1In humans PKU is an inherited metabolic disease which if un

1.In humans, PKU is an inherited metabolic disease, which if untreated leads to death at about the age of six. It is inherited as a recessive trait. What is the probability that two heterozygous parents will have:

a)a PKU child?

b)A heterozygous child?

c)A homozygous non-PKU child?

Explain your answer.

2.Unattached earlobes are inherited as a dominant trait; attached earlobes a recessive trait. Suppose a homozygous man with unattached earlobes marries a homozygous woman has attached earlobes:

a)What type of earlobes will their children have?

b)If these children marry someone with the same genotype as themselves, will all of their children have attached lobes?

Explain your answer.

Solution

1) As PKU (Phenylketouria) is autosomal recessive in nature. We have less probability that child will be affected by this if heterozygous parents are involved. This can be explained as:

AA - normal human, Aa - carrier and aa - PKU diseased

On crossing two heterozygous Parents. Aa X Aa, we get AA, Aa and aa as offspring. i.e 1/4 population is of AA type, 1/4 population is of aa type and 1/2 population is of Aa type.

a)The probability of having a PKU child is 1/4, since only 1 out of 4 childrent has recessive trait and PKU is expressed only when it is completely recessive.

b) The probability of having a Heterozygous child is 1/2, as 2 out of 4 children will have Aa trait, These two children will have normal phenotype but would be a carrier for next generation.

c) The probability of having a Homozygous non-PKU child is again 1/4 with AA trait. This child will be completely normal genotypically and phenotypically.

2) Unattached earlobes being dominant can have two genotypes- \'AA\' and \'Aa\', as in both case dominant character is expressed. while attached earlobes being recessive will have only \'aa\' genotype.

When a Homozygous man with unattached earlobes (AA) marries a homozygous woman with attached earlobes (aa), we get heterozygous child (Aa) which again will have dominant character which is unattached earlobes. Because Unattached earlobes can be expressed in both \'AA\' and \'Aa\' condition.

a) Therefore, the children will have unattached earlobes with genotype \'Aa\'.

b) If \'Aa\' genotype child marries with someone of \'Aa\' genotype, only 1 out of 4 child will have attached lobe (aa-genotype), rest 1 out of 4 will have \'AA\' genotype (unattached earlobe) and 2 out of 4 will also have unattached earlobe \'Aa\' genotype.