Question

Imagine that DNA is composed of equal proportions of six different types of nucleotide, instead of the normal fout. The strucleotides ae A.C.G.1.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 a four-base recognition sequence?

Multiple Choice
A. 425 bp
B. 125 bp
C. 1056 bp
D. 1296 bp
E. 256 bp

Answer 1

The average size of the DNA fragments produced when this new DNA is cut by a restriction endonuclease with a four-base recognition sequence would be 1296 bp.

Step-by-step explanation:

When a restriction endonuclease cuts DNA, it recognizes a specific sequence of bases and cleaves the DNA at that site. In this case, the restriction endonuclease recognizes a four-base recognition sequence. Since the DNA is composed of equal proportions of six different types of nucleotides, the average size of the DNA fragments produced when it is cut by the restriction endonuclease can be calculated.

Since there are six different types of nucleotides, the probability of each base occurring at a given position is 1/6. Therefore, the probability of the four-base recognition sequence occurring is (1/6)^4 = 1/1296. The average size of the DNA fragments can be calculated by dividing the total length of the DNA by the probability of the recognition sequence occurring. In this case, the average size of the DNA fragments would be 1296 base pairs (bp).