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The C. elegans genome is about 100,000kb. If C elegans genomic DNA is digested with XbaI, which recognizes and cleaves the sequence TCTAGA, what is the expected number of DNA fragments generated by this digest?

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Answer:

25000 fragments

Explanation:

A restriction enzyme can cut a random DNA sequence once per every 4^n where n = number of bases in the recognition site of enzyme.

There are 6 bases in Xbal's recognition site (TCTAGA) so it will cut once per every 4^6 = 4096 bases (4kb).

Total genomic size = 100000kb

So expected number of fragments = Total genomic size/Size of one fragment

= 100000/4

= 25000 fragments

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