CALCULATING THE COEFFICIENT OF INBREEDING BY COANCESTRY[1]

For complicated pedigrees (mainly found in animal breeding) the method of Coancestry offers an easier option than path analysis for calculating the Coefficient of Inbreeding. Coancestries are especially useful under systems of very close inbreeding where the number of inheritance pathways may be so great that identifying every one becomes too laborious. The Coefficient of Inbreeding of an individual (F) is equal to the Coancestry (f) between the two parents. In Figure 18:

F_X = f_PQ

Figure 18 Basic Formula

f_{PQ =} 1/4(f_AC + f_AD + f_BC + f_BD)

Supplementary Rules for Overlapping Generations

f_PC = 1/2(f_AC + f_BC)

f_PD = 1/2(f_AD + f_BD)

Also f_PQ = 1/2(f_PC + f_PD)

Rules for Calculating Coancestries of Early Generations

Assume that all coancestries of the first generation are 0. The coancestries of the next generation will depend on the relationship between the prospective parents:

1. Selfing (Monoecious plants and hermaphrodites)

f_AA = 1/2(1 + F_A)

If F_A is unknown or assumed to be 0 then f_AA reduces to 1/2. The coancestry f_AA is used again below in calculating coancestries between other closely-related parents.

2. Offspring and Parent

Since these are from different generations the supplementary rule is used:

f_PA = 1/2(f_AB + f_AA)

If A and B are not related and A is not inbred, then f_AB = 0 and f _AA = 1/2 . The coancestry f_PA then reduces to 1/4.

3. Full Sibs

The basic rule applies here, but the pedigree needs to be redrawn to find the formula.

Figure 19 Modified Diagram for Full Sibs

Since f_AB = f_BA then:

f_PQ = 1/4(2f_AB + f_AA + f_BB)

With no previous inbreeding or genetic relationship between A and B this reduces to f_PQ = 1/4.

d) Half Sibs

Redrafting the pedigree as for full sibs:

Figure 20 Modified Diagram for Half Sibs

Applying the basic rule to this diagram we have:

f_PQ = 1/4(f_AB + f_AC + f_BB + f_BC)

With no previous inbreeding or genetic relationship between A and B or between B and C this reduces to f_PQ = 1/8.

Worked Examples

With coancestries, unlike the previous path analyses, it is usual to start with the early generations and systematically work down through the pedigree until the parents of the chosen individual are reached. To be able to calculate the coancestries between actual parents it is also necessary to calculate the hypothetical coancestries between other members of each generation. Thus, a running tally is kept as you progress down through the generations. This avoids having to double back later to find missing values.

Two of the following four examples were used earlier to calculate coefficients of inbreeding by path analysis:

Example	Figure
1	21	First cousin marriages
2	22	Uncle - niece marriage
3	23	Double first cousin marriage
4	24	Close inbreeding (full sibs)

Example 1

Figure 21 First Cousin Marriages

Find F_Q

The coancestries of the members of the early generations are as follows:

f_AB = f_CD = f_CE = f_CF = f_DF = f_EF = 0. Also F_A = F_B = F_C = F_D = F_E = F_F = 0.

Therefore, f_DE (full sibs) = 1/4(2f_AB + f_AA + f_BB) and since

f_AA = 1/2(1 + F_A) = 0.5 and f_BB = 1/2(1 + F_B) = 0.5

Then f_DE = 1/4(0 + 0.5 + 0.5) = 0.25

f_GH = F_I = 1/4(f_CE + f_CF + f_DE + f_DF)

= 1/4(0 + 0 + 0.25 + 0)

= 0.0625

f_IJ = f_KL = f_KM = f_KN = f_LN = f_MN = 0

and F_J = 0

Therefore, f_LM (full sibs) = 1/4(2f_IJ + f_II + f_JJ)

f_II = 1/2(1 + F_I) = 0.53125 and f_JJ = 1/2(1 + F_J) = 0.5

Therefore f_LM = 1/4(0 + 0.53125 + 0.5) = 0.2578125

Finally, F_Q = f_OP = 1/4(f_KM + f_KN + f_LM + f_LN)

= 1/4(0 + 0 + 0.2578125 + 0)

= 0.064453125 = 6.45%

Example 2

Figure 22 Uncle - niece Marriage

Find F_G

f_AB = f_CE = f_DE = 0

Also F_A = F_B = F_C = F_D = F_E = F_F = 0

f_CD (full sibs) = 1/4(2f_AB + f_AA + f_BB)

Since f_AA = 1/2(1 + F_A) = 0.5 and f_BB = 1/2(1 + F_B) = 0.5

Then f_CD = 1/4(0 + 0.5 + 0.5) = 0.25

Therefore F_G = f_CF (overlapping generations) = 1/2(f_CD + f_CE)

= 1/2(0.25 + 0)

= 0.125 = 12.5%

Example 3

Sometimes it is necessary, as shown earlier, to present a pedigree in alternative forms to facilitate the calculation of coancestries. The following are two different ways of depicting the same pedigree:

Figure 23 Double First Cousin Marriage

Conventional Diagram

Modified Diagram

Find F_K

Using first diagram:

f_AB = f_AC = f_AD = f_BC = f_BD = f_CD = f_EG = f_EH = f_FG = f_FH = 0

Also F_A = F_B = F_C = F_D = F_E = F_F = F_G = F_H = 0

and f_AA = 1/2(1 + F_A) = f_BB = 1/2(1 + F_B) = f_CC = 1/2(1 + F_C) = f_DD = 1/2(1 + F_D) = 0.5

Then f_EF (full sibs) = 1/4(2f_AB + f_AA + f_BB)

= 1/4(0 + 0.5 + 0.5) = 0.25

and f_GH (full sibs) = 1/4(2f_CD + f_CC + f_DD)

= 1/4(0 + 0.5 + 0.5) = 0.25

Using second diagram

Therefore, F_K = f_IJ = 1/4(f_EF + f_EH + f_GF + f_GH)

= 1/4(0.25 + 0 + 0 + 0.25)

= 0.125 = 12.5%

Example 4

Figure 24 Close Inbreeding (full sibs)

Find F_I

f_AB = 0

F_A = F_B = F_C = F_D = 0

f_CD (full sibs) = 1/4(2f_AB + f_AA + f_BB)

f_AA = 1/2(1 + F_A) = 0.5 and f_BB = 1/2(1 + F_B) = 0.5

Therefore f_CD = 1/4(0 + 0.5 + 0.5) = 0.25 = F_E = F_F

f_EF (full sibs) = 1/4(2f_CD + f_CC + f_DD)

f_CC = 1/2(1 + F_C) = 0.5 and f_DD = 1/2(1 + F_D) = 0.5

Therefore f_EF = 1/4(0.5 + 0.5 + 0.5)

= 0.375 = F_G = F_H

f_GH (full sibs) = 1/4(2f_EF + f_EE + f_FF)

f_EE = 1/2(1 + F_E) = 0.625 and f_FF = 1/2(1 + F_F) = 0.625

Therefore, F_I = f_GH = 1/4(0.75 + 0.625 + 0.625)

= 0.5 = 50%