Caballero QG Chp. 1, Problem set Q1 (R)

Chapter 1: Continuous variation

Problem 1

In the cross between two pure lines, a heterozygous hybrid was obtained for 20 biallelic loci that affect a quantitative trait. For these loci, one allele has no effect on the trait and the other increases it by one unit.

(a) How many genotypic classes would be found in the offspring by self-fertilization of the hybrid? Answering this one is straightfoward but the nomenclature of ‘genotypic’ class seems distinct to the Caballero book vs. maybe general use of the term. On page 10 we can see the basic formula for quantifying genotypic classes based upon any number of loci is 2n + 1, where n is the number of loci

n <- 20
g <- 2*n + 1
print(g)
[1] 41

(b) How often would descendents be found with a heterozygous genotype at all 20 loci? Let’s start with a Punnett square of a single locus with two alleles:

We can see from this basic table that half of the offspring will be heterozygous for any one locus. Assuming that all loci are unlinked, we would just need to multiple the expectation for one locus by the number of loci we’re focusing on. So, for all loci being heterozygous we could use the formula $(\frac{1}{2})^n$

h <- (1/2)^20
print(h)
[1] 9.536743e-07

(c) With what probability would descendents with a phenotypic value equal to 10 be found? This one would be a bit more difficult if we had to derive these values from scatch. However, we see, also on page 10, we can use a binomial expectation to determine the expected frequencies of specific genotypes/phenotypes (when we know how individual alleles contribute a phenotype). We have 20 loci, each of which can be AA, Ab, and aa, so 40 draws in total. One allele will increase the value by 1 and the other 0 (so fifty-fifty; 1/2). Finally we want exactly a value of 10, not less, not more. Fortunately we can make this calculation with the function dbinom in base R:

dbinom(10, size=40, prob=0.5) 
0.000770942759118042