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5. If you are digesting a large linear fragment that is 300,000 bp in length. How many fragments would be produced if the DNA
Table 4.1 Recognition Sequences and Cutting Sites of Selected Restriction Endonucleases Enzyme Alul BamHI Bgli Clal ECORI Hae
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Answer #1

Once per every 4^n the restriction enzymes cuts the random DNA.

n - It is the number of bases that the restriction enzyme can recognize.

a) SmaI - Recognise the sequence CCC-GGG that means it has got 6 base recognition site that means they will cut the DNA sequence at a frequency of once per every 4^6, that means 4096 bases.

So the number of fragments produced = 300000 / 4096 = 73.2 = approx. 73 fragments

b) HaeIII - Recognise the sequence GG-CC that shows that it got 4 base recognization site so it will cut the DNA sequence at a frequency of once per every 4 ^4 that is 256 bases.

Thus the number of fragments produced = 300000 / 256 = 1171.8 = approx . 1172 fragments

c) NotI - Recognise the sequence GC-GGCCGC that shows that it has got an 8 base recognition site so it will cut the DNA sequence at a frequency of once per every 4^8 that is 65536

Thus the number of fragments produced = 300000 / 65536 = 4.577 = approx. 5 fragments

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