Past exam of the biological physics course of the University of Oxford / 2018 / 4

Use the source's explicit graded production rules. X is already present at a constant concentration; the input at zero activates it rather than starting accumulation of X. Assume its active level exceeds both relevant X thresholds, put ${\alpha }_x={\alpha }_y={\alpha }_z=\alpha$, let $Y_s={\beta }_y/\alpha$, and take $0<h=K_{yz}/Y_s<1$. Before the input Z is not produced, so use the initial steady value $Z(0)=0$.

For I1 incoherent feed-forward loop, activation of X starts Y production; for I4 incoherent feed-forward loop it stops Y production. ConsequentlyThe two delays are the times for Y to cross its output-regulation threshold. In I1 incoherent feed-forward loop, initially absent Y permits the high output rate, then accumulated Y represses it. In I4 incoherent feed-forward loop, initially abundant Y activates the high output rate, then Y's decay removes that activation. Thus both have production ${\beta }_{\mathrm{high}}$ before their respective delay and ${\beta }_{\mathrm{low}}$ afterwards.

With $Z_s={\beta }_{\mathrm{low}}/\alpha$ and $F={\beta }_{\mathrm{high}}/{\beta }_{\mathrm{low}}$, their common piecewise response, with $d=d_1$ or $d_4$, isThe output is continuous at the production switch. There is an overshoot precisely when $F(1-e^{-\alpha d})>1$; otherwise it rises monotonically to $Z_s$. For a threshold not extremely close to its limiting values and the supplied $F\gg 1$, a rapid pulse followed by relaxation to a lower nonzero level is expected. This can provide fast initial response or a transient strong signal, but it is not perfect adaptation to the original zero output.

The original plots below use the illustrative values $h=1/4$ and $F=10$. At $h=1/2$ the two Y curves are complementary and their output delays and Z curves are identical. These threshold values are examples, not extra source data. “AND” here must be read with the stipulated graded rates: replacing them by an ordinary all-or-nothing AND rule would give a different response.

For the final low steady value $Z_s={\beta }_{\mathrm{low}}/\alpha$, the first crossing of $Z_s/2$ occurs during high production if $F(1-e^{-\alpha d})\ge 1/2$. Solving $F(1-e^{-\alpha t_{1/2}})=1/2$ then gives, for either motif,This half-final response of a graded incoherent feed-forward loop is a first-crossing measure. A simple output produced at the final low rate from zero would instead take $(\ln 2)/\alpha$. This comparison explains the acceleration of the first half-final response, not the time needed to settle after a potentially large overshoot.

The condition matters because the source does not fix $K_{yz}/Y_s$. Define $\delta =1-e^{-\alpha d}$, equal to $h$ for I1 incoherent feed-forward loop and $1-h$ for I4 incoherent feed-forward loop. If the high phase ends before half-final output is reached, solve the second branch of part (a) instead:The expressions agree at the boundary. The $F$-only answer is therefore valid for both motifs when $F\min (h,1-h)\ge 1/2$, as holds for large F and a fixed interior threshold. Merely saying $F\gg 1$ does not cover a threshold that approaches zero or one even faster. If the Y threshold is never crossed, the assumed final low state itself changes and one must reconsider the steady state before using this response-time definition.

At the final steady state, the ligands control activity as well as the amount of each factor. Assuming active X exceeds its thresholds and the high Y level exceeds $K_{yz}$, the normalized output production rates areFor I1 incoherent feed-forward loop, active X makes Y abundant, but Y can repress Z only if its own activating ligand is present. Without $S_y$, Y is inactive and the output remains high. For I4 incoherent feed-forward loop, active X represses Y to below its threshold, so the output is low regardless of $S_y$. If X is inactive, the direct permissive arm is absent and both outputs are zero even when Y is active.

Thus I1 incoherent feed-forward loop can retain two-input combinatorial control at steady state as well as generate the transient in part (a). I4 incoherent feed-forward loop loses steady-state sensitivity to $S_y$ once X has suppressed Y, under these particular assumptions. That gives a functional reason the first motif may be more useful in many regulatory contexts. It is not a proof of an evolutionary frequency: promoter implementation, thresholds, noise, costs and other functions also influence how common a motif is. The argument follows the supplied rate table, including its nonzero ${\beta }_{\mathrm{low}}$, rather than a generic Boolean AND truth table.

Use Y and Z as the variables appearing in the equations; the introductory reference to X and Y is a naming mismatch. With $\gamma >0$, $n>0$, a symmetric equilibrium has $Y=Z=x$ andThe left side increases continuously from zero to infinity because its derivative is $1+(n+1)x^n>0$. Thus there is exactly one positive symmetric solution for every allowed parameter pair, which verifies the implicit steady-state expression.

For $\gamma \gg 1$ and $n\gg 1$, look for one high component and one low component. If $Y\sim \gamma$, the other equation gives $Z=\gamma /(1+Y^n)\sim {\gamma }^{1-n}\ll 1$; that small value in turn gives $Y=\gamma /(1+Z^n)\sim \gamma$. Exchanging the components gives the alternative state:For example the relative correction to the high value is of order ${\gamma }^{-n(n-1)}$, while the low value first has the ordinary denominator correction of relative order ${\gamma }^{-n}$. The sharply repressive Hill function is therefore consistent with two alternatives, one factor high and the other low, which can encode a switch. Mutual repression makes a positive feedback loop: a high Y suppresses Z and thereby relieves repression of Y. Existence of alternative equilibria alone does not establish their stability; the explicit quadratic-Hill case in part (e) also permits that check.

The quadratic-Hill mutual repression switch is explicit: for $n=2$, steady state requires $Y(1+Z^2)=\gamma$ and $Z(1+Y^2)=\gamma$. Subtraction givesThus either $Y=Z=x$, giving the unique positive cubic root $x^3+x-\gamma =0$, or $YZ=1$. In the latter case, $Y+1/Y=\gamma$, henceThe non-symmetric pair is real, positive and distinct exactly when $\gamma >2$. For $0<\gamma <2$ only the symmetric solution exists. Eliminating one variable gives the source's hinted factorization,with no singular denominator for real concentrations. The cubic is strictly increasing since $3Y^2+1>0$, so it contributes only the symmetric state.

One can verify the switch interpretation without a lengthy stability analysis. At symmetry the Jacobian matrix eigenvalues are $-1\pm 2x^2/(1+x^2)$: symmetry is stable for $x<1$ and becomes a saddle for $x>1$, corresponding to $\gamma <2$ and $\gamma >2$. At either asymmetric equilibrium the eigenvalues are $-1\pm 2/\gamma$, both negative when $\gamma >2$. ThereforeAt $\gamma =2$, all three branches meet at $(1,1)$ with a zero linear eigenvalue, the switching threshold rather than two distinct robust memory states. The lower panels of the original figure in solution show representative nullclines on either side.