Hey i have a mid term today at 3 , and this one sample question my prof posted online. If someone can please show me how to get the answers i would really appreciate it:
A plant cell is found to have a Ψs = - 0.7 MPa and a Ψp = 0.5 MPa.
Is this cell turgid? (yes or no) (1 mark)
Yes
b. What is the Ψw of this cell?
-0.2
c. If this cell were placed in a large beaker of solution (at atmospheric pressure and with a Ψs = - 0.3 MPa), would water enter or leave the cell? Assume that no solute can cross the plant plasma membrane in either direction and that the volume of neither body of water changes significantly.
leave(down Ψw gradient)
d. Once the cell in c. has come to equilibrium with the beaker solution (again assuming that no solute crosses the plant plasma membrane in either direction and that the volume of neither body of water changes significantly), what will be the Ψp within the cell?
0.4
(-0.3 MPa = -0.7 MPa + 0.4 MPa)
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The equation is :
Ψw = Ψs + ΨP
Sorry entire thing didn't get posted:
a.So the answer for is the cell turgid?
Yes
B. What is the Ψw of this cell?
-0.2
C. If this cell were placed in a large beaker of solution ( at atmospheric pressure and with a Ψs = - 0.3 MPa. would the water enter or leave the cell?
Assume that no solute can cross the plant plasma membrane in either direction and that the volume of neither body of water changes significantly.
Leaves
D.Once the cell in C. has come to equilibrium with the beaker solution( again assuming no solute can cross the plant plasma membrane in either direction and that the volume of neither body of water changes significantly.What will the Ψp be within the cell?
0.4
(-0.3 MPa = -0.7 MPa + 0.4 MPa)
Okay in understand everything except D....SOMEONE PLEASE HELpp
To find the answers to the questions, you need to understand the concept of water potential (Ψw) and how it is calculated using solute potential (Ψs) and pressure potential (Ψp). Here's how you can solve each question:
a. Is this cell turgid? (yes or no)
To determine if the cell is turgid, you need to compare the solute potential (Ψs) and pressure potential (Ψp). If the solute potential (Ψs) is greater than the pressure potential (Ψp), then the cell is considered turgid. In this case, Ψs is -0.7 MPa and Ψp is 0.5 MPa. Since -0.7 MPa is greater than 0.5 MPa, the cell is turgid. Therefore, the answer is "Yes."
b. What is the Ψw of this cell?
To calculate the water potential (Ψw) of the cell, you need to add the solute potential (Ψs) and pressure potential (Ψp) together. In this case, Ψs is -0.7 MPa and Ψp is 0.5 MPa. Adding them together (-0.7 MPa + 0.5 MPa), the water potential (Ψw) of the cell is -0.2 MPa.
c. If this cell were placed in a large beaker of solution (at atmospheric pressure and with a Ψs = -0.3 MPa), would water enter or leave the cell?
To determine if water would enter or leave the cell, you compare the water potential (Ψw) of the cell and the beaker solution. The water will always move from an area of higher water potential to an area of lower water potential. In this case, the water potential (Ψw) of the cell is -0.2 MPa and the water potential (Ψw) of the beaker solution is -0.3 MPa. Since the water potential is higher in the beaker solution (-0.3 MPa) compared to the cell (-0.2 MPa), water will leave the cell. Therefore, the answer is "leave (down Ψw gradient)."
d. Once the cell in c. has come to equilibrium with the beaker solution, what will be the Ψp within the cell?
When the cell and the beaker solution come to equilibrium, it means that there is no net movement of water between them. At equilibrium, the water potential (Ψw) of the cell and the beaker solution will be equal. The water potential (Ψw) of the beaker solution is -0.3 MPa. Since we know the water potential (Ψw) of the cell is -0.2 MPa and the solute potential (Ψs) is -0.7 MPa, we can calculate the pressure potential (Ψp).
Ψw = Ψs + Ψp
Re-arranging the equation to solve for Ψp:
Ψp = Ψw - Ψs
Substituting the known values, Ψw (-0.3 MPa) and Ψs (-0.7 MPa), into the equation:
Ψp = -0.3 MPa - (-0.7 MPa) = 0.4 MPa
So, the pressure potential (Ψp) within the cell at equilibrium will be 0.4 MPa.