The color of some beans is controlled by three different genes, A, B, and C, making bean color a polygenic treat. If we start with parents that are heterozygous for each gene, then for each individual genotype, probability can be deduced using a simple Punnett Square. Below are Punnett Squares for each individual gene. b B Cc Bb AA Aa BB CC bb Aa Ba Cc a aa CC CC = 25% BB = 25% AA = 25% Which rule for calculating probability (And Rule or Or Rule) would be used to determine the probability of a bean with the genotype AABBCC? Product rule Therefore, to calculate the probability of a bean being AA, BB, and CC, the individual probabilities would need to be multiplied (Note: this is not asking for a number - it's asking you what you need to do mathematically with the individual probabilities to calculate a bean being AA, BB, CC.) The probability of a bean being AABBCC is 1.56 • Report your answer as a percent and round to two decimal places. Given a parent plant with the genotype AaBbCc is mated with a plant with genotype AaBbCc, what is the probability of an F1 offspring with the genotype AABBCC? Express your answer as a percent, but do not include the percent symbol (%). E.g an answer of 0.3333 would be reported as 33.33, not as 1/3 or 0.333 or 33.33% Do not round your answer. 3.125

Human Heredity: Principles and Issues (MindTap Course List)

11th Edition

ISBN:9781305251052

Author:Michael Cummings

Publisher:Michael Cummings

Chapter5: The Inheritance Of Complex Traits

Section: Chapter Questions

Problem 5QP: As it turned out, one of the tallest Potsdam Guards had an unquenchable attraction to short women....

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This my second time posting this question. I already answered it but I want to make sure I have the correct answer. And if is not correct could you please tell me which one is not correct and what is the correct answer will be.

The color of some beans is controlled by three different genes, A, B, and C, making bean color a polygenic treat.
If we start with parents that are heterozygous for each gene, then for each individual genotype, probability can be deduced using a simple Punnett Square. Below are Punnett Squares for each
individual
gene.
b
B
Cc
Bb
AA
Aa
BB
CC
bb
Aa
Ba
Cc
a
aa
CC
CC = 25%
BB = 25%
AA = 25%
Which rule for calculating probability (And Rule or Or Rule) would be used to determine the probability of a bean with the genotype AABBCC? Product rule
Therefore, to calculate the probability of a bean being AA, BB, and CC, the individual probabilities would need to be multiplied
(Note: this is not asking for a number - it's asking
you what you need to do mathematically with the individual probabilities to calculate a bean being AA, BB, CC.)
The probability of a bean being AABBCC is 1.56
• Report your answer as a percent and round to two decimal places.
Given a parent plant with the genotype AaBbCc is mated with a plant with genotype AaBbCc, what is the probability of an F1 offspring with the genotype AABBCC?
Express your answer as a percent, but do not include the percent symbol (%).
E.g an answer of 0.3333 would be reported as 33.33, not as 1/3 or 0.333 or 33.33%
Do not round your answer.
3.125

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