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Recommended: Catalase and enzyme activity
Experimental Design Research Question: Does the rate of reaction (mL/sec) for the enzyme catalase differ in endotherms versus ectotherms? Alternative Hypothesis 1: If the enzyme catalase reacts, then the rate of reaction for the endotherms will have a consistent reaction rate because the endotherms produce more heat for the enzyme to operate efficiently. Alternative Hypothesis 2: If the enzyme catalase reacts, then the rate of reaction for the ectotherms will have a lower reaction rate than the endotherms because the ectotherm body temperature rises and falls to the temperature of the environment. Null Hypothesis: If the enzyme catalase reacts, then there is no statistical difference between the rate of reaction of endotherms and ectotherms. …show more content…
If the p-value is greater than .05, then the null hypothesis would be rejected. The null hypothesis was rejected because the p-value was less than 0.05, it was 0.02. The trends in the data that occurred was the averages of the Chicken Liver (Endotherm) had a higher rate of reaction with 7.90 mL/sec, while the Perch Liver (Ectotherm) had a lower rate of reaction with 6.05 mL/sec. Therefore, the endotherm had higher data points, while the ectotherm had lower data points showing that the endotherm rate of reaction is faster. Variance the measure of variability, which is the spread of the data. The variance for Chicken Liver (Endotherm) is 9.11 mL/sec and Perch Liver (Ectotherm) is 2.14 mL/sec. Since variation shows how close the data points compare to the mean, the endotherm has a larger difference between the variance and the mean than the ectotherm, which shows that the endotherm has less precision in the number set. The trends in the standard deviation of the Chicken Liver (Endotherm) is 3.82 mL/sec and the Perch Liver (Ectotherm) is 1.46 mL/sec. A high standard deviation shows that the data is less reliable since the data is widely spread and a low standard deviation shows that the data are more reliable since the data is closer to the mean. …show more content…
The null hypothesis is rejected since the p-value fell below .05, with .02 therefore there is a statistical difference between the rate of reaction of an enzyme in an ectotherm and endotherm. The data supports the claim since the ectoderm has a lower reaction rate than the endotherms because the ectotherm body temperature rises and falls to the temperature of the environment (Ecto- versus endothermic organisms, 2016). The reason for this is because the endotherm had a higher rate of reaction with 7.90 mL/sec, while ectotherm had a lower rate of reaction with 6.05 mL/sec. Therefore, the endotherm had higher data points, while the ectotherm had lower data points showing that the endotherm rate of reaction is faster. The alternate hypothesis number one was rejected. The alternate hypothesis number one was “If the enzyme reacts, then the rate of reaction for the endotherms will have a consistent reaction rate because the endotherms produce more heat for the enzyme to operate efficiently.” Although the endotherms do produce more heat for the enzyme to react, the rate of reaction does not remain consistent with the same reaction rate throughout the experiment. The average rate of reaction for the endotherms was 7.90 mL/sec and the endotherm had a higher standard deviation which shows it less reliable since the data points are spread unevenly. The standard
This evidence alone suggests that higher increases in substrate concentration causes smaller and smaller increases in enzyme activity. As substrate concentration increases further, some substrate molecules may have to wait for an active site to become empty as they are already occupied with a substrate molecule. So, the rate of the reaction starts to level off resulting in a plateau in the graphs. This means that the reaction is already working at its maximum rate, and will continue working at that rate until all substrates are broken down. The only way the reaction rate would increase, is if more enzyme was added to the solution. This confirms that increases in substrate concentration above the optimum does not lead to greater enzyme activity. Therefore, the rate of reaction is in proportion to the substrate
Is there a difference in the rate of reaction of catalase activity between pinto beans and carrots? Based on our research, we believe that the catalase activity in pinto beans will increase more relative to the catalase activity in whole carrots because pinto beans are higher in protein. We conducted an experiment to test our hypothesis that if we increase the hydrogen peroxide concentration then we will see higher kinetic saturation in pinto beans over whole
In the following experiment, we will attempt to examine the relationship between metabolic rate and environmental temperature in both an ectoderm and an endotherm. I predict that for the ectotherm, the metabolic rate will increase as the outside environment temperature will increase. I also predict that the metabolic rate in the endotherm will remain relatively the same as the outside environment temperature changes. I also make the prediction that the ectotherm will have much lower metabolic rates than the endotherm.
The rate law determines how the speed of a reaction occurs, thus allowing the study of the overall mechanism formation in reactions. In the general form of the rate law, it is A + B C or r=k[A]x[B]y. The rate of reaction can be affected by the concentrations such as A and B in the previous equation, order of reactions, and the rate constant with each species in an overall chemical reaction. As a result, the rate law must be determined experimentally. In general, in a multi-step reaction, there will be one reaction that is slower than the others.
With this information we were able to identify any patterns and similarities. Hypothesis: The higher the temperature of water, potato and H²O², the rate at which the Enzyme will work will be faster therefore producing more oxygen. The reaction will be the same without the catalase (potato). Therefore in both experiments the Enzyme will work more rapidly and produce more oxygen. Aim: To test the hypothesis.
The metabolic rate for the cricket and the cockroach will be different when physical stress and temperature changes are present.The null hypothesis is that the cricket’s metabolic rate will be similar to (Blaptica dubia) cockroach’s rate when physical stress and temperatures changes are present.
Note: Since the P-value for the Equality of Variances is above .05, the Pooled method, or Equal variances is used to compute the t-value. If the P-value had been less than .05, then the Satterthwaite method would have been used to compute the t-value.
Over the observed fifty seconds, there was a consistency among the temperatures. Without a calculated percent error, we are able to assume the average temperature was twenty-six degrees Celsius. There are factors that could have caused error to arise in our data collection. One factor could be that the temperature of the room was not consistent throughout the room. Another factor may have been the performance of the thermometer. The grasp in which the thermometer was held for procedure B may also be a factor.
Investigating the Rate of Reaction between Amylase and Starch. Plan Aim: To be able to The aim of this investigation is to find out whether the volume of amylase affects the rate of reaction between amylase and starch. Prediction: I predict that the greater the volume of amylase then the faster the rate of reaction between the starch and amylase. I predict this because of the lock and key hypothesis.
Objective: The objective of the experiment is to determine what factors cause a change in speed of a reaction. It is also to decide if the change is correlated with the balanced equation of the reaction and, therefore, predictable. To obtain a reaction, permanganate, MnO_4^(1-), must be reduced by oxalic acid, C_2 O_4 H_2. The balanced equation for the reaction is:
- Science Buddies “Carbonation Countdown: The Effect of Temperature on Reaction Time” http://www.scientificamerican.com 29 Aug. 2013. 9 March 2014
The results of this experiment showed a specific pattern. As the temperature increased, the absorbance recorded by the spectrophotometer increased indicating that the activity of peroxidase enzyme has increased.At 4C the absorbance was low indicating a low peroxidase activity or reaction rate. At 23C the absorbance increased indicating an increase in peroxidase activity. At 32C the absorbance reached its maximum indicating that peroxidase activity reached its highest value and so 32 C could be considered as the optimum temperature of peroxidase enzyme. Yet as the temperature increased up to 60C, the absorbance decreased greatly indicating that peroxidase activity has decreased. This happened because at low temperature such as 4 C the kinetic energy of both enzyme and substrate molecules was low so they moved very slowly, collided less frequently and formed less enzyme-substrate complexes and so little or no products. Yet, at 23 C, as the temperature increased, enzyme and substrate molecules
In this lab, it was determined how the rate of an enzyme-catalyzed reaction is affected by physical factors such as enzyme concentration, temperature, and substrate concentration affect. The question of what factors influence enzyme activity can be answered by the results of peroxidase activity and its relation to temperature and whether or not hydroxylamine causes a reaction change with enzyme activity. An enzyme is a protein produced by a living organism that serves as a biological catalyst. A catalyst is a substance that speeds up the rate of a chemical reaction and does so by lowering the activation energy of a reaction. With that energy reactants are brought together so that products can be formed.
If the temperature had been increased in batches of 5oC it would have provided a greater range of results to work with, which would create a high level of data to work with when compiling the mean and plotting the graphs as evidence.
Abstract: Enzymes are catalysts therefore we can state that they work to start a reaction or speed it up. The chemical transformed due to the enzyme (catalase) is known as the substrate. In this lab the chemical used was hydrogen peroxide because it can be broken down by catalase. The substrate in this lab would be hydrogen peroxide and the enzymes used will be catalase which is found in both potatoes and liver. This substrate will fill the active sites on the enzyme and the reaction will vary based on the concentration of both and the different factors in the experiment. Students placed either liver or potatoes in test tubes with the substrate and observed them at different temperatures as well as with different concentrations of the substrate. Upon reviewing observations, it can be concluded that liver contains the greater amount of catalase as its rates of reaction were greater than that of the potato.