Question

A DNA molecule is 160 base pairs long. It has 30% Guanine. How many Adenine bases are present in this DNA molecule?

• A. 48
• B. 64
• C. 96
• D. 192

Hint:

An alternative faster calculation route: - $\%A = 50\% - \%G = 50\% - 30\% = 20\%$. - Total bases = $160 \times 2 = 320$. - Adenine count = $320 \times 0.20 = 64$. Always remember to multiply base pairs by 2 to get individual bases!

Answer 1

The Correct Option is B

Concept: According to Chargaff's Rules of base pairing for double-stranded DNA molecules:
• The total number of purines equals the total number of pyrimidines: $A + G = T + C$.
• Adenine pairs exclusively with Thymine via two hydrogen bonds ($A = T$).
• Guanine pairs exclusively with Cytosine via three hydrogen bonds ($G = C$).
• Consequently, the percentage values satisfy: $\%A = \%T$ and $\%G = \%C$.

Step 1: Compute the total number of individual nucleotide bases.
The double-stranded DNA molecule is 160 base pairs (bp) long. Since every single base pair consists of exactly 2 individual nitrogenous bases, the total count of bases ($N$) contained within the molecule is: \[ N = 160 \times 2 = 320 \text{ individual bases} \]

Step 2: Determine the percentage allocation for all four bases.
We are given that the percentage of Guanine is 30%: \[ \%G = 30\% \] By Chargaff’s rule, Cytosine must have an identical percentage: \[ \%C = \%G = 30\% \] Summing up the Guanine and Cytosine composition gives: \[ \%G + \%C = 30\% + 30\% = 60\% \] The remaining percentage of the DNA molecule must be composed entirely of Adenine and Thymine bases: \[ \%A + \%T = 100\% - 60\% = 40\% \] Since Adenine and Thymine occur in a perfect 1:1 ratio ($\%A = \%T$), we divide this remaining percentage by 2: \[ \%A = \frac{40\%}{2} = 20\% \]

Step 3: Calculate the absolute number of Adenine bases.
Now, we calculate 20% of the total 320 individual bases present in the DNA structure: \[ \text{Number of Adenine bases} = 20\% \text{ of } 320 = \frac{20}{100} \times 320 \] Simplifying the fractions: \[ \text{Number of Adenine bases} = 0.2 \times 320 = 64 \] Thus, there are exactly 64 Adenine bases, matching Option (B).