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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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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Independent Assortment
In genetics, independent assortment refers to how different genes independently separate from each other when reproductive cells develop. This principle is crucial in understanding the genetic variation seen in offspring from the same parents. The idea of independent assortment is rooted in the behavior of chromosomes during meiosis. During this process, homologous chromosomes are separated into different gametes. Because each gene is located on a chromosome, this separation affects how alleles for different traits are distributed.For example, consider our gene pairs, \(A/a\), \(B/b\), and \(C/c\). Since they assort independently, the allele a gamete receives for one gene does not influence the allele it receives for another gene. This leads to multiple combinations in the offspring. Each gene pair behaves as if it is inherited separately from the others.Independent assortment allows for genetic diversity, which is vital for evolution as it contributes to changes within a population over time. The unique combination of genes from independently assorting pairs can result in offspring with a wide array of traits.
Dominant and Recessive Alleles
Genes exist in pairs and can have different versions called alleles. When it comes to how these alleles express themselves, one might wonder why some traits seem to mask others. This is where dominant and recessive alleles come into play.A dominant allele is the version of a gene that will express its trait in the presence of another different allele. It effectively "dominates" and is represented by an uppercase letter (e.g., \(A\), \(B\), \(C\)). A recessive allele, on the other hand, only expresses its trait when paired with another recessive allele. Its presence is recessive to the dominant, and it's denoted by a lowercase letter (e.g., \(a\), \(b\), \(c\)).For instance, if an individual inherits an \(A\) dominant allele from one parent and an \(a\) recessive allele from the other, the \(A\) allele's trait will be expressed. Only when an individual has two recessive alleles (\(aa\)) will the recessive trait be displayed. This concept of dominance and recessiveness is key to predicting offspring traits through genetic probability and helps us to understand the result of certain genetic crosses.
Punnett Squares
Punnett squares are a simple yet powerful tool used in genetics to predict the probable genetic makeup of offspring from a pair of parents. Developed by Reginald Punnett, these diagrams facilitate understanding of how alleles segregate independently.To use a Punnett square, follow these steps:
  • List the alleles for one parent on the top, typically showing all possible gametes they can produce.
  • List the alleles for the other parent on the side, again listing all possible gametes.
  • Fill in the squares by combining the alleles from the row and column where they intersect.
This method determines the probability of offspring inheriting particular genetic combinations. For example, if we're considering two \(Bb\) parents and want to predict offspring genotypes like \(Bb\), we list \(B\) and \(b\) for both parents across the top and side of the square. Filling in the squares gives the probability of each combination, providing a visual representation of potential genetic outcomes.Punnett squares are especially useful in educational settings, helping students visualize genetic probability and understand complex inheritance patterns in a straightforward way.

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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?

Two organisms, \(A A B B C C D D E E\) and aabbccddee, are mated to produce an \(\mathrm{F}_{1}\) that is self-fertilized. If the capital letters represent dominant, independently assorting alleles: (a) How many different genotypes will occur in the \(\mathrm{F}_{2}\) ? (b) What proportion of the \(\mathrm{F}_{2}\) genotypes will be recessive for all five loci? (c) Would you change your answers to (a) and/or (b) if the initial cross occurred between \(A A b b C C d d e e \times a a B B c c D D E E\) parents? (d) Would you change your answers to (a) and/or (b) if the initial cross occurred between \(A A B B C C D D E E \times\)aabbccddEE parents?

Among dogs, short hair is dominant to long hair and dark coat color is dominant to white (albino) coat color. Assume that these two coat traits are caused by independently segregating gene pairs. For each of the crosses given below, write the most probable genotype (or genotypes if more than one answer is possible) for the parents. It is important that you select a realistic symbol set and define each symbol below. (a) dark, short \(\times\) dark, long \(26 \quad 24 \quad 0\) (b) albino, short \(\times\) albino, short \(0 \quad 0 \quad 102 \quad 33\) (c) dark, short \(\times\) albino, short \(16 \quad 0 \quad 16\) (d) dark, short \(\times\) dark, short \(175 \quad 67 \quad 61 \quad 21\) Assume that for cross (d), you were interested in determining whether fur color follows a 3: 1 ratio. Set up (but do not complete the calculations) a Chi-square test for these data [fur color in cross (d)].

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.

Correlate Mendel's four postulates with what is now known about homologous chromosomes, genes, alleles, and the process of meiosis.

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