Question

The haploid human genome contains about 3 × 109 nucleotides. On average, how many DNA fragments would be produced if this DNA was digested with restriction enzyme PstI (a 6-base cutter)? RsaI (a 4-base cutter)? How often would an 8-base cutter cleave?

Answer 1

Answer:

PstI (a 6-base cutter) would produce 732421 fragments.

RsaI (a 4-base cutter) would produce 11718750 fragments

An 8-base cutter would produce 45776 fragments.

Explanation:

The frequency of restriction sites for an enzyme is (1/4)ⁿ. Four is the number of different bases (A,T,C,G) and n is the number of nucleotides the enzyme recognizes.

PstI (a 6-base cutter)

It cuts at a frequency of (1/4)⁶ = 1/4096. This means that every 4096 nucleotides there will be 1 restriction site that PstI recognizes (so the fragments will have a length of 4096 nucleotides).

If the genome has 3x10⁹ nucleotides, then the number of DNA fragments produced will be 3x10⁹ / 4096 = 732421.

RsaI (a 4-base cutter)

It cuts at a frequency of (1/4)4 = 1/256. The fragments will have a length of 256 nucleotides.

If the genome has 3x10⁹ nucleotides, then the number of DNA fragments produced will be 3x10⁹ / 256 = 11718750.

8-base cutter

It cuts at a frequency of (1/4)⁸ = 1/65536.

If the genome has 3x10⁹ nucleotides, then the number of DNA fragments produced will be 3x10⁹ / 65536 = 45776.

As you can see, the shorter the site of recognition of a restriction enzyme, the higher the frequency of cutting, so there will be more fragments generated by digestion.