Skip to main content
Pearson+ LogoPearson+ Logo
Ch. 9 - DNA Structure and Analysis
Klug - Essentials of Genetics 10th Edition
Klug10th EditionEssentials of GeneticsISBN: 9780135588789Not the one you use?Change textbook
All textbooksKlug 10th EditionCh. 9 - DNA Structure and AnalysisProblem 30b
Chapter 9, Problem 30b

Newsdate: March 1, 2030. A unique creature has been discovered during exploration of outer space. Recently, its genetic material has been isolated and analyzed. This material is similar in some ways to DNA in its chemical makeup. It contains in abundance the 4-carbon sugar erythrose and a molar equivalent of phosphate groups. In addition, it contains six nitrogenous bases: adenine (A), guanine (G), thymine (T), cytosine (C), hypoxanthine (H), and xanthine (X). These bases exist in the following relative proportions:
A =T = H and C = G = X
X-ray diffraction studies have established a regularity in the molecule and a constant diameter of about 30 Å. Together, these data have suggested a model for the structure of this molecule.
What base-pairing properties must exist for H and for X in the model?

Verified step by step guidance
1
Step 1: Understand the given base proportions and their equivalences. The problem states that A = T = H and C = G = X in relative amounts, suggesting that these bases form pairs with equal molar ratios, similar to Chargaff's rules in DNA.
Step 2: Recall the base-pairing rules in DNA: adenine (A) pairs with thymine (T) via two hydrogen bonds, and cytosine (C) pairs with guanine (G) via three hydrogen bonds. These pairs maintain a constant diameter of the double helix.
Step 3: Since the molecule has a constant diameter (~30 Å) and the sugar-phosphate backbone is similar to DNA, the new bases hypoxanthine (H) and xanthine (X) must pair in a way that preserves this diameter and the regularity of the structure.
Step 4: Given the equivalences A = T = H and C = G = X, hypothesize that hypoxanthine (H) pairs with adenine (A) or thymine (T) analogously, and xanthine (X) pairs with cytosine (C) or guanine (G) analogously, maintaining complementary pairing and hydrogen bonding patterns.
Step 5: Conclude that the base-pairing properties of H and X must allow them to form stable hydrogen bonds with their complementary bases, preserving the molecular geometry and constant diameter, similar to how A-T and C-G pairs function in DNA.

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above.
Video duration:
1m
Was this helpful?

Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Base Pairing Rules in Nucleic Acids

Base pairing involves specific hydrogen bonding between nitrogenous bases, ensuring complementary pairing and stable double-stranded structures. In DNA, adenine pairs with thymine, and guanine pairs with cytosine, maintaining consistent molecular dimensions. Understanding these rules helps predict how novel bases like hypoxanthine (H) and xanthine (X) might pair to preserve structural regularity.
Recommended video:
Guided course
08:44
Genetics Basics

Chemical Structure and Properties of Nitrogenous Bases

Nitrogenous bases differ in ring structure and functional groups, influencing their hydrogen bonding capabilities. Hypoxanthine and xanthine are modified purines with keto and amino groups that affect their pairing potential. Analyzing their chemical groups helps determine possible complementary partners and the nature of hydrogen bonds they can form.
Recommended video:
Guided course
06:25
DNA Structure

Molecular Geometry and Structural Constraints of Nucleic Acids

The uniform diameter (~30 Å) and regularity observed via X-ray diffraction imply strict spatial constraints on base pairing. Complementary bases must fit within this geometry to maintain the molecule’s stability. This concept guides the prediction of how new bases pair to preserve consistent width and helical structure.
Recommended video:
Guided course
03:49
Ribosome Structure
Related Practice
Textbook Question

One of the most common spontaneous lesions that occurs in DNA under physiological conditions is the hydrolysis of the amino group of cytosine, converting the cytosine to uracil. What would be the effect on DNA structure of a uracil group replacing cytosine?

673
views
Textbook Question

In some organisms, cytosine is methylated at carbon 5 of the pyrimidine ring after it is incorporated into DNA. If a 5-methyl cytosine molecule is then hydrolyzed, what base will be generated?

642
views
Textbook Question

Newsdate: March 1, 2030. A unique creature has been discovered during exploration of outer space. Recently, its genetic material has been isolated and analyzed. This material is similar in some ways to DNA in its chemical makeup. It contains in abundance the 4-carbon sugar erythrose and a molar equivalent of phosphate groups. In addition, it contains six nitrogenous bases: adenine (A), guanine (G), thymine (T), cytosine (C), hypoxanthine (H), and xanthine (X). These bases exist in the following relative proportions:

A =T = H and C = G = X

X-ray diffraction studies have established a regularity in the molecule and a constant diameter of about 30 Å. Together, these data have suggested a model for the structure of this molecule.

Propose a general model of this molecule. Describe it briefly.

541
views
Textbook Question

Newsdate: March 1, 2030. A unique creature has been discovered during exploration of outer space. Recently, its genetic material has been isolated and analyzed. This material is similar in some ways to DNA in its chemical makeup. It contains in abundance the 4-carbon sugar erythrose and a molar equivalent of phosphate groups. In addition, it contains six nitrogenous bases: adenine (A), guanine (G), thymine (T), cytosine (C), hypoxanthine (H), and xanthine (X). These bases exist in the following relative proportions:

A =T = H and C = G = X

X-ray diffraction studies have established a regularity in the molecule and a constant diameter of about 30 Å. Together, these data have suggested a model for the structure of this molecule.

Given the constant diameter of 30 Å, do you think that either (i) both H and X are purines or both pyrimidines, or (ii) one is a purine and one is a pyrimidine?

593
views
Textbook Question

During gel electrophoresis, DNA molecules can easily be separated according to size because all DNA molecules have the same charge-to-mass ratio and the same shape (long rod). Would you expect RNA molecules to behave in the same manner as DNA during gel electrophoresis? Why or why not?

822
views