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Ch. 5 - Chromosome Mapping in Eukaryotes
Chapter 5, Problem 12

Three gene pairs located on separate autosomes determine flower color and shape as well as plant height. The first pair exhibits incomplete dominance, where the color can be red, pink (the heterozygote), or white. The second pair leads to personate (dominant) or peloric (recessive) flower shape, while the third gene pair produces either the dominant tall trait or the recessive dwarf trait. Homozygous plants that are red, personate, and tall are crossed to those that are white, peloric, and dwarf. Determine the F₁ genotype(s) and phenotype(s). If the F₁ plants are interbred, what proportion of the offspring will exhibit the same phenotype as the F₁ plants?

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Identify the genotypes of the parent plants. The red, personate, and tall plant is homozygous dominant for all traits, so its genotype is RRTTSS. The white, peloric, and dwarf plant is homozygous recessive for all traits, so its genotype is rrttss.
Determine the F₁ genotype by crossing the parent genotypes. Since the first trait shows incomplete dominance, the F₁ genotype for color will be Rr, resulting in pink flowers. The other traits are simple dominance, so the F₁ genotype will be Tt for height and Ss for shape.
Identify the F₁ phenotype. The F₁ plants will be pink (Rr), personate (Ss), and tall (Tt).
Consider the F₁ interbreeding. Use a Punnett square to determine the genotypic and phenotypic ratios of the F₂ generation. Each trait is inherited independently, so calculate the probability for each trait separately and then multiply the probabilities.
Calculate the proportion of F₂ offspring with the same phenotype as the F₁ plants. For each trait, determine the probability of the F₂ offspring having the same phenotype as the F₁ (pink, personate, tall) and multiply these probabilities together.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Incomplete Dominance

Incomplete dominance is a genetic phenomenon where the heterozygous phenotype is an intermediate blend of the two homozygous phenotypes. In this case, crossing a red flower (homozygous) with a white flower (homozygous) results in pink flowers (heterozygous). This concept is crucial for understanding how the flower color trait is inherited in the given scenario.
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Variations on Dominance

Gene Interaction

Gene interaction refers to the way different gene pairs influence each other's expression and the resulting phenotype. In this question, three gene pairs affect flower color, shape, and plant height, demonstrating how multiple genes can interact to produce a variety of traits. Understanding this concept is essential for predicting the phenotypes of the offspring from the crosses described.
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Interacting Genes Overview

Punnett Square

A Punnett square is a diagram used to predict the genotypes and phenotypes of offspring from genetic crosses. By organizing the alleles of the parent plants, one can visualize the possible combinations in the F₁ generation and determine the proportions of phenotypes in subsequent generations. This tool is vital for solving the question regarding the F₁ genotypes and the phenotypic ratios of the offspring.
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Chi Square Analysis
Related Practice
Textbook Question
In the cross shown here, involving two linked genes, ebony (e) and claret (ca), in Drosophila, where crossing over does not occur in males, offspring were produced in a 2 + : 1 ca : 1 e phenotypic ratio: These genes are 30 units apart on chromosome III. What did crossing over in the female contribute to these phenotypes?

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Textbook Question
In a series of two-point mapping crosses involving five genes located on chromosome II in Drosophila, the following recombinant (single-crossover) frequencies were observed: pr–adp 29% pr–vg 13 pr–c 21 pr–b 6 adp–b 35 adp–c 8 adp–vg. 16 vg–b. 19 vg–c 8 c–b. 27 Given that the adp gene is near the end of chromosome II (locus 83), construct a map of these genes.
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Textbook Question
In a series of two-point mapping crosses involving five genes located on chromosome II in Drosophila, the following recombinant (single-crossover) frequencies were observed: pr–adp 29% pr–vg 13 pr–c 21 pr–b 6 adp–b 35 adp–c 8 adp–vg. 16 vg–b. 19 vg–c 8 c–b. 27 In another set of experiments, a sixth gene, d, was tested against b and pr: d–b 17% d–pr 23% Predict the results of two-point mapping between d and c, d and vg, and d and adp.
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Textbook Question
Two different female Drosophila were isolated, each heterozygous for the autosomally linked genes b (black body), d (dachs tarsus), and c (curved wings). These genes are in the order d–b–c, with b being closer to d than to c. Shown here is the genotypic arrangement for each female along with the various gametes formed by both: Identify which categories are noncrossovers (NCOs), single crossovers (SCOs), and double crossovers (DCOs) in each case. Then, indicate the relative frequency in which each will be produced.
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Textbook Question
As in Problem 12, flower color may be red, white, or pink, and flower shape may be personate or peloric. For the following crosses, determine the P₁ and F₁ genotypes:
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Textbook Question
In Drosophila, a cross was made between females—all expressing the three X-linked recessive traits scute bristles (sc), sable body (s), and vermilion eyes (v)—and wild-type males. In the F₁, all females were wild type, while all males expressed all three mutant traits. The cross was carried to the F₂ generation, and 1000 offspring were counted, with the results shown in the following table. Phenotype Offspring sc s v 314 + + + 280 + s v 150 sc + + 156 sc + v 46 + s + 30 sc s + 10 + + v 14 No determination of sex was made in the data. Using proper nomenclature, determine the genotypes of the P₁ and F₁ parents.
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