Zoology - Molecular Basis of Inheritance Question with Solution | TestHub

Length of DNA sample$\mathrm{ }=13.6\mathrm{ }\mathrm{n}\mathrm{m}$. Length of$1\mathrm{ }\mathrm{t}\mathrm{u}\mathrm{r}\mathrm{n}=3.4\mathrm{ }\mathrm{n}\mathrm{m}$. Thus No. of turns in the sample$=\frac{13.6\mathrm{n}\mathrm{m}}{3.4\mathrm{n}\mathrm{m}}=4$

No. of nucleotides in 1 turn$= 20$. Thus, No. of nucleotides in the given sample$= 20\times 4=80$

Since cytosine is 10%, By Chargaff's rule,$\mathrm{G}\mathrm{ }=\mathrm{ }\mathrm{C}\mathrm{ }=\mathrm{ }10\mathrm{\%}$while$\mathrm{A}\mathrm{ }=\mathrm{ }\mathrm{T}\mathrm{ }=\mathrm{ }40\mathrm{\%}$

No. of Guanine molecules$=\frac{1}{1}\mathrm{x}\mathrm{ }8=8$. No. of Cytosine$= 8$

No. of Adenine$=\frac{40}{100}\times \mathrm{ }80=32$. No. of Thymine$= 32$

Thus, no. of horizontal steps with GC linkage$= 8$and No. of horizontal steps with AT linkage$= 32$

GC are linked by 3 weak hydrogen bonds. Thus no. of weak hydrogen bonds due to GC linkage$= 8\times 3 =24$

AT are linked by 2 weak hydrogen bonds. Thus no. of weak hydrogen bonds due to AT linkage$=32\times 2=64$

Total no. of weak hydrogen bonds in the given sample$=\mathrm{ }24\mathrm{ }+\mathrm{ }64\mathrm{ }=\mathrm{ }88$