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? .

Ok so this is a punnet square. Use FOIL when crossing dihybrids. First, outside, inside last. So this mean your boxes will be outlined by YT,  Yt, yT, and yt. In the photo, two boxes are shown with YyTT, and there are 16 boxes total. 2/16 is .125, or 12.5% That is the chance of the offspring with that genotype.

The answer is c i just finished taking the test

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