Assignment 7.5 Falconer & Mackay, chapter 7

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