Four different histidine auxotrophs of Neurospora, a eukaryotic mold, are tested fol histidine and histidine precursors. The data are shown in the following table. A plus sign (+) means growth. Which conversion step does Mutant 4 catalyze? Hint: first order the dataset and write out the precursors. Compound SR T Q Tryptophan Matant 1 Mutant 2 Matant 3 + Mutant 4 Rto Q O QtoT O Tto Q O Tto tryptophan O sto T

Four different histidine auxotrophs of Neurospora, a eukaryotic mold, are tested fol histidine and histidine precursors. The data are shown in the following table. A plus sign (+) means growth. Which conversion step does Mutant 4 catalyze? Hint: first order the dataset and write out the precursors. Compound SR T Q Tryptophan Matant 1 Mutant 2 Matant 3 + Mutant 4 Rto Q O QtoT O Tto Q O Tto tryptophan O sto T

Human Heredity: Principles and Issues (MindTap Course List)

11th Edition

ISBN:9781305251052

Author:Michael Cummings

Publisher:Michael Cummings

Chapter10: From Proteins To Phenotypes

Section: Chapter Questions

Problem 10QP: b. Compounds A, B, C, and D are known to be intermediates in the pathway for production of protein...

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a. Compounds A, B, C, and D are known to be intermediates in the pathway for production of protein E. To determine where the block in protein-E production occurred in each individual, the various intermediates were given to each individuals cel Is in culture. After a few weeks of growth with the intermediate, the cells were assayed for the production of protein E. The results for each individuals cells are given in the following table. A plus sign means that protein E was produced after the cells were given the intermediate listed at the top of the column. A minus sign means that the cells still could not produce protein E even after being exposed to the intermediate at the top of the column. Draw the pathway leading to the production of protein E.
The intermediates A, B, C, D, E, and F all occur inthe same biochemical pathway. G is the product of thepathway, and mutants 1 through 7 are all G−, meaningthat they cannot produce substance G. The followingtable shows which intermediates will promote growthin each of the mutants. Arrange the intermediates inorder of their occurrence in the pathway, and indicatethe step in the pathway at which each mutant strain isblocked. A + in the table indicates that the strain willgrow if given that substance, an O means lack of growth.SupplementsMutant A B C D E F G1 + + + + + O +2 O O O O O O +3 O + + O + O +4 O + O O + O +5 + + + O + O +6 + + + + + + +7 O O O O + O +
A pure culture of an unknown bacterium was streaked onto plates of a variety of media. You notice that the colony morphologyis strikingly different on plates of minimal media with glucose compared to that seen on trypticase soy agar plates. How can you explain these differences in colony morphology? Also, describe what happens when a nonsense mutation is introduced into the gene encoding transposase within a transposon and why is it more likely that insertions or deletions will be more detrimental to a cell than point mutations?
The intermediates A, B, C, D, E, and F all occur in the same biochemical pathway G is the product of the pathway, and mutations 1 through 7 are all G –, meaning that they cannot produce substance G. The following table shows which intermediates will promote growth in each of the mutants. Arrange the intermediates in order of their occurrence in the pathway at which each mutant strain is blocked. A “+” in the table indicates that the strain will grow if given that substance, an “o” means lack of growth.
In a process of production of a recombinant protein by E. coli cells, it was observed accumulation of acetate in the culture medium. In this situation, it can be said that: (a) certainly the process in question was being conducted in anaerobiosis (B).Acetate accumulation is advantageous for the process as the acetate formation reaction generates 1 molecule of ATP (c)Knowing that decreasing the temperature of the process causes a reduction in the rate of glycolysis, this could be a strategy to reduce the accumulation of acetate (d).the acetate formed can be re-assimilated by the cell if the glyoxylate pathway is activated at some point in the culture
A number of auxotrophic mutant strains were isolated from wild-type haploid Neurospora crassa. These strains responded to the addition of certain nutritional supplements to minimal culture medium either by growth (+) or no growth (0) The data from this experiment are presented in the table below. Diagram a biochemical pathway, complete with positions of intermediates, that is consistent with the data. Indicate where in the pathway each mutant strain is blocked.
In this problem, just put the order of the intermediates on the pathway, starting with P and ending with Z and explain your reasoning. (So to get you started, notice the class 1 mutants—and there is only mutant in this class, which is mutant #5—won’t grow on minimal medium, but will grow if you give it substance Z. However, giving it substances, p, w, x, or y won’t help. This means that the gene that is mutated lies on the pathway in a place that the Z substances is to its right and all the other substances are to its left. The easiest next one to look at is the line with 2 pluses, and then the line with 3 pluses, and then the line with 4 pluses).
Compounds A, B, C, and D are known to be intermediates in the pathway for production of protein E. To determine where the block in protein-E production occurred in each individual, the various intermediates were given to each individual’s cell in culture. After a few weeks of growth with the intermediate, the cells were assayed for the production of protein E. The results for each individual’s cells are given in the following table. A plus sign means that the protein E was produced after the cells were given the intermediate listed at the top of the column. A minus sign means that the cells still could not produce protein E even after being exposed to the intermediate at the top of the column. a) If an individual who is homozygous for the mutation found in individual 2 and heterozygous for the mutation found in individual 4 mates with an individual who is homozygous for the mutation found in individual 4 and heterozygous for the mutation found in individual 2, what could the…
Give only typing answer with explanation and conclusion For the genotype shown below, which best describes the expression of the B-galactosidase gene. I+ Oc Z+ / F’ Is Constitutive Repressed Inducible None of the above
For 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.
. Seven E. coli mutants were isolated. The activity ofthe enzyme β-galactosidase produced by cells containing each mutation alone or in combination with othermutations was measured when the cells were grown inmedium with different carbon sources.Lactose +Glycerol Lactose GlucoseWild type 0 1000 10Mutant 1 0 10 10Mutant 2 0 10 10Mutant 3 0 0 0Mutant 4 0 0 0Mutant 5 1000 1000 10Mutant 6 1000 1000 10Mutant 7 0 1000 10F′ lac from mutant 0 1000 101/ mutant 3F′ lac from mutant 0 10 102/ mutant 3Mutants 3 + 7 0 1000 10Mutants 4 + 7 0 0 0Mutants 5 + 7 0 1000 10Mutants 6 + 7 1000 1000 10Assume that each of the seven mutations is one andonly one of the genetic lesions in the following list.Identify the type of alteration each mutation represents.a. superrepressorb. operator deletionc. nonsense (amber) suppressor tRNA gene (assumethat the suppressor tRNA is 100% efficient in suppressing amber mutations)d. defective CRP–cAMP binding sitee. nonsense (amber) mutation in the β-galactosidase genef.…
Given the following genotypes, explain, by answering the questions in each number, how the mutation (identified by a (-) superscript) will affect E. coli grown in lactose medium. Will there be a complete set ofgene products? (Yes/No) Will the lac operon be turnedon/off? Will the cell survive? (Yes/No) a. i + p + o + z - y + b. i + p - o + z + y + c. i + p + o - z + y +