Water uptake in plants Questions

Question 1

Three identical potted plants were watered and their mass recorded. Their pots were wrapped in sealed in plastic bags so that only the plant was in the open air. Each plant was then placed in different conditions.
After 6 hours, the bags were removed and the mass of the plants was recorded again. The table shows the results.

Mass in g/Plant A (cool,still air)/Plant B (warm, still air)/Plant C (warm windy)

at start/436/452/448
at end/412/398/332
% change/5.5/a/b

a) Calculate the percentage change in mass for plants B and C.
b) Explain the difference in results.

Answer 1

a) 452 - 398 = 54
54/4.52 = 11.9

a = 11.9%

448 - 332 = 116
116/4.48 = 25.89

b = 25.9%

b) The results show that evaporation of water from the plant was faster in warm air than cool air and even faster in windy air than in still air. This is because evaporation from the stomata is faster in hot and windy conditions.

It is faster because water diffuses down a concetration gradient from high to low. Hot or windy conditions cause the concentration of water directly outside the leaf to decrease, making a steeper concetration gradient.

Question 2

a) A student placed a leafy shoot in a potometer. The radius of the capillary tube was 0.5 mm. The bubble moved 50 mm in 5 minutes. Calculate the rate of transpiration in mm3/min.
b) Explain how this measurement would be different is you repeated this with the same shoot in warmer conditions.
c) Describe how you would carry out this investigation to make sure the results in the two different conditions could be compared fairly.

Answer 2

a) πr²d = π x 0.5² x 50 = 39.25 mm³

Rate = 39.25/5 = 7.85 mm³/min

b) higher rate/faster, because increased temperature increases rate of evaporation of water from leaf
c) all conditions kept the same except temperature; one example is e.g. air movement/light intensity