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The pedigree chart below represents the inheritance of a genetic disorder caused by a dominant allele. If Individual 2 (who is heterozygous for the disorder) has children with Individual 3 (who is unaffected), what is the probability that their child will inherit the disorder?- Pedigree chart: - Individual 1 (affected): Homozygous dominant (AA) - Individual 2 (heterozygous).Heterozygous (Aa)- Individual 3 (unaffected)Homozygous recessive (aa) 50% 25% 100% 75%
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To determine the probability that their child will inherit the disorder, we need to consider the possible combinations of alleles that can result from the cross between Individual 2 (heterozygous, Aa) and Individual 3 (unaffected, aa).Since the disorder is caused by a dominant allele, an individual only needs one copy of the dominant allele (A) to be affected. Therefore, if an individual has at least one copy of the dominant allele, they will have the disorder.In this case, Individual 2 is heterozygous (Aa), meaning they have one copy of the dominant allele (A) and one copy of the recessive allele (a). Individual 3 is homozygous recessive (aa), meaning they have two copies of the recessive allele.When Individual 2 and Individual 3 have children, the possible combinations of alleles are:1. Aa (heterozygous, affected)2. aa (homozygous recessive, unaffected)Since the disorder is caused by a dominant allele, an individual with at least one copy of the dominant allele (A) will have the disorder. Therefore, the probability that their child will inherit the disorder is 50%, as there is a 50% chance that the child will receive the dominant allele (A) from Individual 2.Therefore, the correct answer is
.