Genetic Analysis: An Integrated Approach (2nd Edition)
2nd Edition
ISBN: 9780321948908
Author: Mark F. Sanders, John L. Bowman
Publisher: PEARSON
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Chapter 19, Problem 16P
You have isolated two petite mutants,
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To understand the genetic basis of locomotion in the diploid nematode Caenorhabditis elegans, recessive mutations were obtained, all making the worm “wiggle” ineffectually instead of moving with its usual smooth gliding motion. These mutations presumably affect the nervous or muscle systems. Twelve homozygous mutants were intercrossed, and the F1 hybrids were examined to see if they wiggled. The results were as follows, where a plus sign means that the F1 hybrid was wild type (gliding) and “w” means that the hybrid wiggled.a. Explain what this experiment was designed to test. b. Use this reasoning to assign genotypes to all 12 mutants. c. Explain why the phenotype of the F1 hybrids between mutants 1 and 2 differed from that of the hybrids between mutants 1 and 5
Chapter 19 Solutions
Genetic Analysis: An Integrated Approach (2nd Edition)
Ch. 19 - 17.1 Reciprocal crosses of experimental animals or...Ch. 19 - 17.2 How are some of the characteristics of the...Ch. 19 - 17.3 The human mitochondrial genome encodes...Ch. 19 - What is the evidence that transfer of DNA from the...Ch. 19 - Draw a graph depicting the relative amounts of...Ch. 19 - Prob. 6PCh. 19 - 17.7 What is the evidence that the ancient...Ch. 19 - 17.8 Outline the steps required for a gene...Ch. 19 - 17.9 Consider the phylogenetic tree presented in...Ch. 19 - Most large protein complexes in mitochondria and...
Ch. 19 - What insights have analyses of human mitochondrial...Ch. 19 - You are a genetic counselor, and several members...Ch. 19 - A mutation in Arabidopsis immutans results in the...Ch. 19 - What type or types of inheritance are consistent...Ch. 19 - You have isolated (1) a streptomycin-resistant...Ch. 19 - You have isolated two petite mutants, pet1 and...Ch. 19 - 17.15 Consider this human pedigree for a vision...Ch. 19 - A 50- year - old man has been diagnosed with MELAS...Ch. 19 - 17.17 The first person in a family to exhibit...Ch. 19 - Prob. 20PCh. 19 - Prob. 21PCh. 19 - 17.19 What is the most likely mode of inheritance...Ch. 19 - 17.20 In , the Russian Tsar Nicholaswas deposed,...Ch. 19 - 17.21 The dodo bird (Raphus cucullatus) lived on...Ch. 19 - Cytoplasmic male sterility (CMS) in plants has...Ch. 19 - 17.23 Wolves and coyotes can interbreed in...Ch. 19 - Prob. 27PCh. 19 - Prob. 28P
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- If you expose a culture of human cells (for example,HeLa cells) to 3H-thymidine during S phase, howwould the radioactivity be distributed over a pair ofhomologous chromosomes at metaphase? Would theradioactivity be in (a) one chromatid of one homolog,(b) both chromatids of one homolog, (c) one chromatid each of both homologs, (d) both chromatids ofboth homologs, or (e) some other pattern? Choose thecorrect answer and explain your reasoning.arrow_forwardThree autosomal recessive mutations in yeast, all producing the same phenotype (m1, m2, and m3), are subjected to complementation analysis. Of the results shown below, which, if any, are alleles of one another? Predict the results of the cross that is not shown—that is, m2 * m3. Cross 1: m1 * m24 F1: all wild-type progeny Cross 2: m1 * m34 F1: all mutant progenyarrow_forwardYou are working with two different yeast cultures to study their genetics. But, you are not sure what the mating type of either of the yeast cultures is. (As a reminder, yeast mating types are either a or alpha). As a test, you mix the two cultures together and then observe them under a microscope. If this is the image that you see, what can you conclude about the mating types of the yeast? Explain your answer, including what type of cell morphology you see on this microscope slide.arrow_forward
- An individual is heterozygous for a reciprocal translocation, with the following chromosomes: A • B C D E F A • B C V W X R ST • U D E F R ST • U V W X Q. Explain why the fertility of this individual is likely to be less than the fertility of an individual without a translocation.arrow_forward. Imagine that you have done a cross between twostrains of yeast, one of which has the genotype A B Cand the other a b c, where the letters refer to threerather closely linked genes in the order given. Youexamine many tetrads resulting from this cross, andyou find two that do not contain the expected two Band two b spores. In tetrad I, the spores are A B C, AB C, a B c, and a b c. In tetrad II, the spores are A BC, A b c, a b C, and a b c. How have these unusualtetrads arisen?arrow_forwardWhy have we chosen a human Rab protein with no direct yeast equivalent as an “outgroup” for this study? [at least mention 2 points]arrow_forward
- In the lab, you discover two yeast haploid mutants that cannot produce arginine. You cross them together and the resulting diploid produces arginine. If producing arginine is the wild type phenotype, what can you definitively conclude? A.The haploid strains have identical mutations in the same genes. B.The haploid strains have identical mutations in different genes. C.The haploid strains have mutations in different genes. D.The haploid strains have mutations in the same gene. E.The haploid strains must belong to the complementation group encoding the first enzyme in the biosynthetic pathway.arrow_forwardWhy would a mutation in a somatic cell of a multicellular organism not necessarily result in a detectable phenotype?arrow_forwardThe garden pea (Pisum sativum) is normally a diploid and has 7 chromosomes in one set (in other words, n = 7). Use this information to answer the following questions: a. How many centromeres would be in a cell of a tetraploid pea plant? ______ b. How many total chromatids would be in a cell of a tetraploid pea plant in metaphase of mitosis? _______ c. How many total chromosomes would be in a cell of a pea plant that is monosomic for a single chromosome?arrow_forward
- 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?arrow_forwardIn an electrophoretic gel across which is applied a powerful electrical alternating pulsed field, the DNA of the haploid fungus Neurospora crassa (n = 7) moves slowly but eventually forms seven bands, which represent DNA fractions that are of different sizes and hence have moved at different speeds. These bands are presumed to be the seven chromosomes. How would you show which band corresponds to which chromosome?arrow_forwardTwo diploid species of closely related frogs, which we will callspecies A and species B, were analyzed with regard to the genesthat encode an enzyme called hexokinase. Species A has two distinctcopies of this gene: A1 and A2. In other words, this diploidspecies is A1A1 A2A2. Species B has three copies of the hexokinasegene, which we will call B1, B2, and B3. A diploid individualof species B would be B1B1 B2B2 B3B3. These hexokinase genesfrom the two species were subjected to DNA sequencing, and thepercentage of sequence identity was compared among these genes.The results are shown here. Percentage of DNA Sequence Identity A1 A2 B1 B2 B3A1 100 62 54 94 53A2 62 100 91 49 92B1 54 91 100 67 90B2 94 49 67 100 64B3 53 92 90 64 100…arrow_forward
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