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Ch. 10 - DNA Structure and Analysis
Klug - Concepts of Genetics  12th Edition
Klug12th EditionConcepts of Genetics ISBN: 9780135564776Not the one you use?Change textbook
Chapter 10, Problem 32a

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.

Verified step by step guidance
1
Step 1: Identify the key components of the molecule based on the problem description. The molecule contains a 4-carbon sugar called erythrose, phosphate groups in molar equivalence, and six nitrogenous bases: adenine (A), guanine (G), thymine (T), cytosine (C), hypoxanthine (H), and xanthine (X).
Step 2: Analyze the base pairing rules implied by the relative proportions: A = T = H and C = G = X. This suggests that bases are grouped into two sets with equal molar amounts, indicating possible base pairing between these groups.
Step 3: Consider the structural implications of the X-ray diffraction data showing a regular structure with a constant diameter of about 30 Å. This suggests a helical or double-stranded structure similar to DNA, but with a larger diameter due to the different sugar and additional bases.
Step 4: Propose that the molecule forms a double helix with strands composed of erythrose-phosphate backbones, where bases from one strand pair specifically with bases from the complementary strand. The base pairing likely involves hydrogen bonding between A, T, and H on one side and C, G, and X on the other, maintaining the constant diameter.
Step 5: Summarize the model by describing the molecule as a double-stranded helical structure with erythrose sugars and phosphate groups forming the backbone, and six nitrogenous bases pairing in a specific manner to stabilize the structure and maintain uniform diameter.

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Key Concepts

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

Nucleic Acid Structure and Components

Nucleic acids are polymers made of sugar-phosphate backbones and nitrogenous bases. DNA typically contains a 5-carbon sugar (deoxyribose), phosphate groups, and four bases (A, T, G, C). Understanding the chemical makeup and base composition is essential to model any nucleic acid-like molecule, including variations in sugar type and additional bases.
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Ribosome Structure

Base Pairing and Complementarity

Base pairing involves hydrogen bonding between complementary nitrogenous bases, which stabilizes the nucleic acid structure. In DNA, A pairs with T and G pairs with C. The question’s base proportions (A=T=H and C=G=X) suggest specific pairing rules that must be considered to propose a stable double-stranded structure.
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Complementation

X-ray Diffraction and Molecular Geometry

X-ray diffraction reveals the three-dimensional arrangement and dimensions of molecules, such as the helical structure and diameter of nucleic acids. A constant diameter (~30 Å) indicates a regular, possibly double-helical structure, guiding the spatial arrangement of sugar-phosphate backbones and base pairs in the proposed model.
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Related Practice
Textbook Question

A primitive eukaryote was discovered that displayed a unique nucleic acid as its genetic material. Analysis provided the following information:

A major hyperchromic shift is evident upon heating and monitoring UV absorption at 260 nm.

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Textbook Question

A primitive eukaryote was discovered that displayed a unique nucleic acid as its genetic material. Analysis provided the following information:

Base-composition analysis reveals four bases in the following proportions: Adenine = 8%; Guanine = 37%; Xanthine = 37%; Hypoxanthine = 18%

409
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Textbook Question

A primitive eukaryote was discovered that displayed a unique nucleic acid as its genetic material. Analysis provided the following information:

About 75 percent of the sugars are deoxyribose, while 25 percent are ribose.

Postulate a model for the structure of this molecule that is consistent with the foregoing observations.

405
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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.

What base-pairing properties must exist for H and for X in the model?

531
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?

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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?

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