A specific X-linked recessive allele is lethal. Males that inherit the recessive allele and females that are homozygous recessive will abort before birth. What is the probability that a heterozygous mother will give birth to a child with only one copy of the recessive allele?

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Arclight Arclight · Answer 1 · 2019-08-19T22:34:12+02:00

Answer:

The probability that a heterozygous mother will give birth to a child with only one copy of the recessive allele is [tex]\frac{1}{3}[/tex]

Step-by-step explanation:

Let the recessive allele causing lethal disease be represented by "r"

Let the dominant allele causing no lethal disease be represented by "R"

Genotype of heterozygous mother will be [tex]X_RX_r[/tex]

Genotype of heterozygous mother will be [tex]X_rY[/tex]

A cross between [tex]X_RX_r[/tex] and [tex]X_rY[/tex] will produce following offspring

[tex]X_RX_r[/tex]

[tex]X_RY[/tex]

[tex]X_rX_r[/tex]

[tex]X_rY[/tex]

Thus, out of four offspring only child with genotype [tex]X_rY[/tex] will die

Thus, only three offspring remain alive and out of these three only one child with genotype [tex]X_RX_r[/tex] has one copy of recessive allele.

Thus, the probability that a heterozygous mother will give birth to a child with only one copy of the recessive allele is [tex]\frac{1}{3}[/tex]