Evolution__3rd_Edition

200 PART 2 / Evolutionary Genetics
Linkage disequilibrium is defined
as a deviation from a random
expectation
The butterflies provide an example
We construct a model of haplotype
frequencies over time
Haplotype Frequency in population
A1B1 A1B2 A2B1 A2B2 a = p1q1 + D
b = p1q2 − D
c = p2q1 − D
d = p2q2 + D
The total frequencies add up to one. That is, a + b + c + d = 1. (Also, p 1 q 1 + p 1 q 2 + p 2 q 1 +
p 2 q 2 = 1, and the sum of the two +D and two –D factors is zero.) The important term
to understand is D; it is a measure of “linkage disequilibrium.” Linkage equilibrium
is when D = 0 and means that the alleles at the two loci are combined independently.
The two B alleles would then be found with any one A allele (such as A 1 ) in the same
frequencies as they are found in the whole population. If we take all the A 1 genes, q 1 of
them are with B 1 genes and q 2 with B 2 genes; likewise, q 1 of the A 1 genes are with B 1
genes and q 2 with B 2 . At linkage equilibrium, the frequency of the A 1 B 1 haplotype is
p 1 q 1 . D measures the deviation from linkage equilibrium. If D > 0, A 1 is more often
found with B 1 (and less often with B 2 ) than would be expected if alleles at the two loci
were combined at random a the population contains an excess of A 1 B 1 (and of A 2 B 2 )
haplotypes.
Papilio memnon is an example of high linkage disequilibrium. If Clarke and
Sheppard are correct, the allele T + is almost always combined with the other alleles W 1 ,
F 1 , E 1 , and B 1 rather than with W 2 , W 3 , or W 4 (and equivalent alleles at the other loci).
There is a large excess of the haplotypes T + W 1 F 1 E 1 B 1 , T – W 2 F 2 E 2 B 2 , T – W 3 F 3 E 3 B 3 , etc.,
while haplotypes such as T + W 2 F 2 E 2 B 2 , T + W 1 F 2 E 2 B 2 , or T + W 1 F 1 E 2 B 2 are almost absent.
The linkage disequilibrium in P. memnon, as we have seen, is caused by selection. In
this section, however, we are asking how a set of haplotype frequencies should change
through time in the absence of selection.
Let us return again to the haplotype A 1 B 1 . It has frequency defined as a in one generation.
What will its frequency be in the next generation? (We can use again the notation
a′ as the frequency of A 1 B 1 one generation on.) In the absence of selection, the frequencies
of each gene will be constant, but the frequencies of the haplotypes can be altered
by recombination. The frequency of A 1 B 1 cannot be altered by recombination in double
or single homozygotes: the number of A 1 B 1 haplotypes coming out of an A 1 B 1 /A 1 B 1
individual, or of an A 1 B 1 /A 1 B 2 individual, is the same as the number going in, whether
or not there is recombination. The frequency can only be altered by recombination
in the double heterozygotes A 1 B 1 /A 2 B 2 and A 1 B 2 /A 2 B 1 . When recombination takes
place in an A 1 B 1 /A 2 B 2 individual, the number of A 1 B 1 haplotypes is decreased. When it
takes place in an A 1 B 2 /A 2 B 1 individual, the number of A 1 B 1 is increased. To be exact,
half the genes of an A 1 B 1 /A 2 B 2 double heterozygote are A 1 B 1 ; when recombination
hits between the loci the frequency of A 1 B 1 decreases by an amount − 1 /2. Similarly,
recombination in an A 1 B 2 /A 2 B 1 individual increases the frequency of A 1 B 1 by an
amount + 1 /2.
The frequency of A 1 B 1 /A 2 B 2 heterozygotes in the population is 2ad and of A 1 B 2 /A 2 B 1
is 2bc. The frequency at which the alleles at two loci are recombined per generation is
defined as r. (r can theoretically have any value up to a maximum of 0.5, if the loci are
on different chromosomes; r is between 0 and 0.5 for loci on the same chromosome
depending on how tightly linked they are a see Section 2.8, p. 35.) So:
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