Results:
Part A: Calorimetric analysis
Table 1: Absorbance Values for Standard Solution, Sample X and Sample Y.
Figure 1: Graph of absorbance, A660 against mass of phosphorus (µg).
Calculation of concentration of X and Y:
Using the equation from the graph above, y = 0.0004x,
When the average absorbance of diluted X is 0.015,
0.015 = 0.0004x
x =
= 37.5 µg
Concentration of undiluted sample X = 37.5µg x 10
= 375 µg
5 mL of phosphate solution was taken from the sample X tube, hence
Concentration of phosphate sample X, µg mL =
= 75 µg mL-1
When the average absorbance of diluted sample Y is 0.007,
0.007 = 0.0004 x
x =
= 17.5 µg
Concentration of undiluted sample Y = 17.5µg x 10
= 175µg
5 mL of phosphate solution was taken from the sample Y tube, hence
Concentration of phosphate sample Y =
= 35 µg mL-1
Part B: UV-Vis spectrophotometry
Questions:
Part A: Calorimetric analysis
Table 2: Values for Protein Concentration in mg/mL and Absorbance
Figure 6: A standard graph of absorbance, A540 against protein concentration (mg/mL).
From the graph above, the equation of the straight line is y = 0.0473x,
To calculate the concentration of undiluted sample A, using the equation above,
Concentration of undiluted protein A = 0.27mg/mL
0.27 = 0.0473x
x =
= 5.708 mg mL-1
To calculate the concentration of diluted sample A, using the equation above,
Concentration of diluted protein A = 0.03mg/mL
0.03 = 0.0473x
x =
= 0.6342 mg mL-1
Since this sample has been diluted 10 times, the concentration of the undiluted sample using this value = 0.6342 mg mL-1 × 10
= 6.342 mg mL-1
To calculate the concentration of undiluted sample B, using the equation above,
Concentration of undiluted protein B = 1.15mg/mL
1.15 = 0.0473x
x =
= 24.31 mg mL-1
To calculate the concentration of diluted sample B, using the equation above,
Concentration of undiluted protein B = 0.17mg/mL
0.17 = 0.0473x
x =
= 3.594 mg mL-1
Since this sample has been diluted 10 times, the concentration of the undiluted sample using this value = 3.594 mg mL-1 × 10
= 35.94 mg mL-1
For each sample, two different dilutions are made, because according to the Beer-Lambert’s Law, absorbance is directly proportional to concentration (Stewart & Ebel, 2000). Hence, in order to prove this law, the samples are made into two different concentrations through dilution to see its effects on the absorbance. Moreover, dilution was carry out
For sample A, the undiluted sample would be considered to be more accurate. The discrepancy between the pairs of results can be seen in the difference of absorbance reading. This is because during the process of dilution, the diluted sample could be contaminated with other substances. The presence of foreign particles could affect the absorbance readings hence the calculated diluted concentration would be affected.
For sample B, the diluted sample would be more accurate. This is because the absorbance of the undiluted sample is more than 1. This high absorbance would be the outlier of the plotted graph, affecting the equation which is used to calculate for the concentration. After dilution, the absorbance is lowered, deviation would be smaller, a more appropriate equation could be used to calculate for the concentration, giving a more accurate concentration.
Part B: UV-Vis spectrophotometry
1. Sodium dithionite was used as a reducing reagent which removes the oxygen atoms from haemoglobin molecules and forming deoxyhaemoglobin molecules (Armitrano & Tortora, 2007). Hence, before sodium dithionite was added to the haemoglobin solution, there are three peaks shown in the absorption spectra which are absorbance measured was 0.488 at a wavelength 411nm, 0.058 at a wavelength 540nm, and 0.055 at a wavelength 577nm. However, after sodium dithionite was added, the oxyhaemoglobin became deoxyhaemoglobin where the oxygen is released from the haemoglobin and this result in an absorbance measurement of 0.421 at a wavelength of 428nm. The difference in absorbance and wavelength is due to the structure of the haemoglobin, in oxyhaemoglobin, the haemoglobin is bound to an oxygen atom, while in deoxyhaemoglobin it isn’t (Stewart & Ebel, 2000). Hence, the presence of no bounding atom in the deoxyhaemoglobin causes the haemoglobin group to have high affinity and be more unstable, hence absorbs more energy (light) to excite its electron to a more stable state (Olson, 1976). Where else, oxyhaemoglobin has an oxygen atom attached to the haemoglobin and is already quite stable and is low in affinity, thus absorbing less energy (Olson, 1976). Furthermore, the different wavelength values is due to the colour of the haemoglobin solution before and after adding the sodium dithionite, as different colour will give rise to different wavelength (Stewart & Ebel, 2000).
2. In the absorption spectrum of bovine serum albumin (BSA), there were two peaks with the absorbance 3.498 at the wavelength 229nm and 0.618 at 279nm. The first peak is determined by the peptide bond as peptide bond has a very weak absorption at a wavelength of 210nm and 220nm (Boyer, 2009). As for the second peak, it might contributed by side chains of amino acids (Gore, 2000). Some amino acids which are having aromatic side chains especially tyrosine and tryptophan absorb most of the light in the UV region where the wavelength is lower than 300nm (Gore, 2000). From the protein spectrum obtained above, the wavelength of absorption is at 229nm and 279nm, both are lower than 300nm, it can be said that there is possibility that these peaks are due to the presence of aromatic side chains. Hence, in general term, the peaks in protein absorption spectra is on the basis of amino acid side chains (Gore, 2000). For the RNA spectrum, there were two peaks with wavelength reading of 203nm, absorbance value of 2.504 and 258nm with absorbance reading 1.972. The possible constituent that provides these peaks is either the nitrogenous base pair of pyrimidine or purine (Boyer, 2009). Usually the solution of nucleic acid strongly absorbs light at the region of 260nm (Boyer, 2009). Each purines and pyrimidine has slightly different structures from the others, hence different absorption at slightly different wavelength.
3. (a) Within the protein, the components that are responsible for the absorption maxima observed are the peptide bonds and the aromatic phenyl group consisting of tyrosine, tryptophan and phenylalanine (Boyer, 2009). As for the RNA, the components involve are the pyrimidine and purine bases (Boyer, 2009).
(b) For RNA at 260nm, the absorbance is approximately 0.750 and the extinction coefficient is 25.0 cm-1(mg/mL)-1. Hence, using the Beer-Lambert Law,
A = ɛcl Where, A = Absorbance; ɛ = extinction coefficient
c = Concentration; l = path length (1 cm)
c =
=
= 0.03 mg/mL
(c) Concentration = 1 mg/mL, Absorbance = 0.600. Hence, using the Beer-Lambert Law,
A = ɛcl Where, A = Absorbance; ɛ = extinction coefficient
c = Concentration; l = path length(1 cm)
ɛ =
=
= 0.600 cm-1(mg/mL)-1
(d) mg/mL is used as a concentration unit for protein and nucleic acid, because both molecules are very minute and can only be obtained in a small number. Hence, to use M would need the solution to be in a 1 liter volume which is too big of a ratio compared to mL. By using mg/mL, the molecules can be measured and calculated easily too as there will be less decimal place and zeros to consider.
Conclusion:
The absorbance spectra of haemoglobin, nucleic acids and protein have been obtained. Haemoglobin spectra will be shifted due to the reduction by sodium dithionite solution. Peaks in the spectra of proteins and nucleic acids are dependent on the nature of amino acid side chain and nitrogenous bases respectively. Concentration of inorganic phosphates from sample X and Y, too, have been obtained, which are 75µg mL-1 and 35µg mL-1 respectively..
References:
Armotrano, R., and Tortora G. J. 2007, Components and Origins of Blood, Laboratory Exercises in Anatomy and Physiology with Cat Dissections, Thomson Brooks/Cole, USA, p. 392.
Boyer, R. 2009, Biochemistry Laboratory – Modern Theory and Techniques, Pearson Education, Inc, San Francisco.
Dalziel, K & O’Brien, J.R.P. 1957, Side Reactions in the Deoxygenation of Dilute Oxyhaemoglobin Solutions by Sodium Dithionite, Vol. 67, pp. 119 – 124.
Gore, M. G., (2000), Spectra of Some Important Naturally Occurring Chromophores, Spectrophotometry and Spectrofluorimetry: A Practical Approach, Oxford University Press, New York, pp. 6 – 7.
Livingstone, D. 2002, Data Analysis for Chemists, Oxford University Press, Oxford.
Mikulecky, P., Gilman, R. M., and Brutlag, K., (2009), Chapter 28: Taking the Tour: Recommended AP Experiments, AP Chemistry for Dummies, Wiley Publishing, Inc., Indiana, p. 294.
Olson, J.S. 1976, Spectral differences between the α and β heme groups within human deoxyhaemoglobin, Vol. 73, No. 4, pp. 1140 – 1144.
Sizer, I.W. & Peacock, A.C. 1947, The ultraviolet absorption of serum albumin and of its constituents’ amino acids as a function of pH, PhD thesis, Massachusetts Institute of Technology, Cambridge.
Stewart, K.K. & Ebel, R.E. 2000, Chemical Measurements in Biological Systems, John Wiley & Sons, Inc, New York.
Zijlstra, W. G., Buursma, A., van Assendelft, O. W., (2000), Preparation of Haemoglobin Derivatives, Visible and Near Infrared Absorption Spectra of Human and Animal Haemoglobin, VSP International Science Publishers Netherlands, Netherlands, p. 43.
Pre-lab:
Part B: Spectrophotometric analysis 2: UV-Vis spetrophotometry
1. a)
(i) Molecular weight of glucose = 180 g mol-1
1 mole of glucose = 180g
0.1 mole of glucose = 18g
Weigh 18g of glucose and transfer it to a 1 litre volumetric flask. Fill the volumetric flask with distil water until the half way mark and shake the flask to mix the content. After mixing, add more distilled water until the bottom of the meniscus is at the mark on the volumetric flask. Stopper and mix well. Label the volumetric flask after mixing.
(ii) Molecular weight of ethanol = 46 g mol-1
1 mole of ethanol = 46g
5 mM of ethanol = 0.005M x 46 g mol-1
= 0.230g L-1
Weigh 0.230g of ethanol and transfer it to a 1 litre volumetric flask. Fill the volumetric flask with distil water until the half way mark and shake the flask to mix the content. After mixing, add more distilled water until the bottom of the meniscus is at the mark on the volumetric flask. Stopper and mix well. Label the volumetric flask after mixing.
b)
(i) 1 liter requires 18g of glucose. Hence,
10mL = x 10
= 0.18g
Weigh 0.18g of glucose and transfer it to a 10 mL volumetric flask. Fill the volumetric flask with distill water until the half way mark and shake the flask to mix the content. After mixing, add more distilled water until the bottom of the meniscus is at the mark on the volumetric flask. Stopper and mix well. Label the volumetric flask after mixing.
(ii) 1 liter requires 0.23g of glucose. Hence,
10mL = x 10
= 0.0023g
Weigh 0.0023g of glucose and transfer it to a 10 mL volumetric flask. Fill the volumetric flask with distill water until the half way mark and shake the flask to mix the content. After mixing, add more distilled water until the bottom of the meniscus is at the mark on the volumetric flask. Stopper and mix well. Label the volumetric flask after mixing.
c)
(i) 0.1 mole of glucose at 1mL is being diluted with 99mL water to have a final volume of 100mL. Hence,
C1V1 = C2V2
(0.1) (1) = C2 (99)
C2 = 1.01× 10-3 M
(ii) 5mM of ethanol at 1mL is being diluted with 99mL water to have a final volume of 100mL. Hence,
C1V1 = C2V2
(0.005) (1) = C2 (99)
C2 = 5.05× 10-5 M
2. (a) Concentration of glucose = 26mM
26mM = 26 × 10-3 mol L-1 × 180 g mol-1
= x 180
= x 180g
= 4.68 x 10-3 mg/µL
(b) Amount of glucose present in 30 µL of the cell extract =
4.68 x 10-3 mg/µL x 30 µL
= 0.1404 mg/µL
