Question

For the following: 1) Identify the genotypes of the parents 2) Complete a Punnett Square & 3) Give the genotypic AND phenotypic results of the cross (percentage OR ratio)

Problem:

In humans, male pattern baldness is an X-linked trait.

Show the cross between a male that is not bald and a female that is a carrier.

(Please write or type response, do not use cursive writing due to visual impairment)

Thank you.

Answer 1

The primary gene for baldness is present on the X chromosome and apart from the gene, there are several other factors that affect hair loss. The baldness is more prominent in males and they receive an X chromosome from mother. The baldness is a recessive trait and males have only a single X chromosome. A recessive trait is expressed in homozygous conditions.

1) Genotypes of parents:

Father's genotype will be X⁺Y (X⁺ represent the normal gene) because he is not bald. The baldness is X-linked and males receive their X chromosome from mother. Males have a single X chromosome (hemizygous) so if he receives the X chromosome with a baldness gene they will express baldness trait which is not the case with father. Therefore, father's genotype will be XY because he has a normal X chromosome expressing no baldness.

Mother is a carrier for the baldness which means that she carries one copy of the baldness gene and a normal copy of the gene on two X chromosomes. Her genotype will be X⁺X^b (X^b represents the baldness gene) as she will carry the baldness gene but will not express the baldness trait because of heterozygous condition of the gene (baldness is recessive trait).

2) The punnet square for cross between father and mother:

	X+	Y
X+	X+X+	X+Y
Xb	X+Xb	XbY

3) Genotypic result of the offsprings: As shown in the punnet square 50% of the daughter will be normal (X⁺X⁺) as they carry both normal alleles and 50% daughter will be the carrier for baldness (X⁺X^b) because they carry one copy of normal and one copy of baldness allele.

As shown in punnet square 50% of the son will carry normal allele (X⁺Y) and 50% son will have baldness gene (X^bY).

Phenotypic result of the offsprings: Phenotypically all daughters will be same and express no baldness trait. Phenotypically 50 % son will be express baldness trait and 50% will have normal hair.