PBIO*3110 – Crop Physiology Lecture #13 Fall Semester 2008 Lecture Notes for Tuesday 21 October Canopy Photosynthesis III – Canopy Photosynthesis Calculations Can we calculate the effects of leaf angle and LAI on both canopy photosynthesis and RUE? Learning Objectives 1. Be able to calculate from theory the photosynthetic rates of two crop canopies that differ only in their leaf angles. 2. Know how to calculate crop radiation use efficiency (RUE) in units of g / MJ. Introduction In the previous lecture we saw how higher leaf angles could result in higher canopy photosynthetic rates provided that: a) LAI was high enough that most of the incident PAR was still intercepted despite the higher leaf angles, and
1 b) incident PPFD normal to the leaf was high enough that leaves were operating in the levelling off or plateau phases of the PRI curve. In this lecture we will gain some more practice in performing the canopy photosynthesis calculations, comparing two canopies that differ in only their leaf angles. Then, we will use that information to calculate the whole canopy radiation use efficiency to understand how that component of the yield equation is affected by canopy structure. Equations Recall the following equations that we used in the previous lecture to calculate the whole canopy photosynthetic rate of an idealized crop canopy with a uniform leaf angle: PPFD normal to the leaf surface (PPFDi, with units of µmol photons m ­2 leaf s ­1 ) equals: PPFDi = PPFDo × k = PPFDo × cos(a) For optically black leaves (i.e., PPFDa = PPFDi), gross CO2 assimilation rate per unit leaf area (AG(leaf), with units of µmol CO2 m ­2 leaf s ­1 ) is: A G =
PPFD a × f × A max PFPD a × f + A max Sunlit leaf area in the crop canopy (LAIS, with units of m 2 leaf m ­2 ground) is LAIs = I’A / k = (1 ­ e ­k ∙ LAI ) / k Canopy gross CO2 assimilation rate (AG(canopy), with units of µmol CO2 m ­2 leaf s ­1 ) AG(canopy) = AG(leaf) ∙ LAIs
2 Fig. 1. Example of vertical leaf angle distribution in maize.
3 Fig. 2. An example of a horizontal leaf angel distribution in maize.
4 A Tale of Two Canopies Here is the problem that we will work through together in class: a) Consider two canopies, one with a leaf angle of 30°, and the other with a leaf angle of 60° (see Figs. 1 and 2). They are otherwise identical, with the following characteristics:
· · · · incident PPFD at the top of the canopy (PPFDo) = 2000 µmol m ­2 s ­1
LAI = 5
ф = 0.07 mol CO2 mol ­1 photon
Amax = 50 µmol CO2 m ­2 s ­1 Using the above information, fill in the following table: Parameter Canopy with 30° leaf angle k I’A (%) PPFDi (µmol PPFD m ­2 s ­1 ) AG (leaf) (µmol CO2 m ­2 s ­1 ) LAIs AG (canopy) (µmol CO2 m ­2 s ­1 ) Canopy with 60° leaf angle cos(30°) = 0.87 1 ­ exp(­0.87 × 5) = 0.987 2000 × 0.87 = 17400 (1740×0.07×50)/(1740×0.07+50) = 35.4 0.987/0.87 = 1.13 1.13 × 35.4 = 40.0 What is the percentage difference in AG between the two canopies? Suggestion: for practice, do the calculations above again, only this time using a LAI of 1.0. Put your results in the following table, and again calculate the percentage difference in AG between the two canopies. Parameter Canopy with 30° leaf angle k I’A PPFDi AG (leaf) LAIs AG (canopy)
5 Canopy with 60° leaf angle Radiation Use Efficiency How does canopy leaf architecture impact crop growth rate or, more specifically, the efficiency of conversion of absorbed solar radiation into crop dry matter (i.e., ε in the yield equation). Radiation use efficiency (RUE) is expressed as dry matter accumulation (g DM m ­2 d ­1 ) per unit intercepted PAR (MJ m ­2 d ­1 ). Let’s calculate RUE of the two example canopies above (i.e., LAI of 5.0, leaf angles of 30° and 60°). First, we must convert our canopy photosynthetic rate to a dry matter accumulation rate in g DM m ­2 s ­1 . For now, we will make the simplifying assumption that the new crop dry matter is made of carbohydrate with the stoichiometry of CH2O. That implies 30 g of dry matter are accumulated for every mol of C (or, mol of CO2) fixed. (Be sure you can calculate that!). So, the rate of dry matter accumulation in g DM m ­2 ground s ­1 is AG x 30 g DM mol ­1 CO2 = µmol CO2 m ­2 ground s ­1 x 10 ­6 mol µmol ­1 x 30 g DM mol ­1 CO2 Note that this value has units of g DM m ­2 s ­1 . This is a Crop Growth Rate (CGR), albeit one with a short time interval (one second). Next, we need to determine how much PAR was absorbed the crop canopy in order to produce that dry matter accumulation rate we just calculated, with units of MJ s ­1 We know the incident PPFD, with units of µmol m ­2 s ­1 , and also the fraction of that incident PPFD that was absorbed by the canopy (I’A). So, all we need to know is the amount of energy contained in those absorbed photons. Recall from Lecture #5 that one mol of PPFD contains approximately 217 kJ of energy (217 x 10 3 J). So, energy of the absorbed radiation in J m ­2 s ­1 is: PPFD x I’A x 217 x 10 3 J = = (µmol photon m ­2 ground s ­1 ) x (10 ­6 mol µmol ­1 ) x (217 x 10 3 J mol ­1 photon) Then, divide this value by 10 6 J MJ ­1 to convert the value to MJ m ­2 s ­1 . We can now calculate the RUE as: CGR / PAR(absorbed) = = g DM m ­2 ground s ­1 / MJ m ­2 ground s ­1 = g MJ
6 What is the percent difference in RUE for the two example canopies? How does this compare to the percent difference in the canopy photosynthetic rates calculated above? What accounts for this difference? How does the estimated value of RUE compare with empirical values of RUE? RUE of the canopy with LAI=5 and a 60° leaf angle is 4.04 g/MJ. Maximum values of RUE for maize are in the order of 3.3 g/MJ (see Lecture #2). Note that RUE in this example is calculated on the basis of gross photosynthesis and respiration may be in the order of 30%. Consequently, RUE would be estimated as 0.7 × 4.04 g/MJ = 2.83 g/MJ.
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