Question 10
Which of the following is not in common to bothprokaryotic and eukaryotic promoters?
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A.
They have consensus sequences that proteins bind inorder to help RNA polymerase assemble in a transcriptionalcomplex.
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B.
They bind proteins that assist RNA polymeraseassembly into the transcriptional complex.
Question 12
In the process of intrinsic transcription termination,what RNA structural features are important to theprocess?
A.
The stem loop that forms because of complementarity between basesmaking up two inverted repeat sequences in the terminationsequence.
B.
The C terminal domain of RNA pol II serves as a scaffold upon whichthe cleavage and polyadenylation complex resides, thereforeenabling efficient transcription termination.
C.
A specific sequence of bases at the 3' end of an mRNA is able tocatalyze its own excision, thereby terminating transcription.
D.
The rut sequence in the mRNA that forms a structure that binds Rhoprotein, which causes mRNA release and transcriptiontermination.
2 points Â
Question 13
Imagine that you do this experiment: you create asynthetic gene, based on what you know about the insulin gene inhumans (that is, what you know about the nature of genes in humans,i.e. eukaryotes). You create a strain of E. coli in which thissynthetic gene has been inserted into the bacterial chromosome.After checking for expression of your synthetic gene, you discoverthat no mRNA is produced. Which of the following is a possibleexplanation for this observation?
A.
The gene is lacking the Shine-Delgarno sequence, therefore RNApolymerase won't be able to assemble on the promoter with any ofthe sigma subunits.
B.
The synthetic gene has a eukaryotic promoter that will not berecognized by any of the E. coli sigma-70 subunits, thus RNApolymerase will not bind to the promoter.
C.
The synthetic gene has a prokaryotic promoter that is missing theTATA box or GC box, thus no sigma subunit can bind and recruit RNApolymerase.
D.
TFIID and the other general transcription factors are not able torecognize a prokaryotic promoter, therefore they cannot assemble onthe promoter of the synthetic gene in a bacterial cell.
2 points Â
Question 14
You are studying the expression of a gene in a certaintype of eukaryotic cell. Under normal conditions, you measure thatthe gene is expressed at a certain value. You are able toexperimentally alter a nucleotide pair at position -75 relative tothe transcription start site. In the altered cells, you now measurea reduced rate of gene expression, 50% the value that you measuredfrom normal cells. What is a reasonable interpretation orhypothesis to explain this observation?
A.
You likely altered a base located in a consensus sequence of thepromoter, and the general transcription factors have a loweraffinity for the mutated sequence, therefore transcriptioninitiation complexes form at a reduced rate.
B.
The mutation most likely altered a consensus sequence for sigmasubunit binding, therefore the appropriate sigma subunit is lesslikely to bind to the sequence in the promoter to recruit RNApolII.
C.
The mutation likely created an improved binding site for TFIID andthe other general transcription factors, thus they are less likelyto disassemble to form a new pre-initiation complex.
D.
You likely altered a base in a sequence that RNA polymerase IIrequires for binding, therefore it is less able to recruit generaltranscription factors to the promoter to assemble thepre-initiation complex.
