Biology: Enzyme Lab Data: Rate, pH, and Temperature
Analyze enzyme reaction rates using lab data
Biology: Enzyme Lab Data: Rate, pH, and Temperature
Analyze enzyme reaction rates using lab data
Biology - Grade 9-12
- 1
In a catalase lab, students measured oxygen production from hydrogen peroxide. The reaction produced 12 mL of oxygen in 3 minutes. Calculate the average reaction rate in mL per minute.
Rate equals amount of product formed divided by time.
The average reaction rate is 4 mL per minute because 12 mL divided by 3 minutes equals 4 mL/min. - 2
A reaction produced 0 mL of oxygen at 0 minutes, 5 mL at 1 minute, 9 mL at 2 minutes, and 12 mL at 3 minutes. During which one-minute interval was the reaction fastest?
The reaction was fastest from 0 to 1 minute because oxygen increased by 5 mL during that interval, which is more than the increases in later intervals. - 3
A student tested an enzyme at different pH values. The rates were: pH 3 = 1 unit/min, pH 5 = 4 units/min, pH 7 = 9 units/min, pH 9 = 5 units/min, and pH 11 = 1 unit/min. Identify the enzyme's optimum pH.
The optimum condition is the one with the highest reaction rate.
The enzyme's optimum pH is pH 7 because that is where the reaction rate is highest at 9 units per minute. - 4
Using the pH data from the previous problem, explain why the enzyme activity decreased at pH 3 and pH 11.
Think about how pH affects protein structure.
The enzyme activity decreased at pH 3 and pH 11 because very acidic or very basic conditions can change the shape of the enzyme's active site. When the active site changes shape, the substrate does not bind as well. - 5
A student tested enzyme activity at different temperatures. The rates were: 10°C = 2 units/min, 20°C = 5 units/min, 30°C = 9 units/min, 40°C = 12 units/min, 50°C = 4 units/min, and 60°C = 0 units/min. What temperature produced the highest enzyme activity?
The highest enzyme activity occurred at 40°C because the reaction rate was greatest there at 12 units per minute. - 6
Using the temperature data from the previous problem, explain why the rate increased from 10°C to 40°C.
Before an enzyme denatures, warmer temperatures usually increase molecular motion.
The rate increased from 10°C to 40°C because higher temperature made enzyme and substrate molecules move faster. This caused more frequent successful collisions up to the enzyme's optimum temperature. - 7
Using the temperature data from problem 5, explain why the rate dropped sharply at 50°C and reached 0 units/min at 60°C.
The rate dropped sharply at 50°C and reached 0 units per minute at 60°C because high temperatures can denature the enzyme. Denaturation changes the enzyme's shape so the active site no longer fits the substrate. - 8
Two groups measured catalase activity at pH 7. Group A produced 18 mL oxygen in 3 minutes. Group B produced 24 mL oxygen in 4 minutes. Which group had the higher average reaction rate?
Calculate each rate before comparing the groups.
Both groups had the same average reaction rate. Group A had a rate of 6 mL/min because 18 divided by 3 equals 6, and Group B had a rate of 6 mL/min because 24 divided by 4 equals 6. - 9
A lab report says, "The enzyme worked better at pH 8 than at pH 6." What specific type of data would best support this claim?
The best support would be quantitative reaction rate data showing a higher rate at pH 8 than at pH 6. For example, the report could compare product formed per minute at each pH. - 10
A student measured starch digestion by amylase using iodine color. The solution turned from dark blue-black to light brown in 2 minutes at pH 7 and in 8 minutes at pH 4. Which pH had the faster enzyme activity, and how do you know?
When the same endpoint is reached, less time means a faster rate.
pH 7 had the faster enzyme activity because the starch disappeared in less time. A shorter time to reach the endpoint means a faster reaction rate. - 11
A student collected the following data for product formation: 0 seconds = 0 mg, 20 seconds = 6 mg, 40 seconds = 11 mg, 60 seconds = 15 mg. Calculate the average reaction rate from 0 to 60 seconds in mg per second.
Use the total change in product over the total time.
The average reaction rate from 0 to 60 seconds is 0.25 mg per second because 15 mg divided by 60 seconds equals 0.25 mg/s. - 12
Using the data in problem 11, calculate the reaction rate from 20 seconds to 40 seconds.
The reaction rate from 20 seconds to 40 seconds is 0.25 mg per second because product increased from 6 mg to 11 mg, a change of 5 mg, over 20 seconds. 5 divided by 20 equals 0.25 mg/s. - 13
A student wants to test how temperature affects an enzyme. Identify the independent variable, dependent variable, and two controlled variables for this experiment.
The independent variable is what the student changes, and the dependent variable is what the student measures.
The independent variable is temperature. The dependent variable is enzyme activity or reaction rate. Two controlled variables could be pH and enzyme concentration, although substrate concentration, reaction time, and total volume should also be kept constant. - 14
A data table shows that an enzyme has high activity at pH 2, low activity at pH 7, and no activity at pH 10. Based on this pattern, what type of environment might this enzyme normally work in?
This enzyme might normally work in a very acidic environment because it has high activity at pH 2. An example could be an enzyme that functions in the stomach. - 15
A graph of enzyme activity versus substrate concentration rises quickly at low substrate concentrations and then levels off at high substrate concentrations. Explain why the graph levels off.
Think about the limited number of enzyme active sites.
The graph levels off because most or all enzyme active sites become occupied by substrate at high substrate concentrations. Once the enzyme is saturated, adding more substrate does not greatly increase the reaction rate.