Question

The genome of E. coli has a molecular weight of 2.5 x 109 Da (Da stands for Dalton).
1. Knowing that the average molecular weight of a nucleotide is 300 Da, the distance between two adjacent nucleotides is 0.34 nm, and the number of base pairs per turn of helix is 10, you are asked to calculate.
- The number of base pairs in the E. coli genome
- The length of the E. coli genome
- The number of helical turns in the genome
2. Calculate the number of proteins with an average molecular weight of 60,000 Da that are coded by the E. coli genome assuming that 80% of the E.coli genome contain open reading frames. Assume that the average molecular weight of an amino acid is 100 Da and that every amino acid is coded by 3 nucleotides, and each open reading frame ends by a stop codon (nucleotide triplet) that does not code for any amino acid.

Answer 1

mass of the E.coli genome= 2.5 x 10^9 Da

the average molecular weight of a nucleotide is 300 Da

so total number of nucleotides in E.coli genome = mass of the E.coli genome/ the average molecular weight of a nucleotide

= 2.5 x 10^9 Da / 300 Da

=8333333.333

so number of basepairs = total number of nucleotides in E.coli genome/2 ( 2 nucleotides make up a basepair)

= 4166666.667 bp

= 4.2*10^6 bp

length of the E. coli genome= number of basepairs * length of a basepair

= 4.2*10^6*0.34 nm

= 1428000 nm

The number of helical turns in the genome =  number of basepairs/ number of base pairs in the helix

= 4.2*10^6 bp / 10

= 4.2*10^5