In charlotte-ngs/lbgfs2020: Livestock Breeding and Genomics (Fall 2020)

# 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.

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Problem 2: Matrices in R

In R, matrices are constructed using the function matrix(). This function accepts different options. We want to see, how these options work.

Your Task: Construct matrices using the different options to better understand the meaning of the different options.

Parameter `data`

data: Specify the different matrix elements

(matA <- matrix(data = c(1:9), nrow = 3, ncol = 3))

data: without specifying the matrix elements

(matB <- matrix(nrow = 3, ncol = 3))

data: specifying not all matrix elements

(matC <- matrix(data = c(1,2,3), nrow = 3, ncol = 3))

(matC2 <- matrix(data = c(1,2,3,4), nrow = 3, ncol = 3))

Parameters `nrow` and `ncol`

Leaving out one of both parameters

(matD <- matrix(data = c(1:9), nrow = 3))

(matE <- matrix(data = c(1:9), ncol = 3))

Parameter `byrow`

(matF <- matrix(data = c(1:9), nrow = 3, ncol = 3, byrow = TRUE))

(matG <- matrix(data = c(1:9), nrow = 3, ncol = 3, byrow = FALSE))

Problem 3: Matrix multiplication in R

In R, matrices can be multiplied using the operator %*% or with the functions crossprod() or tcrossprod(). With crossprod() and tcrossprod() vectors and matrices can be multiplied directly. The conversion of vectors to matrices is done automatically inside of these functions. The result will always be a matrix. When doing matrix-vector multiplications with %*% the vector has to be converted first into a matrix using the function as.matrix().

In a first part of this problem, compare the results of the functions crossprod(), tcrossprod() and %*%.

a) Given are the following matrices

matA <- matrix(data = c(1:9), ncol = 3)
matB <- matrix(data = c(2:10), ncol = 3)

Find out which matrix multiplication with %*% corresponds to the following statement?

crossprod(matA,matB)

b) Given is the vector vecB

vecB <- c(-3,16,1)

Multiply the matrix matA with the vector vecB once using %*% and then with the function crossprod().

Hint: a vector can be converted to a matrix using the function as.matrix().

Problem 4: 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)