Question

A man is brashydactylus (very short fingers; rare autosomal dominant) and his wife is not. Both can taste the chemical phenylthiocarbamide (PTC, common autosomal dominant allele) but their mothers could not.

a. Give the genotypes of the couple. If the genes assort independently and the couple has four children, what is the probability of: (Provide a brief explanation or show your work for each of the following answers, You may find a Punnett Square helpful.)
b. all of them being brachydactylus?
c. none of them being brachydactylus?
d. All of them being tasters? e. All of them being nontasters?
f. all of them being brachydactylus taster?
g.. none being brachydactylus taster?
h. at least one being a brachydactylus taster?

Answer 1

The man who is brachydactylus has the genotype Bb, while his wife, who is not, has the genotype bb. The probability of different outcomes for their four children can be calculated using the principles of autosomal inheritance. The probabilities for each outcome are: all children being brachydactylus: 1/16, all children being non-brachydactylus: 1/16, all children being tasters of PTC: 1/16, all children being non-tasters of PTC: 1/16, all children being brachydactylus tasters of PTC: 1/16, none of the children being brachydactylus tasters of PTC: 1/16, and at least one of the children being a brachydactylus taster of PTC: 15/16.

Step-by-step explanation:

The genotype of the man who is brachydactylus and his wife, who is not, can be determined by following the autosomal dominant inheritance pattern. Since brachydactylus is a rare autosomal dominant trait, the man must have the genotype Bb (B for brachydactylus and b for normal finger length). The wife, who does not have the trait, must have the genotype bb.

When the genes assort independently, the probability of all four children being brachydactylus is 1/16. This can be calculated by multiplying the probability of each child inheriting the brachydactylus allele (Bb) from their father, which is 1/2, four times.

The probability of none of the children being brachydactylus is 1/16. This can be calculated by multiplying the probability of each child inheriting the normal finger length allele (b) from their father, which is also 1/2, four times.

The probability of all the children being tasters of PTC is 1/16. This can be calculated by multiplying the probability of each child inheriting the taster allele (T) from their mother, which is 1/2, four times.

The probability of all the children being non-tasters of PTC is 1/16. This can be calculated by multiplying the probability of each child inheriting the non-taster allele (t) from their mother, which is 1/2, four times.

The probability of all the children being both brachydactylus and tasters of PTC is 1/16. This can be calculated by multiplying the probability of each child inheriting the genotype Bb from their father and the genotype Tt from their mother, which is 1/2 × 1/2, four times.

Lastly, the probability of none of the children being brachydactylus tasters of PTC is 1/16. This can be calculated by multiplying the probability of each child inheriting the genotype bb from their father and the genotype tt from their mother, which is 1/2 × 1/2, four times.

At least one of the children being a brachydactylus taster of PTC can be calculated by subtracting the probability of none of the children being brachydactylus tasters from 1, which is 1 - (1/16) = 15/16.