#Question id: 4091
#Unit 3. Fundamental Processes
To avoid being degraded by nuclear exonucleases, nascent transcripts, pre-mRNA-processing intermediates, and mature mRNAs in the nucleus must have their ends protected. Which of the following mechanism are present in the cell to protect mRNA from degradation?
I. The 5′ cap is protected because it is bound by a heterodimeric nuclear cap-binding complex (CBC), which protects it from 5′ exonucleases and also functions in export of the mRNA to the cytoplasm.
II. 5’-5’ phosphodiester bond in the cap of mRNA plays crucial role in protection
III. The 3′ end of a nascent transcript lies within the RNA polymerase and is thus inaccessible to exonucleases
#Question id: 4092
#Unit 3. Fundamental Processes
An experimental setup provided the assay for identifying factors that facilitate transcription in the presence of chromatin. A factor called FACT (facilitates chromatin transcription) functions as
#Question id: 4093
#Unit 3. Fundamental Processes
Which of the following processes is NOT an example of allosteric regulation?
#Question id: 4094
#Unit 3. Fundamental Processes
The accompanying figure depicts the map of a part of E. coli genome harboring the lac operon.
lac mRNA would hybridize to which of the following DNA?
1) lac I; 2) lac 0; 3) lac Z; 4) lac Y; 5) lac A
#Question id: 4095
#Unit 3. Fundamental Processes
Many eukaryotic genes contain a large number of exons. Correct splicing of such genes requires that neighboring exons be ligated to one another; if they are not, exons will be left out. One early proposal suggested that the splicing machinery bound to a splice site at one end of an intron and scanned through the intron to find the splice site at the other end. Such a scanning mechanism would guarantee that an exon was never skipped. This hypothesis was tested with one minigene with a duplicated 3ʹ splice site. Find diagram of the products you expect from minigene if the splicing machinery binds to a 3ʹ splice site and scans toward a 5ʹ splice site.
#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?