the concentration above atmospheric levels, making CO$_2$ less likely to be limiting. (3) Temperature control — heating in winter and ventilation in summer maintain the optimum temperature for photosynthetic enzymes.

(d) Above $25^\circ$C, the enzymes involved in photosynthesis (particularly RuBisCO in the Calvin cycle) begin to denature. The active site changes shape, reducing the enzyme’s ability to catalyse the reaction. Additionally, at higher temperatures, water loss through stomata increases. The guard cells close the stomata to conserve water, but this reduces CO$_2$ entry, further limiting photosynthesis.

---

Integration Tests

IT-1: Photosynthesis and Respiration Balance (with Ecology)

Question: A plant in a sealed container is exposed to light. The net oxygen production rate is measured at different light intensities:

Light intensity (arbitrary units)	0	50	100	200	400
Net O$_2$ production (mg/hr)	-20	-5	10	25	25

(a) Explain the negative values at low light intensity. (b) Calculate the compensation point (light intensity where net O$_2$ production is zero). (c) At what light intensity does photosynthesis become limited by another factor? Explain. (d) If the plant’s respiration rate remains constant at 20 mg O$_2$/hr, calculate the gross photosynthesis rate at light intensity 200.

Solution:

(a) At low light intensities, the rate of photosynthesis is less than the rate of respiration. The plant consumes more O$_2$ in respiration than it produces in photosynthesis, resulting in net O$_2$ consumption (negative values). The plant is a net consumer of O$_2$ and producer of CO$_2$.

(b) The compensation point is where net O$_2$ $= 0$. From the data, this occurs between light intensity 50 (net $= -5$) and 100 (net $= +10$). Interpolating: $50 + 50 \times 5/15 = 50 + 16.7 \approx 67$ arbitrary units.

(c) At light intensity 400, the net O$_2$ production is 25 mg/hr, the same as at 200. The rate has plateaued, meaning light is no longer the limiting factor. CO$_2$ concentration or temperature is now limiting.

(d) Gross photosynthesis $=$ net O$_2$ production $+$ respiration rate $= 25 + 20 = 45$ mg O$_2$/hr.

IT-2: Respiration and Exercise (with Organisation)

Question: An athlete runs a 400 m race in 50 seconds. During the race: (a) Explain why the athlete cannot rely solely on aerobic respiration. (b) The athlete’s breathing rate increases from 15 to 40 breaths per minute and heart rate from 70 to 180 bpm. Calculate the increase in oxygen delivery per minute if tidal volume increases from 0.5 L to 2.5 L per breath. (c) After the race, the athlete continues to breathe heavily for several minutes. Explain the biochemistry of the “oxygen debt”. (d) Calculate the total volume of oxygen debt if the athlete needs an extra 4 L of oxygen after the race to fully recover.

Solution:

(a) A 400 m race is a high-intensity activity. Aerobic respiration cannot supply ATP fast enough to meet the high energy demand of the muscles. The circulatory system cannot deliver oxygen quickly enough for aerobic respiration alone. The muscles supplement with anaerobic respiration, which produces ATP quickly but generates lactic acid as a by-product.

(b) Oxygen delivery per minute $=$ breathing rate $\times$ tidal volume. Before: $15 \times 0.5 = 7.5$ L/min. During race: $40 \times 2.5 = 100$ L/min. Increase $= 100 - 7.5 = 92.5$ L/min.

(c) During the race, anaerobic respiration produced lactic acid in the muscles. This causes muscle fatigue and cramp. After the race, the athlete needs extra oxygen to: (1) Break down the accumulated lactic acid into CO$_2$ and water in the liver (this requires oxygen). This process is called the “oxygen debt” or “excess post-exercise oxygen consumption” (EPOC). The heavy breathing continues until the lactic acid is cleared.

(d) The oxygen debt is 4 L. At the elevated breathing rate of 30 breaths/min (after race) with tidal volume 1.5 L: oxygen intake $= 30 \times 1.5 = 45$ L/min. Normal oxygen demand $\approx 7.5$ L/min. Extra oxygen available $= 45 - 7.5 = 37.5$ L/min. Time to repay debt $= 4/37.5 = 0.107$ min $\approx 6.4$ seconds. This seems too fast — in practice, only a fraction of the elevated oxygen intake goes to repaying the debt, so recovery takes several minutes.

IT-3: Bioenergetics in Ecosystems (with Ecology)

Question: A pond ecosystem has the following energy values (in kJ/m$^2$/year): sunlight reaching pond $= 2,000,000$; energy absorbed by producers $= 20,000$; energy transferred to herbivores $= 2000$; energy transferred to carnivores $= 200$. (a) Calculate the percentage of sunlight energy captured by producers. (b) Calculate the ecological efficiency at each trophic level transfer. (c) Explain why energy is lost between trophic levels. (d) Explain why food chains rarely have more than 4 or 5 trophic levels.

Solution:

(a) Percentage captured by producers $= 20,000 / 2,000,000 \times 100\% = 1\%$.

(b) Producer $\to$ herbivore: $2000/20,000 \times 100\% = 10\%$. Herbivore $\to$ carnivore: $200/2000 \times 100\% = 10\%$.

(c) Energy is lost between trophic levels because: (1) Respiration: Organisms use energy for metabolic processes (movement, maintaining body temperature, digestion). This energy is lost as heat. (2) Excretion and egestion: Not all food is digested and absorbed; some passes through as waste. (3) Incomplete consumption: Not all available biomass is eaten (bones, roots, etc.). (4) Excretion: Some absorbed energy is lost in urine. , only about 10% of energy is transferred to the next trophic level.

(d) With approximately 10% efficiency at each transfer, the energy available decreases exponentially: $100\% \to 10\% \to 1\% \to 0.1\% \to 0.01\%$. After 4—5 levels, the energy available is too small to support a viable population of predators. There is insufficient energy to sustain another trophic level.

Summary

The key principles covered in this topic are linked in the sub-pages above. Focus on understanding the definitions, applying the formulas or frameworks, and evaluating strengths and limitations of each approach.

Worked Examples

Worked examples demonstrating the application of key concepts are covered in the detailed sub-pages linked above.

Common Pitfalls

• Confusing terminology or concepts that appear similar but have distinct meanings.
• Overlooking key assumptions or boundary conditions that limit applicability.