The total number of hydrogen bonds in this 1500 base pair double-stranded B-DNA, where 70% of purines are adenine, is 3450.

Key Concepts

Double-stranded B-DNA has 1500 base pairs, meaning 1500 A-T and G-C pairs held by Watson-Crick hydrogen bonds: 2 per A-T pair and 3 per G-C pair. Purines (A + G) equal pyrimidines (T + C) per Chargaff’s rules. The condition “70% of purines are adenine” means A = 0.7(A + G), so G = 0.3(A + G).

Step-by-Step Calculation

Let total purines per strand = P, so A = 0.7P and G = 0.3P. Then T = A = 0.7P and C = G = 0.3P. Total bases per strand = 3000, so P + (T + C) = 2P = 3000, giving P = 1500.

Thus, A-T pairs = 0.7 × 1500 = 1050 and G-C pairs = 0.3 × 1500 = 450. Hydrogen bonds = (1050 × 2) + (450 × 3) = 2100 + 1350 = 3450.

Detailed Solution

Total base pairs N = 1500. Let fraction of purine sites that are A = 0.7, so fraction A-T pairs f_AT = 0.7, f_GC = 0.3. H-bonds = N × (2 f_AT + 3 f_GC) = 1500 × (2×0.7 + 3×0.3) = 1500 × (1.4 + 0.9) = 1500 × 2.3 = 3450.

In double stranded B-DNA, hydrogen bonds stabilize the helix through Watson-Crick base pairing. This guide solves the exact query: “In a 1500 base pair double stranded B-DNA, 70% of purines are adenine bases. The total number of hydrogen bonds will be _____.” Perfect for CSIR NET Life Sciences preparation.

Understanding Purines and Base Pairing

Purines (adenine, guanine) pair with pyrimidines (thymine, cytosine). A-T forms 2 H-bonds, G-C forms 3. Chargaff’s rules ensure A = T, G = C, purines = pyrimidines.

Solving the Problem

  • Total bp = 1500, so 1500 pairs.
  • Purines per strand = 1500 bases (half of 3000).
  • A = 70% of purines = 1050 per strand.
  • G = 30% = 450 per strand.
  • A-T pairs = 1050, G-C pairs = 450.
  • Total H-bonds = 1050×2 + 450×3 = 2100 + 1350 = 3450.

No options to explain, but common traps: forgetting dsDNA has equal per strand, or confusing % purines total. For CSIR NET, remember average H-bonds per bp = 2.3 here.

Answer: 3450