epistasis and sib pair linkage
Mike Miller
mbmiller at SIRRONALD.WUSTL.EDU
Sun Jun 18 13:09:00 EST 1995
We would like for our nonparametric methods of linkage analysis (e.g.,
affected sib pair methods) to be able to detect disease susceptibility
loci under any model of inheritance--we would like them to be
"model-free." I find it interesting that there can be models where all of
the genetic variance is accounted for by two loci, but there is no
additive genetic variance, no dominance genetic variance, and thus no
genic effect at either locus to be detected by the sib pair method. The
IBD distribution at loci A and B would equal the null distribution (1/4,
1/2, 1/4, for 0, 1, 2).
Suppose we have a disease with population prevalence K, and there are two
alleles _of_equal_frequency_ at each of two loci, and the following table
shows the penetrance of each two-locus genotype:
BB Bb bb
=== === ===
AA 2K 0 2K
Aa K K K
aa 0 2K 0
This can also be shown graphically in an interaction plot with one line
for each A-locus genotype:
P 2K | + x + AA
e | \ / \ /
n | \ / \ /
e | \ / \ /
t K | o___\____o____/___o Aa
r | /\ /\
a | / \ / \
n | / \ / \
c | / \ / \
e 0 | x + x aa
-----------------------
bb Bb BB
Genotype at B locus
It is possible to produce an infinite number of these kinds of models by
varying allele frequencies and penetrances while keeping the mean
penetrance constant across both B-locus genotypes and A-locus genotypes.
In the particular case described above (with equal allele frequencies),
the genetic variance is
2
K
V = ---
G 2
The phenotypic variance is always
V = K (1 - K)
P
So the broad-sense heritability is
K / 2
V / V = -------
G P 1 - K
which can not exceed 1/2 because K can not exceed 1/2. (The penetrance of
some genotypes is 2K.) For small K, heritability is K/2 and therefore
these loci have negligible importance.
Assuming no shared-environmental effect, a study of MZ and DZ twins would
show a correlation for the MZs equal to the broad-sense heritability and a
correlation for the DZs equal to zero. Using the simple method to
estimate heritability of taking either the MZ correlation or double the
difference in correlations, 2(MZ-DZ), whichever is greater, we would find
the correct answer. But if we tried to use a sib pair method to find
genes, we would be stumped.
Much the same logic would apply to a quantitative trait with population
mean equal to K, but there would be no limit on the heritability of the
trait and no formula for phenotypic variance in terms of K. The formula
for V would still hold.
G
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