Evolution__3rd_Edition

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A model of haplotype frequency
changes can be built
Linkage disequilibrium arises with
epistatic fitnesses, such as existed
in the mimetic butterflies
CHAPTER 8 / Two-locus and Multilocus Population Genetics 207
These fitnesses are called multiplicative. An individual’s fitness for its two genotypes
is found by multiplying the fitnesses of each of its one-locus genotypes. The genotypes
are independent, in the sense that the effect of one genotype on survival is independent
of the other locus. An individual with the genotype A 1 /A 1 has a chance of surviving
from age 1 to 6 months of w 11 whether its genotype at the other locus is B 1 /B 1 , B 1 /B 2 , or
B 2 /B 2 .
The next step is to derive a recurrence relation between the frequency of a haplotype
in one generation and in the next. However, we do not need to work through all the
algebra here. In outline the procedure is the same as for the single-locus case, with the
additional factor of recombination. The recurrence relation for haplotype frequency
takes account of the frequency and fitness of all the genotypes that a haplotype is found
in. It also has to add and subtract the number of copies gained and lost by recombination:
we multiply by (1 − r) the frequency of the double heterozygotes containing the
haplotype and by r that of the double heterozygote that can generate it if recombination
occurs. The Mendelian rules are then applied, and the frequency of the haplotype in the
next generation results.
Which kinds of selection cause linkage disequilibrium? The question is important
because, as we have seen, two-locus models are particularly needed when linkage
disequilibrium exists. With multiplicative fitnesses, the haplotype frequencies almost
always go to linkage equilibrium. (Linkage disequilibrium is only possible if both loci
are polymorphic. If one gene is fixed at either locus, D = 0 trivially. The fitnesses, w 11 ,
etc., as written above were frequency independent. A doubly heterozygous equilibrium
then requires heterozygous advantage at both loci: w 11 < w 12 > w 22 , x 11 < x 12 > x 22 ; see
Section 5.12.1, p. 123.) If ever linkage disequilibrium exists between two loci that have
multiplicative fitness relations, that disequilibrium will decay to zero as the generations
pass.
The more interesting case is when the fitnesses of the two loci interact epistatically
(the fitnesses are said to show epistasis). The selection in the mimetic polymorphism of
Papilio memnon is epistatic. Epistatic interaction means that the fitness effects of a
genotype depend on what genotype it is associated with at the other locus.
We can simplify the situation in P. memnon by imagining that one locus controls
whether the butterfly has a tail on its hindwing and one other locus controls coloration.
(In reality, at least four loci influence coloration.) Let T + (presence of tail) be dominant
to T – (absence). At the other locus, C 1 is dominant, and C 1 /C 1 and C 1 /C 2 individuals
have a color pattern that mimics a model species with a tail, whereas C 2 /C 2 individuals
are colored like a model species that has no tail. The relative fitness of each genotype
depends on what the genotype at the other locus is. For example, a T + /T + genotype in
the same butterfly as a C 2 /C 2 will be less fit than a T – /T – with C 1 /C 1 . The fitnesses can be
written as follows (the simplification relative to the earlier fitness matrixes arises
because of dominance, and because there is one term for the fitness of both loci
together rather than one term for each locus):
T + /T + T + /T – T – /T –
C 1 /C 1 w 11 w 11 w 21
C 1 /C 2 w 11 w 11 w 21
C 2 /C 2 w 12 w 12 w 22