1. Gene350 Animal Genetics Lecture 16 1 September 2009

2. Last Time Quantitative inheritance of traits in animals – Heritability – Repeatability – Phenotypic, genetic and environmental correlations

3. Today Inbreeding and relationships – Inbreeding coefficient – Relationship coefficient – Calculations

4. Inbreeding & Relationship Inbreeding mating of related individuals mating of individuals more closely related than the average of the breed animal is inbred if, and only if, its parents are related.

5. Genetic results from inbreeding Aa  1/4 AA: 1/2 Aa: 1/4 aa  1/4 AA: 1/2 (1/4 AA: 1/2 Aa: 1/4 aa): 1/4 aa    1/2 AA : 1/2 aa

6. Genetic result of inbreeding increased homozygocity if there is dominance: aa may be inferior inbreeding will cause decrease in performance.

7. Phenotypic result of inbreeding inbreeding depression decline in performance from mating related individuals result of inbreeding when there is non-additive gene action (dominance).

8. Phenotypic result of inbreeding inbreeding depression more pronounced for lowly heritable traits class of trait inbreeding depression reproductionhigh growthmoderate carcasslow

9. Measurement of inbreeding inbreeding coefficient (F) degree of homozygocity % increase in homozygocity above the average of the breed if F =.25 the animal is 25% more homozygous than the average of the breed

10. Inbreeding coefficient (F) minimum value = 0 maximum value = 1 for livestock rare for F >.6 lab animals F can approach.9 plants with self-fertilization F can approach 1.0.

11. Calculation of Inbreeding coefficient (F) Fx = inbreeding coefficient of individual X  = summation sign n = number of segregations (arrows) between the sire and dam in each separate path F A = inbreeding coefficient of the common ancestor in each path common ancestor - appears in both sire and dam portions of the pedigree.

12. Step 1: Generate an arrow pedigree  D  B --  H  HB X --  XD  DC  C --  L  L.

13. Step 2: List all paths that connect the sire and the dam. B D C Step 3: Identify the common ancestor in each path B D C.

14. Step 4: Calculate the inbreeding coefficient of the common ancestors. F D = 0 Step 5: Count the number of arrows in each path to determine "n". Only one path and it has two arrows. Therefore, n=2.

15. Step 6: Calculate the value of each path connecting the sire and dam value of path = = =

16. Step 7: Add together the value for each path connecting the sire and dam F =.125

17. | G | I --| || H | S ---| ||| I || A --| || M Z ---| || I || S --| ||| A | D ---| || I | A --| | M

18. ISZADISZAD Three generations - parent offspring matings S D S I A D S A D

19. ISAISA S D S A D S I A D S D F S =.25* 1 ½ 1+1 (1+.25) =.3125 S A D F A = 0 2 ½ 2+1 (1+ 0) =.125 S I A D F I = 0 3 ½ 3+1 (1+ 0) =.0625  =.500 F =.50 S D F S =.25* S A D F A = 0 S I A D F I = 0 S D F S =.25* 1 S A D F A = 2 S I A D F I = 0 3 value = ½ 1+1 (1+0) =.25 * Inbreeding of S path I A

20. Relationships Additive relationship is most commonly used measure of relationship Measures fraction of like genes shared by 2 animals Indicates reliability of an animal in predicting the genetic value of the other animal Knowledge of relationships is important in: – selecting animals on the basis of relatives’ records – Arranging matings to avoid high levels of inbreeding – Establishing lines tracing to desirable animals

21. Measurement of relationship relationship coefficient (R XY ) proportion of genes in common between relatives if R XY =.25 the two individuals have 25% more genes in common than two random members of the breed

22. Common relationships An animal is related to itself by 100% The relationship between a parent and its offspring is 50% The halving of relationships and genes occurs with each generation The relationship to each parent is 50%, to each grandparent 25%, to each great- grandparent 12.5%

23. Common relationships Calves by the same bull but different dams are called paternal half-sibs; average relationship is 25% Maternal half-sibs (same mother, different father) 25% related Full sibs (full brothers and sisters) are 50% related. 25% through sire and 25% through dam

24. Common relationships PairRelatives% IFB,CP-O50 KGCAC,DFS50 A,GGP-GO25 HDBA,KGGP-GGO12.5 C,HF.aun.nie25 EG,HF.cou12.5

25. Common relationships DE,FHB/S25 IAE,JH.aun.nie12.5 EI,JH.cou6.25 BC,Citself100 F JC H

26. Common relationships Relationships can be both direct and collateral Direct relatives are in direct line of descent such as parent and progeny or grandparent and grandprogeny A, C, G and K are direct relatives Collateral relatives have one or more common ancestors but are not direct relatives

27. Measurement of relationship relationship coefficient (R XY ) R = relationship between individuals X and Y  = sum together all the values of the paths n= number of arrows connecting X and Y F A = inbreeding coefficient of the common ancestor Fx &F Y = inbreeding coefficients of animals X and Y

28. Step 1: Generate an arrow pedigree  D  B --  H  HB X --  XD  DC  C --  L  L. Calculate relationship between X and D

29. Step 2: List all paths that connect the X with D. XBD XCD Step 3: Identify the common ancestor in each path XBD XCD

30. Step 4: Calculate the inbreeding coefficient of the common ancestors F D = 0 Step 5: Count the number of arrows in each path to determine "n" Two paths, each has two arrows Therefore, n=2 (for each path).

31. Step 6: Calculate the value of each path connecting the sire and dam value of paths ½ n (1+F A ) 1st path ½ 2 (1+0) =.25 2nd path ½ 2 (1+0) =.25

32. Step 7: Add together the value for each path connecting the sire and dam sum =.50 Step 8: divide by value in denominator =.4714

33. | G | I --| || H | S ---| ||| I || A --| || M Z ---| || I || S --| ||| A | D ---| || I | A --| | M

34. ISZADISZAD Z S Z D A S Calculate R ZS Z D S Z D A I S

35.  = 1.125 Z S Z D S Z D A S Z D A I S F S =.25 F A = 0 F I = 0 1 2 3 4 ½ 1 (1+.25) =.625 ½ 2 (1+.25) =.3125 ½ 3 (1+ 0) =.125 ½ 4 (1+ 0) =.0625 =.8216

36. Phenotypic effects of inbreeding increased homozygocity leads to more homozygotes recessive homozygotes tend to be inferior large loss in reproduction and livability some loss in growth and efficiency little loss in carcass merit.

37. Phenotypic effects of inbreeding  litter size  libido  sperm count  conception rate  survival rate  ovulation rate  rebreeding interval  growth.

38. Linebreeding pedigrees constructed so that F <.1 R >.25 to a single common ancestor single common ancestor appears several times > 3 generations back

39. Abdallah Charles Kent Mare Mambino Amazonia One Eye Bellfounder Hambletonion10 Son of Messinger Bishop’s Hambletonion Silver Tail Imp. Messinger F <.05R H10,IM >.25 single common ancestor

40. 16 Tons Leo Tom Lady Gray Little Fanny Joe Reed II Lady Reed Joe Reed F too high more than one common ancestor

41. Linebreeding dangers of linebreeding inbreeding will build up may linebreed to inferior son of outstanding ancestor ideals may change.

42. Linebreeding who should linebreed? Outstanding purebred herd hard to find breeding stock to improve the herd Clearly identified outstanding ancestor Good knowledge of pedigrees and inbreeding.

43. Avoidance of inbreeding need to keep inbreeding at low level F >.1time to think about it F >.2time to worry about it producers should not actively pursue inbreeding.

44. Avoidance of inbreeding avoid mating close relatives nothing closer than cousins keep ratio of males / females high keep replacements from several families avoid bottlenecks (period of reduced census number).