Page 263 - Applied Probability

11. Radiation Hybrid Mapping
250
Interlocus Distances for the Best Order
108.7
17.4
11.9
12.5
37.9
1 —— 2 —— 3 —— 4 —— 5 —— 6
The total map length between locus 1 and locus 6 is b = 188.4 cR under
this order. In the Bayesian analyses, b was increased to 200 cR to determine
a prior interval length of 7×200 = 280 cR. Interlocus distances based on the
5
posterior mode then give a total map length of only 176.5 cR. Apparently,
imposition of a tight prior tends to decrease estimated interlocus distances.
11.11 Problems
1. For m loci in a haploid clone with no missing observations, the ex-
pected number of obligate breaks E[B(id)] is given by expression
(11.2).
(a) Under the correct order, show [1] that
m−1 m−1
2
Var[B(id)] = 2r(1 − r) θ i,i+1 − 2r(1 − r) θ
i,i+1
i=1 i=1
m−2 m−1
!

2
+(1 − 2r) θ i,i+1 θ j,j+1 (1 − θ i+1,j ) ,
i=1 j=i+1
where the breakage probability θ i+1,j = 0 when i+1 = j. (Hint:
Let S i be the indicator of whether a break has occurred between
loci i and i + 1. Verify that
2
E(S i S j )= r(1 − r)θ i,i+1 θ j,j+1 [1 − θ i+1,j (1 − 2r) ]
by considering four possible cases consistent with S i S j = 1. The
first case is characterized by retention at locus i, nonretention
at locus i + 1, retention at locus j, and nonretention at locus
j +1.)
(b) The above expression for Var[B(id)] can be simplified in the
Poisson model by noting that
j−1

= (1 − θ k,k+1 ).
1 − θ i+1,j
k=i+1
Using this last identity, argue by induction that

m−2 m−1

θ i,i+1 θ j,j+1 (1 − θ i+1,j )
i=1 j=i+1