Formula for In a DNA fragment, there are 8 turns, with 40% of the bases are cytosine. What would be the total number of hydrogen bonds present in this DNA fragment?

I need the formula

there are 10 bases per turn approximately
step 1 : so if 8 turns means 8 x10 = 80 ie n turns means n x 10 bases
step 2 : 40% cytosine means 40% of this 80 bases is triple hydrogen bond that is 32 bases; if p percentage triple bond bases means (either cytosine % or guanine % could be given in question) then (n x 10) x p/100
step 3 : these 32 bases have triple bonds rest (80 - 32 =) 48 bases have double bond; (n x 10) - [(n x 10) x p/100]
step 4 : total no. of hydrogen bond is 32 x 3 = 96 added to 48 x 2 = 96 equals 192;
ie

if n is no. of turns and p is % of triple bond bases then total no. of H bonds in the given turn is
[(n x 10) x p/100] x 3 + {(n x 10) - [(n x 10) x p/100]} x 2

if n is no. of turns and p is % of double bond bases (ADENINE & THYMINE) then total no. of H bonds in the given turn is
[(n x 10) x p/100] x 2 + {(n x 10) - [(n x 10) x p/100]} x 3

edited Sep 12, 2014 by pady_1