The coding sequence of M. avium photoreactivation enzyme gene is 3600 bp and the
ID: 10380 • Letter: T
Question
The coding sequence of M. avium photoreactivation enzyme gene is 3600 bp and the G+C content of the DNA is 66% and its genome is 5,480,000 bp. The M avium gene has been chosen because it has the highest activity of any known photreactivation enzyme, but it grows very slowly and is unsuitable for production of large amounnts of the enzyme. The gene will be cloned in E.coli, a rapidly growing bacterium.
What will the average length of the fragments be if the M avium DNA is cut with restriction endonuclease EcoR1 that recognizes sequence 5'-GAATTC-3' ?
How many fragments will be formed from the M.avium DNA if cut by the restriction endonuclease BamH1?
Explanation / Answer
As this holds good for one strand, we double the figure to get the number of bases for both the strands.
we get a 512 bases segment.