The One Per Cent Ceiling: Why Field Photosynthesis Falls Far Short of Its Theoretical Limit — Epoche B2
It is often assumed that because evolution has had billions of years to refine photosynthesis, the process must be close to optimal. This belief is intuitive but wrong. In the field, most crops convert only about one per cent of the solar energy falling on them into new biomass over a growing season, while the theoretical ceiling for so-called C3 plants — the largest group, including wheat and rice, named for the three-carbon compound that is the first stable product of their carbon fixation — is near 4.6 per cent. That 4.6 per cent ceiling is itself the product of mechanistic losses built into the process; the further fall to about one per cent in the field is driven largely by canopy-level and environmental factors: incomplete light interception by young or sparse canopies, light saturation of upper leaves at midday, seasonal and agronomic downtime when no crop is standing, and water and nutrient limitation. This essay traces the chain of mechanistic losses that sets the ceiling, following the accounting of Zhu, Long and Ort (2008). Defining the efficiency Efficiency here means one thing only: the chemical energy stored in new dry matter divided by the solar energy incident on the same ground area over the same period. Because the energy passes through a series of stages, and each stage keeps only a fraction of what the previous one delivered, the overall efficiency $\eta$ is a product of factors rather than a sum of them: $$\eta = f_{\text{abs}} \cdot f_{\text{leaf}} \cdot f_{\text{photon}} \cdot f_{\text{quantum}} \cdot f_{\text{photoresp}} \cdot f_{\text{resp}}$$ Here $f_{\text{abs}}$ is the fraction of the solar spectrum that can be absorbed at all, $f_{\text{leaf}}$ the fraction of that light which the leaf actually captures rather than reflecting or transmitting, $f_{\text{photon}}$ describes energy lost within each absorbed photon, $f_{\text{quantum}}$ the efficiency of using photons to fix carbon, $f_{\text{photoresp}}$ the fraction of fixed carbon retained after photorespiratory losses, and $f_{\text{resp}}$ the fraction of that carbon which survives mitochondrial respiration — the plant burning some of its own sugar to power growth and maintenance. Expanding the original version of this account has revealed one omission worth stating plainly: the leaf-capture term $f_{\text{leaf}}$ was missing from the formula as first written, though the loss it represents is real and is included in the published budgets. Because these factors multiply, each loss compounds the next: a stage that keeps a third of what reaches it destroys two thirds of everything the earlier stages worked to preserve. Where the energy goes Each factor can be given a number, and each number has a reason. Spectral loss: $f_{\text{abs}} = 0.487$. Photosynthesis uses only photosynthetically active radiation (PAR), roughly $400$–$700\,\text{nm}$, because these are the wavelengths chlorophyll and its accessory pigments can absorb. Shorter-wavelength ultraviolet is largely filtered by the atmosphere and is damaging rather than useful; longer-wavelength infrared photons individually carry too little energy to lift an electron across the gap in a chlorophyll molecule, and simply warm the leaf. Integrating the measured solar spectrum at the Earth's surface over $400$–$700\,\text{nm}$ gives about 48.7 per cent of the total energy — so more than half the sunlight is disqualified before any biology begins. Leaf capture: $f_{\text{leaf}} \approx 0.90$. A leaf is not a perfect absorber. Roughly 10 per cent of the PAR striking it is reflected from the waxy surface or passes straight through, which is why a leaf held to the sun glows green rather than appearing black. Of the original 100 units of sunlight, 43.8 remain. Photon energy loss: $f_{\text{photon}} \approx 0.85$. Here the quantum nature of light imposes a tax that has no analogue in ordinary engineering. A photon's energy depends on its wavelength, $E = hc/\lambda$, which for convenience gives $E\,[\mathrm{eV}] \approx 1240/\lambda\,[\mathrm{nm}]$: a blue photon at $450\,\text{nm}$ carries $2.76\,\text{eV}$, a red one at $700\,\text{nm}$ only $1.77\,\text{eV}$. But a photosystem does not care how much energy a photon brought. Absorption raises an electron to an excited state, and within picoseconds that excitation relaxes down to the lowest excited level of the reaction centre — equivalent to a red photon at about $700\,\text{nm}$ — with the surplus released as heat. Every photon is therefore worth the same $1.77\,\text{eV}$ to the chemistry, whatever it was worth on arrival. Averaged over the PAR photons in sunlight, whose mean energy corresponds to roughly $590\,\text{nm}$, this "thermalisation" discards about 15 per cent of the absorbed energy, leaving 37.2 units. The cost of fixing one carbon atom The largest single loss is the next one, and it can be derived rather than merely asserted. Fixing one molecule of $\text{CO}_2$ into carbohydrate in the Calvin–Benson cycle — the sequence of reactions that builds sugar from carbon dioxide — requires 3 molecules of ATP (the cell's energy currency) and 2 of NADPH (its reducing currency, a carrier of electrons). The figure of 2 NADPH follows from simple bookkeeping: carbon in $\text{CO}_2$ is fully oxidised and carbon in carbohydrate, $(\text{CH}_2\text{O})$, is not, and closing that gap takes 4 electrons; each NADPH delivers 2; hence 2 NADPH. Now count photons. Each of those 4 electrons is stripped from water and passed along a chain through two photosystems in series, and each photosystem must absorb one photon to move one electron across its own step. Two photosystems times 4 electrons gives $$8 \text{ photons per } \text{CO}_2\text{ fixed},$$ and this is why the exponent of the argument is two photosystems and not one: photosynthesis splits the job of lifting an electron from water (a very poor electron donor) to NADP$^+$ into two smaller photochemical steps, because no single pigment in the visible range has enough en