The ability to taste the chemical phenylthiocarbamide is an

The ability to taste the chemical phenylthiocarbamide is an autosomal dominant phenotype, and the inability to taste it is recessive. If a taster woman with a nontaster father marries a taster man who, in a previous marriage, had a nontaster daughter, what is the probability that their first child will be: a nontaster girl? a taster girl? a taster boy? the \"autosomal\" pattern indicates that the gene in question is not on a sex chromosome; the sex of the individual is not a factor in the inheritance of the trait A man and his wife are both heterozygous for two independently assorting recessive traits: phenylketonuria (PKU) and cystic fibrosis (Cf). What is the probability that their first child will show at least one of these two disease traits? (Assume that neither gene is on the X/Y chromosomes.]

Solution

6. Phenylthiocarbamide (PTC) tasting is an autosomal dominant phenotype. According to the facts in the given problem, the taster woman marrying taster man are both heterozygous (T/t). The probability of tasters is ¾ (75%) and non-taster is ¼ (25%).

a. Probability of a non-taster girl

p (non-taster/ non-taster girl)* p (girl)= ¼* ½

= 1/8

b. Probability of a taster girl

p (taster/ taster girl)* p (girl)= ¾* ½

= 3/8

c. Probability of a taster boy

p (taster/ taster boy)* p (boy)= ¾* ½

= 3/8

7. Both phenylketonuria (PKU) and cystic fibrosis (CF) are two independent diseases and autosomal recessive.

Supposing, PP= Normal (No PKU)pp= PKU

CC= Normal (No CF)cc= CF

If both parents are heterozygous for PKU (Pp) and CF (Cc) then:

Probability of CF disease trait= ¼ CC, ½ Cc and ¼ cc

Probability of PKU disease trait= ¼ PP, ½ Pp and ¼ pp

Since both disease traits are present in parents, then the probability their first child will have either CF or PKU will be calculated by:

¼ for CF + ¼ for PKU = ½

So, the probability would be ½.