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 15, Problem 22P
You have identified five genes in S. cerevisiae that are induced when the yeast is grown in high salt (NaCl) medium. To study the potential roles of these genes in acclimation to growth in high
a. How will you do this?
b. How would your answer differ if you were working with tomato plants instead of yeast?
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#22) You 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?
Three 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 type
In Caulobacter crescentus development, several proteins are directly involved in generating cell polarity which includes PleC, DivJ, and DivK. These proteins were discovered through genetics, but they are essential genes; which means, cells cannot survive long-term in their absence. However, temperature-sensitive and cold-sensitive mutants have enabled researchers to observe the consequences of the absence of these proteins before the cells die. For each one, answer the following questions: A) What are the consequences (phenotypes) when the function of the protein is lost? Generally these will involve, but are not limited to, positioning of the flagella, pili, and stalk. B) Based on the mechanism of development of cell polarity involving phosphorylation and dephosphorylation of DivK, how do these phenotypes arise?
Chapter 15 Solutions
Genetic Analysis: An Integrated Approach (3rd Edition)
Ch. 15 - 15.1 What purpose do the bla and lacZ genes serve...Ch. 15 - The human genome is 3109 bp in length. How many...Ch. 15 - 15.3 Ligase catalyzes a reaction between the...Ch. 15 - You have constructed four different libraries: a...Ch. 15 - Using the genomic libraries in Problem 4, you wish...Ch. 15 - The human genome is 3109bp. You wish to design a...Ch. 15 - 15.7 Using animal models of human diseases can...Ch. 15 - 15.8 Compare methods for constructing homologous...Ch. 15 - 15.9 Chimeric genefusion products can be used for...Ch. 15 - 15.10 Why are diseases of the blood simpler...
Ch. 15 - Injection of double-stranded RNA can lead to gene...Ch. 15 - Compare and contrast methods for making transgenic...Ch. 15 - 15.13 It is often desirable to insert cDNAs into a...Ch. 15 - 15.14 A major advance in the s was the development...Ch. 15 - 15.15 The bacteriophage lambda genome can exist in...Ch. 15 - 15.16 The restriction enzymes Xho and Sal cut...Ch. 15 - 15.17 The bacteriophage has a single-stranded DNA...Ch. 15 - 15.18 To further analyze the CRABS CLAW gene (see...Ch. 15 - You have isolated a genomic clone with an EcoR I...Ch. 15 - 15.20 You have identified a cDNA clone that...Ch. 15 - 15.21 You have isolated another cDNA clone of the...Ch. 15 - 15.22 You have identified five genes in S....Ch. 15 - You have generated three transgenic lines of maize...Ch. 15 - 15.24 Bacterial Pseudomonas species often possess...Ch. 15 - 15.25 Two complaints about some transgenic plants...Ch. 15 - 15.26 In Drosophila, lossoffunction Ultrabithorax...Ch. 15 - Prob. 27PCh. 15 - The highlighted sequence shown below is the one...Ch. 15 - Vitamin E is the name for a set of chemically...Ch. 15 - The RAS gene encodes a signaling protein that...Ch. 15 - 15.31 You have cloned a gene for an enzyme that...Ch. 15 - 15.32 About of occurrences of nonautoimmune type...Ch. 15 - Describe how having the Cas 9 gene at a genomic...Ch. 15 - 15.34 Would a gene drive system spread rapidly...
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- Schizosaccharomyces pombe, also known as "fission yeast," is a powerful model organism in molecular and cell biology. While performing a genetic screen, you discover an auxotrophic S. pombe strain that is unable to synthesize one or more vitamins. The following table represents the key experiments you performed during your genetic screen. Fill in the table with the outcome of each experiment for your mutant strain (using + for growth and - for no growth). Medium Rich media Minimal media Minimal media + all vitamins Minimal media + all amino acids Growth Wild-type + + + + Mutant + + + > > >arrow_forwardConcerning the Tools of Genetics Box Analysis ofCell-Cycle Mutants in Yeast:a. Describe how you would use replica plating ofmutagenized, haploid yeast cells to identifytemperature-sensitive (ts) mutations in essentialgenes needed for yeast growth and survival.b. Among the many ts mutations you found in part(a), how would you distinguish mutations in genesneeded for cell-cycle progression from those ingenes needed for other aspects of the life of yeasts?c. If you had a large collection of yeast cell-cyclemutants, how would you determine which of themutations are in the same gene and which are indifferent genes?arrow_forwardTo identify genes controlling the cell cycle in budding yeast, a genetic screen was carried out. In this screen, haploid yeast cells were exposed to a DNA damaging agent to introduce random mutations in the genome. By culturing cells at an elevated temperature (e.g. 37 degrees), where many mutated genes lose their function, scientists identified yeast mutants that showed growth defects and arrest at specific stages of the cell cycle (e.g. in mitosis with large buds). In this screen, mutants of the cyclin-dependent kinase were identified, but not mutants of cyclins. Explain the reason for this outcome.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_forwardThe Neurospora octad shown came from a cross between mt and m' strains. a. Is this an MI or an MIl octad or neither? Explain. b. Diagram the production of this octad. c. Is it possible to observe evidence of heteroduplex formation in a Neurospora ascus even if gene conversion did not occur during formation of the octad? Explain.arrow_forwardThe p53 protein was discovered through its association with SV40 T antigen and assumed initially to be an oncoprotein. a. What is the current consensus as the function of p53 and what evidence caused this change in view? b. How does the effect of mutation in the p53 gene differ from the effect of mutation in the RB gene what is the molecular basis for this difference ?arrow_forward
- Yeast cells are eukaryotes, and they require a functional secretory pathway to grow and to maintain cellular organization. But surprisingly, when SRP is removed from yeast cells by deleting the relevant genes, the cells are still alive (although they grow slowly). a) How can yeast survive without SRP? Propose two alternative hypotheses. b) How might electron microscopy of normal and SRP-deficient yeast cells help you to distinguish between these two hypotheses?arrow_forwardThe synthesis of flower pigments is known to be dependent on enzymatically controlled biosynthetic pathways. For the crosses shown here, postulate the role of mutant genes and their products in producing the observed phenotypes: (a) P1: white strain A * white strain B F1: all purple F2: 9/16 purple: 7/16 white (b) P1: white * pink F1: all purple F2: 9/16 purple: 3/16 pink: 4/16 whitearrow_forwardYou have isolated 8 mutants in yeast that fail to grow on minimal media plates but do grow when they are supplemented with Arginine. You know that Arginine is synthesized in a biochemical pathway within wild-type yeast, but you do not know how many gene products it takes for the pathway. You have all of the lines as both a and a cells and mate each strain to each other in pairwise crosses and plate them on minimal media to see if they grow. You obtain the following results with (+) representing growth, and (-) indicating no growth: a 1 5 1 a 4 5 6 7 8 How many genes are represented? O 1 3 7 O Cannot tell from the data a + + + + + • + + i 0 +, + + + • + + 7 + + + + + , . + + + + + m + + + + + + + 2 + + + + + i + -I + + . . + + +arrow_forward
- Assume that there is a double stranded break on DNA double helix of an eukaryotic cell due to X-ray radiation and it is not repaired. In addition, the cell’s Apaf-1 protein is not expressed due to a null mutation in the Apaf-1 gene. Please discuss the effect of not having Apaf-1 expression in the cell with non-repaired double stranded break.arrow_forwardDiscuss how ultra violet light works as a mutagen. This could include: What is UV light and the Mutations commonly introduced by UV light? What are the Repair mechanisms in yeast that fix damage caused by UV light? Describe the phenotypes of Saccharomyces Cerevisiae plates, the first plate is the control yeast, while the second plate has been exposed to low UV light, the third plate has been exposed to high UV light. Normal yeast has round smooth white colonies. Are there any signinifcant differences?arrow_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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