Question

Sucrose (M)	0	0.1	0.2	0.3	0.4	0.5	0.6	0.7
Initial Weight (g)	6.25	6.33	6.38	6.50	6.54	6.40	6.56	6.25
Final Weight (g)	6.76	6.70	6.72	6.57	631	6.04	6.10	5.77
∆ Weight (g)	0.51	0.37	0.34	0.07	-0.23	-0.36	-0.46	-0.48
%∆ Weight	0.08	0.06	0.05	0.01	-0.03	-0.05	-0.07	-0.08

a. Sucrose concentration at which zero percent change in weight is observed is:

b. Determine the water potential (Ψ_w ) of the potato tissue. Show your calculations.

Answer 1

a) The percentage change in weight is calculated using the formula:

% change in weight = final weight – initial weight x 100%   initial weight

We will plot a graph of the percentage change in weight (y-axis) against the concentration of sucrose solution (x-axis).From the graph, we can determine the concentration of sucrose solution which causes no change in weight of the tissue.

Sucrose Conc	%∆ Weight
(in M)
0	0.08
0.1	0.06
0.2	0.05
0.3	0.01
0.4	-0.03
0.5	-0.05
0.6	-0.07
0.7	-0.08

Equation of the graph is: y = -0.2512x + 0.0842

To determine the concentration of sucrose solution which causes no change in weight, we will put %Weight = 0

i.e y = 0

So, putting y=0 and solving for x, the equation becomes : 0 = -0.2512x + 0.0842

0.2512x = 0.0842

   x = 0.0842/0.2512 = 0.335 M

Sucrose concentration at which zero percent change in weight is observed is: 0.335M

b) Water potential of a plant cell is made up of two important components:

w = s + p

w is the overall water potential of a cell.

s is the solute or osmotic potential

p is the pressure potential

In an open solution where there is no turgor pressure, the p is equal to zero. Thus, w = s

To calculate the s of the solution causing no change in weight of the potato tissues using the following formula:

s = -miRT

m = molarity
i = ionization constant = 1 for sucrose
R = gas constant = 8.314 J K ^-1 mol ^-1
T = room temperature in K

Sucrose concentration at which zero percent change in weight is observed,m = 0.335M = 0.335molL^-1

assuming room temperature, T = 22^oC = 295K

s = -(1)(0.335molL^-1)(8.314JK^-1mol^-1)(295K)

w =s = -821.63JL^-1 = -8.2163 bar (1JL^-1 = 0.01bar)

The water potential (Ψw ) of the potato tissue = -8.2163 bar