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The protein calcineurin binds to the protein calmodulin with an association rate of \(8.9 \times 10^{3} \mathrm{M}^{-1} \mathrm{~s}^{-1}\) and an overall dissociation constant, \(K_{\mathrm{d}}\), of 10 \(\mathrm{n} \mathrm{M}\). Calculate the dissociation rate, \(k_{\mathrm{d}}\), including appropriate units.

Short Answer

Expert verified
The dissociation rate, \(k_d\), is \(8.9 \times 10^{-5} \mathrm{s}^{-1}\).

Step by step solution

01

Understand the Relationship

The association rate, the dissociation constant, and the dissociation rate are part of a relationship given by the equation: \( K_d = \frac{k_d}{k_a} \), where \(K_d\) is the dissociation constant, \(k_d\) is the dissociation rate, and \(k_a\) is the association rate.
02

Reorganize the Equation

To find \(k_d\), reorganize the equation: \( k_d = K_d \times k_a \). This allows us to compute the unknown dissociation rate.
03

Convert Units

Convert the dissociation constant \(K_d\) from nanomolar (nM) to molar (M) for consistency, as 1 nM = \(1 \times 10^{-9}\) M. Thus, 10 nM = \(10 \times 10^{-9}\) M = \(1 \times 10^{-8}\) M.
04

Substitute Known Values

Substitute the known values into the equation: \( k_d = (1 \times 10^{-8} \, \text{M}) \times (8.9 \times 10^3 \, \text{M}^{-1} \, \text{s}^{-1}) \).
05

Calculate the Dissociation Rate

Perform the multiplication to find \(k_d\): \( k_d = (1 \times 10^{-8} \, \text{M}) \times (8.9 \times 10^3 \, \text{M}^{-1} \, \text{s}^{-1}) = 8.9 \times 10^{-5} \, \text{s}^{-1} \).
06

Evaluate Units and Result

Ensure that the units cancel out correctly. The result has units of \( s^{-1} \), indicating a dissociation rate. Therefore, \( k_d = 8.9 \times 10^{-5} \, \text{s}^{-1} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Biochemical Kinetics
Biochemical kinetics explores the rate at which biological processes occur. It focuses on how quickly reactions reach equilibrium. This is crucial in understanding how proteins interact within cells. The speed of these interactions impacts cellular functions and overall organism health. Calculating rates such as association and dissociation helps scientists understand these processes.
To measure how proteins bind and separate, scientists often use experimental data. For instance, the association rate measures how fast two molecules bind, while the dissociation rate measures how fast the complex breaks apart. Together, these rates provide a comprehensive view of the dynamic nature of biological interactions.
Protein Interaction Analysis
Protein interaction analysis is the study of how proteins communicate and bond with each other. This field is essential since proteins perform numerous cellular tasks. Understanding these interactions can reveal the molecular mechanisms of diseases.
Studying interactions involves a variety of techniques like surface plasmon resonance and mass spectrometry. These methods help quantify how tightly a protein binds to its partner, which is critical in drug design. By evaluating the strength and rate of these bindings, researchers can manipulate these interactions for therapeutic purposes. Ultimately, protein interaction analysis helps bridge the gap between molecular biology and medicine.
Association and Dissociation Constants
Association and dissociation constants are fundamental concepts in molecular biology. The association constant, often denoted as \( k_a \), reflects how likely a protein will bind to another molecule. In contrast, the dissociation constant, \( K_d \), indicates how readily a complex will disassemble.
\( K_d \) is often used to describe a protein's affinity for a ligand. Lower \( K_d \) values suggest stronger binding, meaning the molecules are more stable together. The dissociation rate, \( k_d \), can be calculated using the formula \( K_d = \frac{k_d}{k_a} \). This calculation is vital in developing inhibitors or activators in drug therapy. Scientists can adjust these constants to control biological processes effectively.
Molecular Biology
Molecular biology involves understanding the molecular mechanisms within cells. It focuses on the interaction and regulation of genetic materials. This field has led to breakthroughs such as gene editing and cloning.
At the heart of molecular biology is the study of macromolecules, including DNA, RNA, and proteins. Proteins, being the workhorses of the cell, are the primary targets in molecular biology studies. By understanding how proteins interact and their kinetics, researchers can elucidate the entire biochemistry of living organisms. Molecular biology has reshaped our comprehension of life and continues to be a cornerstone of modern biological research.

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Most popular questions from this chapter

To fully appreciate how proteins function in a cell, it is helpful to have a threedimensional view of how proteins interact with other cellular components. Fortunately, this is possible using online protein databases and three- dimensional molecular viewing utilities such as JSmol, a free and user- friendly molecular viewer that is compatible with most browsers and operating systems. In this exercise, examine the interactions between the enzyme lysozyme and the Fab portion of the antilysozyme antibody. Use the PDB identifier 1FDL to explore the structure of the IgG1 Fab fragment-lysozyme complex (antibody- antigen complex). To answer the questions, use the information on the Structure Summary page at the Protein Data Bank (www.rcsb.org), and view the structure using JSmol or a similar viewer. a. Which chains in the three-dimensional model correspond to the antibody fragment, and which correspond to the antigen, lysozyme? b. What type of secondary structure predominates in this Fab fragment? c. How many amino acid residues are in the heavy and light chains of the Fab fragment? In lysozyme? Estimate the percentage of the lysozyme that interacts with the antigen- binding site of the antibody fragment. d. Identify the specific amino acid residues in lysozyme and in the variable regions of the Fab heavy and light chains that are situated at the antigen- antibody interface. Are the residues contiguous in the primary sequence of the polypeptide chains?

Protein A has a binding site for ligand \(\mathrm{X}\) with a \(K_{\mathrm{d}}\) of \(3.0 \times 10^{-7}\) ?. Protein \(\mathrm{B}\) has a binding site for ligand \(\mathrm{X}\) with a \(K_{\mathrm{d}}\) of \(4.0 \times 10^{-8} \mathrm{M}\). Calculate the \(K_{\mathrm{a}}\) for each protein. Which protein has a higher affinity for ligand X? Explain your reasoning.

A monoclonal antibody binds to Gactin but not to F-actin. What does this tell you about the epitope recognized by the antibody?

Which of these situations would produce a Hill plot with \(n_{\mathrm{H}}<1.0\) ? Explain your reasoning in each case. a. The protein has multiple subunits, each with a single ligand-binding site. Ligand binding to one site decreases the binding affinity of other sites for the ligand. b. The protein is a single polypeptide with two ligandbinding sites, each having a different affinity for the ligand. c. The protein is a single polypeptide with a single ligand-binding site. As purified, the protein preparation is heterogeneous, containing some protein molecules that are partially denatured and thus have a lower binding affinity for the ligand. d. The protein has multiple subunits, each with a single ligand-binding site. Ligands bind independently to each site, do not affect the binding affinity of other sites, and bind with identical affinities.

Studies of oxygen transport in pregnant mammals show that the \(\mathrm{O}_{2}\) saturation curves of fetal and maternal blood are markedly different when measured under the same conditions. Fetal erythrocytes contain a structural variant of hemoglobin, HbF, consisting of two \(a\) and two \(\gamma\) subunits \(\left(\alpha_{2} \gamma_{2}\right)\), whereas maternal erythrocytes contain \(\mathrm{HbA}\left(\alpha_{2} \beta_{2}\right)\). a. Which hemoglobin has a higher affinity for oxygen under physiologic conditions? b. What is the physiological significance of the different \(\mathrm{O}_{2}\) affinities? When all the BPG is carefully removed from samples of \(\mathrm{HbA}\) and \(\mathrm{HbF}\), the measured \(\mathrm{O}_{2}\)-saturation curves (and consequently the \(\mathrm{O}_{2}\) affinities) are displaced to the left. However, HbA now has a greater affinity for oxygen than does HbF. When BPG is reintroduced, the \(\mathrm{O}_{2}\)-saturation curves return to normal, as shown in the graph. c. What is the effect of BPG on the \(\mathrm{O}_{2}\) affinity of hemoglobin? How can this information be used to explain the different \(\mathrm{O}_{2}\) affinities of fetal and maternal hemoglobin?

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