2 Linkage Disequilibrium
33
Although there are four haplotypes, there is only a single D needed to describe
the extent of LD. That is clear when the relationship between haplotype and allele
frequencies is used. For example, if D is defined by Eq. (2.1), then
f Ab − f A f b = f A − f AB − f A f b = f A (1 − f b ) − f AB = f A f B − f AB = −D.
It is possible, then, to express all the haplotype frequencies in terms of the allele
frequencies and D:
f AB = f A f B + D
f Ab = f A f b − D
f aB = f a f B − D
f ab = f a f b + D.
(2.2)
Notice that, if there are more AB haplotypes than expected at linkage equilibrium
(D > 0), there have to be more ab haplotypes and fewer Ab and aB haplotypes.
Although D is defined by Eq. (2.1), there is an equivalent but different expression:
D = f AB − f A f B = f AB − (f AB + f Ab ) (f AB + f aB )
= f AB (1 − f AB − f Ab − f aB ) − f Ab f aB = f AB f ab − f aB f Ab
(2.3)
Equation (2.2) tells us that, because none of the haplotype frequencies can be
negative, there is a limit on the magnitude of D imposed by the allele frequencies. If
D > 0, D must be no greater than the smaller of f A f b and f a f B , and if D < 0, D must
be larger than −f A f B and −f a f b . That is,
− min (f a f b , f A f B ) ≤ D ≤ min (f A f b , f a f B ) .
(2.4)
One question that arises when computing D for different pairs of loci is whether
a particular value is large or small. For example, does D = 0.006 indicate a small
or large amount of LD for a pair of loci? Inequality (2.4) tells us that the answer
depends on the allele frequencies and suggests that it is useful to express D relative
to its maximum or minimum possible value. We can do this by defining
D =
D
min(f A f b ,f a f B ) if D > 0
=
D
− min(f a f b ,f A f B ) if D < 0
(2.5)
which indicates how close D is to its maximum or minimum value (Lewontin 1964).
In the example with D = 0.006, if f A = 0.4 and f B = 0.01, then D
= 1, while if
f A = 0.4 and f B = 0.3, then D
= 0.0333.
33
Although there are four haplotypes, there is only a single D needed to describe
the extent of LD. That is clear when the relationship between haplotype and allele
frequencies is used. For example, if D is defined by Eq. (2.1), then
f Ab − f A f b = f A − f AB − f A f b = f A (1 − f b ) − f AB = f A f B − f AB = −D.
It is possible, then, to express all the haplotype frequencies in terms of the allele
frequencies and D:
f AB = f A f B + D
f Ab = f A f b − D
f aB = f a f B − D
f ab = f a f b + D.
(2.2)
Notice that, if there are more AB haplotypes than expected at linkage equilibrium
(D > 0), there have to be more ab haplotypes and fewer Ab and aB haplotypes.
Although D is defined by Eq. (2.1), there is an equivalent but different expression:
D = f AB − f A f B = f AB − (f AB + f Ab ) (f AB + f aB )
= f AB (1 − f AB − f Ab − f aB ) − f Ab f aB = f AB f ab − f aB f Ab
(2.3)
Equation (2.2) tells us that, because none of the haplotype frequencies can be
negative, there is a limit on the magnitude of D imposed by the allele frequencies. If
D > 0, D must be no greater than the smaller of f A f b and f a f B , and if D < 0, D must
be larger than −f A f B and −f a f b . That is,
− min (f a f b , f A f B ) ≤ D ≤ min (f A f b , f a f B ) .
(2.4)
One question that arises when computing D for different pairs of loci is whether
a particular value is large or small. For example, does D = 0.006 indicate a small
or large amount of LD for a pair of loci? Inequality (2.4) tells us that the answer
depends on the allele frequencies and suggests that it is useful to express D relative
to its maximum or minimum possible value. We can do this by defining
D =
D
min(f A f b ,f a f B ) if D > 0
=
D
− min(f a f b ,f A f B ) if D < 0
(2.5)
which indicates how close D is to its maximum or minimum value (Lewontin 1964).
In the example with D = 0.006, if f A = 0.4 and f B = 0.01, then D
= 1, while if
f A = 0.4 and f B = 0.3, then D
= 0.0333.