Concept explainers
A classic way to isolate thymidylate synthase—negative mutants of bacteria is to treat a growing culture with thymidine and trimethoprim. Most of the cells are killed, and the survivors are greatly enriched in thymidylate synthase—negative mutants.
a. What
b. What is the biochemical rationale for the selection? (That is, why are the mutants not killed under these conditions?)
c. How would the procedure need to be modified to select mammalian cell mutants defective in thymidylate synthase?
Trending nowThis is a popular solution!
Learn your wayIncludes step-by-step video
Chapter 19 Solutions
Biochemistry: Concepts and Connections
Additional Science Textbook Solutions
Fundamentals of General, Organic, and Biological Chemistry (8th Edition)
General Chemistry: Principles and Modern Applications (11th Edition)
Brock Biology of Microorganisms (14th Edition)
Campbell Biology: Concepts & Connections (9th Edition)
Chemistry: A Molecular Approach (4th Edition)
Microbiology with Diseases by Taxonomy (5th Edition)
- You have an extract of BL21(DE3) cells containing a soluble, 75 kDa recombinant protein that has a histidine tail at its C-terminus. This extract is still in raw form and you need to purify this protein without losing its solubility. Check the option below that represents the best purification methodology in the order of execution. * a) (1) isoelectric precipitation; (2) microfiltration; (3) affinity chromatography; (4) anion exchange chromatography b) (1) centrifugation; (2) microfiltration; (3) isoelectric precipitation; (4) affinity chromatography c) (1) centrifugation; (2) microfiltration; (3) affinity chromatography; (4) ultrafiltration d) (1) microfiltration; (2) jumping out; (3) centrifugation; (4) affinity chromatographyarrow_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_forwardConsider the biochemical pathway shown here. Suppose that a strain of bacteria must synthesize compound 4 to survive and divide. Successful survival and division of bacteria is observed as growth of colonies on an agar plate. This strain of bacteria can grow colonies on minimal medium as long as it is supplemented with compound 1. You are in a lab that has isolated several mutants of this strain. You find that these mutants cannot grow on minimal medium supplemented with compound 1, though they can grow colonies if supplemented with compound 4. Considering what you know about the Beadle-Tatum experiments, which of the following statements would be one that should be true?arrow_forward
- You generate mutants in the metabolic pathway for starlase. You conduct some complementation tests (after testing for dominance of course) and come up with the following results: 1 2 4 6 1 + + + 2 + 3 + 4 + 5 6 a. How many complementation groups are there? [Select] b. You conduct some additional experiments to elucidate the starlase metabolic pathway. Your results are shown below. Use this information alongside information from the complementation table above to place the intermediates in the correct order on the pathway. (HINT: use the complementation groups from the table above to help you consolidate information on the tables below. Reference practice question 3 from today's lecture for help). Addition to minimal medium Mutant None starlase P 1 + + 2 + + 4 + + 5 + 6 Precursor --> [ Select ] [ Select ] [ Select ] --> starlase c. Mutant 4 has a loss-of-function mutation for which enzyme in the starlase synthesis pathway? [ Select ] E1 E2 ЕЗ Е4 Precursor > Intermediate 1→ Intermediate…arrow_forwardThe bacteria that you are growing to express biochemisfunase are resistant to ampicillin, so you keep ampicillin in your media to prevent contamination from other bacterial strains. You find a tube of ampicillin in the lab freezer that says “2000x AMP” on the lid, which indicates that it is 2000 times more concentrated than what you need (i.e., a concentration of 1x ampicillin in your media). How much of the 2000x ampicillin stock would you add to your 1.0-L culture to get a final concentration of 1x ampicillin?arrow_forwardAssume that a series of compounds has been discovered in Neurospora. Compounds A–F appear to be intermediates in a biochemical pathway. Conversion of one intermediate to the next is controlled by enzymes that are encoded by genes. Several mutations in these genes have been identified and Neurospora strains 1–4 each contain a single mutation. Strains 1–4 are grown on minimal media supplemented with one of the compounds A–F. The ability of each strain to grow when supplemented with different compounds is shown in the table (+ = growth; o = no growth). Which biochemical pathway fits the data presented? Media Supplement Strain A B C D E F 1 o o o + + + 2 o o o o + + 3 o o o o + o 4 o o + + + + A) A → B → C → D → E → F B) A → B → C → F → D → E C) F → B → C → D → A → E D) A → B → C → D → F → E E) A → B → F → E → C → Darrow_forward
- (b) Both laboratories used 10 micrograms of protein each in their kinetic assays. Protein concentrations weredetermined by the Bradford protein assay. Assay conditions employed in the two labs (pH, temperature,etc.) were also identical. What would be the most plausible cause for the discrepancy in the Vmax valuesfor the compound I? Explain.Recall that the Bradford assay measures total protein amounts in sample solution based on complexformation between a dye and proteins. Also, the assay solution used in both labs does not contain anyinhibitors.arrow_forwardSelect all that apply from the following to each transformation process : The following are White non-glowing colonies White glowing colonies Blue non-glowing colonies Blue glowing colonies None of the above The transformation are: a. LB + kanamycin plates that have been spread with the E. coli cells transformed with your ligations, what phenotype of the colonies do you expect to obtain . b. LB + kanamycin plates that have been spread with the E. coli cells transformed with water, what phenotype of the colonies do you expect to obtain. c. LB plates that have been spread with the E. coli cells transformed with your pHSG298, what phenotype of the colonies do you expect to obtain. d. LB + kanamycin plates that have been spread with the E. coli cells transformed with PHSG298, what phenotype of the colonies do you expect to obtain. e. LB plates that have been spread with the E. coli cells transformed with your ligations, what phenotype of the colonies do you expect to obtain. f. LB plates…arrow_forward. 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.arrow_forward
- What is the purpose of Southern's blotting technique? Explain in detail the biochemical principle that underpin each step of the method.arrow_forwardThe objective is to study a novel protease P isolated from the digestive tract of an Amazonian insect. This protease can exist into two forms Pi and Pa which have identical amino acid sequences (both of 80 kDa). However, only Pa shows proteolytic activity. To better understand the activation mode of Pi (inactive form) in Pa (active form), the following experiment was done using DIPF. DIPF (diisopropylphosphofluoridate) is a well-known irreversible inhibitor of serine proteases. It reacts with the catalytic serine residue of the active site of proteases as shown below: Enzyme -CH₂OH + CH(CH3)2 O F-P=0 O CH(CH3)2 Diisopropylphospho- fluoridate (DIPF) Enzyme -CH,—O CH(CH3)2 O <=0 O CH(CH3)2 DIP-Enzyme Both proteases Pa and P₁ were incubated with 32P-DIPF for 30 min at 37°C, and then dialysed to remove excess of unreacted radiolabelled reagent. The two proteases were then analyzed in Sodium Dodecyl Sulphate-Polyacrylamide Gel Electrophoresis (SDS-PAGE), with and without 2-mercaptoethanol.…arrow_forwardFor each of the following genotypes, explain how mutation (identified by a (-) will affect the organism grown in lactose medium. Indicate whether a) B-galactosidase will be synthesized or not, b) synthesis of B-galactosidase is inducible (1) or constitutive (C) and c) growth of the organism will occur or not. a. I-P+O+Z+Y+A+ b. I+P-O+Z+Y+A+ c. I+P+O-Z+Y+A+ d. I+P+0-Z+Y+A+/l+P+O+Z-Y-A- e. I-P+0+Z+Y+A+/l+P-O+Z+Y+A+arrow_forward
- BiochemistryBiochemistryISBN:9781319114671Author:Lubert Stryer, Jeremy M. Berg, John L. Tymoczko, Gregory J. Gatto Jr.Publisher:W. H. FreemanLehninger Principles of BiochemistryBiochemistryISBN:9781464126116Author:David L. Nelson, Michael M. CoxPublisher:W. H. FreemanFundamentals of Biochemistry: Life at the Molecul...BiochemistryISBN:9781118918401Author:Donald Voet, Judith G. Voet, Charlotte W. PrattPublisher:WILEY
- BiochemistryBiochemistryISBN:9781305961135Author:Mary K. Campbell, Shawn O. Farrell, Owen M. McDougalPublisher:Cengage LearningBiochemistryBiochemistryISBN:9781305577206Author:Reginald H. Garrett, Charles M. GrishamPublisher:Cengage LearningFundamentals of General, Organic, and Biological ...BiochemistryISBN:9780134015187Author:John E. McMurry, David S. Ballantine, Carl A. Hoeger, Virginia E. PetersonPublisher:PEARSON