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Biological Molecules - Structure of DNA

Grade 12A LevelBiology

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

🔑Concepts

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DNA (Deoxyribonucleic Acid) is a polymer made of monomers called nucleotides. Each nucleotide consists of a phosphate group (PO43−PO_4^{3-}), a pentose sugar called deoxyribose (C5H10O4C_5H_{10}O_4), and a nitrogenous base.

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There are four types of nitrogenous bases divided into two categories: Purines (Adenine (AA) and Guanine (GG)) which have a double-ring structure, and Pyrimidines (Cytosine (CC) and Thymine (TT)) which have a single-ring structure.

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The DNA molecule consists of two polynucleotide strands running in opposite directions, known as being antiparallel (5′→3′5' \rightarrow 3' and 3′→5′3' \rightarrow 5').

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Complementary Base Pairing: Bases on opposite strands are held together by hydrogen bonds. Adenine (AA) always pairs with Thymine (TT) via 22 hydrogen bonds, while Guanine (GG) always pairs with Cytosine (CC) via 33 hydrogen bonds.

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The 'backbone' of the DNA strand is formed by alternating sugar and phosphate groups linked by covalent phosphodiester bonds.

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The two strands twist around a common axis to form a double helix structure, which protects the nitrogenous bases inside.

📐Formulae

[A]=[T][A] = [T]

[G]=[C][G] = [C]

[A]+[G]=[T]+[C][A] + [G] = [T] + [C]

%A+%T+%G+%C=100%\%A + \%T + \%G + \%C = 100\%

💡Examples

Problem 1:

A sample of double-stranded DNA is analyzed and found to contain 22%22\% Cytosine (CC). Calculate the percentage of Adenine (AA) present in the sample.

Solution:

The percentage of Adenine is 28%28\%.

Explanation:

According to Chargaff's rule, [C]=[G][C] = [G]. If C=22%C = 22\%, then G=22%G = 22\%. Together, C+G=44%C + G = 44\%. The remaining percentage for A+TA + T is 100%−44%=56%100\% - 44\% = 56\%. Since [A]=[T][A] = [T], the percentage of Adenine is 56%2=28%\frac{56\%}{2} = 28\%.

Problem 2:

If one strand of DNA has the sequence 5′−ATG CGT−3′5'-ATG CGT-3', what is the sequence of the complementary strand in the 5′→3′5' \rightarrow 3' direction?

Solution:

5′−ACG CAT−3′5'-ACG CAT-3'

Explanation:

First, find the complementary bases: A→TA \rightarrow T, T→AT \rightarrow A, G→CG \rightarrow C, C→GC \rightarrow G, G→CG \rightarrow C, T→AT \rightarrow A. This gives the 3′→5′3' \rightarrow 5' sequence: 3′−TAC GCA−5′3'-TAC GCA-5'. Reversing this to provide the answer in the 5′→3′5' \rightarrow 3' direction gives 5′−ACG CAT−3′5'-ACG CAT-3'.