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

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

Problem 1: Vectors

Given are the following two vectors $v$ and $w$. Compute

the sum $v+w$,
the difference $v-w$ and
the dot product $v\cdot w$.

\begin{center} $v = \left[\begin{array}{c} 3 \ -5 \ 1 \ 9 \ \end{array}\right]$, \hspace{2ex} $w = \left[\begin{array}{c} 1 \ 9 \ -12 \ 27 \ \end{array}\right]$ \end{center}

Problem 2: Vectors in R

Verify your computations from Problem 1 in R. Start by assigning the vectors $v$ and $w$ with the components given in Problem 1. If you are not sure how to assign vectors in R, please have a look at https://bookdown.org/rdpeng/rprogdatascience/r-nuts-and-bolts.html#creating-vectors and its the corresponding video at https://youtu.be/w8_XdYI3reU

Problem 3: Angle between Vectors

Given are two vectors

\begin{center} $a = \left[\begin{array}{c} 8 \ 0 \ 4 \ \end{array}\right]$ and
$b = \left[\begin{array}{c} 2 \ 21 \ x \ \end{array}\right]$ \end{center}

How do we have to choose $x$, such that the vectors are perpendicular? Verify your solution with R

Problem 4: Phenotypes and Genotypes

# milk yield of Delilah
vec_lact_perf_Delilah <- c(5852, 6833, 7984, 7869, 7322, 8216, 8622, 7851)

# milk yield of Rosy
vec_lact_perf_Rosy <- c(6249, 7312)

# breeding values of Delilah
nLactDelilah <- length(vec_lact_perf_Delilah)
l_ebv_Delilah <- list(Milk = -1037, ND = 112)

# breeding values of Rosy
nLactRosy <- length(vec_lact_perf_Rosy)
l_ebv_Rosy <- list(Milk = 471, ND = 122)

Farmer Frank Miller has two cows named Delilah and Rosy. Delilah has completed r nLactDelilah lactations with the following results

knitr::kable(data.frame(Lactation = 1:nLactDelilah, `Milk Yield` = vec_lact_perf_Delilah))

Rosy is a young cow and has completed just r nLactRosy lactation which is shown below.

knitr::kable(data.frame(Lactation = 1:nLactRosy, `Milk Yield` = vec_lact_perf_Rosy))

Your Tasks

a. Compute for both cows the sum and the mean of all lactation results using R. Hint: Have a look at the functions mean() and sum() in R. b. Our farmer wants to know which of the two cows would be a better mother for his breeding herd when looking at the traits milk yield and longevity. The trait longevity is defined as the number of years a cow is able to produce milk. From the breeding association the farmer receives the following predicted breeding values for the two cows. Please explain which of the two cows is the better choice as a mother.

knitr::kable(data.frame(Cows = c("Delilah", "Rosy"), `Milk Yield` = c(l_ebv_Delilah$Milk, l_ebv_Rosy$Milk), Longevity = c(l_ebv_Delilah$ND, l_ebv_Rosy$ND)))