A freshwater alga has an internal osmotic pressure of 0.45 MPa. The external med
ID: 218670 • Letter: A
Question
A freshwater alga has an internal osmotic pressure of 0.45 MPa. The external medium, which is very dilute, has an osmotic pressure of 0.01 MPa.
a. What is ???
b. If the cell is at osmotic equilibrium, what is ???
c. What is ?P?
d. Suppose solutes are added to the external medium to raise the osmotic pressure to 0.12 MPa. Assuming that water comes to equilibrium before any change in internal concentrations occurs, what is the new turgor?
e. The alga then initiates a regulatory mechanism to alter its internal solute concentrations to restore the original turgor. When the original turgor is restored, what is the new ?? Did the alga need to increase or decrease internal solute concentration?
Explanation / Answer
Asnwer-a: It is a case of hypotonic solution.
Answer-b: If a cell is at osmotic equilibrium them their will be no movement of water either from inside or outside. It would attain a condition of isotonic solutions.
Answer-c: The osmostic pressure difference is 0.44 MPa [0.45 - 0.01 MPa].
Answer-d: Turgor pressure can be calculated by considering water potential and osmotic potential. Water potential has Zero value and osmotic potential is 0.12 MPa (given)
Therefore Turgor pressure = Water potential - Osmotic potential
= 0 - 0.12
= -0.12 MPa
Answer-e: In order to restore the original turgor pressure, the new value could be anything above zero. Negative turgor pressure means there is a water deficient condition inside the cell. Zero turgor pressure means cell sap pressure is equals to atmospheri pressure. So, when the value of turgor pressure is above zero, it will show high water concentration.
In order to increase the value of turgor pressure, an alga need to increase the internal solute concentration (similar to the case of hypotonic solutions) so that water from external surrounding can move inside the cell and increases the turgor pressure.
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