Ta có: $L_{ADN}$ = (N/2)x3,4
==> N= 2L/3,4= 2x6000/3,4= 3529 nu
C= L/34= 6000/34= 176,47
%A = %T= (A/N)x100
<==> 15= (A/3529)x100
==> A=T= 529 nu
Ta có: %A + %G=50
==> %G= 50-15=35%
==> 35= (G/3529)x100
==> G=X= 1235nu
H=2A + 3G= (2x529) + (3x1235)= 4763 liên kiết