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`Choose the best matching definition for each term. Multiple Cloning Site: A key feature of a plasmid that allows for efficient insertion of your insert Maltose Binding Protein : common affinity tag used in protein purification due to its affinity to amylose Insect Cells: A common eukaryotic expression system that uses baculovirus to introduce your target gene DpnI: A restriction enzyme that specifically cuts methylated DNA Other Incorrect Match Options: The three bases on DNA that code for an amino acid A cloning vector made up of a circular piece of double stranded DNA A common affinity tag used in protein purification due to its affinity to immobilized metal ions A single polypeptide composed of multiple tandem proteins Electroporation: A technique used for the transformation of bacteria and yeast cells PCR: A technique used to amplify specific DNA sequences. Yeast: A low cost eukaryotic expression system that is difficult to lyse T7 promoter: A key feature of a plasmid that allows for the controlled expression of your target gene Hydrophobic effect: A key interaction between enzymes and substrates that provides favorable entropy to a reaction Fusion protein: A single polypeptide composed of multiple tandem proteins His tag: A common affinity tag used in protein purification due to its affinity to immobilized metal ions Antibiotic Resistance Marker: A key feature of a plasmid that allows for the selection of transformed cells
The effects of an UNCOMPETITIVE inhibitor can be overcome by adding an infinite quantity of substrate. F The effects of an UNCOMPETITIVE inhibitor can be overcome by adding an infinite quantity of substrate. F An UNCOMPETITIVE inhibitor only results in an inhibited VMax leaving the observed KS unchanged. F The effects of a COMPETITIVE inhibitor can be overcome by adding an infinite amount of substrate. T In order to solve for the Ki of an UNCOMPETITIVE inhibitor you can replot the Lineweaver-Burk slope vs. [I]. F In order to solve for the Ki of an COMPETITIVE inhibitor you can replot the Lineweaver-Burk slope vs. [I]. T In order to solve for the Ki of an COMPETITIVE inhibitor you can replot the Lineweaver-Burk Y-intercept vs. [I]. F True NON-COMPETITIVE inhibition only occurs when k-1 >> kP T True NON-COMPETITIVE inhibition only occurs when k-1 << kP. F The Lineweaver-Burk plot for a MIXED TYPE inhibitor approaches completely
vertical as [I]⟶⟶∞∞.
T
The Lineweaver-Burk plot for a PARTIAL MIXED TYPE inhibitor approaches
completely vertical as [I]⟶⟶∞∞.
F
A MIXTED-TYPE inhibitor only results in an apparent KS leaving the observed VMax unchanged. F
Choose the type of inhibition suggested by the following plot: uncompetitive Choose the type of inhibition suggested by the following: VMax is unaffected by [I] Competitive Choose the type of inhibition suggested by the following equation: IC 50 = Ki Non-competitive
Choose the type of inhibition suggested by the following plot: Partial mixed type ` Choose the type of inhibition suggested by the following plot: non- competitive
(KS not changed; v⟶⟶0 as
[I]⟶
Given the following purification table: Fraction Total Protein (mg) Total Activity (Units) Crude Lysate 6000 10, Affinity Chromatography Pooled Fractions
SEC Final Pure Fractions
What is the best estimate for the total fold-purification of your enzyme? Using a size exclusion chromatography column, you run the following standards: Protein or Compound Molecular Weight (Da) Vitamin B12 1, Myoglobin 17, Ovalbumin 44, Gamma-globulin 158, Thyroglobulin 670, And get the following chromatogram (A 280 vs Elution time):
At what elution time would you estimate that your 30,000 Da enzyme would elute from the same column?
using a spectrophotometer monitoring absorption at 340 nm. The initial change in absorbance that you measure is 115 mAU/min (mAU are milli- absorption units, which is a dimensionless measure of total absorption, 1 mAU = 0.001 AU). Given that the extinction coefficient for NADPH at 340 nm is 6220 M-1^ cm-1^ and the pathlength of you cuvette is 0.5 cm. What is the initial velocity of your enzyme catalyzed reaction in [NADH]/min? 37 μM/min Consider the following reaction scheme:
Using a stopped-flow system allows this reaction to be observed during pre- steady state and steady state conditions. These observations result in the following plot when monitoring the [P] 1. What can you conclude about your reaction given these data? k 2 is rate limiting at pre-steady state and k 3 is rate limiting at steady state The effects of a competitive inhibitor cannot be overcome by adding an infinite amount of substrate. False Q18 You are atudying an enzyme catalyzed reaction that you have determined the following constants for:
In the space provided write in which type of inhibition is displayed by inhibitors I, X and Y. Use "C" for competitive, "UC" for uncompetitive, "NC" for non-competitive, "MT" for mixed type, "PMT" for partial mixed type and "ITT" for impossible to tell. I=PTM X=C Y-=UC
The following velocity equation applies in the case of a partial mixed type inhibitor: Of the following equations which ones represent expressions for the inhibited VMax (VMaxi) and apparent KS (KSApp) for a partial mixed type inhibitor?
The active site of an enzyme must be deprotonated to bind a substrate and the substrate is positively charged and remains so at the pH range tested. Given that: KS = 5 μM Ke = 0.1 μM Substrate concentration = 200 μM Select the best expression for the apparent KS (KSApp) for the inhibited reaction. KSApp = [ Select ] Ks * ( 1 + [H+]/Ke ) What velocity relative to the maximum velocity (V/VMax) would you expect for this reaction with the concentration of H+^ = 1 μM (i.e. pH 6.0)? V/VMax = [ Select ] 0.
You are working with what you know is a competitive inhibitor of your enzyme. You measure your reaction in the absence of inhibitor and get the value for KS = 20 μM. You then vary the concentration of inhibitor and measure your KSApp for each concentration of inhibitor. Given that this is a competitive inhibitor you expect to have a linear relationship between KSApp and the inhibitor concentration. However, when you plot your data out it is not linear and can only be fit with a second order polynomial i.e. parabola. This reminds you of a situation discussed in your Enzyme Kinetics course in which:
𝐾𝑆𝐴𝑝𝑝=𝐾𝑆(1+[𝐼]𝐾𝑖) 2 KSApp=KS(1+[I]Ki)
What situation would result in the above relationship? Two identical inhibitor binging sites Using the above equation for KSApp and your best fit equation from the plot (i.e. y = 0.0889x^2 + 2.6667x + 20), determine an estimate for Ki in μM. Ki = [ 15 μM
Indicate whether the IC 50 (i.e., KiApp) increases, decreases, or remains the same in response to a decrease in substrate concentration for the two types of inhibitors listed below. (a) For a Competitive inhibitor, IC 50 [ decreases (b) For an Uncompetitive inhibitor, IC 50 [ INCREASES
You are studying an enzyme catalyzed reaction that you have calculated the following constants for: KM = 6.5 X 10-4^ M Vmax = 300 nmoles x liter-1^ x min- You screen for inhibitors of your enzyme and find an inhibitor that appears to function as a competitive inhibitor. Adding 2 x 10-5^ M substrate and 1 x 10-5^ M inhibitor, you record an initial velocity of 1.5 nmole x liter-1^ x min-1. Calculate the apparent KM (KMApp) in the presence of inhibitor. Provide your answer to two decimal places (three significant figures) in the units of mM. KMApp = mM Calculate the Ki for the inhibitor. Provide your answer to two decimal places (three significant figures) in the units of μM. Ki = μM
