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Chapter 3: Problem 18

Consider three independently assorting gene pairs, \(A / a, B / b,\) and $C / c,\( where each demonstrates typical dominance \)(A-, B-, C-)$ and recessiveness \((a a, b b, c c) .\) What is the probability of obtain ing an offspring that is \(A A B b C c\) from parents that are \(A a B b C C\) and \(A A B b C c ?\)

Short Answer

Expert verified
Answer: The probability of obtaining an offspring with the desired genotype is \(\frac{1}{8}\), or 12.5%.

Step by step solution

01

Determine the probability for the A/a gene pair

Parents have genotypes \(A a\) and \(A A\). For the offspring to have the genotype \(A A\), it must inherit an 'A' allele from each parent. The probability of this happening is 1 (100%) from the \(A A\) parent and 1/2 (50%) from the \(A a\) parent. Therefore, the probability for inheriting an \(A\) allele from each parent is \((1) \times \left(\frac{1}{2}\right) = \frac{1}{2}\).
02

Determine the probability for the B/b gene pair

Both parents have the genotype \(B b\). For the offspring to have the genotype \(B b\), it must inherit 'B' from one parent and 'b' from the other parent (or vice versa). There are two possible combinations for this to happen: (i) Inherit 'B' from the first parent and 'b' from the second parent. The probability of this occurring is \(\left(\frac{1}{2}\right) \times \left(\frac{1}{2}\right) = \frac{1}{4}\) (ii) Inherit 'b' from the first parent and 'B' from the second parent. The probability of this occurring is \(\left(\frac{1}{2}\right) \times \left(\frac{1}{2}\right) = \frac{1}{4}\). Therefore, the overall probability for inheriting the \(B b\) genotype is \(\frac{1}{4} + \frac{1}{4} = \frac{1}{2}\).
03

Determine the probability for the C/c gene pair

Parents have genotypes \(C C\) and \(C c\). For the offspring to have the genotype \(C c\), it must inherit 'C' from the first parent and 'c' from the second parent. The probability for inheriting 'C' from the first parent is 1 (100%) and the probability for inheriting 'c' from the second parent is 1/2 (50%). Therefore, the probability for inheriting the \(C c\) genotype is \((1) \times \left(\frac{1}{2}\right) = \frac{1}{2}\).
04

Calculate the overall probability

We can now calculate the overall probability of obtaining an offspring with the genotype \(A A B b C c\) by multiplying the probability of inheriting each gene pair: \(\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} = \frac{1}{8}\). The probability of obtaining an offspring with the desired genotype is \(\frac{1}{8}\), or 12.5%.

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Most popular questions from this chapter

A geneticist, in assessing data that fell into two phenotypic classes, observed values of \(250: 150 .\) He decided to perform chi- square analysis using two different null hypotheses: (a) the data fit a 3: 1 ratio; and (b) the data fit a 1: 1 ratio. Calculate the \(\chi^{2}\) values for each hypothesis. What can you conclude about each hypothesis?

Two true-breeding pea plants are crossed. One parent is round, terminal, violet, constricted, while the other expresses the contrasting phenotypes of wrinkled, axial, white, full. The four pairs of contrasting traits are controlled by four genes, each located on a separate chromosome. In the \(F_{1}\) generation, only round, axial, violet, and full are expressed. In the \(\mathrm{F}_{2}\) generation, all possible combinations of these traits are expressed in ratios consistent with Mendelian inheritance. (a) What conclusion can you draw about the inheritance of these traits based on the \(\mathrm{F}_{1}\) results? (b) Which phenotype appears most frequently in the \(\mathrm{F}_{2}\) results? Write a mathematical expression that predicts the frequency of occurrence of this phenotype. (c) Which \(\mathrm{F}_{2}\) phenotype is expected to occur least frequently? Write a mathematical expression that predicts this frequency. (d) How often is either \(P_{1}\), phenotype likely to occur in the \(F_{2}\) generation? (e) If the \(F_{1}\) plant is testcrossed, how many different phenotypes will be produced?

Albinism in humans is inherited as a simple recessive trait. Determine the genotypes of the parents and offspring for the following families. When two alternative genotypes are possible, list both. (a) Two parents without albinism have five children, four without albinism and one with albinism. (b) A male without albinism and a female with albinism have six children, all without albinism.

How many different types of gametes can be formed by individuals of the following genotypes? What are they in each case? (a) \(A a B b\) (b) \(A a B B\) (c) \(A a B b C c\) (d) \(A a B B c c\) (e) \(A a B b c c,\) and (f) \(A a B b C c D d E e ?\)

Why was the garden pea a good choice as an experimental organism in Mendel's work?
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