Question

Duchenne’s muscular dystrophy is a recessive X-linked disease (Xd). A cross takes place between a heterozygote normal female and a male with the disorder. A) Determine the genotypes of the parents. B) What would be the phenotypic and genotypic ratios of males and females of the F1 generation?

Answer 1

Muscular dystrophies are a group of genetic diseases that result in progressive muscle weakness and loss of muscle mass. Dystrophin is a protein required for normal functioning of muscles which is mutated in this disorder.

Duchenne Muscular dystrophy is a type of muscular dystrophy which is inherited in an X-linked recessive pattern.

Dystrophin is the largest gene identified in humans located in the short arm of X chromosome.

Due to the mutated Dystrophin gene, no functional dystrophin protein is produced. Dystrophin protein transfers force of muscle contraction from inside of the cell to the cell membrane.

Given that the cross takes place between heterozygous normal female and diseased male.

(A)To determine genotype of the parents.

Normal human genotype is

XX for females

XY for males

The genotypes of parents are:

• Heterozygous normal female : XX°

The (°) represents presence of mutant allele.

Since the female has two X chromosomes and is heterozygous for the given trait, one of the two alleles is mutant and the other is normal.

• The diseased male : X°Y

Human males have one X chromosome and one Y chromosome. If dystrophin gene on the only X chromosome is mutated, the males with the mutated gene will always have the disease, and carrier condition does not exist.

This is also a reason why the disease is more prevalent among males.

(B) Phenotypic and genotypic ratios of males and females in F1 generation:

The cross between parents result in F1 generation.

              P XX° × X°Y

              F1 XX° XY X°X° X°Y

The phenotypes are

XX° heterozygous carrier(normal) female

XY normal male

X°X° homozygous diseased female

X°Y diseased male

Phenotypic ratio = 1:1:1:1

Female carrier. = 25%

Male normal. = 25%

Female diseased = 25%

Male diseased. = 25%

Genotypic ratio = 1:1:1:1

XX° = 25%

XY. = 25%

X°X° = 25%

X°Y. = 25%