Genetic Analysis: An Integrated Approach (3rd Edition)
3rd Edition
ISBN: 9780134605173
Author: Mark F. Sanders, John L. Bowman
Publisher: PEARSON
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Chapter 17, Problem 14P
You have isolated two petite mutants,
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In yeast, LYS5, ADE1, and URA2 are genes required to synthesize lysine,
adenine and uracil, respectively. A MATa lys5 ADE1 URA2 haploid was mated
to a MATA LYS5 ade1 ura2 haploid to make a triply heterozygous diploid. This
diploid was put through meiosis, 100 tetrads were dissected and the spore
colonies transferred to medium lacking either lysine, adenine or uracil to
determine which spores were prototrophic or auxotrophic for those nutrients.
The following results were obtained:
LYS5-ADE1
ADE1-URA2
LYS5-URA2
PD
NPD
Ι
20
22
58
18
20
62
50
8
42
Based on this information, select the statements below that are TRUE.
Select 4 correct answer(s)
A) The LYS5 and URA2 genes are linked.
B)
The genotypes of the four spores in the ADE1-URA2 tetratypes is
ade1 ura2
ade1 URA2
ADE1 ura2
ADE1 URA2
C) The map distance between ADE1 and URA2 is 51 CM.
D) ADE1 and URA2 are likely located on the same chromosome.
E) The map distance between LYS5 and URA2 is 29 CM.
☐ F) LYS5 and ADE1 independently assort…
A heterozygous diploid yeast Aa Bb went through meiosis. What percentage of the haploid spores will have recombinant combinations of alleles?
What if genes A and B are unlinked? Explain
What is genes A and B are linked? Explain
A yeast strain with a mutant spo11- allele has been isolated. The mutant allele is nonfunctional; it makes no spo11 protein. What do you suppose is the phenotype of this mutant strain?
Chapter 17 Solutions
Genetic Analysis: An Integrated Approach (3rd Edition)
Ch. 17 - 17.1 Reciprocal crosses of experimental animals or...Ch. 17 - 17.2 How are some of the characteristics of the...Ch. 17 - 17.3 The human mitochondrial genome encodes...Ch. 17 - What is the evidence that transfer of DNA from the...Ch. 17 - Draw a graph depicting the relative amounts of...Ch. 17 - Prob. 6PCh. 17 - 17.7 What is the evidence that the ancient...Ch. 17 - 17.8 Outline the steps required for a gene...Ch. 17 - 17.9 Consider the phylogenetic tree presented in...Ch. 17 - You are a genetic counselor, and several members...
Ch. 17 - A mutation in Arabidopsis immutans results in the...Ch. 17 - What type or types of inheritance are consistent...Ch. 17 - You have isolated (1) a streptomycin-resistant...Ch. 17 - You have isolated two petite mutants, pet1 and...Ch. 17 - 17.15 Consider this human pedigree for a vision...Ch. 17 - A 50- year - old man has been diagnosed with MELAS...Ch. 17 - 17.17 The first person in a family to exhibit...Ch. 17 - Prob. 18PCh. 17 - 17.19 What is the most likely mode of inheritance...Ch. 17 - 17.20 In , the Russian Tsar Nicholaswas deposed,...Ch. 17 - 17.21 The dodo bird (Raphus cucullatus) lived on...Ch. 17 - Cytoplasmic male sterility (CMS) in plants has...Ch. 17 - 17.23 Wolves and coyotes can interbreed in...Ch. 17 - Prob. 24PCh. 17 - Prob. 25PCh. 17 - Most large protein complexes in mitochondria and...Ch. 17 - As described in this chapter, mothers will pass on...
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- Humans and many other complex organisms are diploid, possessing two sets of genes, one inherited from the mother and one from the father. However, a number of eukaryotic organisms spend most of their life cycles in a haploid state. Many of these eukaryotes, such as Neurospora and yeast, still undergo meiosis and sexual reproduction, but most of the cells that make up the organism are haploid. Considering that haploid organisms are fully capable of sexual reproduction and generating genetic variation, why are most complex eukaryotes diploid? In other words, what might be the evolutionary advantage of existing in a diploid state instead of a haploid state? And why might a few organisms, such as Neurospora and yeast, exist as haploids?arrow_forwardFive independently isolated mutants of yeast all of which are unable to grow in the absence of histidine (his-) were isolated. Each mutant was crossed with all the others. The experimental results of this experiment, using + to indicate growth or 0 to indicate no growth are given below. mutant 1 13 14 15. List the members of each How many genes are definecarrow_forwardTo 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 5arrow_forward
- In a haploid yeast strain, eight recessive mutationswere found that resulted in a requirement for theamino acid lysine. All the mutations were found to revert at a frequency of about 1 × 10−6 except mutations5 and 6, which did not revert. Matings were madebetween a and α cells carrying these mutations. Theability of the resultant diploid strains to grow onminimal medium in the absence of lysine is shown inthe following chart (+ means growth and − means nogrowth.)1 2 3 4 5 6 7 81 − + + + + − + −2 + − + + + + + +3 + + − − − − − +4 + + − − − − − +5 + + − − − − − +6 − + − − − − − −7 + + − − − − − +8 − + + + + − + −a. How many complementation groups were revealedby these data? Which point mutations are foundwithin which complementation groups?The same diploid strains are now induced to undergosporulation. The vast majority of resultant spores areauxotrophic; that is, they cannot form colonies whenplated on minimal medium (without lysine). However,particular diploids can produce rare spores…arrow_forwardThe mutations called bobbed in Drosophila result from variable reductions (deletions) in the number of amplified genes coding for rRNA. Researchers trying to maintain bobbed stocks have often documented their tendency to revert to wild type in successive generations. Propose a mechanism based on meiotic recombination which could account for this reversion phenomenon. Why would wild-type flies become more prevalent in Drosophila cultures?arrow_forwardIn a wild-type fungus, protein E (encoded by the haplosufficient gene E) normally dimerizes to catalyzes a biochemical reaction necessary for the production of a dark pigment. Ed represents a mutant, dominant negative allele of gene E. What is the predicted phenotype of a fungus cell of genotype E*/Ed, and why? O wild type (normal production of the dark pigment), as E is haplosufficient mutant (no pigment production), as no dimers will form in the heterozygous mutant (no pigment production), as the mutant allele Eg is dominant O wild type (normal production of the dark pigment), as dimers of wild-type and mutant protein E will be formed in the heterozygousarrow_forward
- You cross two yeast strains one is an ade auxotroph the other is a pro auxotroph and allow the diploid to sporulate. When you score each spore in the ascus you find the following proportions: 518 PD, 8 NPD, and 225 T. a.) What are the genotypes of each spore in all three types of the tetrads. b) Are these genes linked why or why not? c.) If these genes are unlinked what would you expect the progeny numbers and ratios to be? d.) What is the formula to determine the most accurate distance between these genes? If linked what is the map distance?arrow_forward. A diploid strain of yeast was made by mating a haploidstrain with a genotype w−, x−, y−, and z− with a haploidstrain of opposite mating type that is wild type for thesefour genes. The diploid strain was phenotypically wildtype. Four different X-ray-induced diploid mutantswith the following phenotypes were produced fromthis diploid yeast strain. Assume a single new mutation is present in each strain.Strain 1 w− x+ y− z+Strain 2 w+ x− y− z−Strain 3 w− x+ y− z−Strain 4 w− x+ y+ z+When these mutant diploid strains of yeast go throughmeiosis, each ascus is found to contain only two viablehaploid spores.a. What kind of mutations were induced by X-rays tomake the listed diploid strains?b. Why did two spores in each ascus die?c. Are any of the genes w, x, y, or z located on thesame chromosome?d. Give the order of the genes that are found on thesame chromosomearrow_forwardBaker'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…arrow_forward
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