Answers to Leaf Gas Exchange Problems


1. Stomatal resistances are variable, because stomata can open and close.  Cuticular resistances are thought to be constant until severe water deficits, at which point the cuticle may crack, and this would lower the resistance to water vapor loss.  The diagram should show all the possible flows of water as it can leave a leaf.  Compare a hypostomatous vs amphistomatous leaf.

This is a very involved question, and I would not ask such a difficult one on your test, BUT, I do expect you to understand how this calculation works!!

Check your lecture notes for how to calculate total leaf resistance from one side. For doing two sides, assume that each leaf surface is in parallel. Then, add the inverse of the resistance for each side:

1       +       1         =     1
     rupper        rlower         rtotal

Substitute as follows:

a. For each side of the leaf, the leaf resistance is in series with the boundary layer resistance, so
                            rupper = ruleaf + ruboundary

For the lower surface,
                            rlower = rlleaf + rlboundary

b. Assume that cuticular resistances are embedded within each rl for now.
c. Now add the inverses and cross multiply to get common denominators:

             1                                       1                                     =              1
  [ruleaf + ruboundary]            [rlleaf + rlboundary]                                     rtotal

Cross multiply and get:

[rlleaf + rlboundary]*[ruleaf + ruboundary]                =  rtotal


[rlleaf + rlboundary] + [ruleaf + ruboundary]

d. In this particular case, stomatal conductance of the abaxial (bottom) leaf surface is twice that of the upper (adaxial) surface. In resistance terms, this means that the abaxial stomatal resistance is one half that of the adaxial surface:

                            (rsupper = 2*rslower).

e. From our notes, the one sided leaf resistance (with cuticular resistances still included) is equal to:

ruleaf = 2*rslower * rc / [2*rslower + rc]                     Adaxial surface

rlleaf = rslower * rc / [rslower + rc]                             Abaxial surface

f. Substitute the upper and lower leaf resistances into the equation for rtotal from step c.

[rslower * rc / [rslower + rc] + rlboundary]*[2*rslpwer * rc / [2*rslpwer + rc] + ruboundary]                    =    rtotal


[rslower * rc / [rslower + rc] + rlboundary] + [2*rslpwer * rc / [2*rslpwer + rc] + ruboundary]

Looks horrendous, but this is what you're left with. Now, if the boundary layers are equivalent on both sides (a good assumption) then you might be able to simplify. Same goes if cuticular resistances are the same on both sides (not always a good assumption). However, with spreadsheets, a computer can solve this equation easily.  Anyway, it goes to show that determination of the total leaf resistance is not a trivial task!

2. Use the equation in step e of question 1 and add in boundary layer resistance (remember, it is in series and can be added directly). Thus:

rleaf = rs * rc / [rs + rc] + rb

rleaf= 0.5*2.0/[0.5+2.0] + 1.0 = 0.4 + 1.0 = 1.4 s/cm

3. Our formula for transpiration (E) is: E = VPD*gl
                                                                    BP
Here, VPD/BP = 0.025 kPa/kPa and gl = 400 mmol m-2 s-1

So: E = 0.025*400 = 10 mmol m-2 s-1


4. If gl remains constant, then we have an equation for a straight line. If gl decreases as VPD increases, then we no longer have a linear relationship, and the line will curve downward.  Most stomata tend to close as the humidity goes down, thus preventing excessive water loss from leaves.  It is rare to find a situation where the conductance does not change as VPD does.  I leave it to you to graph these relationships.


5. Yes, if the gl decreased so much that it greatly inhibited the flow of water out of the leaf, even though VPD was high. Could a leaf transpire in air with 100% RH? The answer is yes, but only if it is in the sun. If the leaf was in the sun, it would absorb heat from the sun's radiation, and be slightly above air temperature. This would raise the vapor pressure in the leaf to a value above that of the surrounding air, thus providing a gradient in vapor pressure which would lead to leaf water loss. In the absence of a temperature difference between leaf and air, both the leaf and air would have the same vapor pressure (as may occur at night) and there would be no net exchange of water vapor.
    This also brings up another point not emphasized in class - that water loss from leaves, while mostly diffusional, is also due in part due to pressurized flow out the stomata.  When a leaf does heat up above air temperature, a slight pressure can build up in the leaf, resulting in a mass flow of water vapor out of the leaf.  This may add an additional 2-10% to the transpiration rate above the amount resulting from diffusion alone.  If you were to really get into doing gas exchange, you would make allowances for this in your calculations, but for the purposes of this class, we'll ignore it for now.  If you take Organismal Ecology, you will learn how to deal with this phenomenon.

6. To calculate the water potential of the air, use the formula provided in your notes:

Yw = RTln(e/eo)
       Vw

For e/eo, which is RH in absolute terms, the water potential when the RH is 100% is zero, since the ln of 1 is always zero. If the %RH = 99% at 25oC, then:

Yw= (0.0813)(298)ln(0.99) = -13.52 bars (-1.3 MPa)

                                                         0.018

For 90% RH:(0.0813)(298)ln(0.90) = -141.8 bars (-14.2 MPa)

                                                               0.018

For 75% RH: = -387.2 bars (-38.7 MPa)

I leave it up to you to solve the latter parts of this question.  Simply plug in the appropriate values into the above equation.

7. If the leaf is at 23oC, then its vapor pressure is 2.809 kPa (read off your chart).
If the %RH = 60%, then at an air temperature of 25oC, the vapor pressure would be:

                                               VPair = 0.60*3.167 = 1.900 kPa

The VPD would be:

VPleaf - VPair = 2.809 - 1.900 = 0.909 kPa

Solve our equation for gl:                         gl = [BP*E]/VPD

gl= [101.3*10]/0.909 = 1114 mmol m-2 s-1

If this were done in Boone, where the BP is only 91.3 kPa, simply substitute this into the above equation:

gl= [91.3*10]/0.909 = 1004 mmol m-2 s-1

If you did this in Boone, the new conductance would be 91.3*10/0.909 = 1004 mmol m-2 s-1

This is almost an 11% difference. So, you see how important it is to correct for the barometric pressure in the calculation of leaf conductance.


8. For this problem, you want to solve for leaf temperature.  Take our standard equation from question 7 and rearrange to solve for leaf temperature as follows:

E = gl*(VPD/BP) = gl*(VPl  -  VPa)/BP

(VPl  -  VPa) = E*BP/gl

VPl  = E*BP/gl  + VPa
VPl  = 10*101.3/700 + 0.50*4.243 = 3.569 kPa

This means that leaf temperature is just a hair over 27oC as determined from your vapor pressure chart.  Remember, if you know the vapor pressure of your leaf (which is the saturation vapor pressure at that temperature) you can determine leaf temperature.

If the air temperature suddenly decreased by 5oC, the VPair would still be 2.122 kPa, since we're cooling the air, but not taking any water out of it. This is an important concept to master - changing the air temperature does not change the amount of water vapor in it, unless you go below the dew point, at which point water will condense out of the air.  In our case, dropping the temperature does not saturate the air with water vapor. To do that, we would have had to lower air temperature to just below 19oC (at 18oC the VPsat = 2.063 kPa).

Now gl is only 350 mmol m-2 s-1 and E is just 5 mmol m-2 s-1. Substitute into our equation:

5*101.3/350 + 2.122 = 3.569 kPa

This gives us an estimated leaf temperature exactly the same as before, since the lower E is compensated by the lower leaf conductance. But if leaf conductance stayed the same, at 700, then we would have:

5*101.3/700 + 2.122 = 2.846 kPa

This results in a VPsat between 24oC and 23oC.  The exact leaf temperature can be estimated by linear extrapolation.  Do the following:

First, determine the difference in VP between 24 and 25oC:    2.983 - 2.809 = 0.174 kPa
Next, figure out the difference in VP between our estimate above and that at 23oC: 2.846-2.809 = 37
Now determine the percent of the total difference between the two temperatures that this difference is, and that is also the percent difference in temperature.  That is, 37*100/174 = 21.3% of the total difference.  Therefore, the temperature at which the air would be saturated is 21.3% of the distance from 23oC, or 23.2oC.  Thus our leaf new leaf temperature is 23.2oC.  The reason the leaf temperature decreases is the fact that with less transpiration, there is less cooling due to evaporation and because the air temperature also dropped.


10. At 20oC, the saturation VP is 2.338 kPa, and at 70% RH, it is 0.70*2.338 = 1.637 kPa

If the air heats to 30oC, then the RH is 1.637*100/4.243 = 39% RH

Air at 65% RH at 27oC has a VP of 2.317 kPa. The leaf has a VP of 2.809 kPa. The

VPD is 2.809-2.317 = 0.492 kPa

If the leaf temperature increases by 2oC, the new VPD is 3.167-2.317 = 0.850 kPa


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