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Concept explainers
To analyze:
To detect interactions between proteins using the two-hybrid system, the following results are obtained:
A clone coding gene A gave positive results with clones B and Cwhile clone B gave positive results with clones A, D, and E but not C; and clone E gave positive results only with clone B.
Alternative clone F gave positive results with clone G but not with any of A–E. Can you explain these results?
Introduction:
The two-hybrid system is a method to determine the interaction between two proteins. The system is based on the modular nature of the transcription factor Gal4 from yeast. The two-hybrid systemactivates the downstream reporter gene. It is doneby binding of transcription factor onto the upstream activating sequence (UAS).
For Two-Hybrid system, the transcription factor divides into two parts i.e. DNA binding domains (DBD or BD) and activating domain (AD). Two fusion proteins are prepared –
Both BD and AD have different functions. The domain AD and BD is responsible for the activation of transcriptionand binding to the UASrespectively.
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Chapter 18 Solutions
Genetic Analysis: An Integrated Approach (2nd Edition)
- Determine the coefficient of coincidence and the interferencearrow_forwardPrice et al. [(1999). J. Bacteriol. 181:2358–2362] conducteda genetic study of the toxin transport protein (PA) of Bacillusanthracis, the bacterium that causes anthrax in humans. Withinthe 2294-nucleotide gene in 26 strains they identified five pointmutations—two missense and three synonyms—among differentisolates. Necropsy samples from an anthrax outbreak in 1979revealed a novel missense mutation and five unique nucleotidechanges among ten victims. The authors concluded that thesedata indicate little or no horizontal transfer between differentB. anthracis strains. Question: On what basis did the authors conclude that evidence ofhorizontal transfer is absent from their data?arrow_forwardPrice et al. [(1999). J. Bacteriol. 181:2358–2362] conducteda genetic study of the toxin transport protein (PA) of Bacillusanthracis, the bacterium that causes anthrax in humans. Withinthe 2294-nucleotide gene in 26 strains they identified five pointmutations—two missense and three synonyms—among differentisolates. Necropsy samples from an anthrax outbreak in 1979revealed a novel missense mutation and five unique nucleotidechanges among ten victims. The authors concluded that thesedata indicate little or no horizontal transfer between differentB. anthracis strains. Question: What is meant by ”horizontal transfer”?arrow_forward
- Price et al. [(1999). J. Bacteriol. 181:2358–2362] conducteda genetic study of the toxin transport protein (PA) of Bacillusanthracis, the bacterium that causes anthrax in humans. Withinthe 2294-nucleotide gene in 26 strains they identified five pointmutations—two missense and three synonyms—among differentisolates. Necropsy samples from an anthrax outbreak in 1979revealed a novel missense mutation and five unique nucleotidechanges among ten victims. The authors concluded that thesedata indicate little or no horizontal transfer between differentB. anthracis strains. Question: Which types of nucleotide changes (missense or synonyms)cause amino acid changes?arrow_forwardBy conducting conjugation experiments between Hfr and recipientstrains, Wollman and Jacob mapped the order of many bacterialgenes. Throughout the course of their studies, they identified severaldifferent Hfr strains in which the F-factor DNA had been integratedat different places along the bacterial chromosome. A sample of theirexperimental results is shown in the table:Draw a map that shows the order of genes and the locations ofthe origins of transfer among these different Hfr strains?arrow_forwardIn the Western blot shown here, proteins were isolated from redblood cells and muscle cells from two different individuals. Oneindividual was unaffected, and the other suffered from a diseaseknown as thalassemia, which involves a defect in hemoglobin. Theblot was exposed to an antibody that recognizes β globin, whichis one of the polypeptides that constitute hemoglobin. Equal totalamounts of cellular proteins were added to each lane. Explain these results.arrow_forward
- Suppose you wanted to study genes controlling thestructure of bacterial cell surfaces. You decide tostart by isolating bacterial mutants resistant to infection by a bacteriophage that binds to the cellsurface. The selection procedure is simple: Spreadcells from a culture of sensitive bacteria on a petriplate, expose them to a high concentration ofphages, and pick the bacterial colonies that grow.To set up the selection you could (1) spread cellsfrom a single liquid culture of sensitive bacteria onmany different plates and pick every resistant colony; or (2) start many different cultures, eachgrown from a single colony of sensitive bacteria,spread one plate from each culture, and then pick asingle mutant from each plate. Which methodwould ensure that you are isolating many independent mutations?arrow_forwardStreptococcus pneumoniae cells of genotype str s mtl - are transformed by donor DNA of genotype strr mtl+ and (in a separate experiment) by a mixture of two DNAs with genotypes strr mtl - and str s mtl+. The accompanying table shows the results.a. What does the first row of the table tell you? Why? b. What does the second row of the table tell you? Why?arrow_forwardBelow is an EMSA showing four different reactions, A-D. In each tube there is some combination of labelled DNA probe, Protein X (the protein you are studying), and an antibody for Protein X. Identify which combination of components are found in each of the four reactions and explain how you determined that based on the molecular interactions being studied and your knowledge of gel electrophoresis. It is possible that multiple lanes have the same component(s). A B C D EMSAarrow_forward
- One issue with interrupted-mating experimentssuch as that in Problem 19 is that gene order may beambiguous if the genes are close together. Anothershortcoming is that such experiments do not provide accurate map distances. The reason is that researchers select for the first Hfr marker transferred intothe recipient, but the recovery of F− exconjugantswith a later Hfr marker is complex, depending bothon transfer of the marker into the cell and on crossovers that transfer the marker into the recipientchromosome.To make more accurate maps, bacterial geneticistsoften do Hfr × F− crosses in a different way: Theyselect for exconjugants that contain a late Hfr marker,and then screen for the presence of the earlier markers.This method ensures that all of the markers haveentered the F− cell, so relative gene distances arenow accounted for solely by crossover frequencies.Furthermore, gene order is clarified by considering thecrossovers responsible for each class of exconjugants.As an example,…arrow_forwardIn a transformation experiment, donor DNA was obtained from a prototroph bacterial strain (x* y z'), and the recipient was a triple auxotroph (x y z ). The following transformant classes were obtained: x* y z xy'z 160 130 190 170 x'y 3 xy x'y 5 What general conclusions can you draw about the linkage relationships among the three genes?arrow_forwardAn Hfr strain that is hisE+ and pheA+ was mixed with a strain thatis hisE− and pheA−. The conjugation was interrupted and the percentageof recombinants for each gene was determined by streakingon a medium that lacked either histidine or phenylalanine. Thefollowing results were obtained:Determine the map distance (in minutes) between these twogenes.arrow_forward
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