Question 3 of 25intermediate🔧 ApplyNumerical2 marks

If the length of E. coli DNA is 1.36 mm, can you calculate the number of base pairs in E.coli?

Correct Answer

The number of base pairs in E. coli is 4×104 \times 10⁶ bp (approximately 4 million base pairs).

Given:

  • Length of E. coli DNA = 1.36 mm = 1.36×101.36 \times 10⁻³ m
  • Distance between two consecutive base pairs = 0.34 nm = 0.34×100.34 \times 10⁻⁹ m

Calculation: Number of base pairs = Total length of DNA / Distance between two consecutive base pairs

Number of base pairs = (1.36×101.36 \times 10⁻³ m) / (0.34×100.34 \times 10⁻⁹ m)

Number of base pairs = 4×104 \times 10⁶ bp

Exercise: In-text Question (Page 5) | Q: 3 | (Page 5)
For More Understanding

Explanation

This numerical problem is from NCERT Class 12 Biology Chapter 5 (Principles of Inheritance and Variation) / Chapter on Molecular Basis of Inheritance. The textbook provides the key constant that the distance between two consecutive base pairs in DNA is 0.34 nm. Using this value, students can calculate the number of base pairs by dividing the total DNA length by the distance per base pair. The calculated value of 4×104 \times 10⁶ bp closely matches the actual E. coli genome size of 4.6×104.6 \times 10⁶ bp mentioned elsewhere in the textbook context.

Solution Steps

  1. Step 1: Convert the length of E. coli DNA from mm to metres: 1.36 mm = 1.36×101.36 \times 10⁻³ m

  2. Step 2: Note the distance between two consecutive base pairs = 0.34 nm = 0.34×100.34 \times 10⁻⁹ m

  3. Step 3: Apply the formula: Number of base pairs = Total length / Distance between consecutive bp

  4. Step 4: Calculate: (1.36×101.36 \times 10⁻³) / (0.34×100.34 \times 10⁻⁹) = 4×104 \times 10⁶ bp

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