Textbook Question
Draw the structures of the following nucleotides.
(c) cyclic guanosine monophosphate (cGMP)
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Ch. 23 - Carbohydrates and Nucleic Acids
Wade9th EditionOrganic ChemistryISBN: 9780135213728
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Table of contents
- Ch.1 - Structure and Bonding107
- Ch. 2 - Acids and Bases; Functional Groups115
- Ch.3 - Structure and Stereochemistry of Alkanes100
- Ch.4 - The Study of Chemical Reactions99
- Ch.5 - Stereochemistry93
- Ch.6 - Alkyl Halides; Nucleophilic Substitution118
- Ch. 7 - Structure and Synthesis of Alkenes; Elimination158
- Ch.8 - Reactions of Alkenes171
- Ch.9 - Alkynes91
- Ch.10 - Structure and Synthesis of Alcohols143
- Ch.11 - Reactions of Alcohols104
- Ch. 12 - Infrared Spectroscopy and Mass Spectrometry22
- Ch. 13 - Nuclear Magnetic Resonance Spectroscopy45
- Ch. 14 - Ethers, Epoxides, and Thioethers72
- Ch. 15 - Conjugated Systems, Orbital Symmetry, and Ultraviolet Spectroscopy89
- Ch. 16 - Aromatic Compounds74
- Ch. 17 - Reactions of Aromatic Compounds126
- Ch. 18 - Ketones and Aldehydes193
- Ch. 19 - Amines129
- Ch. 20 - Carboxylic Acids77
- Ch. 21 - Carboxylic Acid Derivatives141
- Ch. 22 - Condensations and Alpha Substitutions of Carbonyl Compounds175
- Ch. 23 - Carbohydrates and Nucleic Acids93
- Ch. 24 - Amino Acids, Peptides, and Proteins54
Chapter 23, Problem 64
Draw the structure of a four-residue segment of DNA with the following sequence. (3′end) G-T-A-C (5′ end)
Verified step by step guidance1
Understand the problem: You are tasked with drawing a four-residue segment of DNA with the sequence G-T-A-C, starting from the 3′ end and ending at the 5′ end. DNA is composed of nucleotides, each containing a sugar (deoxyribose), a phosphate group, and a nitrogenous base (Guanine, Thymine, Adenine, Cytosine).
Step 1: Begin by drawing the backbone of the DNA strand. The backbone consists of alternating deoxyribose sugars and phosphate groups. Each sugar is connected to the next via a phosphodiester bond, which links the 3′ hydroxyl group of one sugar to the 5′ phosphate group of the next sugar.
Step 2: Attach the nitrogenous bases to the 1′ carbon of each deoxyribose sugar. For the given sequence (3′ end G-T-A-C 5′ end), attach Guanine (G) to the first sugar, Thymine (T) to the second sugar, Adenine (A) to the third sugar, and Cytosine (C) to the fourth sugar.
Step 3: Ensure the correct orientation of the strand. The 3′ end of the DNA strand should have a free hydroxyl (-OH) group on the 3′ carbon of the first sugar, and the 5′ end should have a free phosphate group attached to the 5′ carbon of the last sugar.
Step 4: Double-check the structure for accuracy. Verify that the phosphodiester bonds are correctly formed, the bases are properly attached to the sugars, and the strand is oriented from 3′ to 5′ as specified in the problem.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
DNA Structure
DNA is a double helix composed of nucleotides, each containing a sugar, a phosphate group, and a nitrogenous base. The sequence of these bases (adenine, thymine, cytosine, and guanine) encodes genetic information. In a DNA strand, the 3' and 5' ends refer to the orientation of the sugar-phosphate backbone, which is crucial for understanding how strands are synthesized and paired.
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DNA Double Helix Concept 1
Base Pairing
In DNA, base pairing occurs between specific nitrogenous bases: adenine pairs with thymine (A-T) and cytosine pairs with guanine (C-G). This complementary pairing is essential for the stability of the DNA double helix and for accurate replication and transcription processes. Understanding base pairing is vital for drawing correct DNA sequences.
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Directionality of DNA Strands
DNA strands have directionality, indicated by the 5' and 3' ends. The 5' end has a phosphate group attached to the fifth carbon of the sugar, while the 3' end has a hydroxyl group on the third carbon. This directionality is important for processes like DNA replication and transcription, as enzymes synthesize new strands in a 5' to 3' direction.
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1:49DNA Double Helix Example 1