156 + Interview Questions in Gene Expression : Transcription Prokaryotes in Cytogenetics Page 2 InterviewSolution

51.	Under which condition will the yeast express a-alpha mating type?(a) HAT activation(b) TBP activation(c) HDAC inactivation(d) HAT inactivationI had been asked this question by my school principal while I was bunking the class.Origin of the question is Transcription in Eukaryotes : Gene Silencing at Telomere in division Gene Expression : Transcription Prokaryotes of Cytogenetics
Answer» Correct answer is (d) HAT inactivation Easy explanation: HAT acetylates the histone H3 and H4 so it activates the gene while HDAC deacetylates them and INACTIVATES the genes. Under CONDITION when HAT is active but HDAC is inactive we will get an expression of both the telomere gene which is normally REPRESSED. This will lead to a-alpha mating type.

Discussion

52.	Which is the 1^st enzyme in capping?(a) Guanyltransferase(b) RNA 5’ triphosphatase(c) N7G methytransferase(d) Guanul transferaseThis question was posed to me in class test.I want to ask this question from Post-Transcriptional Modification topic in chapter Gene Expression : Transcription Prokaryotes of Cytogenetics
Answer» Correct choice is (d) Guanul transferase For explanation I would say: 1ST the 5’ triphosphate group is removed by the phosphatase. Then the GPP is added to it making it the G cap. Lastly the methyl transferase TRANSFERS the methyl group to the N7 position of the cap GUANINE.

Discussion

53.	If you make a chimeric factor with the DNA binding element of an activator and a functional domain of a repressor, how will this factor behave?(a) It will act like a repressor(b) It will act like an activator(c) It will not be able to bind the DNA so although it binds via the repressing domain it won’t cause repression(d) Repression domain will be non-functionalThis question was posed to me in an online quiz.This interesting question is from Transcription in Eukaryotes : Activators and Repressors in portion Gene Expression : Transcription Prokaryotes of Cytogenetics
Answer» Correct ANSWER is (a) It will act like a repressor For explanation: As activators and REPRESSORS have a domain organization it is seen that if a DNA binding domain is provided it doesn’t AFFECT the activity. As long as the factor binds to the correct cis element it will STILL be a functional repressor.

Discussion

54.	What is the sequence of the heptad repeat of C terminal domain?(a) TSPWSTT(b) YSPTSGS(c) YSPTSPS(d) YSLPSTSI had been asked this question in a national level competition.My enquiry is from Transcription in Eukaryotes : RNA Polymerases in section Gene Expression : Transcription Prokaryotes of Cytogenetics
Answer»

Discussion

55.	When sigma subunit dissociates from the RNA polymerase what happens to the DNA binding clamp of the core?(a) It opens again(b) It closes again(c) It enters a semi open stage(d) It is phophorylated and ubiquitination occursI got this question in final exam.My query is from Transcription in Prokaryotes : Sigma Factor topic in section Gene Expression : Transcription Prokaryotes of Cytogenetics
Answer» RIGHT option is (d) It is phophorylated and ubiquitination occurs Easiest explanation: In the ABSENCE of the sigma SUBUNIT initially the RNA Pol clamp is closed. Sigma makes a WINDOW of open clamp conformation when RNA pol loads on the DNA. When sigma dissociates the clamp closes again AROUND the DNA preventing the pol to fall off.

Discussion

56.	You want to find out which of the consequent gene X or Y is being transcribed. For this you do a hybridization competition analysis and plot the radioactivity curve with increasing concentration of the competitor unlabelled mRNA. If the transcribed gene was X, which observation fits your result?(a) The radioactivity of y decreases while that of x remains the same(b) The radioactivity of x decreases while that of y remains the same(c) The radioactivity of x and y decreases(d) The radioactivity of y decreases while x increasesThis question was addressed to me in an online interview.My enquiry is from Transcription in Prokaryotes : Initiation topic in portion Gene Expression : Transcription Prokaryotes of Cytogenetics
Answer»

Discussion

57.	The catalytic activity resides in which subunit of RNA polymerase?(a) β′ (156 kDa)(b) β (151 kDa)(c) α (37 kDa)(d) σ70 (70 kDa)The question was asked by my college director while I was bunking the class.This intriguing question originated from Transcription in Prokaryotes : Initiation in chapter Gene Expression : Transcription Prokaryotes of Cytogenetics
Answer»

Discussion

58.	U rich RNA that helps in splicing are______________(a) sn RNA(b) snoRNA(c) hnRNA(d) mRNAI got this question during a job interview.Asked question is from Post-Transcriptional Modification in division Gene Expression : Transcription Prokaryotes of Cytogenetics
Answer» Correct choice is (a) SN RNA The explanation is: The U RICH NA are the small nuclear or SNRNA. The snoRNA are needed for ribosomal rRNA PROCESSING in the nucleolus. hnRNA is HETEROGENEOUS nuclear RNA are mRNA is the RNA that is spliced by snRNA.

Discussion

59.	In the case of group 1 RNA, which one of the following can’t attack?(a) GMP(b) GDP(c) GTP(d) ATPThe question was asked in a national level competition.This key question is from Post-Transcriptional Modification topic in division Gene Expression : Transcription Prokaryotes of Cytogenetics
Answer» Right CHOICE is (d) ATP To elaborate: It is seen that in the case of group II AUTO splicing RNA the attack is made by the 3’-OH end of the RIBOSE attached to the G base. It is immaterial WHETHER its GMP, GDP or GTP. HOWEVER, A base is not seen to attack in that case.

Discussion

60.	Which histone(s) are acted upon by HAT?(a) H1, H2(b) H3(c) H2A, H2B(d) H3, H4This question was posed to me during an interview for a job.I want to ask this question from Transcription in Eukaryotes : Chromatin Remodelling in section Gene Expression : Transcription Prokaryotes of Cytogenetics
Answer» Right OPTION is (d) H3, H4 To elaborate: 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.

Discussion

61.	In which of the following chromosome compaction is absent?(a) Cellulomonas flavigena(b) Allium sativum(c) Panthera tigris(d) Saccharomyces cerevisiaeThe question was asked during an interview for a job.This interesting question is from Transcription in Eukaryotes : Chromatin Remodelling in portion Gene Expression : Transcription Prokaryotes of Cytogenetics
Answer» The correct CHOICE is (a) Cellulomonas flavigena The explanation: Cellulomonas flavigena is a prokaryote, and UNLIKE the REST which are eukaryotes they don’t have chromosome compaction and HISTONES.

Discussion

62.	You transfer the DAM gene to the telomere of yeast. What will be your expected result?(a) Methylation at the A sites in specific sequence(b) Methylation at the C sites in specific sequence(c) No methylation at the A sites in specific sequence(d) No methylation at the C sites in specific sequenceThe question was asked at a job interview.I want to ask this question from Transcription in Eukaryotes : Gene Silencing at Telomere topic in section Gene Expression : Transcription Prokaryotes of Cytogenetics
Answer» The correct answer is (c) No methylation at the A sites in specific sequence The BEST EXPLANATION: DAM is a METHYLASE that methylates the A of GATC sequence. When expressed in Yeast it methylated the yeast DNA, however, in telomere the EXPRESSION is blocked by the telomere silencing effect. Thus, we don’t get methylated DNA.

Discussion

63.	Prokaryotes have 1 RNA polymerase. How many RNA polymerases are there in eukaryotes?(a) 2(b) 3(c) 4(d) NoneThe question was posed to me in final exam.I want to ask this question from Transcription in Eukaryotes : RNA Polymerases in chapter Gene Expression : Transcription Prokaryotes of Cytogenetics
Answer» The correct option is (b) 3 To elaborate: 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.

Discussion

64.	Which operon behaves similar to Bacillus subtilis trp operon in terms of attenuation?(a) Lac operon(b) His operon(c) Phe operon(d) Ara operonThis question was addressed to me in my homework.Question is taken from Operons : Trp in section Gene Expression : Transcription Prokaryotes of Cytogenetics
Answer» RIGHT option is (b) His operon To EXPLAIN I would say: 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.

Discussion

65.	If in an experiment you mutate the two trp codons in the attenuator to ala codon what will be the impact on repression?(a) Repression will be normal(b) Reduced repression(c) No repression so the gene would be constitutively turned on(d) Activation will take placeThe question was posed to me in an interview for job.My doubt is from Operons : Trp in division Gene Expression : Transcription Prokaryotes of Cytogenetics
Answer» The correct option is (b) Reduced repression Explanation: The basis of attenuation at the attenuator is the PRESENCE of TWO trp operon which stalls the RNA polymerase when trp is deficient. Mutating the trp to ala ELIMINATES the attenuator’s mode of repression. HOWEVER, there is still an intrinsic mode of repression by binding of a repressor, which is still active.

Discussion

66.	Which of the Ara promoter regulation is controlled by binding of repressors on the O1 regulating unit?(a) P^A(b) P^B(c) P^C(d) P^DI got this question in my homework.The origin of the question is Operons : Lac and Ara topic in chapter Gene Expression : Transcription Prokaryotes of Cytogenetics
Answer» The correct option is (c) P^C To explain I WOULD say: When ARA C is produced in EXCESS it bind to O1 and represses the promoter PC. This is true as o1 LIES closer to PC. However the other promoter P^BAD has a REGULATING unit O2 that acts from a distance by looping out.

Discussion

67.	We know that region 2.4 of sigma factor binds DNA at -10box. What is the function of 2.1?(a) It is a flexible loop with no known function(b) It binds -35box(c) It binds RNA polymerase(d) It binds region 4.1 to give sigma its specific structure and functionI had been asked this question in an interview.My question is taken from Transcription in Prokaryotes : Sigma Factor topic in section Gene Expression : Transcription Prokaryotes of Cytogenetics
Answer» Right CHOICE is (C) It BINDS RNA POLYMERASE The EXPLANATION: Region 2.1 helps to bind the sigma factor to the RNA polymerase. The binding is not absolute as this factor is dissociable. -35box on the other hand is recognized by region 4.2.

Discussion

68.	You transcribe a gene and obtain mRNA. To detect which strand is transcribed you replicate each strand in vitro and use the newly synthesized strand for hybridization with mRNA. If mRNA hybridize with the newly replicated strand from 1^st strand then the gene transcribed was in ________(a) 1^st strand(b) 2^nd strand(c) Same copy in both strands(d) DispersedThe question was posed to me during a job interview.Question is taken from Transcription in Prokaryotes : Sigma Factor topic in section Gene Expression : Transcription Prokaryotes of Cytogenetics
Answer» Right CHOICE is (B) 2^nd strand Best explanation: mRNA produced from 1^st strand and DNA produced by replicating 1^st strand should have the same sequence and they will not hybridize. However, if the gene is transcribed from 2^nd strand then this RESULT is POSSIBLE. RNA pol only transcribes ONE strand at a time so dispersed and both being transcribed are not possible.

Discussion

69.	In an experiment you wrongly added Rifampicin antibiotic to Rifampicin sensitive bacteria. You try to correct your mistake by adding more core subunits (resistant to Rifampicin) to your mixture. What would you expect to see?(a) Normal transcription to restart(b) Transcription will restart but it will be non-specific as you added core enzyme(c) Impaired transcription with more abortive transcripts(d) No transcriptionThe question was posed to me during an interview.The above asked question is from Transcription in Prokaryotes : Sigma Factor in division Gene Expression : Transcription Prokaryotes of Cytogenetics
Answer» Right choice is (a) NORMAL transcription to restart The best I can explain: The sigma factor is not AFFECTED by Rifampicin and it is dissociable. The sigma already present in mixture will add to the Rifampicin RESISTANT core and restore normal transcription of specific genes. No increased abortive transcription will be seen.

Discussion

70.	What is the direction in which the transcript produced by RNA polymerase grows?(a) 3’->5’ direction on 3’->5’ strand(b) 5’->3’ direction on 5’->3’ strand(c) 3’->5’ direction on 5’->3’ strand(d) 5’->3’ direction on 3’->5’ strandI got this question in an interview for job.My question comes from Transcription in Prokaryotes : Initiation in division Gene Expression : Transcription Prokaryotes of Cytogenetics
Answer» Correct option is (d) 5’->3’ DIRECTION on 3’->5’ strand Easy explanation: In the process of RNA synthesis there is a short phase of hybrid DNA-RNA FORMATION. This is possible only when two strands are oppositely ORIENTED. As 3’-OH is necessary for adding new nucleotides, RNA is synthesized from 5’->3’ direction, so the template MUST be 3’->5’ oriented.

Discussion

71.	Which of the following mRNA will not be polyadenylated?(a) Actin mRNA(b) Receptor mRNA(c) Canonical histone mRNA(d) Transport channel mRNAThe question was posed to me by my school principal while I was bunking the class.This interesting question is from Post-Transcriptional Modification in section Gene Expression : Transcription Prokaryotes of Cytogenetics
Answer» Correct option is (c) Canonical HISTONE mRNA For EXPLANATION: Most of the MRNAS are polyadenylated except the major histone mRNAs. These are the canonical mRNA that NEED a rapid TURNOVER and have a unique 3’ end. However, the non-canonical mRNA produced during S phase are polyadenylated.

Discussion

72.	Splicing concensus sequence is ___________(a) Exon/GU–intron–AG/exon(b) Exon/UG–intron–AT/exon(c) Exon/GU–intron–GA/exon(d) Exon/AU–intron–CG/exonI have been asked this question in an interview.The question is from Post-Transcriptional Modification in division Gene Expression : Transcription Prokaryotes of Cytogenetics
Answer» Right choice is (a) Exon/GU–intron–AG/exon For explanation I WOULD say: It has been seen that splicing occurs at this consensus SEQUENCE. However, as there COULD be many more MOTIEF similar to this within the intron there are ADDITIONAL consensus sequences.

Discussion

73.	In the consensus sequence 5′-AG/GUAAGU–intron–YNCURAC–YnNAG/G-3′Y could be ______(a) Base A(b) Base G(c) Base C(d) Base TThis question was posed to me by my school teacher while I was bunking the class.The above asked question is from Post-Transcriptional Modification in portion Gene Expression : Transcription Prokaryotes of Cytogenetics
Answer» The CORRECT choice is (c) Base C The explanation is: Y could be any pyrimidine that RULES out A and G as base. Now we know that RNA doesn’t have T, so the only pyrimidine possible from the given COMBINATION is C.

Discussion

74.	In a Drosophila polytene chromosome just before molting you add green labelled antibody against unphosphhorylated RNA pol II CTD and red labelled antibody against phosphorylated RNA pol II CTD. These give fluorescence on adding the respective substrates. What will be the colour in gene 74 EF and 75 B?(a) Red(b) Green(c) Yellow(d) MagentaThis question was addressed to me by my college director while I was bunking the class.Query is from Transcription in Eukaryotes : RNA Polymerases in division Gene Expression : Transcription Prokaryotes of Cytogenetics
Answer» Correct answer is (a) Red To elaborate: The 74EF and 75B GENES are highly TRANSCRIBED during and before MOLTING. Thus, there WOULD be many RNA pol II with phosphorylated CTD there which will bind the red labelled antibody. Yellow colour is seen when red and GREEN fluorescence mix.

Discussion

75.	Which RNA polymerase can bind CBP?(a) I(b) II(c) III(d) NoneThe question was posed to me in a national level competition.My question comes from Transcription in Eukaryotes : RNA Polymerases topic in section Gene Expression : Transcription Prokaryotes of Cytogenetics
Answer» CORRECT option is (b) II For 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.

Discussion

76.	When uncharged tRNA concentration is low what will you expect as the activity of tryptophan operon?(a) Low(b) Medium(c) High(d) Very highI had been asked this question in a national level competition.The question is from Operons : Trp topic in section Gene Expression : Transcription Prokaryotes of Cytogenetics
Answer» RIGHT option is (d) Very HIGH Best 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.

Discussion

77.	In presence of Thyroid hormone the receptor TR binds to ____________(a) RXR directly(b) mRPD3 directly(c) NcoR directly(d) mRPD3 indirectlyThe question was asked in examination.My question is from Transcription in Eukaryotes : Chromatin Remodelling topic in portion Gene Expression : Transcription Prokaryotes of Cytogenetics
Answer» The correct CHOICE is (a) RXR directly To explain I would say: 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.

Discussion

78.	Which of these operons is anabolic?(a) Lac(b) Ara(c) Trp(d) PheThis question was addressed to me by my college director while I was bunking the class.This intriguing question originated from Operons : Trp topic in section Gene Expression : Transcription Prokaryotes of Cytogenetics
Answer» Correct answer is (c) TRP The explanation is: Trp operon is anabolic as it codes for enzymes that help in the synthesis of the amino acid tryptophan. On the other hand ARA and Lac operons are catabolic as they help in breaking down the RESPECTIVE SUGAR.

Discussion

79.	TFIIB can bind to a promoter element in the core promoter of eukaryotes. This element is __________(a) Inr(b) BRE(c) DPE(d) TATAThis question was posed to me in a national level competition.I need to ask this question from Transcription in Eukaryotes : Activators and Repressors in portion Gene Expression : Transcription Prokaryotes of Cytogenetics
Answer» Correct choice is (b) BRE Explanation: BRE STANDS for TFIIB recognition site. It is here that the ELEMENT can bind. On the other hand TFIID BINDS at the TATA box.

Discussion

80.	When Rho catches up with RNA polymerase it _______(a) Stalls(b) Falls off(c) Backtracks(d) Moves onThe question was asked during a job interview.My doubt stems from Transcription in Prokaryotes : Termination in section Gene Expression : Transcription Prokaryotes of Cytogenetics
Answer» Right option is (b) Falls off To explain I WOULD say: 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.

Discussion

81.	RPA 2 is one of the larger subunits of RNA polymerase in eukaryotes. It is homologous to the prokaryotic _____________(a) Alpha subunit(b) Beta subunit(c) Beta prime subunit(d) Omega subunitThe question was asked in an interview for job.I want to ask this question from Transcription in Eukaryotes : RNA Polymerases topic in section Gene Expression : Transcription Prokaryotes of Cytogenetics
Answer» The correct answer is (b) Beta subunit To ELABORATE: RPA 2 is HOMOLOGOUS to the beta subunit while RPA 1 is homologous to the beta PRIME subunit which is the catalytic subunit.

Discussion

82.	The catalytic unit of RNA polymerases when placed properly during initiation is just over _________(a) -1 site(b) 0 site(c) +1 site(d) – 10 siteI got this question during an interview for a job.This key question is from Transcription in Prokaryotes : Initiation topic in division Gene Expression : Transcription Prokaryotes of Cytogenetics
Answer» Right option is (C) +1 site Explanation: The catalytic subunit is exactly over the first nucleotide of mRNA which is at +1 site. -1 signifies one site upstream to initiation while there is no 0 site. -10 box is recognized by SIGMA FACTOR.

Discussion

83.	Which factor phosphorylates the serine residues in RNA pol II CTD?(a) TFIIA(b) TFIIB(c) TFIIH(d) TFIIDThis question was posed to me during an internship interview.My enquiry is from Transcription in Eukaryotes : RNA Polymerases in division Gene Expression : Transcription Prokaryotes of Cytogenetics
Answer» The correct option is (c) TFIIH To elaborate: TFIIH phosphorylates the serine 2 of the HEPTAD REPEAT that DRIVES the RNA pol into INITIATION. The other factors mentioned are only part of pre-initiation complex and not of transcribing RNA pol.

Discussion

84.	Which of these properties do not agree with trp operon attenuator?(a) It brings about repression of trp operon(b) It consists of one stem loop system(c) It has two codons for tryptophan in sequence(d) Ribosome stalls at the attenuatorI got this question in exam.This question is from Operons : Trp topic in section Gene Expression : Transcription Prokaryotes of Cytogenetics
Answer» Correct option is (b) It consists of one stem loop system The best I can explain: Attenuator does’t just contain one loop but it can form two in PRESENCE of TRYPTOPHAN and an alternate one in absence of the same for B. subtilis and one terminator and ANOTHER anti-terminator loops for E. Coli. The other facts about trp operon are absolutely correct.

Discussion

85.	Which part of sigma factor associates with -10box?(a) 1.2(b) 2.4(c) 4.4(d) 4.2I got this question in final exam.Asked question is from Transcription in Prokaryotes : Initiation in division Gene Expression : Transcription Prokaryotes of Cytogenetics
Answer» Correct answer is (b) 2.4 Easiest EXPLANATION: the 2.4 i.e. (4^th part of 2^nd subunit) associates with the -10BOX in promoter. While 4.2 associates with -35box of the promoter. 1.2 sub UNIT doesn’t associate with DNA; and there is no 4.4 unit in sigma FACTOR.

Discussion

86.	Which of these systems give the best mode for turning trp operon off?(a) Repressor(b) Attenuator(c) Repressor with a downstream poly A tail(d) Repressor with an attenuatorI had been asked this question in unit test.Asked question is from Operons : Trp topic in division Gene Expression : Transcription Prokaryotes of Cytogenetics
Answer» Right answer is (d) Repressor with an attenuator To explain: A repressor provides 70 fold lowering of expression while an attenuator PROVIDED an additional 10 fold lowering of expression. TOGETHER they have a net effect of 700 fold repression over the trp operon.

Discussion

87.	Which of these combinations will result is am operon that can never produce beta galactosidse?(a) I^– + P + O^+ + Z^+ + Y^+ + A^+ / I^+ + P+ O^+ + Z^+ + Y^+ + A^+(b) I^+ + P + O^C + Z^– + Y^+ + A^+ / I^+ + P+ O^+ + Z^– + Y^+ + A^+(c) I^s + P + O^C + Z^+ + Y^+ + A^+ / I^+ + P+ O^C + Z^+ + Y^+ + A^+(d) I^-d + P + O^+ + Z^+ + Y^+ + A^+ / I^– + P+ O^+ + Z^+ + Y^+ + A^+I had been asked this question in a job interview.Enquiry is from Operons : Lac and Ara in portion Gene Expression : Transcription Prokaryotes of Cytogenetics
Answer» Correct choice is (b) I^+ + P + O^C + Z^– + Y^+ + A^+ / I^+ + P+ O^+ + Z^– + Y^+ + A^+ Explanation: This is because although this operon is constitutively turned on, yet it has a mutated LAC Z gene which produces beta galactosidase. This mutation EXISTS in both of the loci so it can’t PRODUCE this enzyme even if the operon is on.

Discussion

88.	Which of these is not a promoter element?(a) TATA box(b) DPE(c) CCAAT box(d) InrThe question was posed to me during an internship interview.The origin of the question is Transcription in Eukaryotes : Activators and Repressors topic in division Gene Expression : Transcription Prokaryotes of Cytogenetics
Answer» The CORRECT choice is (C) CCAAT box Best explanation: The CCAAT box is a part of the upstream activating elements which is upstream to the promoter. It is recognized by CTF or CCAAT RECOGNIZING Transcription Factor. The REMAINING options are part of a core promoter.

Discussion

89.	Promoter proximal elements is positioned ________________(a) Outside the core promoter(b) Downstream to start site(c) Within the enhancer(d) Just before the start siteI have been asked this question in homework.I want to ask this question from Transcription in Eukaryotes : Activators and Repressors in division Gene Expression : Transcription Prokaryotes of Cytogenetics
Answer» The CORRECT OPTION is (b) Downstream to start site For explanation I would say: The PROMOTER PROXIMAL elements are a part of the CORE promoter. They occur downstream to the promoter start site. Enhancers on the other hand are much farther and outside the core promoter.

Discussion

90.	RNA has a proofreading mechanism.(a) True(b) FalseThe question was posed to me in examination.This intriguing question comes from Transcription in Prokaryotes : Termination topic in division Gene Expression : Transcription Prokaryotes of Cytogenetics
Answer» Correct option is (a) True The best I can explain: A lot of ENERGY is invested in transcription so to ensure that a WRONG transcript is not produced RNA has a proofreading MECHANISM. This is seen in the PROCESS of backtracking.

Discussion

91.	What is the 1^st nucleotide of a completely synthesized transcript?(a) A(b) G(c) C(d) TI have been asked this question in an interview.Asked question is from Transcription in Prokaryotes : Elongation in chapter Gene Expression : Transcription Prokaryotes of Cytogenetics
Answer» Correct option is (b) G The BEST EXPLANATION: The 5’ cap can be considered as the 1^st nucleotide of the COMPLETE transcript. And as it is ppGpp the nucleotide is GUANINE or G.

Discussion

92.	Which would be an appropriate method to detect the core promoter regions in a eukaryotic gene?(a) Footprint(b) Sequencing(c) Linker scanning(d) S1 endonucleaseI got this question in semester exam.My question is from Transcription in Eukaryotes : Activators and Repressors in chapter Gene Expression : Transcription Prokaryotes of Cytogenetics
Answer» Correct ANSWER is (c) Linker scanning To explain: In linker scanning method a SMALL segment of the DNA is removed from PROMOTER region and replaced with a RANDOM segment to see its effect on the transcription rate. Thus, as absence of promoter would lead to less transcription this method can be used to detect promoter ELEMENTS.

Discussion

93.	Which of the following is the proper sequence of assembly of factors on the branch poinr A?(a) U1-> U2-> U6(b) U2AF -> U2->BBP(c) BBP->U2AF->U2(d) BBP->U1->U2I had been asked this question in a job interview.My question is taken from Post-Transcriptional Modification in chapter Gene Expression : Transcription Prokaryotes of Cytogenetics
Answer» The correct ANSWER is (C) BBP->U2AF->U2 The explanation is: BBP stands for BRANCH point binding PROTEINS. In the branch site A 1st the U2AF binds followed by BBP then it is replaced by U2. Then U2 can INTERACT with U6 that replaced U1.

Discussion

94.	The 3’- OH end of the branch point A makes the attack in splicing of an intron.(a) True(b) FalseI had been asked this question by my college director while I was bunking the class.My doubt is from Post-Transcriptional Modification in portion Gene Expression : Transcription Prokaryotes of Cytogenetics
Answer» The correct OPTION is (b) False The EXPLANATION: The 2’-OH of A base is seen to make this attack on the 5’ end of the intron at the phosphodiester bond while the original 3’-OH end of the A is STILL in the bonding with the PHOSPHATE. This makes a lariat structure.

Discussion

95.	Which of this is untrue about TBP?(a) It is a part of TFIID(b) It makes a kink in the DNA(c) It is orientation independent(d) It is sequence independentThe question was asked in a job interview.My enquiry is from Transcription in Eukaryotes : Transcription Factors topic in chapter Gene Expression : Transcription Prokaryotes of Cytogenetics
Answer» Correct choice is (d) It is sequence independent Best explanation: TBP specifically recognizes and BINDS to TATA consensus sequence which makes it sequence dependent. Its correct ORIENTATION is HOWEVER determined by other factors like TFIIA and TFIIB. It makes an 80 KINK in the DNA and is a part of TFIID.

Discussion

96.	How many stem loop structures play important role in complete system of tryptophan gene regulation?(a) 3(b) 4(c) 6(d) 8I got this question during an online interview.My question is taken from Operons : Trp in portion Gene Expression : Transcription Prokaryotes of Cytogenetics
Answer» Right option is (C) 6 The 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.

Discussion

97.	A Lac repressor is a tetramer repressed when bound to the inducer. The trp repressor is a ____________________(a) Dimer inactivated when bound to the inducer(b) Dimer activated on inducer binding(c) Tetramer inactivated on inducer binding(d) Tetramer activated on inducer bindingI had been asked this question by my school teacher while I was bunking the class.This is a very interesting question from Operons : Trp topic in division Gene Expression : Transcription Prokaryotes of Cytogenetics
Answer» Correct answer is (b) Dimer activated on INDUCER binding To explain: A trp REPRESSOR is a dimer which can’t bind to the cis ELEMENT unless it is bound to the inducer tryptophan. Only then it undergoes conformational CHANGE such that it then BINDS to the operator.

Discussion

98.	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 terminatorThis question was addressed to me in unit test.This question is from Transcription in Prokaryotes : Termination topic in chapter Gene Expression : Transcription Prokaryotes of Cytogenetics
Answer» The CORRECT answer is (a) No VISIBLE effect To EXPLAIN: 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.

Discussion

99.	What form of helix would you expect to see in double stranded RNA?(a) A form(b) B form(c) C form(d) D formI had been asked this question in an interview for internship.Question is from Transcription in Prokaryotes : Termination topic in section Gene Expression : Transcription Prokaryotes of Cytogenetics
Answer» Right answer is (a) A FORM The explanation is: 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.

Discussion

100.	In a stretch of DNA being transcribed, the region forward to the Elongation is __________(a) Positively supercoiled(b) Negatively supercoiled(c) Uncoiled(d) Forms a secondary structureI have been asked this question in my homework.This question is from Transcription in Prokaryotes : Elongation topic in section Gene Expression : Transcription Prokaryotes of Cytogenetics
Answer» CORRECT choice is (a) Positively supercoiled Best EXPLANATION: The region forward to the RNA polymerase elongation machinery is positively supercoiled DUE to the strain. If the GYRASE doesn’t release this strain the polymerizing machinery will stall. Such regions can’t be uncoiled or folded.

Discussion

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U rich RNA that helps in splicing are______________(a) sn RNA(b) snoRNA(c) hnRNA(d) mRNAI got this question during a job interview.Asked question is from Post-Transcriptional Modification in division Gene Expression : Transcription Prokaryotes of Cytogenetics

In the case of group 1 RNA, which one of the following can’t attack?(a) GMP(b) GDP(c) GTP(d) ATPThe question was asked in a national level competition.This key question is from Post-Transcriptional Modification topic in division Gene Expression : Transcription Prokaryotes of Cytogenetics

Which histone(s) are acted upon by HAT?(a) H1, H2(b) H3(c) H2A, H2B(d) H3, H4This question was posed to me during an interview for a job.I want to ask this question from Transcription in Eukaryotes : Chromatin Remodelling in section Gene Expression : Transcription Prokaryotes of Cytogenetics

Prokaryotes have 1 RNA polymerase. How many RNA polymerases are there in eukaryotes?(a) 2(b) 3(c) 4(d) NoneThe question was posed to me in final exam.I want to ask this question from Transcription in Eukaryotes : RNA Polymerases in chapter Gene Expression : Transcription Prokaryotes of Cytogenetics

Which operon behaves similar to Bacillus subtilis trp operon in terms of attenuation?(a) Lac operon(b) His operon(c) Phe operon(d) Ara operonThis question was addressed to me in my homework.Question is taken from Operons : Trp in section Gene Expression : Transcription Prokaryotes of Cytogenetics

Which of the Ara promoter regulation is controlled by binding of repressors on the O1 regulating unit?(a) P^A(b) P^B(c) P^C(d) P^DI got this question in my homework.The origin of the question is Operons : Lac and Ara topic in chapter Gene Expression : Transcription Prokaryotes of Cytogenetics

Which RNA polymerase can bind CBP?(a) I(b) II(c) III(d) NoneThe question was posed to me in a national level competition.My question comes from Transcription in Eukaryotes : RNA Polymerases topic in section Gene Expression : Transcription Prokaryotes of Cytogenetics

Which of these operons is anabolic?(a) Lac(b) Ara(c) Trp(d) PheThis question was addressed to me by my college director while I was bunking the class.This intriguing question originated from Operons : Trp topic in section Gene Expression : Transcription Prokaryotes of Cytogenetics

When Rho catches up with RNA polymerase it _______(a) Stalls(b) Falls off(c) Backtracks(d) Moves onThe question was asked during a job interview.My doubt stems from Transcription in Prokaryotes : Termination in section Gene Expression : Transcription Prokaryotes of Cytogenetics

Which factor phosphorylates the serine residues in RNA pol II CTD?(a) TFIIA(b) TFIIB(c) TFIIH(d) TFIIDThis question was posed to me during an internship interview.My enquiry is from Transcription in Eukaryotes : RNA Polymerases in division Gene Expression : Transcription Prokaryotes of Cytogenetics

Which part of sigma factor associates with -10box?(a) 1.2(b) 2.4(c) 4.4(d) 4.2I got this question in final exam.Asked question is from Transcription in Prokaryotes : Initiation in division Gene Expression : Transcription Prokaryotes of Cytogenetics

RNA has a proofreading mechanism.(a) True(b) FalseThe question was posed to me in examination.This intriguing question comes from Transcription in Prokaryotes : Termination topic in division Gene Expression : Transcription Prokaryotes of Cytogenetics

What is the 1^st nucleotide of a completely synthesized transcript?(a) A(b) G(c) C(d) TI have been asked this question in an interview.Asked question is from Transcription in Prokaryotes : Elongation in chapter Gene Expression : Transcription Prokaryotes of Cytogenetics

How many stem loop structures play important role in complete system of tryptophan gene regulation?(a) 3(b) 4(c) 6(d) 8I got this question during an online interview.My question is taken from Operons : Trp in portion Gene Expression : Transcription Prokaryotes of Cytogenetics