Photosynthesis: exam questions

Answer: B – the light-dependent reactions occur on the thylakoid membranes and the Calvin cycle occurs in the stroma.

• light energy used to split / photolysis of water (1)
• hydrogen (from the water) combined with carbon dioxide (1)
• to form glucose / a fuel (that stores the hydrogen) (1)
• oxygen is released (into the atmosphere) (1)
• Max 3

• hydrolysis of ATP to ADP and Pi releases energy in one step / quickly (1)
• in small usable amounts / only where it is needed (1)
• photophosphorylation is the addition of phosphate to ADP (1)
• which needs energy / is an endergonic reaction, supplied by light (1)
• Max 3

• similarity: both make ATP / both involve excited electrons passing along an electron transport chain / both use chemiosmosis (1)
• non-cyclic uses photosystems I and II but cyclic uses photosystem I only (1)
• non-cyclic: electrons come from water, cyclic: electrons return to the same chlorophyll (1)
• non-cyclic produces reduced NADP but cyclic does not (1)
• non-cyclic produces oxygen (by photolysis) but cyclic does not (1)
• Max 4. At least one similarity and one difference are needed.

• GP increases and RuBP decreases in the dark (accept figures: GP 100 to 180, RuBP 50 to 10) (1)
• no ATP / reduced NADP is made in the dark (since the light-dependent reactions stop) (1)
• so GP cannot be reduced to GALP, so GP accumulates (1)
• carbon dioxide is still fixed by rubisco for a short time, so RuBP is still used up / GP is still made (1)
• RuBP is not regenerated (from GALP, which also needs ATP) (1)
• Max 4

• carbon dioxide combines with RuBP (1)
• catalysed by rubisco (1)
• to form two molecules of GP (1)
• GP reduced to GALP using ATP and reduced NADP (1)
• GALP used to make glucose / hexose sugars / regenerate RuBP (1)
• (carbon skeletons from) GP / GALP used for amino acids, lipids (accept glycerol / fatty acids) (1)
• sugars used to make polysaccharides / nucleic acids (accept starch, cellulose) (1)
• Max 5

• thylakoids stacked into grana give a large surface area for pigments / photosystems / electron carriers / ATP synthase (1)
• pigments in photosystems absorb a range of wavelengths of light (1)
• stroma contains enzymes / rubisco for the Calvin cycle (1)
• stroma surrounds the grana so ATP / reduced NADP made on thylakoids are used close by (1)
• small thylakoid space allows a proton gradient to be built up quickly (1)
• envelope controls what enters and leaves (1)
• chloroplast has own DNA / ribosomes to make some proteins (1)
• Max 4

• absorption spectrum shows the amount of light absorbed at each wavelength; the action spectrum shows the rate of photosynthesis (1)
• most light absorbed at 450 and 650 nm / least at 550 nm (1)
• rate of photosynthesis is highest where absorption is highest / the two vary together (1)
• so the light absorbed (by pigments) is used in photosynthesis / pigments such as chlorophyll absorb in the blue and red regions (1)
• rate slightly lower than the absorption at 650 nm / not all absorbed light is used in photosynthesis (accept any one comparison with figures) (1)
• Max 4

• Rf = distance moved by pigment ÷ distance moved by solvent front (1)
• 6.2 ÷ 8.0 = 0.78 (accept 0.775) (1)
• 2.4 ÷ 8.0 = 0.30 (1)
• the pigment with Rf 0.78 is more soluble (in the solvent) (1)
• Max 3 (calculation marks: correct working with the wrong answer is credited once)

• named plant (for example Cabomba / Elodea) cut under water, placed in a boiling tube with the cut end upwards (1)
• water bath at a range of temperatures (for example 10 to 40 °C) with a thermometer / set temperature each time (1)
• let the plant equilibrate / adjust to each new temperature (1)
• measure rate as number of bubbles / volume of oxygen per minute (accept volume of gas collected in a capillary / gas syringe) (1)
• light intensity controlled by keeping lamp at a fixed distance and a heat shield / LED lamp used (1)
• carbon dioxide supply controlled (for example same concentration of sodium hydrogencarbonate solution) (1)
• same plant / same length / mass of plant used each time (1)
• repeat at each temperature and calculate a mean (1)
• Max 6

• rate at 10 cm = 4.0 ÷ 10 = 0.40 cm³ min⁻¹ (1)
• rate at 20 cm = 0.10 cm³ min⁻¹ and at 40 cm = 0.025 cm³ min⁻¹ (accept 0.025 or 0.03) (1)
• light intensity is proportional to 1 ÷ distance², so doubling the distance reduces intensity to one quarter (1)
• the rate is proportional to light intensity here, so light is the limiting factor (accept rate falls to a quarter each time the distance doubles) (1)