Biochem Report1 - Spectrophotometry

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Question Exp 1

FIGURE 1.2 Standard curve for estimation of protein concentration

[pic 2]

Protein concentration of A and B

Calculation:

e.g. Absorbance = 0.0476(protein concentration) – 0.002

Absorbance Aundiluted = 0.0476(protein concentration) – 0.002

Protein concentration = (0.27-0.002) ÷ 0.0476

= 5.63 mg/mL

TABLE 1.4 Protein concentration of sample A and B

Unknown

Absorbance540

Protein concentration mg/mL

A (undil)

0.27

5.63

A(dil 1/10)

0.03

0.59

B(undil)

1.15

24.12

B(dil 1/10)

0.17

3.53

Why are two different dilutions of each sample prepared?

To make comparison as which solution is better to use in finding the protein concentration of the samples.

Explanations for discrepancy between the pairs of result for each sample.

There is differences between the pairs of each samples. For both sample A absorbance is well within the standard curve but the final concentration between the diluted and undiluted solution shows that Adiluted has higher protein concentration than Aundiluted. For undiluted sample B the absorbance value extrapolate the standard curve meanwhile absorbance of diluted sample B stay within the range. Final concentration of protein is higher in Bdiluted than in Bundiluted .

Which results for each sample is more accurate?

For sample A and B, the diluted solution is more accurate as Beer-Lambert Law applies only for dilute solutions.

Experiment 2

Question

- Based on the oxy-haemoglobin spectra, the absorbance peaks is at 0.065 for both 541nm and 575nm wavelength. The red color that the oxy-haemoglobin gives are transmitted by wavelength more than 600nm. Based on the deoxy-haemoglobin spectra, it shows that when the wavelength is 557nm the absorbance peak is at 0.055. Deoxy-haemoglobin gave a bluish red color because the wavelength transmitted are shorter than 500nm which gives bluish color to the blood while at the same time wavelength transmitted more than 600nm gives the red color.(Richard-A, 1982)

- The constituent responsible for peak observed in absorption spectra of protein is tyrosine and tryptophan while in absorption spectra of nucleic acid are the heterocyclic bases of nucleic acids.

- a) Components of the compound that responsible for the absorption maxima absorb at (278nm,0.633) of BSA shows the presence of aromatic amino acids such as tyrosine, tryptophan and phenylalanine (Mach, 1992). While for RNA spectra, the absorption peak is at 2.117 of 258nm wavelength indicates the presence of nucleic acid subunits and purines.

b) c = 2.117/25 ×1 mg/mL

c = 0.085 mg/mL

c) ε = 0.633/(1×1)

= 0.633cm-1 (mg/mL)-1

d) mg/mL molar is more often used for concentration of nucleic acid and protein in BSA and RNA because the nucleic acid and protein exist only in small mass hence it is harder to count it by mM solution.

Reference

Boyer R (2006) Spectroscopic analysis of biomolecules, Biochemistry laboratory: modern theory and techniques, Pearson Benjamin Cummings: New Jersey, 207.

Mach H, Middaugh CR, Lewis RV (1992) Statistical determination of the average values of the extinction coefficients of tryptophan and tyrosine in native proteins. Analytical Biochemistry, 200, 74-80.

Paselk RA (1982) Demonstrating color-wavelength relations. Journal of Chemical Education, 59, 383.

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