Concept explainers
Form a small discussion group and decide on the most likely genetic explanation for each of the following situations;
a. A man who has red–green color blindness and a woman who has complete color vision have a son with red–green color blindness. What are the genotypes of these three people, and how do you explain the color blindness of the son?
b. Cross A performed by Morgan and shown in Figure 3.18 is between a mutant male fruit fly with white eyes and a female fruit fly from a pure-breeding, red-eye stock. The figure shows that 1237 F1 progeny were produced, all of them with red eyes. In reality, this isn’t entirely true. Among the 1237 F1 progeny were 3 male flies with white eyes. Give two possible explanations for the appearance of these white-eyed males.
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Genetic Analysis: An Integrated Approach (3rd Edition)
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- Please answer both questions in detail. Correct answers are highlighted.arrow_forward1)se; 12 cM 2)h; 12 cM 3)g; 8 cM 4)se; 8 cMarrow_forwardConsider the following linked traits in fruit flies: V= vermillion eyes (v+ is red eyes) cv = crossveinless wings (cv+ have crossveins) ct = cut wing margins (ct+ have uncut margins) You perform the following cross: v+v+ cv+cv+ ct+ct+ x vv cvcv ctct You then cross an F1 fly with a fly recessive for all three traits: v+v cv+cv ct+ct x vv cvcv ctct These are the offspring you observed from this cross: Red, crossveins, uncut 580 Vermillion, crossveinless, cut 592 Red, crossveinless, cut 89 Vermillion, crossveins, uncut 94 Red, crossveinless, uncut 45 Vermillion, crossveins, cut 40 Red, crossveins, cut 3 Vermillion, crossveinless, uncut 5 Draw a map of these 3 genes on the fly chromosome they inhabit.arrow_forward
- What will be the results of the following crosses, where N is black, n is brown, L is short hair and l is long hair. Make the tables and the corresponding explanation: 1. Crossing a Nnll woman dog with a Nnll dog. 2. Crossing a NnLl woman dog with a NnLl dog 3. Crossing a woman dog nnLl with a dog NNllarrow_forward. a. A mouse cross A/a ⋅ B/b × a/a ⋅ b/b is made, and inthe progeny there are25% A/a ⋅ B/b, 25% a/a ⋅ b/b,25% A/a ⋅ b/b, 25% a/a ⋅ B/bExplain these proportions with the aid of simplifiedmeiosis diagrams.b. A mouse cross C/c ⋅ D/d × c/c ⋅ d/d is made, and inthe progeny there are45% C/c ⋅ d/d, 45% c/c ⋅ D/d,5% c/c ⋅ d/d, 5% C/c ⋅ D/dExplain these proportions with the aid of simplifiedmeiosis diagrams.arrow_forwardThe data set attached summarizes F2 numbers from an F1 cross arising from two, true-breeding Drosophila strains (P generation), which differ with respect to two mutant traits. Here are the hypothesis: Leg length - The wild-type and mutant alleles for leg length are incomplete dominant relative to each other. Justification: The data set includes three phenotypic categories for leg length: wild type (long leg), medium leg, and truncated wings. The presence of three distinct phenotypes suggests an incomplete dominance pattern, where the heterozygous individuals exhibit an intermediate leg length phenotype (medium leg). The absence of purebred short-legged individuals supports the idea that the long leg allele is dominant over the short leg allele. This shows that mode of inheritance is incomplete dominance of the alleles relative to each other. Since the data does not mention any specific differences between males and females, we can assume that the mode of inheritance for the trait is…arrow_forward
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