Animal Breeding and Genetics

1. Animal Breeding and Genetics Instructor: Dr. Jihad Abdallah Prediction of Breeding Values (selection methods for a single trait)

2. Methods for prediction of Breeding Values (methods of selection for a single trait) • Using records on the same animal (phenotypic selection) • Using records on relatives • Combining several sources of information using an index (selection index) • BLUP

3. Using records on the same animal (phenotypic selection): Average of n records on the same animal: Average of n records repeatability For a single record, this reduces to:

4. Accuracy of prediction Accuracy of prediction of the breeding value is the correlation between the predicted breeding value and the true breeding value. It varies from 0 to 1. Accuracy based on average of n records: For a single record, this reduces to:

5. Example: suppose a ewe has three milk records: 150kg in the first record, 100 kg in the second, and 110 kg in the third. Suppose also that the mean of the flock is 100 kg, the heritability is 0.25 and the repeatability is 0.50. Predict the breeding value of this ewe and calculate the accuracy of prediction based on: • Her first lactation record. • The average of her three records.

6. B. Prediction using information on relatives • Using information on sibs (sib-selection) • Using information on progeny (progeny testing) • Using information on pedigree (selection on pedigree)

7. 1) Using information on sibs (half-sibs or full-sibs): - Single record on each of m sibs: Average of records on m sibs Where: r is the additive relationship between the individual and its sibs t is the correlation of phenotypic values between sibs Covariance due to common environmental effects For paternal half-sibs For full-sibs Maximum value of accuracy is 0.5 using half-sibs and 0.71 using full-sibs

8. Examples on prediction of breeding value based on sib information Example 1: a dairy bull has 10 paternal half-sib cows each with a single record. Suppose that the average of milk records on the 10 half-sibs is 6000 kg and the population average is 5000 kg. Assuming that the heritability of milk yield is 0.30, compute the breeding value of the bull from these information and the accuracy of the prediction. Solution:

9. Example 2: a rabbit has carcass weight information on 10 full-sibs each with a single record. Suppose that the average carcass weight of his 10 full-sibs is 600 g above the mean of the population. Assuming that the heritability of carcass weight is 0.40, compute the breeding value of carcass weight for this rabbit from these information and the accuracy of the prediction. Assume that the environmental covariance (c2) is 0.10. Solution: Note that the accuracy of prediction is larger than that obtained from using information on half-sibs. Why?

10. 2) Using information on Progeny: - Single records from p progeny (half-sib or full-sib progeny): Average of records on p progeny Where: r =1/2 t is the correlation of phenotypic values between progeny Covariance due to common environmental effects If the progeny are half-sibs If the progeny are full-sibs

11. - If each progeny has n records: p = number of progeny n = number of records per progeny re = repeatability • NOTES: • Accuracy increases more by increasing the number of progeny than increasing the number of records per progeny • Accuracy using half-sib progeny is larger than that using full-sib progeny • Accuracy can reach 1.0 if the number of half-sib progeny is very large

12. Examples on prediction of breeding value based on progeny test Example 1: a dairy sire has 10 half-sib daughters each with a single record. Suppose that the average of milk records on the 10 daughters is 6000 kg and the population average is 5000 kg. Assuming that the heritability of milk yield is 0.30, compute the breeding value of the sire from these information and the accuracy of prediction. Solution: Note that the accuracy of prediction is larger than that obtained from using information on sibs. Why?

13. Example 2: Solve the same problem if the sire has 40 daughters instead of 10 and their average remained the same. Solution: Note that the accuracy of prediction increased by increasing the number of daughters

14. Example 3: Solve example 2 assuming each daughter has two records instead of one and their average remained the same. Assume repeatability is 0.50 For p = 40, n = 5 accuracy = 0.92

15. 3) Using information on parents Square of the accuracy of breeding value of the sire Square of the accuracy of the breeding value of the dam

16. Example: an animal’s sire has 50 progeny test records and the dam has two repeated measurements . If the heritability of the trait is 0.25 and the repeatability is 0.50, find the accuracy of the animal’s predicted breeding value First we find the square of the accuracy of breeding value of the sire: Second we find the square of the accuracy of breeding value of the dam: Then we find the accuracy of the prediction of breeding value of the animal:

17. C- Combining several sources of information using an index • We desire to maximize selection response in a trait by using several sources of information on relatives. • The breeding value of the trait is A, and the family index used to predict the breeding value is • , where Pj is the family information as a deviation from an appropriate mean and the bj are partial regression coefficients of A on Pj

18. EXAMPLE 1: phenotype of individual and phenotype of a single relative If the relative is a half-sib, VA = 30,and VP=100, then r = 0.25 , t = ¼ (0.30) = 0.075

19. BLUP (Best Linear Unbiased Prediction Makes use of the additive relationship matrix Requires computer software May incrase inbreeding