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 
base pairs.
1200 nucleotides = 1 protein
5 x nucleotides = X no. of proteins
X = 
= 4167
The answer is 4167 proteins.
Answer: Both Meiosis I and II have the same number and arrangement of phases: prophase, metaphase, anaphase, and telophase. Both produce two daughter cells from each parent cell. However, Meiosis I begins with one diploid parent cell and ends with two haploid daughter cells, halving the number of chromosomes in each cell.
Explanation:
Competitive inhibition vs allosteric inhibition
In competitive the substrate and inhibitor bind at the same active site - pretty straightforward. In allosteric regulation (speaking specifically about inhibition here), the inhibitor is binding at a site other than the active site, and changing the enzyme in some way to make it inactive.
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