AP Biology : 2.8 Tonicity and Osmoregulation- Exam Style questions with Answer- FRQ

Question

The paramecium Paramecium aurelia has a contractile vacuole, which it uses to pump excess water out of its cell. P. aurelia was placed in four different salt solutions with concentrations of 0.02 molar, 0.04 molar, 0.08 molar, and 0.10 molar salt. The number of contractions of the contractile vacuole per minute was measured over a 10-minute period.
(a) Describe how the water potential of the surrounding solutions would affect the rate of contraction of the contractile vacuole.
(b) Identify the independent variable and dependent variable in this experiment.
(c) As a follow-up experiment, P. aurelia is placed in a beaker that contains distilled water. Predict what effect, if any, this would have on the rate of contraction of the contractile vacuole.
(d) Justify your prediction from part (c) using your knowledge of water potential.

▶️Answer/Explanation

Ans:

(a) If the water potential of the surrounding solution was higher than
that of the paramecium, more water would enter the paramecium’s
cell and the contractile vacuole would have to pump more often to
remove the excess water from the cell.
(b) The independent variable is the concentration of the salt solution.
The dependent variable is the number of contractions of the
contractile vacuole per minute.
(c) The contractile vacuole would have to pump more times per minute
to remove the excess water.
(d) Distilled water would have a higher water potential than P. aurelia.
So more water would move into the paramecium and the
contractile vacuole would have to work harder to remove the
excess water.

Question

A student conducts an experiment with four different root vegetables (carrots, beets, parsnips, and potatoes). Five cubes of equal sizes and
surface areas are cut from each of the vegetables, and their masses are recorded. Each cube is placed in a beaker that contains a 0.35 molar
sucrose solution for 24 hours. After 24 hours, the cubes were removed from the solutions and weighed, and the percent change in mass for
each cube was calculated. The mean percent change in mass for each vegetable and the standard errors of the mean are shown in the table.

VegetableMean Percent Change in MassStandard Error of the Mean
Carrots+7.5%1.0%
Beets+21.5 %2.5%
Parsnips-15.5 %1.5%
Potatoes-3.5%0.5%

(a) Explain why some of the vegetables would have a positive percent change in mass while others would have a negative percent change in mass during the course of the experiment.
(b) Construct an appropriately labeled graph that shows the mean percent change in mass for each vegetable. Include 95% confidence intervals.

(c) Analyze the data to determine which vegetable has a water potential closest to the water potential of the 0.35 molar sucrose solution used in the experiment. Justify your answer with evidence from the experiment.
(d) It is determined that the sugar content in each vegetable is the major determinant of the vegetable’s water potential. Turnips have a higher sugar content than carrots but a lower sugar content than beets. If the experiment was repeated and included turnip cubes, predict the percent change in mass for the turnip cubes. Justify your prediction with evidence from the experiment.

▶️Answer/Explanation

Ans:

(a) Vegetables with a lower water potential than the surrounding 0.35
molar sucrose solution would gain water and mass. Vegetables with
a higher water potential than the surrounding 0.35 molar sucrose
solution would lose water and mass.
(b)

(c) The vegetable with a water potential that is closest to that of the
0.35 molar sucrose solution would have the smallest percent
change in mass. If a vegetable had the same water potential as the
0.35 molar sucrose solution, its cells would be isotonic to the
sucrose solution, and you would expect to find a 0% change in
mass. The smaller the percent change in mass, the closer the
vegetable’s water potential is to that of the 0.35 molar sucrose
solution. The potato has a water potential that is closest to that of
the 0.35 molar sucrose solution because its percent change in mass
(–3.5%) is the smallest and closest to zero.
(d) Turnips would have a percent change in mass that is greater than
+7.5% but less than +21.5%. Since turnips have a higher sugar
content than carrots, the water potential of the turnip cells would be
less than that of the carrot cells. So the turnips would gain more
water from the surrounding 0.35 molar sucrose solution than the
carrots did (more than 7.5%). Since the turnips have a lower sugar
content than the beets, the water potential of the turnip cells would
be higher than the water potential of the beet cells. So the turnips
would gain less water from the surrounding 0.35 molar sucrose
solution than the beets did (less than 21.5%).

Question

If a person is in the hospital with severe dehydration, often an intravenous infusion of physiological saline (0.9% saline) will be administered.
(a) Describe how the water potential of a person’s cells would be affected by severe dehydration.
(b) Explain why 0.9% saline is used to rehydrate a person with severe dehydration and why distilled water is not used.
(c) During strenuous exercise under extreme heat, some athletes are advised to consume salt tablets. Predict whether consuming a salt tablet would lead to water loss or water conservation in the athlete’s body cells.
(d) Justify your prediction from part (c) using your knowledge of water potential.

▶️Answer/Explanation

Ans:

(a) When people are severely dehydrated, their cells have less water
and the water potential of their cells is lower.
(b) If distilled water were used to rehydrate a person, too much water
would enter the person’s cells and there would be a risk of the
blood cells bursting from the pressure caused by the excess water.
A saline solution of 0.9%, which is the same concentration of
saline that is normally seen in body cells, can rehydrate a severely
dehydrated person without running the risk of bursting the blood
cells.
(c) Consuming a salt tablet would lead to water conservation.
(d) Consuming salt lowers the water potential of the person’s body
cells, so less water would leave the cells as sweat or urine.

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