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
Answer:
highly folded surface of the inner mitochondrial membrane
Explanation:
hope I helped <3
Binary fission enables them to reproduce quickly.
Answer:
the spring time is when they call for mates
Answer:
D
Explanation:
They would be dying off, or leaving to get away from the disease
