required practical 3: Production of a dilution series of a solute to produce a calibration curve with which to identify the water potential of plant tissue

Question 1

PART 1 collecting data: describe a method to produce of a calibration curve with which to identify the water potential of plant tissue (eg. potato)

Answer 1

1. create a SERIES OF DILUTION using a 1 mol
   dm-3 SUCROSE solution (0.0, 0.2, 0.4, 0.6, 0.8,
   1.0 mol dm-3)
2. Use cork borer to cut cylindrical potatos on a white squared tile and use scalpel to trim the ends into IDENTICAL CYLINDERS
3. Blot DRY with a paper towel and measure /
   record INITIAL MASS of each piece using a balance
4. put one chip in each solution and
   leave for a set time (20-30 mins) in a
   water bath at 30oC (degrees celsius)
5. Blot DRY with a paper towel and measure /
   record FINAL MASS of each piece

REPEAT (3 or more times) at each concentration

Question 2

control variables

Answer 2

-volume of solution, eg. 20 cm3
-size, shape and surface area of plant tissue
-source of plant tissue e.g. variety or age
-blot dry to remove excess water for each potato cylinder
-length of time in solution
-temperature of water bath

Question 3

PART 2: processing data

Answer 3

6. Calculate % change in mass = (final - initial mass)/ initial mass
7. Plot a graph with concentration on x axis and percentage change in mass
   on y axis (calibration curve)
   ○ Must show positive and negative regions
8. Identify concentration where line of best fit intercepts x axis (0% change)
   ○ Water potential of sucrose solution = water potential of potato cells

Question 4

explain why % change in mass is calculated. (2)

Answer 4

-enables comparison / shows proportional change
-as plant tissue samples had different initial masses

Question 5

Explain why the potatoes are blotted dry before weighing. (2)

Answer 5

-solution on surface will add to mass (only want to measure water taken up or lost)
-amount of solution on cube varies (so ensure same amount of solution on outside)

Question 6

Explain the changes in plant tissue mass when placed in different concentrations of solute

Answer 6

INCREASE IN MASS
-water moved into cells by osmosis
-as water potential of solution higher than inside cells
DECREASE IN MASS
-water moved out of cells by osmosis
-as water potential of solution lower than inside cells
NO CHANGE
-no net gain/loss of water by osmosis
-as water potential of solution = water potential of cells