web version
Dag Lindgren 2001-08-20, small edit 04-04-29
Coancestry, f, is a quantification of relatedness between individuals, the coancestry of mates becomes the inbreeding, F, of the progeny. Group coancestry, GC, is the average of coancestry values among a group of individuals, and useful for monitoring the degree of relatedness in a group. Other ways of expressing coancestry is Gene Diversity, GD, loss of Gene Diversity (=group coancestry) and Status Number, Ns, which is useful as an effective number. GD=1-GC; Ns=0.5/GC.
All problems are not solved, they stand as challenges for those who want to check that they understood. You can solve them and send the problem and solution to Dag Lindgren for check.
Relevant reading is available on /Derivations_of_coancestry_ based_on_pedigree.doc and Status_Number_reveiw_97.html. and a minicourse which is found on TBT under tutorials.
It is not as trivial as it looks like to get the coancestry values (see Lindgren et al 1996 for a general algorithm).
(Group coancestry = average coancestry = average kinship=average of elements in coancestry matrix=f)
It is difficult to get non-Latin (=non-American) characters, which reproduce well on all screens. Therefore the Latin characters GC is used for Group Coancestry, although it is preferred to use the Greek letter capital theta Θ, and f is used for pair-wise coancestry in spite of that the Greek letter small θ may be preferred when it works. fPQ means coancestry between P and Q.
Among the example files where is also a document with an extract of some of these examples.
Exercises 1.1-1.5
Calculate the group coancestry for the following situations:
1.1 Two unrelated non-inbred trees.
1.2 Two non-inbred trees, which are members of the same full sib family (f = coancestry = 0.25).
1.3 Two unrelated trees which origin from selfing (F = 0.5 and f = 0.5 (1+F)=0.75).
1.4 Two very large unrelated families of equal size with the same mother but two different fathers.
1.5 A, B and C are unrelated and not inbred. Three crosses are made: A*B, B*C and A*C (single-pair mating). The best individual in each full sib family is selected. Calculate group coancestry for a population consisting of these three selected individuals.
1.6 As 1.5, A, B and C are selected plus trees from a wild population, what is the loss of Gene Diversity?
Solutions for 1.1-1.31.1-3. Make coancestry matrixes for the two individuals (tree A and B)
Example 1.1 |
||
| A | B | |
| A | 0.5 | 0 |
| B | 0 | 0.5 |
GC = Group coancestry = (0.5+0.5)/4=0.25
Example 1.2 |
||
| A | B | |
| A | 0.5 | 0.25 |
| B | 0.25 | 0.5 |
GC = 1.5/4=0.375
Example 1.3 |
||
| A | B | |
| A | 0.75 | 0 |
| B | 0 | 0.75 |
GC = 1.5/4=0.375
Solution 1.4: There are n individuals. 0.5n individuals with each father. The n x n coancestry matrix will consist of n self-coancestries (coancestry=0.5); 0.5n2 - n full sibs (coancestry = 0.25) and 0.5n2 half sibs (coancestry = 0.125) In total [0.5n+0.25(0.5n2-n)+0.125*0.5n2]/n2=0.25/n+0.1875, if n is large the first term will vanish.
Solution 1.5: A row in the coancestry matrix will
be
0.5; 0.125; 0.125 ; average 0.25. As all rows has the
same elements the average will be the same for all (GC=0.25).
Solution 1.6: The loss of Gene Diversity compared to the wild population is the same thing as Group Coancestry, thus GC=0.25 (25 %).
Exercises 2.1-2.5
Calculate the status number for the following situations:
2. 1 Two non-inbred trees which are mother and child
2.2 Two unrelated trees of which one is non-inbred and the other origins from selfing
2.3 Two trees, one (C) is the selfed progeny from tree A, the other (D) is the progeny from the cross between A and B. A and B are non inbred and not related.
2.4 A population consisting of one tree from the cross A*B, two trees from the cross A*C and one tree which originates from selfing of C
2.5 A population consisting of one tree as well as its wind-pollinated progeny of size
n. It is assumed that the progeny are true half sibs, thus no full sibs; no selfing and no relatedness between the pollen and the mother or the individual pollen grains. Solutions for 2.1-2.5Example 2.1 |
||
| mother | Child | |
| Mother | 0.5 | 0.25 |
| Child | 0.25 | 0.5 |
GC =1.5/4=0.375; Ns=0.5/0.375=1.333
Example 2.2 |
||
| non-inbred | Selfed | |
| non-inbred | 0.5 | 0 |
| Selfed | 0 | 0.75 |
GC =1.25/4=0.3125; Ns=1.6
Example 2.3 |
||
| C | D | |
| C | 0.75 | 0.25 |
| D | 0.25 | 0.5 |
What is the coancestry between C and D? Let us pick one gene from C and one from D and compare them. The gene from D can be from A (50%chance) or B (50% chance). The gene from C must be from A, so if the gene from D is from B, the genes must be different. Even if both genes compared comes from A they may be different. Let's call them A1 and A2. It is equal chance for gene A1 and A2 to be transmitted to D. A single arbitrary gene from C may equally well be A1 or A2. Thus, given that both genes are from A, they may equally well be A1 or A2, thus the probability for identity is 0.25. Thus C and D are not more related than ordinary full sibs
GC =/4=; Ns=
Solution 2.4
| AB | AC1 | AC2 | Cself | |
| AB | 0.5 | 0.125 | 0.125 | 0 |
| AC1 | 0.5 | 0.25 | 0.25 | |
| AC2 | 0.5 | 0.25 | ||
| Cself | 0.75 |
GC = 0.2656; Ns=1.88
Solution 2.5. The coancestry between mother and progeny is 0.25 and between the half-sibs 0.125. Make a coancestry matrix with size n+1 and calculate the average for all elements.
GC = (0.125n2 + 0.875n + 0.5)/(n+1)2
Exercises 3.1-3.2
3.1 Develop formula to calculate the group coancestry and status number as a function of family size for a full sib family where the parents are unrelated, one parent is from a selfed progeny and the other parent has unrelated parents, start with a grandparents-grandchild crossing figure.
Solution 3.1.
A C D
|| \ /
P Q
Parent P has F=0.5 and parent Q has F=0; fPQ=0.
Consider self-coancestry, which is 0.5 as parents unrelated (fPQ=0), there is no inbreeding in progeny.
Consider the coancestry between full sibs. Genes may be shared as both come from Q, the chance that both come from Q is 0.25 (0.5 in one individual and 0.5 in another = 0.5 x 0.5 = 0.25) and if they do, the chance that it is the same gene is 0.5. Genes may be shared as both come from P, the chance that both come from P is 0.25 and if they do, the chance that it is the same gene is 0.75. Thus the pair-coancestry between sibs will be 0.3125. If a coancestry matrix for the sibs will be drawn it will have 0.5 at the diagonals and 0.3125 elsewhere.
Consider a full sib family of size n
GC = [0.5n + n(n-1)0.25(0.75+0.5)]/n2 =
GC = 0.1875/n + 0.3125
3.2 Develop formula to calculate the group coancestry as a function of family size for a full sib family where parent P has inbreeding of FP=0.1 and parent Q has inbreeding of FQ=0.2 and the coancestry between P and Q is fPQ=0.1. Note that coancestry between parents becomes inbreeding in the progeny.
Example 3.2. SolutionNote that self coancestry of an individual itself is fPP = 0.5(1+FP). GC = [n(0.5+fPQ) + n(n-1)(0.25fPP + 0.5 fPQ + 0.25 fQQ)]/n2 =
= 0.2125/n+0.3375
3.3 Consider the wind-pollinated progeny of a tree. It has size n. The progeny consists of a mixture of selfed seeds and out-crossed seeds. The number of selfed individuals in the progeny is s. The out-crossed seeds are assumed to be half-sibs, but the wind-borne pollens is related with itself with coancestry f=0.01, but not related to the mother.
Example 3.3. Solution.
There are two types of individuals in the wind-pollinated progeny, s selfing full sibs and (n- s) half-sibs.
Cross-coancestry
We can construct a matrix for the coancestry between different members in wind-pollinated progeny:
| Cross-coancestry | s | n- s |
| s | 0.5 | 0.25 |
| n- s | 0.25 | 0.25*0.5 + 0.25*0.01 = 0.1275 |
Take a gene from the outcrossed seed. There is one chance in two that it will be the father gene, if so coancestry is 0. If it is the mother gene it is one chance in two that it will be the same gene as in the selfed seed (as there are two alternative genes in its parent).
The coancestry will be 0.25*0.5 + 0.25*0.01 = 0.1275 for "half-sibs". To realise the later formula, regard that the genes considered might both come from the seed parent or both come from the pollen parent, one chance in 4 for both cases. If both come from pollen parents the coancestry is 0.01 (this implies no full sibs).
Self-coancestry
Self-coancestry will be 0.5(1+F) = 0.75 for selfing full sibs
.The out-crossed half-sibs" will not be inbred, thus their self-coancestry will be 0.5.
Evidently there is no self-coancestry involved when we compare different sib types.
Total contributions to group coancestry
Altogether where are n seeds, which can form relations in n*n ways
Contribution to group coancestry from selfed seeds: S(0.75+( S-1)0.5)
Therefore the group coancestry will be:
GC= [
S(0.75+( S-1)0.5)+ (n-s)(0.5+(n-s-1)*0.1275)+=[0.75s+0.5s2-0.5s+0.5n-0.5s+0.1275n2+0.1275s2-0.255ns-0.1275n+0.1275s+0.5ns-0.5s2]/ n2
=[-0.1225s +0.3725n+0.245ns +0.1275n2+0.1275s2] / n2
(Suggestion; check the results for the extremes n=1 and s=0 or 1)
Exercise
44 A seed orchard has 10 equally fertile unrelated and not inbred clones. The clones are not self-fertile. There is no contaminating pollen. Calculate the expected group coancestry or status number for the following cases:
4.1A. The crop is harvested in a sack and a large number of seeds is taken, consider the seeds in the sack.
Example 4.1A. Solution:
There are 20 founder genes in the sack, equally frequent. If two genes are taken from the sack, the chance that the second will be equal to the first is evidently 0.05.
4.1.B. Same as 4.1.A. What is the Gene Diversity of the seeds in the sack?
Example 4.1B. Solution:
Gene Diversity = 1-0.05 =0.95.
4.1.C. Same as 4.1.A but only five of the ten clones contribute pollen (thus each of these five gives 20% of the pollinations). What is the group coancestry? You may assume the crop is infinite.
Example 4.1C. Solution
5 clones contribute each 15% of the genes and 5 clones each 5% of the genes. The chance that both genes picked from the sack come from the same clone is 5*0.15*0.15 + 5*0.05*0.05=0.125. The chance that it is the same gene is half of that, thus 0.0625.
4.2. Ten seeds taken at random from the sack are considered.
Example 4.2. Solution.
This means the pooling of 20 genes into a gene pool. Two are taken from the gene pool and considered. The two genes may come from the same of the 10 seeds. The chance for that is 1/10. When the genes may be copy of the same gene (probability 1/2), if so they are identical, or that may be from different homologous genes (probability 1/2). As the clones were self-sterile, such genes cannot be identical. The two genes may come from different seeds, the chance for that is 9/10. The probability that genes in different seeds are identical by descent must be 1/20 as there are 20 equally probable alternatives.
GC = 0.1*0.5+0.9/20 = 0.095Ns = 5.3
4.3. One seed is taken from each of the clones.
Example 4.3. Solution.
Let's fill a coancestry matrix for the 100 possible relationships between the ten seeds. Each relationship is an element in a square matrix. As all clones have identical characteristics we do not need to deal with individual characteristics of clones or seeds.
The self-coancestry express the seeds relationship with itself. The seeds are not inbred (no selfing), thus self-coancestry is 0.5 and there are 10 elements such elements, in 90 relations the seeds will be different. The individual relationships are not known, and the male contributions depend partly on chance, thus we will predict a likely value, but it will differ between different samples of ten seeds.
Let us now consider how different seeds are related. They may be halfsibs or unrelated, but can not be full sibs as they can not share mother. We separate three cases: That both compared genes in the different seeds come from the mother, both from the father, or one from the mother and one from the father.
1. Both come from the mother, the probability of that is 1/4, and the probability that they are IBD is 0, because all seeds have different mothers.
2. Both come from the father, the probability is 1/4. Let us compare the paternal gametes in the seeds 1 and 2 from mother 1 and 2. The father of seed 1 could be father 2-10 and the father of seed 2 could be 1 or 3-10. They can share father if it is 3-10, the chance that a father is 3-10 is 8/9 and that both do it is (8/9) 2 = . If they share father they are half-sibs, thus coancestry 0.125.
3. One from the mother, the other from the father, the probability is 0.5. Let us consider seed 1 and 2. If the mother is clone 1, the father to seed 2 could be clone 1 with probability 1/9 and the seeds become half-sibs with coancestry 0.125. It could equally well be the mother of clone 2, thus the probability should be multiplied by 2.
To sum up, the group ancestry will be:
GC=[0.5*10+90*(0+0.25*(8/9) 2 *0.125 + 0.5*2*0.125/9 )]/100
=(5+90*(0.0247 + 0.0139 ))/100=0.0847
Ns=1/(2GC)=5.9
2. Both come from the father, the probability is 1/4. Let us compare the paternal gametes in the seeds 1 and 2 from mother 1 and 2. The father of seed 1 could be father 2-10 and the father of seed 2 could be 1 or 3-10. They can share father if it is 3-10, the chance that a father is 3-10 is 8/9 and that both do it is (8/9) 2 = . If they share father they are half-sibs, thus coancestry 0.125.
3. One from the mother, the other from the father, the probability is 0.5. Let us consider seed 1 and 2. If the mother is clone 1, the father to seed 2 could be clone 1 with probability 1/9 and the seeds become half-sibs with coancestry 0.125. It could equally well be the mother of clone 2, thus the probability should be multiplied by 2.
To sum up, the group ancestry will be:
GC=[0.5*10+90*(0+0.25*(8/9) 2 *0.125 + 0.5*2*0.125/9 )]/100
=(5+90*(0.0247 + 0.0139 ))/100=0.0847
Ns=1/(2GC)=5.9
4.4 A large wind pollinated family is collected from one of the clones. These seeds can be considered a mixture of seeds with fathers outside the seed orchard (thus true half-sibs assuming all pollen parents outside the seed orchard are different and unrelated) and fathers inside the seed orchard (forming nine different full sib families). The maternal clone will not contribute paternally as there was no selfing.
Solution 4.4. This example will be moved to 5.4 when solved and here a simpler version without contamination will be given to conform to the structure!
4.5. The clones are crossed by single pair mating and 5 set of full-sib seeds are produced. An equal but large number of seeds is taken from each cross.
Solution 4.5. A large full sib family has status number 2, so five unrelated large full sibs of equal size has status number 10.olution 4.5. A large full sib family has status number 2, so five unrelated large full sibs of equal size has status number 10. A large full sib family has status number 2, so five unrelated large full sibs of equal size has status number 10.4.6A.
n seeds are taken from each of N unrelated full sib families (thus N*n seeds in total).Solution 4.6A:
GC= [ (0.5N*n)+0.25*n(n-1)*N] /N²n²
= 0.25 (n+1)/N*n
Ns=2N*n /(n+1)
Example 4.6B.: Two seeds are taken from each cross (thus 10 seeds in total).
Solution 4.6B:
Sa = seed from cross a etc
Sa1 |
Sa2 |
Sb1 |
Sb2 |
Sc1 |
Sc2 |
Sd1 |
Sd2 |
Se1 |
Se2 |
|
Sa1 |
0.5 |
0.25 |
||||||||
Sa2 |
0.25 |
0.5 |
||||||||
Sb1 |
0 |
0 |
0.5 |
0.25 |
||||||
Sb2 |
0 |
0 |
0.25 |
0.5 |
||||||
Sc1 |
0 |
0 |
0 |
0 |
0.5 |
0.25 |
||||
Sc2 |
0 |
0 |
0 |
0 |
0.25 |
0.5 |
||||
Sd1 |
0 |
0 |
0 |
0 |
0 |
0 |
0.5 |
0.25 |
||
Sd2 |
0 |
0 |
0 |
0 |
0 |
0 |
0.25 |
0.5 |
||
Se1 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0.5 |
0.25 |
Se2 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0 |
0.25 |
0.5 |
GC = (10*0.25+10*0.5)/100=0.075
Ns = 0.5/0.075 = 6.7
4.7 The clones vary in reproductive success as a different fraction of seeds are harvested from them. The clones are equally fertile on the paternal side, thus equally successful as pollen parents. There is no pollen contamination.
4.7A The ten clones are seed parents to resp. 0, 2, 4, 6, 8, 10, 12, 14, 16 and 28 percent of a large number of seeds (by the way, is this formulation realistic?). For simplicity we allow the reproductive success to be the same for all 10 pollen parents.
Solution 4.7A
Contribution as male is 10% for all clones (thus 5% of the genes).
Contribution as a female, for example for the second
clone, is 2%.
Average contribution as male and female for the second
clone is (0.1+0.02)/2 = 0.06.
The square sum of average contribution from all clones is
[(0.1+0)/2]^2+[(0.1+0.02)/2]^2+...+[(0.1+0.28)/2]^2
=sum(pi^2) = 0.115
If the parents are unrelated and non-inbred, GC and Ns will be
GC = 0.5*sum(pi^2) = 0.5*0.115 = 0.0575
Ns = 0.5/GC = 0.5/0.0575 = 8.7
A weakness with the formulation is that it assumes selfing has
the same share in the reproductively successful pollinations as out-crossing.
4.7B The same as 4.7A, but assume selfing does not occur. The clones are equally successful as pollen parents on other clones, but as the absence of selfing matters, reproductive success will not be equal for all pollen parents .
4.7C Kang (2001) uses the concept "sibling coefficient", what would that be in 4.7A.
Solution 4.7C
Sibling coefficient=1.15
4.8 The ten clones vary in fertility (expected number of offspring), but besides that it is a random (stochastic) variation in number of seeds and success of pollen gametes. The variation in fertility can be described by coefficient of variation in number of offspring per clone = 1. Give the group coancestry as a function of the number of seeds harvested from the seed orchard.
Solution 4.8 to be made
Exercise 5
The complication of pollen contamination is considered. There are 10 clones in the orchard, but some of the pollen parents are found outside the seed orchard. Thus the clones have a lower paternal than maternal fertility. All pollen parents outside the seed orchard are different and unrelated
5.1 The 10 clones in the orchard are equally fertile as seed parents and as pollen parents, but 50 % of the pollen origins from sources outside the seed orchard, thus the clones have a lower paternal than maternal fertility.
5.1.A. What is the group coancestry and status number of a large crop?;
5.1.B. What is group coancestry if 10 seeds are taken at random from the large crop?
Solution 5.1A: Only if none of two genes taken comes from the contaminating pollen there can be coancestry, the chance for that is 0.75*0.75. If both genes comes from the orchard genotypes the probability they are identical is 0.05 (compare 4.1). Thus the reply is 0.75*0.75*0.05=0.0281 and the corresponding status number is 17.8.
Solution 5.1B: This seems to be a classic application of the 0.5/N factor for "increment of inbreeding" (see Falconer!). What Falconer and other call increment attributable to new inbreeding is actually better viewed as increment of group coancestry! Thus if we take 10 seeds at random from a large sac of seeds with GC=0.0281, the group coancestry will be GC=0.05 + 0.95*0.0281 = 0.0767.5.2 Same as 5.1 but only five of the ten clones contribute pollen (thus each of these five gives 10% of the pollinations). What is the status number? You may assume the crop is infinite.
Solution 5.2
Use of the seed orchard manager work-sheet or the formula in Kang et al (submitted): Insert A=1 as there is no fertility variation. Pollen contamination is 50%, thus M=0.25. There are 5 pollen parents (Nf=5), 10 seed parents (Nm=10), and 5 shared seed and pollen parent (Nfm=5). Then, GC is 0.03125 and Ns is 16.0.
Alternatively the problem can be treated like this: There are three sources to the gene pool of the seeds, 5 clones which contributes 10% each; 5 clones which contribute 5% and contaminating pollen which contributes 25%. GC = 0.5[5*0.12 + 5*0.052] + 0*0.25 = 0.03125
5.4 A large wind pollinated family is collected from one of the clones. These seeds can be considered a mixture of seeds with fathers outside the seed orchard (thus true half-sibs assuming all pollen parents outside the seed orchard are different and unrelated) and fathers inside the seed orchard (forming nine different full sib families). The maternal clone will not contribute paternally as there was no selfing.
see 4.4 Solution 5.4, used Excel worksheet seed orchard manager (available at TBT).5.5 There are three clones in a seed orchard: A, B and C. A large wind pollinated family is collected from clone A. The seed parent is thus only clone A. Pollen parents are 10% clone A (thus selfing), 20% clone B (thus a full sib family), 30% clone C (another full sib) and 40% pollen migrants. Clone A and B are unrelated and noninbred. All contaminating pollen with an origin outside the seed orchard are unrelated. Calculate group coancestry of the seed crop harvested from clone A.
Solution 5.5 to be made
Exercise 6
Seeds are harvested from a stand. Initially all trees in the stand are unrelated and not inbred. All pollen is from the stand. A new stand is established with the seeds collected. Later seeds are harvested from that. This goes on to the 10th generation. What will the group coancestry, status number and inbreeding in the stand be?
6.1 Assuming complete random mating. Assume there are 100 trees in the stand in each generation (breeding population is 100). All trees are equally fertile.
Example 6.1. Solution. Initially (in the first generation stand) group coancestry is GC=0.005. The increment of group coancestry in each generation is 0.005 (=1/2N). As this is rather small we can approximate the increase over 9 cycles to generation 10 to 9*0.005=0.045. Thus in generation 10 GC=0.05; Ns=10; F=0.045.
6.2 All trees are equally fertile. We intentionally plant one tree from each mother tree, but the contributions from pollen parents are at random. It may be solved including or excluding selfing.
Example 6.2. Solution to be made.
6.3 Trees vary in fertility (A=2) and there are besides that random variations in the number of successful gametes. We replace the 100 trees in each generation.
Solution Example 6.3. Solution to be made.
6.4: The clones are crossed by single pair mating and 5 sets of full-sib seeds are produced. All parents are non-inbred. Fathers 1 and 2 are halfsibs; Fathers 3 and 4 are fullsibs, else where are no relationships between the parents. An equal but large number of seeds is taken from each cross. Write the coancestry matrix between families!!! Calculate group coancestry!!!
Solution Example 6.4. Solution to be made.
Exercise
77 A plantation consists of 10 half sib families, each represented by 3 members. All trees spread equal amounts of pollen (no selfing). The best phenotype in the plantation is harvested for seeds (assume it is a large number of seeds harvested). What is the group coancestry in this sib family?
Example 7 Solution to be made.
Exercise 8
8. Use of COADEMO (available on TBT-web site)
You have the following genotypes
Genotype |
Breeding value |
F |
Relatedness to the other |
A |
1 |
0.25 |
Full sib to B (father and mother in common with B)1 |
B |
2 |
0.25 |
Full sib to A ( " A) |
C |
3 |
0 |
Half sib to D (same father as D but different mother) |
D |
4 |
0.5 |
Half sib to C (same father as D but different mother)2 |
E |
5 |
0.5 |
Unrelated |
Write the coancestry matrix
Give the penalty constant at which the set of three genotypes that maximises population merit changes
Solution 8.
Yongqi Zheng says it changes at c=12. But, now when I read this, it strikes me that this example is not well specified, relevant information is missing, this is a common phenomenon in science. Make the simplest possible specifications before you solve it.
Exercise 9
9 A*B and C*D are two
unrelated full sib families. A, B, C and D are not inbred.
9.1 What is the group
coancestry for a population consisting of NA*B
members of family A*B and NC*D members of family C*D.
9.2 Let NA*B
=2 and NC*D=3 (thus two members from the first family and three from the
second).
| 9.2 Let NA*B =2 and NC*D=3 (thus two members from the first family and three from the second). The matrix shows the coancestry relations between the trees | ||||||
| 1 | 2 | 3 | 4 | 5 | ||
| 1 | 0,5 | 0,25 | 0 | 0 | 0 | |
| 2 | 0,25 | 0,5 | 0 | 0 | 0 | |
| 3 | 0 | 0 | 0,5 | 0,25 | 0,25 | |
| 4 | 0 | 0 | 0,25 | 0,5 | 0,25 | |
| 5 | 0 | 0 | 0,25 | 0,25 | 0,5 | 0,18 |
9.2 Let NA*B
=2 and NC*D=3 (thus two members from the first family and three from the
second).
Exercise 10
10. Trees from two families are put into a seed orchard. The fertility (flowering, size; ) can vary between families, it is double as high for the first family as the second. A large number of seeds are collected from the orchard.
10.1 to 10.2 are the same as 9.1 and 9.2 added that the fertility of the trees are the same and the group coancestry of the seed orchard crop is asked for.
Solution 10.1 and 10.2 are the same as 9.1 and 9.2.
10.3.There
are 5 trees in the orchard, 2 from one family and 3 from another, thus NA*B
=2 and NC*D=3. The
fertility (flowering, size; the same fertility on male and female side) varies between families, it is double as high for
the first family as the second.
0.6667^2*0.375 + 0.333^2*0.333=0.2??
10.4 There are 5 trees in the orchard, 2 from one family and 3 from another, related, family (half sib), NA*B =2 and NC*B=3. The fertility (flowering, size; the same fertility on male and female side) varies between families, it is double as high for the first family as the second.
0.6667^2*0.375 + 0.333^2*0.333 + 2*0.667*0.333*0.125 = 0.2??
10.5 There are 5 trees in the orchard, 2 from one family and 3 from another, related, family (half sib), NA*B =2 and NC*B=3. The fertility (flowering, size) varies between families and sexes, the female fertility is double as high for the first family as the second, while the male fertility is four times higher.
Solution
10.5: 0.6667*0.8*.375
+0.333*0.2*0.333+(0.667*0.2+0.333*0.8)*0.125 = 0.272
10.6. Similar to 10.5, but let the male and female fertility of each tree be known, but different. The following values apply:
| Tree / gender | female | male |
Complementation and solution to be made...
Exercise 11
There are four clones in a seed orchard (1,2,3 and 4). Clone 1 and 2 are full sibs, otherwise where are no relationships. Their fertility (contribution of successful gametes go the crop) is 0.5; 0.2; 0.2 and 0.1. Calculate the effective number of the seed orchard clones as well as the expected seed crop.
Solution 11: Effective number of clones is: 3,2 and effective number of seed crop is: 2.27 ; look for details in the EXCEL spreadsheet SEEDORCHARDCOA.
Exercise 12
Multigenerational. Repeated full sib mating, for (cf Lindgren et al 199?).
Base
population
|
|
A |
B |
|
A |
0.5 |
0 |
|
B |
0 |
0.5 |
Solution Example 12. Solution to be made.
Tips and comments
The Tree Breeding Tool (TBT) web site is useful. It is accessible from
http://www.genfys.slu.se/staff/dagl/Breed_Home_Page/
All problems are not solved, they stand as challenges for those who want to check that they understood. You can solve them and send the problem and solution to Dag Lindgren for check.
Relevant reading is available on /Derivations_of_coancestry_ based_on_pedigree.doc and Status_Number_reveiw_97.html. and a minicourse which is found on TBT under tutorials.
It is not as trivial as it looks like to get the coancestry values (see Lindgren et al 1996 for a general algorithm).
(Group coancestry = average coancestry = average kinship=average of elements in coancestry matrix=f)It is difficult to get non-Latin (=non-American) characters, which reproduce well on all screens. Therefore the Latin characters GC is used for Group Coancestry, although it is preferred to use the Greek letter Q , and f is used for pair-wise coancestry in spite of that the Greek letter q may be preferred when it works. fPQ means coancestry between P and Q.