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An animal cell with an internal solute concentration of 0.02 OsM is placed in a test tube containing a solution with a solute concentration of 0.10 OsM. What will happen to the cell?
A) nothing-it will stay the same size
B) the cell will shrink as water moves out of the cell
C) the cell will swell as water moves into the cell
D) the cell will burst
E) none of the above

Respuesta :

Answer:

B) the cell will shrink as water moves out of the cell

Explanation:

Water moves from a region of lower to that of higher osmolarity. In the given example, the cell's osmolarity is lower (0.02 OsM) as compared to the outer solution (0.10 OsM). Since the cells' interior is hypotonic to the outer solution, water will move out of the cell towards the solution in the test tube. Exosmosis of water out of the animal cell will make the cell to shrink.

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