This question can be solved by making a punnett square. The genotypes are given:AaBbxaabb.

The potential gametes theAaBbdog can produce areAB,Ab,aB, andab. Theaabbdog can only produce one gamete:ab.

Putting these gametes in our punnett square we can see that we end up with the following potential offspring:AaBb,Aabb,aaBb, andaabb.

Each of these possible offspring carries a different phenotype.AaBbwill carry both dominant traits and be black with long ears.Aabbwill be yellow with long ears.aaBbwill be black with short ears. Finally,aabbwill be yellow with short ears. Each of these gametes is produced in the same ratio, making these phenotypes exist in a 1:1:1:1 probability.

Punnett squares give information about the potential genotype ratios of offspring possible from the cross of two members of the parental generation. The letters represent alleles of various genes, but do not give any information about the allele frequencies. To get information about the allele frequencies, more information about the size and make-up of the population would be needed. The actual cross is between potential gametes produced by the parental generation. Each square shows the potential offspring from these potential gametes.

The shortcut for this problem involves the standard phenotypic ratios for a dihybrid cross. Nine offspring will show both dominant traits. Three will show one dominant trait and the other recessive trait. Three will show the inverse phenotypes, with the opposite dominant trait and recessive trait combination. One offspring will show both recessive traits. Based on these ratios, we can see that three of the sixteen offspring will show the dominant smooth trait and the recessive green phenotype.

We can also solve by using a dihybrid punnett square. The cross described will beAaBbxAaBb, in which theAalleles signify color and theBalleles signify shape.

Consider the color of the peas. In order to have green peas, two recessive alleles must combine in the next generation. According to a punnett square where both sides are heterozygous for the trait, there is only a one in four chance of this taking place. Since smooth is the dominant shape for the peas, a punnett square where each side is heterozygous shows a three in four chance that pea plants will have this shape. By multiplying these two probabilities, we determine that three out of sixteen pea plants will have smooth, green peas.