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O-Level Nutrition and Transport in Flowering Plants: worked solution

4 marks. Full working, one step per line.

Question

Using the data in Table 6.1 (temperatures 10–45 °C and the matching oxygen volumes 9, 15, 23, 31, 34, 32, 19.5, 3 cm3), plot a graph of oxygen volume against temperature and join the points with a smooth best-fit curve.

Worked answer

Step 1 - decide which variable goes on which axis. Temperature is the variable that was changed, so it goes on the x-axis; volume of oxygen is what was measured, so it goes on the y-axis. Step 2 - choose the scales. Temperature runs from 10 °C to 45 °C and the volumes run from 3 cm³ to 34 cm³. Choose easy steps (for example 5 °C per two squares across, and 5 cm³ per two squares up) so the plotted points fill more than half the grid in each direction. Never use awkward steps such as 3 or 7 per square, because reading intermediate values then becomes guesswork. Step 3 - label the axes. Write 'temperature / °C' along the x-axis and 'volume of oxygen / cm³' up the y-axis. The unit must be given as well as the quantity. Step 4 - plot the eight readings, each as a small neat cross: (10, 9) (15, 15) (20, 23) (25, 31) (30, 34) (35, 32) (40, 19.5) (45, 3) Note 19.5 cm³ falls half way between two grid lines, so read that one carefully. Every point should be within half a small square of its true position. Step 5 - draw the curve. Join the points with a single smooth freehand curve, not with ruled lines from dot to dot, and do not extend it beyond the first and last readings. The curve rises steeply from 10 °C, flattens off at a peak of about 34 cm³ near 30 °C, then falls away more and more steeply to 3 cm³ at 45 °C, because above the optimum the enzymes controlling photosynthesis start to denature.

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This question is part of O-Level Nutrition and Transport in Flowering Plants, in O-Level Pure Biology.

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