Lab 4- Column Chromatography and Redox Biochemistry

docx

School

Concordia University *

*We aren’t endorsed by this school

Course

271,221

Subject

Chemistry

Date

May 27, 2024

Type

docx

Pages

Uploaded by MegaSnailMaster1165

Results Part 1: Column Chromatography Table 1. Color of the fractions obtained in Column Chromatography of LDH, Cytochrome C and FMN, and LDH assay results for the clear fractions at 340nm. Fraction Color Rate (Abs/min) at 340 nm LDH activity (μmol/min) % LDH activity #1 Clear 0.04275708 0.007245 5.67% #2 Clear -0.1442514 0.02444 19.1% #3 Clear -0.7537793 0.1277 100% #4 Clear -0.1865322 0.03161 24.8% #5 Light orange -0.02583046 0.004377 3.43% #6 Light orange - - - #7 Orange - - - #8 Brownish orange - - - #9 Light orange - - - #10 Light yellow - - - #11 Light yellow - - - #12 Yellow - - - #13 Bright yellow - - - #14 Yellow - - - #15 Light yellow - - - Part : Reduction using Ascorbate Table 2. Reduction of Cytochrome C and FMN with ascorbate (vitamin C) monitored at the respective wavelengths of 550 nm and 450 nm. Protein Fraction Color Wavelength Absorbance before Ascorbate Absorbance after Ascorbate Flavin Mononucleotide (FMN) #13 Bright Yellow 450 nm 2.0236 1.9819 Cytochrome C #8 Reddish Brown 550 nm 1.0690 1.5765

Calculations  Fraction #1: A = ε x L x C ∆C(mol/min) = ∆A/(ε x L) ∆C(mol/min) = 0.04275708/(6250 x 1) = 6.841 x 10 -6 mol/min NADH (mol/min) = ∆C x Tv = (6.841 x 10 -6 ) x (1.059 x 10 -3 ) = 7.245 x 10 -9 mol/min Enzyme activity =0.007245 μmol/min %LDH activity = (0.007245 μmol/min / 0.1277 μmol/min) x 100% = 5.67 %  Fraction #2: A = ε x L x C ∆C(mol/min) = ∆A/(ε x L) ∆C(mol/min) = 0.1442514/(6250 x 1) = 2.308 x 10 -5 mol/min NADH (mol/min) = ∆C x Tv = (2.308 x 10 -5 ) x (1.059 x 10 -3 ) = 2.444 x 10 -8 mol/min Enzyme activity =0.02444 μmol/min %LDH activity = (0.02444 μmol/min / 0.1277 μmol/min) x 100% = 19.1%  Fraction #3: A = ε x L x C ∆C(mol/min) = ∆A/(ε x L) ∆C(mol/min) = 0.7537793/(6250 x 1) = 1.206 x 10 -4 mol/min NADH (mol/min) = ∆C x Tv = (1.206 x 10 -4 ) x (1.059 x 10 -3 ) = 1.277 x 10 -7 mol/min Enzyme activity =0.1277 μmol/min %LDH activity = 100% because the highest enzyme activity.  Fraction #4: A = ε x L x C ∆C(mol/min) = ∆A/(ε x L) ∆C(mol/min) = 0.1865322/(6250 x 1) = 2.985 x 10 -5 mol/min NADH (mol/min) = ∆C x Tv = (2.985 x 10 -5 ) x (1.059 x 10 -3 ) = 3.161 x 10 -8 mol/min Enzyme activity =0.03161 μmol/min %LDH activity = (0.03161 μmol/min / 0.1277 μmol/min) x 100% = 24.8%  Fraction #5: Column Chromatography and Redox Biochemistry . 2

A = ε x L x C ∆C(mol/min) = ∆A/(ε x L) ∆C(mol/min) = 0.02583046/(6250 x 1) = 4.132 x 10 -6 mol/min NADH (mol/min) = ∆C x Tv = (6.841 x 10 -6 ) x (1.059 x 10 -3 ) = 4.377 x 10 -9 mol/min Enzyme activity =0.004377 μmol/min %LDH activity = (0.004377 μmol/min / 0.1277 μmol/min) x 100% = 3.43% Post-Lab Questions Part 1: 1. In what order did you expect FMN, LDH and Cytochrome C to elute from the column. Explain Why. Gel-Filtration chromatography is conducted to separate the species of proteins or molecules present in the solution according to their size. The column is packed with resin Bio Gel P-200 which contains fine porous beads made of an insoluble polymer. Therefore, the small molecules are capable of entering these porous beads and therefore spend a lot more time in the resin, whereas the larger molecules in the fractionation range of 30-200 kDa, are able to pass through the resin relatively quickly. Thus, this technique allows the large molecules to pass through the column quickly and be the first to elute form the column, whereas the small ones remain in the column chromatography elute the last, since they are in the porous beads longer and travel slower. The Sample mixture provided contained LDH, Cytochrome C and FMN. It was expected that LDH would be the first to elute from the column since it is the largest protein, with a molecular size of 140 kDa and is composed of around 322 residue amino acids. Cytochrome C would be the second and FMN would be the last to emerge out of the column because of their respective sizes of 12kDa and 0.46kDa. 2. Did these three molecules elute in the expected order? If not, suggest a reason why? The three molecules indeed elute in the expected order, with LDH being concentrated in fraction #3, Cytochrome C in fraction #8 and FMN in fraction #13. It was also easy to distinguish the order for cytochrome C and FMN since they are colored reddish brown and yellow respectively. The fractions with the brightest colors were assumed to be the ones with the species being most concentrated in. For LDH, since it is colorless, multiple LDH assays were conducted on the clear fractions and LDH activity was found to be the highest in the fraction #3. 3. What was the purpose of assaying LDH activity in the column fractions? The purpose of assaying LDH activity in the column fractions was to determine which of the fractions had LDH concentrated. And since LDH is colorless, it was difficult to determine its presence in a fraction with naked eye. It was expected that LDH would be the first to be separated since it is the largest protein out of the three, therefore assaying Column Chromatography and Redox Biochemistry . 3

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version

Access to all documents
Unlimited textbook solutions
24/7 expert homework help

LDH was also necessary to confirm whether the order of elution of the three species predicted was indeed attained. 4. What wavelengths did you choose to measure FMN and Cytochrome C reduction? Why? To monitor the reduction of FMN with ascorbate, the absorbance was observed at the wavelength of 450 nm, because according to the absorbance vs wavelength graph of FMN on page 75 of the lab manual, there is a significant difference between the absorbances of oxidized FMN and reduced FMN. This wavelength was also chosen because it demonstrates the maximum difference of absorbances for reduced and oxidized specie and no other wavelength on the graph can demonstrate this difference. This huge difference in absorbances will allow to observe the reduction happening after adding ascorbate as the absorbance should decrease when FMN is reduced. The wavelength for Cytochrome C reduction was set at 550 nm for the same reason. Since at 550 nm, reduced cytochrome C demonstrates a peak, therefore we should observe an elevation in the absorption observed at this wavelength after adding ascorbate. 5. Did Ascorbate reduce Cytochrome C or FMN or both? What evidence you have for this? Explain your results in terms of redox potentials of the species involved. You have to look for the redox potential of all three species. Cytochrome C was reduced with ascorbate and its absorbance before and after adding ascorbate was monitored at the wavelength of 550nm. According to the absorbance vs wavelength graph of Cytochrome C on page 75 of the biochemistry lab manual, cytochrome C should portray an increase in absorbance at 550 nm when it is reduced, and this was proved by the experiment as the absorbance recorded after adding ascorbate was increased by 0.5075 A, and a color change was also observed after adding ascorbate, therefore Cytochrome C was indeed reduced. Looking at the graph of absorbance vs wavelength for FMN, it is indicated that FMN should record a significant decrease in absorbance at 450 nm when it is reduced. This was also portrayed as expected: FMN displayed a decrease of 0.0417 A in absorbance at 450 nm, however no color change was observed. LDH was used to reduce NADH, and this reduction was successful as proved by the LDH activity observed at 340 nm, oxidized NAD+ should show no absorbance at this wavelength. Even though all the species were reduced, some were more reduced than the others and this is due to their redox potentials. Based on the table of standard reduction potentials, the reduction potential of FMN to FMNH 2 and of Cyt C is -0.22 and +0.07 respectively (Berg et al., 529). A negative reduction potential indicates that the oxidized form has lower affinity for electrons, whereas a positive reduction potential indicates the contrary. Therefore, since FMN showed lower absorption differences, and since it has a negative reduction potential, it is concluded that FMN does not like to be reduced, whereas cytochrome C does, since it displayed a fairly large difference in absorbances before and after adding ascorbate and since it has a positive reduction potential. Column Chromatography and Redox Biochemistry . 4

6. Predict the results if you had measured the absorbance at 400nm of all the fractions. If the absorbance of cytochrome C were measured at 400 nm, it would be difficult to observe the difference between reduced and oxidized form of cytochrome C, since both forms demonstrate a similar absorption peak at that wavelength. However, for FMN, a difference in the absorptions can still be detected because there is difference between the absorbances of oxidized and reduced FMN, however it is still not the maximum difference between the peaks therefore it is more accurate to measure the reduction at the wavelengths where there is a signification and maximum difference between the peaks of absorbances of oxidized and their reduced species. Also, for the LDH assay, there is no absorbance at 400 nm, so that would lead to no results. Column Chromatography and Redox Biochemistry . 5

References 1. Biochemistry I- Laboratory Manual Chem 271. Dept. of Chemistry, Concordia University. ISBN 978-1-5251-1274-4. 2. Berg, J. M., Tymoczko, J. L., Gatto, G. J., & Stryer, L. (2015). Biochemistry (8th ed.). W. H. Freeman. Column Chromatography and Redox Biochemistry . 6

Your preview ends here

Eager to read complete document? Join bartleby learn and gain access to the full version