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Course: Advanced Animal Breeding

Course: Advanced Animal Breeding. MS program in Animal Production Faculty of Graduate Studies An-najah National University Instructor: Dr. Jihad Abdallah Lecture 2: Inbreeding and Genetic Relationships. Inbreeding.

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Course: Advanced Animal Breeding

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  1. Course: Advanced Animal Breeding MS program in Animal Production Faculty of Graduate Studies An-najah National University Instructor: Dr. Jihad Abdallah Lecture 2: Inbreeding and Genetic Relationships

  2. Inbreeding • Inbreeding means the mating of individuals that are related to each other by ancestry (have a common ancestor) • We define the inbreeding coefficient, F, as the probability that two alleles at any locus in an individual are identical by descent (IBD), that is have the same common ancestor. • Sometimes the term autozygous is used to refer to IBD alleles and the term allozygous is used to refer to alleles non-IBD. • F is a probability and thus ranges from 0 (non-inbred individual or population) and 1.0 (completely inbred individual or population) • By definition, F in the base population is 0.

  3. Inbreeding in the idealized population • Assume initially an “ideal” population of finite size N, the population size is constant over generations, and mating is at random including self-fertilization (panmictic population). F = 0 in this population. • What is the inbreeding after one generation of panmixia (random mating including selfing)? • At t =1  F1 = (1/2N) • At t =2  F2 = (1/2N)+ (1-1/2N)F1 At any generation t:

  4. Thus inbreeding is cumulative and equals the “new” inbreeding + “old” inbreeding that occurred in the previous generation. • We can call the new inbreeding or the “increment” as ΔF. • ΔF=(1/2N) for the ideal population only: • We can express the inbreeding coefficient in terms of ΔF as: When Inbreeding coefficient is expressed in terms of ΔF, the previous equation is valid for any breeding system and not only the idealized population though only in the idealized population that ΔF=(1/2N) and in general: This is the average inbreeding over all loci of individuals in a population of finite size

  5. We can look at the dispersive process in terms of inbreeding. • The variance of the gene frequency among the sub-populations at any generation t can be written as: • Genotypic frequencies: Genotype A1A1 A1A2 A2A2 Frequency in the whole population

  6. Example: for the following data, find the inbreeding coefficient. • aa Aa AA TOTAL • 42 76 448 566 • Solution: The observed frequency of heterozygotes is 76/566 = 0.1343 p = [2 (448 )+ 76]/1132 = 0.8587, q = (1-0.8587)= 0.1413 Ht = 2p0q0(1-Ft) NOTE: p is an estimate of p0 and q is an estimate of q0  0.1343=2(0.8587)(0.1413)(1-Ft)  Ft = 0.447

  7. Quantifying sub-population differentiation • Sub-population differentiation can be quantified using the overall reduction of heterozygosity. • H0 = 2p0q0 is the frequency of hetrozygotes (heterozygosity) in the base population before subdivision. • Ht = 2p0q0(1-Ft) is the heterozygosity in the collection of sub-populations after t generations of drift. • This leads to:  Measures the amount of differentiation (large values indicate large differentiation) Usually denoted as FST and called fixation index

  8. Example:

  9. Effective population size (Ne) • Real populations depart from the idealized population model in many ways: real populations don’t have the same number of individuals breeding each generation or the same number of breeding males and females, or equal number of offspring for all individuals • When N varies between generations, sexes, or offspring, it must be replaced by an average value which accounts for the population’s history and structure • This average N is called Ne, the effective population size. Ne is a weighted average number of breeding individuals. • Neis defined by Falcconer and Mackay (1996) as the number of individuals that would give the same rate of inbreeding if they breed in the manner of the idealized population. • Ne is generally less than the actual size of the population • ΔF=(1/2Ne) and as before:

  10. Estimation of Ne 1. Exclusion of closely related matings: - Self-fertilization excluded  Ne = N+1/2 (approx.) - With sib-mating excluded  Ne = N+2 (approx.) 2. Different numbers of males and females:  gives a good approximation unless both Nm and Nf are very small (approx) Nm = number of breeding males Nf= number of breeding females Example: a herd of cattle of 4 bulls and 100 cows  Ne = 15.385 4 bulls and 200 cows  Ne = 15.686 Little change by doubling the number of cows

  11. (approx) 3. Unequal numbers in successive generations: Example:N1=100, N2=50, N3=200 Solution:t = 3

  12. 4.Non-random distribution of family size (Vk): • We define family size (k) as the number of progeny (offspring) per parent or pair of parents that survive to become breeding individuals. • In the ideal population, there is random variation of family size and each individual contributes equally to the next generation, so the family size per pair of parents is 2 and the variance of family size is also 2 (family size follows a Poisson dist). • When parents do not have an equal chance of contributing to the next generation, the variance of family size is greater than 2. (approx)  assumes equal number of males and females and constant population size If Vk = 2 as in the ideal population  Ne = N Vk > 2  Ne < N Vk < 2  Ne > N

  13. If males can mate with more than one female  Vkm = variance of family size in males Vkf = variance of family size in males

  14. 5. Minimal inbreeding: • By choice of the individuals to be used as parents, the variance of family size can be reduced below its random amount. • If the individuals are chosen equally from all families (two individuals from each family), then Vk = 0  Ne = 2N-1 • If the number of males and females are not equal (more females), Vk can be made zero by choosing as parents one male from each sire’s progeny and one female from each dam’s progeny  Nm = actual number of males Nf= actual number of females

  15. Inbreeding and Relationships in pedigreed populations • How to calculate the inbreeding coefficient of an individual and the relationship between two individuals in a known pedigree? • Two methods: 1. Path method 2. Tabular method

  16. Path method 1. Inbreeding coefficient (FX) O FX = inbreeding coefficient of individual X i = is the ith common ancestor of the parents of X n = is the number of individuals in any path of relationship between the parents of X counting the parents of X 1 2 3 4 In this example there are 3 common ancestors for the parents of X (O, 1 and 2) X We assume that F= 0 for any individual which has one or both parents missing (unknown)

  17. Calculation of FX in the previous example using the Path method FO = 0 (both parents are unknown) F1 = 0 (one parent is missing and assumed unrelated to O) F2 = 0 (one parent is missing and assumed unrelated to O) FX = 1/32 + 1/32 +4/32 + 4/32 = 10/32 = 0.3125 We can show that F3 = 0.125 and F4 = 0.125

  18. 2. Wright’s coefficient of relationship (RXY): It is the probability that two alleles carried by two individuals are identical by descent, the two individuals having one or more common ancestors. RXY = Wright’s coefficient of relationship between individuals X and Y i = the ithcommon ancestor of X and Y n = the number of arrows connecting X and Y through the ith common ancestor rXY is called the numerator relationshipcoefficient

  19. Calculation of R34 in the previous example using the Path method r34= 1/16 + 1/16 +4/16 + 4/16 = 10/16 = 0.625 We need to calculate F3 and F4 F3 = F4 =(1/2)3(1+0):=0.125 3 and 4 have one common ancestor which is O

  20. Tabular method Rules: rij is the numerator relationship between individuals i and j sj is the sire of individual j, djis the dam of individual j rij = rji riiis the relationship between the individual i and itself is the numerator relationship between the sire of i and the dam of i rij = 0 if both parents of j are unknown NOTE: Individual j is the younger individual. If i and j are in the same generation, then j is the one with more known parents

  21. FO =0, rOO = 1, rO1 = rO2 = 0.5(rOO + 0) = 0.5(1+0)= 0.5 rO3 = rO4 = 0.5(rO1 + rO2) = 0.5(0.5+0.5) = 0.5, rOX = 0.5(rO3 + rO4) = 0.5(0.5+0.5) = 0.5 F1 = 0  r11 =1, r12 =0.5(rO1+0)=0.5(0.5+0)=0.25, r13 = r14 = 0.5(r11+r12) = 0.5(1+0.25) = 0.625,r1X =0.5(r13+r14)=0.5(0.625+0.625) = 0.625 F2 = 0  r22 =1, r23 = r24 = 0.5(r22+r21)=0.5(1+0.25) = 0.625,r2X = 0.5(r23+r24) = 0.5(0.625+0.625) = 0.625 F3= 0.5(r12) = 0.5(0.25) = 0.125  r33 = 1+0.125 =1.125 r34 = 0.5(r13+r23) = 0.5(0.625+0.625)=0.625, r3X = 0.5(r33+r34)= 0.5(1.125+0.625) = 0.875, F4 = 0.5(r12) = 0.5(0.25) = 0.125  r44 = 1+0.125 =1.125, r4X = 0.5(r44+r34) = 0.5(1.125+0.625)=0.875, FX = 0.5(r34) = 0.5(0.625) = 0.3125  rXX = 1+FX = 1+0.3125 =1.3125

  22. Numerator relationships for the previous example: Numerator relationship matrix (A)

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