- Take home problem - solve by March 30 (Answer in Blue):
1. Consider the growth of Arabidopsis plants on complete mineral salts on 10 mL 0.5% agar in a 100 mL air-tight bottle. Calculate the maximum theoretical dry weight increase.
Note Arabidopsis is a C3 plant.
If you assume as stated in the text that air is 0.036% CO2 then there is 3.6 x 10-4 L CO2 per L of air (some of you have air as 0.04% CO2 which is fine and will give a similar answer; I’m not sure of the current level). The compensation point for typical C3 plants is ~ 60 mL/L air or 6 x 10-5L/L. Thus ~ 3 x 10-4 L CO2/L air can be “fixed” by this C3 plant. Therefore we have 3 x 10-4 L CO2 per L air x 0.1 L air in the bottle (the CO2 in air should reach an equilibrium with the 99%+ H2O in the agar but if you subtracted the volume of the agar this is OK and the final number will be very close).
3 x 10-4 L CO2/L air x 0.1 L air = 3 x 10-5 L CO2 in the bottle.
1 mole of an ideal gas is ~ 22.4 L/mole
3 x 10-5 L CO2/22.4 L/mole » 1.3 x 10-6 moles CO2.
Moles CO2 ßà moles CH2O (unless H2O is limiting which is not the case here)
1.3 x 10-5 moles CH2O x 30g/mole » 4 x 10-5 g or 40 mg of growth; small even for Arabidopsis!
2. Calculate the dry weight increase when the agar is supplemented with 1 % sucrose.
10 mL 1% sucrose = 0.1 g or 1 x 10-1 g sucrose
g sucrose ßà g CH2O (minus sucrose derivatives utilized in respiration but since this will add to the CH2O in the closed container and growth up to the compensation point then 1:1 actually). 0.1 g CH2O = 100,000 mg.
3. Calculate how much CO2 would need to added to this closed atmosphere in the bottle for growth only on mineral salts to result in a dry weight increase = to that for plants grown on 1 % sucrose.
0.1 g sucrose => 0.1 g CH2O; 1 x 10-1 g/30 g/mole = 3 x 10-3 moles.
3 x 10-3 moles x 22.4 L/mole = 6.7 x 10-2 L or 67 mL (if this much were added to the bottle it would be ~ 67% CO2!).
6.7 x 10-2 L CO2/3.6 x 10-4 CO2/air = 187 L air
which is » 1,900 times the volume of the 100 mL bottle!