Answer:
correct answer: 14
Step-by-step explanation:
the question is incomplete, so a graph with the complete questionnaire is attached and the requested question is answered
We will start from the exercise requesting the minimum length that a sequence forms with more than 1.4x10ˆ8 permutations
for this we deduce that two base pairs can have:
4 * 4 = 16 permutations
we deduce that 3 base pairs can have:
4 * 4 * 4 = 64 permutations
Thus we multiply successively until we reach that 14 base pairs can obtain: 2.68x10ˆ8 permutations that are random
Thus, the 14bp sequence will be found only once in the genome of the fruit fly.