Using the Punnett square that you created in question number 5, what is the probability that the resulting child will be heterozygous for Huntington’s disease ?


A. 25%

B. 50%

C. 75%

D. 100%

Using the Punnett square that you created in question number 5 what is the probability that the resulting child will be heterozygous for Huntingtons disease A 2 class=

Respuesta :

Huntington’s disease is an autosomal dominant affection. Parental's genotype FF and ff. Cross: FF x ff. F1) 100% Ff, heterozygous individuals (Option D).

What is the Huntington’s disease?

Huntington's disease is caused by a mutation in the gene that codifies the Huntington protein (Htt).

The mutation produces an altered form of the protein leading to the neuron death in certain brain areas.

Huntington's disease characterizes for being,

  • Hereditary, passing from generation to generation. To express the disease the person must have been born with an altered gene.

     

  • Autosomal, affecting men and women equally because the mutated gene is located on an autosomal chromosome.

  • Dominant, meaning that by inheriting only one copy of the altered gene from any of the parents, the receiving person will express the disease. The mutation in the gene dominates over the normal gene copy.

  • Expressed by heterozygosis. Most people affected by the disease are heterozygous, with a normal copy and a mutated copy.

Question number 5

To produce only individuals with the genotype Ff, the parentals genotypes must be homozygous dominant and recessive. This is,

Cross: The cross is done between one homozygous recessive and one homozygous dominant individual

Parentals)  FF   x   ff

Gametes)  F   F     f   f

Punnett square)    F      F

                    f        Ff       Ff

                    f        Ff       Ff

F1) 100% of the progeny is expected to be heterozygous Hh, expressing Huntingnot's disease.

You can learn more about the Huntingnot's disease at

https://brainly.com/question/12572808

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