= Selective biochemical reaction pathways
An <enzyme> accelerates particular <chemical reactions>, with <catalytic selectivity> arising from recognition and from the relative ease of the subsequent chemistry. Strong binding alone does not guarantee fast conversion. Nor does <catalysis> change the <chemical equilibrium> for an unchanged net <chemical reaction>. These distinctions explain how <enzymes> can direct the flow of material without creating an extra source of <free energy>.<Chemistry 2e by Paul Flowers and others>[§12.7, opening paragraphs and Enzyme Structure and Function: alternative mechanisms, regeneration and specificity]{ref}
For an original model of competing routes, let a <substrate> $S$ become either a desired <reaction product> $P$ or another <reaction product> $Q$:
$$
S\xrightarrow{k_P}P,\qquad S\xrightarrow{k_Q}Q.
$$
The labels represent hypothetical <chemical species> with the same elemental composition, not molecular structures. Assume fixed volume, irreversible <first-order reactions>, constant effective <rate constants>, and initially $100\,\mathrm{\mu mol}$ of $S$ with neither <reaction product> present. The model approximates a limited operating regime; real <enzyme kinetics> can include saturation and reverse reactions.
\Image[/biochemical-competing-reaction-routes.png]
{title=Selective acceleration changes the split between competing biochemical <reaction products>}
{description=Original illustrative <chemical reaction network> and limiting product amounts from $100\,\mathrm{\mu mol}$ of hypothetical $S$. Chosen effective <rate constants> are $k_P=0.08$ or $0.32\,\mathrm{min^{-1}}$ and $k_Q=0.02\,\mathrm{min^{-1}}$. The boxes label <chemical species>; they are not molecular structures. The bars are model limits, not experimental data.}
The <reaction-rate equations> give $dn_S/dt=-(k_P+k_Q)n_S$, $dn_P/dt=k_Pn_S$ and $dn_Q/dt=k_Qn_S$, where $n$ denotes <amount of substance>. Consequently,
$$
n_S=n_0e^{-(k_P+k_Q)t},\qquad
n_P=\frac{k_P}{k_P+k_Q}(n_0-n_S),\qquad
n_Q=\frac{k_Q}{k_P+k_Q}(n_0-n_S).
$$
With $k_P=0.08$ and $k_Q=0.02\,\mathrm{min^{-1}}$, the eventual split is $80\,\mathrm{\mu mol}$ of $P$ and $20\,\mathrm{\mu mol}$ of $Q$. At ten minutes, the calculated amounts are approximately $36.8$, $50.6$ and $12.6\,\mathrm{\mu mol}$ for $S$, $P$ and $Q$. Selectively increasing $k_P$ fourfold changes the eventual desired fraction to $0.32/0.34=16/17\approx94.1\%$. Increasing both <rate constants> fourfold instead preserves the $80\%$ split and only shortens the conversion time. Speed and selectivity answer different questions.
The bookkeeping matters independently of that model. In ideal <alcoholic fermentation>, complete conversion of $3\,\mathrm{mmol}$ of <glucose> yields $6\,\mathrm{mmol}$ each of <ethanol> and <carbon dioxide>. Of the initial $18\,\mathrm{mmol}$ of <carbon> <chemical atoms>, $12$ remain in <ethanol> and $6$ enter <carbon dioxide>. This is a chosen material-balance example using the net <stoichiometry> explained under <Saccharomyces cerevisiae>; it gives neither an observed yield nor the mechanism of the individual steps. Equal amounts of the two <reaction products> also do not mean equal <masses>.
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