Biology - Molecular Basis of Inheritance Question with Solution | TestHub

Length of the DNA segment = $340\text{A˚}$

Distance between two consecutive base pairs in B-DNA = $3.4\text{A˚}$

Total number of base pairs = $\frac{340}{3.4}=100$

Total number of nitrogenous bases = $100\times 2=200$

According to Chargaff's rule, $A=T$ and $G=C$.

Given number of thymine bases ($T$) = $40$

Number of adenine bases ($A$) = $40$

Total number of $A$ and $T$ bases = $40+40=80$

Total number of $G$ and $C$ bases = $200-80=120$

Number of guanine bases ($G$) = $\frac{120}{2}=60$

Answer: $60$