**- 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 C_{3}
plant.

If you assume
as stated in the text that air is 0.036% CO_{2} then there is 3.6 x 10^{-4}
L CO_{2} per L of air (some of you have air as 0.04% CO_{2} 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^{-5}L/L. Thus ~ 3 x 10^{-4} L CO_{2}/L
air can be “fixed” by this C3 plant. Therefore
we have 3 x 10^{-4} L CO_{2} per L air x 0.1 L air in the
bottle (the CO_{2} in air should reach an equilibrium with the 99%+ H_{2}O
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 CO_{2}/L air x 0.1 L air = 3 x 10^{-5} L CO_{2} in
the bottle.

1 mole of an
ideal gas is ~ 22.4 L/mole

3 x 10^{-5} L CO_{2}/22.4 L/mole » 1.3 x 10^{-6} moles
CO_{2}.

Moles CO_{2}
ßà moles CH_{2}O (unless H_{2}O
is limiting which is not the case here)

1.3 x 10^{-5}
moles CH_{2}O 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 CH_{2}O
(minus sucrose derivatives utilized in respiration but since this will add to
the CH_{2}O in the closed container and growth up to the compensation
point then 1:1 actually). 0.1 g CH_{2}O = 100,000 mg.

3. Calculate how much CO_{2}
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 CH_{2}O; 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% CO_{2}!).

6.7 x 10^{-2}
L CO2/3.6 x 10^{-4} CO_{2}/air = 187 L air

which is » 1,900 times the volume of the 100 mL bottle!