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Chapter 5: Problem 4

The \(E\). coli nickel-binding protein binds to its ligand, \(\mathrm{Ni}^{2+}\), with a \(K_{\mathrm{d}}\) of \(100 \mathrm{~nm}\). Calculate the \(\mathrm{Ni}^{2+}\) concentration when the fraction of binding sites occupied by the ligand \((Y)\) is (a) \(0.25\), (b) \(0.6\), (c) \(0.95 .\)

Short Answer

Expert verified
To find \\([ ump{Ni}^{2+}]\\), use \\([ ump{Ni}^{2+}] = \frac{YK_d}{1-Y} \\).

Step by step solution

01

Understanding the Binding Equation

To tackle this problem, we need to use the equation for fractional occupancy \( Y \) in terms of the dissociation constant \( K_d \) and the ligand concentration \( [ump{Ni}^{2+}] \): \[ Y = \frac{[ump{Ni}^{2+}]}{K_d + [ump{Ni}^{2+}]} \].This equation describes the fraction of binding sites occupied \( Y \) in terms of the concentration of the ligand and the dissociation constant \( K_d \). We will solve for the concentration of \( [ump{Ni}^{2+}] \) at given occupancy levels.
02

Rearrange the Equation for [ ump{Ni}^{2+}]

We need to rearrange the formula \( Y = \frac{[ump{Ni}^{2+}]}{K_d + [ump{Ni}^{2+}]} \) to solve for \([ump{Ni}^{2+}]\). Multiplying both sides by \(K_d + [ump{Ni}^{2+}]\) and simplifying gives us: \[ Y(K_d + [ump{Ni}^{2+}]) = [ump{Ni}^{2+}], \],which can be further rearranged to:\[ YK_d + Y[ump{Ni}^{2+}] = [ump{Ni}^{2+}]. \]Finally, isolating \([ump{Ni}^{2+}]\) gives:\[ (1-Y)[ump{Ni}^{2+}] = YK_d \]\[ [ump{Ni}^{2+}] = \frac{YK_d}{1-Y} \].

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Key Concepts

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

Ligand Binding
Ligand binding is a fundamental concept in biochemistry. It describes the process where molecules, called ligands, interact with specific proteins or receptors. This interaction often occurs at a binding site characterized by its shape, charge, and hydrophobic or hydrophilic properties that complement the ligand. Ligands can be various types of molecules, such as ions, small molecules, or even other proteins. When a ligand binds to a protein, it can trigger a biological response, which is crucial for many cellular processes. For example, hormones binding to their receptors to initiate signaling pathways.

In the context of our exercise, the ligand in question is the nickel ion \(\mathrm{Ni}^{2+}\). The specific protein in \(E. coli\) binds this ion with a certain affinity, measured by how tightly or weakly it holds onto the \(\mathrm{Ni}^{2+}\). Understanding this binding helps us solve how much of the ligand is required to reach certain levels of protein activation, also known as fractional occupancy.
Dissociation Constant
The dissociation constant, \(K_d\), is a critical parameter in the study of ligand-protein interactions. It provides a quantitative measure of the affinity between a ligand and its binding site on a protein. The \(K_d\) value is defined as the concentration of the ligand at which half of the available binding sites are occupied.

A low \(K_d\) indicates strong binding affinity, meaning the ligand binds tightly to the protein. Conversely, a high \(K_d\) suggests weaker binding affinity. For the \(E. coli\) nickel-binding protein, the \(K_d\) of 100 nm tells us that at this concentration of \(\mathrm{Ni}^{2+}\), half of the protein's binding sites are occupied. This knowledge aids in calculating \(\mathrm{Ni}^{2+}\) concentrations for different levels of site occupancy.
Fractional Occupancy
Fractional occupancy, or \(Y\), refers to the proportion of binding sites on a protein that are occupied by ligand molecules. It is a dimensionless number ranging from 0 to 1, where 0 indicates no occupied sites and 1 indicates full occupation. This concept is central to understanding how effectively a ligand can engage a protein at given concentrations.

The equation \( Y = \frac{[\mathrm{Ni}^{2+}]}{K_d + [\mathrm{Ni}^{2+}]}\) allows us to compute \(Y\) based on the known \(K_d\) and the concentration of the ligand. By rearranging this equation, as shown in the solution, we find the necessary ligand concentration to achieve a specific fractional occupancy. For example, for \(Y = 0.95\), the ligand concentration reflects the scenario where most of the binding sites are occupied, indicating strong binding conditions. This calculation is critical for applications such as drug design, where target saturation is often required for therapeutic efficacy.

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Most popular questions from this chapter

Under appropriate conditions, hemoglobin dissociates into its four subunits. The isolated \(a\) subunit binds oxygen, but the \(\mathrm{O}_{2}\)-saturation curve is hyperbolic rather than sigmoid. In addition, the binding of oxygen to the isolated \(a\) subunit is not affected by the presence of \(\mathrm{H}^{+}, \mathrm{CO}_{2}\), or BPG. What do these observations indicate about the source of the cooperativity in hemoglobin?

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?

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.

An antibody with high affinity for its antigen has a \(K_{\mathrm{d}}\) in the low nanomolar range. Assume an antibody binds an antigen with a \(K_{\mathrm{d}}\) of \(5 \times 10^{-8}\) M. Calculate the antigen concentration when \(Y\), the fraction of binding sites occupied by the ligand, is a. \(0.4\), b. \(0.5\), c. \(0.8\), d. \(0.9\).

Some pathogens have developed mechanisms to evade the immune system, making it difficult or impossible to develop effective vaccines against them. a. African sleeping sickness is caused by a protozoan called Trypanosoma brucei, carried by the tsetse fly. The trypanosome surface is dominated by one coat protein, the variable surface glycoprotein (VSG). The trypanosome genome encodes over 1,000 different versions of VSG. All of the cells in an initial infection feature the same VSG coat on their surfaces, and this is readily recognized as foreign by the immune system. However, an individual trypanosome in the broader population will switch and randomly begin expressing a different variant of the VSG coat. All the descendants of that cell will have the new and different protein on their surface. As the population with the second VSG coat increases, an individual cell will then switch to a third VSG protein coat, and so on. b. The human immunodeficiency virus (HIV) has an error-prone system for replicating its genome, effectively introducing mutations at an unusually high rate. Many of the mutations affect the viral protein coat. Describe how each pathogen can survive the immune response of its host.
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