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ne the initial conditions; allow selection to act; calculate the allelic frequency after selection (qn 1); calculate q (change in allelic frequency from one generation to the ^ next); then calculate equilibrium frequency, q , when q becomes zero; and examine the stability of the equilibrium. In the analysis that follows, we consider a single autosomal locus in a diploid, sexually reproducing species with two alleles and assume that selection acts directly on the phenotypes in a simple fashion (i.e., it occurs at a single stage in the life of the organism, such as larval mortality in Drosophila). After selection, the individuals remaining within the population mate at random to form a new generation in Hardy-Weinberg proportions.

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In table 20.2, we outline the model for selection against the homozygous recessive genotype. The initial population is in Hardy-Weinberg equilibrium. Even with selection acting during the life cycle of the organism, Hardy-Weinberg proportions will be reestablished anew after each round of random mating, although presumably at new allelic frequencies. All selection models start out the same way.They diverge at the point of assigning tness, which depends on the way natural selection is acting. In the model in table 20.2, the dominant homozygote and the heterozygote have

Substituting the expression for the three currents in the nodal equations (equations 32 and 33), we obtain the following relationships: va va vb =0 (37) iS R1 R2 va vb vb =0 R2 R3 (38)

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Equations 37 and 38 may be obtained directly from the circuit, with a little practice Note that these equations may be solved for va and vb , assuming that iS , R1 , R2 , and R3 are known The same equations may be reformulated as follows: 1 1 + R1 R2 1 R2 va + 1 R2 vb = iS (39) vb = 0

We can analyze selection by using our standard modelbuilding protocol of population genetics namely, de-

Schematics, Diagrams, Circuits, and Given Data: I1 = 10 mA; I2 = 50 mA; R1 = Assumptions: The reference (ground) node is chosen to be the node at the bottom of the

indication of how KCL will be applied to determine the nodal equations Note that we have selected to ground the lower part of the circuit, resulting in a reference voltage of zero at that node Applying KCL at nodes 1 and 2 we obtain I1 v1 0 v1 v2 v1 v2 =0 R1 R2 R3 (node 1) (node 2)

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Now we can write the same equations more systematically as a function of the unknown node voltages, as was done in equation 39 1 1 1 + + R1 R2 R3 1 1 R2 R3 v1 + 1 1 R2 R3 v2 = I1 v2 = I2 (node 1) (node 2)

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Genotype AA Initial genotypic frequencies Fitness (W ) Ratio after selection Genotypic frequencies after selection p2 1 p2 p2 W Aa 2pq 1 2pq 2pq W aa q2 1 s q2(1 s) q2(1 s) W 1 sq2 1 W Total 1

With some manipulation, the equations nally lead to the following form: 16v1 06v2 = 10 06v1 + 11v2 = 50 These equations may be solved simultaneously to obtain v1 = 1357 V v2 = 5286 V Knowing the node voltages, we can determine each of the branch currents and voltages in the circuit For example, the current through the 10-k resistor is given by: i10 k v1 v2 = 393 mA = 10,000

indicating that the initial (arbitrary) choice of direction for this current was the same as the actual direction of current ow As another example, consider the current through the 1-k resistor: i1 k = v1 = 1357 mA 1,000

individuals that survive to reproduce, only six aa individuals would survive to reproduce. The total of the three genotypes after selection is 1 sq2. That is, p2 1 2pq sq2 q2(1 s) p2 2pq q2 sq2

In this case, the current is negative, indicating that current actually ows from ground to node 1, as it should, since the voltage at node 1 is negative with respect to ground You may continue the branch-by-branch analysis started in this example to verify that the solution obtained in the example is indeed correct

Comments: Note that we have chose to assign a positive sign to currents entering a node,

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