1-DAV-202 Data Management 2023/24
Previously 2-INF-185 Data Source Integration

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· Dates of project submission and oral exams:
Early: submit project May 24 9:00am, oral exams May 27 1:00pm (limit 5 students).
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Sign up for one the exam days in AIS before June 11.
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· Cloud homework is due on May 20 9:00am.


HWr2

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See also the current and the 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

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 have seen 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

Read the microarray data, and preprocess them (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 how many genes were significant with p-value at most 0.01 for each acid
  • Report 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-vales from tasks B for using Bonferroni correction
pb.adjusted = p.adjust(pb, method ="bonferroni")
# add this adjusted vector to frame a
a <-  cbind(a, 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, acetate 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)