Fundamentals of Biochemistry: Life at the Molecular Level 5th Edition

Published by Wiley
ISBN 10: 1118918401
ISBN 13: 978-1-11891-840-1

Chapter 3 - Nucleotides, Nucleic Acids, and Genetic Information - Exercises - Page 77: 7

Answer

Total number of genome in an haploid organism = 45000-Kb Number of genome in diploid organism = 2*45000 = 90,000 kb Given 21 % G residues Now we know that according to Chargaff's rule, number of A=T and G=C 21% G = 21% 90,000 G = (21/100)*90000 = 18,900 C= (21/100)*90000 = 18,900 A= (100%- G% - C %)/ 2 A= (100 -21-21)/2 A= (100-42)/2 = 29 A=T = (29/100)*90000 =26,100 A= 26,100 T= 26,100

Work Step by Step

Total number of genome in an haploid organism = 45000-Kb Number of genome in diploid organism = 2*45000 = 90,000 kb Given 21 % G residues Now we know that according to Chargaff's rule, number of A=T and G=C 21% G = 21% 90,000 G = (21/100)*90000 = 18,900 C= (21/100)*90000 = 18,900 A= (100%- G% - C %)/ 2 A= (100 -21-21)/2 A= (100-42)/2 = 29 A=T = (29/100)*90000 =26,100 A= 26,100 T= 26,100
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