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A woman who is a carrier (heterozygous) for Huntington’s Disease, which is a dominant disease, marries a man who is also a carrier for the disease. What are the genotypes? What is the probability that their children will have the disease? What is the probability that their children will not have the disease? Show a Punnett square!

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nandana

Answer:

Explanation:

Huntington’s disease is an inherited disorder that causes damage to certain brain cells. It is caused by a dominant allele. In a genetic diagram:

the recessive allele can be shown as h

the dominant allele can be shown as H

Someone who is homozygous (HH) or heterozygous (Hh) for the dominant allele will develop Huntington’s disease.

Ver imagen nandana
coge4eva

Answer: If H represents the trait for Huntington disease and is dominant trait, and h represents a normal trait.

The woman and the man are carriers, therefore their genotype is Hh.

The probability that their children will have the disease is 3/4.

The probability that their children will not have the disease is 1/4.

Explanation: The trait for Huntington disease (H) is dominant over the normal gene (h). A cross between two carriers (heterozygous) adults will yield one who is child homozygous for Huntington disease (HH), two children who are heterozygous for Huntington disease (Hh) and one child homozygous for normal allele (hh).

The two children heterozygous for Huntington (Hh) will appear outwardly as Huntington disease because the trait for Huntington is a dominant gene. In total three of the children will have Huntington disease while one will not have Huntington disease.

See the attached punnet square for illustration

Ver imagen coge4eva

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