Calculate the KM and Vmax for the enzymatic reaction graphed below.
![Wn [s] 005 006 006 002 001 vfumol/min)](http://img.homeworklib.com/questions/29d4ba30-2c40-11eb-bafa-a5d6ac7465d5.png?x-oss-process=image/resize,w_560)
Correct, detailed explanation (with work shown) needed for thumbs up
Vmax is the maximum rate of reaction. It is that point in the y-axis in he graph given, when the enzyme gets saturated (the curve reaches a constant). In the given graph, the corresponding Vmax when the enzyme activity becomes constant is 2umol/min
Vmax = 2 umol/min
Km is the point on x axis at Vmax/2 in y axis. Vmax / 2 = 1. The corresponding point on x axis is 30uM (approximately).
Km = 30 uM
Calculate the KM and Vmax for the enzymatic reaction graphed below. Correct, detailed explanation (with work...
a. What is the Km and Vmax for PFK1 when treated with OmM (represents control for enzymatic activity) or with 5mM AMP Show work on the graph draw lines for Vmax and Km. Control: Vmax (AMP), 0.32 0.28 0.2 Km 0.24 0.20 (s)-FTY20 (20LM): 0.18 Vmax 0.12 0.08 0.04 Km 0.00 Fructose-6-phosphate (mM) b. Fructose-6-phosphate is the substrate of the reaction. Based on answer in "a", what type of regulation occurs with and what site on PFK1 is AMP likely...
Nobody has been able to solve the Km. The Vmax values are
correct. I believe I have my graph wrong. Can someone please help
me solve this.
Using the Lineweaver-Burk Equation 1/Vo Km/Vmax[S]1/Vmax) create a graph of both the Non-Inhibited data and Inhibited data below (on the same graph axes) and calculate the KM and VMax for each line. Vo (umol/L.min) Vo (Hmol/L-min) [Sol (umol/L) Non-Inhibited Data Inhibited Data .00e-06 13.9 7.60 .00e-06 18.0 9.90 ..10e- 05 10e-05 26.0 14.8...
calculate phase difference
find the phase difference
of the phase difference in the space below the table. f = 100 Hz At maximum voltage, the current is (circle one): f = 200 Hz At maximum voltage, the current is (circle one): f = 300 Hz At maximum voltage, the current is (circle one): Maximum Minimum Zero and increasing Zero and decreasing Nonzero and increasing Nonzero and decreasing Other Maximum Minimum Zero and increasing Zero and decreasing Nonzero and increasing Konzero...
21 and 22 please with work shown
21/ Show that for competitive inhibition of an enzymatic reaction, the intercepts on the horizontal axis of a plot of 1/v versus 1/[S) at different inhibitor concentrations are equal to 1/Km (1 + [/K). 48 81 73 0.4 145 (22 [S] (UM) y - no inhibitor (um/min) y - with inhibitor (UM/min) Ti g t be 0.1 0.2 123 wool 0.6 95 0.75 160 108 Renin acts as a specific protease that cleaves...
Need help with number 13! I already asked about number
12. The inverse velocity and inverse substrate concentration relationship for an enzyme-catalyzed reaction is given below V Vmax Vmax S For the hydration of CO2 catalyzed by carbonic anhydrase, it was determined experimentally that (dm s mol 4023.9+ 39.934 at a total enzyme IS] concentration of 2.32 × 10-y mol-dm- What is the value of the Michaelis constant KM for this enzymatic reaction? (B). 9.92x103 mol dm3 (D). 100.8 mol...
The value of Km for the shown data for a hexokinase-catalyzed reaction is with the unit of . The value of Vmax for. the same reaction is with the unit of . Be sure to give the values with the correct number of significant figures. You might have to construct a kinetic plot. For units, choose one answer from (uM, 1/ UM, HM/second, uM x second, mM, 1/mM, second, 1/second, mM/second, mM x second) vo (mM/sec) Glucose concentration (mm) 0.10...
Interpret the data above
1. What is the aporoximate Km and Vmax (units matter)
2.Briefly explain how you found each
3.if you used 0.020 microM enzyme in your studies, what is
kcat in units of s^-1? Show your work units matter
4. Does this enzyme appear to display cooperativity? Explain
how you came up to that conclusion.
Directions: Below are data from 4 separate experiments that you must analyze/evaluate. Using the information from all 4 experiments, you must then propose...
12)Show work step by step
13)Show work step by step
Below is BoX 6-1 if needed
Applying the Michaelis-Menten Equation IV An enzyme is found that catalyzes the reaction X= Y Researchers find that the Kyn for the substrate A is 4 pm, and the keat is 20 min . (a) In an experiment, (X) = 6 mm, and the initial velocity, V, was 480 nm min!. What was the (E) used in the experiment? (b) In another experiment, (E)...
Disregard the work shown it’s not correct, I need correct
answer with proper explanation
Thank you
12. (6 points) If the reaction below initially consisted of just carbon and 1.50 atm of CO2 what are the equilibrium pressures of C02 and CO? Kp 14.1 2CO(g) 1.56 C (s) + CO2(g) 1' ?? 2 x 1 a yv>,14 , i v-?.is,0 Poo = 2a
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