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?

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 dieThus, 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]