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 24: Q15P (page 876)

For the RNA sequence AUUGGCAUCCGAUAA, draw the secondary structure that maximizes its base pairing.

Short Answer

Expert verified

The secondary structure is given below:

Step by step solution

01

Definition of Concept

Polyribonucleotide poly(A) melting curve has a sigmoidal shape. At first, the absorbance increases at a slow rate. As the temperature rises, the absorbance increases exponentially until it reaches a steady-state.

02

Step 2:Draw the secondary structure

RNAs have a variety of secondary structures. Helixes, internal loops, bulges, and hairpin loops may be present.

The RNA sequence AUUGGCAUCCGAUAA could consist of five (5) base pairs.

The base pairing is as follows:

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

How many histone variants occur in human cells? How do they differ from each other? How does the histone variant CENP-A relate to chromosomal structure? Which histone variant is associated with X-chromosome inactivation?

Use the information in Table 11-1 to estimate the half-life, in years, of a phosphodiester bond in DNA.

Polyoma virus DNA can be separated by sedimentation in an ultracentrifuge (Section 5-2E) at neutral pH into three components that have sedimentation coefficients of 20, 16, and 14.5S and that are known as Types I, II, and III DNAs, respectively. These DNAs all have identical base sequences and molecular masses. In 0.15 M NaCl, both Types II and III DNA have melting curves of normal cooperativity and a Tm of 88°C. Type I DNA, however, exhibits a very broad melting curve and a Tm of 107°C. At pH 13, Types I and III DNAs have sedimentation coefficients of 53 and 16S, respectively, and Type II separates into two components with sedimentation coefficients of 16 and 18S. How do Types I, II, and III DNAs differ from one another? Explain their different physical properties.

Which of the DNAs shown in Figure 24-8 would move fastest during agarose gel electrophoresis?

A standard A - T base pair is shown in Fig. 24-1. Under certain conditions, an adenine residue in B-DNA can transiently adopt the syn conformation. Draw the resultingbase pair. (Hint: It has two hydrogen bonds.) Is the helix diameter larger or smaller in this alternate arrangement?
See all solutions

Recommended explanations on Biology Textbooks

Reproduction

Read Explanation

Cellular Energetics

Read Explanation

Cells

Read Explanation

Control of Gene Expression

Read Explanation

Biological Processes

Read Explanation

Responding to Change

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