Open Question: Parents with the genotypes AaBbCc and AaBbCc are crossed, what is the probability that the offspring?

12:14 Publicado por Flechado16

THe 1/2 X 1/2 X 1/2 = 1/8 = 0.125 is correct

You are looking at each independently-assorting trait independently. Using the multiplication rule of probability, probability of 3 independent events occurring simultaneously is the product of their individual probabilities.

In your Punnet Square, you would have an 8X8 square with 64 boxes. 8 of those should have had AaBbCc in them, giving you a 1/8 probability.


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