In the "biochemical assay of b-galactosidase activity", what is o-nitrophenyl-b-D-galactosidase (ONPG)used for?O When ONPG is cleaved by b-galactosidase, we are able to assay b-galactosidase activity.O It provides essential nutrients for bacterial growth.It partially disrupts the cell membrane to allow cellular proteins to diffuse out of the cell.O ONPG cleaves the b-galactosidase that is made by the lac operon so we can see how much activity there isin the cell.ONPG cleaves galactose so we can measure how much lactose there is in the cell.

Introduction The lac operon model was first proposed by F. Jacob and J. Monod. Lac operon an…

Answered: In the "biochemical assay of…

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 9QP: a. Compounds A, B, C, and D are known to be intermediates in the pathway for production of protein...

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Measure the uptake of leucine by epithetial cells of the mouse intestine. Measurements of the rate of update of L-leucine, D-Leucine, and L-valine , with and without Na+ in the assay were perform and yield different results (see table below). A) What can you conclude about the properties and mechanism of leucine transporter? B) Would you expect L-leucine uptake to be inhibited by Ouabain, which is a cardiac glycoside drug treatment?
GTP-y-s serves as an analog of GTP that cannot be hydrolyzed any further. How would a Co-IP experiment differ between G- alpha & G-beta proteins in the presence of GTP-y-a and GDP?
CTP synthetase catalyzes the glutamine-dependent conversion of UTP to CTP. The enzyme is allosterically inhibited by the product, CTP. Mammalian cells defective in this allosteric inhibition are found to have a complex phenotype: They require thymidine in the growth medium, they have unbalanced nucleotide pools, and they have an elevated spontaneous mutation rate. Explain the likely basis for these observations.
. Mutants of Neurospora crassa that lack carbamoyl phosphate syn- thetase I (CPS I) require arginine in the medium in order to grow, whereas mutants that lack carbamoyl-phosphate synthetase II (CPS II) require a pyrimidine, such as uracil. A priori, one would expect the active CPS II in the arginine mutants to provide sufficient carbamoyl phosphate for arginine synthesis, and the active CPS I in the pyrimidine mutants to "feed" the pyrimidine pathway. Explain these observations.
A Leu →Ala mutation at a site buried in the core of the enzyme lysozymeis found to be destabilizing. Explain the observed effect of this mutationon lysozyme stability by predicting how enthalpy (ΔH°), conformationalentropy (ΔS°peptide), and the hydrophobic effect (ΔS°solvent) are expected to change for the mutant compared to wild-type lysozyme. Explain how ΔG°for unfolding is affected by your predicted changes in enthalpy or entropy.
A classic way to isolate thymidylate synthase–negative mutants of bacteriais to treat a growing culture with thymidine and trimethoprim. Most ofthe cells are killed, and the survivors are greatly enriched in thymidylatesynthase–negative mutants.(a) What phenotype would allow you to identify these mutants?(b) What is the biochemical rationale for the selection? (That is, why are themutants not killed under these conditions?)(c) How would the procedure need to be modified to select mammalian cellmutants defective in thymidylate synthase?
Mutations in the Galactose-1-Phosphate Uridyl Transferase Gene Can you use RFLP to identify this mutation?
Why do E. coli cells with a defective lacZ gene fail to show galactoside permease activity after the addition of lactose in the absence of glucose?
b. 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. Denote the point in the pathway in which each individual is blocked.
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.
Wild-type E. coli cells can synthesize all 20 common amino acids, but some mutants, called amino acid auxotrophs, are unable to synthesize a specific amino acid and require its addition to the culture medium for optimal growth. Besides their role in protein synthesis, some amino acids are also precursors for other nitrogenous cell products. Consider the three amino acid auxotrophs that are unable to synthesize glycine, glutamine, and aspartate, respectively. Match each nitrogenous product with the mutant or mutants that would fail to synthesize it. Each mutant may fail to synthesize more than one nitrogenous product, and some nitrogenous products may fail to be synthesized by more than one mutant. glycine auxotrophs glutamine auxotrophs aspartate auxotrophs Answer Bank adenine nucleotides cytosine nucleotides guanine nucleotides uridine nucleotides
Consider the following simple regulatory pathways. Assume the full pathway is shown. A- E- B- F- C- G- D- 1 A H- 2 B || L You identify several null mutations (a complete deletion of the gene). For each mutant (indicated with a - sign), determine whether the final product (I, J, K or L) is inducible, uninducible, or constitutive. 3 C 4 D- [Choose ] [Choose ] [Choose ] [Choose ] [Choose ] [Choose ] [Choose ] E [Choose ] F G I H || J K

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