1. Chapter 4: Relationship and inbreeding  Definitions  Calculation of relationship and inbreeding coefficients  Examples  Segregation of recessive by inbreeding  The tabular method

2. Definition, Relationship Related individuals are individuals with common ancestors, this can for instance be a common father The relationship coefficient a xy is equal to the probability of identical genes in two animals due to common ancestors

3. Definition, Inbreeding Inbreeding occurs after mating of related individuals The degree of inbreeding F x is the probability of identical homozygosity due to common ancestors

4. Formulas: Relationship- and inbreeding coefficient The relationship coefficient a xy =  (½) n (1 + F A ), where n is the number of generations between X and Y from the common ancestor A, level of inbreeding is F A The inbreeding coefficient F x = ½ a mf (m = mother, f = father) a xx = 1+ F x

5. Example: Double first cousin breeding

6. Simple forms of relationship

7. Simple forms of relationship, continued

8. Practical application of relationship and inbreeding coefficients Important in the control of inbreeding on individual basis Inbreeding should normally be avoided, and should not be more than 10% By breeding value estimation the relationships coefficients are important for weighing the information from related animals

9. Segregation of identical homozygotes by full sib breeding

10. Segregation of the recessive by inbreeding: Example The gene frequeny for a recessive disease is q = 0,01 and p = 0,99 Which corresponds to the genotype frequencies f(rr) = q 2 = 0,01 2 = 0,0001 f(Rr) = 2pq = 2  0,01  0,99 = 0,0198 f(RR) = p 2 = 0,99 2 = 0,9801

11. Segregation of the recessive, continued P F1 F2 P(mothers father or mothers mother is heterozygot) = 2(2pq) = 2  0,198 P(x is rr  mothers father or mothers mother is heterozygot) = 1/16

12. Segregation of the recessive, continued The joint probability of rr in offspring from full sib breeding depends on: The segregation of rr among the offspring P(rr) = 2  0,0198  1/16 = 0,0025 This corresponds to an increase in f(rr) of 25 times compared to the original population

13. Segregation of the recessive, continued F = 0,25  25 % increase in the frequency of homozygotes - both recessive and dominant - And a loss of heterozygotes Genotype AA Aaaa Frequency p 2 2pqq 2 +pqF -2pqF+pqF

14. The tabular method - two basic formulas Inbreeding (F) for an animal is equal to half of the relationship between its parents Additive relationship (a XY ) between two animals is equal to half of the relationship between the one animal, X, and the other animal, Y’s, parents, A and B

15. The tabular method : Example

16. The tabular method : Example continued