Aspecific 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?

The probability that a heterozygous mother will give birth to a child with only one copy of the recessive allele is

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

Genotype of heterozygous mother will be

A cross between and will produce following offspring

Thus, out of four offspring only child with genotype will die

Thus, only three offspring remain alive and out of these three only one child with genotype 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

to find the perimeter of the rectangle this is the expression:

x-9+x-9+3x-4+3x-4=138

if we simplify it:

8x-26=138

remember, 138 is what it all adds up to

the length is 11.5cm

the width is 57.5cm

the area is 661.25cm^2

the answers might not be correct use at own risk

12.4, 12.387, 12.26, 12.252, 6.241, 6.2, 6.015, 6.009, 1.411, 1.2, 1.056, 1.007, 0.31,   0.13, 0.5, 0.04

step-by-step explanation:

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