Logout Open In App
Open In App
Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Chapter 5: Problem 3

Termination of a prokaryotic transcript A. is a random process. B. requires the presence of the rho subunit of the holoenzyme. C. does not require rho factor if the end of the gene contains a G-C rich palindrome. D. is most efficient if there is an A-T-rich segment at the end of the gene. E. requires an ATPase in addition to rho factor.

Short Answer

Expert verified
A. Is a random process. B. Requires the presence of the rho subunit of the holoenzyme. C. Does not require the rho factor if the end of the gene contains a G-C rich palindrome. D. Is most efficient if there is an A-T-rich segment at the end of the gene. E. Requires an ATPase in addition to rho factor. Answer: C. Does not require the rho factor if the end of the gene contains a G-C rich palindrome.

Step by step solution

01

Choice A: Is a random process.

Termination of a prokaryotic transcript is not a random process. It is a regulated mechanism that ensures the transcription process is completed properly. So, choice A is not correct.
02

Choice B: Requires the presence of the rho subunit of the holoenzyme.

Rho factor is a protein that is involved in the termination process of a prokaryotic transcript, particularly in rho-dependent termination. However, there is another mechanism called rho-independent termination. Therefore, the presence of the rho subunit is not a requirement for all termination processes. So, choice B is not correct.
03

Choice C: Does not require the rho factor if the end of the gene contains a G-C rich palindrome.

In rho-independent termination, a G-C rich hairpin loop forms in the mRNA, which is followed by a series of U residues. This hairpin formation weakens the interaction between the RNA polymerase and the DNA template, causing termination without the requirement of rho factor. So, choice C is correct.
04

Choice D: Is most efficient if there is an A-T-rich segment at the end of the gene.

An A-T-rich segment at the end of the gene is not directly related to the efficiency of termination in prokaryotic transcription. It is the presence of the U residues in the mRNA that contributes to termination in rho-independent termination. So, choice D is not correct.
05

Choice E: Requires an ATPase in addition to rho factor.

Rho factor itself acts as an ATPase. Therefore, an additional ATPase is not required for the termination process. So, choice E is not correct. Based on the analysis, the correct answer is: C. Does not require rho factor if the end of the gene contains a G-C rich palindrome.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

In eukaryotic transcription, A. RNA polymerase does not require a template. B. all RNA is synthesized in the nucleolus. C. consensus sequences are the only known promoter elements. D. phosphodiester bond formation is favored because there is pyrophosphate hydrolysis. E. RNA polymerase requires a primer.

The synthesis of normal adult hemoglobin (HbA) requires the coordinated synthesis \(\alpha\) -globin and \(\beta\) -globin. \(\beta\) -Thalassemia is a genetic disease leading to a deficiency of \(\beta\) -globin chains and an inability of the blood to deliver oxygen properly. \(\beta\) -Thalassernia can result from a wide variety of mutations. One mutation leading to \(\beta\) -thalassemia occurs at a splice junction. Which of the following statements about removing introns is correct? A. Small nuclear ribonucleoproteins (snRNP) are necessary for removing introns. B. The consensus sequences at the \(5^{\prime}-\) and \(3^{\prime}\) -ends of introns are identical. C. Removal of an intron does not require metabolic energy. D. The exon at one end of an intron must always be joined to the exon at its other end. E. The nucleoside at the end of the intron first released forms a bond with a \(3^{\prime}\) -OH group on one of the nucleotides within the intron.

Protooncogenes produce products that have specific roles in regulating growth and differentiation of normal cells. Mutations can turn these genes into oncogenes whose products are less responsive to normal control. Unmutated protein \(\mathrm{p} 53,\) a tumor suppressor, is a transcription factor, inhibiting some genes and activating others. P53 inhibits genes with TATA sequences and activates genes for DNA repair. The TATA sequence A. occurs about 25 bp downstream from the start of transcription. B. binds directly to RNA polymerase. C. binds transcription factors which bind RNA polymerase. D. binds p53. E. is an enhancer sequence.

In the cellular turnover of RNA, A. repair is more active than degradation. B. regions of extensive base pairing are more susceptible to cleavage. C. a small double-stranded region of mRNA could lead to cleavage of the mRNA if one strand has perfect complementarity with a region of the mRNA. D. the products are always nucleotides with a phosphate at the \(5^{\prime}-\mathrm{OH}\) group. E. all species except rRNA are cleaved.

Fragile \(X\) syndrome is a common form of inherited mental retardation. The mutation in the disease allows the increase of a CGG repeat in a particular gene from a normal of about 30 repeats to \(200-1000\) repeats. This repeat is normally found in the \(5^{\prime}\) untranslated region of a gene for the protein FMR1. FMR1 might be involved in the translation of brain-specific mRNAs during brain development. The consequence of the very large number of \(C G G\) repeats in the DNA is extensive methylation of the entire promoter region of the FMR1 gene. Methylation of bases in DNA usually A. facilitates the binding of transcription factors to the DNA. B. makes a difference in activity only if it occurs in an enhancer region. C. prevents chromatin from unwinding. D. inactivates DNA for transcription. E. results in increased production of the product of whatever gene is methylated.
See all solutions

Recommended explanations on Chemistry Textbooks

Kinetics

Read Explanation

Nuclear Chemistry

Read Explanation

Physical Chemistry

Read Explanation

Inorganic Chemistry

Read Explanation

Organic Chemistry

Read Explanation

Chemical Analysis

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free