We first discuss the permutation property for one SNP. Consider a nuclear family with
two parents, m affected and n unaffected siblings.
Let 𝐺𝑇𝑖 be the number of allele 1 transmitted from parents, and let 𝐺𝑁𝑇𝑖 be the
number of non-transmitted allele 1 from parents, to affected sib i , where i=1,2,…,m
Let 𝐺 ′ 𝑇𝑗 be the number of allele 1 transmitted from parents, and let 𝐺 ′ 𝑁𝑇𝑗 be the
number of non-transmitted allele 1 from parents, to unaffected sib j , where j=1,2,…,n
Table A.1 shows the transmitted and non-transmitted statistics for the affected and
unaffected sibs for the nuclear family.
Since the total number of allele 1 in parents is fixed, 𝐺𝑇1 + 𝐺𝑁𝑇1 = 𝐺𝑇2 + 𝐺𝑁𝑇2 =
⋯ = 𝐺𝑇𝑚 + 𝐺𝑁𝑇𝑚 = 𝐺 ′ 𝑇1 + 𝐺 ′ 𝑁𝑇1 = 𝐺 ′ 𝑇2 + 𝐺 ′ 𝑁𝑇2 = ⋯ = 𝐺 ′ 𝑇𝑛 + 𝐺 ′ 𝑁𝑇𝑛
If m=n
𝐺𝑇1 − 𝐺 ′ 𝑇1 = 𝐺 ′ 𝑁𝑇1 − 𝐺𝑁𝑇1
𝐺𝑇2 − 𝐺 ′ 𝑇2 = 𝐺 ′ 𝑁𝑇2 − 𝐺𝑁𝑇2
.
.
.
𝐺𝑇𝑛 − 𝐺 ′ 𝑇𝑛 = 𝐺 ′ 𝑁𝑇𝑛 − 𝐺𝑁𝑇𝑛
𝑛
n
𝑛
n
′
′
∴ ∑ 𝐺𝑇𝑖 − ∑ 𝐺 𝑇𝑗 = ∑ 𝐺 𝑁𝑇𝑗 − ∑ 𝐺𝑁𝑇𝑖 − − − − − (1)
𝑖=1
𝑚
𝑗=1
𝑗=1
m
𝑖=1
𝑚
m
∑ 𝐺𝑇𝑖 − ∑ 𝐺 ′ 𝑇𝑗 = ∑ 𝐺 ′ 𝑁𝑇𝑗 − ∑ 𝐺𝑁𝑇𝑖 − − − − − (2)
𝑖=1
𝑗=1
𝑗=1
𝑖=1
We define 𝑛𝑇 = number of triads (two parents and one affected offspring) and
𝑛𝑆 = number of discordant sib pair (one affected and one unaffected sibs) in the
nuclear family.
Let 𝑋𝑇 be the sum of difference in allele counts between the transmitted and the nontransmitted alleles over all triads. Let 𝑋𝑆 be the sum of difference in allele counts
between the affected and unaffected sibs over all discordant sib pairs.
𝑚
Therefore , 𝑋𝑇 = ∑(𝐺𝑇𝑖 − 𝐺𝑁𝑇𝑖 )
𝑖=1
𝑚
𝑛
If 𝑚 = 𝑛, 𝑋𝑠 = ∑ 𝐺𝑇𝑖 − ∑ 𝐺 ′ 𝑇𝑗
𝑖=1
𝑚
𝑗=1
𝑛
If 𝑚 ≠ 𝑛, 𝑋𝑠 = 𝑛 ∑ 𝐺𝑇𝑖 − 𝑚 ∑ 𝐺 ′ 𝑇𝑗
𝑖=1
𝑗=1
The PDT statistic for this nuclear family is calculated as
1
𝑛𝑇 +𝑛𝑆
(𝑋𝑇 + 𝑋𝑆 )
After permuting the transmitted and non-transmitted alleles, the statistics for 𝑋𝑇 and
𝑋𝑆 can be calculated as:
m
𝑋(𝑇) = ∑(𝐺𝑁𝑇𝑖 − 𝐺𝑇𝑖 )
i=1
𝑚
n
𝑋(𝑆) = ∑ 𝐺𝑁𝑇𝑖 − ∑ 𝐺 ′ 𝑁𝑇𝑗
𝑖=1
j=1
We investigate the relationship between 𝑋𝑇 + 𝑋𝑆 and 𝑋(𝑇) + 𝑋(𝑆) in the following
three cases.
(Case 1) The numbers of affected and unaffected sibs are equal. (m=n)
𝑚
m
𝑋𝑇 = ∑(𝐺𝑇𝑖 − 𝐺𝑁𝑇𝑖 ) = − ∑(𝐺𝑁𝑇𝑖 − 𝐺𝑇𝑖 ) = −𝑋(𝑇)
𝑖=1
𝑚
i=1
𝑛
𝑛
𝑛
𝑋𝑆 = ∑ 𝐺𝑇𝑖 − ∑ 𝐺 ′ 𝑇𝑗 = ∑ 𝐺𝑇𝑖 − ∑ 𝐺 ′ 𝑇𝑗
𝑖=1
𝑗=1
𝑚
𝑖=1
𝑗=1
𝑛
n
n
′
𝑋(𝑆) = ∑ 𝐺𝑁𝑇𝑖 − ∑ 𝐺 𝑁𝑇𝑗 = ∑ 𝐺𝑁𝑇𝑖 − ∑ 𝐺 ′ 𝑁𝑇𝑗
𝑖=1
𝑛
j=1
𝑛
𝑋𝑆 = ∑ 𝐺𝑇𝑖 − ∑ 𝐺 ′ 𝑇𝑗
𝑖=1
𝑖=1
𝑗=1
j=1
n
𝑛
𝑛
n
𝑓𝑟𝑜𝑚 (1) ∑ 𝐺 ′
′
𝑁𝑇𝑗 − ∑ 𝐺𝑁𝑇𝑖 = − (∑ 𝐺𝑁𝑇𝑖 − ∑ 𝐺 𝑁𝑇𝑗 )
=
𝑗=1
𝑖=1
𝑖=1
j=1
= −𝑋(𝑆)
∴ 𝑋𝑇 + 𝑋𝑆 = −(𝑋(𝑇) + 𝑋(𝑆) )
(Case 2) The number of the affected sibs is greater than the number of the
unaffected sibs. (m>n)
In this case, 𝑋𝑇 and 𝑋(𝑇) have the same form as in case 1. Therefore, we only
discuss 𝑋𝑆 and 𝑋(𝑆) here.
𝑚
𝑛
𝑋𝑆 = 𝑛 ∑ 𝐺𝑇𝑖 − 𝑚 ∑ 𝐺 ′ 𝑇𝑗
𝑖=1
𝑗=1
𝑚
𝑛
𝑋(𝑆) = 𝑛 ∑ 𝐺𝑁𝑇𝑖 − 𝑚 ∑ 𝐺 ′ 𝑁𝑇𝑗
𝑖=1
𝑗=1
𝑚
𝑛
𝑋𝑆 = 𝑛 ∑ 𝐺𝑇𝑖 − 𝑚 ∑ 𝐺 ′ 𝑇𝑗
𝑖=1
𝑗=1
𝑛
𝑛
𝑚
𝑛
= 𝑛 ∑ 𝐺𝑇𝑖 − 𝑛 ∑ 𝐺 ′ 𝑇𝑗 + 𝑛 ∑ 𝐺𝑇𝑖 − (𝑚 − 𝑛) ∑ 𝐺 ′ 𝑇𝑗
𝑖=1
𝑗=1
𝑛
𝑖=𝑛+1
𝑛
𝑗=1
𝑚
𝑚
′
= 𝑛 (∑ 𝐺𝑇𝑖 − ∑ 𝐺 𝑇𝑗 ) + ∑ (𝐺𝑇𝑖 − 𝐺
𝑖=1
𝑗=1
𝑛
𝑛
′
𝑇1 ) + … +
𝑖=𝑛+1
∑ (𝐺𝑇𝑖 − 𝐺 ′ 𝑇𝑛 )
𝑖=𝑛+1
𝑚
𝑛
= 𝑛 (∑ 𝐺𝑇𝑖 − ∑ 𝐺 ′ 𝑇𝑗 ) + ∑ ∑(𝐺𝑇𝑖 − 𝐺 ′ 𝑇𝑗 )
𝑖=1
𝑗=1
𝑛
𝑖=𝑛+1 𝑗=1
𝑛
𝑚
𝑛
′
= 𝑛 (∑ 𝐺 𝑁𝑇𝑗 − ∑ 𝐺𝑁𝑇𝑖 ) + ∑ ∑(𝐺 ′ 𝑁𝑇𝑗 − 𝐺𝑁𝑇𝑖 ) − − − 𝑎𝑐𝑐𝑜𝑟𝑑𝑖𝑛𝑔 𝑡𝑜 (1)
𝑗=1
𝑚
𝑖=1
𝑖=𝑛+1 𝑗=1
𝑛
𝑋(𝑆) = 𝑛 ∑ 𝐺𝑁𝑇𝑖 − 𝑚 ∑ 𝐺 ′ 𝑁𝑇𝑗
𝑖=1
𝑗=1
𝑛
𝑛
𝑚
𝑛
′
= 𝑛 ∑ 𝐺𝑁𝑇𝑖 − 𝑛 ∑ 𝐺 𝑁𝑇𝑗 + 𝑛 ∑ 𝐺𝑁𝑇𝑖 − (𝑚 − 𝑛) ∑ 𝐺 ′ 𝑁𝑇𝑗
𝑖=1
𝑛
𝑗=1
𝑖=𝑛+1
𝑛
𝑚
𝑗=1
𝑚
= 𝑛 (∑ 𝐺𝑁𝑇𝑖 − ∑ 𝐺 ′ 𝑁𝑇𝑗 ) + ∑ (𝐺𝑁𝑇𝑖 − 𝐺 ′ 𝑁𝑇1 ) + … + ∑ (𝐺𝑁𝑇𝑖 − 𝐺 ′ 𝑁𝑇𝑛 )
𝑖=1
𝑗=1
𝑖=𝑛+1
𝑖=𝑛+1
𝑛
𝑛
𝑚
𝑛
′
= 𝑛 (∑ 𝐺𝑁𝑇𝑖 − ∑ 𝐺 𝑁𝑇𝑗 ) + ∑ ∑(𝐺𝑁𝑇𝑖 − 𝐺 ′ 𝑁𝑇𝑗 )
𝑖=1
𝑛
𝑗=1
𝑖=𝑛+1 𝑗=1
𝑛
𝑚
𝑛
𝑋𝑆 = 𝑛 (∑ 𝐺 ′ 𝑁𝑇𝑗 − ∑ 𝐺𝑁𝑇𝑖 ) + ∑ ∑(𝐺 ′ 𝑁𝑇𝑗 − 𝐺𝑁𝑇𝑖 )
𝑗=1
𝑖=1
𝑛
𝑖=𝑛+1 𝑗=1
𝑛
𝑚
𝑛
′
= − {𝑛 (∑ 𝐺𝑁𝑇𝑖 − ∑ 𝐺 𝑁𝑇𝑗 ) + ∑ ∑(𝐺𝑁𝑇𝑖 − 𝐺 ′ 𝑁𝑇𝑗 )} = −𝑋(𝑆)
𝑖=1
𝑗=1
𝑖=𝑛+1 𝑗=1
∴ 𝑋𝑇 + 𝑋𝑆 = −(𝑋(𝑇) + 𝑋(𝑆) )
(Case 3) The number of the affected sibs is less than the number of the
unaffected sibs. (m<n)
In this case, 𝑋𝑇 and 𝑋(𝑇) have the same form as in case 1. Therefore, we only
discuss 𝑋𝑆 and 𝑋(𝑆) here.
𝑚
𝑛
𝑋𝑆 = 𝑛 ∑ 𝐺𝑇𝑖 − 𝑚 ∑ 𝐺 ′ 𝑇𝑗
𝑖=1
𝑗=1
𝑚
𝑛
𝑋(𝑆) = 𝑛 ∑ 𝐺𝑁𝑇𝑖 − 𝑚 ∑ 𝐺 ′ 𝑁𝑇𝑗
𝑖=1
𝑚
𝑗=1
𝑛
𝑋𝑆 = ∑ 𝐺𝑇𝑖 − 𝑚 ∑ 𝐺 ′ 𝑇𝑗
𝑖=1
𝑗=1
𝑚
𝑚
𝑚
𝑛
′
= 𝑚 ∑ 𝐺𝑇𝑖 − 𝑚 ∑ 𝐺 𝑇𝑗 + (𝑛 − 𝑚) ∑ 𝐺𝑇𝑖 − 𝑚 ∑ 𝐺 ′ 𝑇𝑗
𝑖=1
𝑚
𝑗=1
𝑚
𝑖=1
𝑚
𝑗=𝑚+1
𝑚
= 𝑚 (∑ 𝐺𝑇𝑖 − ∑ 𝐺 ′ 𝑇𝑗 ) + ∑(𝐺𝑇𝑖 − 𝐺 ′ 𝑇𝑚+1 ) + … + ∑(𝐺𝑇𝑖 − 𝐺 ′ 𝑇𝑛 )
𝑖=1
𝑗=1
𝑖=1
𝑖=1
𝑚
𝑚
𝑚
𝑛
′
= 𝑚 (∑ 𝐺𝑇𝑖 − ∑ 𝐺 𝑇𝑗 ) + ∑ ∑ (𝐺𝑇𝑖 − 𝐺 ′ 𝑇𝑗 )
𝑖=1
𝑗=1
𝑚
𝑖=1 𝑗=𝑚+1
𝑚
𝑚
𝑛
= 𝑚 (∑ 𝐺 ′ 𝑁𝑇𝑗 − ∑ 𝐺𝑁𝑇𝑖 ) + ∑ ∑ (𝐺 ′ 𝑁𝑇𝑗 − 𝐺𝑁𝑇𝑖 ) − −𝑎𝑐𝑐𝑜𝑟𝑑𝑖𝑛𝑔 𝑡𝑜 (2)
𝑗=1
𝑖=1
𝑚
𝑖=1 𝑗=𝑚+1
𝑛
𝑋(𝑆) = 𝑛 ∑ 𝐺𝑁𝑇𝑖 − 𝑚 ∑ 𝐺 ′ 𝑁𝑇𝑗
𝑖=1
𝑗=1
𝑚
𝑚
𝑚
𝑛
= 𝑚 ∑ 𝐺𝑁𝑇𝑖 − 𝑚 ∑ 𝐺 ′ 𝑁𝑇𝑗 + (𝑛 − 𝑚) ∑ 𝐺𝑁𝑇𝑖 − 𝑚 ∑ 𝐺 ′ 𝑁𝑇𝑗
𝑖=1
𝑗=1
𝑚
𝑖=1
𝑚
𝑗=𝑚+1
𝑚
𝑚
′
= 𝑚 (∑ 𝐺𝑁𝑇𝑖 − ∑ 𝐺 𝑁𝑇𝑗 ) + ∑(𝐺𝑁𝑇𝑖 − 𝐺 𝑁𝑇𝑚+1 ) + … + ∑(𝐺𝑁𝑇𝑖 − 𝐺 ′ 𝑁𝑇𝑛 )
𝑖=1
𝑗=1
𝑚
𝑚
′
𝑖=1
𝑖=1
𝑚
𝑛
= 𝑚 (∑ 𝐺𝑁𝑇𝑖 − ∑ 𝐺 ′ 𝑁𝑇𝑗 ) + ∑ ∑ (𝐺𝑁𝑇𝑖 − 𝐺 ′ 𝑁𝑇𝑗 )
𝑖=1
𝑗=1
𝑚
𝑖=1 𝑗=𝑚+1
𝑚
𝑚
𝑛
′
𝑋𝑆 = 𝑚 (∑ 𝐺 𝑁𝑇𝑗 − ∑ 𝐺𝑁𝑇𝑖 ) + ∑ ∑ (𝐺 ′ 𝑁𝑇𝑗 − 𝐺𝑁𝑇𝑖 )
𝑗=1
𝑖=1
𝑚
𝑖=1 𝑗=𝑚+1
𝑚
𝑚
𝑛
= − {𝑚 (∑ 𝐺𝑁𝑇𝑖 − ∑ 𝐺 ′ 𝑁𝑇𝑗 ) + ∑ ∑ (𝐺𝑁𝑇𝑖 − 𝐺 ′ 𝑁𝑇𝑗 )} = −𝑋(𝑆)
𝑖=1
𝑗=1
𝑖=1 𝑗=𝑚+1
∴ 𝑋𝑇 + 𝑋𝑆 = −(𝑋(𝑇) + 𝑋(𝑆) )
Because
1
𝑛𝑇 +𝑛𝑆
is a constant,
in Cases 1,2and3.
1
𝑛𝑇 +𝑛𝑆
(𝑋𝑇 + 𝑋𝑆 ) = −
1
𝑛𝑇 +𝑛𝑆
(𝑋(𝑇) + 𝑋(𝑆) ) is satisfied
Therefore, if we permute the transmitted and non-transmitted alleles simultaneously
from both parents to all the children, it simply results in a sign change in the PDT
statistic. The argument is still true when we have different numbers of affected and
unaffected siblings.
Since there can be LD among SNPs, we need to permute haplotypes to preserve the
LD structures among SNPs when analyzing multiple SNPs. We then discuss the
relationship between the permutation of haplotypes and the single-SNP PDT statistics
for two SNPs.
𝑎
We use ( ) to denote a haplotype for two SNPs, where a is the allele at the first
𝑏
SNP, and b is the allele at the second SNP.
Also consider a nuclear family with two parents, m affected and n unaffected siblings.
1
Let 𝐻𝑇𝑖 be the number of haplotype ( ) transmitted from parents, and let 𝐻𝑁𝑇𝑖 be
1
1
the number of non-transmitted haplotype ( ) from parents, to affected sib i , where
1
i=1,2,…,m
1
Let 𝐻 ′ 𝑇𝑗 be the number of haplotype ( ) transmitted from parents, and let 𝐻 ′ 𝑁𝑇𝑗
1
1
be the number of non-transmitted haplotype ( ) from parents, to unaffected sib j ,
1
where j=1,2,…,n
1
Table A.2 shows the transmitted and non-transmitted statistics for haplotype ( ) for
1
affected and unaffected sibs for the nuclear family.
1
Since the number of haplotype ( ) in parents is fixed, 𝐻𝑇1 + 𝐻𝑁𝑇1 = 𝐻𝑇2 + 𝐻𝑁𝑇2 =
1
⋯ = 𝐻𝑇𝑚 + 𝐻𝑁𝑇𝑚 = 𝐻 ′ 𝑇1 + 𝐻 ′ 𝑁𝑇1 = 𝐻 ′ 𝑇2 + 𝐻 ′ 𝑁𝑇2 = ⋯ = 𝐻 ′ 𝑇𝑛 + 𝐻 ′ 𝑁𝑇𝑛
If m=n
𝐻𝑇1 − 𝐻 ′ 𝑇1 = 𝐻 ′ 𝑁𝑇1 − 𝐻𝑁𝑇1
𝐻𝑇2 − 𝐻 ′ 𝑇2 = 𝐻 ′ 𝑁𝑇2 − 𝐻𝑁𝑇2
.
.
.
𝐻𝑇𝑛 − 𝐻 ′ 𝑇𝑛 = 𝐻 ′ 𝑁𝑇𝑛 − 𝐻𝑁𝑇𝑛
𝑛
n
𝑛
n
′
′
∴ ∑ 𝐻𝑇𝑖 − ∑ 𝐻 𝑇𝑗 = ∑ 𝐻 𝑁𝑇𝑗 − ∑ 𝐻𝑁𝑇𝑖 − − − − − (1)
𝑖=1
𝑚
𝑗=1
m
𝑗=1
m
𝑖=1
𝑚
∑ 𝐻𝑇𝑖 − ∑ 𝐻 ′ 𝑇𝑗 = ∑ 𝐻 ′ 𝑁𝑇𝑗 − ∑ 𝐻𝑁𝑇𝑖 − − − − − (2)
𝑖=1
𝑗=1
𝑗=1
𝑖=1
Let 𝑋𝑇 be the sum of the difference in the haplotype counts between the transmitted
and the non-transmitted haplotypes overall all triads. Similarly, let 𝑋𝑆 be the sum of
the difference in haplotype counts between the affected and unaffected sibs over all
discordant sib pairs.
𝑚
Therefore , 𝑋𝑇 = ∑(𝐻𝑇𝑖 − 𝐻𝑁𝑇𝑖 )
𝑖=1
𝑚
𝑛
If m = n, 𝑋𝑠 = ∑ 𝐻𝑇𝑖 − ∑ 𝐻 ′ 𝑇𝑗
𝑖=1
𝑗=1
𝑚
𝑛
If m ≠ n, 𝑋𝑠 = 𝑛 ∑ 𝐻𝑇𝑖 − 𝑚 ∑ 𝐻 ′ 𝑇𝑗
𝑖=1
𝑗=1
1
The PDT statistic for haplotype ( ) for this nuclear family is calculated as 𝐷(1) =
1
1
1
𝑛𝑇 +𝑛𝑆
(𝑋𝑇 + 𝑋𝑆 )
After permuting the transmitted and non-transmitted haplotypes, then
m
𝑋(𝑇) = ∑(𝐻𝑁𝑇𝑖 − 𝐻𝑇𝑖 )
i=1
𝑚
n
𝑋(𝑆) = ∑ 𝐻𝑁𝑇𝑖 − ∑ 𝐻 ′ 𝑁𝑇𝑗
𝑖=1
𝑃(1) =
1
j=1
1
(𝑋 + 𝑋(𝑆) )
𝑛𝑇 + 𝑛𝑆 (𝑇)
Using the similar procedures for the single-SNP situation, we can show that 𝐷(1) =
1
−𝑃(1)
1
1
2
2
The same procedure can also be applied to haplotypes ( ) , ( ) , ( ) and the same
2
1
2
conclusion can be made.
𝑎
Define 𝐷(𝑎) is the PDT statistic for haplotype ( ) at two SNPs
𝑏
𝑏
𝑎
𝑃(𝑎) is the PDT statistics for haplotype ( ) after permuting the transmitted
𝑏
𝑏
and non-transmitted haplotypes , where a=1,2 and b=1,2
Let the PDT statistics for allele 1 at the first and second SNPs be 𝐷1 and D1′ ,
respectively. Similarly, let the PDT statistics for allele 2 at the first and second SNPs
be 𝐷2 and D2′ , respectively. It can be shown that
𝐷1 = 𝐷(1) + 𝐷(1) = −𝑃(1) − 𝑃(1) = − (𝑃(1) + 𝑃(1) ) = −𝐷(1)
1
2
1
2
1
2
𝐷2 = 𝐷(2) + 𝐷(2) = −𝑃(2) − 𝑃(2) = − (𝑃(2) + 𝑃(2) ) = −𝐷(2)
1
2
1
2
1
2
𝐷1′ = 𝐷(1) + 𝐷(2) = −𝑃(1) − 𝑃(2) = − (𝑃(1) + 𝑃(2) ) = −𝐷(1′ )
1
1
1
1
1
1
𝐷2′ = 𝐷(1) + 𝐷(2) = −𝑃(1) − 𝑃(2) = − (𝑃(1) + 𝑃(2) ) = −𝐷(2′ )
2
2
2
2
2
2
Thus, if we permute the transmitted and non-transmitted haplotypes, it simply results
in simultaneous sign changes in the PDT statistics for the two SNPs.
We then check if the results are still true when a haplotype has recombination.
Assume we consider a nuclear family with one affected and one unaffected siblings
and only the affected sib has a haplotype with recombination from the mother.
We only consider the case where the genotypes at the two SNPs in the mother are
both heterozygous. If at least one SNP is homozygous, the observation of haplotype in
the affected sib is the same as the situation of no recombination. Then the same proof
from the condition of no recombination can be applied.
1 2
Therefore, we consider the condition that the mother has haplotype ( ) ( ). Then,
1 2
we would like to show that 𝐷(1) + 𝐷(1) = −𝑃(1) − 𝑃(1) is still true.
1
2
1
2
𝑎
Let 𝑇𝐴𝑎𝑏 be the number of haplotype ( ) transmitted from parents and 𝑁𝑇𝐴𝑎𝑏 be
𝑏
𝑎
the number of non-transmitted haplotype ( ) from parents to the affected child.
𝑏
𝑎
Let 𝑇𝑈𝑎𝑏 be the number of haplotype ( ) transmitted from parents and 𝑁𝑇𝑈𝑎𝑏 be
𝑏
𝑎
the number of non-transmitted haplotype ( ) from parents, to the unaffected child.
𝑏
Table A.3 summarizes the transmitted and non-transmitted statistics for haplotype
𝑎
( ) for the affected and unaffected siblings.
𝑏
1
For haplotype ( )
1
1
If there is no recombination in the mother, then the number of haplotype ( ) in
1
parents is fixed, 𝑇𝐴11 + 𝑁𝑇𝐴11 = 𝑇𝑈11 + 𝑁𝑇𝑈11
1
However, since haplotype ( ) has recombination in the mother and only the affected
1
child has the haplotype with recombination from the mother, the number of haplotype
1
1
( ) in the affected child is the number of haplotype ( ) in the unaffected child
1
1
minus 1. Therefore, 𝑇𝐴11 + 𝑁𝑇𝐴11 + 1 = 𝑇𝑈11 + 𝑁𝑇𝑈11
∴ 𝑇𝐴11 − 𝑇𝑈11 = 𝑁𝑇𝑈11 − 𝑁𝑇𝐴11 − 1 − − − − −(1)
Let 𝑋𝑇 1 be the difference in the haplotype counts between the transmitted and non( )
1
transmitted haplotypes. Similarly, let 𝑋𝑆 1 be the difference in the haplotype counts
( )
1
between the affected and the unaffected siblings.
Therefore , 𝑋𝑇 1 = 𝑇𝐴11 − 𝑁𝑇𝐴11 , 𝑋𝑆 1 = 𝑇𝐴11 − 𝑇𝑈11
( )
1
( )
1
In this case, nT =1 and nS =1
1
The PDT statistic for haplotype ( ) for this nuclear family is calculated as 𝐷(1) =
1
1
1
2
(𝑋𝑇 1 + 𝑋𝑆 1 )
( )
1
( )
1
After permuting the transmitted and non-transmitted labels, then
𝑃𝑇 1 = 𝑁𝑇𝐴11 − 𝑇𝐴11
( )
1
𝑃𝑆 1 = 𝑁𝑇𝐴11 − 𝑁𝑇𝑈11
( )
1
Therefore, 𝑋𝑇 1 = 𝑇𝐴11 − 𝑁𝑇𝐴11 = −(𝑁𝑇𝐴11 − 𝑇𝐴11 ) = −𝑃𝑇 1
( )
1
( )
1
𝑋𝑆 1 = 𝑇𝐴11 − 𝑇𝑈11 = 𝑁𝑇𝑈11 − 𝑁𝑇𝐴11 − 1 = −𝑃𝑆 1 − 1
( )
1
( )
1
− − − 𝑏𝑦 (1)
1
For haplotype ( )
2
1
Similar to haplotype ( ), we can show that 𝑇𝐴12 + 𝑁𝑇𝐴12 = 𝑇𝑈12 + 𝑁𝑇𝑈12 + 1
1
∴ 𝑇𝐴12 − 𝑇𝑈12 = 𝑁𝑇𝑈12 − 𝑁𝑇𝐴12 + 1 − − − − −(2)
Also define 𝑋𝑇 1 = 𝑇𝐴12 − 𝑁𝑇𝐴12 , 𝑋𝑆 1 = 𝑇𝐴12 − 𝑇𝑈12
( )
2
( )
2
1
The PDT statistic for haplotype ( ) for this nuclear family is calculated as 𝐷(1) =
2
2
1
2
(𝑋𝑇 1 + 𝑋𝑆 1 )
( )
2
( )
2
After permuting the transmitted and non-transmitted haplotypes, then
𝑃𝑇 1 = 𝑁𝑇𝐴12 − 𝑇𝐴12
( )
2
𝑃𝑆 1 = 𝑁𝑇𝐴12 − 𝑁𝑇𝑈12
( )
2
Therefore, 𝑋𝑇 1 = 𝑇𝐴12 − 𝑁𝑇𝐴12 = −(𝑁𝑇𝐴12 − 𝑇𝐴12 ) = −𝑃𝑇 1
( )
2
( )
2
𝑋𝑆 1 = 𝑇𝐴12 − 𝑇𝑈12 = 𝑁𝑇𝑈12 − 𝑁𝑇𝐴12 + 1 = 1 − 𝑃𝑆 1
( )
2
∴ 𝐷(1) + 𝐷(1) =
1
2
=
( )
2
− − − 𝑏𝑦 (2)
1
(𝑋
+ 𝑋𝑆 1 + 𝑋𝑇 1 + 𝑋𝑆 1 )
( )
( )
( )
nT + nS 𝑇(11)
1
2
2
1
(−𝑃𝑇 1 − 𝑃𝑆 1 − 1 – 𝑃𝑇 1 + 1 − 𝑃𝑆 1 )
( )
( )
( )
( )
nT + nS
1
1
2
2
=−
1
(𝑃
+ 𝑃𝑆 1 + 𝑃𝑇 1 + 𝑃𝑆 1 ) = −𝑃(1) − 𝑃(1)
( )
( )
( )
nT + nS 𝑇(11)
1
2
1
2
2
The argument is still true when we have different numbers of affected and unaffected
siblings. The procedure can also be extended to multiple SNPs and it can be shown
that permuting transmitted and non-transmitted haplotypes at multiple SNPs results in
simultaneous sign changes for all of the SNPs.
Figures
Figure 1.A. Independent haplotypes are labeled with different colors. Haplotypes
transmitted from the mother to the affected sib have a crossover between the two
SNPs.
Tables
Table A.1
Transmitted
Non
Transmitted
Affected
Affected
Affected
Unaffected
Unaffected
Unaffected
sib 1
sib 2
sib m
sib 1
sib 2
sib n
𝐺𝑇1
𝐺𝑇2
…
𝐺𝑇𝑚
𝐺 ′ 𝑇1
𝐺 ′ 𝑇2
…
𝐺 ′ 𝑇𝑛
𝐺𝑁𝑇1
𝐺𝑁𝑇2
…
𝐺𝑁𝑇𝑚
𝐺 ′ 𝑁𝑇1
𝐺 ′ 𝑁𝑇2
…
𝐺 ′ 𝑁𝑇𝑛
Affected
Affected
Affected
Unaffected
Unaffected
𝐻𝑇1
𝐻𝑇2
…
𝐻𝑇𝑚
𝐻 ′ 𝑇1
𝐻 ′ 𝑇2
…
𝐻 ′ 𝑇𝑛
𝐻𝑁𝑇1
𝐻𝑁𝑇2
…
𝐻𝑁𝑇𝑚
𝐻 ′ 𝑁𝑇1
𝐻 ′ 𝑁𝑇2
…
𝐻 ′ 𝑁𝑇𝑛
Table A.2
Transmitted
Non
Transmitted
Table A.3
Affected
Transmitted
Non Transmitted
Unaffected
𝑇𝐴𝑎𝑏
𝑇𝑈𝑎𝑏
𝑁𝑇𝐴𝑎𝑏
𝑁𝑇𝑈𝑎𝑏
Unaffected