CBSE · Class 12 · Biology · Molecular Basis of InheritanceA double-stranded DNA fragment has 10,000 base pairs (bp). If the proportion of Adenine is 20%, calculate the total number of Adenine, Thymine, Guanine, and Cytosine nucleotides present in this DNA fragment. Also, calculate the total length of the DNA double helix in nanometers.
Step-by-Step Solution
Step 1: Calculate Total Nucleotides\nGiven:
- Total base pairs (bp) = $10,000$
- Since each base pair consists of 2 nucleotides, total number of nucleotides $= 10,000 \times 2 = 20,000$.
Step 2: Apply Chargaff's Rules\nAccording to Chargaff's rules for double-stranded DNA:
- Percentage of Adenine (A) = Percentage of Thymine (T) $= 20%$
- Therefore, Percentage of Guanine (G) + Percentage of Cytosine (C) $= 100% - (20% + 20%) = 100% - 40% = 60%$
- Since G = C, percentage of Guanine = $30%$ and Cytosine = $30%$.
Step 3: Calculate Actual Number of Each Nitrogenous Base
- Number of Adenine (A) $= 20% \text{ of } 20,000 = \frac{20}{100} \times 20,000 = 4,000$
- Number of Thymine (T) $= 20% \text{ of } 20,000 = 4,000$
- Number of Guanine (G) $= 30% \text{ of } 20,000 = \frac{30}{100} \times 20,000 = 6,000$
- Number of Cytosine (C) $= 30% \text{ of } 20,000 = 6,000$
Step 4: Calculate Total Length of DNA Double Helix
- Distance between two consecutive base pairs in a DNA double helix $= 0.34 \text{ nm}$ (or $3.4 \text{\AA}$).
- Total length $= \text{Total base pairs} \times \text{Distance between adjacent base pairs}$
- Total length $= 10,000 \times 0.34 \text{ nm} = 3,400 \text{ nm}$.
Final Answer Summary:
- Adenine = $4,000$
- Thymine = $4,000$
- Guanine = $6,000$
- Cytosine = $6,000$
- Total length of DNA = $3,400 \text{ nm}$
💡 Study Guide: This question tests core syllabus concepts from Molecular Basis of Inheritance. For formulas, key summaries, and mock exam reference guides, read the full Molecular Basis of Inheritance Revision Notes.