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PBG 650 Advanced Plant BreedingPowerPoint Presentation

PBG 650 Advanced Plant Breeding

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PBG 650 Advanced Plant Breeding

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PBG 650 Advanced Plant Breeding

Module 5: Quantitative Genetics

- Genetic variance: additive and dominance

- The variance of a variable X is:
V(X) = E[(Xi- X)2] = E(Xi2) - X2

- The covariance of variable X and variable Y is:
Cov(X,Y) = E[(X - X)(Y - Y)] = E(XY) - XY

- The variance of a constant is zero
V(c) = 0 V(c+X) = V(X)

- The variance of the product of a variable and a constant is the constant squared times the variance of the variable
V(cX) = c2V(X)

- The variance of a sum of random variables is the sum of the variances plus twice the covariance between the variables
V(X + Y) = V(X) + V(Y) + 2Cov(X,Y)

P = G + E

G = A + D + I

P = A + D + I + E

Gijkl = + (i +j + ij) + (k +l + kl) + Iijkl

Because there are no

covariances among

the components

Variance of breeding values

(No adjustment for the mean is necessary because the mean of breeding values is zero)

When p=q=1/2

σA2 =(1/2)a2

When d=0

σA2 = 2pqa2

Variance of dominance deviations

(No adjustment for the mean is necessary because the mean of dominance deviations is zero)

When p=q=1/2

σD2 =(1/4)d2

When d=0, σD2 = 0

For a single locus

(It can be shown that the Cov(A,D) = 0)

P=MP=midparent value

M

Mean(X) = (ΣfiXi) = q2(0) + 2pq(1) + p2(2) = 2p(q+p) = 2p

= p2(22) + 2pq(12) +q2(02) – (2p)2= 2pq

M

= p2(2)(P+a) + 2pq(1)(P+d) + q2(0)(P-a)– (2p)(P+a(p-q)+2pqd)

= 2pq[a+d(q-p)] = 2pq

Same result with scaled values (a, d, -a) or the adjusted genotypic values:

= p2(2)(a-M) + 2pq(1)(d-M) +q2(0)(-a-M)-(2p)(0) 2pq

p=0.6 q=0.4

- Options for estimating variances
- Use formulas with known values of a and d
- Calculate breeding values and dominance deviations, and estimate their variances
- Regress observed values on number of Z1 alleles

Example from Falconer & Mackay

P = (6+14)/2 = 10

a=14-10=4

d=12-10=2

p=0.6 q=0.4

σG2 = 0.16(-5.76)2+0.48(0.24)2+0.36(2.24)2-02 = 7.1424

σA2 = 0.16(-4.32)2+0.48(-0.72)2+0.36(2.88)2-02 = 6.2208

σD2 = 0.16(-1.44)2+0.48(0.96)2+0.36(-0.64)2-02 = 0.9216

p=0.6 q=0.4

Mean(X) = 0.16(0) + 0.48(1) + 0.36(2) = 1.20

= 0.16(02) + 0.48(12) +0.36(22) – (1.20)2

= 0.4800 = 2pq

= 0.16(0)(6) + 0.48(1)(12) + 0.36(2)(14)– (1.20)(11.76) =1.7280=2pq

The result is the same if we use the adjusted genotypic values:

= 0.16(0)(-5.76) + 0.48(1)(0.24) + 0.36(2)(2.24) – (1.20)(0) =1.7280

genotypic value

breeding value

a = 3.6

1

Z1Z2

0

Z2Z2

2

Z1Z1

Excel

- With no dominance, all genetic variance is additive and maximum genetic variance occurs when p=q=0.5
- With complete dominance
- maximum additive genetic variance occurs when the unfavorable allele has a frequency of q=0.75
- maximum dominance variance occurs when q=0.5
- maximum genetic variance occurs when q2=0.5 (q=0.71)

Total genetic variance increases with selfing!!

Hallauer, Carena and Miranda, 2010