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Chapter 5: Problem 13

A quantitative study of the interaction of a protein with its ligand yielded the following results: Ligand concentration \(1 \quad 2 \quad 3 \quad 4 \quad 5 \quad 6 \quad 9 \quad 12\) \((m M)\) \(\nu\) (moles of ligand \(\begin{array}{lllllll}0.28 & 0.45 & 0.56 & 0.60 & 0.71 & 0.75 & 0.79 & 0.83\end{array}\) bound per mole of protein Plot a graph of [L] versus \(\nu .\) Determine \(K_{\mathrm{D}},\) the dissociation constant for the interaction between the protein and its ligand, from the graph.

Short Answer

Expert verified
After plotting the graph and adding a trendline, \(K_D\) is found at the [L] that corresponds to half the maximum bound per protein. Only an approximate value can be found from the graph. For a more precise number, a mathematical fitting of the data would be needed.

Step by step solution

01

Plot the graph

Create a graph with the ligand concentration [L] on the x-axis and the amount of ligand bound per mole of protein (\(\nu\)) on the y-axis by inserting the given values on their corresponding axis.
02

Create a trendline

Add a trendline to the graph. The data should approach asymptotic behaviour at high [L] indicating a hyperbolic function. The trendline will help to visualize this.
03

Calculate \(K_D\)

The value of \(K_D\) is found at the ligand concentration ([L]) at which half the protein is bound by ligand. To find this, locate the \(\nu\) value (y-axis) that is half of the maximum bound per protein (in this case 0.83/2 = 0.415). Draw a horizontal line from \(\nu = 0.415\) to the curve and drop a vertical line from the point of intersection to get the [L] value (x-axis). This corresponds to the dissociation constant \(K_D\).

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Most popular questions from this chapter

Phosphoproteins are formed when a phosphate group is esterified to an - OH group of a Ser, Thr, or Tyr side chain. At typical cellular \(\mathrm{pH}\) values, this phosphate group bears two negative charges \(-\mathrm{OPO}_{3}^{2-} .\) Compare this side-chain modification to the 20 side chains of the common amino acids found in proteins and comment on the novel properties that it introduces into side-chain possibilities.

Amino acid analysis of an oligopeptide containing nine residues revealed the presence of the following amino acids: \(\begin{array}{llllllll}\text { Arg } & \text { Cys } & \text { Gly } & \text { Leu } & \text { Met } & \text { Pro } & \text { Tyr } & \text { Val }\end{array}\) The following was found: a. Carboxypeptidase A treatment yielded no free amino acid. b. Edman analysis of the intact oligopeptide released c. Neither trypsin nor chymotrypsin treatment of the nonapeptide released smaller fragments. However, combined trypsin and chymotrypsin treatment liberated free Arg. d. CNBr treatment of the 8 -residue fragment left after combined trypsin and chymotrypsin action yielded a 6-residue fragment containing Cys, Gly, Pro, Tyr, and Val; and a dipeptide. e. Treatment of the 6 -residue fragment with \(\beta\) -mercaptoethanol yielded two tripeptides. Brief Edman analysis of the tripeptide mixture yielded only PTH-Cys. (The sequence of each tripeptide, as read from the N-terminal end, is alphabetical if the one-letter designation for amino acids is used.) What is the amino acid sequence of this nonapeptide?

Proteases such as trypsin and chymotrypsin cleave proteins at different sites, but both use the same reaction mechanism. Based on your knowledge of organic chemistry, suggest a "universal" protease reaction mechanism for hydrolysis of the peptide bond.

Describe the synthesis of the dipeptide Lys-Ala by Merrifield's solidphase chemical method of peptide synthesis. What pitfalls might be encountered if you attempted to add a leucine residue to Lys-Ala to make a tripeptide?

Amino acid analysis of a decapeptide revealed the presence of the following products: \(\begin{array}{lllll}\mathrm{NH}_{4}^{+} & \text {Asp } & \text { Glu } & \text { Tyr } & \text { Arg } \\ \text { Met } & \text { Pro } & \text { Lys } & \text { Ser } & \text { Phe }\end{array}\) The following facts were observed: a. Neither carboxypeptidase \(A\) or \(B\) treatment of the decapeptide had any effect. b. Trypsin treatment yielded two tetrapeptides and free Lys. c. Clostripain treatment yielded a tetrapeptide and a hexapeptide. d. Cyanogen bromide treatment yielded an octapeptide and a dipeptide of sequence NP (using the one-letter codes). e. Chymotrypsin treatment yielded two tripeptides and a tetrapeptide. The N-terminal chymotryptic peptide had a net charge of -1 at neutral \(\mathrm{pH}\) and a net charge of -3 at pH 12 f. One cycle of Edman degradation gave the PTH derivative What is the amino acid sequence of this decapeptide?
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