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19-C. The protein bovine serum albumin can bind several molecules of the dye methyl orange. To measure the binding constant K for one dye molecule, solutions were prepared with a fixed concentration \(\left( {{X_0}} \right)\) of dye and a larger, variable concentration of protein (P). The equilibrium is Reaction 19-18, with X 5 methyl

orange.

Experimental data are shown in cells A16-D20 in the spreadsheet on the next page. The authors report the increase in absorbance \((\Delta A)\) at 490 nm as P is added to X. X and PX absorb visible light, but P does not. Equilibrium expression 19-20 applies and (PX) is given by Equation 19-21. Before P is added, the absorbance is \({\varepsilon _X}{X_{0 - }}\). The increase in absorbance when P is added is

The spreadsheet uses Solver to vary K and \(\Delta E\) in cells B10:B11 to minimize the sum of squares of differences between observed and calculated \(\Delta A\) in solutions with different amounts of P. Cell E16 computes (PX) from Equation 19-21, which is Equation A on line 6 of the spreadsheet. Cells F16 and G16 find (X) and (P) from mass balances. Cell H16 computes \(\Delta {A_{calc}} = \Delta E(PX)\)which is Equation B on line 7.

To estimate a value of K in cell B10, suppose that 50% of X has reacted in row 20 of the spreadsheet. The total concentration of X is\({X_0} = 5.7\mu M\). If half is reacted, then \((X) = (PX) = 2.85\mu M\) and\((P) = {P_0} - (PX) = 40.4 - 2.85 = 37.55\mu M\). The binding

constant is \(K = (PX)/(P)(X)) = (2.85\mu M)/(37.55\mu M)(2.85\mu M)) = \)\(2.7 \times 1{0^4}\) which we enter as our guess for K in cell B10. We estimate \(\Delta \varepsilon \)in cell B11 by supposing that 50% of X has reacted in row 20. In Equation B on line 7, \(\Delta A = \Delta \varepsilon (PX)\).The measured

value of \((\Delta A)\)in row 20 is 0.0291 and we just estimated that \((PX) = \)\(2.85\mu M\). Therefore, our guess for \(\Delta \varepsilon \)in cell B11 is \(\Delta \varepsilon = \Delta A/\)\((PX) = (0.0291)/(2.85\mu M) = 1.0 \times 1{0^4}\)

Your assignment is to write formulas in columns E through J of the spreadsheet to reproduce what is shown and to find values in cells E17:J20. Then use Solver to find K and \(\Delta \varepsilon \)in cells B10:B11 to minimize \(\Sigma {\left( {{A_{oths\;}} - {A_{calc\;}}} \right)^2}\)in cell I21.

Short Answer

Expert verified

The value of\({\rm{\Delta }}\varepsilon \)and K which is in cell B10:B11 will minimize \({\rm{\Sigma }}{\left( {{A_{{\rm{oths\;}}}} - {A_{{\rm{calc\;}}}}} \right)^2}\)in cell I21. We get the spreadsheet as,

Step by step solution

01

Find K:

The value of\({\rm{\Delta }}\varepsilon \)and K which is in cell B10:B11 will minimize \({\rm{\Sigma }}{\left( {{A_{{\rm{oths\;}}}} - {A_{{\rm{calc\;}}}}} \right)^2}\)in cell I21. We get the spreadsheet as,

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Most popular questions from this chapter

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Scatchard plot for binding of estradiol to albumin. Data in the table come from a student experiment to measure the binding constant of the radioactively labeled hormone estradiol (X)to the protein, bovine serum albumin (P).Estradiol(7.5nM)was equilibrated with various concentrations of albumin for 30minat37°C.A small fraction of unbound estradiol was removed by solid phase microextraction (Section24-4) and measured by liquid scintillation counting. Albumin is present in large excess, so its concentration in any given solution is essentially equal to its initial concentration in that solution. Call the initial concentration of estradiol [X]0and the final concentration of unbound estradiol [X]. Then bound estradiol is[X]0―[X]and the equilibrium constant is

X+P⇌PX   K=[PX][X][P]=[X]0-[X][X][P]

which you can rearrange to
localid="1663648487221" [X]0[X]=K[P]+1

A graph of [X]0/[X]versus [P]should be a straight line with a slope of K.The quotient [X]0/[X]is equal to the counts of radioactive estradiol extracted from a solution without albumin divided by the counts of estradiol extracted from a solution with estradiol. (b) What fraction of estradiol is bound to albumin at the first and last points?

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The protein bovine serum albumin can bind several molecules of the dye methyl orange. To measure the binding constant K for one dye molecule, solutions were prepared with a fixed concentration (x0)of dye and a larger, variable concentration of protein (P). The equilibrium is Reaction 19-18, with X 5 methyl

orange.

Experimental data are shown in cells A16-D20 in the spreadsheet on the next page. The authors report the increase in absorbance∆Aat 490 nm as P is added to X. X and PX absorb visible light, but P does not. Equilibrium expression 19-20 applies and [PX] is given by Equation 19-21. Before P is added, the absorbance is. The increase in absorbance when P is added is

The spreadsheet uses Solver to vary K and ∆Ein cells B10:B11 to minimize the sum of squares of differences between observed and calculatedin solutions with different amounts of P. Cell E16 computes [PX] from Equation 19-21, which is Equation A on line 6 of the spreadsheet. Cells F16 and G16 find [X] and [P] from mass balances. Cell H16 computes Δ´¡calc=Δ·¡[PX]which is Equation B on line 7.

To estimate a value of K in cell B10, suppose that 50% of X has reacted in row 20 of the spreadsheet. The total concentration of X is. If half is reacted, then[X]=[PX]=2.85μ²Ñand[P]=P0−[PX]=40.4−2.85=37.55μ²ÑThe binding

constant is K=[PX]∧P][X])=[2.85μ²Ñ]∧37.55μ²Ñ][2.85μ²Ñ])=2.7×104which we enter as our guess for K in cell B10. We estimatein cell B11 by supposing that 50% of X has reacted in row 20. In Equation B on line 7, Δ´¡=Δε[PX].The measured

value ofin row 20 is 0.0291 and we just estimated that. Therefore, our guess for localid="1668328314124" Δεin cellB11isΔε=Δ´¡/[PX]=(0.0291)(2.85μ²Ñ)=1.0×104in cell B11 is

Your assignment is to write formulas in columns E through J of the spreadsheet to reproduce what is shown and to find values in cells E17:J20. Then use Solver to find K andin cells B10:B11 to minimizeΣAoths−Acalc2in cell I21.

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