☜
Restriction Sites Plus 4

Restriction Calculations 4 — Advanced II

 
Calculate the number of fragments a given restriction enzyme would be expected to generate from a piece of DNA of given length but unknown sequence. In some cases the DNA is circular, and in others it is linear.
Assume that each base — A, T, G or C — occurs equally frequently in the DNA. Give your answer to the nearest whole number of fragments.

Number of fragments    
 
First, work out the frequency of occurrence of the restriction site as 1-in-x bases, as explained in the example for the Intermediate level calculation. Then take the size of the DNA in kb (kilobases) and multiply by 1000 to get the size in bases. Divide this by x and round to the nearest whole number. Finally check if the DNA is circular or linear, and in the latter case add 1. The following is a specific example:
  • Take the case of a restriction enzyme with a recognition site, AGCT, cleaving linear DNA, 1 kb in length (assuming 50% GC).
  • The frequency of occurrence of AGCT in the DNA is 1-in-256 bases.
  • Dividing 1x1000 bp by 256 gives 4 as the nearest whole number.
  • Add 1, because the DNA is linear (compare cutting a rubber band with cutting a shoe lace). This gives a total of 5 restriction fragments.