A man has six fingers on each hand and six toes on each foot. His wife and their daughter have the normal number of digits. Remember that extra digits is a dominant trait. What fraction of this couple's children would be expected to have extra digits?

Verified step by step guidance

1. Identify the genotypes of the parents. The man with six fingers and toes has the dominant trait, so his genotype must include at least one dominant allele (D). We represent his genotype as Dd, assuming he is heterozygous since the wife and daughter do not exhibit the trait. The wife has the normal number of digits, which is a recessive trait, so her genotype is dd.
2. Set up a Punnett square using the genotypes of the parents. Place Dd on one side (for the man) and dd on the other side (for the wife).
3. Fill in the Punnett square. The top row will have D from the father and d from the mother, and the bottom row will have d from the father and d from the mother. This results in the following squares: Dd, Dd, dd, dd.
4. Analyze the Punnett square to determine the genotypes of potential offspring. From the square, you can see that there are two Dd offspring and two dd offspring.
5. Calculate the probability of the offspring having extra digits. Since the trait for extra digits is dominant, the offspring with a Dd genotype will show the trait. There are two Dd outcomes out of four possible outcomes, so the fraction of children expected to have extra digits is 1/2.