Enzyme activity (Effect of temperature) Research Question How does temperature affect the activity Of the enzyme catalase? Hypothesis The higher temperature would increase Catalase activity leading to a faster reaction. Teme degrees 4 10 22 410 65+ Results Time to float disa Triall 5.85 3.06 4.00 1/29/24 Mini Lab #1 10.49 9.43 Trial a 5.70 2.86 5.26 7.86 11.36 Average 5.775 2.96 4.63 9.175 10.395 Rate Class average 5

Enzyme activity (Effect of temperature) Research Question How does temperature affect the activity Of the enzyme catalase? Hypothesis The higher temperature would increase Catalase activity leading to a faster reaction. Teme degrees 4 10 22 410 65+ Results Time to float disa Triall 5.85 3.06 4.00 1/29/24 Mini Lab #1 10.49 9.43 Trial a 5.70 2.86 5.26 7.86 11.36 Average 5.775 2.96 4.63 9.175 10.395 Rate Class average 5

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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For this question you will need to work all of your answers on scratch paper. Once completed, upload an image of your answers showing all work and clearly identifying your answers. Use the Browse My Computer button to upload an image. It is known that the activity of penicillin declines over several weeks when it is stored at room temperature in the absence of stabilizers. The following concentrations of active penicillin was found as several weeks passed. This data shows the decline in effectiveness of the penicillin due to a decomposition reaction. Time in weeks 0 5.0 10.0 15.0 20.0 25.0 30.0 Concentration of Penicillin in Molarity 0.076 0.046 0.020 0.010 0.0050 0.0020 0.0010 a. Show all your work, what is the average rate of the reaction? b. Show all your work, what is the instantaneous rate from 5.0 to 20.0 weeks? c. Show all your work, what is the instantaneous rate from 0.0 to 10.0 weeks? d. Show all your work, what is the instantaneous rate from 15.0 to 30.0 weeks?
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
Experiment 1: Determination of the Rate, Order, and Rate Constant Table 3: Data for Trial 1 Water displaced in mL Time (seconds) 2 mL 49 4 mL 84 6 mL 116 8 mL 145 10 mL 174 12 mL 202 14 mL 230 16 mL 253 18 mL 280 20 mL 313 22 mL 350 Table 4: Data for Trial 2 Water displaced in mL Time (seconds) 2 mL 78 4 mL 144 6 mL 175 8 mL 205 10 mL 255 12 mL 308 14 mL 330 16 mL 409 18 mL 463 20 mL 515 22 mL 575 Table 5: Data for Trial 3 Water displaced in mL Time (seconds) 2 mL 53 4 mL 100 6 mL 139 8 mL 178 10 mL 213 12 mL 247 14 mL 282 16 mL 319 18 mL 363 20 mL 412 22 mL 463 help me with this part 1. Should the rate constant (k) be the same for all three trials in this experiment? Explain your answer.…
8. Data on the speeds of a Michaelis-Menten enzyme reaction v (in micromolar/minute) as a function of the substrate concentration S (in micromolar) has been collected. Your goal is to find a model that fits this data. a) What type of transformation do you need to do for the data so that the transformed data looks like a linear function? b) Assume that the appropriate transformation on the data has been performed and that a linear regression line has been fitted to the transformed data and that its equation is given by Y = 0.501147.X +0.454986 Compute the parameters Vmax and KM and write the model. and KM = v(S) = micromolar minute c) What is the context domain of this function? d) Predict the speed of the reaction (in micromolar/minute) when the concentration is 12.45 micromolar. Round your answer to two decimal places. v(S) = micromolar minute e) What is the maximal speed of the reaction? micromolar minute
Question 15 Plots A and B show enzyme kinetics data in the absence and presence of inhibitor. Which response describes the effect of the inhibitor? B Vo (M¹) 2960 10 A [S] (mm) 6 8 10 Apparent Km > Km; Apparent Vmax > Vmax Apparent Km > Km; Apparent Vmax Vmax -1 1.2 1 0.8 0.6 0.4 0.2 0 -0.2 1 1/[S] (mM¹¹) Analyse the plot and calculate the values. This question is harder than it seems. 2
1. An enzyme-catalyzed reaction has a KM of 1 mM and a Vmax of 5 nM s-1. What is the reaction velocity when the substrate concentration is 0.25 mM? 2. If an enzyme-catalyzed reaction has a velocity of 2 mm/min and a Vmax of 10 mm/min when the substrate concentration is 0.5 mM, what is the KM?
What subclass of enzymes catalyzes the following reaction? Select the single best answer. H H 0 | | H HN-C-CO, + 0,c-c-c-c-co, CH3 H H u་མང་ HH NH3 HC–C-CO, + 0,c-c———CO, HHH O Transaminase X $ O Kinase Decarboxylase O Protease
I need help in question no 2 On the graph paper, plot a graph of reaction rate vs. temperature by using the Y-axis for reaction rate and the X-axis for temperature
This question has multiple parts. Work all the parts to get the most points. For the reaction below: OH O H*, catalyst OH a Draw the hemiacetal intermediate. 0 + 11*** // [F ? ChemDoodle Cengage Learning Cengage Technical Support Previou Email Instruc
Suppose that you have an enzyme concentration at the molarity calculated in question 2 and the product that you monitor has a value of x = 6200 M1 cm1. You ran eight reactions and found the following: Reaction # [substrate] (μM) Slope (Abs/min) 0 0.000 2345678 10 0.163 20 0.288 60 0.586 100 0.740 200 0.920 500 1.078 1000 1.143 M-4.45x10-M Calculate the values of Kcat and KM for the reaction (remember to include units). Show your plot and your calculations.
Determine the values of KM and Vmax. For the Vmax obtained, calculate the turnover number assuming that 1 X 10-4 mol of enzyme were used. Substrate concentration (mol L-1) Velocity (mM min-1) 2.500 0.588 1.000 0.500 0.714 0.417 0.526 0.370 0.250 0.256
Part a can you please help ?