Alkaline Phosphatase And Vmax Lab

In this lab or experiment, the aim was to determine the following factors of enzymes: (1) the effects of enzymes concentration the catalytic rate or the rate of the reaction, (2) the effects of pH on a particular enzyme, an enzyme known and referred throughout this experiment as ALP (alkaline phosphate enzyme) and lastly (3) the effects of various temperatures on the reaction or catalytic rate. Throughout the experiment 8 separate cuvettes and tubes are mixed with various solutions (labeled as tables 1,3 & 4 in the apparatus/materials sections of the lab) and tested for the effects of the factors mentioned above (concentration, pH and temperature). The tubes labeled 1-4 are tested for pH with pH paper and by spectrophotometer, cuvettes 1a-4a was tested for concentration and cuvettes labeled 1b-4b was tested for temperature in four different atmospheric conditions (4ºC, 23ºC, 32ºC and 60ºC) to see how the enzyme solution was affected by the various conditions. After carrying out the procedures the results showed that the experiment followed the theory for the most part, which is that all the factors work best at its optimum level. So, the optimum pH that the enzymes reacted at was a pH of 7 (neutral), the optimum temperature that the reactions occurs with the enzymes is a temperature of 4ºC or

In the experiment we used Turnip, Hydrogen Peroxide, Distilled Water, and Guaiacol as my substances. On the first activity, Effect of Enzyme concentration of Reaction Rate for low enzyme concentration, we tested three concentrations of the turnip extract, and hydrogen peroxide. For the Turnip Extract I used 0.5 ml, 1.0 ml, and 2.0 ml. For hydrogen peroxide we used 0.1 ml, 0.2 ml, and 0.4 ml. We used a control to see the standard, and used a control for each enzyme concentration used. The control contains turnip extract and the color reagent, Guaiacol. We prepared my substrate tubes separately from the enzyme tubes. My substrate tube

The practical was carried out to investigate the effect of pH on the reaction of the enzyme acid phosphatase.

From Figure 5, Vmax = 0.0172 and Km = 0.3144. At ½ Vmax (Figure 4), the rate of reaction increases steadily with an increase in substrate concentration. At this point, the rate of product formation is partially limited by substrate availability. As the curve approaches Vmax, the enzyme catalyzed saturation increases. In effect, the enzyme activity increases the rate of product formation up to a point whereby the reaction rate remains constant as more substrate is added (saturation).

Lab six requires students to observe the effects of pH and enzyme concentration on catecholase activity. Enzymes are organic catalysts that can affect the rate of a chemical reaction depending on the pH level and the concentration of the enzyme. As pH comes closer to a neutral pH the enzyme is at its greatest effectiveness. Also at the absorbance of a slope of 0.0122 the enzyme is affected greatly. The pH effect on enzymes can be tested by trying each pH level with a pH buffer of the same pH as labeled as the test tube and 1mL of potato juice, water, and catechol. This is all mixed together and put in the spectrophotometer to test how much is being absorbed at 420nm. As the effect on enzyme concentration can be tested almost the same way. This part of the exercise uses different amounts of pH 7-phosphate buffer and potato juice, and 1mL of catechol mixed together in a test tube. Each substance is put in the spectrophotometer at a wavelength set tot 420nm. The results are put down for every minute up to six minutes to see how enzyme concentration affects reaction rate. The results show that the pH 8 (0.494) affects the enzyme more than a pH of 4 (0.249), 6 (0.371), 7 (0.456), and 10 (0.126). Also the absorbance is greatest at a slope of 0.0122 with test tube C that has more effect on the reaction rate, than test tube A, B, and D.

In the first part of the enzyme lab, we mixed a substrate and an indicator with an enzyme. There was also a neutral buffer in each of the chemical mixtures. The neutral buffer regulated the pH to around 7. We got a color palette and once we mixed each together, we observed and saw a change in the color of the substance. The darker and more brown the substance got, the more oxygen produced by the reaction. Our results showed that amount of oxygen produced increased about 10% a minute until it sort of equilibrated at 4 minutes and didn’t change to the fifth minute mark. If we were to change anything we did in the experiment, we would make our comparisons to the chart more precise. Overall we thought it

In both experiments, there was one environment that the enzyme worked best it. The graphs show that in the salt experiment 4% salt concentration was the optimum environment for the enzyme, and pH 6 buffer solution was the optimum pH environment for the enzyme. In the other environments the enzyme did not work as fast, therefore not creating as much product. The hypothesis was that the enzyme would work the most efficiently in the most neutral environment. The data does support this hypothesis; therefore we would accept the hypothesis. The neutral state for pH is pH 7, so the enzyme would be most active at exactly pH 7. With pH 6 buffer solution, the enzyme is working, but after pH 7, the enzyme begins to denature at becomes less effective. For the salt experiment, the hypothesis was that the enzyme would be more productive in the solution that had the least salt concentration. As the salt concentration increase, the enzyme would be being to denature. In this experiment, the hypothesis was incorrect, so it had to be

Where kn represents the individual rate constants, E the enzyme, S the substrate, and P the product. When almost none of the product reverts to substrate the rate of the product formation i.e. the catalytic rate, V, is equal to k3 x [ES]. V is defined as the number of moles of product formed per second. Reaction rates are described by the Michaelis-Menton equation, shown in equation 4 (Boon, 2007).

First, it was downloaded the spreadsheet from eLearn provided by the instructor titled as “Experiment I data”. The first thing, it was calculated the slope for each Time point and Response set of data. It was noticed that inhibitor concentration [I] was constant, and substrate [S] concentration was increasing. So, it was used the Response vs. Time data to allowed to measure the initial velocity [Vo] of the enzyme converting substrate to product. It was typed the formula “=slope(“ in the shift cell called “slope”. When it was typed the formula, it was selected the Y-values which is the Response column dragged down until the last response. Then, put a comma to separate the X-values which is the Time column followed by a closed parenthesis. Since, slope should be calculated for all of the others, it can be save time by doing

The purpose of this lab experiment was to determine the relationship between temperature and the rate of enzymatic activity in yeast cells. In the lab, the temperature was the independent variable. The temperatures consisted of 6°C, 24°C, 34°C, 46°C. The dependent variable in the experiment was the rate of enzymatic activity in yeast cells. The temperatures were tested by using a LabQuest and pressure probe that tested pressure inside the plastic test tube. LabQuest graphed the data and created a line of best fit that was used to determine the slope of the graph. The slope of the graph represents the rate of enzymatic activity. The slope was found for each temperature in 2 different trials. Then, the rate of enzyme activity (kPa/sec) for each temperature in the 2 trials were averaged. These averages were used to develop a graph that shows the relationship between temperature and the rate of enzyme activity. According to the results of the experiment, as the temperature increases, the rate of enzymatic activity decreases. Each enzyme has its own optimal temperature in which it can function efficiently.

Since the catalytic activity of ATPase seems to follow first order kinetics, the rate constant in s-1 for ATPase activity can be ascertained from the activity assay and then the following relationship can be used with known molecular weight of the ATPase enzyme in mg mmol-1 to calculate Vmax with the units of µmol mg-1 min-1:

An enzyme is a protein that acts as a catalyst which reduces the activation energy needed for a chemical reaction. Without the presence of enzyme, cell reactions would take so long that they would detectable. During a reaction, in the presence of an enzyme, the substrate first creates a complex with the enzyme. While the substrate is a part of the complex, it’s converted into the product. Then, finally, the complex dissociates from the molecule which allows the release of the enzyme and formed product. An enzyme’s activity depends on a variety of conditions which includes the pH level and temperatures. Phosphorylase is an enzyme that catalyze the addition of a

For this experiment we determine the Michaelis constant (Km) and the maximum velocity (Vmax) and inhibition of alkaline phosphate. Using the different graph created and different methods of calculation I was able to determine the Km and Vmax for the different mixtures, and also what type of inhibitor it was. For mixtures 1-5, the Vmax was 4.04 and the Km was 0.019. For mixtures 6-10, the Vmax was 3.82 and the Km was 0.023. Using the slope from the graph of Lineweaver-Burk for reactions 6-10 I was able to determine Ki, which was 171.53, by calculating the [I] using C1V1=C2V2 scheme, which was 0.012. By plotting in the same graph the Lineweaver-Burk graph for reaction 1-5 and 6-10, I was able to determine that it was competitive inhibitor and

When the substrate concentration has been increased, there are more substrate molecules that collide with the enzyme molecules (RSC, section 4). When this happens, more of the product will be formed (RSC, section 4). When a pH change occurs, there will also be an effect in the enzyme. When the pH changes from the optimal pH level to higher or below the level, there will be a decrease in the reaction rate (RSC, section 4). This occurs because there will be more enzyme molecules that have active sites that are not fitting with the shape of the substrate (RSC, section 4).

The hypothesis is as the substrate concentration has an increase so will the reaction of velocity if the amount of enzyme is kept constant.