Pearson Edexcel · GCSE · Biology

    Course version: 1BI0 · Version from Sept 2016

    Demonstrate an understanding of rate calculations for transpiration

    Full specification wording

    Topic 6 – Plant structures and their functions: 6.13 Demonstrate an understanding of rate calculations for transpiration

    Transpiration is the loss of water vapour from leaves, mainly through stomata.

    Checked against the Pearson Edexcel 1BI0 specification (opens in a new tab) on .

    • Point 12 of 15

    Study notes

    From the checked teaching notes for this specification point.

    The idea

    Transpiration is the loss of water vapour from leaves, mainly through stomata. Its rate is a measured quantity, not a guess: you find how much water the shoot takes up or loses over a set time and divide the change by that time. A potometer gives a distance moved by an air bubble; if the bubble travels 30 mm in 5 minutes, the rate is 30 ÷ 5 = 6 mm per minute. If a mass changes by 2.4 g in 8 minutes, the rate is 2.4 ÷ 8 = 0.3 g per minute. Always state the unit and match it to what was measured (distance, volume or mass per unit time). Rate calculations let you compare treatments fairly, for example a shoot in moving air against one in still air, because each result is expressed per minute.

    What earns the marks

    • Rate of transpiration is calculated as the change in the measured quantity (distance moved by an air bubble, volume of water taken up, or mass lost) divided by the time taken.
    • The unit of the answer must combine the measured quantity with time, for example mm min⁻¹, cm³ min⁻¹ or g min⁻¹.
    • A worked example: a bubble moving 30 mm in 5 minutes gives 30 ÷ 5 = 6 mm min⁻¹.
    • Comparing rates between treatments is valid only when each is expressed per unit time, so conditions such as temperature, humidity, light and air movement can be linked to changes in rate.
    • Reading a potometer scale correctly and timing accurately are needed before any division is carried out.

    Common mistakes and how to put them right

    • Multiplying the distance by the time instead of dividing.

      Put it right: Rate is change ÷ time, so 30 mm in 5 min is 6 mm min⁻¹, not 150.

    • Giving a bare number with no unit.

      Put it right: Attach the correct combined unit such as mm min⁻¹ so the answer is meaningful.

    • Using total time in the wrong place, for example time ÷ distance.

      Put it right: Keep the measured change as the numerator and time as the denominator.

    Exam tips

    • Write the formula down first: rate = change in measured quantity ÷ time taken, then substitute the numbers.
    • Check the units of both quantities before dividing and build the answer unit from them.
    • If asked to compare two shoots, calculate both rates per minute before drawing any conclusion.

    Try the lesson

    The start of the Topic 6 – Plant structures and their functions: 6.13 Demonstrate an understanding of rate calculations for transpiration lesson, made from checked teaching notes for this specification point.

    Have a go first

    A shoot loses 24 cm³ of water by transpiration over 6 hours. What is the mean rate of transpiration in cm³ per hour?

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