1. Which of the following will increase transcription?
a) Shielding positive charge of DNA
b) Shielding positive charge of histones
c) Shielding promoter for polymerase binding
d) Shielding termination region
Explanation: Histone binds to the negatively charged DNA due to its positive charges. On DNA binding those regions of the DNA are not available for transcription. Thus, shielding the positive charge would loosen this binding and enable more transcription.
2. Which histone(s) are acted upon by HAT?
a) H1, H2
b) H3
c) H2A, H2B
d) H3, H4
Explanation: These are the two histone residues that are having lysine side chains available for acetylation. On, HAT action they loosely bind to DNA allowing transcription.
3. In a substrate X catabolic gene promoter you replace the normal histones by your experimental histones – ones with arginine side chains in place of histidine. What will be the effect on the concentration of substrate X against the control?
a) Concentration will increase
b) Concentration will decrease
c) Concentration will remain constant
d) Concentration will fluctuate and no definite result will be obtained
Explanation: Arginine side chains can’t be acetylated, so there is no regulation. They are always positively charged and never bind loosely so no gene product is produced. As these gene products are catabolic in their absence the substance X is not broken down so its concentration remains constant.
4. You take a crude cell lysate and treat it with DNase. Now wash it and add protease. Use specific restriction endonuclease to get your desired fragment and run it on a gel. You see a single band. Which of the following will be true about the fragment?
a) It is actively transcribed
b) Only part of it is transcribed
c) No part of it transcribed
d) Undefined transcriptional nature
Explanation: DNase would only get access to uncondensed DNA. Uncondensed DNA is transcriptionally active. As the DNase would have cleaved the fully or partially transcriptionally active DNA, and our fragment is intact, we can say it was not at all being transcribed.
5. Which of the following is not a common mode of histone modification in eukaryotes?
a) Methylation
b) Phosphortlation
c) Sulphonation
d) Ubiquitinylation
Explanation: The positively charged N-termini of histone is modified by the phosphorylation, ubiquitinylation, methylation and also acetylation processes in order to shield its positive charges.
6. Which of the following association acts as transcriptional activator?
a) Myc- Max
b) Mad- Max
c) Map- Mad
d) MapK- Myc
Explanation: When Max associates with Myc the gene is transcriptionally active, while when Max associates with Mad the gene is inactivated. This is because Mad binds to SIN3A which brings about a HDAC.
7. In an experiment you perform the following- you attach a flag epitope to HDAC, you over express a mutated Mad1-pro gene within the experimental cell, you immunoprecipitate the protein complex using flag epitope and run that on a non-denaturing SDS PAGE and western blot. Now you add antibodies for HDAC, MAD1 and SIN3A stripping after each addition. What will you observe?
a) Bands at same position every time
b) 3 bands at different positions one at a time
c) Bands at same position on 2 occasions
d) No band
Explanation: As Mad1-pro can’t bind to SIN3A due to the mutation it will not precipitate with the HDAC-SIN3A complex. Due to non-denaturing nature of the gel both the proteins migrate together to the same distance so the band appears at the same position.
8. In presence of Thyroid hormone the receptor TR binds to ____________
a) RXR directly
b) mRPD3 directly
c) NcoR directly
d) mRPD3 indirectly
Explanation: In presence of thyroid hormone the receptor binds to RXR directly. In absence however, it binds to NcoR directly, which binds to SIN3 and mRPD3 which is an HDAC, this binding inactivates the genes.
9. If the events in post transcriptional modification are listed as-
i) Polyadenylation ii) capping iii) splicing
What would be the correct sequence?
a) i->ii->iii
b) iii->ii->i
c) i->iii->ii
d) ii->i->iii
Explanation: Post transcriptional modification is very strictly synchronized process and no next step takes place unless the preceding one is complete. The sequence is 1st adding cap, then Polyadenylation and then splicing.
10. Which RNA transcript would be capped by the capping enzymes in vitro mixture of RNA?
a) Pol II transcripts
b) Pol I and Pol III transcripts
c) Pol I transcripts
d) Pol I, Pol II and Pol III transcripts
Explanation: In vivo, only pol II transcripts are capped, however in vitro when the different pol transcripts are mixed with the capping enzymes all of them are capped showing no specificity.