Calvin cycle: what this calculator is modeling
The Calvin cycle (light-independent reactions) is the pathway where plants use ATP and NADPH
(made by the light reactions) to convert CO2 into carbohydrate.
For planning and teaching, the most useful part is the carbon and energy accounting:
how many moles of CO2, ATP, and NADPH are associated with a given carbohydrate target.
Key simplification used
Standard textbook bookkeeping (net synthesis):
per CO2 fixed the Calvin cycle consumes approximately
3 ATP and 2 NADPH.
This already includes the costs of the regeneration steps.
Stoichiometry summary used by the calculator
Where these numbers come from (general formulas)
Let the product contain C carbon atoms per molecule (for example: G3P has C = 3, glucose has C = 6).
Each CO2 contributes one carbon, so:
\[
n_{\mathrm{CO_2}} = C \cdot n_{\mathrm{product}}
\]
Using the standard per-CO2 energy costs:
\[
n_{\mathrm{ATP}} = 3 \cdot n_{\mathrm{CO_2}}, \qquad
n_{\mathrm{NADPH}} = 2 \cdot n_{\mathrm{CO_2}}
\]
For example, if the target is 1 mol glucose (C = 6), then
\(n_{\mathrm{CO_2}} = 6\) mol, \(n_{\mathrm{ATP}} = 18\) mol, and \(n_{\mathrm{NADPH}} = 12\) mol.
How the calculator handles the three modes
Mode 1 Sugar target → requirements
You enter a carbohydrate type and an amount (mol or grams). The calculator converts grams to moles (if needed),
then computes CO2, ATP, and NADPH requirements using the formulas above.
Mode 2 CO2 fixed → sugar potential
You enter CO2 fixed and choose which product to express the output as.
The calculator converts CO2 to “product-equivalent moles” by dividing by the carbon count:
\[
n_{\mathrm{product,theoretical}} = \frac{n_{\mathrm{CO_2}}}{C}
\]
Mode 3 Per turn (per CO2)
A “turn” here means fixing 1 CO2. The calculator reports:
\(3\) ATP and \(2\) NADPH per turn, plus optional scaling to any number of turns you enter.
Efficiency factor: modeling losses (0–100%)
Real photosynthesis is not 100% efficient (photorespiration, respiration, limitations in light, CO2, and temperature).
The calculator uses an efficiency factor \( \eta \) between 0 and 1 to scale the theoretical result.
How efficiency is applied in each direction
From sugar target → requirements
If only a fraction \( \eta \) of the theoretical output is realized, you need to “aim higher” in inputs:
\[
n_{\mathrm{CO_2,required}} = \frac{n_{\mathrm{CO_2,theoretical}}}{\eta}, \quad
n_{\mathrm{ATP,required}} = \frac{n_{\mathrm{ATP,theoretical}}}{\eta}, \quad
n_{\mathrm{NADPH,required}} = \frac{n_{\mathrm{NADPH,theoretical}}}{\eta}
\]
From CO2 fixed → sugar produced
The net carbohydrate potential is reduced:
\[
n_{\mathrm{product,net}} = \eta \cdot n_{\mathrm{product,theoretical}}
\]
If \( \eta = 1 \), you recover the textbook stoichiometry. If \( \eta = 0.80 \), you model a 20% loss.
Worked examples
Example A: glucose target
Target: 2 mmol glucose, assume \( \eta = 1 \).
\[
C = 6,\quad n_{\mathrm{product}} = 2\ \mathrm{mmol}
\]
\[
n_{\mathrm{CO_2}} = C\cdot n_{\mathrm{product}} = 6 \cdot 2 = 12\ \mathrm{mmol}
\]
\[
n_{\mathrm{ATP}} = 3\cdot n_{\mathrm{CO_2}} = 3 \cdot 12 = 36\ \mathrm{mmol}
\qquad
n_{\mathrm{NADPH}} = 2\cdot n_{\mathrm{CO_2}} = 2 \cdot 12 = 24\ \mathrm{mmol}
\]
Example B: CO2 fixed → G3P equivalent
CO2 fixed: 30 mmol. Express as G3P (C = 3). Assume \( \eta = 0.90 \).
\[
n_{\mathrm{product,theoretical}} = \frac{n_{\mathrm{CO_2}}}{C} = \frac{30}{3} = 10\ \mathrm{mmol}
\]
\[
n_{\mathrm{product,net}} = \eta\cdot n_{\mathrm{product,theoretical}} = 0.90 \cdot 10 = 9\ \mathrm{mmol}
\]
Example C: per turn (per CO2)
For 1 CO2 fixed (one “turn”):
\[
n_{\mathrm{ATP}} = 3,\qquad n_{\mathrm{NADPH}} = 2
\]
For 12 turns, multiply both by 12: \(36\) ATP and \(24\) NADPH.