12 Istrative purposes, and it becone\"?$???D? Apene at ione laas i.the e m man w

Question

12 Istrative purposes, and it becone"?$???D? Apene at ione laas i.the e m man wo ldewical coples of the same gene from L The probabtity of this happening is the isbreeding corno.eester,3,to the dan, Dyte aceter, could be transmitsed to X ehrough 5 and to X through D. ifso, Xould be homormpout-with The formula for calculating inbreeding coellicients is as folees Nnter the ano-nanning from the sres,though where This generacion will be taken care ef when the total pathways have been summed and then multipled by Xldivided by 2) F, refers to the inbreeding coefficient of individualx E is the Greek symbol meaning to sm or add all paths. s the power to which one-halt muat be ralised, depending upon the number of anow connecting the sire and dam through the conmor f, is the Inbreeding coefficient of the common ancesto The mumber of arrows connecting the sire and dan with the common ancestor is tiwa, and this is the n in the previeus fermula. The commen ancestor is not inbred, se the calculatioss of the inbreeding eetkint proceeds by leting f the common ancestor is not inbred, the formule to use in calculating the Inbrreding cicent Halt-aib mating The follewing pedigree and arow dagram show a haifsb mating the sire and the dam of individual having the same sire. 1 sh mating, except hat an additional path and common ancestor are Involved. Note Pedigree Arrow diagram Sraightened pathway put in the amow dagrem The only common ancester in this pedgree is individual 2, because he appears in the pedigree ofb the sire and dam of individual X. The anrow dagm through the sire and only one through th. ?? The pethway may now be ?-ghtened out for thows that s only one pathway from I te X

Explanation / Answer

1. Inbreeding coefficient (f) = (1/2)N

Where N => Number of nodes = 2

f = (1/2)2 = 1/4 =0.25

f = 1/4 =0.25

2. Paths are

XS356DX

XS456DX

XS35DX

XS45DX

N1=N2=3

f = (1/2)N1 +(1/2)N2 = (1/2)3 + (1/2)3 = 1/8 + 1/8 =1/4

f = 1/4 = 0.25

3. Paths are

XS1462DX

XS1452DX

XS1DX

XS2DX

N1=N2=4

f = (1/2)N1 +(1/2)N2 = (1/2)4 + (1/2)4 = 1/16 + 1/16 =1/8

f = 1/8 = 0.125

4. Paths are

XS1493DX

XSDX

N1=N2=5

f = (1/2)N1 +(1/2)N2 = (1/2)5 + (1/2)5 = 1/32 + 1/32 =1/16

f = 1/16 = 0.0625

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