The first hypothesis was unsupported as Tube 4 (positive control) had the most succinate (.2mL) but ended with the lowest % transmittance out of the three sample we were examining (Tubes 2-4). We believe this discrepancy to be due to the variability of the mitochondrial solution. If the Tube 4 mitochondrial suspension had less concentrated enzyme than the solution for Tube 3 and Tube 2 then Tube 4 would be biologically incapable of matching their rate of reduction. It's worth mentioning that Tube 2 and 3 had a %T change of 23.1% and 23.2% respectively while Tube 4 exhibited a change of only 20.4%; further supporting the idea that Tube 4, despite its increase in succinate concentration, wa lacking in other enzymes/proteins necessary for the …show more content…
In the second experiment I knocked over the 24 degrees celsius test tube so the starting volume was 2.5mL in contrast to the other tubes 1 mL initial volume; this exposure may have affected the outcome for this sample. The third experiment was both correct and incorrect as the monosaccharides, fructose and galactose, demonstrated wildly different CO2 production. Sucrose produced about 11mL of CO2 while galactose produced only 0.5mL. The disaccharides of sucrose, maltose, and lactose demonstrated a similar trend where sucrose produced significant amounts of CO2 whole maltose and lactose lagged behind in production. However, fructose still produced significantly more CO2 (1.5mL) than the most productive disaccharide sucrose. This suggest that while the energy barrier of two reactions vs one reaction between monosaccharides and disaccharides might be a contributor, it is more likely that the availability of specialized enzyme is more important. For example, in the data, it is more likely that there is a sucrose enzyme in higher concentration/availability than a galactose enzyme given the substantial difference in CO2
The mole is a convenient unit for analyzing chemical reactions. Avogadro’s number is equal to the mole. The mass of a mole of any compound or element is the mass in grams that corresponds to the molecular formula, also known as the atomic mass. In this experiment, you will observe the reaction of iron nails with a solution of copper (II) chloride and determine the number of moles involved in the reaction. You will determine the number of moles of copper produced in the reaction of iron and copper (II) chloride, determine the number of moles of iron used up in the reaction of iron and copper (II) chloride, determine the ratio of moles of iron to moles of copper, and determine the number of atoms and formula units involved in
What is its melting point? 1,405 K: 1,132 C: 2,070 F. boiling point? 4,404 K: 4,131 C: 7,468 F
The purpose of this lab was to identify unknown bacteria cultures using various differential tests, and my unknown bacteria is #17. The identification of these unknown cultures was accomplished by separating and differentiating possible bacteria based on specific biochemical characteristics. Whether the tests performed identified specific enzymatic reactions or metabolic pathways, each was used in a way to help recognize those specifics and identify the unknown cultures. The differential tests used to identify the unknown cultures were Gram stain, Catalase, Mannitol Salt Agar (MSA), Blood Agar, Novobiocin, Coagulase, and DNAse (Alachi, 2007).
Make three exposures using given technical factors on a phantom knee in PA position . Include saline bags in exposures 1 and 2 to demonstrate patient soft tissue thickness.
Begining by labeling 7 different 2.0 ml tubes 0 thru 6 for each compound. Then add 1ml of extract to tube 0. Then add 0.5 ml of DMSO to tubes 1 thru 6. Now make a 1:2 serial dilution from 0(pure extract) to 6(1:16)
Measure the initial width, length, and thickness of the steel specimen using a Dial Caliper. Relieve pressure in Amatrol T9014 and adjust the height of the bottom platform to insert steel specimen. Insert one pin into the bottom platform to hold the steel specimen into the fixture. Slide two locking bars down the steel specimen. Adhere one locking bar to the bottom of the specimen and one at the top, lock them in place using the attached thumb screws. Insert the Linear Vernier Caliper in the top locking bar and zero out the caliper, allowing it to rest on the bottom locking bar. Compress the hydraulic cylinder until the indicator reads a force of zero. Lock the Linear Vernier Caliper in place by tightening the top thumb screw. [1] Compress
Hypothesis: The purpose of this experiment was to investigate the question will different food sources affect the level of activity of detoxification enzymes in bean beetles? The class alternate hypothesis is different food sources will affect the level of activity of the detoxification enzymes in bean beetles. The null hypothesis is the different food sources will not have any effect on the level of activity of the detoxification enzymes in bean beetles. Experimental design: The independent variables in this experiment were the types of beans (bean 1 was mung beans and bean 2 was adzuki beans) and enzymes assays used.
The type of sugar affects the rate of cellular respiration because each sugar is classified as either a monosaccharides, disaccharides or polysaccharides. The data from this experiment was collected by the amount of carbon dioxide produced from the type of sugar that was used. The data was then analyzed using a line graph. Data was also collected in class averages. There were three sugars in this experiment, glucose, lactose, and fructose. An example of a monosaccharides would be glucose and fructose. The slope for glucose should be about 283.07. The slope for fructose should be about 269.77. Second, an example of a disaccharides would be lactose. The slope for a disaccharide in this experiment should be about 67.055. The data shows that
2. (5 pts) List and explain the names and affiliations of the various characters/stakeholders in this story – I’m looking for us to use the story to map out the complexities that are generally associated with solving public health puzzles – the stakeholders you list and explain here should apply to many of the cases we consider going forward.
Oxidation State: Oxidation state or oxidation number (O.N.) is the apparent charge assigned to an atom of an element in a molecule or in an ion. Valency: Valency is the combining capacity of an atom with other atoms. It depends upon the number of electrons in the outermost shell. Q. 9: Calculate the oxidation number of Nitrogen in HNO3.
However, Tube E contained the same amount of glucose, but did not have the highest absorption value or quickest generation time. This is most likely due to the fact that Tube E did not contain phosphate buffer. Without the buffer, the acid produced by the bacteria’s fermentation may have lowered the pH of the solution
This can be explained through the transport mechanisms used for both disaccharides across a cellular membrane. Sucrose is broken down into fructose and glucose by the enzyme invertase and then transported into the cell via facilitated diffusion (Lagunas 1993). Meanwhile, maltose is not broken down into its monosaccharides until after it is transported into the cell via active transport, a process which requires an energy input (Lagunas 1993). Facilitated diffusion is a much easier process for the cell than active transport. The only reason maltose does not fall considerably behind sucrose in the rate of fermentation is because it is made of two glucose molecules, the primary input of glycolysis. Meanwhile, sucrose is made of one glucose molecule and one fructose molecule, so as these molecules enter the cell, fructose must first be converted to a usable form before it enters glycolysis. This collection of reactions was what kept the production of carbon dioxide between the maltose and sucrose substrates similar (Figure
However, with the additional fructose, it was hypothesized to cause a faster rate of fermentation of the yeast, yet the standard protocol with fructose was shown to be significantly faster with a rate of 0.3 while fructose was 0.00001. However, the results do agree with the background information found since it has been hypothesized that the yeast may be destroyed by the toxicity because fructose decreased the tolerance for ethanol in yeast (Francisco Javier De la Torre-Gonzalez Jose). It also supports the experiment performed by Zinnai, Venturi and others, they found that metabolization of glucose is preferred over fructose (Zinnai, Venturi, Sanmartin, Quartacci, & Andrich, 2013). Our results adds onto previous experiments and supports their findings which lead the conclusions found one step closer to being defined and completely explained since it is important to do further research after only one experiment to ensure that it is accurate and precise. Nonetheless, both experiments do support the conclusion that yeast concentration is positively correlation with production of CO2 since even though with fructose, the rate was significantly slower, it was still
The results from the experiment support the hypothesis that all six sugar samples would produce carbon dioxide as a result of cellular respiration. The carbon dioxide produced can be correlated with the energy being produced during the cellular respiration because it is a by-product of the decomposition reaction. The experiment proved my second theory wrong that splenda would produce the least amount of carbon dioxide of the sugars. According the the graph in Figure 02, the lowest producing sugar was in fact, lactose. Glucose was the highest carbon dioxide producing sugar. Sucrose was the second highest producing sugar and Splenda the third highest. The experiment supported my last theory that of all the sugars, glucose would produce the most carbon dioxide being that its rate of energy production was about 18 times that of lactose. This is because glucose is a simple sugar that is directly used in the glycolysis cycle which leads to the following energy producing steps. The other sugars, sucrose, lactose, and splenda are disaccharides that require
Baker’s yeast is a staple in any human diet because of its usefulness in making bread. As yeast respires, CO2 air bubbles form within the bread, giving it its fluffy interior (Strains of Bread Making Yeast). In this experiment, the respiration rate of S. cerevisiae was tested by placing yeast with different carbon sources within a test tube, and using a CO2 detector to measure the carbon dioxide levels as the yeast cells respired. CO2 can be used to measure cellular respiration because of the 1:1 ratio between oxygen and carbon dioxide (Measuring Respiration). Glucose, a carbon source that could enter glycolysis directly, should be the one that has the highest respiration rate. In comparison, sucrose and lactose must have their disaccharide glycosidic linkages broken before entering glycolysis, and glycerol must go through many steps before being able to enter glycolysis (Principles of Biology).