Ten mutant mouse strains with white coats were tested by crossing each with all the other strains. The results are shown in the table below. A "-" symbol indicates the offspring were white (failure to complement), and a "+" symbol indicates that the offspring showed wild-type brown coats (complementation). Which of the other strains have a mutation in the same gene as strain 1? a b с d None of the other strains have a mutation in the same gene as strain 1. All of the other strains have a mutation in the same gene as strain 1. Selected answer will be automatically saved. For keyboard navigation, press up/down arrow keys to select an answer. Strains 2, 4, 5, 6, 8, 9, and 10 have a mutation in the same gene as strain 1. Mutant Strains 3 and 7 have a mutation in the same gene as strain 1. 1 2 + 3+ 4 +++ - 5 + + + - 6+-++- 7++++ - 8 + + + + + + − 10 + -++--+ + + 1 2 3 4 5 6 7 8 9 10 Mutant -: offspring are mutant +: offspring are wild-type

Ten mutant mouse strains with white coats were tested by crossing each with all the other strains. The results are shown in the table below. A "-" symbol indicates the offspring were white (failure to complement), and a "+" symbol indicates that the offspring showed wild-type brown coats (complementation). Which of the other strains have a mutation in the same gene as strain 1? a b с d None of the other strains have a mutation in the same gene as strain 1. All of the other strains have a mutation in the same gene as strain 1. Selected answer will be automatically saved. For keyboard navigation, press up/down arrow keys to select an answer. Strains 2, 4, 5, 6, 8, 9, and 10 have a mutation in the same gene as strain 1. Mutant Strains 3 and 7 have a mutation in the same gene as strain 1. 1 2 + 3+ 4 +++ - 5 + + + - 6+-++- 7++++ - 8 + + + + + + − 10 + -++--+ + + 1 2 3 4 5 6 7 8 9 10 Mutant -: offspring are mutant +: offspring are wild-type

Biology (MindTap Course List)

11th Edition

ISBN:9781337392938

Author:Eldra Solomon, Charles Martin, Diana W. Martin, Linda R. Berg

Publisher:Eldra Solomon, Charles Martin, Diana W. Martin, Linda R. Berg

Chapter11: The Basic Principles Of Heredity

Section: Chapter Questions

Problem 12TYU: Genes A and B are 6 map units apart, and A and C are 4 map units apart. Which gene is in the middle...

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A yeast geneticist irradiates haploid cells of a strain that is an adenine-requiring auxotrophic mutant, caused by mutation of the gene ade1. Millions of the irradiated cells are plated on minimal medium, and a small number of cells divide and produce prototrophic colonies. These colonies are crossed individually with a wildtype strain. Two types of results are obtained:(1) prototroph × wild type : progeny all prototrophic(2) prototroph × wild type : progeny 75% prototrophic, 25% adenine-requiring auxotrophsa. Explain the difference between these two types of results.b. Write the genotypes of the prototrophs in each case.c. What progeny phenotypes and ratios do you predict from crossing a prototroph of type 2 by the original ade1auxotroph?
In a series of two-point map crosses involving five genes locatedon chromosome II in Drosophila, the following recombinant (single- crossover) frequencies were observed:pr–adp 29pr–vg 13pr–c 21pr–b 6adp–b 35adp–c 8adp–vg 16vg–b 19vg–c 8c–b 27 In another set of experiments, a sixth gene (d) was testedagainst b and pr, and the results were d - b = 17% andd - pr = 23%. Predict the results of two-point mapsbetween d and c, d and vg, and d and adp.
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You have already localized the genes to the same chromosome by deletion mapping, and now decide that the best way to accomplish the mapping is to conduct two simultaneous three-point testcross experiments. The genes you are investigating are as follows: N = round leaves, n = notched leaves; H = smooth stems, h = hairy stems; R = purple flowers, r = red flowers; B = grey seeds, b = black seeds; and Y = green pods, y = yellow pods. Earlier experiments you have done already established that gene B is in the middle of this gene cluster, so you design both three-point test crosses to include that gene. Cross #1 is designed as RrHhBb x rrhhbb while cross #2 is NnBbYy x nnbbyy. The results of both crosses are given in the table below. Based on the information given, determine the arrangement of these five genes including the position of each allele in the heterozygous fly and the distances between each pair of genes. (Hint: treat each experiment separately, knowing that gene B is in the…
When true-breeding mice with brown fur and short tails (BBtt) were crossed to true-breeding mice with white fur and long tails (bbTT), all the F1 offspring had brown fur and long tails. The F1 offspring were then crossed to mice with white fur and short tails (bbtt). The fur gene and the tail length gene are 7 cM apart. Part a Which of the following F2 offspring are recombinant? (Select all that apply.) a) White fur, short tail b) Brown fur, long tail c) You need more information to answer this question. d) White fur, long tail e) Brown fur, short tail Part b Using the information given in Part a, if there were 400 mice in the F2 generation, how many do you expect would have white fur and long tails? a) 25 b) 186 c) 14 d) 372 e) 28 f) 75
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