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+PPXK=[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?

Short Answer

Expert verified

The fraction of estradiol is bound to albumin at the first and end point are 0.21 and 0.77 respectively.

Step by step solution

01

Define radioactivity.

It is property that is exhibited by certain type of matter of emitting energy and sub atomic particles spontaneously.

02

Calculate fraction of estradiol bound to albumin at first and end points.

Fraction of free estradiol at first point=11.26=0.79

Fraction of free estradiol at end point=14.36=0.23

Fraction of estradiol bound to albumin at first point=1-0.79=0.21

Fraction of estradiol bound to albumin at end point=1-0.23=0.77

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Most popular questions from this chapter

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 absorbanceAat 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 ΔAcalc=ΔE[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]=2.85μMand[P]=P0[PX]=40.42.85=37.55μMThe binding

constant is K=[PX]P][X])=[2.85μM]37.55μM][2.85μM])=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, ΔA=Δε[PX].The measured

value ofin row 20 is 0.0291 and we just estimated that. Therefore, our guess for localid="1668328314124" Δεin cellB11isΔε=ΔA/[PX]=(0.0291)(2.85μM)=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 minimizeΣAothsAcalc2in cell I21.

What is the advantage of a time-resolved emission measurement with Eu3+versus measurement of fluorescence from organic chromophores?

The end of Box 19-2 states that “the advantage of up conversion for biomedical probes is that low energy near-infrared (800 to 1000 nm) incident radiation stimulates little background emission from the complex biological matrix that can be highly fluorescent under visible radiation.” Suggest why near-infrared radiation stimulates less emission than visible radiation and why this behavior is useful.

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.

Explain how signal amplification is achieved in enzyme linked immunosorbent assays.

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