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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