Archives for October 2006

Introduction

Acids have a pH between 0 and 7, while bases have a pH between 7 and 14. A solution with a pH of 7 is said to be neutral; it is neither an acid nor a base. When given an acidic solution, it is possible add basic solution in order to neutralize it. To tell if the solution has been neutralized, an indicator such as phenolphthalein is used. The indicator will make the solution change color when it has become basic. In this experiment, a 10% solution of unknown molarity HCl was titrated with 1.00 * 10^-1 M NaOH in order to neutralize the HCl. The molarity of the HCl was then able to be calculated knowing the molarity of NaOH, the volume of NaOH used, and the volume of HCl.

Experimental

First, 25.00 ml of unknown sample of acid was delivered to a 250 mL volumetric flask using a volumetric pipette. The volumetric flask was then filled with de-ionized water to the mark in order to complete the 10% solution. A buret was then filled with 1.00 * 10^-1 M NaOH solution and its starting point was recorded. Next, 25.00 mL of the acid solution was delivered to 300 mL Erlenmeyer flask and 3 drops of phenolphthalein were added. The Erlenmeyer flask was put under the buret and the NaOH solution was dispensed into the Erlenmeye flask until the indicator turned the solution pink for around 15 seconds. The final recording on the buret was recorded, and the process was performed 3 times in order to reduce error.

Results

Molarity of NaOH solution: 1.00 * 10^-1 M

Identification of acid: I

Mean Molarity of Undiluted acid solution, M: 1.01 M

Calculations

In order to find the volume of titrant used in mL, I simply subtracted the initial buret reading from the final buret reading. To convert that volume in mL to L, I divided by 1000, as there are 1000 mL in 1 L. To find the molarity of diluted acid solution, I used the equation M₁V₁ = M₂V₂. M₁ equals the molarity of the NaOH solution used (1.00 * 10^-1 M), V₁ equals the volume of titrant used in L, V₂ equals the volume of acid solution used (0.02500 L), and M₂ is the molarity of the acid solution, which was solved for. To find the molarity of the undiluted acid solution, I knew that a 10% solution was used, so I multiplied the molarity of the diluted solution by 10 to get the molarity of the undiluted solution. Lastly, to find the mean molarity of undiluted solution, I added the molarity of undiluted acid solution from the 3 trials and then divided by 3.

Discussion/Conclusions

The final result of the molarity of the undiluted acid came very close to a whole number, which should mean my results are valid. It is also a very plausible number for the molarity of a solution. The volume of titrant used in my last two trials is nearly exact, so I must have performed them very well. The first trial is probably slightly off, as it was my first time doing a titration. On my first trial I added drops of NaOH too quickly towards the end. It is necessary to add the final drops and half drops very carefully towards the end, as it is a fine line between neutralizing the solution and making it basic. Overall, it was a very tedious experiment that relied on precision in order to achieve viable results.

Introduction

When two aqueous solutions are mixed together, they often react chemically and form products. The chemical reaction may be visible as a change in color from the reactants to the products, the release of gas in the product, or the formation of a precipitate. A precipitate forms when one of the resulting products is aqueous and the other product is insoluble. The insolubility means that the product will be a solid, which usually settles to the bottom of the resulting aqueous solution.

It is possible to determine how much precipitate will form before actually combining the reactants. Using stoichiometry, one can figure out how much precipitate theoretically will be produced. In the lab there are outside factors, which can affect how much precipitate is actually formed. Only under ideal conditions would the actual amount of precipitate formed match the theoretically amount of precipitate formed.

In this lab, strontium iodate monohydrate was synthesized using the following equation: Sr(NO₃)₂ + 2KIO₃ —-> Sr(IO₃)₂ + 2KNO₃. With the knowledge of the starting amount of Sr(NO₃)₂ and KIO₃ used, it was possible to figure out how much strontium iodate monohydrate should be theoretically produced. Then after finding out how much strontium iodate monohydrate actually was produced, the percent yield was found.

Experimental

First, approximately 40.00 ml of 5.00 x 10^-2 M Sr(NO₃)₂ and 50.00 mL of 1.00 x 10^-1 M KIO₃ were put into separate graduated cylinders. They were then combined in a beaker sitting in ice. This was to prevent the strontium iodate monohydrate from becoming soluble in water, as its solubility in water goes up with its temperature. The mixture was then stirred for about 10 minutes, so that all the precipitate could form. The precipitate and supernate were then poured into a vacuum filter, and the beaker was rinsed with ice cold distilled water to get all the precipitate out. The filter paper used in the vacuum filter was first weighed before filtration, then once the filtered precipitate was dry, the filter paper and precipitate was weighed. This whole procedure was performed twice.

Results

Volume of Sr(NO₃)₂ solution used: 39.99 mL (1^st run), 40.25 ml (2^nd run)

Molarity of Sr(NO₃)₂ solution used: 5.00 x 10^-2 M

Volume of KIO₃ solution used: 49.98 mL (1^st run), 49.50 mL (2^nd run)

Molarity of KIO₃ solution used: 1.00 x 10^-1 M

Mass of product, watch glass, and filter paper: 32.04 g (1^st run), 27.32 g (2^nd run)

Mass of watch glass and filter paper: 31.33 g (1^st run), 26.47 g (2^nd run)

Mass of product: 0.71 g (1^st run), 0.85 g (2^nd run)

Mass of Sr(NO₃)₂: 0.68 g (1^st run), 0.82 g (2^nd run)

Number of moles of Sr(NO₃)₂ used: 0.00200 moles (1^st run), 0.00201 moles (2^nd run)

Number of moles of KIO₃ used: 0.00500 moles (1^st run), 0.00495 moles (2^nd run)

Limiting reagent: Sr(NO₃)₂

Theoretical yield of product, moles: 0.00200 moles (1^st run), 0.00201 moles (2^nd run)

Theoretical yield of product, g: 0.875 g (1^st run), 0.880 g (2^nd run)

Percent yield: 78.% (1^st run), 93.% (2^nd run)

Mean percent yield: 86.%

Calculations

In order to find the mass of the precipitate, took the mass of the product, watch glass, and filter paper minus the mass of the watch glass and filter paper. To find the moles of the reactants used, I used the equation Molarity = Moles/Liters and rearranged it to the equation Moles = Molarity x Liters. Before subbing the volume in, I had to convert mL to L by dividing by 1000. To find the limiting reagent, I first had to look at the balanced equation of the chemical reaction, which was Sr(NO₃)₂ + 2KIO₃ —-> Sr(IO₃)₂ + 2KNO₃. Since I knew for every 2 moles of KIO₃ used, 1 mole of Sr(NO₃)₂ was used, I could substitute the actual number of moles of each that were used in the experiment to find the limiting reagent. If KIO₃ was the limiting reagent, then 0.00250 moles of Sr(NO₃)₂ would be needed for the 0.00500 moles of KIO₃, but there were only 0.00200 moles of Sr(NO₃)₂ used, so that made it the limiting reagent.

In order to find the percent yield, I first took the number of moles of Sr(NO₃)₂ used because from the balanced equation, I knew that there would be an equal number of moles of Sr(IO₃)₂ produced. I then found the molar mass of Sr(IO₃)₂, which is 437.43 g, then multiplied by 0.00200 moles, which is the number of moles of Sr(IO₃)₂ theoretically produced, to find the number of grams of Sr(IO₃)₂, theoretically produced (0.875 g). I then took mass of the precipitate strontium iodate monohydrate produced (0.71 g) and needed to find the mass of Sr(IO₃)₂ produced. So I found the percent of Sr(IO₃)₂ that makes up strontium iodate monohydrate by taking the molar mass of Sr(IO₃)₂ (437.43 g) and divided by the molar mass of strontium iodate monohydrate (455.44 g) to get 96.046%. I then multiplied the mass of the product by this to find the mass of Sr(IO₃)₂ actually formed (0.71 g x 0.96046 g= 0.68 g). I then took 0.68 g and divided by the mass of Sr(IO₃)₂ theoretically produced (0.875 g) and multiplied by 100 to find get the percent yield of 78.%. I repeated this for the values found in the second run. For the mean percent yield, I took the percent yields found for each run, added, then up, and divided by 2.

Discussion/Conclusions

My results were fairly close to what they should have been. A mean percent yield of 86.% is probably good considering all the factors that can cause the percent yield to be less than 100%. The Sr(IO₃)₂ could get too warm and wash away in the water, the precipitate could not completely form when mixing the reactants, and some precipitate could become stuck in the beaker and not wash out. Those are factors that are not all easily controlled, so overall my percent yield of 86.% seems plausible when looking at those factors that could affect the results.

If I were to repeat this experiment, I would probably take more time in letting the reactants mix and form the precipitate. That way I could be sure almost all the precipitate actually formed. I would also keep the wash bottle in colder conditions to make sure none of the precipitate washed away.

Abstract

Enzymes are catalysts which lower the activation of chemical reactions, thus making them happen more rapidly. In this experiment, different amount of enzyme and substrate were put in a test tube, then were observed using a spectrophotometer to see how fast the reacted to produce product. It was found that as the concentration of enzyme was increased, the speed of reaction increased. Likewise, it was found that as the concentration of substrate was increased, the speed of reaction increased. As there is more enzyme, it is able to react with more substrate at once, therefore increasing the rate of reaction. When there is more substrate, there will be just as much enzyme, also making the rate of the reaction increase. Thus, higher levels of enzyme or substrate mean there will be a higher turnover rate of product.

Introduction

A catalyst is a substance that lowers the activation energy of a chemical reaction (Zubay et al., 1995). In order words, this means that a catalyst makes chemical reactions happen faster than they normally do. Enzymes are catalysts that speed up reactions in living cells. For example, the enzyme carbonic anhydrase makes the chemical reaction CO₂ + H₂O —> H₂CO₃ happen 10⁷ times faster than it normally does. In this example, the enzyme reacted with CO₂. The substance which an enzyme catalyzes is called a substrate. Enzymes bind to substrates in order to speed the reaction in turning the substrate to a product. When there is a little amount of substrate, there will be a small amount of enzyme, but as the level of substrate increases, the level of enzyme increases. In this experiment, the speed at which enzymes and substrates react was observed.

Materials and Methods

For the first experiment, different enzyme concentrations were tested to see how fast they reacted. A Spec20 spectrophotometer was first turned on and set to a wavelength of 340 nm. A reaction mixture consisting of 0.850 ml distilled water, 1.000 ml buffer stock, 0.100 ml 15 mM NAD+, and 1.000 ml ethanol was then put into the test tube using a micropipettor. This mixture was agitated to mix it together, and the spectrophotometer was then calibrated using this test tube. Once calibrated, 0.050 ml ADH enzyme was added to the test tube mixture and agitated. The test tube was then quickly put into the spectrophotometer and its absorbance readings were recorded. The readings were recorded every 15 seconds for 3 minutes. This process was repeated for 2 more reaction mixtures with adjustments to the amount of distilled water and AHD enzyme included. The second reaction mixture had 0.100 ml ADH enzyme and 0.800 ml distilled water, and the third one had 0.025 ml ADH enzyme and 0.875 ml distilled water.

For the second experiment, different substrate concentrations were tested to see their reaction rates. Once again, a Spec20 spectrophotometer was turned on and set to a wavelength of 340 nm. A reaction mixture consisting of 1.85 ml distilled water, 1.0 ml buffer stock, 100 ul NAD+, and 0 ul ethanol stock was put into a test tube using a micropipettor. The mixture was agitated, then put into the spectrophotometer and the spectrophotometer was calibrated. Once 50 ul of ADH was added to the reaction mixture, the test tube was agitated and quickly put into the spectrophotometer. Absorbance readings were then recorded every 15 seconds for 2 minutes. Another reaction mixture with the same amounts of each substance was made, observed, and recorded. This process was repeated for five more reaction mixtures with differences in the amount of distilled water and ethanol stock used. The amounts of distilled water and ethanol stock for these reaction mixtures were 1.80 ml distilled water and 50 ul ethanol stock, 1.75 ml distilled water and 100 ul ethanol stock, 1.65 ml distilled water and 200 ul ethanol stock, 1.35 ml distilled water and 500 ul ethanol stock, and 0.85 ml distilled water and ethanol stock.

Results

Table I

Table II

The concentration of enzyme had an effect of the reaction rate. As the concentration of the enzyme went up, the velocity went up (Table II, Figure 2). The initial slope also became steeper as the concentration of the enzyme increased (Table II, Figure 1).

The concentration of substrate affected the rate of reaction. As the concentration of substrate went up, the velocity increased steeply then evened out (Figure 4). From Figure 4 (v vs. [S]), the estimated Vmax is 0.0620 umol/min and the estimated Km is 0.0140 mM. From Figure 3 (1/v vs. 1/[S]), the estimated Vmax is 0.0568 umol/min and the estimated Km is 10 mM. The actual Vmax and Km recorded from SigmaPlot were 0.06566 umol/min and 0.01522 mM. The Vmax obtained from each graph are fairly close, but the Km are not very close at all. The Vmax and Km from the v vs. [S] graph are very similar to the actual Vmax and Km obtained from SigmaPlot.

Discussion

Shono and co-workers (1995) observed the rate of reaction versus the concentration of a substrate different from the one used in this experiment. Their graphs resulting from their experiment are very similar to the graph resulting from this experiment. The “velocity vs. substrate concentration” graphs follow the almost exact same curve, but the levels of concentration were higher in Shono’s experiment, resulting in higher rates of reaction. In the “1/v vs. 1/[S]” graphs, Shono’s line seems to have about the same slope, but it crosses the x-axis at a much lower value than it did in the graphs for this experiment.

This can be attributed to error in procedure of the experiment, which caused outlier values the line to be skewed. It was not known these outlier values should have been disregarded until computing calculations for Km and Vmax using the “1/v vs. 1/[S]” graph and comparing to other similar graphs. The Vmax recorded from Figure 3 was fairly close to the Vmax found from Figure 4. The Km from Figure 3, 10 mM, is not very close, however. That value is way larger than the Km found from Figure 4, 0.0140 mM, and the actual Km calculated by SigmaPlot, 0.01522 mM.

From the experiment, the results showed that with a higher concentration of enzyme, the higher rate of reaction. When catalyzing a reaction, the enzyme binds to the substrate (Bolsover et al., 1997). If there is a higher concentration of enzyme, this means that there will be more enzyme to bind to substrate at once, therefore making the turnover rate of substrate to product higher. This is why a higher concentration of enzyme produces a higher turnover rate. The turnover rate begins to slow down and stop as the amount of substrate runs out, and that is why the absorbance rates began to even out in Figure 1. There was a limited amount of substrate that could be converted into product.

As there was a higher concentration of substrate, the rate of reaction increased and then leveled off, as shown in Figure 4. The Michaelis-Menten graph assumes that the enzyme and substrate are in equilibrium (Zubay et al., 1995). Therefore, as there is a higher concentration of substrate, there will be an equally high concentration of enzyme to react with the substrate. As there is a higher concentration of each, the rate of reaction increases. The curve evens off, however. This is because each substrate has a maximum velocity at which it can convert from substrate to product. The enzyme can not catalyze the substrate to turnover faster than this.

The affinity for the enzyme and substrate in this experiment was fairly high. The Km, 0.01522 mM, is a low number. This means the enzyme and substrate reacted very quickly to produce product.

Literature Cited

Bolsover, S.R., J.S. Hyams, S. Jones, E.A. Shepard, and H.A. White. 1997. From Genes to Cells. (Wiley-Liss, NY). 424 p.

Shono, M., M. Wada, T. Fujii, 1995. Partial Purification of a Na+ -ATPase from the Plasma Membrane of the Marine Alga Heterosigma akashiwo. Plant Physiol 108: 1615-1615.

Zubay, Geoffrey, William M. Parson, and Dennis E. Vance. 1995. Principles of Biochemistry. (Wm. C. Brown, Dubuque, Iowa) 863 p.

Growing up, there are times when we feel the urge to test the rules set before us. It does not matter whether it is a parent, teacher, or authority setting standards to adhere by. When we are told do to something, we instinctively want to do the opposite of what we are told. It is part of human nature. Rules and restrictions are things that our primitive instincts tell us to bend and break. For the most part, we are able to control these urges. We know that there are consequences involved with breaking rules. But every once in a while, we decide to test our luck. We think that we can get away without abiding by what we have been taught. First we must muster up some courage and confidence, and then when the time is right, we cross the line between what is “right” and what is “wrong”. There is a certain excitement or euphoria felt when we do something we are not supposed to do. This thrill of breaking rules overcomes us and we feel invincible. There is nothing that can stop us. With careful planning, it is possible to get away with breaking rules unscathed, but then there are times when we are caught red handed. There are not many feelings worse than when we are found out. We go straight from an extreme high to a complete low. The spotlight shines on us and there is nothing we can do to escape punishment. Though we know how bad it feels to be caught doing wrong, we keep testing limits because of the excitement we get. It is a very thin line we walk, but we keep walking on that thin line. I like the poem “Running on Empty” by Robert Phillips because it does such a good job at portraying this risk taking attitude.

In the poem, a teenager wants to take his father’s car out for a joyride. The father decides to let him borrow the car, but the teenager is to make sure the car’s gas tank is always half full. The son agrees, but when the fuel gauge reaches below half full, he keeps driving. He is overcome with an excitement of defiance against his father. He feels an adrenaline rush from going against what he is told. The car is still running, so he feels like he could drive forever. He and the car are invincible, nothing can stop them. The teenager’s high is only momentary as the car suddenly runs out of gas and he is stranded in the middle of nowhere at night time. He is in a state of a shock that the car actually stopped. He had run on empty for so long and his father seemed wrong about needing to keep the gas tank half full. In the morning he is able to refill the car and go home.

The poem does an excellent job of relaying the emotion felt by the teenager. Lines 4 though 16 are written as one long sentence describing the teenager driving. The sentence builds up to the moment when he runs out of gas. This creates an image to the reader of the car speeding down the highway, going and going until it abruptly stops. There is also repetition of some words in this sentence, which portrays the excitement and intensity felt by the teenager. Line 5, “The fuel gauge dipping, dipping…”, and line 9, “…mile after mile, faster and faster…”, show the intensity felt by the teenager. The repetition almost seems like his heartbeat rising. In lines 15-16, “…the wind screaming past like the Furies,” a simile is used. This also gives the reader an idea of how fast and exhilarating the experience of driving the car must have been. Lines 17-22 are short sentences, which contrast the long sentence describing the driving. Phillips does this in order to portray the shock felt by the teenager.

When I read this poem, I can almost feel the teenager’s exhilaration of breaking his father’s advice. I have had times in my life similar to that of the teenager in the poem where I did not listen to my parents and regretted not listening to them in the end. I have felt his same thrill of defiance, but then also his same feeling of shock when caught in the act of rebellion. The one long sentence with repetition of words really makes me feel like I am in that car with my heart racing. I can also feel the teenager’s shock and disbelief when the car runs out of gas. I can relate to this poem all too well, and that is why I like it so much.

The shuttle system for St. Joseph’s University currently begins its runs in front of the Sourin Residence Hall at five minutes and thirty-five minutes after the hour. The R5 Septa line stops at the Overbrook train station on the hour and on the half hour. For this reason, I believe the shuttle system should begin its runs later in order to somewhat coordinate its schedule with the train schedule. If I want to take the shuttle to the train station, currently I have to wait about fifteen to twenty minutes between the time the shuttle arrives at the Overbrook station and the time the train arrives. Alternatively, I usually choose to walk to the train station instead because then I have an extra ten minutes before I need to leave Sourin to catch the train. I do not need to start walking towards the train station until fifteen minutes or forty-five minutes after the hour. When it becomes colder out, I can see problems with both having to walk to the train station and having to wait for the train to come if I take the shuttle.

I understand that the reason the shuttle currently comes at five and thirty-five minutes after the hour is because that gives students time to get from class to Sourin and then take the shuttle wherever they need to go. Most classes end at fifty minutes after the hour, so fifteen minutes gives students adequate time to get from almost anywhere on the campus to Sourin by five minutes after the hour. Other classes end at times like fifteen minutes, thirty minutes, and forty-five minutes after the hour. The current shuttle times also give students coming from these classes a fair amount of time to arrive at Sourin before the shuttle comes.

If the shuttle were to come at fifteen minutes and forty-five minutes after the hour, then students going to the train station would not have to wait as long for the train to come. Another positive outcome from this change would be that students coming from classes would have an extra ten minutes before the shuttle comes. This would give them time to talk to a teacher after class, stop at the library to print a paper, or finish any tasks they need to do before the shuttle comes. They would not be in as much of a rush to make it to Sourin in time. I believe this change in the shuttle schedule would benefit all students and allow them to make better use of this convenience.

Introduction

For this lab, seven small experiments were performed. First, solid Mg was combined with a HCl solution. The reactants started to bubble and give off a colorless gas was given all while the Mg was dissolving. Next a Pb(NO₃)₂ solution was mixed with a KI solution. These two clear, colorless liquids produced a transparent yellow liquid and yellow sediment at the bottom of the test tube. After that, CuSO₄ ∙ H₂O, a fine blue crystal solid, was heated to leave behind a white powdery solid. Next a HCl solution and NaOH solution with phenolphthalein were combined. When the HCl solution was added to the NaOH, its color changed from purple to clear. Following that experiment, Cu was heated in the presence of O₂. The shiny, copper colored metal turned dark gray and lost its luster. After that, a CuSO₄ solution was combined with solid Fe to produce rust. Lastly, a FeCl₃ solution (yellow in color) was mixed with a NaOH solution (colorless in color) to yield a cloudy orange liquid.

Abstract

Sugars catabolize through the process of glycolysis. Glycolysis causes the sugar to undergo phosphorylation and ferment, which yields CO₂. In this experiment, different sugar solutions were mixed with a yeast solution. The yeast solution caused the sugar solutions to undergo glycolysis and produce CO₂. Glucose, fructose, and mannose all produced CO₂, yet galactose did not. Mannose and fructose followed very similar curves of time versus the production of CO₂, while glucose followed a different curve. The ability for the sugar to undergo glycolysis was dependent on its ability to accept a phosphate group during phosphorylation. Therefore the sugar molecules most easily able to accept a phosphate group produced the most CO₂.

Introduction

Everyone knows that yeast makes dough rise, but exactly why does this happen? Sugars are broken down through the process of glycolysis (Black, 1999). In order to begin this catabolizing (breaking down) process, the sugar must first gain a phosphate group, which is called phosphorylation. Often the phosphate group is gained from ATP. Once the sugar has undergone these processes, it will begin to ferment and yield a byproduct of CO₂. Buchner (1897) performed experiments using a yeast “juice” and different sugars such as can sugar, glucose, fructose, and maltose. When these two ingredients were combined, they reacted to produce CO₂ for days. When the yeast juice alone was heated, it would also produce CO₂. He also noticed that some sugars did not undergo the fermentation process with the yeast juice to produce CO₂. In this study, different sugar solutions were combined with a yeast solution in order to see how fast and if they react to produce CO₂.

Materials and Methods

A yeast solution along with three sugar solutions and laboratory instruments and supplies were distributed. The three sugar solutions included mannose, fructose, and glucose. A fourth sugar solution had to be concocted by measuring 1 g of raw galactose on a top loading scale and mixing it with 10 ml of distilled water measured in a graduated cylinder. The galactose and water were mixed in a test tube. All of the sugar solutions were 10% solutions.

Using a micropipettor, 2 ml of yeast solution and 2 ml of the first sugar solution, mannose, were measured and mixed together in a test tube. After being combined, a Pasteur pipette was filled with 1 ml of the new solution. “Play-Doh” was put on the tip of the pipette so that CO₂ would not be able to escape and also to create vacuum so the solution would not fall out of the pipette. The pipette was then placed upside down in the test tube with the remaining solution of yeast and mannose. The pipette was then observed every 5 minutes to see if any CO₂ had accumulated at the top. Any progress was recorded. This process was repeated with the rest of the sugar solutions. A control sample was also tested, using distilled water instead of a sugar solution. All of this was done at a room temperature of about 20º C.

Results

Table I:

Graph I:

Out of the samples that did produce CO₂, the mannose sample took the longest to begin producing CO₂ (Figure 1). It took 25 minutes until it started. Its biggest jumps in CO₂ production were between the 20 minute and 25 minute, and 25 minute and 30 minute marks. This was when right when it began producing CO₂. Its CO₂ production increased steadily until the 65 minute mark when production started to even out.

The fructose sample took 20 minutes to begin producing CO₂, which was second fastest out of the samples that did produce CO₂. The fructose had its highest CO₂ production in the first three time intervals it began producing CO₂. After that it steadily increased and only seemed to slightly begin to even out production of CO₂. It produced the most CO₂ out of all the samples.

The glucose sample started producing CO₂ the earliest out of all the samples, at the 15 minute mark. It increased production of CO₂ steadily until the 45 minute mark when production suddenly spiked in production for 15 minutes. The CO₂ production looked like it may have been starting to even out when recording stopped.

Galactose and the control sample did not produce any CO₂ at all. They remained at 0 mL of CO₂ production throughout the whole experiment.

Discussion

Prescott and co-workers (1999) took a look at the chemical reactions between microorganisms and carbohydrates. They noted that the sugars glucose, fructose, and mannose are all catabolized, or broken down, through the process of phosphorylation, which is process of adding a phosphate group to a molecule (Black, 1999). Most commonly ATP adds the phosphate group to the sugar molecule, and then it is able to enter glycolysis, which is the process of breaking down a sugar. Once in the process of glycolysis, the sugar will begin to ferment. One common byproduct of fermentation is CO₂.

In this experiment, glucose, fructose, and mannose were the only sugar solutions that produced CO₂. As described by Prescott and co-workers (1999), those are the only sugars that are able to be broken down through phosphorylation. The yeast must have had ATP in it, which would have added a phosphate to the sugar molecules. After gaining the phosphate, the sugars began to break down, ferment, and produce CO₂.

Prescott and co-workers (1999) also noted that in order for galactose to be catabolized, it must first go through a three-step process to be converted into a type of glucose. In the experiment, the galactose solution did not react with the yeast to produce any CO₂. This is because the galactose was not converted, so therefore it could not gain a phosphate and begin the process of phosphorylation to break down, undergo glycolysis, ferment, and produce CO_2.

Lastly, the control group did not produce any CO₂ either. That is because there was nothing for it to react with and it was at room temperature. If the yeast was heated, it would have given off CO₂ (Buchner, 1897).

Literature Cited

Buchner, E. 1897. “Alcoholic Fermentation without Yeast Cells”.

Prescott et al., 1999. “Catabolism (breakdown) of Carbohydrates”.

Black, 1999. “Glycolysis and Fermentation”.

“Running on Empty” – Robert Phillips

This poem relates to the theme of memories and struggles of childhood because it is about a young man’s rebellion against his father. The father tells his son to always keep the car’s gas tank half full. The son ignores his advice and drives until the gas tank is almost empty for the thrill of defiance. The car eventually runs out of gas and he is stranded for the night. The need to defy one’s parents is part of growing up. It is part of the struggle between listening to one’s parents and making one’s own decisions. The poem is part of the writing trend of “New Narrative” because the poem tells a story and it does not follow any rhyming pattern or meter.

“I was stolen…” – Charles Simic

This poem relates to the theme of memories and struggles of childhood because it is about a child’s struggle to find their identity. The child switches between a gypsy lifestyle and a high class lifestyle. This contrast in settings could confuse a child and leave them without a sense of who they are. This poem shows the struggle of finding one’s identity while growing up. The poem fits into the writing trend of “Language Poetry” because it contains very simple sentence structure and has no structure of rhyme or meter.