ANSC 313 Lecture Notes - Lecture 10: Genotype, Zygosity
Breeding values!
Calcula&ons*of*breeding*values*and*EPD's*
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**********V*=*A*+*D*
V*=*genotypic*value*
A*=*breeding*value*
D=*dominance*devia&on*
If*heterozygote*V*is*not*half*way*between*the*two*homozygotes*
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Genotypic*value*(V)*
The*devia&on*of*the*phenotype*from*the*average*of*the*homozygotes*(m)*
Mean*=*9v*
A1A1*
12*units*
V11*=*12*-*m*=*3*
a*
A1A2*
10*units*
V12*=*10*-*m*=*1*
d*
A2A2*
6*units*
V22*=*6*-*m*=*-3*
-a*
Incomplete*dominance*opera&ng*at*this*case*
0<*d*<*a*
To*alter*phenotype*
* •* Improve*environment*(*beQer*barn,*feed)
* •* Change*genotype
Wish*to*measure*how*much*changing*the*genotype*would*help*
Genotypic*value*=*V*
Breeding*value*=*A*
(Addi&ve*gene&c*component)*
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An*animal*breeding*value*is*directly*propor&onal*to*how*many*favourable*alleles*It*
carried*AND*transmits*to*offspring*
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Ex.*
Lacta&on*genotype*in*dairy*caQle*is*associated*with*milk*yield*but*the*genotype*cant*
Document Summary
If heterozygote v is not half way between the two homozygotes. The devia&on of the phenotype from the average of the homozygotes (m) V11 = 12 - m = 3 a. V12 = 10 - m = 1 d. Wish to measure how much changing the genotype would help. An animal breeding value is directly propor&onal to how many favourable alleles it carried and transmits to o spring. Lacta&on genotype in dairy caqle is associated with milk yield but the genotype cant. But at worst llllll be assessed directly since we don"t know which genes are involved (yet!) Each l allele adds about 100 litres of milk. If all cows in a herd or breed are: llllll: all o spring are at best llllll. Therefore for all bulls ( genotype llllll or llllll) Detect no improvements in calves regardless of bull. Bu some bulls make the calves worse than their dams.
