1. What form of helix would you expect to see in double stranded RNA?
a) A form
b) B form
c) C form
d) D form
Explanation: Double stranded RNA generally forms A form helix while double stranded DNA forms B form helix. The other forms also occur in DNA under special conditions.
2. A ssRNA shows inverted tandem repeat. What would be its secondary structure?
a) Linear
b) Hair pin
c) Stem-loop structure
d) Crucifix
Explanation: A tandem repeat has some bases between the palindrome regions resulting in the formation of the loop. The remaining complementary sequences of the palindrome forms the stem. In case of dsRNA similar sequence would form a crucifix while absence of intersecting sequence will form hairpin.
3. To study the functioning of intrinsic terminator you increase the number of G-C base pairs in the stem and decrease the U bases from the tail region. What would be its effect on termination?
a) No visible effect
b) Efficiency will decrease
c) Efficiency will increase
d) The RNA polymerase will stall at the terminator
Explanation: The increase in number of G-C base pairs in the stem region stabilizes the stem, and decreasing in number of U in the tail destabilizes it. These two have a compensating effect on terminator efficiency.
4. Which of these interactions is the weakest?
a) rC-dG
b) rG-dC
c) rA-dT
d) rU-dA
Explanation: The interaction between A and T or U base is through 2 hydrogen bonds only. So this is weaker than G-C which mediates by 3 hydrogen bonds. Further RNA DNA interactions are weaker making rU-dA the weakest of interaction which is used in termination of transcription.
5. You modify your gene DNA sequence by introducing an attenuator within the transcribed region. You isolate the labeled gene product and perform SDS PAGE. You observe two bands one higher and another slightly lower. Which of this band corresponds to properly modified sequence?
a) Higher
b) Lower
c) Both the bands as they show differently folded protein
d) This is an experimental artifact
Explanation: An attenuator will lead to premature transcription termination. This will result in a truncated protein that will migrate faster in the gel and form the lower band. Protein folding doesn’t affect its migration in SDS page.
6. Rho protein, that is necessary for transcription termination, is a ________
a) Homotetramer
b) Heterotetramer
c) Heterohexamer
d) Homohexamer
Explanation: Rho is a 46KDa hexameric protein with 6 identical subunits. Thus, it is a holohexamer. A four subunit protein with all identical subunit would be a holotetramer, not identical subunit would be heterotetramer respectively.
7. Rho shows base preference although it doesn’t have any concensus cis binding sequence. Which region will be expected to bind Rho better?
a) G and C rich region
b) A and T rich region
c) Region rich in C and poor in G
d) Region rich in A and poor in T
Explanation: Rho binds best to regions having high C content as much as 41% and low G content as much as 14%. As in RNA there is no compulsory base pairing the amount of G is not necessary equal to the amount of c.
8. Rho dependent termination mechanism doesn’t reguire___________
a) ATP
b) Stem loop structure
c) G-C rich stem
d) Sigma factor
Explanation: Sigma factor plays no role in termination, rather it dissociates long before the RNA pol terminates. Rho however requires ATP for its rolling function and the stem loop structure formation. This stem loop structure is strengthened by G-C rich regions in the stem.
9. When Rho catches up with RNA polymerase it _______
a) Stalls
b) Falls off
c) Backtracks
d) Moves on
Explanation: Rho protein helps in termination of transcription. It binds to the growing transcript rolls up and catches up with the polymerase when the loop starts to form. This dissociates the complex from the DNA leading to the formation of free Rho, DNA, RNA transcript and polymerase.
10. You can see non-cannonical base pairing in________
a) dsDNA
b) ssDNA
c) dsRNA
d) ssRNA
Explanation: Along with canonical base pairing with the normal base pairs according to Watson-Crick rules, dsRNA also forms non-cannonical base pairs. However, dsDNA can’t do so. ssRNA and ssDNA are called so because they are single stranded so there is no base pairing at all.