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Rate of Respiration

Extracts from this document...

Introduction

Aim The purpose of this investigation is to discover whether different respiratory substrates will affect the rate of respiration of yeast. I will investigate this by measuring the amount of C02 evolved during anaerobic respiration. As C02 is a waste product produced during respiration, measuring the volume of c02 produced will allow me to evaluate which of the sugars are best metabolized during respiration (of yeast). Background information Yeast Yeasts are a form of eukaryotic microorganisms that are used heavily in industry to be used for fermentation of alcohols and baking. Yeast digest the sugars using extracellular enzymes to break down the compounds, thus yeast are known as saprophytes, which the products will be transported into the cell of the yeast by facilitated diffusion to be respired. The yeast synthesises the enzymes required that will be specific to the substrate that they will be digesting. For example, domestic yeast is sold in as a sucrose solution, and the cell membrane of the yeast contains a high concentration of sucrose enzymes, thus the yeast adapts its digestive enzymes to its environments to digest specific sugars. Yeast can respire both through aerobic and anaerobic respiration, depending on the availability of oxygen present within the surrounding area. If oxygen is not present then fermentation occurs which converts sugars i.e. glucose into ethanol and CO2 (Glucose --> Ethanol and CO2), which is the anaerobic process. This is then respired out into surroundings. However if there is oxygen available then the sugar is fully metabolised and produces water and CO2 (glucose + oxygen --> CO2 + water), in this process energy is released for the cell to respire. In both cases of respiration CO2 is produced as a waste product, hence I have chosen it as a measure to calculate the rate of rate of respiration (based on the volume of C02 evolved). When oxygen is provided to a yeast cell then respiration follows in a series of steps, in the experiment this will only occur in small amounts as the supply of oxygen will be limited. ...read more.

Middle

1.3 1.4 1.4 1.4 1.4 0 0.1 0.1 0.1 0.2 0.2 0.3 0.3 0.3 0.3 0 7 8.4 9 9.1 9.2 9.2 9.3 9.3 9.3 0.0 0.0 0.0 0.0 0.0 0.1 0.1 0.1 0.1 0.2 0.0 0.0 0.0 0.0 0.2 0.2 0.2 0.2 0.2 0.3 0.0 0.0 0.0 0.0 0.3 0.3 0.3 0.5 0.5 0.5 0.0 0.0 0.0 0.2 0.2 0.2 0.3 0.3 0.4 0.4 0.0 0.8 2.6 3.2 3.3 3.7 4.1 4.1 4.1 4.1 0.0 2.1 2.4 2.6 2.6 2.6 2.6 2.6 2.6 2.6 0.0 4.3 4.8 4.8 5.8 5.1 5.1 5.2 5.4 5.4 0.0 3.3 3.8 3.9 4.0 4.1 4.1 4.1 4.1 4.1 0.0 0.3 1.6 2.4 2.4 2.4 2.4 2.4 2.4 2.4 0.0 1.8 2.6 3.5 4.7 4.7 4.7 4.7 4.7 4.7 0.0 2.4 2.4 2.7 3.1 3.3 3.3 3.3 3.3 3.3 0.0 0.4 0.8 0.8 1.4 1.6 1.6 1.8 1.8 1.8 0.0 5.9 7.1 8.3 9.2 9.9 11.3 11.7 11.7 11.9 0.0 3.1 4 5.7 6.4 7.9 8.1 8.2 8.5 8.6 0 7.8 12.1 15.2 15.2 15.2 15.2 15.2 15.2 15.2 0 6.9 10.6 14.9 14.9 14.9 14.9 14.9 14.9 14.9 0 1.3 2.2 2.6 3.2 3.6 5.1 5.3 5.5 5.5 0.0 0.9 1.3 2.4 3.9 4.0 4.5 5.1 5.6 5.6 0 1.8 2.4 2.8 4.2 4.2 5.7 5.9 6.5 8 0 0.5 1.5 1.5 2.5 2.6 3.8 3.8 4.6 5.5 0 0.1 1.1 1.7 1.8 1.8 1.8 1.8 1.8 1.8 0 0 0 0.2 0.4 0.8 0.8 0.8 0.8 0.8 0 0.4 1.2 1.5 2.1 2.3 2.9 3.2 3.4 3.7 0 1.5 1.9 2.2 2.9 3 3 3 3 3 0 0.4 0.6 0.6 0.6 0.6 0.6 0.6 0.6 0.6 0 0 0 0 0.3 0.4 0.4 0.4 0.4 0.4 0.0 0.0 0.0 0.2 0.2 0.2 0.2 0.2 0.2 0.2 YEAST ONLY 0.0 1.0 1.3 1.3 1.3 1.3 1.3 1.3 1.3 1.3 0.0 4 4.1 4.9 6.8 7.9 8.5 9.2 10.4 10.4 0.0 2.7 2.8 3.1 3.1 3.1 3.1 3.1 3.1 3.1 0.0 0.2 0.2 0.2 0.2 0.2 0.2 ...read more.

Conclusion

This can lead to the assumption that some yeast have more or less numbers of an enzyme than another yeast organism, thus one yeast organism may be adapted to digest maltose better than another organism as it has a higher concentration of maltase enzymes. This will cause a false positive value in the amount of C02 evolved. This will affect all reactions systematically due to the fact that I have used the yeast from the same batch of production This means that the yeast have been bred the same and should be genetically identical. Thus any genetic variation in the yeast (such as more carrier proteins for a specific substrate in the membrane) would have affected all results with a falsely positive value. Also, referring to my error bar chart, the profile of the graph does not change if the bars are increased assuming that this limitation has affected the results and thus fructose still produces the greatest volume of C02 and remains my conclusion remains valid. Respiration I assumed that all respiration that occurred in the conical flash during my experiments were anaerobic. However, the conical flask increased the surface area for the top portion of yeast/substrate solution, which could have respired aerobically as it had access to oxygen. Thus more C02 would have been produced as aerobic respiration produces a greater yield of ATP in comparison to anaerobic respiration As I have conducted all experiments using the same type of conical flask (i.e. same capacity, shape), then all results will be affected if any aerobic respiration had occurred. This will cause a falsely positive value of the volume of C02 produced as aerobic respiration produces more ATP thus more C02. However the amount of aerobic respiration for each reaction is varied, thus referring to my bar graph, I can see that if I increase the graph profile, the profile of the graph does not change (fructose remains substrate that produces the greatest volume of C02) and thus my conclusion remains valid. ?? ?? ?? ?? Faisal Faruque 13.11 Biology Coursework ...read more.

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