Cytogenetics Questions and Answers – Transcription in Prokaryotes : Initiation


This set of Cytogenetics Multiple Choice Questions & Answers (MCQs) focuses on “Transcription in Prokaryotes : Initiation”.

1. Which protein mentioned below can reverse central dogma?
a) Ribosome
b) Restriction Endonuclease
c) Reverse Transcriptase
d) RNA Polymerase
View Answer

Answer: c
Explanation: 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.

2. 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)
View Answer

Answer: a
Explanation: β′ and β subunits are mainly associated with the transcription start site where β′ subunit mainly confers the catalytic activity. The sigma subunit provides specificity while alpha subunit interacts with regulatory proteins.

3. 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’ strand
View Answer

Answer: d
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.

4. In an experiment you use RNA polymerase without its sigma factor for transcription. What will be the result that you observe?
a) More transcription
b) Less transcription
c) More specific transcription
d) More random transcription
View Answer

Answer: d
Explanation: Sigma subunit is responsible for recognizing the transcription start site. In its absence the RNA pol will transcribe at non-specific random sites. Sigma plays no role in the rate of transcription so there is no chance in transcription rate in its absence.

5. In an experiment you mutate the consensus sequence at the -10 box. You observe the rate of transcription reduces. Now you make complementary mutation to the sigma factor. What will you observe?
a) Further decrease in transcription rate
b) Increase in transcription rate
c) Original transcription rate
d) Increase in rate of random transcription
View Answer

Answer: c
Explanation: Complementary mutation in sigma factor helps to restore its -10 box recognition capacity which restores normal transcription rate. However, to improve the rate further both the -10 and -35 box sequences should be made more like the consensus.

6. You wanted to isolate active RNA polymerase (with a part of the transcript associated with it ) from cellular mix. For this, you perform an immunoprecipitation with anti sigma antibody. Then you check the precipitate. What will you observe?
a) RNA polymerase with transcript
b) RNA polymerase without transcript
c) Only antibody
d) DNA with associated RNA polymerase and a transcript
View Answer

Answer: b
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.

7. Which part of sigma factor associates with -10box?
a) 1.2
b) 2.4
c) 4.4
d) 4.2
View Answer

Answer: b
Explanation: the 2.4 i.e. (4th part of 2nd 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.

8. 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 increases
View Answer

Answer: b
Explanation: As the unlabelled mRNA against X will compete with the labeled one for hybridization the intensity of radioactivity will gradually decrease with increased concentration. But as it will not bind to Y gene the Y labeled mRNA faces no competition and its intensity remains same.

9. Which of these is the 1st event to take place during transcription initiation?
a) Formation of a closed initiation complex
b) Formation of open initiation complex
c) Formation of absorptive transcript
d) Promoter clearance
View Answer

Answer: b
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.

10. 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 site
View Answer

Answer: c
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.

11. In an experiment you mutate the C terminal domain of alpha subunit of the RNA polymerase. What will you expect to see?
a) Transcription is absent
b) Transcription is at random sites
c) Transcription is less
d) Transcription is more
View Answer

Answer: c
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.

12. Mutations as far as 100 base pairs upstream to +1 site will also affect the transcription rate.
a) True
b) False
View Answer

Answer: a
Explanation: 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.

13. In an experiment you add increasing amount of sigma factor to a mixture of DNA fragment and core polymerase in vitro. You also add [14C]ATP and [γ-32P]ATP in the mixture and check the rate of incorporation. Results show that the incorporation of both of this labels increase on adding more sigma factor. What will you conclude from your observation?
a) Sigma confers specificity and increases transcription rate
b) Sigma increases initiation and elongation rate
c) Sigma increases initiation of transcription rate
d) Sigma causes faster elongation
View Answer

Answer: b
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.

14. In an experiment, you wanted to see which mutation will have worse effect on transcription. In the 1st set you deleted 94 amino acids from the C-terminus and in the 2nd set you replace the 265th C-terminal arginine to cystine. Which in your opinion will have the worst effect?
a) 94 amino acid mutant
b) 265th amino acid mutant
c) Both will have similar negative effect
d) No visible effect on rate of transcription
View Answer

Answer: c
Explanation: This is because the 265th arginine is most important for recognition of the up element. Thus, its deletion alone can have a similar negative effect on transcription to the deletion of 94 amino acids from C terminal end.

15. Which cation is placed in the catalytic subunit of RNA polymerase?
a) Zn2+
b) Mn2+
c) Mg2+
d) Fe3+
View Answer

Answer: c
Explanation: 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.

Sanfoundry Global Education & Learning Series – Cytogenetics.

To practice all areas of Cytogenetics, here is complete set of 1000+ Multiple Choice Questions and Answers.

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Manish Bhojasia, a technology veteran with 20+ years @ Cisco & Wipro, is Founder and CTO at Sanfoundry. He lives in Bangalore, and focuses on development of Linux Kernel, SAN Technologies, Advanced C, Data Structures & Alogrithms. Stay connected with him at LinkedIn.

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