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Correct answer is (c) Sulphonation

The BEST 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.

Right choice is (c) TYPE III

The explanation is: The Type III of CLASS III promoter is a non-classical class III promoter which resembles type II. It has a TATA box, PSE and an upstream DSE. There is no 4^th type of class III promoter and type I and II don’t resemble class II.

The CORRECT answer is (b) 5

For explanation I would say: The beta, beta prime, TWO alpha and one omega SUBUNIT homology to eukaryotic RNA POLIS common in all three RNA polymerases. These are RPA 1-5 respectively. The rest subunits are non-essential or dispensable.

Right choice is (a) U5

To explain I would say: The C complex is the final STRUCTURE before the attack. Then U6 has replaced U1 and U2 has replaced U2AF and U4 dissociates. Only U5 is STILL there stabilizing the structure by binding the two ends to CLOSE.

The correct CHOICE is (b) False

Easiest EXPLANATION: The modifications are carried out as the transcript is produced. Like when the transcript is 25-30 NUCLEOTIDES long it is capped and there has also been evidence showing that SPLICING also takes place during TRANSCRIPTION.

The correct option is (b) SIR2

To elaborate: The SIR or silent INFORMATION regulator PROTEINS are responsible for repression at the telomere by deacetylating the histones. While SIR3and SIR4 BIND to the N-termini the SIR2 protein which is attracted by SIR 3-4 is having the HDAC ACTIVITY.

Right answer is (b) ALPHA

Explanation: In yeast mating TYPE alternates with every GENERATION, so if a parent is ‘a’ then bud will be ‘alpha’ and VICE versa. Yeast with a-alpha mating type is sterile while there is no rho mating type of yeast.

Correct choice is (d) Primer extension

The explanation is: While primer extension method can be used to locate the 5’ end of the RNA this doesn’t do any HELP in studying the assembly of PRE initiation complex- basically as that’s upstream to even TRANSCRIPTION start site. The remaining methods are however very USEFUL.

The CORRECT CHOICE is (c) Ara C

Best explanation: As ara C level rises it itself binds beside its promoter preventing the RNA polymerase to bind to the same there by having a NEGATIVE FEEDBACK control. Ara BAD promoter lacks this type of control and is regulate din different way.

The correct ANSWER is (c) To protect the transcript from exonuclease

Explanation: There are several free exonuclease in the nucleus and the 5’ cap protects the mRNA from them. In its absence, VALUABLE INFORMATION will be lost from the transcript and abnormal proteins would result after translation. RNA normally font STICK to DNA and absence of ATPase activity rules out 2^nd option.

Right answer is (b) Sigma INCREASES initiation and ELONGATION rate

The best explanation: [γ-^32P]ATP INCORPORATION marks more initiation while [^14C]ATP incorporation indicates elongation. As on adding sigma factor both incorporations seems to increase, sigma must increase both initiation and elongation rate.

The correct OPTION is (a) True

The explanation is: Splicing REQUIRES the 5’ end GU and the 3’ end AG with a branch point A. How big is the intron or the intermediate SEQUENCES play no role in splicing.

Correct choice is (b) 3’->5’ exonuclease

Explanation: The 3’ END of the RNA is susceptible to chewing by 3’->5’ exonucleases. So, the poly A TAIL acts as a safeguard which EVEN if chewed partly will protect the valuable information of the RNA. This in PLACE of inhibition rather promotes the next steps LIKE transcription and export.

Correct CHOICE is (b) 5

The BEST I can explain: when RNA TRANSCRIPT is only 30 bases long it is still within the initiation RANGE. Then the 2nd SERINE residue is phosphorylared only.

The CORRECT CHOICE is (c) Enhancer to TAF

The best I can explain: As enhancer elements work from a distance the CONNECTING DNA is often looped and the enhancer connects to the RNA pol bound TAF through a mediator. A silencer may connect to TAF in a similar way.

Correct answer is (c) Attenuation of trp

To explain: 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.

The correct ANSWER is (b) I^+ + P + O^C + Z^+ + Y^+ + A^+ / I^+ + P+ O^+ + Z^+ + Y^+ + A^+

The explanation: O^C is an operon that can’t bind to the repressor. Consecutively it is always turned on. I^S produces a repressor which binds tightly to operon and never lets it turn on- This is the CASE of constitutively turning off.

The correct CHOICE is (C) Negative effect on transcription

Easy EXPLANATION: RAP1 is necessary for silencing at the TELOMERE. As silenced gene can’t be expressed so it has a negative effect. The effect can be even called very negative as the silencing is very strong.

The CORRECT choice is (c) TFIIS

Explanation: TFIIS is the transcription factor that HELPS to overcome the transcriptional stalling of RNA POLYMERASES and also to help the polymerase to BACKTRACK.

Right ANSWER is (d) pol II -> pol III , pol I unaffected

Easiest EXPLANATION: RNA polymerase II is inactivated by very small level of alpha amanitin as low as 1 microgram PER ml while on the other hand pol III is affected at 10 MICROGRAMS per ml concentration. Polymerase I remain unaffected EVEN at very high alpha amanitin concentration.

The correct ANSWER is (a) Function normally

Easiest explanation: The set of GENE product PRODUCED from the wild type I+ loci will be sufficient to induce repression in both the OPERONS. THUS, the system will not be impaired or non-functional.

The correct answer is (b) Helix turn helix

Best explanation: Region 4.2 of sigma subunit has a helix turn helix MOTIF that HELPS it to bind to the major groove of DNA. The SIMILAR motif has been SEEN in region 3 as well. Zinc fingers and leucine zipper are examples of other DNA BINDING motif.

Right option is (d) It prevents sigma to BIND to DNA

For EXPLANATION: The 1^st region forms a flap LIKE STRUCTURE that BLOCKS the DNA binding sites. This helps to prevent unbound sigma to bind to DNA. That will only block the DNA binding site and lead to no transcription in absence of RNA pol.

Correct CHOICE is (a) Immediate early

The best explanation: Sigma factor in HOLOENZYME ensures transcription of immediate early gene FIRST SOON after viral INFECTION. This is followed by delayed and late gene transcription. CoreEnzyme shows no such specificity.

Correct answer is (c) Experimental artifact

To explain: These short oligomers are ABORTIVE transcripts produced by the RNA POLYMERASE before it leaves the promoter. There is no lagging strand in RNA and SINCE the experiment is repeated several TIMES with caution the other justifications are rules out.

The CORRECT answer is (d) rU-dA

For explanation I would say: 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.

Correct ANSWER is (c) Reverse Transcriptase

The explanation is: Reverse transcriptase helps in cDNA synthesis from RNA, which is opposite to the CENTRAL dogma, where RNA is synthesized from DNA. Ribosome and RNA POLYMERASES OBEY central dogma, while Restriction Endonuclease is unrelated as it simply cleaves DNA at a specific site.

Right option is (b) U4, U6

The explanation is: U4 and U6 snRNA are found in the cytoplasm. They can associate to form a common RNP with the other PROTEINS when they can move to the NUCLEUS. In the nucleus, it interacts with U5 to form the tripartite complex.

Correct answer is (c) AAUAAA—–CA—-UUUUGUGU

Explanation: The cleavage sequence has a CONSENSUS of AAUAAA sequence towards the 5’ and a G/U box towards the 3’ end and a CA DINUCLEOTIDE in the middle. The cleavage is just after 3’ to the CA.

Correct answer is (d) AU box

Explanation: The TATA box and Pribnow box are PART of the promoter so they are not EVEN transcribed into RNA. The GU box on the other hand is cleaved off during polyadenylation while the AU box is retained even after polyadenylation.

The CORRECT answer is (b) GDP

The best explanation: The 5’ cap has a guanine RESIDUE which MAKES Gpp(pN)n where N is any NUCLEOTIDE. This is also called the 5’ G cap. Other bases like A, T, C or U don’t participate in capping.

Right option is (B) Linear REPEATED sequences

Explanation: RAP1 has no sequence specification. It binds to the repeated sequences at the TELOMERIC region in the linear DNA lacking nucleosomes. Then, it loops over and spreads the repression to the entire TELOMERE.

Right ANSWER is (d) PROMOTER PROXIMAL elements

The best I can explain: While upstream ACTIVATING ELEMENT is orientation independent, enhancers and silencers are position as well as orientation independent. On the other hand promoter proximal elements are orientation dependent.

Correct choice is (a) RNA pol I

The 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.

The correct option is (d) Delta subunit

To elaborate: While σ^70 in E. coli has an extra 245 amino acid that helps it to LOOSEN the polymerase from non-specific SITE, σ^43 in BACILLUS subtilis lacks it. Thus, it needs another subunit of core polymerase- delta subunit to perform this function.

Right option is (b) σ^43

Best explanation: σ^43 is the normally active sigma factor for Bacillus subtilis while σ^70 is the normally active sigma factor for E. COLI. σ^32 is expressed under STRESSED CONDITIONS like heat shock when it competes with σ^70 for RNA pol core.

The correct choice is (a) INCREASE in UV absorption

Easy EXPLANATION: MELTING is the separation of the two strands of DNA which CAUSES an increase in the amount of UV RAYS absorbed by the DNA. This is known as Hyperchromic shift. It has no effect on florescence of DNA.

The CORRECT option is (b) Streptolydigin

The best I can explain: Streptolydigin prevents elongation while Rifampicin prevents initiation. Penicillin and Tetracyclin ACT on the RIBOSOME and not on RNA polymerase although EVEN they ultimately inhibit protein SYNTHESIS.

Correct option is (c) CATALYST

For explanation: CBP ACTS as a catalyst which concentrates the ENZYME activity at the pol II transcripts so that they are selectively capped at their 5’ ENDS.