- 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 ~ 380 ppm or 0.038% CO₂ then there is 3.8 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). The compensation point for typical C3 plants is ~ 50 - 60 ppm or ~6 x 10^-5L/L air. 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 at ~ pH 7 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 can be converted into 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 ug of growth*; small even for Arabidopsis!

*There will also some dry matter accumulation due to uptake of minerals particularly incorporation of N into protein but this won't change the total number much. 

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 = ~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 CO2/3.8 x 10^-4 CO₂/air = 176 L air

This is ~1,800 times the volume of the 100 mL bottle!