In charlotte-ngs/lbgfs2021: Livestock Breeding and Genomics (ETHZ) - Fall Semester 2021

# knitr::opts_chunk$set(echo = FALSE, results = 'asis')

Problem 1: Breeding Values For a Monogenic Trait

We assume that the absorption of cholesterol is determined by a certain enzyme. The level of enzyme production is determined by a single bi-allelic locus $E$. The genotype frequencies and the genotypic values for the two dairy cattle populations Original Braunvieh and Brown Swiss are given in the following table.

l_maf <- list(ob = 0.25, bs = 0.1)
l_a <- list(ob = 15, bs = 29)
l_d <- list(ob = 3, bs = 0)
tbl_geno_freq_value <- tibble::tibble(Variable = c(
  "$f(E_1E_1)$",
  "$f(E_1E_2)$",
  "$f(E_2E_2)$",
  "$a$",
  "$d$"),
  `Original Braunvieh` = c(
    l_maf$ob^2, 
    2*l_maf$ob*(1-l_maf$ob),
    (1-l_maf$ob)^2,
    l_a$ob,
    l_d$ob
  ),
  `Brown Swiss` = c(
    l_maf$bs^2,
    2*l_maf$bs * (1-l_maf$bs),
    (1-l_maf$bs),
    l_a$bs,
    l_d$bs
  ))
knitr::kable(tbl_geno_freq_value, booktabs = TRUE, escape = FALSE)

Hints

• Assume that allele $E_1$ is the allele with the positive effect on the enzyme level
• Assume that the Hardy-Weinberg Equilibrium holds in both populations

Your Task

Compute the breeding values for all three genotypes in both populations.

Your Solution

Problem 2: Quantitative Genetics

In a population the following numbers of genotypes were counted for a given genetic locus called $A$.

dfGenotypeFreq <- data.frame(Genotypes = c("$A_1A_1$", "$A_1A_2$", "$A_2A_2$"),
                             Numbers   = c(24, 53, 23),
                             stringsAsFactors = FALSE)
knitr::kable(dfGenotypeFreq)

a) Compute the genotype frequencies

b) Compute the allele frequencies

c) Compute the population mean $\mu$ under the following assumptions

• the difference between the genotypic values of the homozygous genotypes is $20$ and
• the genotypic value of the heterozygous genotype is $2$.

Your Solution

charlotte-ngs/lbgfs2021 documentation built on Dec. 19, 2021, 3:01 p.m.