Youve discovered a new virus which has a 10 kb double strand

You’ve discovered a new virus which has a 10 kb double stranded DNA genome consisting of 65% A/T. How many times would you expect the enzyme HaeIII (GG/CC) to cut this genome? Round off to the nearest whole integer.

Solution

Genome of virus = 10kb = 10,000bp

How to calculate number of restriction sited in genome is as follows:

If the genome have approximate 1:1 ration of A/T :G/C (Chargaff’s rule), it means 50% G/C, 50% A/T

Enzymes HaeIII (GG/CC) have target site of 4pb

Therefore HaeIII (GG/CC) will cut the DNA once in every 256pb

[Enzyme will cut after A^a bp

where A= number of bases in DNA i.e A, T, G,C (4)

a= number of bases in the restriction site (4) (HaeIII (GG/CC))

Therefore 4⁴=256]

The number of cuts made by HaeIII = Total number of bp in genome/ 256 (vary enzyme to enzyme)

=10,000/256

=39

HaeIII will cut the genome about 39 times approximately if genome follows Chargaff\'s rules.

But in the question viral genome have 65% A/T

Therefore G/C= 100-65=35%

So Number of cuts in genome with G/C 35% =39*35/50=27

Therefore HaeIII will cut viral genome (65% A/C), 27 times