Exercises
Six Exercises, with Worked Solutions
Graded from a quick estimate to an open argument. Hints are inline; the full solutions are below, so cover them if you want the problem first.
The problems
1. Colour tuning
Using the free-electron model, how much longer would the conjugated chain need to be to shift absorption from 470 nm to 590 nm, with N kept fixed? Then explain why real red-shifted channelrhodopsins achieve this through the protein environment instead.
Hint: λ scales as L², so L must grow by √(590/470) ≈ 1.12.
2. Photon energy budget
A 470 nm photon carries about 2.6 eV. Convert this to kJ/mol and compare it with the barrier of more than 1 eV to thermal trans→cis isomerization in the ground state. Why does the molecule almost never isomerize in the dark?
Hint: 1 eV per molecule ≈ 96.5 kJ/mol.
3. Timescales
Place these on a logarithmic time axis: photon absorption (~1 fs), passage through the conical intersection (~100s of fs), channel opening (~1 ms), membrane charging τ (~15 ms), a behavioural response (~1 s). How many orders of magnitude separate the first and last?
Hint: Count powers of ten between 10⁻¹⁵ s and 10⁰ s.
4. Channel count
Repeat the estimate in Part 3 for a small neuron with C = 30 pF and R = 500 MΩ. Why are small neurons easier to drive with light?
Hint: The threshold current drops to about 40 pA.
5. Photon flux
Typical optogenetic experiments use about 1 to 10 mW/mm² of blue light at the target. Calculate the photon flux in photons per second per µm² at 5 mW/mm² and 470 nm.
Hint: Divide intensity by the energy per photon, hc/λ.
6. Discussion
Is optogenetics an example of quantum biology? Argue both sides: the trigger is a quantum event, but the outcome is a classical, many-channel average. Where would you draw the line?
Hint: There is no single right answer here — state your criterion and apply it consistently.
Worked solutions
1. Colour tuning
Since λ ∝ L² at fixed N, the new length is L′ = L √(590/470) ≈ 1.12 L. Starting from 15.4 Å, L′ ≈ 17.3 Å, about 1.9 Å or 1.3 extra bonds. But a longer chain means more π electrons, which pulls N up too, and the molecule would no longer fit the protein pocket. Nature instead keeps retinal fixed and moves charges around it: weakening the counter-ion near the nitrogen spreads the positive charge along the chain and narrows the gap.
2. Photon energy budget
\[ E_{\text{photon}} = 2.6\ \text{eV} \times 96.5\ \tfrac{\text{kJ/mol}}{\text{eV}} \approx 250\ \text{kJ/mol} \]
That is more than twice the ground-state barrier. In the dark, the chance that thermal motion supplies a barrier Ea of at least 1 eV is set by the Boltzmann factor, with kBT ≈ 0.0267 eV at 310 K:
\[ e^{-E_a/k_B T} \le e^{-1/0.0267} = e^{-37} \approx 10^{-16} \]
Thermal isomerization is therefore extremely rare, which is what makes retinal a clean, low-noise light switch. Visual rhodopsin isomerizes spontaneously only about once in centuries per molecule.
3. Timescales
From about 10⁻¹⁵ s (absorption) to about 1 s (behaviour) is fifteen orders of magnitude. For scale: if absorption took one second, the behavioural response would take about 30 million years.
4. Channel count for a small neuron
The rheobase is Imin = uth / R = 20 mV / 500 MΩ = 40 pA. At 3.5 fA per channel, that is about 11,000 open channels, roughly a quarter of the large neuron's 40,000. The time constant is unchanged, τ = RC = (500 MΩ)(30 pF) = 15 ms, as the RmCm argument predicts. Small cells are easier to drive because their high input resistance converts a small current into a large voltage.
5. Photon flux
Energy per photon:
\[ E = \frac{hc}{\lambda} = \frac{1240\ \text{eV}\!\cdot\!\text{nm}}{470\ \text{nm}} \approx 2.64\ \text{eV} \approx 4.2\times10^{-19}\ \text{J} \]
The intensity is 5 mW/mm² = 5 × 10⁻³ W per 10⁶ µm² = 5 × 10⁻⁹ W/µm². Dividing:
\[ \Phi = \frac{5\times10^{-9}\ \text{W}/\mu\text{m}^2}{4.2\times10^{-19}\ \text{J}} \approx 1.2\times10^{10}\ \text{photons}\ \text{s}^{-1}\,\mu\text{m}^{-2} \]
Each channel, though, is a tiny target: retinal's absorption cross-section is about 2 × 10⁻¹⁶ cm² = 2 × 10⁻⁸ µm². The rate of absorptions per channel is Φσ ≈ (1.2 × 10¹⁰)(2 × 10⁻⁸) ≈ 200 per second. With a channel cycle of about 10 ms, that is enough to keep a large fraction of channels open at once, which is why light levels near 1–10 mW/mm² are standard.
6. Discussion — guidance, not an answer
For: the trigger is irreducibly quantum, with a non-adiabatic transition through a conical intersection whose efficiency the protein has tuned. Against: by the time anything biologically relevant happens, the signal is a classical average over tens of thousands of channels, and no quantum coherence survives. A useful distinction is between biology that uses quantum events — true of all photochemistry, including vision and photosynthesis — and biology whose function depends on sustained quantum coherence, which remains debated. State where you draw the line and why.