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Chapter 5: Problem 1

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.

Short Answer

Expert verified
Protein B has a higher affinity for ligand X with a \(K_{a}\) of \(2.5 \times 10^{7} \: M^{-1}\).

Step by step solution

01

Understanding Kd and Ka

The dissociation constant, denoted as \(K_{d}\), measures the affinity of a protein for its ligand. It is inversely related to the association constant, \(K_{a}\). A lower \(K_{d}\) indicates a higher affinity, while \(K_{a}\) is calculated as the reciprocal of \(K_{d}\), \(K_{a} = \frac{1}{K_{d}}\).
02

Calculate Ka for Protein A

Protein A has a \(K_{d}\) of \(3.0 \times 10^{-7} \: M\). To find \(K_{a}\), compute:\[ K_{a} = \frac{1}{K_{d}} = \frac{1}{3.0 \times 10^{-7}} \]Calculating this gives:\[ K_{a} = 3.33 \times 10^{6} \: M^{-1} \]
03

Calculate Ka for Protein B

Protein B has a \(K_{d}\) of \(4.0 \times 10^{-8} \: M\). To find \(K_{a}\), compute:\[ K_{a} = \frac{1}{K_{d}} = \frac{1}{4.0 \times 10^{-8}} \]Calculating this gives:\[ K_{a} = 2.5 \times 10^{7} \: M^{-1} \]
04

Compare Affinities

Protein A has a \(K_{a}\) of \(3.33 \times 10^{6} \: M^{-1}\), while Protein B has a \(K_{a}\) of \(2.5 \times 10^{7} \: M^{-1}\). Since a higher \(K_{a}\) indicates a higher affinity, Protein B has a higher affinity for ligand X than Protein A.

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Key Concepts

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

Dissociation Constant (Kd)
The dissociation constant, represented as \( K_d \), is a key concept in understanding protein-ligand interactions. It is an indicator of the stability of the complex formed between a protein and its ligand. In simple terms, \( K_d \) helps us understand how easily a ligand can leave a protein once it is bound. A lower \( K_d \) value means that the ligand binds more tightly to the protein and is less likely to dissociate, resulting in a higher affinity of the protein for the ligand. On the other hand, a higher \( K_d \) value indicates a weaker binding, as the ligand more readily dissociates from the protein. :
  • In this exercise, Protein A has a \( K_d \) of \( 3.0 \times 10^{-7} \) M, suggesting a weaker binding compared to Protein B.
  • Protein B, with a \( K_d \) of \( 4.0 \times 10^{-8} \) M, forms a stronger complex with the ligand.
Association Constant (Ka)
The association constant, denoted as \( K_a \), is the direct counterpart to \( K_d \). While \( K_d \) focuses on the dissociation of a ligand from a protein, \( K_a \) measures how likely the protein and ligand are to associate or bind. The relationship between \( K_a \) and \( K_d \) is mathematically inversed; thus, \( K_a = \frac{1}{K_d} \).

Practically Calculated Examples:

  • For Protein A, \( K_a \) is calculated as \( \frac{1}{3.0 \times 10^{-7}} = 3.33 \times 10^{6} \) M-1.
  • For Protein B, \( K_a \) is computed as \( \frac{1}{4.0 \times 10^{-8}} = 2.5 \times 10^{7} \) M-1.
These calculations reveal that the higher the \( K_a \), the greater the affinity of the protein for its ligand. Protein B displays a higher \( K_a \), which suggests a stronger propensity for forming complexes with ligand \( X \).
Affinity
Affinity refers to the strength of binding between a protein and its ligand. A higher affinity means the protein and ligand bind more tightly and remain in complex more favorably. This concept is fundamentally linked to both \( K_d \) and \( K_a \).

Key Takeaways on Affinity:

  • A higher \( K_a \) suggests a higher affinity, whereas a lower \( K_d \) also indicates a higher affinity.
  • For practical understanding, consider the exercise conclusion: Protein B, having a \( K_a \) of \( 2.5 \times 10^{7} \) M-1 and a lower \( K_d \) value, shows a higher affinity than Protein A.
  • This means that Protein B is more effectively able to bind and hold onto ligand \( X \) compared to Protein A.
The interplay of these constants helps scientists and researchers determine the most suitable bindings for biological purposes, such as drug design and understanding metabolic pathways.

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Most popular questions from this chapter

When a vertebrate dies, its muscles stiffen as they are deprived of ATP, a state called rigor mortis. Using your knowledge of the catalytic cycle of myosin in muscle contraction, explain the molecular basis of the rigor state.

A team of biochemists uses genetic engineering to modify the interface region between hemoglobin subunits. The resulting hemoglobin variants exist in solution primarily as \(a \beta\) dimers (few, if any, \(\alpha_{2} \beta_{2}\) tetramers form). Are these variants likely to bind oxygen more weakly or more tightly? Explain your answer.

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?

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) \(0.95 .\)

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?
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