convert 7.85 x 10^-8 km to um
Convert 5.0 um to meters. O 5.0*10*9m O 5.0 x 103 m O 5.0 x 106m O 5.0 x 10-6 m
Question 8 5. Convert 85 MPH into feet per second. 6. Convert 120 km/hr into feet per second. 7. A cubic equation; x3 + 2.40x2-7.14x = 0. Solve for x values( 225 N 8. Use vector addition to find the resultant. Angle 150 N Angles in degrees. 5 5 9. Use the component method to find the resultant. (Last problem we did in class) 35 lbs Angles in degrees. 25 lbs 10. Why is finding a resultant important, in design?....10...
Part A An enzyme that follows Michaelis-Menten kinetics has a KM value of 10.0 uM and a kcat value of 201 s-1. At an initial enzyme concentration of 0.0100 uM, the initial reaction velocity was found to be 1.07 x 10- uM/s. What was the initial concentration of the substrate, [S], used in the reaction ? Express your answer in micromolar to three significant figures. ► View Available Hint(s) PO ALO O O ? [S]; = MM UM
1.3 Convert: (a) 16.3 m to mm (b) 16.3 m to km (c) 4 x 10-6 uF (microfarad) to pF (picofarad) (d) 2.3 ns to us (e) 3.6 x 10 v to MV (f) 0.03 mA (milliamp) to LA
a. An enzyme has a Vmax of 100 umol/min and a Km of 40 uM. When substrate concentration is 40 uM what is the initial reaction rate? b. An enzyme with a Vmax of 100 umol/minute and a Km of 10 uM was reacted with a irreversible active site specific inhibitor. After reaction with the inhibitor, the enzyme was assayed using a 2 mM concentration of substrate, and it gave a reaction rate of 20 umol/min. What percentage of the...
Values for an Enzyme and a substrate are: Km=4 uM and kcat=20/min. For an experiment where [S]=6mM, it was found that Vo=480nM/min. What was the enzyme concentration? Give your answer in nM. Using the same kcat and Km as the previous question, if [Et]=0.5uM gives a Vo=5 uM/min, what wat the [S]? Give your answer in uM. reaction is run with the kcat=20/min and the Km=4uM. Use the enzyme concentration from question 1 above. A very strong inhibitor is added creating...
Convert the following measurement. mol mol 5.6 x 10-8 8 cm kg.m x I ?
5. The Km of an enzyme of an enzyme-catalyzed reaction is 6.5 uM. What substrate concentration will be required to obtain 55% of Vmax for this enzyme? (10 pts)
In the presence of 1.4 uM of a competitive inhibitor, how will the kinetics of an enzyme change? KM = 10 UM, Vmax = 2 nM/min, K; = 1 UM Km will be: In the presence of 1.4 uM of a competitive inhibitor, how will the kinetics of an enzyme change? KM = 10 UM, Vmax = 2 nM/min, K; = 1 UM Vmax will be:
Convert the following measurement. mol mol - 8 8.9 x 10° = g.cm kg.m x 6 ?