Partial Enzyme Lab Report Submi x+ Chapter 4 Competency 3XA cloud.scorm.com/ScormEnginelnterface/defaultui/player/legacy.html?configuration=uJ3nrCSmzNPWgXiwoJ4YwRLimO8D6cRHCEe163Lap90fEcUCaNJhdd-I..I AppsGoogle I MyMasteryPatho Welcome to LSU M. P MasteringBiology E FRENCHALEKS - Adaptive L.Pt Periodic Table - Pta.Other bookmarksE Reading listChapter 4 Competency 3POWERED BY THE RUSTICI ENPREVIOUSNEXTCLOSE ITEMRETURN TO LMSPage 7 of 26Your Learning Points 3Osiagwu, Chiemeka KA reaction is carried out by mixing together a solution of K,CO, and a solution of Fe(NO,),.K,Co, (aq) + Fe(NO,), (aq) - FeCo, (s) + 2KNO, (aq)Assume the purple particles below each represent 1 mole of particles. Construct the balanced netionic equation by dragging the appropriote components into the gray boxes. Drag each unreactedspectator ion along with its balancing coefficient into the blue box that represents the aqueoussolution.Spectator ionsBalanced Net lonic Equation+FeCo, (s)9:39 PMA 4)10/9/2021O Type here to search73°F近出

Given : K2CO3 + Fe(NO3)2 --> FeCO3 + 2KNO3

Answered: A reaction is carried out by mixing…

Chemistry for Today: General, Organic, and Biochemistry

9th Edition

ISBN:9781305960060

Author:Spencer L. Seager, Michael R. Slabaugh, Maren S. Hansen

Publisher:Spencer L. Seager, Michael R. Slabaugh, Maren S. Hansen

Chapter20: Enzymes

Section: Chapter Questions

Problem 20.26E

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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?
Enzyme X can act on either substrate A or B; the Km on A < Km on B. Vmax for A > Vmax on B.Sketch 1 plot for 1/v vs. 1/[S] plot for Enzyme X in the presence of A and in the presence of B.
please helb.
Calculate the specificity constant for an enzyme-catalyzed reaction if the total enzyme concentration is 15 nM, Vmax is 9 M/s, and Km is 150 M. A. Has the enzyme achieved catalytic perfection? (Y or N)
Q1: the following table gives the results of enzyme-catalyzed reaction. The enzyme concentration [E] = 0.00012 mmol/L Use the information of substrate concentration [S] and rate Vo for both experiments, with inhibitor and without inhibitor to answer the following questions (NOTE: [E] = 0.00012 mmol/L) [S], mmol/L 0.710 0.400 0.310 0.098 0.066 0.040 Experiment 11 without inhibitor VO, mmol/L-min 0.200 0.180 0.160 0.120 0.100 0.070 Experiment 2 with inhibitor [1] = 0.033 mmol/L VO, mmol/L-min 0.180 0.150 0.110 0.070 0.050 0.040 1- Calculate Km and Vmax for both experiments using Michaelis-Menten equation. [S] vs Vo 2- Calculate Km and Vmax for both experiments using Lineweaver-Burk plot, 1/[S] vs 1/vo 3- Calculate Kcat for the enzyme (without inhibition) 4- Based on the values of Km and Vmax, with inhibitor and without inhibitor, determine the type of the Inhibition (competitive, non-competitive or uncompetitive)
Q1: the following table gives the results of enzyme-catalyzed reaction. The enzyme concentration [E] = 0.00012 mmol/L Use the information of substrate concentration [S] and rate Vo for both experiments, with inhibitor and without inhibitor to answer the following questions (NOTE: [E] = 0.00012 mmol/L) [S], mmol/L 0.710 0.400 0.310 0.098 0.066 0.040 Experiment 1 without inhibitor VO, mmol/L-min 0.200 0.180 0.160 0.120 0.100 0.070 Experiment 2 with inhibitor [1] = 0.033 mmol/L VO, mmol/L-min 0.180 0.150 0.110 0.070 0.050 0.040 1- Calculate Km and Vmax for both experiments using Michaelis-Menten equation. [S] vs Vo 2- Calculate Km and Vmax for both experiments using Lineweaver-Burk plot, 1/[S] vs 1/vo 3- Calculate Kcat for the enzyme (without inhibition) 4- Based on the values of Km and Vmax, with inhibitor and without inhibitor, determine the type of the Inhibition (competitive, non-competitive or uncompetitive) Note:- show all your work to answer the above questions. Upload your answers…
File Edit Dashboard. app.101edu.co 2 View S History HO ** # Bookmarks H- 3 E D Profiles Tab Aktiv Chemistry Choose the best classification based on the chain length for the monosaccharide shown. 80 CHO H OH CH₂OH $ 4 2 R Window Help a F 27 28 X G the pineal gland stimulates pro x Glycogen: Mammalian Energy X % 5 P I T G W Question 18 of 20 A 6 MacBook Air F6 Y H & 7 F7 U * *00 8 A) triose B) tetrose C) pentose D) hexose E) heptose FB 1 ( 9 K ☎ a s C The structure for D-fructose is X F9 O O OT C P PI * DOCX
3. An organic molecule needed to make an enzyme function correctly is called a(n) subunit cofactor coenzyme O apoenzyme
4. Based on the graph below, What is the approximate value for K, and the V for this enzyme? A M max 14C peptide MM/min to + H 1000 1500 500 [Ac-CoA], UM
← M Gmail k > C + app.aktiv.com YouTube Maps Lab Rpt 1-Protein Draw the product of the reaction shown below. Ignore inorganic byproducts. 1. NaBH4 2. H3O+ Drawing CO Calendar o Probl Atoms, and F Draw or ta
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