Chapter 26.3, Problem 26.7P

Interpretation Introduction

Interpretation:

The full structure of the DNA dinucleotide C – T has to be drawn and 5’ and 3’ ends has to be identified.

Concept Introduction:

Composition of nucleic acid: Nucleic acid is a polymer of nucleotides. Each nucleotide has three parts: a sugar, a nitrogenous base, and a phosphate group. Two nucleotides are joined by phosphate diester linkage where a free phosphate on 5’ carbon of one nucleotide and a free –OH group on 3’ carbon of another nucleotide.

Sugar: In both DNA and RNA, sugar portion is found. In DNA, the sugar is D-ribose, where at 2’hydroxyl group is absent and in RNA, the hydroxyl group is present at 2’.

Nitrogenous bases: Five types of nitrogenous bases (has unique one-letter code A, G, T, U, and C) are derived from two parent compounds called purine and pyrimidine. The purine derivatives are Adenine and Guanine are two fused nitrogen containing rings. The pyrimidine derivatives are Thymine, Cytosine, and Uracil are only one nitrogen containing six-membered ring.  Adenine, Guanine, Thymine, and Cytosine are the nitrogenous bases present in DNA. Adenine, Guanine, Cytosine and Uracil are the nitrogenous bases present in RNA.

Nucleotide: (Nucleoside + phosphate)

Nucleotides are the building blocks of nuclei acids; monomers of DNA and RNA polymers. At carbon-5’ of the ribose sugar, a phosphate group is added which is collectively known as nucleotide. Phosphate groups can be added to any of the nucleotide to form diphosphate or triphosphate.

Numbering the atoms in sugar and base rings:

In order to distinguish the atoms in the sugar of a nucleoside and atoms of a base ring, numbers without prime is used for atoms in the base ring and numbers with prime used for the atoms in the sugar ring.