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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![The enzyme BURT was found to function at VO=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?](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F0c97a642-79d5-4404-a525-e17b193e9e6a%2F21248d59-8860-4c37-96e3-473e1063316e%2Fsv4oe9f_processed.jpeg&w=3840&q=75)
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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 140Problems 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?
- :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.4. 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.
- 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 %23A 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 2The 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 33.8
- For 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-1The 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 80How 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)