Midday Photosynthesis Collapse in Irrigated Wheat at Anthesis
A wheat crop (cv. HD-3086) on a sandy loam in a semi-arid block is at 50% anthesis in the second week of March. The block received 60 mm of rain 12 days ago and has not been irrigated since. On a clear day (max 31 °C, RH 34%, wind 12 km/h) a porometer records flag-leaf stomatal conductance falling from 0.38 to 0.09 mol m-2 s-1 between 09:30 and 13:00, net photosynthesis falling from 21 to 8 µmol m-2 s-1, leaf water potential dropping from -1.0 to -2.0 MPa and recovering to -1.3 MPa by 18:00. Intercellular CO2 falls from 290 to 190 µmol mol-1 over the same window. Gravimetric soil moisture is 11% at 0-15 cm and 17% at 15-45 cm, against a field capacity of 22% for both layers. The farmer blames a 'nutrient problem' because the tips of the older leaves are yellowing, and wants 25 kg N/ha broadcast as urea at midday.
1. Is the farmer's 'nutrient problem' the primary constraint on photosynthesis at 13:00? Justify your answer using the gas-exchange and soil-moisture data, and explain what the yellowing of the older leaf tips most likely indicates.
Read the worked answer
The primary constraint is water, not nutrition. The diagnostic pair is A and Ci: net photosynthesis falls by 62% while intercellular CO2 also falls, from 290 to 190 µmol mol-1. A fall in Ci during a fall in A means the supply side of the pathway is throttled - the chloroplasts are being starved of CO2 because stomata have closed. If the limitation were biochemical (inactivated Rubisco, damaged electron transport), A would fall while Ci rose or stayed high, because CO2 would accumulate inside the leaf behind closed stomata.
The supporting evidence is consistent: stomatal conductance drops by 76%, leaf water potential reaches -2.0 MPa and recovers to -1.3 MPa by evening (a reversible midday water deficit, not permanent damage), and the subsoil is only at 17% against a field capacity of 22%, i.e. 77% of field capacity, so there is no subsoil reserve to buffer the drying topsoil (11% = 50% of field capacity).
Yellowing of the tips of older leaves is consistent with nitrogen deficiency (N is mobile and moves out of old leaves), but it must be confirmed by tissue analysis before any fertilizer decision; other causes such as sulphur or root-zone constraints should be ruled out. In any case, spreading urea at midday on a dry surface is poor practice: urea needs moisture to hydrolyse, and the ammonia released can volatilize at 31 °C before it is incorporated, while nitrogen applied at anthesis cannot do much for a yield already largely set by the flag leaf.
2. Describe the measurements or experiments you would use to separate stomatal from mesophyll (biochemical) limitation, and to establish whether the midday depression has caused lasting photoinhibition.
Read the worked answer
Paired A-Ci response curves at 09:30 and 13:00. Fit Vcmax and Jmax from the curves. If these parameters are essentially unchanged and only the operating point on the curve has shifted, the limitation is stomatal; if Vcmax/Jmax fall, mesophyll capacity is impaired.
The limitation index l = (Ca - Ci)/Ca. An increase at 13:00 relative to 09:30 quantifies stomatal limitation; a decrease, or high Ci with low A, points to non-stomatal limitation.
Small rewatering test. Irrigate a marked strip and re-measure after 24-48 h. Recovery of A within one or two days demonstrates that the depression was water-driven and reversible.
Chlorophyll fluorescence. Measure Fv/Fm before dawn. A value near 0.78-0.80 indicates healthy photosystem II; a depression that recovers by the next dawn is dynamic photoinhibition (protective), whereas a depression that persists for several days is chronic photodamage.
Leaf temperature relative to air temperature, and the difference (Tleaf - Tair), to confirm that transpirational cooling has been lost.
Do not rely on any single measurement: the A-Ci curve is the decisive test, the rewatering and fluorescence tests tell you whether the damage is permanent, and together they determine whether the correct response is irrigation, shading, or accepting the loss.
3. Calculate how much irrigation water the 0-45 cm layer needs to be returned to field capacity, and state when and how the water should be applied if application efficiency is 70%.
Read the worked answer
Work layer by layer in depth of water (percentage x layer depth):
- 0-15 cm: deficit = 22 - 11 = 11 percentage points; 11% x 15 cm = 1.65 cm = 16.5 mm.
- 15-45 cm: deficit = 22 - 17 = 5 percentage points; 5% x 30 cm = 1.50 cm = 15.0 mm.
- Total net requirement = 3.15 cm = 31.5 mm.
Gross requirement at 70% application efficiency = 31.5 / 0.70 = 45 mm (450 m3/ha). For a 0.4 ha block this is about 180 m3, roughly 4.5 hours with a 40 m3/h tubewell.
Judgement points: anthesis to early grain filling is the most water-sensitive stage, so this is not a water to defer. Apply in the evening or at night to cut evaporation and to avoid irrigating into the hottest part of the day, and split the 45 mm into two runs, because a sandy loam takes roughly 1.5 cm/h and a single 45 mm application will pond and run off. Check that the field is not crusted and that the tail end is not over-irrigated. After this irrigation, schedule the next one at about 50-60% depletion of available water rather than waiting for visible wilting, and keep a simple IW/CPE record to build a local rule.