Respuesta :
Answer: The possible genotype for their child is Hh. All the children will be heterozygous for Huntington (Hh).
The possible phenotype for their child is Huntington disease.
Explanation: If H represents the trait for Huntington and h represents normal trait; and if Huntington trait (H) is dominant over normal trait (h), therefore the genotype of the mother who is homozygous for Huntington disease is HH and the genotype of the man who is homozygous normal is hh.
A cross between the man and woman will produce offsprings who are all heterozygous for Huntington disease Hh. Phenotypically, the offsprings will manifest as Huntington disease.
See the attached punnet square for more information
A dominant allele codes for Huntington's disease, while the recessive allele codes for normal phenotype. The genotype of the progeny will be heter0zyg0us Hh, and they will express the disease.
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Huntington's disease is an autosomal and dominant affection, caused by a mutation in the gene that codifies for the Huntingtin protein (Htt).
Let us say that
- H is the dominant allele that codes for the disease
- h is the recessive allele that codes for normal phenotype
Cross: mother h0m0zyg0us dominant with father h0m0zyg0us recessive
Parentals) HH x hh
Gametes) H H h h
Punnett square) H H
h Hh Hh
h Hh Hh
F1) 100% of the progeny is expected to be heter0zyg0us, Hh, and express the disease.
The only possible genotype for children is to be heter0zyg0us for the trait and express the disease.
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