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

0 comentarios:
Publicar un comentario