SLIDE 5 5
( )
IBD that were alleles diff d transmitte CA that prob paths both down allele identical transmited CA that prob.
2 1 2 1
CA
F +
A note about the common ancestor (CA)
A C D E F I C D E F I B Path 1 Path 2
Segment 1 Segment 2 Probability x1 x1 ½ × ½ = ¼ x1 x2 ¼ x2 x1 ¼ x2 x2 ¼ Genotypes: X1 X1 or X2 X2
¼ + ¼ = ½
Source 1 for IBD: same allele passed down both segments Source 2: It is also possible that X1 and X2 are IBD
(¼)FCA + (¼)FCA = (½)FCA
x1 & x2 = 1/4 x2 & x1= 1/4
- I. Individual inbreeding in a pedigree sense :
Step 3: Compute the probability of the path (in our case 2 paths, each with n = 6 segments).
A C D E F I C D E F I B Path 1 Path 2
( ) ( )
CA n
F
i
+
− 1
2 / 1
1 Probability of a path:
( ) ( ( ) )
slide last from i.e., segments; CA for two y Probabilit CA CA to connected not segments 2
for y Probabilit 2
F 2 1 2 1 2 1 + ×
− n
( ) ( )( )
CA n
F + ×
−
1 2 1 2 / 1
2