Quantitative Genetics

1
Quantitative Genetics Plant Breeding Populations
are created by breeders to serve as sources of
new cultivars. These populations are describable
by their genotypic and allelic frequencies (note
that as plant breeders we describe by their
phenotypic frequencies many times but need to
understand the underlying genetics) Recommend
Breeding for Quantitative Traits in Plants
Rex Bernardo (Dep. Agronomy Plt. Genetics,
Univ.of Minn.) published by Stemma
Press, Woodbury , MN
2
Quantitative Genetics Plant Breeding Consider a
diploid species with 2 alleles A1 and
A2 Genotype n Geno. freq. (f)
A1A1 240 P110.4 A1A2 240 P120.4
A2A2 120 P220.2 600
Consider A1 as the allele and A2 as the
allele Gene Action
3
Quantitative Genetics Plant Breeding Consider a
diploid species with 2 alleles A1 and
A2 Genotype n Geno. freq. (f)
A1A1 240 P110.4 A1A2 240 P120.4
A2A2 120 P220.2 600
Consider A1 as the allele and A2 as the
allele Gene Action
Dominance Genotype Additive Partial Complete Ov
erdom. A1A1 5 5 5
5 A1A2 3 4 5
6 A2A2 1 1 1 1
4
Quantitative Genetics Plant Breeding Consider a
diploid species with 2 alleles A1 and
A2 Genotype n Geno. freq. (f)
A1A1 240 P110.4 A1A2 240 P120.4
A2A2 120 P220.2 600 Genotypic freq
Allelic frequencies
Consider A1 as the allele and A2 as the
allele Gene Action
P11 240 / 600 .4 P12 240 / 600 .4 P22
120 / 600 .2 Sums to 1.0
p A1 P11 (.5)(P12) / 1200 .6 (240
240 240) / 1200 .6 q A2 P22 (.5)(P12)
/ 1200 .4 (120 120 240) / 1200 .4
5
Quantitative Genetics Plant Breeding Hardy-Weinb
erg Equilibrium (HWE) states that a populations
of a diploid species, considering 2 alleles or 1
allele versus all other alleles at that locus,
can be described by the expansion of (a b)2 or
in our terms (p q)2 p2 2pq q2, thus we
can add Consider a diploid species with 2
alleles A1 and A2 Genotype n Geno.freq.
(f) f after random mating A1A1 240 P110.4
A1A2 240 P120.4 A2A2 120 P220.2 Assumption
s and empirical proof of HWE (next slides)
p2P11 (.5)(P12)/1200 .62 .36 2pq 2.6.4
0.48 q2 P22 (.5)(P12)/1200 .42 .16
6

• Quantitative Genetics Plant Breeding
• HWE states that genotypic and allelic freq. will
  not change in a cross pollinating species under
  the following assumptions
• random mating
• no selection is practiced (naturally or
  otherwise)
• no differential migration
• mutation rates are equal, i.e. A?a a?A
• species in question is diploid
• usually considering 1 gene and 2 alleles or 1
  allele of an allelic series against all other
  alleles in the series

7
Quantitative Genetics Plant Breeding Empirical
proof of concept of HWE Consider a population
derived by crossing AA with aa AA x aa F1 Aa
(self) F2 .25 AA .5 Aa .25 aa
8
Quantitative Genetics Plant Breeding Empirical
proof of concept of HWE Consider a population
derived by crossing AA with aa AA x aa F1 Aa
(self) F2 .25 AA .5 Aa .25 aa
But what if the species is X-pollinated? Then we
put AA and aa in a crossing block and allow
pollination to occur and HWE states that in the
next generation, analogous to the F2 ( skips the
F1) in the selfed scenario Freq. of A p .5
a 1-p .5 p2 2p(1-p) (1-p)2 .52
2(.5)(.5) .52 .25 AA .5 Aa .25 aa
9
Quantitative Genetics Plant Breeding Empirical
proof of concept of HWE Consider a population
derived by crossing AA with aa
But what if the species is X-pollinated? Then we
put AA and aa in a crossing block and allow
pollination to occur and HWE states that in the
next generation, analogous to the F2 in the
selfed scenario Freq. of A p .5 a 1-p
.5 p2 2p(1-p) (1-p)2 .52 2(.5)(.5) .52
.25 AA .5 Aa .25 aa
Why? AA Aa aa AA x AA
4 AA x aa 4 aa x AA
4 aa x aa
4 4 8 4
.25 .5 .25
10
Quantitative Genetics Plant Breeding Empirical
proof of concept of HWE Consider a population
derived by intercrossing P1 BB P2 BB P3
Bb P4 Bb
11
Quantitative Genetics Plant Breeding Empirical
proof of concept of HWE Consider a population
derived by intercrossing P1 BB P2 BB P3
Bb P4 Bb
What is the freq. of p B and (1-p) b? 6/8
.75 B p 2/8 .25 b (1-p) HWE .752
2(.75)(.25) .252 .5675 BB .375 Bb .0625 bb
12
Quantitative Genetics Plant Breeding Empirical
proof of concept of HWE Consider a population
derived by intercrossing P1 BB P2 BB P3
Bb P4 Bb
What is the freq. of p B and (1-p) b? 6/8
.75 B p 2/8 .25 b (1-p) HWE .752
2(.75)(.25) .252 .5675 BB .375 Bb .0625 bb
Proof were in HWE p B .5675 .5 (.375) /
1 .75
13
Empirical proof of HWE for this population of 4
parents BB Bb bb BB1 x BB1 4 BB1 x BB2 4 BB1
x Bb3 2 2 BB1 x Bb4 2 2 BB2 x BB1 4 BB2 x
BB2 4 BB2 x Bb3 2 2 BB2 x Bb4 2 2 Bb3 x BB1
2 2 Bb3 x BB2 2 2 Bb3 x Bb3 1 2 1 Bb3 x
Bb4 1 2 1 Bb4 x BB1 2 2 Bb4 x BB2 2
2 Bb4 x Bb3 1 2 1 Bb4 x Bb4 1 2 1
Totals 36 24 4 64 .5625 .375 .0625
14
Quantitative Genetics Plant Breeding Lets go
back to a previous slide where we had a
population involving a diploid species with 2
alleles, A1 and A2, and plants in the following
freq. Genotype n freq. (f)
A1A1 240 P110.4 A1A2 240 P120.4
A2A2 120 P220.2
HWE p2 2p(1-p) (1-p)2 .36 A1A1 .48 A1A2
.16 A2A2
Can we devise empirical proof for this population?
15
Set up the table as below each individual has an
equal opportunity to be fertilized by any other
individual with each cross producing 4 possible
genotypes, i.e. a perfect population for each
plant to plant cross as shown earlier
16
p22p(1-p)(1-p)2 .62 (2)(.6).4) .42
.36 A1A1.48A1A2.16A2A2
17
Quantitative Genetics and Plant
Breeding Extension of HWE Recall the last
assumption as I listed the assumptions for HW to
be in effect was considering 1 gene and 2
alleles or 1 allele of an allelic series against
all other alleles in the series. However, HW can
be extended to describe the allelic and genotypic
frequencies encountered in a population with an
allelic series at a particular locus. What is an
allelic series? Consider a locus with 3 alleles
A1, A2, and A3
18
Quantitative Genetics and Plant Breeding HWE with
an allelic series Consider a locus with 3
alleles A1, A2, and A3, such that A1A1 A2A2
A3A3 A1A2 A1A3 A2A3 Obs. freq. 0.15 0.00 0.20 0
.25 0.35 0.05 If Pij freq. of the AiAj genotype,
then the freq. of the ith allele pi (i.e.
the freq. of Ai) or, pi freq. of AiAi
(1/2) (freq. of all heteroz. with Ai)
or, pi Pii ½ ?iltj Pij
19
Quantitative Genetics and Plant Breeding Consider
a locus with 3 alleles A1, A2, and A3, such
that A1A1 A2A2 A3A3 A1A2 A1A3 A2A3 Obs.
freq. 0.15 0.00 0.20 0.25 0.35 0.05 So, if pi
Pii ½ ?iltj Pij then for this population,
allelic freq. are A1 p1 P11 ½ (P12
P13) .15 ½ (.25 .35) .45 A2 p2 P22
½ (P12 P23) 0 ½ (.25 .05) .15 A3
p3 P33 ½ (P13 P23) .2 ½ (.35 .05)
.40 Note sum to 1.0
20
Quantitative Genetics and Plant Breeding Consider
a locus with 3 alleles A1, A2, and A3, such
that A1A1 A2A2 A3A3 A1A2 A1A3 A2A3 Obs.
freq. 0.15 0.00 0.20 0.25 0.35 0.05 So, allelic
freq. across these genotypes are .45 A1 .15 A2
.40 A3 and at HWE, AFTER ONE GENERATION OF
RANDOM MATING, genotypic freq. will be described
by (p1 p2 p3)2p12 2 p1p2 2 p1p3 p22
2 p2p3 p32 .452 A1A1
21
Quantitative Genetics and Plant Breeding Consider
a locus with 3 alleles A1, A2, and A3, such
that A1A1 A2A2 A3A3 A1A2 A1A3 A2A3 Obs.
freq. 0.15 0.00 0.20 0.25 0.35 0.05 So, allelic
freq. across these genotypes is .45 A1 .15 A2
.40 A3 and at HWE, AFTER ONE GENERATION OF
RANDOM MATING, genotypic freq. will be described
by (p1 p2 p3)2p12 2 p1p2 2 p1p3 p22
2 p2p3 p32 .452 A1A1 (2)(.45)(.15)
A1A2
22
Quantitative Genetics and Plant Breeding Consider
a locus with 3 alleles A1, A2, and A3, such
that A1A1 A2A2 A3A3 A1A2 A1A3 A2A3 Obs.
freq. 0.15 0.00 0.20 0.25 0.35 0.05 So, allelic
freq. across these genotypes is .45 A1 .15 A2
.40 A3 and at HWE, AFTER ONE GENERATION OF
RANDOM MATING, genotypic freq. will be described
by (p1 p2 p3)2p12 2 p1p2 2 p1p3 p22
2 p2p3 p32 .452 A1A1 (2)(.45)(.15)
A1A2 (2)(.45)(.4) A1A3
23
Quantitative Genetics and Plant Breeding Consider
a locus with 3 alleles A1, A2, and A3, such
that A1A1 A2A2 A3A3 A1A2 A1A3 A2A3 Obs.
freq. 0.15 0.00 0.20 0.25 0.35 0.05 So, allelic
freq. across these genotypes is .45 A1 .15 A2
.40 A3 and at HWE, AFTER ONE GENERATION OF
RANDOM MATING, genotypic freq. will be described
by (p1 p2 p3)2p12 2 p1p2 2 p1p3 p22
2 p2p3 p32 .452 A1A1 (2)(.45)(.15)
A1A2 (2)(.45)(.4) A1A3 .152 A2A2
24
Quantitative Genetics and Plant Breeding Consider
a locus with 3 alleles A1, A2, and A3, such
that A1A1 A2A2 A3A3 A1A2 A1A3 A2A3 Obs.
freq. 0.15 0.00 0.20 0.25 0.35 0.05 So, allelic
freq. across these genotypes is .45 A1 .15 A2
.40 A3 and at HWE, AFTER ONE GENERATION OF
RANDOM MATING, genotypic freq. will be described
by (p1 p2 p3)2p12 2 p1p2 2 p1p3 p22
2 p2p3 p32 .452 A1A1 (2)(.45)(.15)
A1A2 (2)(.45)(.4) A1A3 .152 A2A2
(2)(.15)(.4) A2A3
25
Quantitative Genetics and Plant Breeding Consider
a locus with 3 alleles A1, A2, and A3, such
that A1A1 A2A2 A3A3 A1A2 A1A3 A2A3 Obs.
freq. 0.15 0.00 0.20 0.25 0.35 0.05 So, allelic
freq. across these genotypes is .45 A1 .15 A2
.40 A3 and at HWE, AFTER ONE GENERATION OF
RANDOM MATING, genotypic freq. will be described
by (p1 p2 p3)2p12 2 p1p2 2 p1p3 p22
2 p2p3 p32 .452 A1A1 (2)(.45)(.15)
A1A2 (2)(.45)(.4) A1A3 .152 A2A2
(2)(.15)(.4) A2A3 .42 A3A3
26
Quantitative Genetics and Plant Breeding Consider
a locus with 3 alleles A1, A2, and A3, such
that A1A1 A2A2 A3A3 A1A2 A1A3 A2A3 Obs.
freq. 0.15 0.00 0.20 0.25 0.35 0.05 .45 A1 .15
A2 .40 A3 and .452 A1A1
(2)(.45)(.15) A1A2 (2)(.45)(.4) A1A3 .152
A2A2 (2)(.15)(.4) A2A3 .42 A3A3 .2025A1A1
.135A1A2 .36A1A3 .0225A2A2 .12A2A3
.16A3A3 Is this population in HWE? Recalculate
allelic freq. A1 p1 P11 ½ (P12 P13)
.2025 ½ (.135 .36) .45 A2 p2 P22 ½
(P12 P23) .0225 ½ (.135 .12) .15 A3
p3 P33 ½ (P13 P23) .16 ½ (.36 .12)
.40
YES
27
Quantitative Genetics and Plant Breeding Consider
a locus with 3 alleles A1, A2, and A3, such
that A1A1 A2A2 A3A3 A1A2 A1A3 A2A3 Obs.
freq. 0.15 0.00 0.20 0.25 0.35 0.05 And we could
do this with any number of alleles in a series at
a single locus, e.g. 5 alleles (p1 p2 p3
p4 p5)2 Note that we are determining HW for a
given locus or gene and not for every possible
gene or trait. Demonstrates however what occurs
in an idealized population for every gene
28
Quantitative Genetics and Plant Breeding In this
idealized HW population, we can describe the
genotypic mean for a given gene Consider a
single locus, A, with alleles A1 and
A2 Genotypic mean can be expressed as MMP
a(p-q) 2dpq where Mpopulation
mean MPmidparent value or mean of
homozygotes avalue of the homozygote
A1A1 - A2A2 p freq. of A1 q
freq. of A2 d value of the heterozygote
29
Quantitative Genetics and Plant Breeding In this
idealized HW population, we can describe the
genotypic mean for a given gene Consider a
single locus, A, with alleles A1 and
A2 Genotypic mean can be expressed as MMP
a(p-q) 2dpq or graphically A1A1 A1A2 A2A2
a d -a also described
as MPa MPd MP-a
30
Quantitative Genetics and Plant Breeding IF
aperformance of either homozygous genotype (
for A1A1 and for A2A2) minus the MP LET the
performance of A1A1 20 and A2A2 14 then a
20 ((2014)/2) 20 17 3 NOW d value of
the heterozygote MP d and if we rearrange to
MP d then d degree of dominance SUCH THAT
d A1A2 17 3 THEREFORE in this example A1
is completely dominant to A2 SO IF d a
then gene action is dominant d lt a then gene
action is partial dominance (incldg additive)
d gt a then gene action is overdominant
31
Quantitative Genetics and Plant Breeding Simple
example to demonstrate MMP a(p-q) 2dpq LET p
q 0.5 MMP a(p-q) 2dpq 17 3 (.5 - .5)
(2)(3)(.5)(.5) 17 3 (0) 1.5 18.5 Or
M for contribution of the gene only 18.5 MP
1.5
32
Quantitative Genetics and Plant Breeding Simple
example to demonstrate MMP a(p-q) 2dpq LET p
q 0.5 MMP a(p-q) 2dpq 17 3 (.5 .5)
(2)(3)(.5)(.5) 17 3 (0) 1.5 18.5 Or
M for contribution of the gene only 18.5 MP
1.5 HW proof (recall HWp2 2pq q2) .52
(2)(.5)(.5) .52 .25 or .25 A1A1 .5 A1A2
.25 A2A2 AND SINCE A1A1 3 .25 (3) A1A1 .5
(3) A1A2 .25 (-3) A2A2 .75 A1A1 1.5 A1A2
(-.75) A2A2 ? 1.5
33
Quantitative Genetics and Plant Breeding Simple
example to demonstrate MMP a(p-q) 2dpq LET p
q 0.5 MMP a(p-q) 2dpq 17 3 (.5 .5)
(2)(3)(.5)(.5) 17 3 (0) 1.5 18.5 Or
M for contribution of the gene only 18.5 MP
1.5 HW proof (recall HWp2 2pq q2) .52
(2)(.5)(.5) .52 .25 or .25 A1A1 .5 A1A2
.25 A2A2 AND SINCE A1A1 3 .25 (3) A1A1 .5
(3) A1A2 .25 (-3) A2A2 .75 A1A1 1.5 A1A2
(-.75) A2A2 ? 1.5
34
Quantitative Genetics and Plant Breeding Simple
example to demonstrate MMP a(p-q) 2dpq LET p
q 0.5 MMP a(p-q) 2dpq 17 3 (.5 .5)
(2)(3)(.5)(.5) 17 3 (0) 1.5 18.5 Or
M for contribution of the gene only 18.5 MP
1.5 HW proof (recall HWp2 2pq q2) .52
(2)(.5)(.5) .52 .25 or .25 A1A1 .5 A1A2
.25 A2A2 AND SINCE M for A1A1 20 .25 (20)
A1A1 .5 (20) A1A2 .25 (14) A2A2 5 A1A1 10
A1A2 3.5 A2A2 ? 18.5
35
Quantitative Genetics and Plant Breeding Simple
example to demonstrate MMP a(p-q) 2dpq LET p
q 0.5 MMP a(p-q) 2dpq 17 3 (.5 .5)
(2)(3)(.5)(.5) 17 3 (0) 1.5 18.5 Or
M for contribution of the gene only 18.5 MP
1.5 HW proof (recall HWp2 2pq q2) .52
(2)(.5)(.5) .52 .25 or .25 A1A1 .5 A1A2
.25 A2A2 AND SINCE M for A1A1 20 .25 (20)
A1A1 .5 (20) A1A2 .25 (14) A2A2 .75 A1A1
1.5 A1A2 (-.75) A2A2 E 18.5
36
Quantitative Genetics and Plant Breeding For a
polygenic trait without epistasis M ? MP
a(p-q) 2dpq or ? Mpi ? ai(pi qi) 2
??dipiqi AND there is no way to measure a, d,
p,or q for individual loci of a polygenic trait.