### Re: What is the difference between linkage equilibrium & linkage disequilibrium

Date: Thu Dec 21 11:35:11 2000
Posted By: Mark Woelfle, Faculty, Molecular Biology, Vanderbilt University
Area of science: Genetics
ID: 976036598.Ge
Message:
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Many phenotypic traits result from the interaction of alleles at more than
one gene locus and the environment; that is, the trait is multifactorial.
When more than one locus is involved in a trait, the loci can be either
connected or independent. If the alleles at each locus assort randomly into
gametes, one would expect that the frequency of a given allele at one locus
would have no effect on the frequency of an allele at the second locus
involved in the trait; that is, they are independent from one another. Such

Linkage equilibrium occurs in a population with a trait involving N loci
when the frequency of any genotype is equal to the product of the individual
frequencies at each of the N loci. For example, if a trait is determined by
2 loci, let's say A and B, each with two alleles (A/a and B/b) and at each
locus an allele appears in the population with a frequency. At one locus, we
can call these freq(A), freq(a) and at the other, freq(B) and freq(b). There
there are four possible gamete types: AB, aB, Ab, and ab and with a
population in linkage equilibrium, we would expect:

freq(AB) = freq(A) x freq(B)
freq(aB) = freq(a) x freq(B)
freq(Ab) = freq(A) x freq(b)
freq(ab) = freq(a) x freq(b).

In other words, D, the coefficient of linkage disequilibrium =

Freq(AB) x Freq(ab) - Freq(Ab) X Freq(aB)

As you can see, if A assorts independently of B, then

Freq(AB) x Freq(ab) = Freq(Ab) X Freq(aB) AND D = 0.

If A is linked to B, then Freq(AB) will not equal Freq(A) X Freq(B), etc to
a point in the extreme, where A always goes with B, and a always goes with
b, such that A and B are tightly linked and thus

D = 1.

In this case, the population is in complete linkage disequilibrium.

Hope this helps,

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