Question

Assume that the average protein in

Answer 1

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 $10^{6}$ base pairs.

1200 nucleotides = 1 protein

5 x $10^{6}$ nucleotides = X no. of proteins

X = $\frac{5\times 10^{6} \times 1}{1200}$ = 4167

The answer is 4167 proteins.