Question

Chargaffs results yielded a molar ratio of 1.45 for A to G in hen DNA, 1.29 for T to C, 1.06 for A to T, and 0.910 for G to C. The ratio of purines to pyrimidines is 0.990. Calculate the mole fractions of A, G, T and C in hen DNA Mole fraction of A Mole fraction of G Mole fraction of T : Mole fraction of C

Answer 1

Given that Purines: pyrimidine is 0.990. The other given ratios are as follows: A:G=1.45 T:C=1.29 A:T=1.09 G:C=0.910 So, A+G=0.99C+0.99T.....(1) Also, from the given ratios, A =1.45G T=1.29C or C=0.771 Substitute the value of A and C in equation (1), A+G= 0.99C+0.99T (1.45G+G) = 0.99(0.77T)+0.99T 2.456 = 1.53T 1.53 G = " T 2.45 G=0.62T

As known, A+G+C+T=1....(2) A:T=1.09 A = 1.091 Substitute the values of A, C and C in terms of T in equation (2), A+G+C+T=1 1.09T +0.62T+0.77T+T=1 3.48T =1 TS 1 3.48 = 0.287 Calculate the mole fractions of A, G and C using the value of T as follows: G=0.62T G=0.62 (0.287) = 0.178 A=1.09T A = 1.09 (0.287) = 0.313

C=0.771 C=0.77 (0.287) = 0.221 Therefore, the mole fractions are as follows: A = 0.313 G=0.178 C=0.221 T= 0.287