# Investigating the Effect of Increasing Exercise in Aerobic Respiration

by sdurmus (student)

Sercan Durmuş 10/A 71

Investigating the Effect of Increasing Exercise in Aerobic Respiration

Research Question:

How does increasing exercise affect the amount of CO2 produced by the reaction of human cells in aerobic respiration by keeping the solutions in the flasks and the time for blowing the air into the flasks same for all the trials and using the same person for the whole experiment?

Hypothesis:

If the time period of an exercise is increased, the enzymes present in human cells will work more to produce more energy needed to complete the exercise. Therefore, the amount of CO2 produced by human cells will increase as a byproduct from the aerobic respiration.

Variables:

Independent: different time periods for the exercise (resting, 1 minute and 2 minutes of exercise) calculated by a stopwatch.

Dependent: the amount of CO2 released as a byproduct of aerobic respiration that is measured by adding NaOH solution to the solution in the flask

Controlled: the person, the solution for each flask, the time period for blowing air into the flasks for each trial, temperatures of solutions, atmosphere, room temperature, the place that is used for doing exercise

Materials: …

Procedure: …

Results:

Data Collection:

Note: To convert the NaOH added value to CO2 produced, the value from NaOH should be multiplied by 10. For example, for the Flask #1; 64 ml x 10 = 640 μmol, for Flask #2; 68 ml x 10 = 680 μmol

Table 1: The results for the effect of 2 minutes of exercise on CO2 production

Table 2: The results for the effect of 1 minute of exercise on CO2 production

Calculations:

Table 3: Average values for different conditions

The calculations for the new uncertainty value:

The formula for calculating the average uncertainty is

1/ (sqrt (the number of terms))*uncertainty” (Formula)

Before Exercise:

1/ (sqrt (5))*10

=4.47 μmol

After 1 minute of exercise:

1/ (sqrt (2))*10

=7.07 μmol

After 2 minutes of exercise:

1/ (sqrt (3))*10

=5.77 μmol

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