population genetics

Population Genetics Chpt 4 End

Chapter 4 End Problems

Problem 1

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u <- 80
u.s <- 110
u.prime <- 90
S <- u.s - u  #selection differential
R <- u.prime - u  # response to selection
h2 <- R/S  #realized heritability
h2
## [1] 0.3333333

Problem 2

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u <- 140
u.s <- 165
h2 <- 0.4
S <- u.s - u  #selection differential
R <- h2*S
u.prime <- u + R
u.prime
## [1] 150

If females only have u.s=165 and we assume males u=140 then the average u.s = (165+140)/2 = 152.5

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u <- 140
u.s <- 152.5 
h2 <- 0.4
S <- u.s - u  #selection differential
R <- h2*S
u.prime <- u + R
u.prime
## [1] 145

Problem 3

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u <- 60
u.s <- 48
h2 <- 0.12
S <- u.s - u  #selection differential
R <- h2*S
u.prime <- u + R
u.prime
## [1] 58.56

Problem 4

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u.s <- 35
u.prime <- 25 
h2 <- 0.4


#S <- u.s - u  #selection differential
#R <- h2*S
#u.prime <- u + R  or R <- u.prime - u

#Rearrange to get:

#R <- h2*(u.s-u)
#u.prime - u <- h2*(u.s-u)
#solve for u:

u <- (u.prime-h2*u.s)/(1-h2)
u
## [1] 18.33333

Problem 5

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u <- 110
h2 <- 0.25
#S <- u.s - u  #selection differential
S <- 80
u.s <- S + u
u.s
## [1] 190
R <- h2*S
R
## [1] 20
u.prime <- u + R
u.prime
## [1] 130

Problem 6

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u <- 100
#S <- u.s - u  #selection differential
S <- 32
u.s <- S + u
u.s
## [1] 132
#From Table IX p292 half sib covariance equals the additive genetic variance divided by 4 so

sigma2.a <- 12*4
sigma2.p <- 240
h2 <- sigma2.a/sigma2.p
h2
## [1] 0.2
R <- h2*S
R
## [1] 6.4
u.prime <- u + R  
u.prime 
## [1] 106.4

Problem 7

$$\displaystyle\sigma^2_g + \sigma^2_e = 240$$

$$\displaystyle\sigma^2_e = 90$$ so

$$\displaystyle\sigma^2_g = 240 - 190 = 50$$

Broad sense heritability =
$$\displaystyle\frac{ \sigma^2_g }{ \sigma^2_g + \sigma^2_e } = \frac{50}{240} = 0.208$$

Problem 8

additive variance: $$\displaystyle \sigma^2_a$$ dominance variance: $$\displaystyle \sigma^2_d$$ interaction or epistatic variance: $$\displaystyle \sigma^2_i$$ total phenotypic variance of the whole population: $$\displaystyle \sigma^2_p$$ total genotypic variance: $$\displaystyle \sigma^2_g$$

Broad sense heritability $$\displaystyle h^2= \frac{\sigma^2_g}{\sigma^2_p} = \frac{\sigma^2_a +\sigma^2_d +\sigma^2_i }{\sigma^2_p} = 0.7$$

Narrow sense heritability $$\displaystyle h^2= \frac{\sigma^2_a}{\sigma^2_p}$$

If $$\displaystyle \sigma^2_d +\sigma^2_i = 0$$ then Broad sense heritability and narrow sense heritability are the same and so the maximum narrow sense is 0.7. Heritability can be zero, so the range is 0 - 0.7.

Problem 9

Narrow sense heritability $$\displaystyle h^2= \frac{\sigma^2_a}{\sigma^2_p}$$

Rearrange: $$\displaystyle \sigma^2_a = h^2\sigma^2_p$$

Jones: 0.24 x 4= 0.96 = $$\displaystyle\sigma^2_a$$

Smith: 0.12 x 4 = 0.48 = $$\displaystyle\sigma^2_a$$

Problem 10

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x <- c(21, 15, 9)
y <- log(x)
y
## [1] 3.044522 2.708050 2.197225
u.x.asterisk <- (x[1] + x[3])/2
a.x <- x[1] - u.x.asterisk
d.x <- x[2] - u.x.asterisk

 u.x.asterisk; a.x; d.x;
## [1] 15
## [1] 6
## [1] 0
u.y.asterisk <- (y[1] + y[3])/2
a.y <- y[1] - u.y.asterisk
d.y <- y[2] - u.y.asterisk

u.y.asterisk; a.y; d.y;
## [1] 2.620874
## [1] 0.4236489
## [1] 0.08717669
exp(u.y.asterisk); exp(a.y); exp(d.y);
## [1] 13.74773
## [1] 1.527525
## [1] 1.091089

Problem 11

See Box E

$$\displaystyle \sigma^2_a=2pq[a+(q-p)d]^2$$

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a.A <- 0.6
d.A <- 0.2
p.A <- 0.5
q.A <- 1-p.A

sigma2.A <- 2*p.A*q.A*(a.A+(q.A-p.A)*d.A)^2
sigma2.A
## [1] 0.18
a.B <- 0.4
d.B <- 0
p.B <- 0.7
q.B <- 1-p.B

sigma2.B <- 2*p.B*q.B*(a.B+(q.B-p.B)*d.B)^2
sigma2.B
## [1] 0.0672
sigma2.A + sigma2.B
## [1] 0.2472

$$\displaystyle h^2 = \frac{0.2472}{\sigma^2_p}$$

Problem 12

Problem 13

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p <- 0.3
q <- 1-p
a <- 0.1
d <- 0.08

sigma2.a <-2*p*q*(a+(q-p)*d)^2
sigma2.a
## [1] 0.00731808
sigma2.d <- (2*p*q*d)^2
sigma2.d
## [1] 0.00112896
sigma2.g <- sigma2.a + sigma2.d
sigma2.g
## [1] 0.00844704

Problem 14

d > a implies overdominance and an equilibrium at $$\displaystyle \sigma^2_a=0$$ which implies:

$$\displaystyle a + (q-p)d = 0$$

$$\displaystyle a + (1-p-p)d = 0$$

$$\displaystyle a + d - 2pd = 0$$

$$\displaystyle a + d = 2pd$$

$$\displaystyle p=\frac{a+d}{2d}=\frac{0.1+0.12}{2(0.12)}=0.91667$$

Problem 15

Genotype Response Proportion Mean of $F_2$ $\sigma^2_G$
$AA$ $\mu^*+a$ 1/4 $1/4(\mu^*+a)$ $1/4(\mu^*+a-\mu^*)^2$
$AA^{‘}$ $\mu^*+d$ 1/s $1/2(\mu^*)$ $1/2(\mu^*-\mu^*)^2$
$A^{‘}A^{‘}$ $\mu^*-a$ 1/4 $1/4(\mu^*-a)$ $1/4(\mu^*-a-\mu^*)^2$
Totals: $\mu^*$ $1/2a^2$

Genetic variance in the $F_2$ for n loci is $\sigma^2_G=n\frac{a^2}{2}$

$$\sigma^2_{Total}=\sigma^2_g + \sigma^2_e$$

So for the F2 $\sigma^2_2=n\frac{a^2}{2} + \sigma^2$ or solve for $a^2$:

$$a^2=2\frac{(\sigma^2_2-\sigma^2)}{n}$$

$R=2na$ is $R^2=4n^2a^2$ or solve for $a^2$:

$$a^2=\frac{R^2}{4n^2}$$

$$\frac{R^2}{4n^2}=2\frac{(\sigma^2_2-\sigma^2)}{n}$$

Solve for n: $n=\frac{R^2}{8(\sigma^2_2-\sigma^2)}$

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