MP Board · Class 12 · Biology · Molecular Basis of InheritanceA double-stranded DNA fragment has a total of 10,000 base pairs. Analysis of the DNA shows that the proportion of Adenine is 20%. (i) Calculate the total number of Adenine, Thymine, Guanine, and Cytosine bases present in this DNA fragment. (ii) Calculate the total length of this DNA fragment in nanometers (nm), given that the distance between two consecutive base pairs is 0.34 nm.

Step-by-Step Numerical Solution:

Given data:

• Total number of base pairs (bp) = 10,000
• Total number of nucleotides = $10,000 \times 2 = 20,000$
• Percentage of Adenine (A) = 20%

Part (i): Calculation of individual nitrogenous bases

1. According to Chargaff's rules, the amount of Adenine (A) equals Thymine (T), and the amount of Guanine (G) equals Cytosine (C). $$%A = %T = 20%$$

2. The total sum of all four bases is 100%: $$%A + %T + %G + %C = 100%$$ $$20% + 20% + %G + %C = 100%$$ $$40% + %G + %C = 100%$$ $$%G + %C = 100% - 40% = 60%$$

3. Since $%G = %C$, we get: $$%G = \frac{60%}{2} = 30%$$ $$%C = 30%$$

4. Now, calculating the absolute number of each base out of 20,000 total nucleotides:

   • Number of Adenine (A): $$20% \text{ of } 20,000 = \frac{20}{100} \times 20,000 = 4,000$$
   • Number of Thymine (T): $$20% \text{ of } 20,000 = \frac{20}{100} \times 20,000 = 4,000$$
   • Number of Guanine (G): $$30% \text{ of } 20,000 = \frac{30}{100} \times 20,000 = 6,000$$
   • Number of Cytosine (C): $$30% \text{ of } 20,000 = \frac{30}{100} \times 20,000 = 6,000$$

Verification: Total bases = $4,000 + 4,000 + 6,000 + 6,000 = 20,000$ (Correct).

Part (ii): Calculation of the total length of the DNA fragment

1. Total number of base pairs = 10,000
2. Distance between two consecutive base pairs = $0.34 \text{ nm} = 0.34 \times 10^{-9} \text{ m}$
3. Total length of DNA = $\text{Total number of base pairs} \times \text{Distance between two adjacent base pairs}$ $$\text{Total Length} = 10,000 \times 0.34 \text{ nm}$$ $$\text{Total Length} = 3,400 \text{ nm}$$ (or $3.4 \times 10^{-6} \text{ meters}$)