Aerobic cellular respiration is the release of energy from organic compound from organic compounds by metabolic chemical oxidation in the mitochondria within each cell. Cellular respiration involves a series of enzyme-mediated reactions. The equation below shows the complete oxidation of glucose. Oxygen is required for this energy-releasing process to occur.
C6H12O6 + 6O2 -> 6CO2 + 6H2O + 686 kilocalories of energy/mole of glucose oxidized PURPOSE This lab provided process of cellular respiration and how it is affected by temperature in both germinating and dormant pea seeds. Cellular respiration is an ATP-producing catabolic process in which the electron receiver is an inorganic molecule. It is the release of energy from organic compounds by chemical oxidation in the mitochondria within each cell.Carbohydrates, proteins, and fats can all be metabolized, but cellular respiration usually involves glucose: C6H12O6 + 6O2 > 6CO2 + 6H2O + 686 Kcal of energy/mole of glucose oxidized. Cellular respiration involves glycolysis, the Krebs cycle, and the electron transport chain. Glycolysis is a catabolic pathway that occurs in the cytosol and partially oxidizes glucose into two pyruvate (3-C). The Krebs cycle occurs in the mitochondria and breaks down a pyruvate (Acetyl-CoA) into carbon dioxide.These two cycles both produce a small amount of ATP by substrate-level phosphorylation and NADH by transferring electrons from substrate to NAD+.
The Krebs cycle also produces FADH2 by transferring electrons to FAD. The electron transport chain is located at the inner membrane of the mitochondria and accepts energized electrons from enzymes that are collected during glycolysis and the Krebs cycle, and couples this exergonic slide of electrons to ATP synthesis or oxidative phosphorylation.This process produces most of the ATP. Cellular respiration can be measured in two ways: the consumption of O2 (how many moles of O2 are consumed in cellular respiration) and production of CO2 (how many moles of CO2 are produced in cellular respiration). PV = nRT is the formula for the inert gas law, where P is the pressure of the gas, V is the volume of the gas, n is the number of molecules of gas, R is the gas constant, and T is the temperature of the gas in degrees K. This law shows several important things about gases.If temperature and pressure are kept constant then the volume of the gas is directly proportional to the number of molecules of the gas.
If the temperature and volume remain constant, then the pressure of the gas changes in direct proportion to the number of molecules of gas. If the number of gas molecules and the temperature remain constant, then the pressure is inversely proportional to the volume. If the temperature changes and the number of gas molecules is kept constant, then either pressure or volume or both will change in direct proportion to the temperature.Materials Materials are necessary for the lab: 2 thermometers, 2 shallow baths, tap water, ice, paper towels, masking tape, germinating peas, non-germinating (dry) peas, glass beads, 100 mL graduated cylinder, 6 vials, 6 rubber stoppers, absorbent and non- absorbent cotton, KOH, a 5-mL pipette, silicon glue, paper, pencil, a timer, and 6 washers.Hypothesis The respirometer with only germinating peas will consume the largest amount of oxygen and will convert the largest amount of CO2 into K2CO3 than the respirometers with beads and dry peas and with beads alone. The temperature of the water baths directly effects the rate of oxygen consumption by the contents in the respirometers (the higher the temperature, the higher the rate of consumption).Temp (°C)| Time (min)| Beads Alone| Germinating Peas| Dry Peas & Beads Alone| | | Reading at time x| Difference| Reading at Time x| Difference| Corrected Difference | Reading at time x| Difference| Corrected Difference | 21°C| 0| .
9mL| 0| . 9mL| 0| 0| . 9| 0| 0| 21°C| 5| . 9mL| 0| . 85mL| .
05| . 05| . 87| . 02| . 02| 21°C| 10| .
95mL| . 05| . 82mL| . 08| . 13| . 87| . 02| . 07| 21°C| 15| .
95mL| . 05| . 79mL| . 11| .
16| . 86| . 03| . 08| 21°C| 20| . 95mL| .
05| . 74mL| . 16| . 21| . 86| . 03| . 08| 10°C| 0| .
95mL| 0| . 92mL| 0| 0| . 91mL| 0| 0| 10°C| 5| . 94mL| . 01| . 88mL| . 04| .
03| . 90mL| . 1| 0| 10°C| 10| . 92mL| . 03| . 85mL| . 07| .
04| . 87mL| . 04| . 01| 10°C| 15| . 93mL| .02| . 83mL| . 09| .
07| . 86mL| . 05| .
03| 10°C| 20| . 93mL| . 02| .
83mL| . 12| . 10| . 85mL| . 06| . 04| Results 1. In this activity, you are investigating both the effect of germination versus non-germination and warm temperature versus cold temperature on respiration rate.
Two hypothesis. Germinating peas should consume more oxygen than non-germinating peas. Peas germinating at warm temperatures should consume more oxygen than peas germinating at cold temperatures 2. This activity uses a number of controls.
What conditions must be kept constant?.? Water baths held at constant temperature? Volume of KOH is the equal in every tube? Equilibration time is identical for all respirometers. 3. Describe and explain the relationship between the amount of oxygen consumed and time? The amount of oxygen consumed was greatest in germinating peas in warm water. The oxygen consumption increased over time in germinating peas. 4. Condition| Show Calculations| Rate in mL O / minute| Germinating peas at 10oC| 2.
3-1. 5=. 8/5| . 16mL O2 /minute| Germinating peas at room temperature| 4.
6-3. 1/5| . 3mL O2 /minute| Dry peas at 10oC| (. 1)/5=| . 2 mL O2 /minute| Dry peas at room temperature| (. 2-0 )/5=| . 04 mL O 2 /minute| 6. Why is it necessary to correct the readings from the peas with the readings from the beads? The beads carried out no cellular respiration.
The peas did. Changes in atmospheric pressure could have caused changes in respiration rate and correcting the readings provided the most accurate results under the given conditions. 7. Explain the effect of germination versus non-germination on pea seed respiration. Germination causes a higher rate of respiration than the non-germinating peas.
9. What is the purpose of KOH in this experiment?KOH removes carbon dioxide formed during cellular respiration. 10. Why did the vial have to be completely sealed around the stopper. The stopper was completely sealed to prevent water from entering the respirometer. 11.
If you used the same experimental design to compare the rates of respiration of a 25g. reptile and a 25 g. mammal at 10oC what results would you expect? Explain your reasoning. The mammal would carry out a higher rate of cellular respiration.
This is because the mammal maintains a constant temperature that is higher than the temperature of the cold blooded reptiles that will have a temperature of 10 C. 2. If respiration in a small mammal were studied at both room temperature 21 o C and 10oC what results would you predict? Explain your reasoning. The rate of cellular respiration would be higher at 21 degrees C because the 10 degrees C temperature could cause the overall body of the mammal temperature to drop the most. 13. Explain why water moved into the respirometers’ pipettes. Water moved into the pipettes because oxygen was being consumed and allowed water to move only partially into the pipette. 14.
Design an experiment to examine the rates of cellular respiration in peas that have been germinating for 0, 24, 48, and 72 hours. What results would you expect? Why? I would use the same format using respirometers to measure the cellular respiration rate of the peas. The peas that had been germinating for 72 hours would have a higher respiration rate because they have a higher energy demand.
Conclusion: The lab demonstrated many important things relating to cellular respiration. It showed that the rates of cellular respiration are greater in germinating peas than in non-germinating peas.It also showed that temperature and respiration rates are directly proportional; as temperature increases, respiration rates increase as well. Because of this fact, the respirometers placed in the water at 10 oC displayed a lower rate of cellular respiration than the respirometers placed in the room temperature water.
The non-germinating peas consumed far less oxygen than the germinating peas. This is because, though germinating and non-germinating peas are both alive, germinating peas require a larger amount of oxygen to be consumed so that the seed will continue to grow and survive.In the lab, CO2 made during cellular respiration was removed by the potassium hydroxide (KOH) and created potassium carbonate (K2CO3). It was necessary that the carbon dioxide be removed so that the change in the volume of gas in the respirometer was directly proportional to the amount of oxygen that was consumed. The result was a decrease in gas volume within the tube, and a related decrease in pressure in the tube. The respirometer with just the glass beads served as a control group that did not undergo cellular respiration.
Numerous errors could have occurred throughout the lab.The temperature of the baths may have been allowed to fluctuate, which would change the temperature in the vials. The amounts of peas, beads, KOH, and cotton may have varied from vial to vial.
Air may have been allowed to creep into the vial via a leaky stopper or poorly sealed pipette. The vials may have not properly equilibrated, and students could have read the pipettes either too soon or too late. Students may have misread pipettes. KOH could have come into contact with the sides of the vials when it was being dropped onto the cotton. Mathematical inaccuracies may have occurred when completing the table.