Past exam of the mathematics course of the University of Cambridge / 2001 / iii / Paper 52 / 5

Let $u$ be the allele frequency of $A$, with random mating and sufficiently large population size to neglect genetic drift. The Hardy-Weinberg proportions before selection are $u^2,2u(1-u),(1-u{)}^2$. Write the three genotype fitnesses as $w_{AA},w_{Aa},w_{aa}$, with $w_{Aa}$ smaller than either homozygote value. After one viability-selection generation,ConsequentlyPut $a=w_{AA}-w_{Aa}>0$, $b=w_{aa}-w_{Aa}>0$. The underdominant allele-frequency dynamics has stable fixation states zero and one, separated by the unstable thresholdBelow it $A$ decreases to allele fixation of $a$; above it $A$ increases to fixation. Exactly at the threshold the deterministic frequency remains balanced, but any perturbation selects a side. This is underdominance, or heterozygote disadvantage, rather than balancing selection from heterozygote advantage.

In a continuous-time weak-selection or selection-rate convention, the corresponding equation is $\dot{u}=u(1-u)[(a+b)u-b]$. Equal homozygote fitness gives $\dot{u}=su(1-u)(2u-1)$ for a positive selection strength $s$. The next part uses this explicitly stated continuous-time convention; $s$ and migration then have the same time units.

Use a globally coupled underdominant metapopulation with equal habitat sizes:Symmetry preserves a common frequency $x$ in the initially $AA$ habitats and $y$ in the initially $aa$ habitats. With $q=m/N$, the equations reduce exactly toThe maximum positive selection rate is $f(u_m)=s/(6\sqrt{3})$ at $u_m=(3+\sqrt{3})/6$. Comparing it with the worst outward migration flux gives a sufficient barrier condition $\sigma \max (q,1-q)<f(u_m)/u_m$. A slightly stronger optimized condition follows from maximizing the per-capita restoring term $f(u)/u=s(1-u)(2u-1)$ over $u>1/2$: its maximum is $s/8$ at $u=3/4$.

Indeed, on $x=3/4$, with $0\le y\le 1/4$, $\dot{x}\ge 3s/32-(3/4)\sigma (1-q)>0$. On $y=1/4$, with $3/4\le x\le 1$, $\dot{y}\le -3s/32+(3/4)\sigma q<0$. The other two edges point inward as well. The invariant-rectangle criterion for underdominant coexistence therefore gives the explicit robust conditionsStarting from the specified pure habitats, the populations then remain in $x\ge 3/4$, $y\le 1/4$. Their global frequency lies between $3q/4$ and $q+(1-q)/4$, so neither allele fixes. This is a sufficient condition, not a claim that the bound is sharp. If $m=0$ or $m=N$, the population is already fixed; if $\sigma =0$, any mixture of the two pure habitat types persists.

The exactly balanced case illustrates why robustness and the prescribed trajectory differ. When $q=1/2$, symmetry forces $y=1-x$ andFor $\sigma <s/2$ its heterogeneous limiting state isHere $f^{\prime }(x_{\ast })=f^{\prime }(y_{\ast })=-s+3\sigma$, so the full Jacobian matrix has diagonal entries $f^{\prime }(x_{\ast })-\sigma /2$ and off-diagonal entries $\sigma /2$. Its two full-system eigenvalues are $-s+3\sigma$ and $-s+2\sigma$: the symmetric underdominant two-cluster equilibrium is robustly attracting only for $\sigma <s/3$. For larger migration the exactly symmetric initial condition can still avoid fixation, eventually approaching the spatially uniform frequency $1/2$ when $\sigma \ge s/2$, but the symmetry-broken mean-frequency mode is unstable. A finite population with genetic drift need not maintain deterministic coexistence indefinitely.