Q36-6TYU.

Question

A plant cell with a ΨS of -0.65 MPa maintains a constant

volume when bathed in a solution that has a ΨS of -0.30 MPaand is in an open container. The cell has a

(A) ΨP of +0.65 MPa.

(B) Ψ of -0.65 MPa.

(C) ΨP of +0.35 MPa.

(D) ΨP of 0 MPa.

 

Step-by-Step Solution

Verified
Answer


(A)    The statement "ΨP of +0.65 MPa" is false.

(B)    The statement "Ψ of -0.65 MPa" is false.

(C)    The statement "ΨP of +0.35 MPa" is true

(D)    The statement "ΨP of 0 MPa." is false.


1Step 1: Equation of water potential

The water potential equation is expressed as Ψ = ΨS + ΨP, where ΨS is the solute potential and is the water potential (osmotic potential), while ΨP denotes the pressure potential. 

 

A solution's solute potential (S) is inversely proportional to molarity. The solute potential is another term because solutes affect the osmotic potential in the osmosis direction. The solutes found in plants are typically made up of mineral ions and sugars.

 

2Step 2: Explanation of option '(A)'

Ψ = ΨS + ΨP-0.30 MPa= (-0.65 MPa)+ ΨP0.35 MPa=ΨP

Thus, the correct answer is ΨP= 0.35 MPa instead of +0.65 MPa because ΨS is given as -0.30 MPa

Therefore, the given statement is false.


3Step 3: Explanation of option '(B)'

It is given that the Ψ is water potential in a container open to the atmosphere is - 0.35 MPa and plant cell with a ΨS of -0.65 MPa.


Thus, Ψ is 0.35 MPa.

Therefore, the given statement is false.

 

4Step 4: Explanation of option '(C)'

Ψ = ΨS + ΨP-0.30 MPa= (-0.65 MPa)+ ΨP0.35 MPa=ΨP


Thus, the ΨP= +0.35 MPa.

Therefore, the given statement is true.

5Step 5: Explanation of option '(D)'

Ψ = ΨS + ΨP-0.30 MPa= (-0.65 MPa)+ ΨP0.35 MPa=ΨP


Thus, the given option is incorrect because ΨP= 0.35 MPa instead of ΨP= 0 MPa.

Therefore, the given statement is false.