Solutions to Real Quiz 8 of Dr. Z.’s Dynamical Models in Biology class

Name: Dr. Z.

1. Recall that the Hardy-Weinberg rule (where the fraction of the population that have genotype AA is u, the fraction that have genotype Aa is v (and hence the fraction of the population that has genotype aa is w = 1 − u − v) is:

(u, v) → (u^{2} + vu + \frac{1}{4}v^{2}, −2vu − 2u^{2} + 2u − \frac{1}{2}v^{2} + v)

(i) (5 points) If right now, \frac{1}{2} of the population have genotupe AA, \frac{1}{3} of the polpulation have genotype Aa, what is the fraction of aa genotypes (ia) Right now? (ib) In ten generations? (ii) (5 points) Prove that the above equilibrium (after ten generations) is semi-stable. Sol. of 1.(i)a: w = 1 − u − v = 1 − \frac{1}{2} − \frac{1}{3} = \frac{1}{6}. So right now \frac{1}{6} of the population has genotype aa. Sol. of 1.(ib):

(\frac{1}{2}, \frac{1}{3}) → ((\frac{1}{2})^{2} + \frac{1}{2} · \frac{1}{3} + \frac{1}{4}(\frac{1}{3})^{2}, −2\frac{1}{3} \frac{1}{2} − 2(\frac{1}{2})^{2} + 2u − \frac{1}{2}(\frac{1}{3})^{2} + \frac{1}{3}) = (\frac{4}{9}, \frac{4}{9})

So after one generation it is (\frac{4}{9}, \frac{4}{9}). But this now stays for ever after, in particular after ten generations. So w = 1 − \frac{4}{9} − \frac{4}{9} = \frac{1}{9}. Ans. to 1.(ib) After ten generations \frac{1}{9} of the population would have genotype aa. So. to 1(ii) The Jacobian is:

JAC(u, v) [ \begin{matrix}2u + v & u + v/2 \\ −2v − 4u + 2 & −2u − v + 1 \end{matrix} ]

Plugging in u = \frac{4}{9} and v = \frac{4}{9} we have

JAC(u, v) [ \begin{matrix}2 · \frac{4}{9} + \frac{4}{9} & \frac{4}{9} + (\frac{4}{9})/2 \\ −2\frac{4}{9} − 4\frac{4}{9} + 2 & −2\frac{4}{9} − \frac{4}{9} + 1 \end{matrix} ]

[ −\frac{2}{3} −\frac{1}{3}^{\frac{4}{3} \frac{2}{3}} ] .

The characteristic equation is

(\frac{4}{3} − λ)(−\frac{1}{3} − λ) − (\frac{2}{3})(−\frac{2}{3}) = 0

Simplifying

λ^{2} − λ = 0 .

Factorizing:

(λ − 1)λ = 0 .

So the eigenvalues are 1 and 0. Since the absolute values are ≤ 1 but one of them has absolute value 1, (\frac{4}{9}, \frac{4}{9}) it is semi-stable.