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

since your problem is x^4 = 81, what you have to undeo is the "to the fourth power" part.

the inverse (opposite operation) of "to the fourth power" is "to the fourth root".

  (x^4) ^ ¼   = 81 ^ ¼  

  81 = 3^4     so (3 ^ 4) ^ ¼   =   3 ^ 1   = 3

by hand: (3 ^ 4) ^ ¼   =   3 ^ (4 x ¼) = 3 ^ 1   = 3   (when you take a power to a power, multiply the exponents.)

or, do it by calculator. one way of finding the fourth root is to use ^ (1/4)

step-by-step explanation:

me! what did you put for this question?

step-by-step explanation: