Q.An artificial cell made of selectively permeable membrane immersed in a beaker (in the figure). Read the values and answer the following questions?
a. Draw an arrow to indicate the direction of water movement
b. Is the solution outside the cell isotonic, hypotonic or hypertonic?
c. Is the cell isotonic, hypotonic or hypertonic?
d. Will the cell become more flaccid, more turgid or stay in original size?
e. With reference to artificial cell state, the process is endosmosis or exosmosis? Give reasons
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Start your 14-day free trial to unlock the full solution →Step 1. Read the given values. Inside the artificial cell: Ψw = 8, Ψs = 0, Ψp = -8. In the solution outside the cell: Ψw = 0, Ψs = -2, Ψp = 0 (same arbitrary pressure units on both sides).
Step 2. (a) Direction of water movement. Water always moves, across a selectively permeable membrane, from the region of higher water potential (Ψw) to the region of lower water potential. Here Ψw(cell) = 8 is greater than Ψw(outside) = 0, so the arrow must point OUTWARD — from inside the artificial cell, through the membrane, into the surrounding beaker solution.
Step 3. (b) Tonicity of the outside solution. The outside solution has a solute potential of Ψs = -2, i.e. it contains dissolved solutes, whereas the cell's contents show Ψs = 0 (essentially solute-free). Since the outside solution has the more negative solute potential (the higher effective solute concentration), it is HYPERTONIC relative to the cell's contents.
Step 4. (c) Tonicity of the cell. By the same comparison, the cell's contents (Ψs = 0) are more dilute than the outside solution (Ψs = -2), so the cell is HYPOTONIC relative to the outside solution.
Step 5. (d) Change in the cell's state. Because water is leaving the cell (Step 2) and the cell is hypotonic to its surroundings, the cell will lose water/volume and become MORE FLACCID — it will not become more turgid, and it will not stay at its original size. …
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