How To Calculate Breeding Value

BREEDING VALUE: Problem no 1

If A1 = 10 & A2 = -10 then calculate breeding values of A1A1, A1A2, and A2A2.

Breeding value A1A1 = 10 + 10 = 20

Breeding value A1A2 = 10 -10 = 0

Breeding value A2A2 = -10 -10 = -20

BREEDING VALUE: Problem no 2

If in a random mating population having 2 alleles at a single locus, the genotype value and frequency for A1A1 is 320 and 0.25, A1A2 is 300 and 0.5 and A2A2 is 280 and 0.25 respectively then calculate the breeding value for each genotype if p=q=0.5.

Population mean = sum of frequency x values

280 x 0.25 + 300 x 0.5 + 320 x 0.25 = 300

Genetic value = deviation from population mean

e.g. A2A2 = 280 – 300 = -20

Breeding value = sum of average effect of alleles

Average effect of A1

A1 allele will meet A1 at frequency p and A2 at frequency q

A1A1 x p + A1A2 x q = 20 x 0.5 + 0 x 0.5 = 10

Average effect of A2

A2 allele will meet A1 at frequency p and A2 at frequency q

A2A1 x p + A2A2 x q = 0 x 0.5 + 20 x 0.5 = -10

Breeding value of A1A1 = average effect of A1+ average effect of A1

=10+10= 20

Breeding value of A1A2 = average effect of A1+ average effect of A2

=10-10= 0

Breeding value of A2A2 = average effect of A2+ average effect of A2

=-10-10= -20

BREEDING VALUE: Problem no 3

If in a random mating population having 2 alleles at a single locus, the genotype value and frequency for A1A1 is 320 and 0.25, A1A2 is 310 and 0.5 and A2A2 is 280 and 0.25 respectively then calculate the breeding value for each genotype if p=q=0.5.

Population mean = sum of frequency x values

280 x 0.25 + 310 x 0.5 + 320 x 0.25 = 305

Genetic value = deviation from population mean

e.g. A2A2 = 280 305 = -25

Breeding value = sum of average effect of alleles

Average effect of A1

A1 allele will meet A1 at frequency p and A2 at frequency q

A1A1 x p + A1A2 x q = 15 x 0.5 + 5 x 0.5 = 10

Average effect of A2

A2 allele will meet A1 at frequency p and A2 at frequency q

A2A1 x p + A2A2 x q = 5 x 0.5 + 25 x 0.5 = -10

Breeding value of A1A1 = average effect of A1+ average effect of A1

=10+10= 20

Breeding value of A1A2 = average effect of A1+ average effect of A2

=10-10= 0

Breeding value of A2A2 = average effect of A2+ average effect of A2

=-10-10= -20

BREEDING VALUE: Problem no 4

If in a random mating population having 2 alleles at a single locus, the genotype value and frequency for A1A1 is 320 and 0.01, A1A2 is 300 and 0.18 and A2A2 is 280 and 0.81 respectively then calculate the breeding value for each genotype if p=0.1 and q=0.9.

Population mean = sum of frequency x values

280 x 0.81 + 300 x 0.18 + 320 x 0.01 = 284

Genetic value = deviation from population mean

e.g. A2A2 = 280 -284 =  – 4

Breeding value = sum of average effect of alleles

Average effect of A1

A1 allele will meet A1 at frequency p and A2 at frequency q

A1A1 x p + A1A2 x q = 36 x 0.1 + 16 x 0.9 = 18

Average effect of A2

A2 allele will meet A1 at frequency p and A2 at frequency q

A2A1 x p + A2A2 x q = 16 x 0.1 + 4 x 0.9 = -4

Breeding value of A1A1 = average effect of A1+ average effect of A1

=18+18= 36

Breeding value of A1A2 = average effect of A1+ average effect of A2

=10-2= 16

Breeding value of A2A2 = average effect of A2+ average effect of A2

=-2-2= -4

BREEDING VALUE: Problem no 5

Given that the yearling weight of a heifer is 320 kg in a herd with a mean of 250 kg, predict her breeding value and its accuracy if the heritability of yearling weight is 0.45.

From equation,

â_{i =} b(y_i-µ)

â = 0.45(320-250)

= 3150 kg

Also Study: Inbreeding in Humans | Inbreeding in Animals | Inbreeding in Plants