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dedylja [7]
3 years ago
8

both parents are heterozygous for the Huntington's allele. what is the chance that their child will not have Huntington's diseas

e
Biology
1 answer:
iris [78.8K]3 years ago
4 0

Answer:

25% or 1/4

Explanation:

The cross in this question involves a gene coding for the possession or not of Huntington's disease in humans. The dominant allele is denoted by (H) while the recessive allele is denoted by (h). The possession of the dominant allele by an individual means the possession of the Huntington's disease.

In a cross between two heterozygous parents i.e. combination of both dominant and recessive alleles (Hh). Four possible offsprings will be produced with genotypes HH, Hh, Hh and hh.

The genotypes HH, Hh and Hh will phenotypically possess the Huntington's disease while the hh genotype will not possess the Huntington's disease. Hence, the chance of having a child without the disease is 1/4 or 25%.

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