The enzyme BURT was found to function at V0=250 uM/min with a Km=15.0 uM when [S]=10.0 uM. What is the Vmax under these conditions? answer is 625 um/min but how?
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- 2. Using the following data, determine the Vmax and Km of the enzyme catalyzed reaction: [S] (M) 2.5 X 10-6 4.0 X 106 1 X 105 2 X 10'5 Vo (uM/min) 28 40 70 95 [S] (M) 4 X 105 1 X 10 2 X 10³ 1 X 10² Vo (uM/min) 112 128 139 140:The following data were obtained in a study of an enzyme known to follow Michaelis-Menten kinetics Vo (mmol/min) S (mmol/L) 0.8 3 4 217 325 433 488 1,000 :The Km for this enzyme is approximately 647 mM 4 O mM 1 O mM 1,000 Ⓒ mM 3 O mm 6 Oa. Using the graph paper below, to plot the enzyme velocity versus the substrate concentration in the absence of ibuprofen, estimate the Km and Vmax for this enzyme.
- Problems on enzyme reaction kinetics 1. Initial rates of enzyme-catalyzed reaction for various substrate concentrations are listed below: Rate Substrate concentration mol/(L min)x10 177 (mol/L) 0.0041 173 0.00095 0.00052 125 106 0.000103 80 0.000049 0.0000106 0.0000051 67 43 a) Evaluate Vmax and Km by a Lineweaver-Burk plot.For an enzyme that follows Michaelis –Menten kinetics, k1=1X106 M-1 sec-1, k-1= 2x103 sec-1 and k2=2X 102 sec-1. a. What is the Km for the enzyme ? If the enzyme concentration is 10 nanomoles in 1 ml, what is Vmax for the enzyme ? What is the catalytic efficiency for this enzyme?A research group discovers a neW version oI happyasee, which they nappyase", that catalyzes the chemical reaction HAPPY = SAD The researchers begin to characterize the enzyme. In the first experiment, with [E] at 4 nM, they find that the Vmax is 1.6 µM s-1. Based on this experiment, what is the kcat for happyase*? kcat 400 S In another experiment, with [E,] at 1 nM and [HAPPY] at 30 µM, the researchers find that Vo is equal to 300 nM s-1. What i the measured Km of happyase* for its substrate HAPPY? Km 10 x10–6 Incorrect Further research shows that the purified happyase* used in the first two experiments was actually contaminated with a reversible inhibitor called ANGER. When ANGER is carefully removed from the happyase* preparation, and the two experiments repeated, the measured Vmax increases to 4.8 µM s-, and the measured Km is now 15 µM. S For the inhibitor ANGER, calculate the values of a' and a. a' 2 Incorrect 2
- In a 250µl reaction containing 0.3 nmol of a Michaelis enzyme with Km = 4.4 x 10 Mand saturating substrate concentration, product was formed at an initial velocity of 6.3 x 10-3 M min-1. What is kcat for this enzyme? Give your answer in 3 sigfigs in scientific notation, for example 1.20 e Click Save and Submit to save and submit. Click Save All Answers to save all answers. O Type here to search 787 Moddy co PrtScn F8 Home F9 DII F6 %23For an enzyme -catalyzed reaction, KM=10.0 mmol dm-3, vmax=0.250 mmol dm²³ s-'; [E]o= 2.3 x 10-6 mmol dm-3. Calculate the catalytic efficiency (=kb/KM) of the enzyme. 0.920 dm³ mol-l s-1 3.1 x 107 dm³ mol-1 s-1 1.1 x107 dm³ mol-l s-1 0.250 dm³ mol-l s-14. Suppose the data shown below are obtained for an enzyme catalyzed reaction. [S] (MM) v (mmol/ml/min) 0.1 3.33 0.2 5.0 0.5 7.14 0.8 8.0 1.0 8.3 2.0 9.09 a) Plot the data and determine Km and Vmax (please include the plot). b) Assuming the enzyme was used in these reactions at a concentration of 10-6 M, then calculate the turnover number for the enzyme.
- The kinetics of an enzyme are measured as a function of [S] in the presence and absence of 2 mM I. Compute Km and Vmax in the absence and presence of I S] (uM) V (uM/min) without I with I 3 10.4 4.1 5 14.5 6.4 10 22.5 11.3 30 33.8 22.6 90 40.5 33.8How does Kcat influence the velocity of MM enzymes ? A. Linear relationship B. Inverse relationship C. No effect D. Only effects velocity if total enzyme concentration is low v = Vmax [S] KM + [S] Turnover rate kcat = Vmax [ET] Speed (substrates product/sec)The image below contains a Michaelis Menten plot for an enzyme catalysed reaction. Based on this plot, estimate the Vmax and Km. Velocity, uM/sec 40 30 20 10 0 0 ENZYME CATALYSED REACTION 20 60 Substrate concentration, mM 40 Vmax= 39 umol/sec; Km = 14 mM O Vmax = 39 umol/sec; Km = 20 mM O Vmax = 14 umol/sec; Km = 39 mM O Vmax = 20 umol/sec; Km = 14 mM 80