Solve & explain briefly

Solve & explain briefly The value of osmotic pressure becomes equal to diffusion pressure deficit when a cell is placed a typertonic solution (2) hypotonic solution (3) isotonic solution (4) both (2) and (3)

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The correct option is (1)


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Osmotic pressure: Itis defined as the pressure which is developed in a solution when it is separated from its pure solvent by a semi-permeable membrane. Osmotic relations of plant cells : (a) A typical plant cell has a permeable elastic wall, a semi-permeable membrane and an osmotically active solution called cell sap.
(b) If a cell is placed in a hypotonic solution, water enters the cell as a result of endosmosis. It is because the direction of movement of water is from-higher water potential to lower water potential.
© It results in turgor pressure which forces the protoplasm to press against the cell wall in turn, exerts an opposite pressure equal in magnitudes to the turgor pressure. This pressure is called wall pressure.
(d) The wall pressure goes on increasing till the cell becomes fully turgid i.e., WP = TP.
(e) The net force with which water is drawn into the cell is called diffusion pressure deficit. (DPD).
(f) The value of DPD is equal to the difference between OP of the cell sap and the WP which opposes the entry of water.
Thus, DPD = OP - WP (WP = TP)
so DPD = OP -TP
(g) If a cell is placed in hypertonic solution, water will move out of the cell because of the higher chemical potential of the cell sap, this is exosmosis.
(h) The vacuoles shrinks in size. The turgor pressure and wall pressure becomes zero, and the cell becomes flaccid. The value of TP of such a cell is zero, therefore DPD equals to its OP.
DPD = OP-TP (TP = 0)
DPD = OP - 0 DPD = OR
(i) When a plasmolysed cell is placed in water, water rqoves into a cell as solution hypotonic to cell sap.
(j) The values of OP begins to fall and TP increases till
OP equals to TP. In such a conditon DPD becomes zero.
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