The genome of T. aureus bacteria is heavily methylated to prevent it from being

Question

The genome of T. aureus bacteria is heavily methylated to prevent it from being digested by restriction enzymes made by the bacteria. This bacteria grows optimally around 72 degree C. Using the DNA polymerase from this bacteria, I use PCR to amplify the gene, "X" from wildtype and methylase-deficient strains of R coli, and obtain a 2.5Kb product from the methylase-deficient strain. MOST LIKELY, would I expect to see any product from the methylase-containing wildtype E. coli strain? If so, what size would this product be? What specific piece of evidence supports your conclusion? Explain.

Explanation / Answer

Yes, If the gene "X" is present in the wildtype then PCR will amplify the gene and you will get the product. The product size will be same (2.5Kb) if no mutation has occured in gene obtained from methylase deficient strain of E. coli or depends on the primers used to amplify the gene. As it is known that methylation prevents the foreign genetic material from being digested by restriction enzymes and during PCR methylation of daughter strand does not occur.

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