10: Transpiration, Transport and Support in Plants - Problems Flashcards
The diagram below shows a weight potometer user to study the transpiration of a leafy shoot. The leafy shoot was left in the laboratory for 4 hours. During this time, the change in reading of the balance is 1.5g and the change in reading of the measuring cylinder is 2cm3.
What can be deduced from the difference between the readings of the balance and the measuring cylinder? Explain the significance of this difference to the growth of the plant. (3)
The change in the reading of the measuring cylinder (i.e. water uptake) is larger than that of the electronic balance (i.e. water loss).
This indicates that the plant has a net uptake of water.
The water retained in the plant is used for various processes, e.g. photosynthesis and formation of new cells.
The diagram below shows a weight potometer user to study the transpiration of a leafy shoot. The leafy shoot was left in the laboratory for 4 hours. During this time, the change in reading of the balance is 1.5g and the change in reading of the measuring cylinder is 2cm3.
Explain the difference in results if the experiment was repeated with all the leaves of the shoot removed. (3)
The change in the readings of the electronic balance and the measuring cylinder would become smaller.
As transpiration mainly takes place through the stomata in the leaves, removal of the leaves greatly reduces the water loss of the plant by transpiration.
This, in turn, reduces the transpiration pull, which facilitates the uptake of water by the roots. Therefore, water uptake of the plant is also reduced.
DSE Bio 2013 IB Q6
The following set-up can be used to determine the transpiration rate of a leafy shoot.
Explain why the lower end of the shoot should be cut under water when setting up the experiment. (1)
This is to avoid blockage of xylem by the air bubble formed during cutting.
DSE Bio 2013 IB Q6
The following set-up can be used to determine the transpiration rate of a leafy shoot.
State one assumption for using this set-up to measure the transpiration rate. (1)
The rate of transpiration is the same as the rate of water absorption / the water absorbed is used for transpiration only.
DSE Bio 2013 IB Q6
The following set-up can be used to determine the transpiration rate of a leafy shoot.
Explain the change in transpiration rate if the fan placed near the shoot is switched on. (4)
The transpiration rate will increase.
As the air current sweeps away the water vapour around the leafy shoot,
the concentration gradient of water vapour between the atmosphere and the air space in the leaves becomes steeper.
Therefore, water vapour diffuse out to the atmosphere at a faster rate.
DSE Bio 2013 IB Q6
The following photomicrograph shows the appearance of the surface of a leaf during daytime.
In terms of sub-cellular structure, state two differences between cell A and cell B. (2)
Cell B contains chloroplasts but cell A does not.
The cell wall of cell B has uneven thickness while that of cell A is even.
DSE Bio 2013 IB Q6
The following photomicrograph shows the appearance of the surface of a leaf during daytime.
Discuss the functional significance of the change in size of C at night. (3)
The size of C will be reduced
to reduce transpiration / water loss,
when the need for gas exchange decreases in the absence of photosynthesis.
The diagram below shows the percentage saturation of air with water vapour near a stoma.
The cell wall of the guard cells has uneven thickness. Explain how this feature related to the canton of the guard cells. (4)
The uneven thickness of the cell wall of guard cells is important in controlling the opening and closing of stomata.
When the guard cells take up water from the surrounding cells, the guard cells become turgid.
The thinner outer side of the cell wall expands more than the thicker inner side.
Therefore, the guard cells bend and the stoma opens.
The diagram below shows the percentage saturation of air with water vapour near a stoma.
Explain in which direction, X or Y, does water vapour diffuse faster out of the leaf. (3)
Direction X.
The lines are closer in direction X,
which indicates that the concentration gradient of water vapour in direction X is steeper. Therefore, water vapour diffuses out at a higher rate.
A scientist put the roots of plants in solutions with different concentrations of phosphate ions. The rate of absorption of phosphate ions bay the roots was measured. He repeated the experiment by using roots treated with cyanide, which is a chemical that can inhibit respiration. The graph shows the results of the two experiments.
Describe the results of experiments I and I when the concentration of phosphate ions in the solution is below 20 units. (2)
In experiment I, the rate of absorption of phosphate ions by the roots increases as the concentration of phosphate ions in the solution increases.
In experiment II, there is no absorption of phosphate ions by the roots.
A scientist put the roots of plants in solutions with different concentrations of phosphate ions. The rate of absorption of phosphate ions bay the roots was measured. He repeated the experiment by using roots treated with cyanide, which is a chemical that can inhibit respiration. The graph shows the results of the two experiments.
Account for the result of experiment II when the concentration of phosphate ions in the solution increases from 0 to 40 units. (4)
The root cells treated with cyanide
cannot release energy by respiration.
When the concentration of phosphate ions in the solution is lower than that in the plant root from 0 units to 20 units, the root cells cannot absorb phosphate ions from the solution against a concentration gradient by active transport.
When the concentration of phosphate ions in the solution becomes higher than that of the root cells at above 20 units, phosphate ions diffuse into the root cells down a concentration gradient.
The diffusion rate increases as the concentration of phosphate ions in the solution increases.
A scientist put the roots of plants in solutions with different concentrations of phosphate ions. The rate of absorption of phosphate ions bay the roots was measured. He repeated the experiment by using roots treated with cyanide, which is a chemical that can inhibit respiration. The graph shows the results of the two experiments.
Explain the shape of the curve in experiment I when the concentration of phosphate ions in the solution is higher than 20 units. (2)
The curve levels off. In the cell membrane of the root cells, the number of carriers for transporting phosphate ions is limited.
The curve levels off when all the carriers are occupied by phosphate ions.
The electron micrograph below shows part of the transverse section of a phloem tissue.
Identify cells X and Y with a reason. (3)
Cell X: sieve tube
Cell Y: companion cell
A nucleus is present in cell Y but absent in cell X.
The electron micrograph below shows part of the transverse section of a phloem tissue.
State the roles of cells X and Y. (2)
Cell X is responsible for the transport of organic nutrients in the plant.
Cell Y supports the metabolism of cell X.
A potted plant has one leaf enclosed in a plastic bag containing carbon dioxide with radioactive carbon. The plant was put under sunlight for two hours. Then, the radioactivity in different parts of the plant was measured.
Explain why the radioactivity is detected in the shoot, fruit. and root of the plant. (2)
Carbon dioxide containing radioactive carbon is converted to carbohydrates during photosynthesis.
Then the carbohydrates are transported to different parts of the plant including the shoot tips, fruits and roots by phloem.
A potted plant has one leaf enclosed in a plastic bag containing carbon dioxide with radioactive carbon. The plant was put under sunlight for two hours. Then, the radioactivity in different parts of the plant was measured.
Explain how the experiment shows that translocation of organic food in the plant is bidirectional. (1)
Radioactivity was detected both above and below the leaf supplied with carbon dioxide containing radioactive carbon.
A potted plant has one leaf enclosed in a plastic bag containing carbon dioxide with radioactive carbon. The plant was put under sunlight for two hours. Then, the radioactivity in different parts of the plant was measured.
Suggest a reason why the radioactivity in the shoot tip is much lower than that of the fruit. (2)
The organic food transported to the fruit will be stored, so radioactivity accumulates in the fruit.
Meanwhile, the organic food transported to the shoot tip is used in respiration to release energy for growth, so radioactivity in the shoot tip is lower than that of the fruit.
DSE 2019 IB Q10
The graph shows the change in the rate of transpiration and the change in stem diameter of a plant over 24 hours.
Describe the relationship between the rate of transpiration and the stem diameter. (1)
The diameter of the stem decreases as the rate of transpiration increases / the diameter of the stem increases as the rate of transpiration decreases.
DSE 2019 IB Q10
The graph shows the change in the rate of transpiration and the change in stem diameter of a plant over 24 hours.
The change in stem diameter is related to the diameter of the xylem vessels. With reference to the way in which water is transported along the stem, explain the relationship between the rate of transpiration and stem diameter. (2)
water is transported up the stem by transpiration pull.
This force increases with transpiration rate and pulls the xylem vessel’s walls
inwards, thus reducing the diameter of the xylem as well as that of the stem.
DSE 2019 IB Q10
Describe and explain two adaptive features of xylem vessels as a structure for water transport. (4)
It is a hollow tube,
to allow the passage of water and minerals with low resistance.
It has a lignified wall,
to withstand the negative pressure of the transpiration pull and prevent the collapse of xylem vessels.
DSE 2012 IB Q3
The electron micrograph shows part of the stem of a plant containing two cell types, P and Q.
Based on the photograph, state the difference between cell types P and Q in the structure indicated by the arrow heads. (1)
The cell wall of cell type Q is much thicker than the cell wall of cell type P.
DSE 2012 IB Q3
The electron micrograph shows part of the stem of a plant containing two cell types, P and Q.
Explain how these cells contribute to the support of the plant. (4)
When there is ample supply of water,
cell type P provides turgidity to the plant.
Cell type Q has a thickened cell wall,
which provides rigidity to the plant.
The diagram blow shows two stem tissues obtained from a herbaceous plant and a woody plant. The stems were immersed in a concentrated salt solution.
Suggest and explain the appearance of the stems after 24 hours. (7)
The herbaceous stem will bend downwards,
while the woody stem will remain straight.
The herbaceous stem is mainly supported by the turgidity of thin-walled cells in the stem.
When the herbaceous stem is put in concentrated salt solution, the water potential of the bathing solution is lower than the water potential of the thin-walled cells in the stem, so there is a net movement of water molecules from inside the thin-walled cells in the stem to the solution through the differentially permeable cell membrane by osmosis.
The cells become flaccid and cannot provide support to the stem to resist bending.
Cells in the woody stem have thick and lignified cell walls.
The rigidity of these cells provides strong support to the stem to resist bending.
The diagram blow shows two stem tissues obtained from a herbaceous plant and a woody plant. The stems were immersed in a concentrated salt solution.
Explain the appearance of the stems if the experiment was repeated using distilled water instead of concentrated salt solution. (4)
Both stems will remain straight.
As the water potential of distilled water is higher than that of the herbaceous stem, there is a net movement of water molecules from the bathing solution to the thin-walled cells in the herbaceous stem through the differentially permeable cell membrane by osmosis.
The cells become turgid and give support to the stem to resist bending.
Meanwhile, cells in woody stems have thick and lignified cell walls and their rigidity of these cells provides strong support to the stem to resist bending.
