Peter D. Fields Academia, Research, and whatnot

Caballero QG Chp. 1, Problem set Q2 (R)

Chapter 1: Continuous variation

Problem 2

The following table shows the number of sternopleural bristles in 50 individuals of Drosophila melanogaster.

24 24 21 21 21 20 19 19 18 24
23 25 22 24 22 22 18 20 22 28
23 27 23 20 23 21 26 27 27 28
26 26 24 26 27 26 20 22 21 18
18 21 17 21 18 20 23 22 34 22

Knowing that the genetic variance of the trait is VG = 4, deduce the value of the environmental variance (VE) and the phenotypic (CVP) and genotypic (CVG) coefficients of variation. A good place to start here is creating an array of the phenotypes for bristle number:

b <- c(24,24,21,21,21,20,19,19,18,24,23,25,22,24,22,22,18,20,22,28,23,27,23,20,23,21,26,27,27,28,26,26,24,26,27,26,20,22,21,18,18,21,17,21,18,20,23,22,34,22)
str(b)
num [1:50] 24 24 21 21 21 20 19 19 18 24 ...

The solution to this problem requires solving for a few variables from equation (1.2), VP = VG + VE. Let’s start by getting the mean of the bristle phenotype:

mb <- mean(b)
print(mb)
[1] 22.68

We also need to estimate the phenotypic variance:

vb <- var(b)
print(vb)
[1] 11.3649

Great! Now we can easily get our VE value since it’s just the result of subtracting VG from VP, or:

ve <- vb - 4
print(ve)
[1] 7.364898

Getting (CVP) is similarly easy since we now have VP and the mean phenotypic value. Specifically, we see in the text on page 12 that CV2P = $(\frac{V_P}{\bar{P}^2})$. Solving for CVP we have:

cvp <- (sqrt(vb))/mb
print(cvp)
[1] 0.1486414

Finally, and similar to CVP, we can estimate CVG as:

cvg <- (sqrt(4))/mb
print(cvg)
[1] 0.08818342

All done with chapter 1!