Bioinformatics Questions and Answers – Global Gene Regulation

This set of Bioinformatics Multiple Choice Questions & Answers (MCQs) focuses on “Global Gene Regulation”.

1. Which of the given statements is incorrect about global gene regulation?
a) One way to obtain useful information about a genome is to determine which genes are induced or repressed in response to a phase of the cell cycle
b) Sets of a gene whose expression rises and falls under the same condition are likely to have a related function
c) Sets of a gene whose expression rises and falls under the same condition are likely to have dissimilar functions
d) Cell cycle is a developmental phase, or a response to the environment
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2. Which of the given statements is incorrect about Microarray (or microchip) analysis?
a) It is a new technology in which all of the genes of an organism are represented by oligonucleotide sequences spread out in an 80 x 80 array on microscope slides
b) The oligonucleotide sequences cannot be synthesized directly on the slide
c) The oligonucleotides are collectively hybridized to a labeled cDNA library prepared by reverse-transcribing mRNA from cells
d) The amount of label binding to each oligonucleotide spot reflects the amount of mRNA in the cell
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3. Once a set of genes that are co-regulated has been found, the promoter regions of these genes may be analyzed for conserved patterns that represent sites of interaction with specific transcription factors.
a) True
b) False
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4. Which of the given statements is incorrect about Microarray Analysis?
a) It is designed to detect global changes in transcription in a genome
b) It provides information about the levels of protein products of the genes
c) The proteins are first separated in a column on the basis of size and then across a second dimension on a slab on the basis of charge
d) Labeled protein samples may also be extracted from treated cells and separated by two-dimensional gel electrophoresis
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5. In cluster analysis of microarray data– If X_i is the log odds value for gene X at time i, then for two genes X and Y and N observations, a similarity score is calculated. S(X,Y) is also known as the Pearson correlation coefficent. X_offset and Y_offset can be the mean of the observations on X or Y, respectively, in which case is the standard deviation, or else X_offset and Y_offset can be set to zero when a reference state is used. Which of the following best represents it?
a) S(X,Y) = \(\frac{1}{N-2}\) ∑_i=1,N . (X_i – X_offset) (Y_i + Y_offset)/ϕ_xQ_Y
b) S(X,Y) = \(\frac{1}{N}\) ∑_i=1,N . (X_i – X_offset) (Y_i – Y_offset)/ϕ_xQ_Y
c) S(X,Y) = \(\frac{1}{N-1}\) ∑_i=1,N . (X_i + X_offset) (Y_i + Y_offset)/ϕ_xQ_Y
d) S(X,Y) = \(\frac{1}{N}\) ∑_i=1,N+2 . (X_i + X_offset) (Y_i – Y_offset)/ϕ_xQ_Y
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6. In cluster analysis of microarray data– A node is created between the _________ scoring pair, and the gene expressed profiles of these two genes are averaged and the joined elements are weighted by the _________ of elements they contain.
a) lowest, frequency
b) average, sequence
c) lowest, number
d) highest, number
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7. In cluster analysis of microarray data– For n genes, the process is repeated ________ times until a single element remains.
a) n²
b) n
c) n^-1
d) n^-4
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8. The hierarchical clustering method generates a similarity score [S(X,Y)] for all gene combinations, places the scores in a matrix, joins those genes that have the highest score, and then continues to join progressively less similar pairs.
a) True
b) False
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9. In Self-organizing maps a choice is made of a number of clusters by which to organize the data.
a) True
b) False
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10. SVMs (Support vector machines) are a binary classification method to discriminate one set of data points from another.
a) True
b) False
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11. In SVMs (Support vector machines) Data points are log-transformed and normalized as in method A, where for N observations of a gene i, the log transform X_i of the expression level E_i and reference level R_i is?
a) X_i = \(\frac{Log (E_i/R_i)}{\sqrt{\sum_{j=1,N} Log_{z-2} (E_j/R_j)}}\)
b) X_j = \(\frac{Log (E_j/R_i)}{\sqrt{\sum_{j=1,N} Log_z (E_j/R_j)}}\)
c) X_i = \(\frac{Log (E_i/R_i)}{\sqrt{\sum_{j=1,(N-1)} Log_z (E_j/R_j)}}\)
d) X_i = \(\frac{Log (E_i/R_i)}{\sqrt{\sum_{j=1,N} Log_z (E_j/R_j)}}\)
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Sanfoundry Global Education & Learning Series – Bioinformatics.

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