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Consider the following cross:
Parent 1: YyTt
Parent 2: YyTt
Using the rules of probability, determine the probability that the offspring will show YyTT genotype?
We can divide this question to two parts: Yy and TT. First, the parents have Yy(1) and Yy(2). The offspring may have four types: Y1Y2, Y1y2, Y2y1, y1y2. So the probability of getting Yy is 1/2. The same, the probability of getting TT is 1/4. So the final probability of YyTT is (1/2)*(1/4)=1/8
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