Mathematical problems!
Posted: Thu May 27, 2010 12:37 pm
Hi. Im new to this chromatography and have some problem to solve some questions! I will be grateful if you just take a look at the questions and give me answer or some tips!
1: How does the retention time of a substance if the flow rate of the mobile phase is increased?
2: Which parameter would remain unaffected if you use a temperature gradient instead of an isothermal run? 1) elutionorder, 2) retention times, or 3) peak heights?
3: Extinction coefficient of hemoglobin (55 kDa Mw) is at 577 nm, 1.46 x 104 M-1 cm-1. Whitch length of a cuvette would be to have an absorbance of 1.05 when measured by a blood test containing hemoglobin 105g / L?
4: You have access to 8 ml of a polyclonal antibody to use in an immunoassay. For this you need 5 ml with a concentration of antibody at 1 mg / ml. You decide antibody concentration photometrically when you know that 1 mg of IgG / ml exhibits an absorbance of 1.26 in a 1cm cuvette at 280 nm. The results of your measurements show that the absorbance at 280 nm of your solution is 0.875. What is the purity of% with respect to IgG in your original antibody solution? If you find that the antibody concentration is less than what the analysis requires what can you do to overcome this problem so that you can use the antibody in your analysis?
5: The analysis of our sample which we received at ammoniumsulfatutfällningen with SDS-PAGE, we found two proteins of interest A and B in the second pellet.
Determine the molar mass of protein A (28mm) and B (54mm). Walking distance and molecular weight of standards and samples are; serum albumin (67kDa, 11mm), ovalbumin (45kDa, 23 mm), carbonic anhydrase (32 kDa, 34), trypsininhibitor (21.5 kDa, 46 mm) and lysosym (14.4 kDa , 59 mm).
6: 3rd The fraction of the interesting proteins A and B being analyzed with SEC-chromatography because we are interested in the native molar mass of each protein. The bottom line obtained from the gel filtration was Vo = 5.2 ml, Vt = 13.3 ml and Ve for A = 9.9 ml, B = 10.3 ml, tyroglobulin (669 kDa) = 7.7 ml, catalase ( 232 kDa) = 9.4 ml, alcohol dehydrogenase (150 kDa) = 9.8 ml, serum albumin (67 kDa) = 10.0 ml, ovalbumin (43 kDa) = 10.7 ml and RNase (13.7 kDa) = 12 , 0 ml.
a. What is the molar mass of A and B with this method?
b. How are the results of this method compared to those obtained with SDS-PAGE. Provide a reasonable explanation for deviations.
c. Give some way to determine Vt.
1: How does the retention time of a substance if the flow rate of the mobile phase is increased?
2: Which parameter would remain unaffected if you use a temperature gradient instead of an isothermal run? 1) elutionorder, 2) retention times, or 3) peak heights?
3: Extinction coefficient of hemoglobin (55 kDa Mw) is at 577 nm, 1.46 x 104 M-1 cm-1. Whitch length of a cuvette would be to have an absorbance of 1.05 when measured by a blood test containing hemoglobin 105g / L?
4: You have access to 8 ml of a polyclonal antibody to use in an immunoassay. For this you need 5 ml with a concentration of antibody at 1 mg / ml. You decide antibody concentration photometrically when you know that 1 mg of IgG / ml exhibits an absorbance of 1.26 in a 1cm cuvette at 280 nm. The results of your measurements show that the absorbance at 280 nm of your solution is 0.875. What is the purity of% with respect to IgG in your original antibody solution? If you find that the antibody concentration is less than what the analysis requires what can you do to overcome this problem so that you can use the antibody in your analysis?
5: The analysis of our sample which we received at ammoniumsulfatutfällningen with SDS-PAGE, we found two proteins of interest A and B in the second pellet.
Determine the molar mass of protein A (28mm) and B (54mm). Walking distance and molecular weight of standards and samples are; serum albumin (67kDa, 11mm), ovalbumin (45kDa, 23 mm), carbonic anhydrase (32 kDa, 34), trypsininhibitor (21.5 kDa, 46 mm) and lysosym (14.4 kDa , 59 mm).
6: 3rd The fraction of the interesting proteins A and B being analyzed with SEC-chromatography because we are interested in the native molar mass of each protein. The bottom line obtained from the gel filtration was Vo = 5.2 ml, Vt = 13.3 ml and Ve for A = 9.9 ml, B = 10.3 ml, tyroglobulin (669 kDa) = 7.7 ml, catalase ( 232 kDa) = 9.4 ml, alcohol dehydrogenase (150 kDa) = 9.8 ml, serum albumin (67 kDa) = 10.0 ml, ovalbumin (43 kDa) = 10.7 ml and RNase (13.7 kDa) = 12 , 0 ml.
a. What is the molar mass of A and B with this method?
b. How are the results of this method compared to those obtained with SDS-PAGE. Provide a reasonable explanation for deviations.
c. Give some way to determine Vt.