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The correct answer is (B) Formation of OPEN INITIATION complex

The explanation: When the RNA pol is attached to DNA STRANDS that have still not melted the complex so formed is called open initiation complex. When the DNA melts it is known as open initiation complex. This follows the formation of abortive transcript and promoter clearance.

Correct option is (d) Exons not to be SKIPPED

Best explanation: BINDING of SR proteins on the exons ensure that the exon is not skipped. It doesn’t BIND to INTRON and in some weak exons it helps in skipping the exon thus helping in ALTERNATE splicing.

Correct OPTION is (C) No part of it TRANSCRIBED

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.

The CORRECT ANSWER is (c) BANDS at same position on 2 occasions

Easy 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.

Correct option is (B) Affinity chromatography

Easy explanation: An affinity chromatography using oligo T sephadex beads in the column can be USED to detect proper 3’ processing. If the RNA is properly processed the POLY A TAIL will adhere to the oligo T and will not COME out of the column without elution.

Correct answer is (B) CPG islands

The explanation: This is because prokaryotic methylases would METHYLATE the C BASES of the CPG region. Then it there is a de-amination of C base, it will resemble a T base which is not as easily distinguishable as U by UDG. So, this region is very mutagenic.

The correct choice is (C) Non-denaturing SDS PAGE

To ELABORATE: A denaturing SDS PAGE would BREAK the interactions between the FACTORS which is being studied. Thus, a non-denaturing SDS PAGE is used. A native page on the other hand would be less clear while zymography is UNRELATED.

Correct option is (c) When REPRESSOR binds to I2 and I1

The best I can explain: When repressor binds to I1 and I2 then RNA polymerase faces no hindrance in binding to the promoter and the genes are turned on. On the other HAND when induces the repressor bind to O2 and I1 which leads to the looping out and turning off.

Correct answer is (b) Lower

Easy 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.

Correct option is (a) True

To explain I would say: The Fas elements are LOCATED as far as -60 to -150 base PAIRS upstream to +1 site. Although these are not true promoter elements, mutations in the Fas site also effects TRANSCRIPTION.

Right answer is (b) RNA entry channel

Explanation: As the RNA is SYNTHESIZED there is an RNA exit channel but there is no NEED of RNA entry channel. Clamp and flap on the other HAND help to STABILIZE the POLYMERASE structure.

The correct answer is (a) Dna

To ELABORATE: DUE to SPLICING the mature mRNA is SHORTER than the DNA with the introns in DNA looping out when they hybridize. Thus the loops have DNA.

The CORRECT choice is (d) Telomeres

For explanation I would say: TELOMERE GENES are mostly silenced so in case of any deletion mutation the deletion will not be detected. HOWEVER, loss of bases from anywhere else in the chromosome which transcribe RNA could be disastrous for the cell.

Correct answer is (d) TFIIE

Best explanation: While TFIIH is actually PERFORMING the kinase action, TFIIE stimulates TFIIH to phosphorylate the CTD. TFIIA and TFIIF play no ROLE in PHOSPHORYLATION.

Correct OPTION is (b) TBP

To explain I WOULD say: TBP or TATA binding protein is the PART of TFIID complex that recognizes and BINDS the TATA consensus SEQUENCE in the promoter (cis element) of eukaryotic genes. The remaining options are also related to eukaryotic transcription but they have protein-protein interactions.

Right option is (b) False

Explanation: ALTHOUGH these 5 subunits are common to all the RNA POLYMERASES YET they share some subtle differences- the CTD of the beta prime subunit of pol II has heptad repeat. ALSO the alpha subunits in RNA pol II are quite different from that of pol I and III.

Correct CHOICE is (C) dsRNA

Easy 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.

Correct option is (c) Stem-loop structure

To EXPLAIN I WOULD say: 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.

Correct choice is (c) PRESENCE of DNA – RNA hybrid

The explanation is: The Uridine analogue can base pair with the RNA pol enzyme when there is a presence of a complementary STRAND (here DNA). THUS, its incorporation and cross linking will confirm the presence of RNA-DNA hybrid. We use labeled NTP to detect the other PROCESSES mentioned.

Right OPTION is (c) Mg^2+

The explanation is: Magnesium ion is MANDATORY for the catalytic ACTION as it shields the NEGATIVE charge of the phosphate groups of rNTPs and also promotes the nucleophilic attack by the 3’-OH group. Manganese, zinc and iron are part of other biologically important enzymes.

Correct answer is (C) Transcription is less

Explanation: The C terminal domain of the alpha subunit is responsible for BINDING to the UP elements of the promoter. Such binding helps to enhance transcription so in ABSENCE of the binding transcription will be SLOWER. However, still transcription will OCCUR due to the presence of core promoter.

Correct OPTION is (d) AAUAAA weakly but stabilization occurs after G/U binding

For EXPLANATION: CPSF binds to AAUAAA but the interaction is weak. This interaction is strengthened by CStF binding to G/U box and further stabilization occurs at CFI and CFII binding.

Correct option is (a) Band is OBSERVED in both cases

The explanation is: There will not be any change in the intensity of the 1^st band as GT is not binding to GSK tag. The 2^nd case also doesn’t change the intensity of the band as GT BINDS to phosphorylated –GSK-RNA pol II, so no GT BOUND unlabelled RNA exists to compete with the labeled band and decrease its intensity.

Correct choice is (d) PAP

Explanation: PAP BINDS to CPSF and CStF by protein-protein INTERACTION and doesn’t actually bind to the DNA. However, the other factors are cis elements BINDING to SPECIFIC sequences.

The correct OPTION is (d) POL I, Pol II and Pol III transcripts

The 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.

The CORRECT option is (C) Shielding promoter for polymerase binding

To elaborate: 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.

Right OPTION is (a) True

Easy explanation: As the cell goes on dividing in each DIVISION love GENES at the telomere is lost for the primer RNA at the telomere. THUS the DNA keeps shortening. If it shortens beyond a limit the cell undergoes apoptosis.

Correct answer is (b) You add TFIID, TFIIB, TFIIF and polymerase

To explain I would say: TFIID is absolutely necessary for any NUCLEATION to take PLACE at all. The complex formed by this composition is almost functional as TFIIA is not playing that important role. HOWEVER, TFIIB and TFIID absence HUGELY EFFECTS the formation of pre-IC.

Correct option is (c) 2->7->5

The explanation is: The phosphorylation is at the 2^nd serine residue with the HELP of the TFIIH, and then during elongation the phosphorylation is at the 5^th serine residue. Between these two there is a short PHASE of phosphorylation at the 7^th serine.

The CORRECT answer is (d) No RNA pol II will precipitate

The best 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.

Right option is (b) Negative feedback

Easiest explanation: Trp binding to the TRAP makes a trp-TRAP COMPLEX which binds to the RNA and stabilizes the TERMINATOR loop. Hence, TRYPTOPHAN’s PRESENCE has a negative feedback effect.

The correct answer is (d) CSA

For explanation: When the RNA polymerase encounters a bulky LESION the NER PROTEIN CSA heps the RNA to backtrack giving the NER mechanism a CLEARANCE of 100bp to act. XPC and TFIIH are also needed in NER but have different function. DNA G is a primase needed in prokaryotic replication.

The correct answer is (b) RNA polymerase without transcript

Best explanation: Sigma FACTOR dissociates after INITIATION so anti sigma antibody will only PULL down RNA polymerase prior to promoter clearance. Thus, the POLYMERASES so obtained have no transcript ASSOCIATED.

Right CHOICE is (C) A

The explanation is: SPLICING requires a branch point A that can attach the 5’ end with the intron making a lariat structure. This doesn’t require ENERGY molecules like ATP.

Right answer is (c) CONCENTRATION will remain constant

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.

Correct answer is (b) False

The explanation: Although often repressors DIRECTLY associate with RNA POLYMERASES it can also act by the means of a mediator as ENHANCERS do. So, they could be close or far from the transcription start site.

Correct answer is (d) HIGHLY specialized GENES.

The explanation is: The highly specialized genes are likely to have a TATA box within their promoter. On the other hand the rest genes have TATAless PROMOTERS.

The correct CHOICE is (b) Segment 2-3

For explanation I would say: In the absence of tryptophan when the RNA polymerase stalls at segment 1 of the ATTENUATOR it HINDERS the formation of loop 1-2. The alternate loop so FORMED is loop 2-3 which is the ATTENUATION loop.

The CORRECT option is (d) σ^E

Explanation: Rse A BINDS σ^E and PREVENTS it from activating transcription. Under extreme conditions this binding is broken by cleavage of c terminal cytosolic tail of transmambrane Rse A, THUS RELEASING σ^E for transcribing necessary genes.

Correct answer is (b) Activate, inactivate

Easy explanation: Binding of SR proteins to ESE or exon splicing ENHANCERS promotes splicing where that exon is retained. hnRNA is a competitive inhibitor of SR proteins that tend to bind ESE as well, thus it INACTIVATES.

Correct option is (d) U6 : U2

For explanation I would SAY: In the spliceosome complex we SEE that U1 is replaced by U6 and not U2. U2AF on the other hand is replaced by U2. Other matches but d are correct.

Right answer is (a) F

The EXPLANATION is: RNA polymerase bound TFIIF is necessary for recruiting TFIIH and TFIIE, TWO IMPORTANT transcription factors that trigger the INITIATION finally.

The correct answer is (d) B. subtlis

The EXPLANATION is: 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.

Right ANSWER is (d) Anti-TRP with Trp-TRAP

Explanation: Anti- trp is an auto-regulating molecule that is sensitive to different levels of tryptophan. When tryptophan conc. is high enough to FORM trp-TRAP complex yet lower than necessary it binds to the Trp-TRAP complex THUS preventing repression.