Biology Cell Respiration Lab

Identify the two hypotheses being tested in this activity

1. is the rate of respiration higher for germinated peas or non-germinated peas? 2. is the rate of respiration higher for germinated peas in room temperature water or germinated peas in cool water?

Explain your reasoning.

CB organisms don't need to be at a certain temp so they wouldn't be carrying out CR as quickly as the WB organisms that need to use E to get themselves up to a higher temp so to get that E they would be carrying out CR must faster.

Did cell repsiration occur in the peas? How do you know?

Cell respiration did occur in the peas. I know this because as time went by the food coloring went down. This is an indication of the levels of Oxygen in the test tube going down, which means as more oxygen was consumed

This activity uses a number of controls. What conditions must remain constant? Why?

Conditions that must remain constant are pressure and time. Pressure must be controlled because since volume and pressure are inversely proportional, meaning that as one goes up the other goes down. Time must also be kept at constant because if you do one for a longer time than the other, the rate of respiration might be different than it would've if you used the same times for both. Volume of seeds (number of moles of seeds used) 15 germinated & 15 dormant + glass beads because volume of dormant is less than volume of germinated

If respiration in a small mammal were studied at both room temperature (21C) and 10C, what results would you predict? Explain your reasoning.

If CR in a small mammal were studied at room temp and 10C, I would expect their rates of cellular respiration to be higher at room temp because in the cooler, 10C environment, the enzymes used to carry out CR are slowed down, causing the process itself to slow down and be higher at room temp.

If you used the same experimental design to compare the rates of a 25 g cold blooded organism and a 25 g warm blooded organism, both at 10C, what results would you expect?

Results I'd expect from this experiment are that the warm-blooded organism would carry out cellular respiration at a much higher rate because they maintain a higher temperature than 10C so their body would need to work really hard and use energy to keep themselves at their set temperature, while the cold blooded organism would stay at the 10C and not change, so they wouldn't need to carry out cellular respiration nearly as much.

design an experiment to examine the rates of cellular respiration with peas that have been germinating for different lengths of time: 0, 12, 24, and 36 hours. What results would you expect and why?

Set up four test tubes similarly to this experiment. Put absorbent cotton sprayed with KOH in the bottom of each and on top of the absorbent cotton place a piece of non-absorbent cotton. Next write 0, 12, 24, and 36 on one of each of the test tubes so that they can correspond with the length of time the peas have been germinating. Place the same number of peas (25) into each test tube, and to insure the same volume for each add glass beads until all of the test tubes have the same volume. Fill the test tube with a 0 with non-germinating peas and glass beads to make up for the lack of volume, in the test tube labeled 12 add 25 peas that have been germinating for 12 hours and the needed amount of glass beads to create the same volume for each test tube. Do the same for test tubes 24 and 36, adding peas that have been germinating for 24 hours into test tube 24 and peas that have been germinating for 36 hours into test tube 36. Results that I would expect are that the peas that have been germinating for 36 hours to have the highest rates of cellular respiration because they have been growing for the longest and therefore have more cells carrying out mitosis and need more E for it to all be carried out.

What is the effect of germination on the rate of cell respiration in peas? Why do you think this?

The affect of germination on the rate of cell respiration in peas is that in peas that are germinated, the rate of cell respiration is higher because the cells are growing/going through mitosis which requires energy/ATP in order to be carried out which is generated through the process of cellular respiration. In non-germinating peas, they aren't going through mitosis and therefore don't need as much ATP, so their rate of cellular respiration is much less.

What is the effect of temperature on the rate of cell respiration in the peas? Why do you think this is?

The effect of temperature on the rate of cell respiration in the peas is that the higher the temperature, the higher the rate of cell respiration because the cool water causes the enzymes to slow down, causing the reactions of cellular respiration to slow down, and therefore the rate of respiration to be slower.

What are the main differences between a cold blooded and warm blooded organism?

The main differences between a cold blooded and warm blooded animal is that warm blooded animals try to keep their bodies at a set temperature, so if it is cold their bodies will generate heat and if it is hot their bodies will cool down. On the other hand, cold blooded animals are the temperature of their environment and aren't at a set temperature.

What is the purpose of the KOH in the experiment?

The purpose of the KOH in the experiment is to react with the CO2 and form Potassium Carobnate (K2CO3) so only the pressure of the O2 is measured.

Describe the relationship between the amount of O2 consumed and time.

The relationship between the amount of O2 consumed and time is that as more time goes by, more Oxygen is consumed. This means their relationship is a directly proportional relationship-- the more time passed, the more oxygen consumed and the less time passed- the less oxygen consumed.

Why did the vial need to be completely sealed around the stopper?

The vial needed to be completely sealed around the stopper because it insured that the measurement of the pressure in the test tube to be accurate. If it hadn't been tightly sealed, the air inside of the test tube would escape, causing the pressure of the oxygen to be impossible to be measured.