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There are two highly inbred strains of laboratory mice whose adult body weights are very different. Assume that the mouse's body weight is under the control of three pairs of contrasting genes: A vs. a, B vs. b and D vs. d. Assume further that each capital letter gene is responsible for contributing \(5.0\) grams to the total body weight, and that lowercase letters contribute \(2.5\) grams to total body weight. The average weight of mice in Strain I is 30 grams, while that of Strain II mice is 15 grams. (a) What are the most likely genotypes of these two strains? (b) Suppose Strain I and Strain II are crossed. What will be the phenotype of the \(\mathrm{F}_{1}\) progeny?

Short Answer

Expert verified
The most likely genotypes for Strain I and Strain II mice are AABBDD and aabbdd, respectively. If Strain I and Strain II are crossed, the phenotype of the F₁ progeny will be approximately 11.25 grams.

Step by step solution

01

Determine the total contributions from each gene pair

First, we need to find out the total weight contribution from all three gene pairs. As each capital letter gene (A, B, and D) contributes 5 grams and each lowercase letter gene (a, b, and d) contributes 2.5 grams, we have: - AABBDD: 5 + 5 + 5 + 5 + 5 + 5 = 30 grams - aabbdd: 2.5 + 2.5 + 2.5 + 2.5 + 2.5 + 2.5 = 15 grams
02

Identify the possible genotypes for Strain I and Strain II

Now, we have to identify the genotypes of Strain I and Strain II based on their average weights. As mentioned earlier, the weights are: - Strain I: 30 grams ⇒ AABBDD - Strain II: 15 grams ⇒ aabbdd
03

Determine the phenotype of the F1 progeny

Finally, we need to determine the phenotype of the F1 progeny based on the genotypes of their parents. When Strain I (AABBDD) and Strain II (aabbdd) are crossed, the genotype of the F1 progeny will be: - AaBbDd To calculate the phenotype of the F₁ progeny, we have to add up the weight contributions from these heterozygous gene pairs: - Aa: (5 + 2.5) / 2 = 3.75 grams - Bb: (5 + 2.5) / 2 = 3.75 grams - Dd: (5 + 2.5) / 2 = 3.75 grams Now, add up the weights from all gene pairs: Phenotype of F₁ progeny: 3.75 + 3.75 + 3.75 = 11.25 grams So, the phenotype of the F₁ progeny will be approximately 11.25 grams.

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Key Concepts

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

Inbred Strains
Inbred strains are groups of genetically similar organisms that arise through continuous breeding within a closed population. This means that any genetic variations within the group are minimized over generations. Because of this controlled breeding, the genetic makeup or genotype of each organism within the strain becomes nearly identical.

In laboratory settings, inbred strains are incredibly valuable. They provide researchers with a consistent genetic baseline to study the effects of specific genes or external factors on certain traits, such as body weight in mice. In our exercise, each strain of mice has a distinct and fixed genotype that determines their average adult body weight. Strain I is an example of a genetically homogenous group with heavy body weight, while Strain II comprises a lighter counterpart. Understanding these genetic consistencies allows researchers to predict offspring characteristics accurately when two strains are crossed.
Phenotype Calculation
A phenotype refers to the observable physical or biochemical characteristics of an organism, as determined by its genotype. Calculating the phenotype involves understanding the genetic structure and how each gene contributes to an observable trait, such as body weight.

In our example, body weight depends on three pairs of contrasting genes. Each uppercase gene contributes more to the body weight than its lowercase counterpart. By knowing the exact genetic contributions, we can calculate the total weight.
  • For Strain I (\( AABBDD \)), each gene contributes 5 grams, making the phenotype 30 grams.
  • For Strain II (\( aabbdd \)), each gene contributes 2.5 grams, totaling 15 grams.
This precise calculation highlights the critical role of genetics in determining physical attributes and helps predict the phenotype of any potential offspring.
Genotype Analysis
Genotype analysis involves determining the genetic composition of an organism with respect to a particular trait. In this scenario, genotype analysis helps us understand what genes are present and contributes to the observed phenotype.

For our inbred mice strains, the genotype of Strain I is identified as \( AABBDD \), where all contributing genes are capital letters, resulting in higher body weight. Strain II is noted as \( aabbdd \), featuring only recessive lower case genes, hence a lower body weight. When these strains are crossed, their offspring (the \( F_1 \) progeny) inherit one allele from each parent, leading to a heterozygous genotype, \( AaBbDd \).
  • Each gene pair combines the effects of both upper and lower cases, averaging the contributions to the resulting phenotype.
Understanding how to analyze and predict the impact of such genetic crosses is vital in fields like genetics and breeding programs, shaping our ability to manage and influence inheritance patterns.

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Most popular questions from this chapter

The checkered pattern of pigeons is controlled by a dominant gene \(C\); plain color is determined by the recessive allele c. Red color is controlled by a dominant gene \(\mathrm{B}\), and brown color by the recessive allele b. Complete a Punnett square for a dihybrid cross involving a homozygous checkered red bird and a plain brown bird. For this cross, show the expected phenotypes, genotypes, genotypic frequencies and phenotypic ratios for the \(\mathrm{F}_{2}\) generation.

Consider that in horses a single pair of genes controls gait. Some horses can only exhibit a trotting gait, while others exhibit only a pacing gait. In an attempt to determine the pattern of inheritance displayed by this trait, a horse breeder crossed a stallion ( 8 ) that came from a long line of pacers to several mares ( 9 ) that were trotters. All of the foals were trotters. Which gait is under the control of the dominant gene? Using the letters \(\mathrm{T}\) and \(\mathrm{t}\) to represent the dominant and recessive genes, respectively, identify the genotypes of the stallion, the mares and the foals.

In peas, tall (D) is dominant to dwarf (d) and yellow cotyledons (G) is dominant to green (g). If a tall, homozygous, yellow pea plant is crossed with a dwarf, green pea plant, what will be the phenotypic results in the \(\mathrm{F}_{1}\) and \(\mathrm{F}_{2} ?\)

What are the possible gametes that can be formed from the following genotypes, assuming all the gene pairs segregate independently? What are the gamete frequencies? (a) \(\mathrm{A} \mathrm{aBBCc}\) (b) DdEEffGg (c) \(\mathrm{MmNnOo}\)

Normal-length fur in rabbits is controlled by the dominant allele \(\mathrm{R}\), and a short type of fur, called "rex," is determined by the recessive allele \(\mathrm{r}\). The dominant allele B controls black fur color; the recessive allele b controls brown. (a) Diagram a dihybrid cross between a homozygous rabbit with normal-length black fur and a rex rabbit with brown fur. What are the phenotypic ratios resulting from this cross? (b) What proportion of the normal, black rabbits in the \(\mathrm{F}_{2}\) generation of this cross can be expected to be homozygous for both pairs of genes? (c) What would be the expected phenotypic and genotypic results of a backcross between a member of the \(\mathrm{F}_{1}\) generation and a fully recessive rex, brown parent?

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