Question

The AseI restriction enzyme recognizes the sequence ATTAAT. Which of the following prokaryotic genomes will on average yield thelargest fragments after cutting with this enzyme?

	a.	The AseI enzyme will not cut prokaryotic DNA
	b.	S. aureus (G+C content = 34%)
	c.	E. coli (G+C content = 50%)
	d.	Rhodobacter sphearoides (G+C content = 68%)
	e.	The G+C content does not affect fragment size when the recognition sequence has no G or C bases.

Answer 1

Restriction enzymes can cut at specific sites of DNA and give different sized fragments.

In this question, it has been asked which prokaryotic genome would give rise to largest fragments upon cutting with enzyme AseI.

Lets look at available options.

Option A says that the enzyme AseI will not cut prokaryotic DNA. This is false as restriction enzymes generally cannot discriminate between prokaryotic and eukoryotic DNA,

Option E says G+C content does not affect fragment size when the recognition sequence has no G or C bases. This is also false because the restriction digestion frequency depends on two parameters and not just one. The two parameters are

1) GC or AT content of the restriction site

2). GC or AT content of the genome.

How?

Expected cut frequency of any enzyme on a DNA strand is given by following formula

=1/(0.5*GC)^a * (0.5* AT)^b

where GC - ratio of GC content in genome

          AT - ratio of AT content in genome

          a - Number of G and C in restriction site

            b - Number of A and T in restriction site.

Now using the above formula lets calculate the restriction cut frequency of the enzyme AseI for 3 different genomes with different GC content.

Option B, S.aureus (G+C content = 34%)

= 1/ (0.5* 0.34)^0 * (0.5 * 0.66)^6

=1/ 0.00129

=775 (Approximately).

Thus AseI cuts approximately at every 775 bases and hence you will find band size of 775 bases.

Option C, E.coli (G+C content = 50%)

= 1/ (0.5* 0.5)^0 * (0.5 * 0.5)^6

= 1/ 0.0002

= 5000

Thus AseI cuts approximately at every 5000 bases and hence you will find band size of 5000 bases.

Option D, Rhodobacter sphearoides (G+C content = 68%)

= 1/ (0.5* 0.68)^0 * (0.5 * 0.32)^6

=1/ 0.00001677

= 59630

Thus AseI cuts approximately at every 59630 bases and hence you will find band size of 59630 bases.

Hence the right answer is

Option D, Rhodobacter sphearoides (G+C content = 68%) - This on average yield the largest fragments (59630 bases) after cutting with this enzyme(AseI).