IGCSE Biology 0610 Paper 4: Photosynthesis, Plant Transport and Transpiration

Level: Year 10-11 (IGCSE) | Subject: Chemistry 0620 | Time: 75 minutes | Total: 100 marks | Parts: 31

Paper at a glance

Paper	Time	Marks	Calculator
Paper 4 (Extended, structured)	1 h 15 min	75	Not allowed

Questions

Question 1 (16 marks) – The photosynthesis equation and leaf structure

(a) Write the balanced equation for photosynthesis and name two raw materials and two products. (3 marks)

(b) Name the organelle in which photosynthesis takes place and the pigment that absorbs light energy. (2 marks)

(c) Explain the role of light energy in photosynthesis. (3 marks)

(d) Describe the position and role of the palisade layer and the spongy mesophyll in a leaf. (3 marks)

(e) State two functions of stomata in gas exchange in a leaf. (2 marks)

(f) Explain why a plant cannot photosynthesise at night but still respires. (3 marks)

Question 2 (17 marks) – Measuring the rate of photosynthesis and limiting factors

(a) Describe a method for measuring the rate of photosynthesis using an aquatic plant. (3 marks)

(b) Explain why the rate of bubble production rises when the lamp is moved closer. (3 marks)

(c) At a lamp distance of 10 cm there are 40 bubbles per minute, at 20 cm 10 and at 40 cm 2.5. Explain the change in rate when the distance rises from 10 cm to 20 cm. (4 marks)

(d) Explain why the rate no longer rises when light intensity is increased beyond a high value. (4 marks)

(e) Describe how you could show that carbon dioxide concentration is a limiting factor for photosynthesis. (3 marks)

Question 3 (17 marks) – Stomata, guard cells and microscopic measurement

(a) The eyepiece is x10 and the objective is x40. The image of a stoma measures 8.0 mm. Calculate the total magnification and the actual length of the stoma in micrometres. (4 marks)

(b) A field of view of area 0.25 mm2 contains 45 stomata. Calculate the number of stomata per mm2. (3 marks)

(c) Describe how guard cells control the opening and closing of a stoma. (3 marks)

(d) Explain why stomata close when water is in short supply and how this affects photosynthesis. (4 marks)

(e) Explain why leaves usually have more stomata on the lower surface than on the upper surface. (3 marks)

Question 4 (17 marks) – Xylem, the water column and mineral uptake

(a) Describe how the structure of xylem vessels is adapted for the transport of water. (3 marks)

(b) Explain how transpiration pull moves water from the roots to the leaves. (3 marks)

(c) Water rises 2.4 m from the roots in 8.0 hours. Calculate the average rate of rise in m/h and in cm/h. (4 marks)

(d) Explain the roles of cohesion and adhesion in the water column. (4 marks)

(e) Explain how roots take up mineral ions from the soil. (3 marks)

Question 5 (16 marks) – Transpiration, the potometer and environmental factors

(a) Describe how a potometer is used to measure the rate of transpiration. (3 marks)

(b) A shoot loses 12.0 g of water in 4.0 hours and has a leaf area of 0.060 m2. Calculate the water loss per hour and the water loss per m2 per hour. (3 marks)

(c) Explain why wind increases the rate of transpiration. (3 marks)

(d) Explain why a rise in humidity slows down transpiration. (4 marks)

(e) Explain why a leaf still loses water even though its surface has a cuticle. (3 marks)

Question 6 (17 marks) – Photosynthesis, transpiration and horticultural applications

(a) A plant absorbs 0.48 g of carbon dioxide each hour. Calculate the moles of carbon dioxide absorbed per hour and the mass of oxygen produced in the same time. (Relative molecular mass of CO2 = 44, of O2 = 32) (4 marks)

(b) Explain two reasons why a greenhouse may be heated and lit at night. (3 marks)

(c) A plant transpires at 2.4 g/h at 25 C and at 4.8 g/h at 35 C. Calculate the ratio and explain the difference and the risk involved. (4 marks)

(d) Explain why the rate of photosynthesis may fall around midday on a sunny day. (3 marks)

(e) Explain why a plant may still grow poorly even in strong light. (3 marks)

Answers and marking notes

Question 1 – The photosynthesis equation and leaf structure

(a) 6CO2 + 6H2O -> C6H12O6 + 6O2 (one mark); the raw materials are carbon dioxide and water (one mark); the products are glucose and oxygen (one mark). [3 marks]

(b) Chloroplast (one mark); chlorophyll (one mark). [2 marks]

(c) Chlorophyll absorbs light energy and converts it into chemical energy (one mark); that energy is used to split water molecules and release oxygen (one mark); the energy ends up stored in the chemical bonds of glucose (one mark). [3 marks]

(d) The palisade layer lies just below the upper epidermis, with closely packed cells containing many chloroplasts (one mark); it is the main site of photosynthesis (one mark); the spongy mesophyll lies below it and has large air spaces that allow gases to diffuse (one mark). [3 marks]

(e) They allow carbon dioxide to diffuse into the leaf (one mark); they allow oxygen and water vapour to diffuse out of the leaf (one mark). [2 marks]

(f) There is no supply of light energy at night (one mark); chlorophyll cannot carry out the light-dependent stage, so photosynthesis stops (one mark); respiration does not need light energy and therefore continues day and night (one mark). [3 marks]

Question 2 – Measuring the rate of photosynthesis and limiting factors

(a) Place pondweed in a measuring cylinder of sodium hydrogencarbonate solution (one mark); shine a lamp from a fixed distance (one mark); count the bubbles produced per minute as a measure of the rate (one mark). [3 marks]

(b) Moving the lamp closer increases the light intensity (one mark); chlorophyll absorbs more light energy (one mark); the light-dependent stage speeds up, more oxygen is produced and more bubbles appear (one mark). [3 marks]

(c) The distance is doubled (one mark); light intensity is inversely proportional to the square of the distance, so it falls to one quarter of its previous value (one mark); the rate falls from 40 to 10 bubbles per minute (one mark); the fall matches the change in light intensity (one mark). [4 marks]

(d) Light intensity is no longer the limiting factor (one mark); carbon dioxide concentration or temperature has become the limiting factor (one mark); the products of the light-dependent stage are already plentiful but the light-independent stage is limited by carbon dioxide (one mark); the rate therefore stays constant (one mark). [4 marks]

(e) Keep the plant at a high light intensity (one mark); raise the carbon dioxide concentration in steps and measure the rate each time (one mark); if the rate rises as carbon dioxide rises, it is a limiting factor (one mark). [3 marks]

Question 3 – Stomata, guard cells and microscopic measurement

(a) Total magnification = 10 x 40 = 400 (one mark); actual length = 8.0 mm / 400 (one mark); = 0.020 mm (one mark); = 20 micrometres (one mark). [4 marks]

(b) Density = 45 / 0.25 (two marks); = 180 stomata per mm2 (one mark). [3 marks]

(c) Guard cells take up water, swell and become turgid (one mark); their outer walls are thinner so the cells bend outwards (one mark); the pair separates and the stoma opens, and the stoma closes when the cells lose water and become flaccid (one mark). [3 marks]

(d) When water is short the guard cells lose water and the stomata close (one mark); this reduces water loss by transpiration and conserves water (one mark); closure also cuts the entry of carbon dioxide (one mark); less carbon dioxide means a lower rate of photosynthesis (one mark). [4 marks]

(e) The lower surface receives less direct sunlight (one mark); it is cooler, so water is lost more slowly (one mark); gas exchange can still occur while water loss is kept low (one mark). [3 marks]

Question 4 – Xylem, the water column and mineral uptake

(a) The cells are dead and form hollow tubes (one mark); their walls contain lignin, which is rigid and stops the tubes collapsing (one mark); the cells join end to end into a continuous tube, which reduces resistance to water flow (one mark). [3 marks]

(b) Transpiration from the stomata lowers the water potential of the mesophyll cells (one mark); water moves from the xylem into those cells, creating an upward pull (one mark); the pull is transmitted down the continuous water column and draws water up from the roots (one mark). [3 marks]

(c) Rate = 2.4 / 8.0 (one mark); = 0.30 m/h (one mark); = 30 cm/h (one mark); this is an average, and the actual rate varies with the rate of transpiration (one mark). [4 marks]

(d) Cohesion between water molecules keeps the column from breaking (one mark); adhesion between water molecules and the lignin wall holds the water against the wall (one mark); adhesion reduces the tendency of the column to fall back (one mark); together with transpiration pull these forces maintain a continuous water column (one mark). [4 marks]

(e) The concentration of mineral ions in the soil is usually lower than inside the root hair cells (one mark); uptake is therefore by active transport against the concentration gradient and uses energy (one mark); the energy comes from respiration in the root cells (one mark). [3 marks]

Question 5 – Transpiration, the potometer and environmental factors

(a) Insert a leafy shoot into a potometer filled with water and check that the apparatus is airtight (one mark); record how far the air bubble or water column moves along the capillary in a set time (one mark); calculate the rate from the distance moved per unit time (one mark). [3 marks]

(b) Water loss per hour = 12.0 / 4.0 = 3.0 g (one mark); per m2 per hour = 3.0 / 0.060 (one mark); = 50 g (one mark). [3 marks]

(c) Wind removes the layer of moist air around the leaf (one mark); this increases the water vapour concentration difference between the inside and outside of the leaf (one mark); diffusion is faster and transpiration increases (one mark). [3 marks]

(d) The water vapour concentration in the air rises (one mark); the difference in water vapour concentration between the inside and the outside of the leaf falls (one mark); diffusion depends on that concentration difference (one mark); a smaller difference means slower transpiration (one mark). [4 marks]

(e) The cuticle only reduces water loss and cannot stop it completely (one mark); stomata must open for gas exchange (one mark); water can still be lost as water vapour through the stomata (one mark). [3 marks]

Question 6 – Photosynthesis, transpiration and horticultural applications

(a) Moles = 0.48 / 44 = 0.0109 mol (one mark); carbon dioxide and oxygen are produced in a 1:1 molar ratio (one mark); moles of oxygen = 0.0109 mol (one mark); mass = 0.0109 x 32 = 0.349 g (one mark). [4 marks]

(b) Extra light lets photosynthesis continue (one mark); heating keeps enzymes close to their optimum temperature so reactions run faster (one mark); it also prevents low night temperatures from slowing growth and flowering (one mark). [3 marks]

(c) Ratio = 4.8 / 2.4 = 2 times (one mark); a higher temperature gives water molecules more kinetic energy and faster diffusion (one mark); guard cells may lose too much water and close the stomata (one mark); closed stomata reduce the entry of carbon dioxide so the rate of photosynthesis falls (one mark). [4 marks]

(d) Temperatures are high at midday so transpiration is rapid (one mark); excessive water loss makes the plant close its stomata (one mark); less carbon dioxide enters and the rate of photosynthesis falls (one mark). [3 marks]

(e) Photosynthesis is also limited by other factors such as carbon dioxide concentration, temperature or mineral supply (two marks); excess light energy can also damage chlorophyll (one mark). [3 marks]

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