In the following cross: AABbDd x AaBbDD   calculate the probability of getting an offspring with the following genotypes: (by using rules of probability)   AABBDd and AaBbDd

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Calculating the probabilities of this cross is not difficult; however, it will be somewhat time-consuming because the cross is asking you to incorporate three different traits. With only one trait, there are four boxes in your Punnett square. Two traits necessitates 16 boxes. Punnett squares with 64 boxes come from three-trait crosses.

The first thing to do is figure out what gametes each parent can produce. Let's call the first parent listed (AABbDd) "parent 1." Parent 1 can produce the following gametes. ABD, ABD, AbD, AbD, ABd, ABd, Abd, abd.

Parent 2 can produce the following gametes. ABD, aBD, AbD, abD, ABD, aBD, AbD, abD.

Place those gametes along the top and side of your Punnett square and fill in the square.

Despite the fact that there are 64 boxes, only 12 unique allele combinations result from those two particular parents. 8 of the boxes will be AaBbDd. 8 out of 64 boxes means there is a 12.5% chance of producing a child that is heterozygous for all three traits. 4 of the 64 boxes are AABBDd. This means that there is a 6.25% chance of producing a child that is homozygous dominant for the first two traits and heterozygous for the final trait.

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