Cytogenetics Questions and Answers – Transcription in Eukaryotes : RNA Polymerases

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

1. Prokaryotes have 1 RNA polymerase. How many RNA polymerases are there in eukaryotes?
a) 2
b) 3
c) 4
d) None
View Answer

Answer: b
Explanation: 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.
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2. You isolate the nucleolus from the nucleus and performed a density gradient centrifugation with the extract. Then you check the different for the RNA polymerase activity of the different segment. Which RNA polymerase would you expect to find in abundance?
a) RNA pol I
b) RNA pol II
c) RNA pol III
d) RNA pol IV
View Answer

Answer: a
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.

3. Which RNA polymerase can bind CBP?
a) I
b) II
c) III
d) None
View Answer

Answer: b
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.
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4. What is the sequence of the heptad repeat of C terminal domain?
a) TSPWSTT
b) YSPTSGS
c) YSPTSPS
d) YSLPSTS
View Answer

Answer: c
Explanation: A heptad repeat sequence is YSPTSPS where there are 3 phosphoryable serine residues at position 3, 5 and 7.

5. You designed an antibody against phosphorylated CTD of RNA polymerase II. Then you use a DNA fragment with lac operon and try to immunoprecipitate the RNA pol II. What will be the expected result?
a) Moderate concentration of RNA pol II will precipitate
b) RNA pol II that precipitated will be contaminated with pol III
c) High amount of RNA pol II will precipitate
d) No RNA pol II will precipitate
View Answer

Answer: d
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.
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6. If you consider a transcribing RNA pol II with a transcript as small as 30 bases, which serine residue in its CTD will be phosphorylated?
a) 2
b) 5
c) 7
d) None
View Answer

Answer: b
Explanation: when RNA transcript is only 30 bases long it is still within the initiation range. Then the 2nd serine residue is phosphorylared only.

7. Which factor phosphorylates the serine residues in RNA pol II CTD?
a) TFIIA
b) TFIIB
c) TFIIH
d) TFIID
View Answer

Answer: c
Explanation: 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.
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8. In what sequence are the serine residues in RNA pol II phosphorylated?
a) 5->2->7
b) 2->5->7
c) 2->7->5
d) 7->2->5
View Answer

Answer: c
Explanation: The phosphorylation is at the 2nd serine residue with the help of the TFIIH, and then during elongation the phosphorylation is at the 5th serine residue. Between these two there is a short phase of phosphorylation at the 7th serine.

9. In an experiment you add growing concentration of the alpha amanitin. What will be the sequence in which RNA polymerases are affected if at all?
a) Pol I -> pol III , pol II unaffected
b) pol I->pol I ->pol III
c) pol III->pol II -> pol I
d) pol II -> pol III , pol I unaffected
View Answer

Answer: d
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.
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10. 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 subunit
View Answer

Answer: b
Explanation: RPA 2 is homologous to the beta subunit while RPA 1 is homologous to the beta prime subunit which is the catalytic subunit.

11. How many of eukaryotic RNA polymerase subunits are common for all three polymerases?
a) 3
b) 5
c) 10
d) 7
View Answer

Answer: b
Explanation: 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.

12. All the RNA polymerases share 5 common subunits, which are present in all the three polymerases and are all alike. State whether the statement is true or false.
a) True
b) False
View Answer

Answer: b
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.

13. 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) Magenta
View Answer

Answer: a
Explanation: 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.

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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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