Assume that the average protein in E.coli consists of a chain of 400 amino acids. Now, a single amino acid present in the protein requires a codon to code for it. A codon consists of 3 nucleotides. This shows that a protein of length 400 amino acids will require, 400 codons to code for it, that is, 1200 (3 x 400) nucleotides are required. We know that the length of the E.coli genome on average is 5 x [tex]10^{6}[/tex] base pairs.
1200 nucleotides = 1 protein
5 x [tex]10^{6}[/tex] nucleotides = X no. of proteins
X = [tex]\frac{5\times 10^{6} \times 1}{1200}[/tex] = 4167
The answer is 4167 proteins.
