Answer:
4210.5 proteins
Explanation:
(a)The amount of nucleotide pairs in the DNA molecule calculated using
=molecular weight of DNA/molecular weight of a single pair
molecular weight of DNA
= 3.1 × 10^9g/mol
molecular weight of a single pair
= 0.66 ×10^3g/mol
= (3.1 × 10^9g/mol)/(0.66 × 10^3g/mol)
= 4.7 10^6pairs
Lets, Multiply the number of pairs with the length per pair
Take, length per pair = 0.34 nm/pair
= (4.7 × 10^6pairs)(0.34 nm/pair)
= 1.6 × 10^6nm
= 1.6 mm
= (approx 2mm or 0.002 mm).
Therefore,
the DNA is
(1.6 mm)/(0.002 mm)
= 800
The DNA is 800 times times longer than the cell.
This indivates Tat DNA must be tightly coiled to fit into the cell.
(b) the amount of DNA molecule nucleotide pairs has given
= 4.7 × 10^6 nucleotide pairs,
From solution (a), there must be one-third this number of triplet codons
= (4.7 × 10^6)/3
= 1.6 × 10^6codons
If an individual protein has an average of 380 amino acids, each needs a codon,
the number of proteins that can be coded by E. coli DNA
= (1.6 × 10^6codons) × (1 amino acid/codon)
---------------------------------------------
(380 amino acids/protein)
= 4210.5 proteins
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Here is the complete question
The genetic information contained in DNA consists of a linear sequence of coding units known as codons. Each codon consists of three adjacent DNA nucleotides that corresponto a single amino acid in a protien.
The E.coli DNA molecule contains 4.70 x 10^6 base pairs. Determine the number of codons that can be present.
Assuming that the average protein in E.coli consists of a chain of 400 amino acids, calculate the maximum number of protiens that can be coded by an E.coli DNA molecule
