P=G+E
P=G+E
If we replicate a genotype in a large number of environments
P=G+E
If we replicate a genotype in a large number of environments
When a single locus is under consideration,
G=A+D
In a population in HWE:
Genotype | Freq. | Value |
---|---|---|
A1A1 | p2 | +a |
A1A2 | 2pq | d |
A2A2 | q2 | −a |
In a population in HWE:
Genotype | Freq. | Value |
---|---|---|
A1A1 | p2 | +a |
A1A2 | 2pq | d |
A2A2 | q2 | −a |
According to Formula (1) in the STAT Note:
E(f(X))=k∑i=1f(xi)Pr(X=xi)M=p2a+2pqd−q2a=a(p−q)+2pqd
In a population in HWE:
Genotype | Freq. | Value |
---|---|---|
A1A1 | p2 | +a |
A1A2 | 2pq | d |
A2A2 | q2 | −a |
According to Formula (1) in the STAT Note:
E(f(X))=k∑i=1f(xi)Pr(X=xi)M=p2a+2pqd−q2a=a(p−q)+2pqd where M is the deviation from the midpoint of the homozygotes.
Booroola (B) locus influences fecundity in Merino sheep
Genotype | Mean litter size |
---|---|
B1B1 | 2.66 |
B1B2 | 2.46 |
B2B2 | 1.48 |
Booroola (B) locus influences fecundity in Merino sheep
Genotype | Mean litter size |
---|---|
B1B1 | 2.66 |
B1B2 | 2.46 |
B2B2 | 1.48 |
First, determine midpoint:
then, determine a and d:
And finally calculate the population mean
Booroola (B) locus influences fecundity in Merino sheep
Genotype | Mean litter size |
---|---|
B1B1 | 2.66 |
B1B2 | 2.46 |
B2B2 | 1.48 |
Note that this is the deviation from the midpoint, 2.07.
When multiple loci are acting independently and contributing to overall value, the mean is:
M=k∑i=1ai(pi−qi)+2k∑i=1piqidi
Summation over all the loci
k is the number of loci affecting the genotypic value of a trait
Parents pass on alleles, not genotypes
The average effect of an allele depends on
Thus, an average effect can be unique for each population because of differences in allele freq between populations.
Parents pass on alleles, not genotypes
The average effect of an allele depends on
Thus, an average effect can be unique for each population because of differences in allele freq between populations.
Mean deviation from the population mean of individuals who received that allele from one parent
The other allele is received at random
The probability an A1 allele combines with an A2 allele to form a heterozygote, with value d, is q.
The upshot of this is that
Thus, the mean value of individuals that received A1 is pa+qd
The probability an A1 allele combines with an A2 allele to form a heterozygote, with value d, is q.
The upshot of this is that
Thus, the mean value of individuals that received A1 is pa+qd
α1=pa+dq−M=pa+dq−(a(p−q)+2pqd)=q(a+d(q−p))
α=α1−α2=q(a+d(q−p))−(−p(a+d(q−p)))=a+d(q−p)
Genotype | Mean litter size |
---|---|
B1B1 | 2.66 |
B1B2 | 2.46 |
B2B2 | 1.48 |
Genotype | Mean litter size |
---|---|
B1B1 | 2.66 |
B1B2 | 2.46 |
B2B2 | 1.48 |
Mean of individuals receiving a B1 allele (others coming from random) is: pa+qd
Averge effect of B1:
Genotype | Mean litter size |
---|---|
B1B1 | 2.66 |
B1B2 | 2.46 |
B2B2 | 1.48 |
Genotype | Mean litter size |
---|---|
B1B1 | 2.66 |
B1B2 | 2.46 |
B2B2 | 1.48 |
α=a+d(q−p)=0.59+0.39×(0.75−0.25)=0.785
Genotype | Mean litter size |
---|---|
B1B1 | 2.66 |
B1B2 | 2.46 |
B2B2 | 1.48 |
α=a+d(q−p)=0.59+0.39×(0.75−0.25)=0.785
α=α1−α2α2=α1−α=0.589−0.785=−0.196
Genotype | Mean litter size |
---|---|
B1B1 | 2.66 |
B1B2 | 2.46 |
B2B2 | 1.48 |
Genotype | Mean litter size |
---|---|
B1B1 | 2.66 |
B1B2 | 2.46 |
B2B2 | 1.48 |
M=a(p−q)+2pqd=0.59×(0.85−0.15)+2×0.85×0.15×0.39=0.512α=a+d(q−p)=0.59+0.39×(0.15−0.85)=0.317
Genotype | Mean litter size |
---|---|
B1B1 | 2.66 |
B1B2 | 2.46 |
B2B2 | 1.48 |
M=a(p−q)+2pqd=0.59×(0.85−0.15)+2×0.85×0.15×0.39=0.512α=a+d(q−p)=0.59+0.39×(0.15−0.85)=0.317 As population mean increases, average effect of the allele substitution decreases.
α1=q(a+d(q−p))
When d=0, α1=q(a+d(q−p))=(1−p)a=a−pa
Average effect of an allele plotted against its frequency in the population.
α1=q(a+d(q−p))α=a+d(q−p)α1=qα
α1=q(a+d(q−p))α=a+d(q−p)α1=qα
α=a+d(q−p)=a+d(1−p−p)=(a+d)−2d×p
P=G+E
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