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Biomolecules - Nucleic Acids (DNA and RNA)

Grade 12CBSEChemistry

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Nucleic acids are long-chain polymers of nucleotides, also known as polynucleotides. They are responsible for the storage and transfer of genetic information.

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A nucleotide consists of three chemical components: a pentose sugar, a nitrogenous base, and a phosphoric acid group.

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In DNA, the sugar is β-D-2-deoxyribose\beta\text{-D-}2\text{-deoxyribose}, whereas in RNA, the sugar is β-D-ribose\beta\text{-D-ribose}.

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Nitrogenous bases are categorized into Purines (Adenine AA and Guanine GG) and Pyrimidines (Cytosine CC, Thymine TT, and Uracil UU).

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DNA contains the bases AA, GG, CC, and TT. RNA contains AA, GG, CC, and UU (Uracil replaces Thymine).

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A nucleoside is formed by the attachment of a base to the 1′1' position of sugar. A nucleotide is formed when the 5′5'-OH group of the nucleoside is esterified with phosphoric acid.

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Nucleotides are joined together by phosphodiester linkages between 5′5' and 3′3' carbon atoms of the pentose sugar.

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DNA has a double-stranded helical structure where strands are antiparallel and held together by hydrogen bonds. AA pairs with TT via two hydrogen bonds (A=TA=T), and GG pairs with CC via three hydrogen bonds (G≡CG \equiv C).

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Biological functions include Replication (DNA creating copies of itself) and Protein Synthesis (Transcription and Translation).

📐Formulae

Nucleoside=Sugar+Base\text{Nucleoside} = \text{Sugar} + \text{Base}

Nucleotide=Sugar+Base+Phosphoric Acid\text{Nucleotide} = \text{Sugar} + \text{Base} + \text{Phosphoric Acid}

Chargaff’s Rule: [A]+[G]=[T]+[C]\text{Chargaff's Rule: } [A] + [G] = [T] + [C] (for double-stranded DNA)

Linkage: 3′,5′-phosphodiester linkage\text{Linkage: } 3', 5'\text{-phosphodiester linkage}

💡Examples

Problem 1:

If a double-stranded DNA has 20%20\% of Cytosine, calculate the percentage of Adenine in the DNA.

Solution:

According to Chargaff's rule, [C]=[G][C] = [G] and [A]=[T][A] = [T]. Given [C]=20%[C] = 20\%, therefore [G]=20%[G] = 20\%. The total C+G=20%+20%=40%C + G = 20\% + 20\% = 40\%. The remaining 100%−40%=60%100\% - 40\% = 60\% must be A+TA + T. Since [A]=[T][A] = [T], then [A]=60%2=30%[A] = \frac{60\%}{2} = 30\%.

Explanation:

In double-stranded DNA, the concentration of Purines equals the concentration of Pyrimidines due to specific base pairing.

Problem 2:

Write the complementary strand for the DNA sequence: 5′-ATGCCTGA-3′5'\text{-ATGCCTGA-}3'.

Solution:

3′-TACGGACT-5′3'\text{-TACGGACT-}5' or 5′-TCAGGCAT-3′5'\text{-TCAGGCAT-}3'.

Explanation:

DNA strands are antiparallel and follow base-pairing rules (AA with TT, GG with CC). The sequence is read in the 5′5' to 3′3' direction.

Problem 3:

Differentiate between a nucleoside and a nucleotide.

Solution:

A nucleoside contains only a nitrogenous base and a pentose sugar (e.g., Adenosine). A nucleotide contains a nitrogenous base, a pentose sugar, and a phosphate group (e.g., Adenylic acid).

Explanation:

The addition of a phosphate group to a nucleoside at the C5′C5' position converts it into a nucleotide.

Nucleic Acids (DNA and RNA) Class 12 Notes & Examples