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Thin layer Chromatography lab report Introduction
Thin layer Chromatography lab report Introduction
Thin layer Chromatography lab report Introduction
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The amino acids are stained purple and from this we can measure relative distance traveled (RF) to determine which amino acids are present in each sample. The more hydrophobic the amino acid the the further up the paper it appears this is why leucine is above alanine. Alanine is more hydrophobic than glutamate which is at the bottom of the paper chromatography. The paper chromatography proved that only sample A had everything required for the GPT reaction to occur. This was indicated by the presence of both glutamate and alanine. As alanine wasn’t added to this sample it must have been produced by the reaction. This reaction contained the GPT enzyme in the homogenate solution and the substrates sodium L-glutamate and sodium L-pyruvate. The …show more content…
This is seen since there is no reaction between L-leucine and L-pyruvate to produce a second amino acid. If this had been a successful reaction it would be seen on the paper chromatography as two purple spots at different heights, one corresponding to the substrate L-leucine and one due to the product amino acid. The thin layer chromatography also would have had a second yellow spot if the reaction had proceeded as there would have been both the substrate and product -keto acid present. Reaction C substitutes sodium L-glutamate for sodium D-glutamate. This does not react to create alanine as seen on the paper chromatography therefore we can say that GPT is very specific since it can distinguish between these enantiomers. The L-glutamate is found animals but D-glutamate is not so the GPT enzyme has evolved to only react with L-glutamate. We can also see on the thin layer chromatography that no reaction occurred here as there is only the presence of the substrate -keto acid, pyruvate shown in yellow. Reaction D contains a phosphate buffer instead of an amino acid which gives no purple spots of the paper chromatography. This is because the there was no amino acids present in the substrate to react with the ninhydrin reagent and there was also no reaction to produce amino
In experiment I “Watching the Reaction,” none of the runs reacted. 1A and 2A were still dark purple and light purple (respectively) at the beginning of the experiment and 10 minutes after. 3A had a translucent top and a dark purple bottom. There was a clear distinction of color in run 3A, much like combining water with oil. Since the tubes were rinsed multiple times with distilled water, dirty
Living organisms undergo chemical reactions with the help of unique proteins known as enzymes. Enzymes significantly assist in these processes by accelerating the rate of reaction in order to maintain life in the organism. Without enzymes, an organism would not be able to survive as long, because its chemical reactions would be too slow to prolong life. The properties and functions of enzymes during chemical reactions can help analyze the activity of the specific enzyme catalase, which can be found in bovine liver and yeast. Our hypothesis regarding enzyme activity is that the aspects of biology and environmental factors contribute to the different enzyme activities between bovine liver and yeast.
The affects of pH, temperature, and salt concentration on the enzyme lactase were all expected to have an effect on enzymatic activity, compared to an untreated 25oC control. The reactions incubated at 37oC were hypothesized to increase the enzymatic activity, because it is normal human body temperature. This hypothesis was supported by the results. The reaction incubated to 60oC was expected to decrease the enzymatic activity, because it is much higher than normal body temperature, however this hypothesis was not supported. When incubated to 0oC, the reaction rate was hypothesized to decrease, and according to the results the hypothesis was supported. Both in low and high pH, the reaction rate was hypothesized to decrease, which was also supported by the results. Lastly, the reaction rate was hypothesized to decrease in a higher salt concentration, which was also supported by the results.
The control for both curves was the beaker with 0% concentration of substrate, which produced no enzyme activity, as there were no substrate molecules for...
b) Comprehensive diagnostic chemistry panel with significantly increased amylase (1626 with normal being 300-1100 U/L), total
While the tube for specimen Cb turned a tannish white in the lower half of the tube while the top stayed the lavender inoculated tube color. Do to this evidence I determined that both specimens Ca and Cb cannot use the process Casein hydrolysis or Casein coagulation due to lack of soft or firm curds in both tubes. Since there was no casein curds formed, I concluded that specimens Ca and Cb also cannot perform the process of proteolysis. My conclusion is supported by the fact that there was no clearing of the medium. I have also determine that neither of my organisms can make the enzymes rennin, proteolytic or even proteases. I know my specimens cannot produce proteases due to the fact that there was no blue coloring in the tubes which means that the byproduct Ammonia was not produced to increase the pH. Since neither of my specimens can make these enzymes, I concluded that my specimens cannot break down lactose or casein. Although I did learn that specimen Cb can reduce litmus due to the evidence that the lower part of the tube turned a tannish white color with a purple ring at the top. This color change from a purple to a white means that the litmus was reduced turning it clear and leaving the white of the milk to show. Finally I know that specimen Ca cannot reduce litmus due to the fact that the tube had no change in
As the solution pH can influence the stability of NaClO-NH3 blend and the elimination of SO2, NOx, the impact of the pH of NaClO-NH3 blend solution on the instantaneous removal as well as the duration time was investigated, and the final pH after reaction was also detected and shown in Fig. 5. It can be seen that the variation of solution pH has a negligible effect on the desulfurization, but the elevated pH has a great promotion on the NOx removal, the efficiencies are significantly increased from 36% to 99% for NO2 in the pH range of 5–12 and from 19% to 65% for NO when the pH is between 5 and 10, after where, both of them are constant. Hence, the optimal pH of the NaClO-NH3 solution for the
Methionine represents the first limiting amino acid in broiler nutrition, thus different sources are available to balance diets based of corn and soybean. Bioavailability is different for each methionine source because of its rate of absorption and metabolic pathways. A broiler experiment was conducted to determine the relative bioavailability of Hydroxyl Methyl Analog Calcium (HMA-Ca) relative to DL-Methionine(DL-Met). The experiment was conducted at at Lavinesp (Unesp, Jaboticabal). It was used 1890 male broiler Cobb 500 of 21 days old, they were weighted and distributed homogeneously in a complete randomized design with 13 treatments and 7 replicates each. All birds fed either a basal diet deficient in sulphur amino acids, digestible methionine and cysteine (dig Met+Cys), or the basal diet with four levels of HMA-Ca (0.063, 0.183, 0.302 and 0.540%) and DL-Met (0.054, 0.156, 0.259 and 0.463%) to achieve increasing levels of dig Met+Cys. For the analysis, 5% of significance was considered and procedures of non-linear model were used by SAS. Exponential regression determinates bioavailability of HMA-Ca relative to DL-Met by calculating the relation of the slope of HMA-Ca relative to DL-Met
The purpose of this experiment was to discover the specificity of the enzyme lactase to a spec...
EDTA, the chelating agent that binds with magnesium, had a high absorbency and strong color change to red. The correct cofactor was copper which with the chelating agent of PTU and citric acid which both bind strongly to copper which keeps it from binding with the enzyme. This was determined because in the trails, both PTU and citric acid had low absorbency and were clear or roughly clear in color. The catechol in each tube, which was the control for this experiment, allowed the cofactor that would be used in this reaction to be singled out. The way each chelating agent would affect the different cofactors displayed which was not needed for the reaction and which cofactors were needed for the reaction. An inconsistency that may have affected the data would be if the calibration tube malfunctioned in balancing the spectrophotometer to zero. There also could be errors if the calibration tube wasn’t used before each tube was tested in the spectrophotometer. The relationship of the cofactor and amount of enzyme activity would be that if the cofactor is inhibited or not, the enzyme activity would be higher if the cofactor is not inhibited but lower if it was inhibited by the chelating
Abstract: Enzymes are catalysts therefore we can state that they work to start a reaction or speed it up. The chemical transformed due to the enzyme (catalase) is known as the substrate. In this lab the chemical used was hydrogen peroxide because it can be broken down by catalase. The substrate in this lab would be hydrogen peroxide and the enzymes used will be catalase which is found in both potatoes and liver. This substrate will fill the active sites on the enzyme and the reaction will vary based on the concentration of both and the different factors in the experiment. Students placed either liver or potatoes in test tubes with the substrate and observed them at different temperatures as well as with different concentrations of the substrate. Upon reviewing observations, it can be concluded that liver contains the greater amount of catalase as its rates of reaction were greater than that of the potato.
According to the graph on amylase activity at various enzyme concentration (graph 1), the increase of enzyme dilution results in a slower decrease of amylose percentage. Looking at the graph, the amylose percentage decreases at a fast rate with the undiluted enzyme. However, the enzyme dilution with a concentration of 1:3 decreased at a slow rate over time. Additionally, the higher the enzyme dilution, the higher the amylose percentage. For example, in the graph it can be seen that the enzyme dilution with a 1:9 concentration increased over time. However, there is a drastic increase after four minutes, but this is most likely a result of the error that was encountered during the experiment. The undiluted enzyme and the enzyme dilution had a low amylose percentage because there was high enzyme activity. Also, there was an increase in amylose percentage with the enzyme dilution with a 1: 9 concentrations because there was low enzyme activity.
Enzymes are necessary for life to exist the way it does. Enzymes help our bodies carry out chemical reactions at the correct speed. Catalase is one such enzyme, “Catalase is a common enzyme found in nearly all living organisms exposed to oxygen (such as bacteria, plants, and animals). It catalyzes the decomposition of hydrogen peroxide to water and oxygen”.\(Wikipedia). In other words catalase speeds up the breaking down of hydrogen peroxide, which is a byproduct of reactions in our body. Hydrogen peroxide is very common in our body but, “If it were allowed to build up it would kill us”(Matthey).This shows how necessary enzymes such as catalase to life. Without enzymes reactions that take place in our body could be affected greatly. In our
= Before conducting the experiment I would conduct a simple test for the protein by placing a sample of the albumen into a test tube and add biurett reagent. This contains copper (II) sulphate and sodium hydroxide.
In this type of chromatography stationary phase retains the anion. Here A+ will be positively charged which is masked by a negatively charged molecule B-. Mobile phase will have anions (say C-) which need to be separate. As the idea here is also ion exchange. Positive part (A+) attached to agarose is already attached to negative part(B-) So to remove B we need to add another positive part (D+). It will help to remove B and in addition of C- to A+. Then desired anion (C-) will be separated by buffer solution. By using anion exchange chromatography we can separate proteins like aspartic acid and glutamic