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| Description: (a) The probability of A1 IBD equals the likelihood of four transmission events, each with a probability of 1/2. (b) The probability of A2 IBD also requires four transmission events, each with a probability of 1/2. The likelihood of either allele IBD is F = 2(1/2)4. (c) With two common ancestors, there are four alleles (A1 , A2 , A3 , and A4 ) that can be IBD. For this first-cousin mating, the probability for each allele IBD is the same: F = (1/2)6. For all shared alleles combined, F = 4 (1/2)6 = 1/16(a) The probability of A1 IBD equals the likelihood of four transmission events, each with a probability of 1/2. (b) The probability of A2 IBD also requires four transmission events, each with a probability of 1/2. The likelihood of either allele IBD is F = 2(1/2)4. (c) With two common ancestors, there are four alleles (A1 , A2 , A3 , and A4 ) that can be IBD. For this first-cousin mating, the probability for each allele IBD is the same: F = (1/2)6. For all shared alleles combined, F = 4 (1/2)6 = 1/16 Probability of inheriting A, ll Half siblings lll l Probability ofA, IBD =(%)“ = 7% Probability of 2 inheriting A2 m lProbab'lva'BMMl F= Probabil' of eltherA, orA2 IBD wizdx?mwiéfg Probability ofA , allele IBD in lV-I is F=Gr=b e same holds forA A and A4 2’ .3” Probability of any allele (A, A, A, A4) lBD in [W is F = 4( b" = r‘s 3 Picture Stats: Views: 63 Filesize: 411.71kB Height: 2783 Width: 1136 Source: https://biology-forums.com/index.php?action=gallery;sa=view;id=48908 |