1. When uncharged tRNA concentration is low what will you expect as the activity of tryptophan operon?
a) Low
b) Medium
c) High
d) Very high
Explanation: When uncharged tRNA concentration is low this is because of lack of trp. This inactivates anti-trap gene and thus there is no attenuation at all in trp gene. Thus activity is high.
2. How many stem loop structures play important role in complete system of tryptophan gene regulation?
a) 3
b) 4
c) 6
d) 8
Explanation: 3 stem loop for the attenuation and termination in trp gene itself and another 3 for attenuation and termination of anti-trp gene. Overall there are 6 stem loop structures which regulate the entire functioning
3. Which of this mode of trp regulation is not possible in eukaryotes?
a) Anti-TRAP Trp-TRAP
b) Anti-Terminator stabilization
c) Attenuation of trp
d) Trp-TRAP binding
Explanation: Attenuation of trp is based on ribosomal stalling at two trp codons which leads to the formation of attenuation loop. This is only possible for co-transcriptional translation which is absent in eukaryotes.
4. In which microorganism will you find attenuation by alternate loop formation due to ribosomal stalling?
a) S. aureus
b) E. coli
c) S. typhimurium
d) B. subtlis
Explanation: This is a mechanism found in Bacillus subtilis; where other microorganisms like E. coli have a different mechanism. E. coli has the anti–TRAP trp-TRAP mechanism.
5. Which operon behaves similar to Bacillus subtilis trp operon in terms of attenuation?
a) Lac operon
b) His operon
c) Phe operon
d) Ara operon
Explanation: His operon, similar to B. subtilis trp operon, has a streach of 6 His codons in its attenuator which stalls the ribosome and helps in attenuation.
6. Prokaryotes have 1 RNA polymerase. How many RNA polymerases are there in eukaryotes?
a) 2
b) 3
c) 4
d) none
Explanation: Eukaryotes have three RNA polymerases specialized in different types of RNA transcription. While RNA pol I mainly deals with rRNA, RNA pol II with mRNA and RNA pol III deals with tRNA and 5S rRNA.
7. You isolate the nucleolus from the nucleus and performed a density gradient centrifugation with the extract. Then you check the different for the RNA polymerase activity of the different segment. Which RNA polymerase would you expect to find in abundance?
a) RNA pol I
b) RNA pol II
c) RNA pol III
d) RNA pol IV
Explanation: Nucleolus is the centre for rRNA production. As RNA pol I is mainly concerned with rRNA transcription we will expect to find this in abundance. Cytosolic extract on the other hand will be abundant in RNA pol II and III.
8. Which RNA polymerase can bind CBP?
a) I
b) II
c) III
d) None
Explanation: CBP or C terminal binding protein binds to the C terminal tail of heptad repeats when it is phosphorylated. As RNA pol II is the only polymerase with a C terminal tail, only RNA pol II will bind the hepad repeat
9. What is the sequence of the heptad repeat of C terminal domain?
a) TSPWSTT
b) YSPTSGS
c) YSPTSPS
d) YSLPSTS
Explanation: A heptad repeat sequence is YSPTSPS where there are 3 phosphoryable serine residues at position 3, 5 and 7.
10. You designed an antibody against phosphorylated CTD of RNA polymerase II. Then you use a DNA fragment with lac operon and try to immunoprecipitate the RNA pol II. What will be the expected result?
a) Moderate concentration of RNA pol II will precipitate
b) RNA pol II that precipitated will be contaminated with pol III
c) High amount of RNA pol II will precipitate
d) No RNA pol II will precipitate
Explanation: Lac operon is a prokaryotic gene not recognized by eukaryotic RNA polymerase. Thus it will not be transcribing. Non-Transcribing RNA pol II doesn’t have phosphorylated residue in CTD, so it can’t be immunoprecipitated with the given antibody.