Answer:
The correct answer is (0.2)^4.
Explanation:
Based on the given information, the given genomic DNA is A-T rich, and the base pair composition is C-G = 40 % and A-T = 60%. Therefore, as per Chargaff's rule, Each C and G will possess 20%, and each A and T will possess 30%.
Therefore, the probability of each base will be,
C = 20/100 = 0.2
G = 20/100 = 0.2
A = 30/100 = 0.3
T = 30/100 = 0.3
Now the expected frequency of the Tsp E1 restriction sites 5'-CCGG-3' will be 0.2 * 0.2 * 0.2 * 0.2 = (0.2)^4.
