1/V, min/umol 5.50 5.00 4.50 4.00 3.50 answer. 3.00 2.50 2.00 0.00 y = 0.9474x + 2.6649 0.50 1.00 1.50 1/[S], UM -1 y = 0.9997x + 2.032 2.00 2.50 3.00 Looking at the double reciprocal plot for an enzyme in the absence of inhibitor and in the presence of two concentrations of inhibitor, what would be the Vmax for the uninhibited enzyme? (bottom graph) Equation is given. Choose the one best
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![5.50
1/V, min/umol
5.00
4.50
4.00
y = 0.9474x + 2.6649
y = 0.9997x + 2.032
0.00
1.00
2.00
2.50
3.00
1/[S], uM -1
Looking at the double reciprocal plot for an enzyme in the absence of inhibitor and in
the presence of two concentrations of inhibitor, what would be the Vmax for the
uninhibited enzyme? (bottom graph) Equation is given. Choose the one best
answer.
3.50
3.00
2.50
2.00](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F07b00a25-4bbc-4603-8563-821c16d74641%2Fda338a2f-f237-4970-8031-a1f3332c891b%2Fheuf5u8_processed.jpeg&w=3840&q=75)
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- An enzyme-catalyzed reaction has a KM of 20.0 mmol L-1 and Vmax of 17.0 pmol s-1. When a mixed inhibitor is added, the apparent KM is 50.0 mmol L-1 and the apparent Vmax is 5.20 pmol s-1. Calculate α.If a michaelis-type enzyme has an apparent Km of 2*10^-8 and apparent Vmax of 1*10^-4 moles/min in the presence of 1*10^-3 M uncompetitive inhibitor (Ki=1*10^-7 M), what is the true Km of the enzymeThe enzyme serine hydroxymethy ltransferase (SHMT) catalyses the conversion of serine into glycine. The fo llowing table gives the initial rates, vo, for the SHMT-catalysed reaction of the substrate serine at var ious concentratio ns of serine, lSI.[S]/(mmol dm-3) 10 20 30 40vo(μmol dm-3 s-1) 1.63 2.94 4.10 4.95Use the data to determine the values of the MichaelisMenten constant, the maximum velocity of the reaction, and the maximum turnover number of the enzyme.
- The enzyme β-methylaspartase catalyzes the deamination of β-methylaspartate. For this aspartate reaction in the presence of the inhibitor hydroxymethylaspartate (3.8 M), determine KM and whether the inhibition is competitive or noncompetitive (KI = 1.0 M). [S], M V w/o inhibitor, M/s V w/ inhibitor, M/s 1x10-4 0.0259 0.0098 5x10-4 0.0917 0.040 1.5x10-3 0.136 0.086 2.5x10-3 0.150 0.120 5x10-3 0.165 0.142 In the ABSENCE of inhibitor: The Lineweaver-Burke equation is 1V=1V= __________ (1[S])(1[S]) + __________, and the KM is __________ M. In the PRESENCE of inhibitor: The Lineweaver-Burke equation is 1V=1V= ____________ (1[S])(1[S]) + ___________, and the KM is ___________ M. The type of inhibition is ____________. Round-off all answers to two (2) significant figures.For the following aspartase reaction in the presence of the inhibitor hydroxymethylaspartate, determine Km and whether the inhibition is competitive or noncompetitive. You have to plot thegraph on the graph paper and also by using excel.[S] V, No Inhibitor V, Inhibitor Present(molarity) (arbitrary units) (same arbitrary units) 1 x 10-4 0.026 0.0105 x 10-4 0.092 0.0401.5 x 10-3 0.136 0.0862.5 x 10-3 0.150 0.1205 x 10-3 0.165 0.142A particular enzyme has a ΔΔG‡ of -22.1 kJ mol-1 at 37.0 °C. Calculate the rate enhancement of this enzyme. (R = 8.3145 J mol-1 K-1)
- The enzyme β-methylaspartase catalyzes the deamination of β-methylaspartate. For this aspartate reaction in the presence of the inhibitor hydroxymethylaspartate (3.8 M), determine KM and whether the inhibition is competitive or noncompetitive (KI = 1.0 M). In the ABSENCE of inhibitor: The Lineweaver-Burke equation is 1V=1V= __________ (1[S])(1[S]) + __________, and the KM is __________ M. In the PRESENCE of inhibitor: The Lineweaver-Burke equation is 1V=1V= ____________ (1[S])(1[S]) + ___________, and the KM is ___________ M. The type of inhibition is ____________. Round-off all answers to two (2) significant figures.A particular enzyme-catalyzed reaction has an apparent Vmax = 9.00 nmol s-1 and α' = 3.00 when 2.00 µmol L-1 inhibitor X is present and uncompetitively inhibiting the reaction. Calculate Vmax for the uninhibited reaction in nmol s-1.Proline racemase catalyzes the conversion between L-proline and D-proline. The Km and kcat for this reaction are 0.15 M and 550/sec respectively. If the enzyme concentration is 1.45 X 10-5 mmole/ml what is the Vmax of this reaction?
- The equation of the double reciprocal plot is y = 0.5294 x + 1.4960. What is the value of vmax (in M/s)? The substrate concentration is given in units of molarity (M) and reaction velocity has units of molarity per second (M/s). (Report to three significant figures)The Lineweaver - Burk plot (Figure 1) shows an enzyme-catalyzed reaction in the absence and presence of 0.1µM inhibitor (ketoconazole). O Estimate Vmax and Km in the absence and presence of the inhibito.. (ii) Determine the type of inhibition shown by the inhibitor. Explain. 0.1- 0.08 0.06 - With inhibitor 0.04 0.02 Without inhibitor 0.4 -0.2 -0.02 0.2 0.4 0.6 0.8 1.0 1/{S\(&M-1) 1/vo (pmol-11 min)1. A Lineweaver-Burk Plot is shown below. 30 25 Curve A y = 3.1207x + 2.4978 20 15 Curve B y = 1.0003x + 2.3602 10 5 -3 1 5 7 11 1/[Catechol] (mM1) With these curves, determine the following enzyme parameters. Show all pertinent solutions. a. Km of Curve A and Curve B b. Vmax of Curve A and Curve B c. Assuming that one of these curves corresponds to the kinetics of one enzyme and one substrate, which curve represents the effect of an inhibitor? Why do you say so? d. What type of inhibition is exhibited by your answer in question c? Why do you say so? 1/V, (units of activity 1)