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.
Step-by-step 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.