Genetics: From Genes to Genomes
6th Edition
ISBN: 9781259700903
Author: Leland Hartwell Dr., Michael L. Goldberg Professor Dr., Janice Fischer, Leroy Hood Dr.
Publisher: McGraw-Hill Education
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Chapter 6, Problem 30P
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
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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.
A. What conclusions can you make about the alleles of mutant strains 1, 2, and 3 and their relationships with each other?
B. What phenomenon is occurring in the cross between mutant strains 1 and 3?
After crossing the F1 generation of the cross between mutant strains 1…
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…
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.
A. What conclusions can you make about the alleles of mutant strains 1, 2, and 3 and their relationships with each other?
B. What phenomenon is occurring in the cross between mutant strains 1 and 3?
Chapter 6 Solutions
Genetics: From Genes to Genomes
Ch. 6 - Griffith, in his 1928 experiments, demonstrated...Ch. 6 - Griffith, in his 1928 experiments, demonstrated...Ch. 6 - During bacterial transformation, DNA that enters a...Ch. 6 - Nitrogen and carbon are more abundant in proteins...Ch. 6 - If 30 of the bases in human DNA are A, a what...Ch. 6 - Which of the following statements are true about...Ch. 6 - Imagine you have three test tubes containing...Ch. 6 - What information about the structure of DNA was...Ch. 6 - A portion of one DNA strand of the human gene...Ch. 6 - When a double-stranded DNA molecule is exposed to...
Ch. 6 - A particular virus with DNA as its genetic...Ch. 6 - The underlying structure of DNA is very simple,...Ch. 6 - Prob. 13PCh. 6 - Bacterial transformation and bacteriophage...Ch. 6 - The CAP protein is shown bound to DNA in Fig....Ch. 6 - In Meselson and Stahls density shift experiments...Ch. 6 - When Meselson and Stahl grew E. coli in 15N medium...Ch. 6 - If you expose human tissue culture cells for...Ch. 6 - Draw a replication bubble with both replication...Ch. 6 - a. Do any strands of nucleic acid exist in nature...Ch. 6 - As Fig. 6.21 shows, DNA polymerase cleaves the...Ch. 6 - The bases of one of the strands of DNA in a region...Ch. 6 - Replicating structures in DNA can be observed in...Ch. 6 - Indicate the role of each of the following in DNA...Ch. 6 - Draw a diagram of replication that is occurring at...Ch. 6 - Figure 6.18 depicts Watson and Cricks initial...Ch. 6 - Researchers have discovered that during...Ch. 6 - A DNA synthesizer is a machine that uses automated...Ch. 6 - Bacterial cells were coinfected with two types of...Ch. 6 - A yeast strain with a mutant spo11- allele has...Ch. 6 - Imagine that you have done a cross between two...Ch. 6 - The Neurospora octad shown came from a cross...Ch. 6 - From a cross between e f g and e f g strains of...Ch. 6 - In Step 6 of Fig. 6.27, the resolvase enzyme...Ch. 6 - Figure 6.31shows four potential outcomes of...Ch. 6 - Each of the substrates for site-specific...Ch. 6 - Prob. 37PCh. 6 - Suppose that you could inject a wild-type mouse...Ch. 6 - C31 is a type of bacteriophage that infects...Ch. 6 - Cre is a recombinase enzyme encoded by a gene in...Ch. 6 - Like Cre/loxP recombination, site-specific...
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- A yeast geneticist irradiates haploid cells of a strain thatis an adenine-requiring auxotrophic mutant, caused bymutation of the gene ade1. Millions of the irradiatedcells 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 ofresults.b. Write the genotypes of the prototrophs in each case.c. What progeny phenotypes and ratios do you predictfrom crossing a prototroph of type 2 by the original ade1auxotroph?arrow_forwardFor a haploid fungus, the starting point in the biosynthesis of the amino acid arginine is Compound X, which is always present in and absorbed from the environment. The arginine biosynthetic pathway is: Enzyme A Enzyme B Enzyme Ç Compound X It is know that genes encoding enzymes A and C are on two different chromosomes. Compound Y Compound Z- Arginine A mutant strain of genotype a (lacking only enzyme A) is crossed to a mutant strain of genotype c (lacking only enzyme C) to generate a diploid strain. Sporulation (i.e. meiosis) is subsequently induced in the resulting diploid strain. What proportion of the spores (haploids formed by sporulation) is expected to grow on medium without arginine but supplemented with Compound Y? O 100% 50% 0% 25%arrow_forwardWild-type strains of the haploid fungus Neurospora canmake their own tryptophan. An abnormal allele td renders the fungus incapable of making its own tryptophan.An individual of genotype td grows only when its medium supplies tryptophan. The allele su assorts independently of td; its only known effect is to suppress the tdphenotype. Therefore, strains carrying both td and su donot require tryptophan for growth.a. If a td ; su strain is crossed with a genotypically wildtype strain, what genotypes are expected in the progenyand in what proportions?b. What will be the ratio of tryptophan-dependent totryptophan-independent progeny in the cross of part a?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_forwardThree haploid fungal mutants that require compound W for growth were isolated. Each mutant contains a recessive allele in a single gene. Three compounds (A, B and C) in the biosynthetic pathway to W are known, but their order in the pathway is unknown. Each compound is tested for its ability to support the growth of each of the three mutants. Phenotypes of all of the three mutants are shown in the following table (“+" indicates growth, "-" indicates no growth). A C W Mutant 1 Mutant 2 Mutant 3 What would be the phenotype of a haploid mutant that contains both mutant alleles in mutant 2 and 3? Phenotype refers to growth or absence of growth on compounds A, B, C and WN. O Like mutant 1 O Like mutant 2 Like mutant 3 O Like wild typearrow_forwardYou have identified five genes in S. cerevisiae that are induced when the yeast are grown in a high-salt (NaCl) medium. To study the potential roles of these genes in acclimation to the growth in high-salt conditions, you wish to examine the phenotypes of loss- and gain-of-function alleles of each. How will you do this?arrow_forward
- Why have we chosen a human Rab protein with no direct yeast equivalent as an “outgroup” for this study (image given)?arrow_forwardIn your laboratory, you have an F − strain of E. coli that is resistant to streptomycin and is unable to metabolize lactose, but it can metabolize glucose. Therefore, this strain can grow on media that contain glucose and streptomycin, but it cannot grow on media containing only lactose. A researcher has sent you two E. coli strains in two separate tubes. One strain, let’s call it strain A, has an F factor that carries the genes that are required for lactose metabolism. On its chromosome, it also has the genes that are required for glucose metabolism. However, it is sensitive to streptomycin. This strain can grow on media containing lactose or glucose, but it cannot grow if streptomycin is added to the media. The second strain, let’s call it strain B, is an F − strain. On its chromosome, it has thegenes that are required for lactose and glucose metabolism. StrainB is also sensitive to streptomycin. Unfortunately, when strains A and B were sent to you, the labels had fallen off the…arrow_forwardYou 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
- 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?arrow_forwardIn 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…arrow_forwardYou receive four strains of yeast in the mail, and the accompanying instructions state that each strain contains a single copy of transgene A. You grow the four strains and determine that only three strains express the protein product of transgene A. Further analysis reveals that transgene A is located at a different position in the yeast genome in each of the four strains. Provide a hypothesis to explain this result.arrow_forward
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