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vodka [1.7K]
3 years ago
7

Assume that long earlobes in humans are an autosomal dominant trait that exhibits 30% penetrance. A person who is heterozygous f

or long earlobes mates with a person who is homozygous for normal earlobes. What is the probability that their first child will have long earlobes (State probability as a decimal and round to the nearest hundredth i.e. 0.01 with no extra spaces)?
Mathematics
1 answer:
Bess [88]3 years ago
3 0

Answer:

0.75

Step-by-step explanation:

Autosomal dominant: A pattern of inheritance in which an affected individual has one copy of a mutant gene and one normal gene on a pair of autosomal chromosomes

Heterozygous just means that a person has two different versions of the gene (one inherited from one parent, and the other from the other parent).

Being homozygous for a particular gene means you inherited two identical versions.

The trait is autosomal dominant so the characters pass into the next generation in a large ratio. the person who is heterozygous for the character have two type of allele which represents the trait for long earlobes and also for short earlobe.So a person is paired with homozygous individuals who have pure character for short earlobes.The percentage of their first child with long earlobes would be 75 percent. This is due to the dominance of the character in a generation.

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She ordered 80 carnations, 50 roses, and 70 daisies.

*HOPE THIS HELPED!!! :D)

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