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!
Written on April 3rd , 2021 by P. Fields