Each of two parents has the genotype green divided by bluegreen/blue, which consists of the pair of alleles that determine eye coloreye color, and each parent contributes one of those alleles to a child. Assume that if the child has at least one greengreen allele, that color will dominate and the child's eye coloreye color will be greengreen. a. List the different possible outcomes. Assume that these outcomes are equally likely. b. What is the probability that a child of these parents will have the blue divided by blueblue/blue genotype? c. What is the probability that the child will have greengreen eye coloreye color? a. List the possible outcomes. A. green divided by green comma green divided by blue comma and blue divided by bluegreen/green, green/blue, and blue/blue B. green divided by blue and blue divided by greengreen/blue and blue/green C. green divided by green and blue divided by bluegreen/green and blue/blue D. green divided by green comma green divided by blue comma blue divided by green comma and blue divided by bluegreen/green, green/blue, blue/green, and blue/blue b. The probability that a child of these parents will have the blue divided by blueblue/blue genotype is nothing. (Round to two decimal places as needed.) c. The probability that the child will have greengreen eye coloreye color is nothing. (Round to two decimal places as needed.)
Each of two parents has the genotype green divided by bluegreen/blue, which consists of the pair of alleles that determine eye coloreye color, and each parent contributes one of those alleles to a child. Assume that if the child has at least one greengreen allele, that color will dominate and the child's eye coloreye color will be greengreen. a. List the different possible outcomes. Assume that these outcomes are equally likely. b. What is the probability that a child of these parents will have the blue divided by blueblue/blue genotype? c. What is the probability that the child will have greengreen eye coloreye color? a. List the possible outcomes. A. green divided by green comma green divided by blue comma and blue divided by bluegreen/green, green/blue, and blue/blue B. green divided by blue and blue divided by greengreen/blue and blue/green C. green divided by green and blue divided by bluegreen/green and blue/blue D. green divided by green comma green divided by blue comma blue divided by green comma and blue divided by bluegreen/green, green/blue, blue/green, and blue/blue b. The probability that a child of these parents will have the blue divided by blueblue/blue genotype is nothing. (Round to two decimal places as needed.) c. The probability that the child will have greengreen eye coloreye color is nothing. (Round to two decimal places as needed.)
Chapter8: Sequences, Series,and Probability
Section8.7: Probability
Problem 4ECP: Show that the probability of drawing a club at random from a standard deck of 52 playing cards is...
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Question
Each of two parents has the genotype
green divided by bluegreen/blue,
which consists of the pair of alleles that determine
eye coloreye color,
and each parent contributes one of those alleles to a child. Assume that if the child has at least one
greengreen
allele, that color will dominate and the child's
eye coloreye color
will be
greengreen.
a. List the different possible outcomes. Assume that these outcomes are equally likely.
b. What is the probability that a child of these parents will have the
blue divided by blueblue/blue
genotype?c. What is the probability that the child will have
greengreen
eye coloreye color?
a. List the possible outcomes.
green divided by green comma green divided by blue comma and blue divided by bluegreen/green, green/blue, and blue/blue
green divided by blue and blue divided by greengreen/blue and blue/green
green divided by green and blue divided by bluegreen/green and blue/blue
green divided by green comma green divided by blue comma blue divided by green comma and blue divided by bluegreen/green, green/blue, blue/green, and blue/blue
b.
The
probability that a child of these parents will have the
blue divided by blueblue/blue
genotype is
nothing.
(Round to two decimal places as needed.)
c.
The
probability that the child will have
greengreen
eye coloreye color
is
nothing.
(Round to two decimal places as needed.)
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