Past exam of the biological physics course of the University of Oxford / 2017 / 1

Start both systems from zero and compare their times to reach half of their own steady concentrations. Without regulation,In the strong logic approximation for gene regulation, production is initially fully on and the concentration rises towards the repression threshold. The regulated level is pinned near $Y_{\mathrm{st}}\simeq K$, so it reaches $K/2$ before switching:The faster relative response comes from allowing high initial production while keeping the final abundance low. Negative autoregulation can reduce response delays and sensitivity to production fluctuations, and can reduce unnecessary accumulation of a protein. It does not suppress every possible noise source at every frequency.

A perfectly discontinuous on/off right-hand side has a threshold-switching ambiguity rather than an ordinary smooth equilibrium at $K$. Here threshold pinning means the limiting rapid-switching or steep smooth-response idealization. A continuous repression law resolves that idealization. Comparing half of the same prescribed concentration, rather than half of each model's final concentration, would be a different response-time comparison.

For the strong Hill autorepression response, in the requested approximation, $\dot{Y}=\beta K^n{Y}^{-n}-\alpha Y$. Put $z=Y^{n+1}$. Thenso the formal zero-initial-concentration solution isSetting $Y(T_{1/2})=Y_{\mathrm{st}}/2$ givesThe formal ratio decreases with positive $n$; for $n=1$ it is about $0.208$, and for $n=2$ about $0.0642$. Its large-$n$ asymptote is $2^{-(n+1)}/[(n+1)\log 2]$. This is a nonlinear half-response time, not simply the near-equilibrium relaxation time $1/[(n+1)\alpha ]$.

Approximation qualification. The omitted initial layer is precisely the one the source permits us to ignore: the approximate derivative diverges at $Y=0$, whereas the full Hill equation has initial slope $\beta$. The approximation is not uniform as $n\to \infty$ at fixed $R=\beta /(\alpha K)$. Strong repression already at half of the final concentration needs $R^{1/(n+1)}/2\gg 1$. The actual Hill equation obeys $\dot{Y}\le \beta$, so $T_{1/2}\ge Y_{\mathrm{st}}/(2\beta )$. Its steep-switch limit approaches the logic-response scale in a; the formal vanishing large-$n$ result above cannot be extrapolated past its regime of validity.

For the parameter sensitivity of autorepressed expression, unregulated production has $Y_{\mathrm{st}}=\beta /\alpha$, so its logarithmic sensitivity is one. In the strong Hill approximation, $Y_{\mathrm{st}}=(\beta K^n/\alpha {)}^{1/(n+1)}$, with $K,\alpha ,n$ held fixed. HenceThe ideal logic level $Y_{\mathrm{st}}\simeq K$ instead gives $S\simeq 0$, because changing production primarily changes the switching duty fraction, not the threshold. Finite Hill repression has a residual production sensitivity.

The exact Hill steady-state relation makes the comparison continuous. Write $s=(Y_{\mathrm{st}}/K{)}^n$; then $\beta =\alpha Y_{\mathrm{st}}(1+s)$ and $d\beta /d{Y}_{\mathrm{st}}=\alpha [1+(n+1)s]$. ThusIt tends to one for weak repression and to $1/(n+1)$ for strong repression. The calculation concerns a local relative parameter change, not the response to an arbitrary large production perturbation.

Use steady or initial-rate conditions with effectively clamped substrate concentrations. For either substrate $X\in \{S,R\}$, put $U_X=[AX]$ and $V_X=[A{X}^{\ast }]$. The intermediate balances areThe product flux is thereforeThe free enzyme concentration is common to the two competing pathways, so it cancels even when enzyme sequestration changes its magnitude. ConsequentlyEach post-binding factor is the ratio of probabilities of proceeding rather than dissociating at that stage. The expression assumes positive common forward rates and no significant product back reaction. Without the steady/intermediate-balance or initial-rate assumption, the ratio can depend on intermediate populations and observation time; binding and dissociation constants alone would not determine a unique transient ratio.

With equal substrate abundance and association rates, let $a=k_{-}^S<b=k_{-}^R$ and $c=l_{-}^{S\ast }<d=l_{-}^{R\ast }$. ThenBoth factors increase with their forward rate: for example ${\partial }_m[(a+m)/(b+m)]=(b-a)/(b+m{)}^2>0$. Thus maximum discrimination requires enough waiting at each stage for the less stable incorrect complex to be rejected. In the slow-forward limit,The two rejection stages multiply their discrimination. Fast modification or fast product formation bypasses the corresponding dissociation test and drives that factor towards one.

There is a kinetic proofreading fidelity-throughput tradeoff: taking either forward rate literally to zero gives zero product flux, not a useful finite-output optimum. If dissociation rates are well separated, a practical high-yield discrimination window is $k_{-}^S\ll m\ll k_{-}^R$ and $l_{-}^{S\ast }\ll V_A\ll l_{-}^{R\ast }$. Correct complexes then usually pass both stages while incorrect complexes usually dissociate; its fidelity is not the mathematical infimum obtained with vanishing forward rates. The irreversible modification represents a driven, energy-consuming step, permitting successive nonequilibrium checks rather than merely repeating one equilibrium binding comparison. No finite positive optimum can be selected from minimization of $F$ alone without a speed or yield requirement.