Page 179 - Applied Probability

8. The Polygenic Model
164
To calculate the covariance between the trait values u and v of two
relatives, we exploit the bilinearity property
n
n

Cov(u, v)=
i=1 j=1 c i c j Cov(u i ,v j ).
To make further progress, let z ij be the indicator function of the
event that the genes expressed in patch i of the first person and
patch j of the second person are identical by descent. The probabil-
ity Pr(z ij =1) = φ is just the X-linked kinship coefficient between
the two relatives. This probability can be either unconditional or con-
ditional. If it is conditional, then in QTL mapping we condition on
the observed X-linked marker types for the two individuals and their
relatives. In any case, by conditioning on the z ij , prove that
n
2
Cov(u, v)= c i φ.
i=1
Given that only one gene is expressed in the same patch of the same
person, show that the trait variance for a single person is
n n 2
2
Var(u)= c i (1 − φ)+ c i φ.
i=1 i=1
Because φ = 1 for a male and φ = 1 for a non-inbred female, it is
2
trivial to show when the c i are positive that the male trait variance
n 2 1 n 2 1 n 2
( i=1 i ) exceeds the female trait variance 2 i=1 i 2 i=1 i ) .
c
c
c + (
This makes sense; the process of X inactivation smooths the contri-
butions from different alleles and decreases the variance.
In practice, we need to select parameters that can be estimated from
pedigree data. Assuming that there are no inbred females, one way
2
2
i=1 i ) and σ =
of achieving this is to let σ =( n c 2 2 n c . Then,
a f i=1 i
except for female variances, all trait variances and covariances can
2
be expressed as σ φ. For a female trait variance, we amend this ex-
a
2
pression by adding σ (1 − φ), with the understanding that φ =1/2.
f
2
The extra term σ can easily be included in the random environment
f
portion of any multivariate normal model.
12. In the hypergeometric polygenic model, verify that the number of
positive polygenes a non-inbred person possesses follows the binomial
distribution (8.12). Do this by a qualitative argument and by checking
analytically the reproductive property
2n 2n 2n
2n 1 2n 1 2n 1

=
τ g 1 ×g 2 →g 3
2 2 2
g 1 g 2 g 3
g 1 g 2
for polygene transmission under sampling without replacement.