Archives for October 2009

Introduction

Tyrosinase is an enzyme involved with that catalysis of monophenols and catechols. Specifically in mammals, tyrosinase catalyzes two steps in the biosynthesis of melanin pigments from tyrosine. The pigment produced from this reaction is used in eyes, hair, and skin. In this laboratory experiment, the kinetics of mushroom tyrosinase is observed by monitoring the oxidation of L and D-3,4-dihydroxyphenyl alanine (Dopa). A crimson colored complex forms from due in part to the oxidoreductase and copper containing functionality of the tyrosinase. The K_M and V_max for tyrosinase can be calculated from resulting data obtained by monitoring the kinetics of the tyrosinase-DOPA solution with a UV-vis spectrophotometer. The enzymatic activity of tyrosinase can then be inhibited and followed via inhibitors such as thiourea and cinnamic acid.

Experimental

During the first week of the experiment, the enzyme kinetics of tyrosine in the presence of L-Dopa and D-Dopa were observed using a UV-vis spectrophotometer at 475 nm. To begin, six solutions were prepared using varying amounts of phosphate buffer and L-Dopa, but an unwavering amount of tyrosinase. The buffer-L-Dopa solutions were prepared in 1 mL cuvettes, and the tyrosinase, kept on ice, was added immediately before subjecting the solutions to UV-vis spectrophotometry. The cuvettes were inverted using paraffin as a cover, in order to mix the enzyme and substrate together, and thus begin the reactions, which was of kinetic interest. The UV-vis was used to monitor the kinetics for one minute. The recorded data could then used to determine the K_M and V_max of tyrosinase. This procedure was then repeated, only using D-Dopa in lieu of L-Dopa.

During the second week of the experiment, the enzyme kinetics of tyrosinase were observed in the presence of L-DOPA and the inhibitors thiourea and cinnamic acid, and were again monitored using a UV-vis spectrophotometer. As in the aforementioned procedure used during the first week of the experiment, the enzyme kinetics of tyrosinase in the presence of varying amounts of L-Dopa and phosphate buffer was monitored using a UV-vis spectrophotometer. For the next two trials, varying amounts of inhibitor was added along with the L-Dopa and phosphate buffer. The inhibitors used were thiourea and trans-cinnamic acid. Again, the enzyme kinetics were followed using a UV-vis spectrophotometer and by comparing the Michaelis-Menten and Lineweaver-Burk plots of the trial without inhibitor to the trials with inhibitor, it could be deciphered as to what class of inhibitors were being dealt with.

Data

L-Dopa Week 1

K_M = 479.0208
V_max = 0.1321

D-Dopa Week 1

K_M = 6616.7
V_max = 0.4886

L-Dopa Week 2

K_M = 512.77
V_max = 0.1529

Thiourea

* Samples 5 and 6 discounted

K_M = 283.6
V_max = 0.1181

Trans-Cinnamic Acid

* Samples 3, 5, and 6 discounted

K_M = 279.8
V_max = 0.1172

Results

In order to find the K_M and V_max, the raw data was first graphed as absorbance versus time. The slopes elicited from the linear regression of these plots were representative of velocity in terms of A/min. These velocities were then converted to umol/min using the equation A = Elc. Absorbance was divided by 3600 M^-1 cm^-1 and multiplied by 1 cm to give M/min, which was then converted to moles/min by multiplying by 0.001 L, the volume of the cuvette, and finally this value was converted to umol/min by multiplying by 10⁶ umol/mol.

The concentration of the substrates was found by taking the known mg/mL concentration, dividing by the formula weight of the molecule to obtain mol/L, then multiplying by 10⁶ umol/mol to obtain units in uM. These values were then multiplied by the percentage they comprised of the mL solution. The reciprocal values were graphed, 1/V versus 1/[S], with the y-intercept being equal to 1/V_max and the x-intercept being equal to -1/K_M.

As far as the results go, the K_M and V_max for L-Dopa are both significantly lower than that of D-Dopa found during the first week. This shows that tyrosinase exhibits stereoselectivity, otherwise the values would be exactly the same. It should have been expected that L-Dopa would have a higher K_M and V_max than that of D-Dopa however, because the naturally occurring Dopa molecule has L configuration. It seems more likely that the naturally occurring molecule would have fast enzyme kinetics than the synthesized molecule.

In regards to the inhibitors, the produced strikingly similar K_M and V_max values, both of which are lower than that of the reaction without either inhibitor. This suggests that both thiourea and cinnamic acid are uncompetitive inhibitors. The V_max values for the runs with the inhibitors is around 0.12 for each, which is fairly close to that of the run without inhibitor, 0.15, but because the K_M values for the inhibitor runs are nearly half that of the K­_M for the trial without inhibitor, I am not sure how to decipher that. Having equal V_max values could potentially make the inhibitors competitive, but the K_M values should be greater, not lower, than that of the enzyme kinetics without inhibitor.

Conclusion

There is undoubtedly some error in the raw data which affected the K_M and V_max values for all the trials. I had to cut out a lot of data points in order to obtain linear regression lines with reasonable R² values for the original absorbance versus time graphs, the slope of which was the velocity of the reaction. Even then, I still had to cut out more points for the Michaelis-Menten and Lineweaver-Burk plots in order to have reasonable looking graphs and values. Because the K_M and V_max values are not as expected, I would have to say the results are inconclusive. The inhibitors had nearly identical K_M and V_max values and had to be classified as uncompetitive. I would have expected the inhibitors to be competitive or noncompetitive, just because for a laboratory experiment I doubt the professor would have us analyze an uncompetitive inhibitor; it does not show much significance. The K_M and V_max for L-Dopa compared to D-Dopa from week one also seem odd; I would have expected L-Dopa to have the higher enzymatic activity.

I am guessing most of the error came from not being able to keep the solutions cold. There was a lot of waiting around in order to use the UV-vis spectrophotometer, and once the solutions of buffer and L-dopa were concocted, there was no real way to keep them chilled. This is probably what interfered with the ability to obtain valid data. It was also somewhat difficult to add the tyrosinase to each cuvette at the very same time, as for some solutions the droplet would gravitate down into the solution before the rest, so this cause error in the UV-vis spectrophotometer readings as well.

Questions

Question 1

Sodium azide – NaN₃

Non-competitive inhibitor because of its ability to bind to copper.

Sodium cyanide – NaCN

Non-competitive inhibitor because of its ability to bind to copper.

L-phenylalanine – HO₂CCH(NH₂)CH₂C₆H₅

Competitive inhibitor because of its inability to bind to copper.

8-hydroxyquinoline – C₉H₇NO

Competitive inhibitor because of its inability to bind to copper.

Tryptophan – C₁₁H₁₂N₂O₂

Competitive inhibitor because of its inability to bind to copper.

Diethyldithiocarbamate – S₂CN(C₂H₅)₂^–

Non-competitive inhibitor because of its ability to bind to copper.

Cysteine – HO₂CCH(NH₂)CH₂SH

Non-competitive inhibitor because of its ability to bind to copper.

Thiourea – CH₄N₂S

Non-competitive inhibitor because of its ability to bind to copper.

4-chlororesorcinol – C₆H₃(OH)₂Cl

Competitive inhibitor because of its inability to bind to copper.

Phenylacetate – CH3COOC₆H₅

Competitive inhibitor because of its inability to bind to copper.

Question 2

a. Increase

b. Decrease

c. Decrease

d. Increase

e. Increase

f. No change

Purpose

To determine the relationship between force, displacement, potential energy, kinetic energy, and work by using a force sensor to pull a spring and also to push a cart.

Hypothesis

The work done on the spring will be greatest at its furthest displacement and that the greater the work done on the cart, the greater its acceleration will be.

Labeled Diagrams

See attached sheet.

Data

Part One

Stretch

Part Two

Weight of cart: 5.29 N

Mass of cart: 0.54 kg

Graphs

See attached sheets.

Questions

1. In Part 1 you did work to stretch the spring. The graph of force vs. distance depends on the particular spring you used, but for most springs will be a straight line. This corresponds to Hooke’s law, of F=-kx, where F is the force applied by the spring when it is stretched a distance x. k is the spring constant, measured in N/m. What is the spring constant of the spring? From your graph, does the spring follow Hooke’s law? Do you think that it would always follow Hooke’s law, no matter how far you stretched it? Why is the slope of your graph positive, while Hooke’s law has a minus sign?

The spring constant is 81.668 N/m. From the graph, it does appear that the spring follows Hooke’s law as it produced a fairly straight line. I think the spring would follow Hooke’s law until it is all the way stretched out and cannot be stretched any more, or if breaks. The slope of the graph is positive as it is showing force applied on the spring. Hooke’s law shows the force applied by the spring, so that would be in the opposite direction in which it is pulled, thus being negative.

2. The elastic potential energy stored by a spring is given by ∆PE = ½ kx², where x is the distance. Compare the work you measured to stretch the spring to 10 cm, 20 cm, and the maximum stretch to the stored potential energy predicted by this expression. Should they be similar?

The ∆PE for my intervals was 0.20 J at 7 cm, 0.80 J at 14 cm, and 2.16 J at 23 cm. The ∆PE increased as the displacement is increased. This should be expected, as the spring becomes harder to pull the more it is stretched out.

3. In Part 2 you did work to accelerate the cart. In this case the work went to changing the kinetic energy. Since no spring was involved and the cart moved along a level surface, there is no change in potential energy. How does the work you did compare to the change in kinetic energy. Here, since the initial velocity is zero, ∆KE = ½ mv² where m is the total mass of the cart and any added weights, and v is the final velocity. Record you values in the data table.

The work done, 0.04796 J, is greater than the ∆KE, which is 0.027 J.

Conclusion

Lab Summarized

The overall goal of the lab was to investigate the relationship between work, potential energy, and kinetic energy. The goal was achieved using a spring and force sensor along with a motion detector to determine the work done on the spring when pulling it. Using the motion detector, displacement was determined, which could then be used to determine the spring constant from Hooke’s law. The force and acceleration were also collected using the force sensor and motion detector. Graphs produced of force versus position could be integrated to find the work done on the spring over certain intervals. The slope of the linear fit of this graph could also be used to produce the spring constant. Finally, the elastic potential energy stored in the spring could be determined from the aforementioned data.

During the second part of the experiment, the force sensor was used to push and thus accelerate a cart on the frictionless track toward a motion detector. The measured weight of the cart and final velocity could then be used to determine the change in kinetic energy of the cart. The work applied could again be determined by taking an integral over the time period in which the cart was pushed and accelerated with the force sensor.

The data collected for part one seems fairly conclusive. The determined spring constant of 81.668 N/m is comparable to known spring constants. The values for work extrapolated by integrating the graphs of force versus position are extremely close to the calculated values for work, or potential energy. The values of 0.1779 and 0.20 J, 0.8080 and 0.80 J, and 2.152 and 2.16 J are nearly identical, which shows part one of the experiment was performed rather well (or luckily). In regards to part two of the experiment, the work done on the cart found by integrating the graph of force versus position, 0.04796 J, is almost twice as large as the calculated change in kinetic energy, 0.027 J. The only way that this could be accounted for is if the wrong interval was used for the integral on the graph. Looking at the graph, it seems like the integral taken should have been from the start to the top of the peak, and not the whole peak.

Equations

W = F * s

W = ∆PE + ∆KE

∆PE = ½ kx²

∆KE = ½ mv²

F = -kx

Unlike the well know and oft studied chemistry of double bonds between carbons, the chemistry of boron-boron double bonds is for the most part unexplored. It is believed that boron should behave similarly to carbon due to its relativity to the element on the periodic table. Anions containing boron double bonds, specifically [R₂BBR­­₂]^2-, have in the past been predicted to be possible structures of interest to synthesize in the laboratory, however such efforts have failed for the most part.

It was then proposed to explore neutral diborenes, even though they in theory should be highly reactive compounds due to their triplet ground states and two one-electron π-bonds according to molecular orbital theory. The electron deficiency in this structure could however be stabilized by the addition of Lewis base ligands. The stabilizing ability of different ligand groups were assessed, including CO and NHC, which were chosen based on their strong electron donating capabilities. The ligand group that ultimately experimentally produced an actual neutral diborene was :C{N(2,6-PRⁱ₂C₆H₃)CH}₂. Previous work from using this ligand group for stabilizing carbenes suggested that this would be a potential stabilizing ligand for a diborene.

This compound, R(H)B=B(H)R, where R is the aforementioned ligand group, was synthesized beginning with RBBr₃ and KC₈ in diethyl ether. Two products were isolated from this reaction, including the desired diborene R(H)B=B(H)R. It was shown that a ratio of 1:5.4 of RBBr₃ to KC₈ yielded the highest percentage of R(H)B=B(H)R (12%). Any excess amount of RBBr₃ over this ratio resulted in a decrease of R(H)B=B(H)R and thus in increase of the other product, R(H)₂B-B(H)₂R.

A few methods were utilized in order to determine the chemical makeup of these products. NMR resonances of RBH₃, R(H)₂B-B(H)₂R, and R(H)B=B(H)R were respectively reported to be -35.38, -31.62, and +25.30 ppm. The ¹¹B signal of R(H)B=B(H)R produced a quartet, while the other two compounds elicited singlets. This alone could suggest double bond character between borons.

X-ray structural analysis shows a bond distance of 1.828 Å for R(H)₂B-B(H)₂R. This number seems to be on point with calculated B-B bond lengths for similar structures such as the CO-ligated analogue (1.819 Å) and an activated m-terphenyl based diborate (1.83 Å). Crystallization of R(H)B=B(H)R reveals B-C bond distances of 1.547 Å, which is marginally shorter than that of the other molecules. In addition to this, it is calculated that the angles between the C₃N₂ carbene rings and the core are strikingly different than that of the other compounds used and produced. Finally, the B=B bond distance in R(H)B=B(H)R was measured to be much shorter than the B-B distance reported in R(H)₂B-B(H)₂R, again implying a double bond.

DFT computations were also used to support the nature of R(H)B=B(H)R. The analysis was performed on the simplified model, where R=:C(NHCH)_2­. The experimental bond lengths for the non-simplified model seem to be in concordance with the computed B-B and B-C bond lengths, and well as the B-B-C bond angle calculated from the simplified model analyzed using DFT. The bond character of these bonds was also delved into via HOMO representations of the compounds among other computational techniques.

In conclusion, the authors of the paper were able to successfully prove that they had synthesized and characterized the first neutral diborene compound. They also ventured into the nature of the elusive boron-boron double bond. Though it was not necessarily expected that this phenomenon could feasibly be synthesized due to the expected reactivity of the boron-boron double bond, these chemists found a way to isolate the compound. In context to the larger field of chemistry, I suppose that the authors could determine other possible ligand groups that would produce a stable neutral diborene. They could also venture into increasing the percent yield, as 12% is on the low side. Finally, they could explore other group 13 elements, such as Al and Ga to see if they can replicate similar double bond properties.

Purpose

To determine the relationship between force, mass, and acceleration using a cart attached to a pulley with varying weights.

Hypothesis

If the mass of the weights attached to the pulley is increased, the force exerted on the cart and the acceleration of the cart will also increase.

Labeled Diagrams

See attached sheet.

Data

Graphs

See attached sheets.

Questions

1. Is the graph of force vs. acceleration for the cart a straight line? If so, what is the value of the slope?

Yes, the graph produces nearly a straight line; the correlation for a linear fit is 0.9787. The value of the slope is 0.6129 N/(m/s^2).

2. What are the units of the slope of force vs. acceleration graph? Simplify the units of the slope to fundamental units (m, kg, s). What does the slope represent?

The units of the slope of force vs. acceleration are N/(m/s^2). This simplifies to kg. The slope represents the mass of the pulley.

3. What is the total mass of the system (both with and without extra weight) that you measured?

The total mass of the system without the extra weight was 0.6759 kg and the mass of the system with the extra weight (300 g) was 0.9916 kg, which seems to make sense. 0.9916 kg is almost exactly 300 g more than 0.6759 kg.

4. How does the slope of your graph compare (percent difference) with the total mass of the system that you measured?

The slope of the graph without any added weights was 0.6129 kg, which is a 9.78% difference. The slope of the graph with the added weights was 0.8939 kg, which is a 10.36% difference.

5. Are the net force on an object and the acceleration of the object directly proportional?

Yes, as the net force is increased, the acceleration is also increased.

6. Write a general equation that relates all three variables: force, mass, and acceleration.

F = ma

Conclusion

Lab Summarized

The overall goal of the lab was to determine and show the relationship between force, mass, and acceleration. The goal was achieved using a cart and pulley system with varying weights to measure force and acceleration. The forces and accelerations collected were then graphed against each other the construct a linear fit line, whose slope showed the mass of the system (the cart, sensor, and any added weights). This value could then be compared to the mass calculated from the force of the free hanging system. The force and acceleration from each trial run could also be analyzed to show any relationship between the two values.

The data collected seemed to show a direct correlation between force and acceleration. Thus, the stated hypothesis was confirmed that if the force was increased, the acceleration would also increase. The compared values for the masses of the cart systems were about 10% different in each case. For the trial without any added weight, the calculated value of 0.6759 kg is 9.78% different from the extrapolated value of 0.6129 kg. In regards to the trial with the added weight, the calculated value of 0.9916 kg is 10.36% different than the extrapolated value of 0.8939 kg. This error could have been caused by a number of factors. For instance, the air resistance from the weight on the pulley dropping could have caused error, and any possible friction from the track could have attributed to this, too. It could also be thought that if the pulley did not drop straight downward, i.e. it was swaying at all, this would have further error. Lastly, if the rope was not completely taught when the system was put in motion, that could have caused error as well.

Equations

F = ma

a = 9.8 m/s^2

N/(m/s^2) = kg