Q.55 A double helical DNA molecule is composed of 32 mol % of adenosine. The mol % of cytosine in
this DNA molecule is _____.
The mol% of cytosine in this double helical DNA molecule is 18.
Chargaff’s rules dictate equal mol% of adenine (A) to thymine (T), and guanine (G) to cytosine (C), in double-stranded DNA.
Given 32 mol% adenosine (A), apply these base-pairing equalities to find C.
Step-by-Step Solution
Adenosine refers to adenine (A) residues in DNA context. Thus, mol% A = 32%.
Mol% T = mol% A = 32%, since A pairs with T via two hydrogen bonds.
A + T + G + C = 100%, so G + C = 100% – 32% – 32% = 36%; and since G = C, mol% C = 36%/2 = 18%.
Option Analysis
This fill-in-the-blank question (common in CSIR NET Life Sciences) expects a numeric answer without listed options, but typical distractors include:
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32%: Matches A but ignores G=C pairing.
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36%: Equals G+C total, not C alone.
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64%: Incorrectly assumes A+T only (ignores G=C).
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18%: Correct, as derived from (100 – 2×32)/2.
Chargaff’s Rules Explained
First Parity Rule: In dsDNA, %A = %T and %G = %C globally, validating Watson-Crick model.
Second Parity Rule: Holds per strand, ensuring bidirectional complementarity.
Purines (A+G) equal pyrimidines (T+C), stabilizing the helix.
For CSIR NET (Unit 3: Inheritance Biology), memorize: Given one base, compute pairs sequentially.
Exam Application
Such problems test base composition without sequencing data. Practice: If C=20%, then G=20%, A=T=30% each.
Slug: double-helical-dna-32-mol-adenosine-cytosine
