Name___________________________
Microarray Unit Questionnaire
1. A “dye swap” design in microarray analysis …
a) … is designed as a biological replication, in which each sample is labeled
alternately either with red or green dye and hybridized against the array.
b) … allows the researcher to assess variation in the print quality between one array
and the next.
c) … statistically counts as one experiment, not two.
d) … is a method, with which the researcher can test the quality of the cDNA
synthesis reaction.
e) … allows for comparison of the rate of incorporation of the green
fluorophore versus the red fluorophore into each cDNA copy.
2. Microarray data are often transformed into logarithmic values …
a) … because logarithmic transformation is a mathematical way to avoid biologically
impossible “negative gene expression” values. These values can then be flagged as
technical artifacts.
b) … because logarithmic values of log2 transforms fractions smaller than 1
into negative numbers of the same absolute value for genes with the same
magnitude (up or down) “fold-change”.
c) … because tests to assess the statistical difference between treatment and
control samples will not provide a p-value, unless the data are displayed as
logarithms.
d) … because the ratio between two dyes (red/green or green/red) mathematically
cannot be displayed in non-logarithmic values.
e) … because it is the mean (as opposed to the median) that is used in the statistical
analysis.
3. The “null hypothesis” in microarray data …
a) … states that a gene has the same expression in the treatment as in the
control group.
b) … states that two genes have different expression levels in the treatment and the
control group.
c) … assumes that green and red dye in a dye swap analysis incorporate into the
same cDNAs of the probe at different rates.
d) … states that the M value for the same gene in treatment versus control groups
does not equal 0.
e) … states that the mean of all repeated measurements of the same gene as it
appears on an M/A plot lies either above or below the 0 line.
4. Assume that the statistical analysis of several repeated data points for the
same gene from several arrays results in a p-value of >0.05. This means that…
a) … the alternative hypothesis can be rejected, which means that the control and
treatment group are indeed different in their gene expression.
b) … according to the data, either the treatment or the control group does not
contain the gene in question in its DNA.
c) … the null hypothesis can be rejected, which means that the control and treatment
group are indeed different in their gene expression.
d) … the null hypothesis cannot be rejected, which means the control and
treatment group are not different from each other in their gene expression.
e) … statistically, the question whether or not the gene from the control and
treatment group are different in their expression cannot be answered.
5. If microarrays are to be evaluated statistically it is important to have at least
one replicate, …
a) … because from one single slide analysis no biological data could be obtained.
b) … but a dye swap design would not provide replicated data.
c) … which, furthermore, has to include a technical replicate, not jus a biological
replicate, if the data are to have any statistical value.
d) … because if you have only one value for each gene you cannot perform
any statistics on its value.
e) … because each replicate gives a value for either the control or the treatment
group (red or green).
6. In microarray analysis thousands of data points are analyzed resulting in
thousands of p-values after statistical analysis has been performed. This
results in “multiple testing” problems because…
a) … due to the large number of genes analyzed, the p-value for each gene is less
likely to reflect true statistical (in-)significance for the gene that it is assigned to.
b) … due to the large number of genes analyzed, the likelihood that among the
p-value assigned several are false positives is close to 100%.
c) … due to the large number of genes analyzed, the likelihood that among those pvalue are false positives is equal to or less than 5% (0.05).
d) …only one p-value is calculated to reflect the significance for all represented
genes on the array.
e) … each p-value has to be re-calculated as many times as there are genes on the
array.
7. The Bonferroni correction method…
a) … calculates p-values not by the use of t-statistics but by assigning ranks to each
gene, and dividing the rank by the total sample size.
b) … is basically a method that due to the constraints of multiple testing issues of
thousands of genes considers p-values as significant even when they are greater
than 0.05.
c) … is basically a method that due to the constraints of multiple testing
issues of thousands of genes considers p-values as significant only when they
are much smaller than 0.05.
d) … does not rely on a p-value but compares each gene to each other on the array
in pairwise comparisons to eliminate the problems of multiple testing.
e) … calculates a more stringent p-value because it can apply a greater degree of
freedom due to the thousands of replicated data points on each array.