**1-DAV-202 Data Management 2023/24**

Previously 2-INF-185 Data Source Integration

# HWr2

See also the current and the previous lecture.

- Do either tasks A,B,C (beginners) or B,C,D (more advanced). You can also do all four for bonus credit.
- In your protocol write used R commands with brief comments on your approach.
- Submit required plots with filenames as specified.
- For each task also include results as required and a short discussion commenting the results/plots you have obtained. Is the value of interest increasing or decreasing with some parameter? Are the results as expected or surprising?
- Outline of protocol is in
`/tasks/r2/protocol.txt`

## Contents

### Task A: sign test

- Consider a situation in which players played
*n*games, out of which a fraction of*q*were won by*A*(the example in the lecture corresponds to*q=0.6*and*n=10*). - Compute a table of p-values for combinations of
*n=10,20,...,90,100*and*q=0.6, 0.7, 0.8, 0.9*. - Plot the table using
`matplot`(*n*is x-axis, one line for each value of*q*). **Submit**the plot in`sign.png`.**Discuss**the values you see in the plot / table.

Outline of the code:

```
# create vector rows with values 10,20,...,100
rows = (1:10) * 10
# create vector columns with required values of q
columns = c(0.6, 0.7, 0.8, 0.9)
# create empty matrix of pvalues
pvalues = matrix(0, length(rows), length(columns))
# TODO: fill in matrix pvalues using binom.test
# set names of rows and columns
rownames(pvalues) = rows
colnames(pvalues) = columns
# careful: pvalues[10,] is now 10th row, i.e. value for n=100,
# pvalues["10",] is the first row, i.e. value for n=10
# check that for n=10 and q=0.6 you get p-value 0.3769531
pvalues["10","0.6"]
# create x-axis matrix (as in HWr1, part D)
x=matrix(rep(rows, length(columns)), nrow=length(rows))
# matplot command
png("sign.png")
matplot(x, pvalues, type="l", col=c(1:length(columns)), lty=1)
legend("topright", legend=columns, col=c(1:length(columns)), lty=1)
dev.off()
```

### Task B: Welch's t-test on microarray data

Run the code below which reads the microarray data and preprocesses them (the last time we worked with preprocessed data). We first transform it to log scale and then shift and scale values in each column so that median is 0 and sum of squares of values is 1. This makes values more comparable between experiments; in practice more elaborate normalization is often performed. In the rest, work with table *a* containing preprocessed data.

```
# read the input file
input = read.table("/tasks/r2/acids.tsv", header=TRUE, row.names=1)
# take logarithm of all the values in the table
input = log2(input)
# compute median of each column
med = apply(input, 2, median)
# shift and scale values
a = scale(input, center=med)
```

Columns 1,2,3 are control, columns 4,5,6 acetic acid, 7,8,9 benzoate, 10,11,12 propionate, and 13,14,15 sorbate.

Write a function `my.test` which will take as arguments table *a* and 2 lists of columns (e.g. 1:3 and 4:6) and will run for each row of the table Welch's t-test of the first set of columns versus the second set. It will return the resulting vector of p-values, one for each gene.

- For example by calling
`pb <- my.test(a, 1:3, 7:9)`we will compute p-values for differences between control and benzoate (computation may take some time) - The first 5 values of
`pb`should be

> pb[1:5] [1] 0.02358974 0.05503082 0.15354833 0.68060345 0.04637482

- Run the test for all four acids.
**Report**for each acid how many genes were significant with p-value at most 0.01.- See Vector arithmetic in HWr1
- You can count
`TRUE`items in a vector of booleans by`sum`, e.g.`sum(TRUE,FALSE,TRUE)`is 2.

**Report**also how many genes are significant for both acetic and benzoate acids simultaneously (logical "and" is written as`&`).

### Task C: multiple testing correction

Run the following snippet of code, which works on the vector of p-values `pb` obtained for benzoate in task B.

```
# adjusts vectors of p-values from tasks B for using Bonferroni correction
pb.adjusted = p.adjust(pb, method ="bonferroni")
# add both p-value columns pb and pb.adjusted to frame a
a <- cbind(a, pb, pb.adjusted)
# create permutation ordered by pb.adjusted
ob = order(pb.adjusted)
# select from table five rows with the lowest pb.adjusted (using vector ob)
# and display columns containing control, benzoate and both original and adjusted p-value
a[ob[1:5],c(1:3,7:9,16,17)]
```

You should get an output like this:

control1 control2 control3 benzoate1 benzoate2 benzoate3 PTC4 0.5391444 0.5793445 0.5597744 0.2543546 0.2539317 0.2202997 GDH3 0.2480624 0.2373752 0.1911501 -0.3697303 -0.2982495 -0.3616723 AGA2 0.6735964 0.7860222 0.7222314 1.4807101 1.4885581 1.3976753 CWP2 1.4723713 1.4582596 1.3802390 2.3759288 2.2504247 2.2710695 LSP1 0.7668296 0.8336119 0.7643181 1.3295121 1.2744859 1.2986457 pb pb.adjusted PTC4 4.054985e-05 0.2594379 GDH3 5.967727e-05 0.3818152 AGA2 8.244790e-05 0.5275016 CWP2 1.041416e-04 0.6662979 LSP1 1.095217e-04 0.7007201

Do the same procedure for acetate p-values and **report** the result (in your table, report both p-values and expression levels for acetate, not bezoate). **Comment** on the results for both acids.

### Task D: volcano plot, test on data generated from null hypothesis

Draw a volcano plot for the acetate data.

- The x-axis of this plot is the difference between the mean of control and the mean of acetate. You can compute row means of a matrix by
`rowMeans`. - The y-axis is -log10 of the p-value (use the p-values before multiple testing correction).
- You can quickly see the genes that have low p-values (high on y-axis) and also big difference in the mean expression between the two conditions (far from 0 on x-axis). You can also see if acetate increases or decreases the expression of these genes.

Now create a simulated dataset sharing some features of the real data but observing the null hypothesis that the mean of control and acetate are the same for each gene.

- Compute vector
*m*of means for columns 1:6 from matrix*a* - Compute vectors
*sc*and*sa*of standard deviations for control columns and for acetate columns respectively. You can compute standard deviation for each row of a matrix by`apply(some.matrix, 1, sd)`. - For each i in 1:6398, create three samples from the normal distribution with mean
`m[i]`and standard deviation`sc[i]`and three samples with mean`m[i]`and deviation`sa[i]`(use the`rnorm`function). - On the resulting matrix, apply Welch's t-test and draw the volcano plot.
- How many random genes have p-value at most 0.01? Is it roughly what we would expect under the null hypothesis?

Draw a histogram of p-values from the real data (control vs acetate) and from the random data (use function `hist`). **Describe** how they differ. Is it what you would expect?

**Submit** plots `volcano-real.png`, `volcano-random.png`, `hist-real.png`, `hist-random.png`
(real for real expression data and random for generated data).