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.

Solution

The breeding values are computed as shown in the following table.

tbl_bv <- tibble::tibble(Genotype = c(
   "$E_1E_1$",
  "$E_1E_2$",
  "$E_2E_2$"),
  `Breeding Value` = c(
    "$BV_{11} = 2q\\alpha$",
    "$BV_{12} = (q-p)\\alpha$",
    "$BV_{22} = -2p\\alpha$"
  ))
knitr::kable(tbl_bv, booktabs = TRUE, escape = FALSE)

with $\alpha = a + (q-p)d$. The values for $a$ and $d$ are given in the task and the allele frequencies $p$ and $q$ can be computed from the given genotype frequencies.

$$p = f(E_1) = f(E_1E1) + {1\over 2}f(E_1E_2)$$ and $q = 1-p$

For the two populations we get

l_alpha <- list(ob = l_a$ob + (1-2*l_maf$ob)*l_d$ob,
                bs = l_a$bs + (1-2*l_maf$bs)*l_d$bs)
tbl_allele_freq_alpha <- tibble::tibble(Variable = c(
  "$p$",
  "$q$",
  "$\\alpha$"),
  `Original Braunvieh` = c(
    l_maf$ob,
    1-l_maf$ob,
    l_alpha$ob),
  `Brown Swiss` =  c(
    l_maf$bs,
    1-l_maf$bs,
    l_alpha$bs))
knitr::kable(tbl_allele_freq_alpha, booktabs = TRUE, escape = FALSE)

The breeding values for the two breeds are given in the following table

tbl_table_bv_result <- tibble::tibble(Genotype = c(
   "$E_1E_1$",
  "$E_1E_2$",
  "$E_2E_2$"),
  `Breeding Value` = c(
    "$BV_{11}$",
    "$BV_{12}$",
    "$BV_{22}$"
  ),
  `Original Braunvieh` = c(
    2 * (1-l_maf$ob) * l_alpha$ob,
    (1-2*l_maf$ob) * l_alpha$ob,
    -2 * l_maf$ob * l_alpha$ob
  ),
  `Brown Swiss` =  c(
    2 * (1-l_maf$bs) * l_alpha$bs,
    (1-2*l_maf$bs) * l_alpha$bs,
    -2 * l_maf$bs  * l_alpha$bs))
knitr::kable(tbl_table_bv_result, booktabs = TRUE, escape = FALSE)

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))

\pagebreak

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)

Solution

The statement crossprod(matA,matB) corresponds to

t(matA) %*% matB

Alternatively there is the function tcrossprod(). Find out which matrix multiplication is executed by

tcrossprod(matA, matB)

Solution

matA %*% t(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().

Solution

matA %*% as.matrix(vecB)

crossprod(t(matA), vecB)

\pagebreak

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)

a) Compute the genotype frequencies

Solution

nTotNrInd <- sum(dfGenotypeFreq$Numbers)
vGenoTypeFreq <- dfGenotypeFreq$Numbers / nTotNrInd
cat(paste("genotype-frequency", dfGenotypeFreq$Genotypes[1]), ": ", vGenoTypeFreq[1])
cat(paste("genotype-frequency", dfGenotypeFreq$Genotypes[2]), ": ", vGenoTypeFreq[2])
cat(paste("genotype-frequency", dfGenotypeFreq$Genotypes[3]), ": ", vGenoTypeFreq[3])

b) Compute the allele frequencies

Solution

vGenFreqP <- vGenoTypeFreq[1] + 0.5*vGenoTypeFreq[2]
vGenFreqQ <-  vGenoTypeFreq[3] + 0.5*vGenoTypeFreq[2]
cat("allele frequency for A1: ", vGenFreqP)
cat("allele frequency for A2: ", vGenFreqQ)

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 $2$.

Solution

nDeltaHom <- 20
### # additive value A
nAddValue <- nDeltaHom / 2
nDom <- 2
### # population mean
nMu <- (vGenFreqP-vGenFreqQ) * nAddValue + 2 * vGenFreqP * vGenFreqQ * nDom
cat("Population mean: ", nMu, "\n")

charlotte-ngs/lbgfs2020 documentation built on Dec. 20, 2020, 5:39 p.m.