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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 obtaining 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 genotype \(AABbCc\) from these parents is 0.

Step by step solution

01

Find the probability of obtaining an \(AA\) allele from the parents

To get an \(AA\) allele, the offspring has to inherit the dominant allele \(A\) from both parents. The first parent has a genotype of \(Aa\), which means the probability of passing on the \(A\) allele is 1/2 (since there's 1 dominant allele and 2 alleles in total for this gene pair). The second parent has a genotype of \(AA\), which means the probability of passing on the \(A\) allele is 1 (since there are 2 dominant alleles and 2 alleles in total for this gene pair). Multiply the probabilities from each parent to get the probability of obtaining an \(AA\) offspring: \(P(AA) = \frac{1}{2} * 1 = \frac{1}{2}\)
02

Find the probability of obtaining a \(Bb\) allele from the parents

To get a \(Bb\) allele, the offspring has to inherit one dominant allele \(B\) and one recessive allele \(b\) from the parents. Both parents have genotypes of \(Bb\) for this gene pair, which means the probability of passing on the \(B\) allele is 1/2, and the probability of passing on the \(b\) allele is also 1/2. To obtain a \(Bb\) offspring, it can either inherit \(B\) from the first parent and \(b\) from the second, or inherit \(b\) from the first parent and \(B\) from the second. Therefore, we have to add the probabilities of these two cases: \(P(Bb) = (\frac{1}{2} * \frac{1}{2}) + (\frac{1}{2} * \frac{1}{2}) = \frac{1}{4} + \frac{1}{4} = \frac{1}{2}\)
03

Find the probability of obtaining a \(Cc\) allele from the parents

To get a \(Cc\) allele, the offspring has to inherit the recessive allele \(c\) from both parents. The first parent has a genotype of \(CC\) and the second parent has a genotype of \(Cc\). Since the first parent does not have a \(c\) allele, the probability of obtaining a \(Cc\) offspring from these parents is 0. \(P(Cc) = 0\)
04

Calculate the probability of obtaining an \(AABbCc\) offspring from the parents

To find the probability of obtaining an \(AABbCc\) offspring, we need to multiply the probabilities we found in steps 1-3: \(P(AABbCc) = P(AA) * P(Bb) * P(Cc) = \frac{1}{2} * \frac{1}{2} * 0 = 0\) The probability of obtaining an offspring with the genotype \(AABbCc\) from these parents is 0.

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Most popular questions from this chapter

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?

Mendel crossed peas with round, green seeds with peas having wrinkled, yellow seeds. All \(\mathrm{F}_{1}\) plants had seeds that were round and yellow. Predict the results of testcrossing these \(F_{1}\) plants.

In a cross between a black and a white guinea pig, all members of the \(F_{1}\) generation are black. The \(F_{2}\) generation is made up of approximately \(3 / 4\) black and \(1 / 4\) white guinea pigs. Diagram this cross, and show the genotypes and phenotypes.

In an intra-species cross performed in mustard plants of two different species (Brassicajuncea and Brassica oleracea), a tall plant \((T T)\) was crossed with a dwarf (tt) variety in each of the two species. The members of the \(\mathrm{F}_{1}\) generation were crossed to produce the \(\mathrm{F}_{2}\) generation. Of the \(\mathrm{F}_{2}\) plants, Brassica juncea had 60 tall and 20 dwarf plants, while Brassica oleracea had 100 tall and 20 dwarf plants. Use chi-square analysis to analyze these results.

A plant breeder observed that for a certain leaf trait of maize that shows two phenotypes (phenotype 1 and phenotype 2), the \(\mathrm{F}_{1}\) generation exhibits 200 plants with phenotype 1 and 160 with phenotype 2. Using two different null hypotheses and chi-square analysis, compute if the data fits (a) a 3: 1 ratio, and (b) a 1: 1 ratio.
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