Imagine that DNA is composed of equal proportions of six different types of nucleotide, instead of the normal four. The six nucleotides are A, C, G, T, X, and Y. In this new type of DNA molecule, A pairs with T, C pairs with G, and X pairs with Y. What would be the average size of the DNA fragments produced when this new DNA is cut by a restriction endonuclease with an eight-base recognition sequence?

Question

Biology

Wyatt Gay

31 August, 03:05

Imagine that DNA is composed of equal proportions of six different types of nucleotide, instead of the normal four. The six nucleotides are A, C, G, T, X, and Y. In this new type of DNA molecule, A pairs with T, C pairs with G, and X pairs with Y. What would be the average size of the DNA fragments produced when this new DNA is cut by a restriction endonuclease with an eight-base recognition sequence?

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Nixon · Answer 1

The average size of the DNA fragments produced when this new DNA is cut by a restriction endonuclease with an eight-base recognition sequence would be 1679616.

Explanation:

Restriction enzymes can be described as the enzymes which are specific to certain DNA sites and cut the DNA from that particular site.

Normally, the average size of a DNA fragment cut by a restriction enzyme is calculated by the formula 4^{no. of base recognition sequence}.

As the number of bases has changed to 6 and the restriction enzyme has eight base recognition site then the formula will be:

6^{no. of base recognition sequence}

i. e

6^{8} = 1679616