Quiz 5 Question Pool Fall 2023

docx

School

Temple University *

*We aren’t endorsed by this school

Course

3096

Subject

Biology

Date

Feb 20, 2024

Type

docx

Pages

2

Uploaded by ChefRamPerson3860

Report
Quiz Question Pool 5 1) a. What is the charge on an acidic amino acid at pH 7? (1 point) -1 like glutamic acid and aspartic acid because R group loses a proton to solvent b. What is the charge on a basic amino acid at pH 7? (1 point) +1 because of the protonation of the R group 2) What is the charge on a protein when: a. the pH is equal to the protein’s pI? (1 point) neutral b. the pH is above the protein’s pI? (1 point) negative c. the pH is below the protein’s pI? (1 point) positive 3) If you have a negatively charged ion-exchange column such as CM-cellulose and you are using it to isolate a particular protein, would you like the pH to be above or below that protein’s pI? WHY? (3 points) The pH should be below the protein’s pI because it would make the protein positive in charge 4) What are 2 ways to elute proteins from an ion-exchange column? Briefly, how do they work? (4 points) - A salt gradient: charged salt ions compete with bound proteins for the charged resin functional groups. Proteins with few charged groups will elute at low salt concentrations, whereas proteins with many charged groups will have greater retention times and elute at high salt concentrations. -   a pH gradient is chosen that approaches the protein of interest's pI. Proteins will elute when the pH gradient reaches their pI, because they will no longer carry a net charge that allows them to interact with the column resin. 5) What are the 2 steps in two-dimensional (2D) gel electrophoresis and on what basis are proteins separated in each? (4 points) isoelectric focusing and SDS-PAGE; separated by isoelectric point and protein's molecular weight For questions 6-13, see the Help menu of “Protein Purification” ( http://www.agbooth.com/pp_ ajax/ )! 6) Early in purification, two low resolving steps often used are: (2 points)  ammonium sulfate precipitation heat denaturation 7) Traditionally, one unit of enzyme activity is defined as: (2 points) that which causes the conversion of 1 mmol of substrate per minute under the (optimal) conditions of measurement 8) The number of units of enzyme activity per milligram of total protein is the Specific activity of the enzyme. (1 point) 9) Enzyme yield is defined as: (2 points) (units of enzyme after purification)/(units of enzyme in the original preparation) x100
10) We will consider our purification to be complete when the apparently pure protein yields one spot when tested by 2D gel electrophoresis (1 point) 11) DEAE cellulose or CM cellulose are only effective in the pH range of 5 to 9 . The starting buffer should be of reasonably low ionic strength, because the affinity of proteins for the ion exchange resins INCREASES or DECREASES as ionic strength increases. (4 points) 12) When choosing a gel type for optimal fractionation of complex proteins the gel pore size is such that the desired protein is partially excluded from the gel beds (1 point) 13) The O.D. 280 is monitored in each of the eluted fractions because: (2 points) to estimate the protein content of individual fractions contained The following questions are from previous exams: 1 Question 14. Assume that you are trying to analyze what you believe to be a pure protein. You carry out the following procedures: A. When you elute from an ion exchange column, you get a single peak of material. On what basis is this separation occurring? (1 point) Separation of charge B. When this material is run on Sephadex, it elutes as two peaks, one with an apparent molecular weight of 80 kD, and the other with an apparent molecular weight of 40 kD. On what basis is this separation occurring? (1 point) size C. If you electrophorese the protein in the presence of SDS, it migrates as two bands with apparent molecular weights of 40 and 80 kD. On what basis is this separation occurring? (1 point) mass To your surprise, if you treat the sample with β -mercaptoethanol and then electrophorese in SDS, you now get a single band migrating with an apparent molecular weight of 40 kD. D. What was in your original sample? Explain your reasoning. Remember, you need to account for all of the previous results. (2 points)
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
  • Access to all documents
  • Unlimited textbook solutions
  • 24/7 expert homework help