#Question id: 4096
#Unit 3. Fundamental Processes
The lambda repressor binds as a dimer to critical sites on the bacteriophage lambda genome to keep the lytic genes turned off, which allows the bacteriophage lambda genome to be maintained as a silent resident in the bacterial genome. Each molecule of the repressor consists of an N-terminal DNA-binding domain and a C-terminal dimerization domain. Upon induction (for example, by irradiation with ultraviolet light), the genes for lytic growth are expressed, bacteriophage lambda progeny are produced, and the bacterial cell lyses to release the viral progeny. Induction is initiated by cleavage of the lambda repressor at a site between the DNA-binding domain and the dimerization domain. In the absence of bound repressor, RNA polymerase initiates transcription of the lytic genes, triggering lytic growth.
A. Binding as monomers will be sufficiently weak that they do not compete with the binding of RNA polymerase. As a result, the genes for lytic growth will be turned on.
B. Binding as monomers will be sufficiently strong that they will compete with the binding of RNA polymerase. As a result, the genes for lysogenic growth will be turned on
C. The affinity of the dimeric lambda repressor for its binding site is the sum of all the interactions made by each DNA-binding domain. An individual DNA-binding domain will make just half the contacts and provide just half the binding energy as the dimer.
D. The affinity of the dimeric lambda repressor for its binding site is the sum of all the interactions made by each DNA-binding domain. An individual DNA-binding domain will make just double the contacts and provide just double the binding energy as the dimer.
Given that the number (concentration) of DNA-binding domains is unchanged by cleavage of the repressor, which of above outcomes will be possible?
#Question id: 4097
#Unit 3. Fundamental Processes
The 3' end of most eukaryotic mRNAs is defined by the addition of a polyA tail - a processing reaction called polyadenylation. Choose correct set of proteins involved in the polyadenylation
#Question id: 4098
#Unit 3. Fundamental Processes
A merodiploid with a mutant repressor gene (Is) is shown below. The mutant repressor therefore binds irreversibly to both operators and will render
#Question id: 4099
#Unit 3. Fundamental Processes
Five E. coli strains have been identified, each of which has a different mutation that disrupts the normal regulation of a particular operon. For each mutant strain, the mutation has been mapped to the promoter or the operator region; however, the exact sequence changes are not known for these mutations. It is known that the normal promoter/operator consists of a single binding site for a positively acting transcription factor located just upstream of the promoter itself. Short DNA fragments containing the promoter and the operator were subcloned from each of the five mutant strains and from the wild type, purified, and radiolabeled. These fragments were then incubated under conditions of DNA excess with either purified regulatory factor or RNA polymerase or with both polymerase and regulatory factor.
The resulting protein-DNA complexes were separated by electrophoresis, and the radioactive DNA fragments were detected by exposure to x-ray film, giving the results shown below. Electrophoresis is from top to bottom; the largest complexes run slowest.
Based on above experiment, match the following
|
Effect |
Mutant |
|
1. One of the mutations increases the affinity of the polymerase for the promoter. Transcription of the operon is not stimulated by the regulatory factor in this mutant. |
A. Mutant 5 |
|
2. One of the mutations maps to the operator. Transcription of the operon is not stimulated by the regulatory factor in this mutant. Which mutant is most likely to show this effect |
B. Mutant 2 |
|
3. One of the mutations is known to result from a small deletion between the operator and the promoter. The polymerase and the regulatory factor is each able to bind to the mutated DNA sequence, but are unable to form the three components complex. Transcription of the operon is not stimulated by the regulatory factor in this mutant. Which mutant shows the properties that might be expected for such a change? |
C. Mutant 4 |
|
D. Mutant 3 |
|
|
E. Mutant 1 |
#Question id: 4100
#Unit 3. Fundamental Processes
some pre-mRNAs are spliced by an alternative, low-abundance form of the spliceosome. This rare form contains some components common to the major spliceosome, but it contains other unique components such as
#Question id: 4101
#Unit 3. Fundamental Processes
In order to ensure that only fully processed mature mRNAs are allowed to be exported to cytosol, pre-mRNAs associated with snRNPs are retained in the nucleus. To demonstrate this, an experiment was performed where a gene coding a pre-mRNA with a single intron was mutated either at the 5’ or 3’ splice sites or both the splice sites.
Given below are a few possible outcomes, Choose NOT FEASIBLE OUTCOME
A. Pre-mRNA having mutation at both the splice sites will be retained in the nucleus because of the presence of bound snRNPs.
B. Pre-mRNA having mutation at both the splice sites will be exported to cytosol because of the absence of bound snRNPs.
C. Pre-mRNA mutated at either 3’ or 5’ splice sites will be retained in the nucleus because of the presence of bound snRNPs.
D. Pre-mRNA mutated at either 3’ or 5’ splice sites will be exported to cytosol because of the absence of bound snRNPs.