PRE-LAB Buffer preparation. Do calculations (show work) to determine how much of

Question

PRE-LAB Buffer preparation. Do calculations (show work) to determine how much of each reagent is needed for preparation of the Eluent Buffer 0.020M sodium citrate (provided asthe 1. Make 500 ml of Eluant Buffer Eluent buffer is 0.02M (NHSO, dihydrate: M.W.. 294.10 g/mol) and 0.001 M ethylenediaminetetraacetic acid (EDTA), (provided as the disodium salt dihydrate, commonly acronymned Na EDTA 2HO: M.W., 372.24) 1 p Mass of (NH)SO. to use: Mass of sodiumcitrate to use: Mass of EDTA to use: 2. Imagine you have a mixture of two proteins, one 17 kda and the other 50 kda that you ran through a gel tration column. Which protcin would clhute firs Esplain your rationale, in your own words, based on the principles of how gel filtration works to separate protein mixtures. 1 pt

Explanation / Answer

(NH4)2SO4 molecular weight = 132.14 gm

1000ml 1(M) (NH4)2SO4 contains 132.14 gm

1000ml, 0.02(M) (NH4)2SO4 contains 132.14 X 0.02/ 1=2.6428gm(NH4)2SO4

500ml, 0.02(M) (NH4)2SO4 contains 2.6428X 500/1000=1.3214 gm (NH4)2SO4.

sodium citrate molecular weight: 294.10g / mol

1000ml 1(M) sodium citrate contains 294.10 gm

1000l, 0.020( M) contains 294.10 X 0.020/1=5.882gm

500ml, 0.02 (M) sodium citrate contains = 5.882 X 500/ 1000=2.941 gm of sodium citrate.

Molecular weight of EDTA = 372.24

1000ml, 1(M) EDTA contains 372.24 gm of EDTA

1000ml, 0.001 EDTA contains 372.24 X 0.001/1=0.37224 g of EDTA

thus 500 ml of 0.001 (M) EDTA will contains 0.37224 X 500/ 1000 =0.18612 gm EDTA

2. Size exclusion chromatography is used to separate protein based on their molecular weight. SIze exclusion columns are packed with resin beads. As large protein moleculae can not enter the pores of the resin beads they are passed through faster through the column. Whereas, smaller proteins get trapped inside the beads and thus requires longer time to elute from the column. As a result a 50Kd protein will elute faster than 17KDa protein.

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