Answer to Question #105174 in Genetics for David

Question

In humans there is a disease called Phenylketonuria (PKU) which is caused by a recessive allele. People with this allele have a defective enzyme and cannot break down the amino acid phenylalanine. This disease can result in mental disability or death. Let "E" represent the normal enzyme. Also in humans is a condition called galactose intolerance or galactosemia, which is also caused by a recessive allele. Let "G" represent the normal allele for galactose digestion. In both diseases, normal dominates over recessive. If two adults were heterozygous for both traits (EeGg), what are the chances of having a child that is completely normal? Has just PKU? Has just galactosemia? Has both diseases?

Answer 1

According to 3rd Mendel law in dihybrid crosses 9:3:3:1 ratios are found.

That means that 9 people out of 16 will have both dominant traits = healthy, 3 people with dominant E and recessive g = galactosemia, 3 people with recessive e and dominant G = PKU, and 1 person out of 16 will have both recessive alleles = both diseases.

Chances: normal – 9/16 – 56%

PKU – 3/16 – 19%

Galactosemia – 3/16 – 19%

Both – 1/16 – 6%