Question

Question 15. The genome of the bacteriophage Qb consists of about 4000 nucleotides of single-stranded RNA. Inside the E. coli host, replication of this genome requires a RNA-dependent RNA polymerase made up of phage and bacterial proteins. Interestingly, the replicase binds to a region in the center of the RNA genome, yet must start copying the 3 0 end of the RNA template to produce a new strand of RNA in the 5 0 to 3 0 direction. Researchers hypothesized that the presence of a predicted pseudoknot in the Qbgenome allowed the replicase to gain access to the 3 0 end of the RNA. To test their hypothesis, they measured the replication efficiency in vitro using wild-type replicase and different versions of the Qb RNA containing mutations in a region key to the predicted pseudoknot formation. They specifically focused on an eight-nucleotide region from the center of the RNA that was complementary to the 3 0 -terminal hairpin. The wild-type and mutant sequences and replication data are given below.

Wild type 5 0 -UAAAGCAG-3 0 GUUUCGUC
Mutant A 5 0 -UAAACGAG-3 0 GUUUCGUC

Mutant B 5 0 -UAAAGCAG-3 0 GUUUGCUC

Mutant C 5 0 -UAAACGAG-3 0 GUUUGCUC

In vitro replication efficiency (relative to wild type)

Wild type: 100%

Mutant A: 1.6%

Mutant B: 0.6%

Mutant C: 42%

A. Predict why the replication efficiency so low in Mutant A.

B. Predict why the replication efficiency is restored to almost half of wild-type levels in Mutant C?

C. Do you think the results support or refute the hypothesis that the presence of a pseudoknot affects replication?

Answer 1

ANSWER A :-

1. The type of mutation associated with the mutants is nucleotide inversion mutations with respect to the terminal hairloop and the complementary sequence in the center of the structure. (RNA)
2. The mutation is located near the stop codon and more the proximity to the stop codon, more will be the impact in terms of replication which indicates that the replication process will be affected due to mismatching although in some cases, the Frameshift mutations will be negligible. There is a pausing effect which can be seen in terms of presence of inversion mutation. If this effect is seen, the replication efficieny will be affected as can be seen in mutation A (complementary sequence located in the center) and mutation B (mutation in the hairpin). Although both the mutation lead to pausing effect as the base pairs are not read leading to inability of replicase to get access to the stop codon sequence. Hence, the replication efficiency is low as compared to that of the wild type (1.6%)

ANSWER B :-

1. In the mutant C, nucleotide inversion mutations leads to the base pairing between the complementary strand sequence and the hairpin sequence to be read thereby initiating the replicase activity to terminate towards the stop codon.

5'- UAAACGAG -3'

3'- GUUUGCUC -5'

• This forms the reading frame although there is shift in the position of the Guanine and Cytosine residues at the 5th and the 6th nucleotide respectively which alters the mechanism and reduces the efficiency to half of the original wild-type. So, this is the way in which the replication efficieny is stated to be reduced.
• It can also be stated that Mutants A and B do not show presence of complementary binding in order to initiate replication as seen in mutant C where the potential is revived mainly due to possibility of having both the mutant sequences from the Mutant A and Mutant B.

ANSWER C :-

1. Although this mechanism does not illustrate quite enriched details about the role of pseodoknoy in initiating replication by binding of replicase, but it provides insights about the sequences associated with the pseodoknot and the role of mutations in reducing the replication efficiency as can be stated to have been the output in most of the cases.

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