Question

18. You have been hired by a dairy company to assist in their cow breeding program. They have a population of cows from which they have calculated the phenotypic and genotypic variance values for milk fat content. With this information, they want you to find the broad sense heritability of milk fat content in their herd. Variance Type Phenotypic Additive Genetic Dominance Epistatic Value for Cows 82.3 33.4 24.7 15.9 b. They now want to alter the milk fat content produced by their herd. Their population has an average content of 3.7, but they want to select for a higher content by using a breeder population with a mean of 4.6. What would one round of selective breeding do to the mean milk fat content in the next generation?

Answer 1

Ans a: The broad sense heritability is equal to the ratio of total genetic variance and total phenotypic variance. The total genetic variance is the sum of additive, dominance and epistatic variance. So the value of total genetic variance is equal to 33.4+24.7+15.9 which is equal to 74.

So thr broad sense heritability is equal to 74/82.3 which is equal to 0.899.

b: The broad sense heritability is also equal to the ratio of selection response and selection differential. Now selection differential is equal to 4.6-3.7 and selection response is equal to x-3.7, where x is the milk fat content in next generation.

So putting the values we get:

0.899 = (x-3.7)/(4.6-3.7) and this is equal to 0.899*0.9+3.7 which is equal to 4.5091 or 4.51.

Hence in the next generatio the fat content in the milk is imcreased to 4.51 from 3.7.