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Search results “Determining km of an enzyme catalyzes”

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Views: 104640 Shomu's Biology

04:44

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You need to know the 3 parts of the graph: a) initially linear since many free enzymes available at the lower concentration of substrate. b) rate begins to slow as substrate cannot always find a free enzyme (some are being used by other substrate) c) Saturated -- the enzymes are always in use -- increasing substrate will no longer increase rate. Adding inhibitors will change these graphs -- but this is for another assessment statement.
Views: 106758 Richard Thornley

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Description
Views: 89217 Arpan Parichha

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Views: 6443 Priyanka Jain

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This video shows you how to calculate the Km and Vmax for an enzyme catalyzed reaction. Urease is used in this example. Urease catalyzes the hydrolysis of urea to carbon dioxide and ammonia. Before you calculate Km and Vmax, you will have to calculate reaction rates at different substrate concentrations. Details on how to calculate reaction rates can be found here https://rahulpatharkar.000webhostapp.com/2018/12/enzyme-kinetics
Views: 37 Rahul Patharkar

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A quick discussion of how substrate concentration affects reaction rate for enzyme-catalyzed reactions.
Views: 18234 theolderthefox

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Views: 226314 Moof University

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Extra Tutorial Problems - Enzyme Kinetics Question 1: Consider an industrially important enzyme, which catalyzes the conversion of a substrate to form a much more valuable product. An initial rate analysis for the reaction in solution yields the following Michaelis-Menten parameters: Vmax = 0.60 µM/s; Km = 80 µM. What will be the substrate concentration at one-half the maximum velocity of the reaction?
Views: 852 Dr Helen

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A number of important catalytic constants for enzyme reactions: Km, v max, kcat and kcat/vmax. Each convey different information about an enzyme's activity!
Views: 22743 biochemistry rocks

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Views: 16473 Zachary Tan

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How to calculate reaction rate from your absorbance data!
Views: 27327 kirk kawagoe

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Enzyme Kinetics Question 1: The enzyme-catalyzed conversion of a substrate at 25C has a Km of 0.035 M. The rate of the reaction is 1.15 x 10-3 M s-1 when the substrate concentration is 0.110 M. What is the maximum velocity of this reaction?
Views: 1319 Dr Helen

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#iitutor #Biology #SubstrateConcentration https://www.iitutor.com Factors that influence the enzyme that have the ability to alter or reduce their effectiveness include: Temperature, pH, Substrate concentration, Cofactors, Inhibitors (deactivate enzymes to purposely slow their activity temporarily or permanently). The concentration of substrate affects enzyme activity. The rate of an enzyme controlled reaction is affected by the concentration of the substrate. An increase in substrate concentration will increase the rate of reaction until all enzyme active sites are occupied. The higher the substrate concentration, the greater the rate of enzyme reaction, until all available enzyme active sites are being used to catalyse reactions. When all enzymes have a substrate each, the reaction rate plateaus. This is because the reaction rate is working at its maximum capacity. Increasing the substrate concentration beyond the saturation point will not increase the rate of reaction, since all enzymes are working at their maximum turnover rate and will have to be reused to act on the additional substrate. The only way to increase the reaction rate would be to increase the enzyme concentration.
Views: 14778 iitutor.com

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:)
Views: 19492 yr82011

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A presentation that will show you how to calculate the rate of a reaction from experimental data.
Views: 17967 Paul Scott

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Project Title: Development of e-contents on foundation course on analytical biochemistry and separation techniques Project Investigator: Dr. Charmy Kothari Module: Substrate kinetics of enzyme (determination of Km and VMAX)
Views: 3847 Vidya-mitra

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Ever wonder why some folks handle their alcohol better than others? This introductory video to Michaelis-Menten dynamics explains the enzyme behaviour behind why you're left feeling pukey after a few drinks. This animation was done as part of degree requirements for the University of Toronto's Master's in Science in Biomedical Communications. More examples of my work can be found on my website: ursulavisuals.com
Views: 877 Ursula Florjanczyk

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Moof's Medical Biochemistry Video Course: http://moof-university.thinkific.com/courses/medical-biochemistry-for-usmle-step-1-exam For Related Practice Problems with Worked Video Solutions on Enzymes, visit courses.moofuniversity.com. In this video, I basically do an example problem relating to Michaelis-Menton Kinetics, the Hyperbolic Equation, and the Lineweaver-Burk (Double Reciprocal) Plot. For a suggested viewing order of the videos, information on tutoring, personalized video solutions, and an opportunity to support Moof University financially, visit MoofUniversity.com, and follow Moof University on the different social media platforms. Don't forget to LIKE, COMMENT, and SUBSCRIBE: http://www.youtube.com/subscription_center?add_user=MoofUniversity SUPPORT MOOF UNIVERSITY: http://www.moofuniversity.com/support-moof/ BUY A T-SHIRT https://shop.spreadshirt.com/moofuniversity/ INFORMATION ABOUT TUTORING AND PERSONALIZED VIDEO SOLUTIONS: http://www.moofuniversity.com/tutoring/ INSTAGRAM: https://instagram.com/moofuniversity/ FACEBOOK: https://www.facebook.com/pages/Moof-University/1554858934727545 TWITTER: https://twitter.com/moofuniversity
Views: 90427 Moof University

14:07
In this video I do a problem where we determine the Vmax and Km for a normal uninhibited enzyme, and then determine the same the values when it's inhibited. The values are then plotted on a graph, and the type of inhibition is determined from the graph.
Views: 48808 FortuneFavorsPrep

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Temperature: As temperature increases so to does the kinetic energy of the enzyme and substrate molecules which randomly collide. The frequency of collisions increases as the temperature increases thus initially increasing the rate of reaction. This occurs up to a maximum rate of reaction and the temperature at which the maximum rate of reaction is reached is referred to as the optimum temperature. Beyond the optimum temperature, increasing temperature increases the kinetic energy of the molecules to the point that the three-dimensional shape of the enzyme can be lost. Thus the shape of its active site changes and can no longer bind to the substrate, reducing the rate of reaction beyond the optimum temperature. pH: All enzymes have an optimum pH at which they have their maximum rate of reaction. Increasing pH values are as a result of the addition of hydroxide ions (OH-) while decreasing pH values are as a result of the addition of hydrogen ions (H+). The addition of either OH- or H+ can change the charges of the amino acids that make up the polypeptide chains of the enzyme. This can alter the bonding in the protein and change the shape of the active site resulting in denaturation, or alternatively it can prevent the substrate from binding with the enzyme's active site. Therefore either increasing or decreasing pH from the optimum value results in decreasing rate of enzyme activity. Substrate concentration: When discussing substrate concentration it is assumed that enzyme concentration remains constant. When substrate concentration is increased the rate of enzyme activity increases as the enzyme's active sites are gradually filled through the increased number of collisions with the substrate (due to increased substrate concentration). However, there comes a point when all of the enzyme's active sites are bound with substrates and the enzyme's are working at their maximum rate of reaction. Any increase in substrate concentration beyond this point will not result in any additional increase in rate of enzyme activity.
Views: 145583 Stephanie Castle

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Moof's Medical Biochemistry Video Course: http://moof-university.thinkific.com/courses/medical-biochemistry-for-usmle-step-1-exam Questions Answered in This Video: - What are competitive inhibitors, and what is mechanism by which they act? - How do competitive inhibitors affect the values of KM and VMAX? - How do competitively inhibited reactions look on the hyperbolic graph and Lineweaver-Burk plot? Don't forget to LIKE, COMMENT, and SUBSCRIBE: http://www.youtube.com/subscription_center?add_user=MoofUniversity INFORMATION ABOUT TUTORING: http://www.moofuniversity.com/tutoring/ TO SUPPORT MOOF UNIVERSITY WITH A FINANCIAL CONTRIBUTION: http://www.moofuniversity.com/support-moof/ INSTAGRAM: https://instagram.com/moofuniversity/ FACEBOOK: https://www.facebook.com/pages/Moof-University/1554858934727545 TWITTER: https://twitter.com/moofuniversity Video Content Summary: Competitive inhibitors "compete" with substrate for binding the enzyme's active site. Free enzyme will bind either the substrate OR the competitive inhibitor. It cannot bind both at the same time. If the enzyme binds the substrate, the enzyme-substrate complex forms, and the enzyme can convert the substrate into product. If, however, a competitive inhibitor binds the active site, the enzyme-competitive inhibitor complex forms, which cannot proceed towards products, for the simple reason that the competitive inhibitor impedes the ability of the substrate to bind to the enzyme. Since a competitive inhibitor blocks the substrate from binding, the competitive inhibitor essentially lowers the affinity of enzyme for the substrate. Thus increasing the KM. Despite this, VMAX can still be reached with a high enough substrate concentration. If the substrate concentration is sufficiently high, the substrates will out-compete the competitive inhibitor for binding at the active site, thus effectively overcoming the effects of the inhibitor. The effects of a competitive inhibitor on an enzyme-catalyzed reaction are depicted in a variety of ways in the video, and it is shown what happens to the hyperbolic graph and the Lineweaver-Burk plot, otherwise known as the double reciprocal plot.
Views: 15020 Moof University

05:12
Views: 3474 Shomu's Biology

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Contact me at [email protected] Facebook friend me at https://www.facebook.com/kevin.g.ahern Highlights Enzymes II 1. Steps in an enzyme catalyzed reaction can be summarized as E+S = ES = ES* = EP = E+P 2. We are interested in the rate of formation of product. Consequently, it is important that we measure the rate of formation of product early in a reaction before the product accumulates and the reverse reaction begins to occur significantly. Such a velocity, V0, is determined by ([P]/time. 3. If one lets a reaction go for a long time, it will reach equilibrium. At equilibrium, the relative concentration of products and reactants do not change. Initial velocities of reactions are therefore measured so as to avoid allow the product to accumulate and favor the reverse reaction. 4. Pre-steady state conditions occur very early in a reaction. During these conditions, the concentration of free enzyme and the ES complex are changing rapidly. Trying to measure the rate of product formation under these conditions will not give an accurate rate of formation of product. 5. Instead, we measure V0 under conditions where the concentrations of [E] and [ES] are relatively constant and before product begings to accumulate. These conditions are known as steady state. 6. Measurements in pre-steady state conditions, however, are very helpful for determining mechanism(s) of reaction, among other things. Such conditions are measured using an instrument called a stopped-flow device. 7. Measurements of kinetics of reactions involve widely varying concentrations of substrate - from very low to very high. At very low concentrations, the enzyme is frequently not bound to substrate and therefore idle. Reaction rates at these concnetrations are low. 8. At high substrate concenrations, the enzyme is saturated with substrate - as soon as the product is released, another substrate binds. Thus the enzyme is always active and the reaction rate is at the maximum level - called Vmax. 9. A plot of V0 versus [S] for such a reaction will have a hyperbolic curve for most enzymes, with the curve approaching the theoretical maxium velocity of Vmax. Enzymes yielding a hyperbolic plot for V0 vs. [S] are said to follow Michaelis Menten Kinetics. 10. Vmax is a quantity that depends on the amount of enzyme used to determine it. If you double the amount of enzyme, you can expect to double the Vmax. 11. Km is a measure of the affinity of an enzyme for its substrate. It is determined by first finding Vmax, and then Vmax/2. The substrate concentration required to get a reaction to Vmax/2 is the Km. Km, is therefore a substrate concentration and it is independent of the amount of enzyme used - it is a constant for an enzyme. 12. The value of Km is inversely related to an enzyme's affinity for substrate - Low Km corresponds to high affinity. High Km corresponds to low affinity. 13. Some enzymes are affected by binding to their substrates. These enzymes do not exhibit Michaelis Menten kinetics and yield (typically) a sigmoidal plot for V0 vs [S]. 14. Since Vmax varies with the amount of enzyme used for kinetic measurements, one can obtain a measurement of the reaction that is independent of the enzyme concentration if one simply divides Vmax by the concentration of enzyme used. This yields Kcat. 15. The units of Kcat are 'per time'. The time used usually is seconds. A value of 1000 per second means each enzyme molecule is producing 1000 molecules of product per second. 16. Kcat/Km is a measure of the effectiveness or efficiency of an enzyme. The most efficient enzymes will have a large ratio - high speed, but low concentration needed to get to Vmax/2. 17. The most efficient enzymes approach, but do not exceed, a common maximal value. These enzymes are called perfect enzymes and they typically have reaction rates limited only by the rate with which the substrate can diffuse through water to the active site for reaction. 18. Only a few enzymes are perfect. 19. Cofactors/coenzymes are non-amino acids bound to or used by enzymes to help catalyze reactions. 20. A Lineweaver Burk plot is a kinetic plot using the same data as a V0 vs [S] plot, but the data and the axes it is plotted on are inverted (reciprocals). The plot is also called a double reciprocal plot because it plots 1/V0 vs 1/[S]. Such a plot yields a straight line for an enzyme obeying Michaelis Menten kinetics with the intercecpt of the Y-axis corresponding to 1/Vmax and the intercept of the X-axis corresponding to -1/Km
Views: 2759 Kevin Ahern

17:14
Views: 65733 Shomu's Biology

14:43
using a Lineweaver Burk plot to analyse enzyme data This work is licenced under the Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License. To view a copy of this licence, visit http://creativecommons.org/licenses/by-nc-sa/3.0/ or send a letter to Creative Commons, 171 Second Street, Suite 300, San Francisco, California 94105, USA.
Views: 117422 Peter Klappa

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Views: 134759 AK LECTURES

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In this video I have explained how to calculate Km and Vmax of an enzyme in Lineweaver Burk double reciprocal plot. The Lineweaver–Burk plot was widely used to determine important terms in enzyme kinetics, such as Km and Vmax, before the wide availability of powerful computers and non-linear regression software. The y-intercept of such a graph is equivalent to the inverse of Vmax; the x-intercept of the graph represents −1/Km. It also gives a quick, visual impression of the different forms of enzyme inhibition. You Can Subscribe to my Channel for REGULAR UPDATES by clicking on SUBSCRIBE button above! You can follow me on my BLOG by clicking the link below http://drmungli.blogspot.com/ You can follow my Facebook page Biochemistry Made Easy by Dr Prakash Mungli, MD by clicking the link below. Here I post USMLE step-1 style MCQs and you can participate in discussion. https://www.facebook.com/drmungli
Views: 29375 Dr.Mungli

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Sarina Bang leads you through an exploration of enzyme kinetics, including ways to differentiate reversible competitive inhibitors, irreversible competitive inhibitors, and noncompetitive inhibitors. She also helps you understand Michaelis-Menten kinetics and Lineweaver-Burk plots, as discussed in First Aid for the USMLE Step 1. Learn more about our Express Videos series, featuring over 80 hours of high yield videos to help you learn First Aid for the USMLE Step 1, by visiting https://www2.usmle-rx.com/content/first-aid-step-1-express-video
Views: 524 USMLE-Rx

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Views: 987 Priyanka Jain

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Moof's Medical Biochemistry Video Course: http://moof-university.thinkific.com/courses/medical-biochemistry-for-usmle-step-1-exam Questions Answered in This Video: - What are noncompetitive inhibitors, and what is mechanism by which they act? - How do noncompetitive inhibitors affect the values of KM and VMAX? - How do noncompetitively inhibited reactions look on the hyperbolic graph and Lineweaver-Burk plot? Don't forget to LIKE, COMMENT, and SUBSCRIBE: http://www.youtube.com/subscription_center?add_user=MoofUniversity INFORMATION ABOUT TUTORING: http://www.moofuniversity.com/tutoring/ TO SUPPORT MOOF UNIVERSITY WITH A FINANCIAL CONTRIBUTION: http://www.moofuniversity.com/support-moof/ INSTAGRAM: https://instagram.com/moofuniversity/ FACEBOOK: https://www.facebook.com/pages/Moof-University/1554858934727545 TWITTER: https://twitter.com/moofuniversity Video Content Summary: Noncompetitive inhibitors, unlike competitive inhibitors, do not bind at the active site. They bind a site elsewhere on the enzyme. Thus, they can bind free enzyme OR the enzyme-substrate complex. So, the enzyme-substrate complex, the enzyme-noncompetitive inhibitor complex, and the enzyme-substrate-noncompetitive inhibitor complex can all potentially form. Of course, as long as the noncompetitive inhibitor is bound, the enzyme will not be able to convert substrate into product. Since a noncompetitive inhibitor binds at a site other than the active site and does not have an impact on whether or not substrate can bind, the affinity of enzyme for substrate is not changed, and the KM value remains unchanged. However, the effect of a noncompetitive inhibitor cannot be overcome by increasing substrate concentration, and as long as there are noncompetitive inhibitors present, there will be a smaller number of functional enzymes, relative to an uninhibited case. Less functional enzymes means lower rate of catalysis and a lower VMAX. The effects of a noncompetitive inhibitor on an enzyme-catalyzed reaction are depicted in a variety of ways in the video, and it is shown what happens to the hyperbolic graph and the Lineweaver-Burk plot, otherwise known as the double reciprocal plot.
Views: 9036 Moof University

10:12
Moof's Medical Biochemistry Video Course: http://moof-university.thinkific.com/courses/medical-biochemistry-for-usmle-step-1-exam Questions Answered in This Video: - What are uncompetitive inhibitors, and what is mechanism by which they act? - How do uncompetitive inhibitors affect the values of KM and VMAX? - How do uncompetitively inhibited reactions look on the hyperbolic graph and Lineweaver-Burk plot? Don't forget to LIKE, COMMENT, and SUBSCRIBE: http://www.youtube.com/subscription_center?add_user=MoofUniversity INFORMATION ABOUT TUTORING: http://www.moofuniversity.com/tutoring/ TO SUPPORT MOOF UNIVERSITY WITH A FINANCIAL CONTRIBUTION: http://www.moofuniversity.com/support-moof/ INSTAGRAM: https://instagram.com/moofuniversity/ FACEBOOK: https://www.facebook.com/pages/Moof-University/1554858934727545 TWITTER: https://twitter.com/moofuniversity Video Content Summary: Uncompetitive inhibitors are a bit weird, as they do not bind specifically at the active site of an enzyme in the way a competitive inhibitor does, nor do they specifically bind at a site other than the active site in the way a noncompetitive inhibitor does. Simply, they bind the enzyme-substrate complex exclusively. Specifically, in order for an uncompetitive inhibitor to work, it can only bind the enzyme once the substrate has bound the enzyme. The depiction of this is hopefully clear to some extent in my video. Uncompetitive inhibitors can NOT bind free enzyme. Uncompetitive inhibitors deplete the enzyme-substrate complex by converting it into enzyme-substrate-uncompetitive inhibitor complex, which, of course, cannot proceed towards products. This depletion of the enzyme-substrate complex shifts the free enzyme + substrate reaction towards enzyme-substrate complex, which, in essence, increases the affinity of the enzyme for the substrate, thus decreasing the KM value. In addition, the depletion of the enzyme-substrate complex lowers the number of functional or effective enzyme, thus lowering the VMAX. It is worth noting, as well, that the factors by which both the KM and VMAX change are equal. This is important when considering the effects on the graphs. The effects of an uncompetitive inhibitor on an enzyme-catalyzed reaction are depicted in a variety of ways in the video, and it is shown what happens to the hyperbolic graph and the Lineweaver-Burk plot, otherwise known as the double reciprocal plot.
Views: 11483 Moof University

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Views: 422 mrscolosia

09:08
Enzymes exhibit interesting kinetic profiles. Most of them follow Michaelis-Menten kinetics, which assumes rapid establishment of the substrate binding equilibrium, followed by relatively slow transformation into products (the steady-state appoximation). Learn more in this webcast.
Views: 51281 Michael Evans

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Email me at [email protected] Friend me on Facebook at kevin.g.ahern Highlights Enzymes II 1. To account for the amount of enzyme in a reaction, Kcat (also called turnover number) is used. Kcat is equal to Vmax/[Enzyme]. Because the concentration of enzyme is taken into account in this equation, Kcat does NOT vary with the amount of enzyme used and is therefore a constant for an enzyme. Kcat is equal to the number of molecules of product made per enzyme per unit time. A Kcat of 5/second means that that enzyme makes five molecules of product per molecule of enzyme per second. 2. A very important number that does NOT vary according to the quantity of enzyme used (that is to say that it is a constant for a given enzyme) is the Km (the Michaelis constant). Km turns out to be the concentration of substrate required to get an enzymatic reaction to half maximum velocity. Km is a constant for any given enzyme and provides a measure of an enzyme's "affinity" for its substrate. An enzyme with a high Km has a low affinity for its substrate. An enzyme with a low Km has a high affinity for its substrate. Note that Km is NOT Vmax/2. Instead, it is the substrate concentration required to get a reaction to Vmax/2. 3. The active site of an enzyme is the place on the enzyme where the substrate binds and where the reaction is catalyzed. Temperature affects an enzymatic reaction in two ways. First, temperature is a factor in the Gibbs Free Energy, which determines the energy of a reaction. Second, temperature can denature an enzyme and thus negatively impact an enzyme's activity. 4. A very important number for an enzyme iis the Km (the Michaelis constant). Km turns out to be the concentration of substrate required to get an enzymatic reaction to half maximum velocity. Km provides a measure of an enzyme's "affinity" for its substrate. An enzyme with a high Km has a low affinity for its substrate. An enzyme with a low Km has a high affinity for its substrate. Note that Km is NOT Vmax/2. Instead, it is the substrate concentration required to get a reaction to Vmax/2. 5. A "perfect enzyme" is one that has a maximum value for Kcat/Km. These enzymes have maximaly velocities and minimal Km values (high affinity). Perfect enzymes are slowed only by the rate with which the substrate can diffuse to the active site of the enzyme. 6. Determining Vmax from a plot of V versus S is not easy. Consequently, an alteration of this plot is done to make the calculation simpler. The most common alteration is known as a Lineweaver-Burk (double-reciprocal) plot. In it, a double reciprocal plot is performed - 1/V versus 1/S. When this is plotted for an enzymatic reaction, a line is produced, with the x-intercept (place where the line intersects the x-axis) equaling -1/Km and the y-intercept (place where the line intersects the y-axis) equaling 1/Vmax. 7. Inhibition of enzymes occurs competitively when the inhibitor of the enzyme resembles the substrate and competes with the substrate for binding to the enzyme. Because they are competing with each other for the same site on the enzyme, the substrate can "overcome" the inhibitor at high concentrations (because reactons are set up with a fixed amount of inhibitor). Thus, although it will require more substrate in the presence of an inhibitor to get the same enzyme reaction velocity as when there is no inhibitor, the Vmax of a competitive inhibition is unchanged from the uninhibited reaction. 8. Note above that "it will require more substrate in the presence of an inhibitor to get the same enzyme reaction velocity as when there is no inhibitor" so the Km will change. It requires MORE substrate to get the reaction with the inhibitor to half maximum velocity (compared to uninhibited). Therefore, the Km for a reaction undergoing competitive inhibition INCREASES, meaning the affinity of the enzyme for substrate DECREASES. 9. In non-competitive inhibition, the inhibitor binds to a site on the enzyme that is NOT related to the substrate. Therefore, substrate and inhibitor are NOT competing for the enzyme. This means then that increasing the substrate concentration does NOT affect the inhibitor's ability to inhibit the enzyme. It also means that a fixed percentage of the enzyme is always inactivated by a non-competitive inhibitor. In non-competitive inhibition, the apparent Km is the same as the uninhibited reaction, whereas the Vmax decreases.
Views: 1140 Kevin Ahern

51:51
Contact me at [email protected] Facebook friend me at https://www.facebook.com/kevin.g.ahern Highlights Enzymes II 1. The active site of an enzyme is the place on the enzyme where the substrate binds and where the reaction is catalyzed. Temperature affects an enzymatic reaction in two ways. 1) in the Gibbs Free Energy, which determines the energy of a reaction. 2) temperature can denature an enzyme and thus negatively impact its activity. 2. There are two models for the mechanism of enzyme action. The "lock and key" model proposes that enzymes act like a "lock" that only certain keys (substrates) fit. This model describes the binding of substrates, but is not helpful (or accurate) for the mechanism of catalysis. 3. The "induced fit" model proposes that enzymes change in response to binding of substrate and that change is partly responsible for the catalysis that occurs on the substrate. The induced fit model says that enzymes change substrates and that substrates change enzymes. 4. The "tension" that an enzyme experiences during its binding of a substrate helps to cause the substrate to be modified (reacted). It is important to note that after catalysis occurs, the product is released and the enzyme is returned to its original state. 5. Some enzymes have their ability to catalyze a reaction affected by the presence of another molecule. If that molecule is the substrate, one obtains a sigmoidal plot for V vs [S] like that of hemoglobin binding to oxygen. This type of plot is is evidence that the enzyme's activity is affected by the substrate. When the activity of an enzyme is affected by binding a small molecule, the enzyme is described as allosteric. Allosterism specifically means that binding of a small molecule to an enzyme affects the enzyme's activity. 6. A very important number for an enzyme iis the Km (the Michaelis constant). Km turns out to be the concentration of substrate required to get an enzymatic reaction to half maximum velocity. Km provides a measure of an enzyme's "affinity" for its substrate. An enzyme with a high Km has a low affinity for its substrate. An enzyme with a low Km has a high affinity for its substrate. Note that Km is NOT Vmax/2. Instead, it is the substrate concentration required to get a reaction to Vmax/2. 7. A "perfect enzyme" is one that has a maximum value for Kcat/Km. These enzymes have maximaly velocities and minimal Km values (high affinity). Perfect enzymes are slowed only by the rate with which the substrate can diffuse to the active site of the enzyme. 8. Determining Vmax from a plot of V versus S is not easy. Consequently, an alteration of this plot is done to make the calculation simpler. The most common alteration is known as a Lineweaver-Burk (double-reciprocal) plot. In it, a double reciprocal plot is performed - 1/V versus 1/S. When this is plotted for an enzymatic reaction, a line is produced, with the x-intercept (place where the line intersects the x-axis) equaling -1/Km and the y-intercept (place where the line intersects the y-axis) equaling 1/Vmax. 9. Inhibition of enzymes occurs competitively when the inhibitor of the enzyme resembles the substrate and competes with the substrate for binding to the enzyme. Because they are competing with each other for the same site on the enzyme, the substrate can "overcome" the inhibitor at high concentrations (because reactons are set up with a fixed amount of inhibitor). Thus, although it will require more substrate in the presence of an inhibitor to get the same enzyme reaction velocity as when there is no inhibitor, the Vmax of a competitive inhibition is unchanged from the uninhibited reaction. 10. Note above that "it will require more substrate in the presence of an inhibitor to get the same enzyme reaction velocity as when there is no inhibitor," so the Km will change. It requires MORE substrate to get the reaction with the inhibitor to half maximum velocity (compared to uninhibited). Therefore, the Km for a reaction undergoing competitive inhibition INCREASES, meaning the affinity of the enzyme for substrate DECREASES.
Views: 2109 Kevin Ahern

04:23
This video shows you how to calculate reaction rates with Excel. The enzyme urease is used in this example which catalyzes the hydrolysis of urea to carbon dioxide and ammonia. Details of how reaction rates can be used to calculate the Km and Vmax can be found at https://rahulpatharkar.000webhostapp.com/2018/12/enzyme-kinetics
Views: 24 Rahul Patharkar

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Youtube Channel (Hindi) - https://www.youtube.com/channel/UCmIchtf7_PvcAOfT5p5f6eQ Unacademy - https://unacademy.com/user/Sethi Disclaimer The information provided on this channel is a public service with the understanding that Gate Chemistry makes no warranties, either expressed or implied, related to completeness, accuracy, reliability or suitability of the information. The channel disclaims liability for errors and omissions that may have crept in its content.
Views: 7014 Gate chemistry

01:26
Applications and skills: Deduction and interpretation of graphs of enzyme activity involving changes in substrate concentration, pH and temperature.
Views: 7175 Mike Sugiyama Jones

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Video from Dr Nisha Mishra KCMT
Views: 4622 KCMT College

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Derives the rate expression for an enzyme reaction with a substrate to make a product using the rate-determining step approximation. Made by faculty at the University of Colorado Boulder, Department of Chemical & Biological Engineering. Check out our Kinetics/Reactor Design playlists: https://www.youtube.com/user/LearnChemE/playlists?view=50&flow=list&shelf_id=7 Check out our website for screencasts organized by popular textbooks: http://www.learncheme.com/screencasts/kinetics-reactor-design
Views: 10795 LearnChemE

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Which of the following statements BEST describes the Michaelis-Menton constant, Km? Answer: E) It is numerically equal to the substrate concentration required to reach half maximal velocity for an enzyme-catalyzed reaction Script: Ok question 4, which of the following statements best describe the Michaelis-Menton constant (Km)? SO we drew out this little chart, the one we see in our lecture slides. We have the reaction rate on the Y axis, and solute concentration of the X axis. And so, looking at this chart we see that Km is found by finding Half of Vmax, and find at which point this intersection occurs, then travel down. So Km would be by definition numerically equal to the substrate concentration required to reach half maximal velocity for and enzyme-catalyze do reaction. So let's look at the option we have on the board. Letter A) It is a measure of enzyme efficiency. Now Km itself is not measuring enzyme effiecency. We said its equal to substrate concentration, you know looking at our concentration of substrate, thus has nothing to do with enzyme efficiency. So we can cross that one out. B, it is a measure of rate of catalytic process. So we saw from the chart there is on the Y axis the rate of reaction, but Km is not the rate of catalytic process. So we can scratch that one out. Now C, it has units of concentration. Now Km does have units of concentration (concentration of substrate) but it's not what Km stands for. It is just its units, but it's not what it stands for. So we look at D, It is numerically equal to the affinity between the enzyme and its substrate. Now this is a great answer, but it's not the best answer. So D, is a correlation, the actual equation when we look at the Mechaelis-Menton equation. It does not say that the higher affinity, the less substrate. It is just a correlation we can make off of this chart. So then the last answer is E, which as we said earlier, it is numerically equal to the substrate concentration required to reach half maximal velocity for an enzyme-catalyzed reaction. As we can see this is the exact definition of what Km is, D is a great answer but it is not the best answer because it is a correlation not the direct definition of Km.
Views: 163 Human Biochemistry

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Why Km does not change with the enzyme concentration.
Views: 6462 Peter Klappa

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1. Contact me at [email protected] / Friend me on Facebook (kevin.g.ahern) 2. Download my free biochemistry book at http://biochem.science.oregonstate.edu/biochemistry-free-and-easy 3. Take my free iTunes U course at https://itunes.apple.com/us/course/biochemistry/id556410409 4. Check out my free book for pre-meds at http://biochem.science.oregonstate.edu/biochemistry-free-and-easy 5. Course video channel at http://www.youtube.com/user/oharow/videos?view=1 6. Check out all of my free workshops at http://www.youtube.com/playlist?list=PLlnFrNM93wqyTiCLZKehU1Tp8rNmnOWYB&feature=view_all 7. Check out my Metabolic Melodies at http://www.davincipress.com/metabmelodies.html 8. My courses can be taken for credit (wherever you live) via OSU's ecampus. For details, see http://ecampus.oregonstate.edu/soc/ecatalog/ecourselist.htm?termcode=all&subject=BB 9. Course materials at http://oregonstate.edu/instruct/bb350 Lecture Highlights Highlights Enzymes II 1. A "substrate" is a molecule bound by an enzyme which it catalyzes a reaction upon. Substrates bind specific binding sites on enzymes that resemble their structure. An "active site" of an enzyme is a site on an enzyme where the reaction it catalyzes occurs. 2. There are two models for enzyme action relevant for our consideration. The "lock and key" model proposes that enzymes act like a "lock" that only certain keys (substrates) fit. This model works well for describing the binding of substrates, but is not helpful (or accurate) for describing the mechanism of catalysis. 3. The "induced fit" model of enzyme action proposes that enzymes change in response to binding of substrate and that change is at least partly responsible for the catalysis that occurs on the substrate. Thus, the induced fit model says that enzymes change substrates (by catalysis) and that substrates change enzymes (enabling catalysis). 4. It is important to note that after catalysis occurs, the product is released and the enzyme is returned to its original state. 5. As one increases the amount of substrate for an enzymatic reaction, the velocity of the reaction (concentratioin of product made per time) increases. If one uses more enzyme, one produces a faster velocity. 6. An enzymatic reaction's maximum velocity (Vmax) is the limit (maximum) of a plot of the velocity versus the substrate concentration. Enzymatic reactions reach maximum velocity when the enzyme is saturated with substrate. Plots of enzyme velocities versus substrate concentration are called hyperbolic. 7. Some enzymes have their ability to catalyze a reaction affected by the presence of another molecule. If that molecule is the substrate, one obtains a sigmoidal plot like that of hemoglobin binding to oxygen. This type of plot is is evidence that the enzyme's activity is affected by the substrate. When the activity of an enzyme is affected by binding a small molecule, the enzyme is described as allosteric. Allosterism specifically means that binding of a small molecule to an enzyme affects the enzyme's activity. 8. A very important number that does NOT vary according to the quantity of enzyme used (that is to say that it is a constant for a given enzyme) is the Km (the Michaelis constant). Km turns out to be the concentration of substrate required to get an enzymatic reaction to half maximum velocity (slide 12). Km is a constant for any given enzyme and provides a measure of an enzyme's "affinity" for its substrate. An enzyme with a high Km has a low affinity for its substrate. An enzyme with a low Km has a high affinity for its substrate. Note that Km is NOT Vmax/2. Instead, it is the substrate concentration required to get a reaction to Vmax/2. 9. Another important parameter of enzymes is called Kcat (also called turnover number). Kcat is equal to Vmax/[Enzyme]. Because the concentration of enzyme is taken into account in this equation, Kcat does NOT vary with the amount of enzyme used and is therefore a constant for an enzyme. Kcat is equal to the number of molecules of product made per enzyme per unit time. A Kcat of 5/second means that that enzyme makes five molecules of product per molecule of enzyme per second. 10. Determining Vmax from a plot of V versus S is not easy. Consequently, an alteration of this plot is done to make the calculation simpler. The most common alteration is known as a Lineweaver-Burk (double-reciprocal) plot. In it, a double reciprocal plot is performed - 1/V versus 1/S. When this is plotted for an enzymatic reaction, a line is produced, with the x-intercept (place where the line intersects the x-axis) equaling -1/Km and the y-intercept (place where the line intersects the y-axis) equaling 1/Vmax.
Views: 5326 Kevin Ahern

01:05:51
MIT 5.07SC Biological Chemistry, Fall 2013 View the complete course: http://ocw.mit.edu/5-07SCF13 Instructor: JoAnne Stubbe In this classroom lecture, Professor Stubbe focuses on enzymes as catalysts. She describes the theory and mechanics of catalysis and explains why enzymes are so important. License: Creative Commons BY-NC-SA More information at http://ocw.mit.edu/terms More courses at http://ocw.mit.edu
Views: 25565 MIT OpenCourseWare

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Views: 11460 Benjamin Himme

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A short video discussing the Michaelis-Menten enzyme reaction mechanism and its properties.
Views: 12405 Pedro Mendes

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Minilecture on how to generate a Michaelis Menten Graph and Enzyme Inhibition
Views: 8210 Linda Columbus