Doubled Haploids versus RIL's

 
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Doubled Haploids versus RIL's

Reference: Haldane, J.B.S., and C.H. Waddington. 1931. Inbreeding and Linkage. Genetics 16: 357-374.

Haldane and Waddington (1931) showed that the final proportion of crossover zygotes was
R =
   2r   
1 + 2r
where R is the final proportion of crossover zygotes and r is the recombination between two loci.

Example: Suppose the A and B loci are linked in coupling phase linkage with r = 0.2. Then the proportion of AAbb and aaBB RIL’s when the lines are fully inbred (n=) is

R =
   2(0.2)   
 = 0.286
1 + 2(0.2)

This means that there are:

0.357AABB + 0.143 aaBB + 0.143 A Abb + 0.357 aabb when r = 0.2 for RIL’s.

The proportion of each genotype in the F2 generation are determined from a Punnett Square using the gametic array:

{
1 - r
AB +
 r 
aB +
 r 
Ab +
1 - r
 ab } 2
2
2
2
2

For Doubled Haploid lines we can just use the gametic array. The gametic array is the same as the zygotic array because the gametic frequency is the zygotic frequency. Then the zygotic array for doubled haploids is

1 - r
AABB +
 r 
aaBB +
 r 
AAbb +
1 - r
aabb
2
2
2
2

Then the array of doubled haploid lines when r = 0.2 is:
0.4 AABB, 0.1 aaBB, 0.1 A Abb, and 0.4 aabb. When we compare the array of the RIL’s to that of the doubled haploid lines, we can see that the proportion of recombinant lines is greater for the RIL’s. The RIL’s have 14.3% of each recombinant class while the doubled haploid lines have 10% of each recombinant class. The RIL’s have more opportunity for recombination than the doubled haploid lines. The doubled haploid lines can only have observable recombination when the F1 plant undergoes meiosis. In the case of the RIL’s, there are AB/ab and Ab/aB double heterozygotes in each stage of the inbreeding process. In the case of RIL’s there is observable recombination in the F1, F2, F3, F4 etc. generations. The proportion of AB/ab plants in the F2 generation is
(1 - r)2
2
At each stage of inbreeding, the proportion of doubly heterozygous plants decreases, which decreases the amount of observable recombination.

The advantage to DH lines is that inbred lines can be rapidly developed for mapping experiments or for developing new cultivars. The disadvantage to using DH lines in a breeding program is that in the case of repulsion-phase linkage, there would be less opportunity to break up unfavorable linkage relationships. Use of DH lines in a breeding program would be superior to use of RIL’s if the genes were linked in coupling because coupling phase linkage relationships would be more likely to be retained for DH populations.

Linkage and Self-fertilization

Assume coupling phase linkage. The two parental lines have the genotypes AABB and aabb. Let r = observed recombination. The F1 is AB/ab. The gametes produced by the F1 are in the proportions

1 - r
AB:
 r 
aB:
 r 
Ab:
1 - r
ab
2
2
2
2

Haldane and Waddington write the gamete array as:

(1 - ß)AB: ß Ab: ß aB: (1 - ß) ab.

We will assume that the proportion of crossover events is the same in both sexes, then using Haldane and Waddington’s notation, ß = d. Haldane and Waddington give the proportions of each genotypic class as:

2Cn + 2Dn + 4En + Fn + Gn = 2.

We will use the usual convention that the sum of all genotypic frequencies equals unity, then

Cn + Dn + 2En + Fn/2 + Gn/2 = 1.

When we divided the genotypic array of 2Cn + 2Dn + 4En + Fn + Gn = 2 by a factor of 2, then we had to divide the gametic array of (1 - ß)AB: ß Ab: ß aB: (1 - ß) ab by a factor of 2 as well. We will now show that Haldane and Waddington’s ß = r. When the sum of the genotypic frequencies equals unity, then the gamete array is:

(1 - ß)
AB:
 ß 
Ab:
 ß 
aB:
(1 - ß)
ab =
1 - r
AB:
 r 
aB:
 r 
Ab:
1 - r
ab
2
2
2
2
2
2
2
2

From the above expression we can see that ß = r.

In the nth generation of self-fertilization, the 10 genotypic classes occur in the following proportions:

Table 1. Genotypic classes and genotypes produced in the nth generation of self-fertilization.

Class Genotypes
 Cn AABB and aabb
 Dn AAbb and aaBB
 En AABb, AaBB, Aabb, aaBb
 Fn AB/ab
 Gn Ab/aB

Haldane and Waddington (1931) state that 2Cn + 2Dn + 4En + Fn + Gn = 2. This expression can be written equivalently as Cn + Dn + 2En + Fn/2 + Gn/2 = 1. We can write the array as:

Cn AABB + Cn aabb + Dn AAbb + Dn aaBB + En AABb + En AaBB + En Aabb + En aaBb + Fn AB/ab + Gn Ab/aB = 2.

Or equivalently we can write:

Cn /2 AABB + Cn/2 aabb + Dn/2 AAbb +Dn/2 aaBB + En/2 AABb + En/2 AaBB + En/2 Aabb +En/2 aaBb + Fn/2 AB/ab + Gn/2 Ab/aB = 1.

Recombination can only be observed when crossovers occur for double heterozygotes. There are two types of double heterozygotes. Fn is the proportion of parental double heterozygotes in generation n. Gn is the proportion of double heterozygotes that are formed when recombination gametes Ab and aB unite.

In the F2 generation, we can derive the proportion of each genotype using a Punnett Square.

Table 2. Gametic and genotypic frequencies of two loci linked in coupling phase.

 
1-r
 AB
2
 r 
 aB
2
 r 
Ab
2
1-r
 ab
2
1-r
 AB
2
(1-r)2
 AABB
4
(1-r)r
 AaBB
4
(1-r)r
 AaBb
4
(1-r)2
 AB/ab
4
 r 
 aB
2
(1-r)r
 AaBB
4
 r2 
 aaBB
4
 r2 
 Ab/aB
4
(1-r)r
 aaBb
4
 r 
 Ab
2
(1-r)r
AABb
4
 r2 
 Ab/aB
4
 r2 
 AAbb
4
(1-r)r
 Aabb
4
1-r
 ab
2
(1-r)2
 AB/ab
4
(1-r)r
 aaBb
4
(1-r)r
 Aabb
4
(1-r)2
 aabb
4

Now
1 -r
AB +
 r 
aB +
 r 
Ab +
1 - r
ab = 1
2
2
2
2

This means that the sum of the four gametic frequencies equals 100%. To derive the results of the Punnett Square, which provides the proportions of each genotype, we use the algebraic equation:
{
(1 -r)
AB +
 r 
aB +
 r 
Ab +
(1 -r)
ab
}
2
 = 1
2
2
2
2

From the Punnett Square we can see that the expected proportions of the AABB class is
(1-r)2
4

Likewise, the expected proportion of the aabb class is
(1-r)2
4
in the F2 generation.

C2 is the proportion of the AABB and aabb genotypes. The following proportions in the F2 are derived from the Punnett Square. If we use the relationship that Cn + Dn + 2En + Fn/2+ Gn/2 = 1, then the proportion of the AABB class is (1 -r)2/4 and the proportion of the aabb class is (1 -r)2/4.

C2 =
(1-r)2
+
(1-r)2
=
(1-r)2
4
4
2
= the sum of the proportions of AABB and aabb classes.

D2 =
  r 2
+
  r 2
=
  r 2
4
4
2
= the sum of the proportions of AAbb and aaBB classes.

2E2 = 4
{
(1 -r)r
+
(1 -r)r
}
4
4
= the sum of the proportions of AABb, AaBB, Aabb, and aaBb classes.

And E2 = (1 - r)r.

 F 2
= 2
{
(1 -r)2
}
2
4
= the proportion of AB/ab and F2 = (1 - r)2.

 G 2
= 2
{
 r 2
}
=
 r 2
2
4
2
= the proportion of Ab/aB and G2 = r2.

We know that the sum of all the genotypic classes is 100% when we express the proportions as percent. When we express the proportions as fractions, the sum of all genotypic classes equals unity. In the F2 generation:

1 =
(1 -r)2
[1/2 AABB + ½ aabb] +
  r 2
[1/2AAbb + 1/2aaBB]
2
2

+ 2(1 -r)r [1/4 AABb + 1/4 AaBB + 1/4 Aabb + 1/4 aaBb]

+
(1 -r)2
[AB/ab] +
  r 2
[Ab/aB]
2
2

(1 -r)2
(2) +
  r 2
(2) +2r(1 -r)+
(1 -r)2
+
  r 2
4
4
2
2

= C2 + D2 + 2E2 +
  F 2
+
  G 2
 = 1
2
2

In the F3 generation: selfing AABB goes to 100% AABB; selfing aabb goes to 100% aabb for the C3. Selfing AAbb goes to 100% AAbb; selfing aaBB goes to 100% aaBB for the D3.
For the E3, selfing AABb goes to 1/4[AABB + 2AABb + AAbb];
              selfing AaBB goes to 1/4[AABB +2AaBB + aaBB};
              selfing Aabb goes to 1/4[AAbb + 2Aabb + aabb];
              selfing aaBb goes to 1/4[aaBB + 2aaBb +aabb}.

The array in the F3 generation for the AB/ab class is derived from a Punnett square:

Table 3. Gametic and genotypic frequencies for two loci linked in coupling phase.

 
1-r
AB
2
 r 
aB
2
 r 
Ab
2
1-r
ab
2
1-r
AB
2
(1-r)2
AABB
4
(1-r)r
AaBB
4
(1-r)r
AABb
4
(1-r)2
AB/ab
4
 r 
aB
2
(1-r)r
AaBB
4
  r2
aaBB
4
  r2
Ab/aB
4
(1-r)r
aaBb
4
 r 
Ab
2
(1-r)r
AABb
4
  r2
Ab/aB
4
  r2
AAbb
4
(1-r)r
Aabb
4
1-r
ab
2
(1-r)2
AB/ab
4
(1-r)r
aaBb
4
(1-r)r
Aabb
4
(1-r)2
aabb
4

The array in the G3 class is derived from selfing the Ab/aB class which results in the following Punnett square:



Table 4. Genotypic frequencies produced in the F3 generation for the G3 class.

 
1 - r
Ab
2
 r 
AB
2
 r 
ab
2
1 - r
aB
2
1 - r
Ab
2
(1 -r)2
AAbb
4
(1 -r)r
AABb
4
(1 -r)r
Aabb
4
(1 -r)2
Ab/aB
4
 r 
AB
2
(1 -r)r
AABb
4
  r 2
AABB
4
  r 2
AB/ab
4
(1 -r)r
AaBB
4
 r 
ab
2
(1 -r)r
Aabb
4
  r 2
AB/ab
4
  r 2
aabb
4
(1 -r)r
aaBb
4
1 - r
aB
2
(1 - r)2
Ab/aB
4
(1 - r)r
AaBB
4
(1 - r)r
aaBb
4
(1 - r)2
aaBB
4

The C3 class consists of AABB and aabb individuals. We derive the proportion of AABB and aabb individuals in the F3 generation from selfing the C2, E2, F2, and G2 classes. We can see that

C3 = C2 + 1/2E2 +
(1 -r)2
F2 +
  r 2
G2
4
4

The explanation for the above equation is that 100% of the AABB individuals in the F2 generation that were self-fertilized produced AABB offspring and this explains the C2 portion of the equation. The E2 class consists of AABb, AaBB, Aabb, and aaBb individuals and one-fourth of the AABb individuals when self-fertilized produce AABB progeny, also one-fourth of the AaBB individuals produce AABB progeny, so the sum of 1/4E2 plus 1/4E2 equals 1/2E2. If we consider the Punnet Square that shows the results of self-fertilizing AB/ab individuals, we can see that a proportion of (1 -r)2/4 of these progeny will be AABB in the F3 generation. If we look at the Punnet Square that shows the results of self-fertiliztion of Ab/aB individuals we can see that a proportion of r2/4 of these progeny will have the AABB genotype. Where C3 is the proportion of either AABB or aabb individuals and 2C3 is the proportion of AABB and aabb individuals summed together. We can determine D3 using the same reasoning that we used to derive C3:

D3 = D2 + ½E2 +
  r 2
F2 +
(1 -r)2
 G2
4
4

Now we can determine E3, F3, G3.

E3 = 1/2E2 +
(1 -r)r
F2 +
(1 -r)r
G2
2
2

F3
=
(1 -r)2
F2 +
  r 2
 G2
2
4
4

G3
=
(1 -r)2
G2 +
  r 2
F2
2
4
4


The same genetic relationships exist with each generation of self-fertilization, so we can generalize the above equations to relate genotypic frequencies in the parental generation to the genotypic frequencies of the progeny. We have the following equations and can use a computer spreadsheet, such as Excel, to derive the genotypic frequencies for any generation of self-fertilization.

Cn = Cn-1 + [1/2]En-1 + [(1 - r)2 /4]Fn-1 + [r2/4]Gn-1

Dn = Dn-1 + [1/2]En-1 + [r2/4]Fn-1 + [(1 - r)2/4]Gn-1

En = 1/2En-1n-1 +
[2r(1 -r)]
Fn-1 +
[2r(1 -r)]
 Gn-1
4
4

  Fn
=
(1 -r)2
Fn-1 +
  r 2
 Gn-1
2
4
4

  G n
=
(1 -r)2
Gn-1 +
  r 2
 Fn-1
2
4
4

The above equations were derived by Haldane and Waddington’s (1931). We can use these formulas to calculate the genotypic frequencies for each generation of inbreeding as the r value is varied (coupling phase linkage). If we want to find the genotypic frequencies for the case of repulsion phase linkage we can use values of r > 0.5. For example, when r = 0.6 in the above equations, this is equivalent to r = 0.1 for the case of repulsion phase linkage. We can see this because when r = 0.6 for coupling phase linkage, then 1 - r = 1 - 0.6 = 0.1 for repulsion phase linkage. What we have done is substitute 1 - r for r in the gamete frequencies.

For coupling phase linkage the gametic array in the F2 generation is:

(1 - r)
AB:
 r 
Ab:
 r 
aB:
(1 - r)
ab.
2
2
2
2

For repulsion phase linkage the gametic array in the F2 generation is:

 r 
AB:
(1 - r)
Ab:
(1 - r)
aB:
 r 
ab.
2
2
2
2


Table 5. Proportion of each genotypic class for two linked loci in the F2 generation as the intensity of linkage (r) is varied. (click here to see Table 5.)

Table 6. Proportion of each genotypic class for two linked loci in the F3 generation as the intensity of linkage (r) is varied. (click here to see Table 6.)

Table 7. Proportion of each genotypic class for two linked loci in the F4 generation as the intensity of linkage (r) is varied. (click here to see Table 7.)

Table 8. Proportion of each genotypic class for two linked loci in the F5 generation as the intensity of linkage (r) is varied. (click here to see Table 8.)

Table 9. Proportion of each genotypic class for two linked loci in the F6 generation as the intensity of linkage (r) is varied. (click here to see Table 9.)

Table 10. Proportion of each genotypic class for two linked loci in the F7 generation as the intensity of linkage (r) is varied. (click here to see Table 10.)

Table 11. Proportion of each genotypic class as the number of generations of self-fertilization approaches infinity and the intensity of linkage (r) is varied. (click here to see Table 11.)

Table 12. Proportions of each genotypic class for doubled haploid lines as the intensity of linkage (r) is varied. (click here to see Table 12.)

Proportion of Recombinant Gametes in Successive Generations of Self-fertilization

Assume coupling-phase linkage, then the proportion of recombinant gametes produced by self-fertilization of a doubly-heterozygous F1 is r/2 Ab + r/2 aB = r. We will define recombinant gametes as gametes with the haplotypes of Ab or aB. We can determine the proportion of these types of gametes produced by self-fertilization of F2 plants by determining the proportion of Ab or aB gametes produced by each F2 genotype, multiplied by the proportion of each F2 genotype and then summing across genotypes.

F2 plants with the AABB genotype will not produce any Ab or aB gametes. F2 plants with the AaBB genotype will produce the gamete array of 1/2AB and 1/2aB gametes. Any crossovers between the homologous chromosomes will not be apparent because this genotype is heterozygous at only one locus. Among AB/ab genotypes, a proportion of r will be recombinant types. The results for all ten F2 genotypes is summarized in the following table.

Table 13. Proportion of Ab or aB gametes produced in the F3 generation for two loci linked in coupling phase.
F2 Genotype
Proportion of F2 plants
Proportion of Ab or aB gametes per F2 genotype
Proportion of Ab or aB gametes in the F3 generation
AABB
(1 -r)2
4
0
0
AaBB
(1 -r)r
2
 1 
2
(1 -r)r
4
AABb
(1 -r)r
2
 1 
2
(1 -r)r
4
AB/ab
(1 -r)2
2
r
(1 -r)2r
2
aaBB
 r 2
4
1
 r 2
4
Ab/aB
 r 2
2
(1 -r)
r2(1 -r)
2
aaBb
(1 -r)r
2
 1 
2
(1 -r)r
4
AAbb
 r 2
4
1
r2
4
Aabb
(1 -r)r
2
 1 
2
(1 -r)r
4
aabb
(1 -r)2
4
0
0

Now we can sum the last column to determine the proportion of recombinant gametes produced in the F3 generation as a result of self-fertilization of F2 plants.

R2 =
(1 -r)r
 +
(1 -r)r
+
(1 -r)r
 +
(1 -r)r
 +
r2
 +
r2
 +
(1 -r)2r
 +
r2(1 -r)
4
4
4
4
4
4
2
2
=
3r - 2r2
2

Where R2 is the proportion of recombinant gametes produced from self-fertilizing F2 plants. We can check the formula by putting r =0.5 and the result is that R2 = r = 0.5. This result shows that when two loci are independent, the proportion of recombinant gametes stays constant with advanced generations of self-fertilization.

Let R3 equal the cumulative proportion of Ab and aB gametes produced by self-fertilization of F3 plants. The development of the formula for R3 is complicated because we must determine the proportion of Ab or aB gametes produced from F3 plants multiplied by the proportion of each F3 genotype. Only double heterozygous plants produce observable recombinant gametes. Gametes that result from crossovers of AB/ab plants increase the proportion of recombinant gametes, but gametes that result from crossovers of Ab/aB plants decrease the proportion of recombinant gametes.

R3 =
3r - 2r2
+
(1 -r)4r
+
  r 5
-
(1 -r)3r2
-
(1 -r)3r2
2
4
4
4
4
=
3r - 2r2
+
(1 -r)4r
+
  5 5
-
(1 -r)2r3
2
4
4
2

The
3r - 2r2
 portion of the above formula is the proportion of
2
recombinant gametes that were previously developed in the F2
generation.
(1 -r)4
 is the proportion of AB/ab plants in the
4
F3 generation which resulted from AB/ab plants in the F2 that were recreated in the F3 generation. These plants contribute a proportion of r recombinant gametes to the F4 population, so
the result is
(1 -r)4r
4

The
  r 5
 portion of the R3 formula is the proportion of AB/ab F3
4
plants that were the result of Ab/aB F2 plants that produced AB or ab gametes to form AB/ab F3 plants. These AB/ab F3 plants produced a proportion of r Ab or aB recombinant gametes. We have r X r4/4 = r5/4. Lastly we lose a portion of recombinant gametes when Ab/aB F3 plants produce parental gametes which are AB or ab. When AB/ab F2 plants produce Ab/aB progeny and these F3 plants recombine to produce AB or ab gametes, this is a loss of recombinant gametes. Another loss of recombinant gametes is when Ab/aB F2 plants are self-fertilized to produce Ab/aB F3 progeny and then produce AB/ab gametes, this is a loss of recombinant gametes. We can summarize these results in a table.

Table 14. Proportion of Ab and aB gametes produced in the F4 generation for two loci linked in coupling phase.

F2 genotype
F3 genotype
F3 proportion
Gametic proportions
from F3 plants
AB/ab
AB/ab
(1 -r)2/2
r
Ab +
r
aB +
(1 - r)
AB +
(1 - r)
ab
2
2
2
2
Ab/aB
AB/ab
r2/2
r
Ab +
r
aB +
(1 - r)
AB +
(1 - r)
ab
2
2
2
2
AB/ab
Ab/aB
r2/2
r
AB +
r
ab +
(1 - r)
Ab +
(1 - r)
aB
2
2
2
2
Ab/aB
Ab/aB
(1 -r)2/2
r AB + r ab + (1 - r) Ab + (1 - r) aB
2 2 2 2


Using the formulas:
R2 =
3r - 2r2
; R3 =
3r - 2r2
+
(1 -r)2r
+
 r5
-
(1 -r)2r2
;and R =
  2r  
2
2
4
4
2
1 + 2r

we can determine the proportion of Ab and aB gametes produced by the F1, F2, F3 and completely homozygous RIL generations.

Haldane and Waddington (1931) show that for completely inbred lines:
C
=
   1   
and D
=
  2r  
 
1 + 2r
1 + 2r

We can see that C + D = 1

Another way of saying this is that when lines are completely inbred, there are no heterozygous loci for either the A or B locus. The only genotypes when n = 8 are the AABB, aabb, AAbb, or aaBB types. The proportion of AABB and aabb genotypes equals 1/(1 + 2r). The proportion of AAbb and aaBB genotypes is 2r/(1 + 2r).

Table 15. Cumulative proportion of Ab and aB gametes produced by F1, F2, F3, and completely homozygous inbred plants are r, R2, R3, and R, respectively.

r R2 R3 R
____________________
0.1 0.14 0.16 0.17
0.2 0.26 0.28 0.29
0.3 0.36 0.37 0.38
0.4 0.44 0.44 0.44
0.5 0.5 0.5 0.5
____________________

We can see that the proportion of Ab and aB gametes produced by F2 plants (R2) is greater than the proportion of recombinant gametes produced by F1 plants (r), but there is little additional increase in the proportion of recombinant gametes after the F2 generation. This result is explained by the rapid decrease in doubly heterozygous genotypes after the F2 generation. Observable recombination can only occur in doubly heterozygous genotypes.

The results have implications for applied breeders and biotechnologists. Applied breeders should recognize that progress in breaking up linkage relationships between favorable and unfavorable genes will not be greatly enhanced by developing lines that are more inbred than F3- derived lines. In breeding beyond the F3 generation is an advantage to breeders not because deleterious linkage relationships will be broken, but because the additive genetic variance among lines will continue to increase with continued self-fertilization.

Biotechnologists should recognize that when using co-dominant markers to identify linkage relationships between markers, there is little advantage to developing lines that are more inbred than F3-derived lines. For a given number of lines, the probability of observing recombinant genotypes does not improve very much as lines are inbred beyond the F3.

When dominant markers are used, the advantage to developing F6-derived lines rather than F3-derived lines, is that the increased inbreeding reduces the proportion of heterozygous loci. Heterozygous loci provide little information about linkage relationships when markers at both loci are completely dominant.

Copyright 2000©, Ted Helms
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