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Genetic Analysis: An Integrated Approach (2nd Edition)
2nd Edition
ISBN: 9780321948908
Author: Mark F. Sanders, John L. Bowman
Publisher: PEARSON
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Textbook Question
Chapter 6, Problem 16P
Suppose you have an
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Given the following complementation chart for holes in Monstera, give me the biochemical (phenotype) pathway.
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When human hemoglobin undergoes a mutation, the
mutant protein usually does not replace all of the normal HbA in
the red blood cells or erythrocytes of the individual. The erythro-
cytes contain mixtures of varying amounts of both HbA and the
mutant protein depending on the mutation and the individual. Hb
Yakima is a mutant human Hb with an Asp-(B99)His mutation.
The diagram on the right shows that Hb Yakima was separated
by DEAE-cellulose chromatography from HbA with a 0 – 0.1 M
linear gradient of NaCl buffered to pH 8.3. Why is chromatog-
raphy carried out at pH 8.3? If the isoelectric point of HbA is 6.85,
what is the change in total charge caused by the mutation?How
does the change in charge explain the chromatography elution
profile of the Hb Yakima/HbA mixture?
1,5
-Hb-A
Hb -Yakima
1.0
0.5-
20
40
60
80
00
Fraction number
O.D578 nm
For each of the E. coli strains containing the lacoperon alleles listed, indicate whether the strain isinducible, constitutive, or unable to expressβ-galactosidase and permease.a. I+ o+ Z− Y+/ I+ ocZ+ Y+b. I+ o+ Z+ Y+/ I− ocZ+ Y−c. I+ o+ Z− Y+/ I− ocZ+ Y−d. I−P− o+ Z+ Y−/ I+ P+ ocZ− Y+e. Iso+ Z+ Y+/ I− o+ Z+ Y−
Chapter 6 Solutions
Genetic Analysis: An Integrated Approach (2nd Edition)
Ch. 6 - For bacteria that are F+, Hfr, F', and F-, perform...Ch. 6 - The flow diagram identifies relationships between...Ch. 6 - Conjugation between an Hfr cell and an F-cell does...Ch. 6 - Bacteria transfer genes by conjugation,...Ch. 6 - Explain the importance of the following features...Ch. 6 - Prob. 6PCh. 6 - Describe what is meant by the term site-specific...Ch. 6 - What is a prophage, and how is a prophage formed?Ch. 6 - How is the frequency of cotransduction related to...Ch. 6 - Describe the differences between genetic...
Ch. 6 - Among the mechanisms of gene transfer in bacteria,...Ch. 6 - What is lateral gene transfer? How might it take...Ch. 6 - Lateral gene transfer is thought to have played a...Ch. 6 - Prob. 14PCh. 6 - Prob. 15PCh. 6 - 16. Suppose you have an lysis mutant that maps to...Ch. 6 - Five Hfr strains from the same bacterial species...Ch. 6 - An interrupted mating study is carried out on Hfr...Ch. 6 - An Hfr strain with the genotype cys+leu+met+strS...Ch. 6 - A triple-auxotrophic strain of E. coli having the...Ch. 6 - Penicillin was first used in the 1940 s to treat...Ch. 6 - An attribute of growth behavior of eight...Ch. 6 - Synthesis of the amino acid histidine is a...Ch. 6 - The phage P1 is used as a generalized transducing...Ch. 6 - Prob. 25PCh. 6 - Five rII partial-deletion mutants are mapped and...Ch. 6 - A 2013 CDC report identified the practice of...Ch. 6 - Hfr strains that differ in integrated F factor...
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- Genomic DNA from a family where sickle-cell disease is known to be hereditary, is digested with the restriction enzyme MstII and run in a Southern Blot. The blot is hybridised with two different 0.6 kb probes, both probes (indicated in red in the diagram below) are specific for the β-globin gene (indicated as grey arrow on the diagram below). The normal wild-type βA allele contains an MstII restriction site indicated with the asterisk (*) in the diagram below; in the mutated sickle-cell βS allele this restriction site has been lost. What size bands would you expect to see on the Southern blots using probe 1 and probe 2 for an individual with sickle cell disease (have 2 βS alleles)? Probe 1 Probe 2 (a) 0.6kb 0.6kb and 1.2kb (b) 0.6kb and 1.8kb 0.6kb, 1.2kb and 1.8kb (c) 1.2kb 0.6kb (d) 1.8kb 1.8kb a. (a) b. (b) c. (c) d. (d)arrow_forwardFor E. coli strains with the lac genotypes show below, use a plus sign (+) to indicate the synthesis of β-galactosidase and permease and a minus sign (–) to indicate no synthesis of the proteins.arrow_forwardThe cellulase gene of Bacillus licheniformis was successfully cloned into the pET21a vector and expressed in Escherichia coli BL21. The pET21a vector consists of ampicillin resistant gene. To screen for the successful transformants, E. coli BL21 was cultivated on LB agar containing ampicillin (100 pg/mL) and 0.5% (w/v) carboxymethylcellulose, and incubated at 37 0C for 6 hours. After that the agar plate was stained with Congo red solution for 15 minutes and washed twice with sodium chloride solution and the observation is as shown in Figure 3. Answer the following: (i.) Briefly explain why ampicillin was added to the LB agar. (ii) Indicate the function of carboxymethylcellulose in the LB agar. (iii) Conclude how the researchers were able to identify the E. coli BL21 that carried the cellulase gene.arrow_forward
- In E. coli, four Hfr strains donate the following markers,shown in the order donated:Strain 1: M Z X W CStrain 2: L A N C WStrain 3: A L B R UStrain 4: Z M U R BAll these Hfr strains are derived from the same F+ strain.What is the order of these markers on the circularchromosome of the original F+?arrow_forwardYou are interested in studying resistance to heavy metals and have selected the yeast Saccharomyces cerevisea to conduct your studies. You have recovered a deletion mutant that does not tolerate high concentrations of zinc (grows poorly in zinc containing media ) and have designated the mutant pgz-1 (for poor growth in zinc ). (a) What is the advantage to the type of mutant used in this work? What class of mutagen was likely use to generate pgz-1? ( b) Do you expect the PGZ gene to be expressed in your mutant? Explain.arrow_forwardSchizosaccharomyces 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_forward
- Cap, EA1, and Sap are all genes/proteins of interest in this study. For each gene, what gene product is encoded and where is the gene (the literal DNA sequence) located physically in the cell? I need help fimiding this in the artticle and answer as short as possible https://www.ncbi.nlm.nih.gov/pmc/articles/PMC106848/arrow_forwardCTP synthetase catalyzes the glutamine-dependent conversion of UTP to CTP. The enzyme is allosterically inhibited by the product, CTP. Mamma- lian cells defective in this allosteric inhibition are found to have a complex phenotype: They require thymidine in the growth medium, they have unbal- anced nucleotide pools, and they have an elevated spontaneous mutation rate. Explain the likely basis for these observations.arrow_forwardA fluctuation test was carried out to determine the rate of mutation to Azetidine resistance (a toxic proline analog) in S. typhimurium. Twenty tubes of rich medium were each inoculated with a few wild-type cells and the cultures were grown to 10° cells / ml. A 0.1 ml sample of each culture was then plated on each plate (a total of 20 plates) to detect AztR mutants. The results are shown in the following table. Calculate the mutation rate of S. typhimurium to AztR. Culture # # AztR mutants Culture # # AztR mutants 11 12 3 4. 13 14 15 4. 30 303 97 69 14 16 17 18 19 20 10 19arrow_forward
- Let’s suppose you make a transposon library of the cellulose-secreting bacterium Komagataeibacter xylinus, with the goal of finding mutants that produce higher than normal amounts of cellulose, which would be useful industrially. However, despite your best efforts you are unable to isolate any transposon mutants that make more cellulose than the wild-type strain.Why might this have failed? List as many reasons as you can think of.arrow_forwardFor the lac genotypes of Escherichia coli shown in the following Table 1, predict the expression of beta-galactosidase (Z) and permease (Y) is inducible or noninducible or constitutive. Explain your answer. Table 1 Genotype I-P+O+Z+Y+ (i) (ii) I+P+OCZ+Y+ (iii) ISP+O+Z+Y+ (iv) I+P+O+Z+Y-//I+P-O+Z+Y+ (v) ISP+OcZ+Y+//I-P+O+Z+Y- Condition No lactose No lactose lactose lactose No lactosearrow_forwardAfter mutagenesis of wild type Vibrio fisheri, you isolate two different mutant strains (A and B) that, unlike the wild type cells, fail to luminesce when grown to high density in a flask with appropriate medium. Curiously, however, when you inoculate both mutant strains in the same flask, you observe that the mixed (A+B) culture begins to emit light after growing dense. a) What gene/functions are likely affected in each of the two mutants? b) How does this explain their phenotypes?arrow_forward
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