A) 20
B) 35
C) 15
D) 2
Correct Answer: A
Solution :
The frequency of cutting DNA sequence in a random for a given restriction enzyme is once per every 4", where n is the number of bases in the restriction enzymes recognition sequence. The "4" derives from the fact that there are four different possible nucleotides that maybe inserted at anyone position (G, A, T, or C). |
\[Taq\]Has a four-base recognition site, so it will cut once per every 44 or 256 bases. |
Number of restriction sites \[=\frac{Number\,of\,based\,in\,DNA\,molecule\,}{\operatorname{Average}\,length\,of\,restriction\,fragment}\] |
\[=\frac{5000}{256}=19.5\approx 20\] |
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