Genetics: From Genes to Genomes, 5th edition
Genetics: From Genes to Genomes, 5th edition
5th Edition
ISBN: 9780073525310
Author: Leland H. Hartwell, Michael L. Goldberg, Janice A. Fischer, Leroy Hood, Charles F. Aquadro
Publisher: McGraw-Hill Education
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Chapter 5, Problem 46P

A research group has selected three independent Trp− haploid strains of Neurospora, each of which cannot grow in the absence of the amino acid tryptophan. They first mated these three strains with a wild-type strain of opposite mating type, and then they analyzed the resultant octads. For all three matings, two of the four spore pairs in every octad could grow on minimal medium (that is, in the absence of tryptophan), while the other two spore pairs were unable to grow on this minimal medium.

a. What can you conclude from this result?
In the matings of mutant strains 1 and 2 with wild type, one of the two topmost pairs in some octads had spores that could grow on minimal medium while the other of the two topmost pairs in the same octads had spores that could not grow on minimal medium. In the mating of mutant strain 3 with wild type, either all the spores in the two topmost pairs could grow on minimal medium or all could not grow on minimal medium.
b. What can you conclude from this result?
The researchers next prepared two separate cultures of each mutant strain; one of these cultures was of mating type A and the other of mating type a. They mated these strains in pairwise fashion, dissected the resultant octads, and determined how many of the individual spores could grow on minimal medium. The results are shown here.
Mating % of octads with x number of spores viable on minimal medium
x = 0 2 4 6 8
1 × 2 78 22 0 0 0
1 × 3 46 6 48 0 0
2 × 3 42 16 42 0 0
c. For each of the three matings in the table, how many of the 100 octads are PD? NPD? T?
d. Draw a genetic map explaining all of the preceding data. Assume that the sample sizes are sufficiently small that none of the octads are the result of double crossovers.
e. Although this problem describes crosses in Neurospora, it does not help in this case to present the matings in the table as ordered octads. Why not?
f. Why in this problem can you obtain gene ↔ centromere distances from the crosses in the table, even though the data are not presented as ordered octads?
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The figure below shows the life cycle of the fungus Neurospora. The adult stage of the Neurospora is a multicellular haploid.  b) Neurospora has an arginine amino acid synthesis pathway shown below. Suppose I take the strain above that only grows with arginine supplements and cross it to a different mutant Neurospora strain that grows with arginine and citrulline supplements but not with ornithine supplements. Assuming gens A, B, and C are unlinked and there is only one mutation per stain: What percentage of the progeny will grow on ornithine? What percentage on citrulline? What percentage on arginine?
From a cross between e+ f+ g+ and e− f − g− strains ofNeurospora, recombination between these linkedgenes resulted in a few octads containing the followingordered set of spores:e+ f+ g+e+ f+ g+e+ f − g+e+ f − g+e− f − g−e− f − g−e− f − g−e− f − g−a. Where was recombination initiated?b. Did crossing-over occur between genes e and g?Explain.c. Why do you end up with 2 f+ : 6 f − but 4 e+: 4 e−and 4g+: 4g−?d. Could you characterize these unusual octads as MIor MII for any of the three genes involved?Explain.
Baker's yeast, Saccharomyces cerevisiae, is a single-celled, diploid fungus (which is, of course, a eukaryote, that is capable of both meiosis and sexual reproduction). Wild type yeast can normally grow on solid or liquid minimal medium; you isolate three mutant strains which are no longer capable of growing on minimal medium alone, however, they can grow on medium supplemented with adenine. All three yeast strains are homozygous for the underlying alleles. When you cross mutant strain 1 and mutant strain 2, the offspring cannot grow on minimal medium alone and require adenine supplementation; when you cross mutant strain 1 and mutant strain 3, the offspring can grow on minimal medium alone and do not require adenine. After crossing the F1 generation of the cross between mutant strains 1 and 3, you count and determine the phenotypes of 1,000 colonies (here a colony is equivalent to an individual):  563 colonies that can grow on minimal medium alone; 437 colonies that require adenine…

Chapter 5 Solutions

Genetics: From Genes to Genomes, 5th edition

Ch. 5 - Albino rabbits lacking pigment are homozygous for...Ch. 5 - In corn, the allele A allows the deposition of...Ch. 5 - If the a and b loci are 40 cM apart and an AA BB...Ch. 5 - Write the number of different kinds of phenotypes,...Ch. 5 - A DNA variant has been found linked to a rare...Ch. 5 - Figure 5.7a on p. 134 shows chromosomes during...Ch. 5 - Figure 5.7b shows bivalents in mouse primary...Ch. 5 - Cinnabar eyes cn and reduced bristles rd are...Ch. 5 - In Drosophila, the autosomal recessive dp allele...Ch. 5 - From a series of two-point crosses, the following...Ch. 5 - Map distances were determined for four different...Ch. 5 - In the tubular flowers of foxgloves, wild-type...Ch. 5 - In Drosophila, the recessive allele mb of one gene...Ch. 5 - A snapdragon with pink petals, black anthers, and...Ch. 5 - In Drosophila, three autosomal genes have the...Ch. 5 - Drosophila females heterozygous for each of three...Ch. 5 - Male Drosophila expressing the autosomal recessive...Ch. 5 - a. In Drosophila, crosses between F1 heterozygotes...Ch. 5 - A true-breeding strain of Virginia tobacco has...Ch. 5 - Prob. 30PCh. 5 - The following list of four Drosophila mutations...Ch. 5 - Do the data that Mendel obtained fit his...Ch. 5 - Two genes control color in corn snakes as follows:...Ch. 5 - A mouse from a true-breeding population with...Ch. 5 - Neurospora of genotype a c are crossed with...Ch. 5 - A cross was performed between one haploid strain...Ch. 5 - A cross was performed between a yeast strain that...Ch. 5 - Two crosses were made in Neurospora involving the...Ch. 5 - Indicate the percentage of tetrads that would have...Ch. 5 - Prob. 42PCh. 5 - Prob. 43PCh. 5 - Prob. 44PCh. 5 - a. In ordered tetrad analysis, what is the maximum...Ch. 5 - A research group has selected three independent...Ch. 5 - A single yeast cell placed on a solid agar will...Ch. 5 - Prob. 48PCh. 5 - In Drosophila, the yellow y gene is near the end...Ch. 5 - Neurofibromas are tumors of the skin that can...Ch. 5 - Two important methods for understanding the...
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