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:
D
Step-by-step equation:
-10 + 30 = 20
It's like walking ten meters to the left, then walking thirty meters to the right. The result is walking 20 meters to the right. It's the same with force.
It is muscle and nerve tissue
