Question

Based on the notations we’ve learned in class and being used in this exam, which of the following genotypes is a correct way to write out an individual having Wilson’s disease and red green color blindness? SELECT ALL

		ATP7B/atp7b XAXa
		ATP7B/atp7b XAY
		atp7b /atp7b XaY
		atp7b /atp7b XaXa

Answer 1

Wilson disease is caused by autosomal recessive trait where a mutation occurs in The ATPB7 gene.

As it is autosomal recessive , both parents must carry one mutated allele. Here ATPB7 is normal gene and atp7b is mutated gene.

In heterozygous condition , where one normal and one mutant gene wilson disease will not occur. It occur in only homozygous recessive condition where both the mutant allele is present.

Red green colour blindness is a x linked recessive trait. The allele will present in only X chromosome.

Female have two X chromosome so present of two allele will show this trait. If one X chromosome has normal allele and the other has affected allele then the female will be carrier and not produce the disease.

In case of male if he gets the affected X chromosome he will develop the colourblindness because he has only one X chromosome and the other is Y chromosome.

The normal X chromosome is denoted as XA and the affected as Xa.

Option A: ATP7B/atp7b it suggests that one of the parrent has the mutant allele but other has normal allele. But wilson disease is a autosomal recessive trait where both parents need to pass the mutant gene. Hence this child will not produce the disease. It is heterozygous condition .

XAXa it suggests that the offspring is female and has one affected allele. So she will not has colour blindness. Hence this option is wrong.

Option B : ATP7B/atp7b as above one parent has normal gene so the offspring will not has the disease. It is heterozygous condition.

XAY it suggest that the offspring is male and has normal X chromosome so he will not have the colour blindness.

Hence this option is wrong.

Option C: atp7b/atp7b , look here both the parents pass the mutant gene. And in autosomal recessive if both parent pass the mutant gene , it will develop the disease in the offspring. It is the homozygous recessive condition.

XaY , look it is a male offspring and has the affected allele . As it is male and the X chromosome has the affected allele it will develop the colour blindness.

So these option is correct.

Option D: atp7b/atp7b , as avove both the gene are mutant so the offspring will have the wilson disease. It is homozygous recessive condition.

XaXa , look it is a female offspring. Both of her X chromosome have the affected allele. In female if both the X chromosome have the affected allele it will generate colourblindness.

So these option is also correct.

So the right answer is

atp7b/atp7b ;XaY : male child with wilson disease and red green colourblindness.

atp7b/atp7b; XaXa: female child with wilson disease and red green colour blindness.