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Assignment 7.5 Falconer & Mackay, chapter 7. Sanja Franic VU University Amsterdam 2012. 2 alleles, A and B 3 situations: q=.2 q=.5 q=.8 mean phenotypic value per genotype: Question: What are the average effects of the two alleles, and the average effect of gene substitution? . AA. - PowerPoint PPT Presentation
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Assignment 7.5Falconer & Mackay, chapter 7
Sanja FranicVU University Amsterdam 2012
- 2 alleles, A and B
- 3 situations:q=.2q=.5q=.8
- mean phenotypic value per genotype:
Question: What are the average effects of the two alleles, and the average effect of gene substitution?
AA
154
188
122
Genotypic value
Genotype AB BB
Genotype frequency
2pq
p2q2
- 2 alleles, A and B
- 3 situations:q=.2q=.5q=.8
- mean phenotypic value per genotype:
Question: What are the average effects of the two alleles, and the average effect of gene substitution?
Answer:
AA
154
188
122
Genotypic value
Genotype AB BB
Genotype frequency
2pq
p2q2
a= a + d (q - p)aA = -paaB = qa
a=33d=-1
- 2 alleles, A and B
- 3 situations:q=.2q=.5q=.8
- mean phenotypic value per genotype:
Question: What are the average effects of the two alleles, and the average effect of gene substitution?
Answer:
Situation 1)
a= a + d (q - p) = 33.6aA = -pa = -26.88aB = qa = 6.72
AA
154
188
122
Genotypic value
Genotype AB BB
Genotype frequency
2pq
p2q2
AA
154 188122
Phen. valueGenotype AB BB
Genotype freq .32 .64.04
a= a + d (q - p)aA = -paaB = qa
a=33d=-1
- 2 alleles, A and B
- 3 situations:q=.2q=.5q=.8
- mean phenotypic value per genotype:
Question: What are the average effects of the two alleles, and the average effect of gene substitution?
Answer:
Situation 1)
a= a + d (q - p) = 33.6aA = -pa = -26.88aB = qa = 6.72
Situation 2)
a= a + d (q - p) = 33aA = -pa = -16.5aB = qa = 16.5
AA
154
188
122
Genotypic value
Genotype AB BB
Genotype frequency
2pq
p2q2
AA
154 188122
Phen. valueGenotype AB BB
Genotype freq .32 .64.04
a= a + d (q - p)aA = -paaB = qa
a=33d=-1
AA
154 188122
Phen. valueGenotype AB BB
Genotype freq .5 .25.25
- 2 alleles, A and B
- 3 situations:q=.2q=.5q=.8
- mean phenotypic value per genotype:
Question: What are the average effects of the two alleles, and the average effect of gene substitution?
Answer:
Situation 1)
a= a + d (q - p) = 33.6aA = -pa = -26.88aB = qa = 6.72
Situation 2)
a= a + d (q - p) = 33aA = -pa = -16.5aB = qa = 16.5
Situation 3)
a= a + d (q - p) = 32.4aA = -pa = -6.48aB = qa = 25.92
AA
154
188
122
Genotypic value
Genotype AB BB
Genotype frequency
2pq
p2q2
AA
154 188122
Phen. valueGenotype AB BB
Genotype freq .32 .64.04
a= a + d (q - p)aA = -paaB = qa
a=33d=-1
AA
154 188122
Phen. valueGenotype AB BB
Genotype freq .5 .25.25
AA
154 188122
Phen. valueGenotype AB BB
Genotype freq .32 .04.64
- aA: mean deviation from the population mean of individuals who inherited allele A from one parent, the other allele having come at random from the population
- so aA is a deviation from the population mean. What is the population mean?
m = a(p – q) + 2dpq
- in our example (situation 1):
m = a(p – q) + 2dpq = 33(.8-.2) + 2*(-1)*.8*.2 = 19.48
(on a scale where 0 is midpoint!)
- our scale:
m = 19.48 + midpoint = 19.48 + 155 = 174.48
AA
154
188
122
Genotypic value
Genotype AB BB
Genotype frequency
2pq
p2q2
m
aBaA
AA (q)
B (p)
AA (q)
AB (p)
q*(-a)
p*d-qa + pd aA= -qa + pd -
m aA = -p[a+ d(q - p)]
aA = -pa = -26.88 -> 174.48-26.88 = 147.68aB = qa = 6.72 -> 174.48+6.72 = 181.2
Situation 1)
aA = -pa = -26.88 -> 174.48-26.88 = 147.68aB = qa = 6.72 -> 174.48+6.72 = 181.2
f(B)=.8f(A)=.2
Situation 2)
aA = -pa = -16.5 -> 154.5-16.5 = 138aB = qa = 16.5 -> 154.5+16.5 = 171
f(B)=.5f(A)=.5
Situation 3)
aA = -pa = -6.48 -> 134.88-6.48 = 128.4aB = qa = 25.92 -> 134.88+25.92 = 160.8
f(B)=.2f(A)=.8
AA
154
188
122
Genotypic value
Genotype AB BB
Genotype frequency
2pq
p2q2
m
aBaA
AA
188
122
Genotypic value
Genotype AB BB
Genotype frequency
2pq
p2q2
m
aBaA
AA
154
188
122
Genotypic value
Genotype AB BB
Genotype frequency
2pq
p2q2
m
aBaA
A simplified example involving height
A simplified example involving height, d=0:
A simplified example involving height, d=0:
AA
180
200
160
Genotypic value
Genotype AB BB
Genotype frequency
2pq
p2q2
A simplified example involving height, d=0:
AA
180
200
160
Genotypic value
Genotype AB BB
Genotype frequency
2pq
p2q2
A simplified example involving height, d=0:
Situation 1) q=.2 AA
180
200
160
Genotypic value
Genotype AB BB
Genotype frequency
2pq
p2q2
A simplified example involving height, d=0:
Situation 1) q=.2 AA
180
200
160
Genotypic value
Genotype AB BB
Genotype frequency
.32 .64.04
A simplified example involving height, d=0:
Situation 1) q=.2 AA
180
200
160
Genotypic value
Genotype AB BB
Genotype frequency
.32 .64.04
A simplified example involving height, d=0:
Situation 1) q=.2 AA
180
200
160
Genotypic value
Genotype AB BB
Genotype frequency
.32 .64.04
160
180
200
cm
A simplified example involving height, d=0:
Situation 1) q=.2
Population mean:
m = a(p – q) + 2dpq
a = 20d=0p=.8q=.2
m = 12 cm
(on a scale with midpoint=0!)
On actual scale:
m = 180cm + 12cm = 192cm
AA
180
200
160
Genotypic value
Genotype AB BB
Genotype frequency
.32 .64.04
160
180
200
cm
a a
A simplified example involving height, d=0:
Situation 1) q=.2
Population mean:
m = a(p – q) + 2dpq
a = 20d=0p=.8q=.2
m = 12 cm
(on a scale with midpoint=0!)
On actual scale:
m = 180cm + 12cm = 192cm
AA
180
200
160
Genotypic value
Genotype AB BB
Genotype frequency
.32 .64.04
160
180
200
cm
m
192
A simplified example involving height, d=0:
Situation 1) q=.2
Population mean:
m = a(p – q) + 2dpq
a = 20d=0p=.8q=.2
m = 12 cm
(on a scale with midpoint=0!)
On actual scale:
m = 180cm + 12cm = 192cm
a= a + d (q - p) = a + 0 = 20
aA = -pa = -.8*20 = -16aB = qa = .2*20 = 4
AA
180
200
160
Genotypic value
Genotype AB BB
Genotype frequency
.32 .64.04
160
180
200
cm
m
192
A simplified example involving height, d=0:
Situation 1) q=.2
Population mean:
m = a(p – q) + 2dpq
a = 20d=0p=.8q=.2
m = 12 cm
(on a scale with midpoint=0!)
On actual scale:
m = 180cm + 12cm = 192cm
a= a + d (q - p) = a + 0 = 20
aA = -pa = -.8*20 = -16aB = qa = .2*20 = 4
AA
180
200
160
Genotypic value
Genotype AB BB
Genotype frequency
.32 .64.04
160
180
200
cm
m
192
aBaA
A simplified example involving height, d=0:
Situation 1) q=.2
a = 20d=0p=.8q=.2
m = 192cm
a= 20aA = -16aB = 4
AA
180
200
160
Genotypic value
Genotype AB BB
Genotype frequency
.32 .64.04
160
180
200
cm
m
192
aBaA
A simplified example involving height, d=0:
Situation 1) q=.2
a = 20d=0p=.8q=.2
m = 192cm
a= 20aA = -16aB = 4
bvAA=2aA=-32
bvAB=aA+aB=-12
bvBB=2aB=8
AA
180
200
160
Genotypic value
Genotype AB BB
Genotype frequency
.32 .64.04
160
180
200
cm
m
192
bvAA bvBB
A simplified example involving height, d=0:
Situation 1) q=.2
a = 20d=0p=.8q=.2
m = 192cm
a= 20aA = -16aB = 4
bvAA=2aA=-32
bvAB=aA+aB=-12
bvBB=2aB=8
So what does this mean?
The mean deviation from the population mean of offspring of short (AA) individuals mated at random with a sample from the population will be -16. The corresponding deviation for offspring of tall (BB) individuals will be only 4. Why?
In random mating with the population, allele frequencies come into play. E.g. if I randomly mate short (AA) individuals with a sample from the population, I will be mating them with only 4% other short individuals, 32% intermediate individuals, and 64% tall individuals.
AA
180
200
160
Genotypic value
Genotype AB BB
Genotype frequency
.32 .64.04
160
180
200
cm
m
192
aBaA
+
+
= (on average) 176 cm
+
Why? Because the short individual is such an outlier (i.e., is so far removed from the mean), and he/she always provides half the genes for the offspring. Thus, the short individual will have a large effect on ‘pulling’ the offspring mean downwards.
= (on average) 176 cm
A simplified example involving height, d=0:
Situation 1) q=.2
a = 20d=0p=.8q=.2
m = 192cm
a= 20aA = -16aB = 4
bvAA=2aA=-32
bvAB=aA+aB=-12
bvBB=2aB=8
So what does this mean?
The mean deviation from the population mean of AA individuals mated at random with a sample from the population will be -16. The corresponding deviation for BB individuals will be only 4. Why?
In random mating with the population, allele frequencies come into play. E.g. if I randomly mate short (AA) individuals with a sample from the population, I will be mating them with only 4% other short individuals, 32% intermediate individuals, and 64% tall individuals. Still, on average, their offspring will be relatively short (176cm). Why?
Because the alleles of the short individual, being so far removed from the mean, affect the mean genotypic value of the progeny by a lot (by ‘pulling’ it away from the population mean). Thus, allele A is said to have a large (average) effect (aA).
AA
180
200
160
Genotypic value
Genotype AB BB
Genotype frequency
.32 .64.04
160
180
200
cm
m
192
aBaA
Back to our example (7.5)…
Situation 1)
aA = -pa = -26.88 -> 174.48-26.88 = 147.68aB = qa = 6.72 -> 174.48+6.72 = 181.2
f(B)=.8f(A)=.2
Situation 2)
aA = -pa = -16.5 -> 154.5-16.5 = 138aB = qa = 16.5 -> 154.5+16.5 = 171
f(B)=.5f(A)=.5
Situation 3)
aA = -pa = -6.48 -> 134.88-6.48 = 128.4aB = qa = 25.92 -> 134.88+25.92 = 160.8
f(B)=.2f(A)=.8
AA
154
188
122
Genotypic value
Genotype AB BB
Genotype frequency
2pq
p2q2
m
aBaA
AA
188
122
Genotypic value
Genotype AB BB
Genotype frequency
2pq
p2q2
m
aBaA
AA
154
188
122
Genotypic value
Genotype AB BB
Genotype frequency
2pq
p2q2
m
aBaA
Situation 1)
aA = -pa = -26.88 -> 174.48-26.88 = 147.68aB = qa = 6.72 -> 174.48+6.72 = 181.2
f(B)=.8f(A)=.2
Situation 2)
aA = -pa = -16.5 -> 154.5-16.5 = 138aB = qa = 16.5 -> 154.5+16.5 = 171
f(B)=.5f(A)=.5
Situation 3)
aA = -pa = -6.48 -> 134.88-6.48 = 128.4aB = qa = 25.92 -> 134.88+25.92 = 160.8
f(B)=.2f(A)=.8
Note that here dominance IS present, so there are deviations from linearity (unlike in the height [d=0] example). However, d is small in the present example (d=-1), so the two situations are fairly similar. In general, however, dominance does complicate things.
AA
154
188
122
Genotypic value
Genotype AB BB
Genotype frequency
2pq
p2q2
m
aBaA
AA
188
122
Genotypic value
Genotype AB BB
Genotype frequency
2pq
p2q2
m
aBaA
AA
154
188
122
Genotypic value
Genotype AB BB
Genotype frequency
2pq
p2q2
m
aBaA
