Question

Question 10

Which of the following is not in common to both prokaryotic and eukaryotic promoters?

 

 

A.

They have consensus sequences that proteins bind in order to help RNA polymerase assemble in a transcriptional complex.

 

B.

They bind proteins that assist RNA polymerase assembly into the transcriptional complex.

Question 12

In the process of intrinsic transcription termination, what RNA structural features are important to the process?

A.
The stem loop that forms because of complementarity between bases making up two inverted repeat sequences in the termination sequence.

B.
The C terminal domain of RNA pol II serves as a scaffold upon which the cleavage and polyadenylation complex resides, therefore enabling efficient transcription termination.

C.
A specific sequence of bases at the 3' end of an mRNA is able to catalyze its own excision, thereby terminating transcription.

D.
The rut sequence in the mRNA that forms a structure that binds Rho protein, which causes mRNA release and transcription termination.

2 points  

Question 13

Imagine that you do this experiment: you create a synthetic gene, based on what you know about the insulin gene in humans (that is, what you know about the nature of genes in humans, i.e. eukaryotes). You create a strain of E. coli in which this synthetic gene has been inserted into the bacterial chromosome. After checking for expression of your synthetic gene, you discover that no mRNA is produced. Which of the following is a possible explanation for this observation?

A.
The gene is lacking the Shine-Delgarno sequence, therefore RNA polymerase won't be able to assemble on the promoter with any of the sigma subunits.

B.
The synthetic gene has a eukaryotic promoter that will not be recognized by any of the E. coli sigma-70 subunits, thus RNA polymerase will not bind to the promoter.

C.
The synthetic gene has a prokaryotic promoter that is missing the TATA box or GC box, thus no sigma subunit can bind and recruit RNA polymerase.

D.
TFIID and the other general transcription factors are not able to recognize a prokaryotic promoter, therefore they cannot assemble on the promoter of the synthetic gene in a bacterial cell.

2 points  

Question 14

You are studying the expression of a gene in a certain type of eukaryotic cell. Under normal conditions, you measure that the gene is expressed at a certain value. You are able to experimentally alter a nucleotide pair at position -75 relative to the transcription start site. In the altered cells, you now measure a reduced rate of gene expression, 50% the value that you measured from normal cells. What is a reasonable interpretation or hypothesis to explain this observation?

A.
You likely altered a base located in a consensus sequence of the promoter, and the general transcription factors have a lower affinity for the mutated sequence, therefore transcription initiation complexes form at a reduced rate.

B.
The mutation most likely altered a consensus sequence for sigma subunit binding, therefore the appropriate sigma subunit is less likely to bind to the sequence in the promoter to recruit RNA polII.

C.
The mutation likely created an improved binding site for TFIID and the other general transcription factors, thus they are less likely to disassemble to form a new pre-initiation complex.

D.
You likely altered a base in a sequence that RNA polymerase II requires for binding, therefore it is less able to recruit general transcription factors to the promoter to assemble the pre-initiation complex.

Answer 1

Answer:

Q-10): (" They have a -10 and -35 consensus sequence to which proteins bind that assist RNA polymerase in assembling the transcription initiation complex. ")

Because -10 and -30 consensus sequences are present only in prokaryotic promoter.

Q-12):"A" Correct:

• In intrinsic termination or Rho independent termination mRNA at 3 end make a stem loop structure which is generally 7-20 base pair long and is rich in GC sequence which forms 3 hydrogen bond (strongest).
• This hair pin loop is followed by multiple uracil residues which are attached to adenine on DNA.
• The bonding between uracil and adenine is weakest and got broken due to the formation of hairpin loop structure which has multiple GC.

Q-13):"B" Correct:

• The sunthetic gene has the eukaryotic promoter that will not be recongnized by any of the E.coli sigma -70 subunuts, thus RNA polymerase will not bind to the promoter.

Q-14):"D" Correct:

• you likely altered a base located in a consensus sequence of the promoter and the general transcription factors have a lower affinity for the mutated sequence, therefore transcription initiation complexes form at a reduced site

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