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Under the proper environmental conditions, the salt-loving archaeon Halobacterium halobium synthesizes a membrane protein \(\left(M_{\mathrm{r}} 26,000\right)\), known as bacteriorhodopsin, which is purple because it contains retinal (see Fig, 10-20). Molecules of this protein aggregate into "purple patches" in the cell membrane. Bacteriorhodopsin acts as a light- activated proton pump that provides energy for cell functions. X-ray analysis of this protein reveals that it consists of seven parallel \(a\)-helical segments, each of which traverses the bacterial cell membrane (thickness \(45 \AA\) ). Calculate the minimum number of amino acid residues necessary for one segment of \(a\) helix to traverse the membrane completely. Estimate the fraction of the bacteriorhodopsin protein that is involved in membrane-spanning helices. (Use an average amino acid residue weight of 110 .)

Short Answer

Expert verified
Approximately 30 residues are needed per helix, and about 89% of bacteriorhodopsin is in membrane-spanning helices.

Step by step solution

01

Calculate the Number of Amino Acids for One Helical Segment

Each alpha helix is composed of amino acids adding a rise of approximately 1.5 Å per residue. We need to find how many residues are required to span the cell membrane's 45 Å thickness. This is done by dividing the membrane thickness by the rise per residue:\[\text{Number of residues per segment} = \frac{45 \text{ Å}}{1.5 \text{ Å per residue}} = 30 \text{ residues}\]
02

Determine the Total Number of Residues for All Helical Segments

Since there are seven helical segments, each requiring 30 residues to span the membrane, the total number of residues involved in membrane-spanning helices can be calculated:\[\text{Total residues in helices} = 7 \times 30 = 210 \text{ residues}\]
03

Calculate the Total Number of Residues in the Protein

The molecular weight of the bacteriorhodopsin protein is 26,000. Given that the average molecular weight of an amino acid residue is 110, we can calculate the total number of residues in the entire protein:\[\text{Total residues in protein} = \frac{26,000}{110} \approx 236 \text{ residues}\]
04

Estimate the Fraction of Protein Involved in Helices

To find the fraction of the bacteriorhodopsin protein involved in **membrane-spanning** helices, divide the total number of residues in helices by the total number of residues in the protein:\[\text{Fraction involved in helices} = \frac{210}{236} \approx 0.89\]This means that about 89% of the protein is involved in membrane-spanning helices.

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Key Concepts

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

Bacteriorhodopsin
Bacteriorhodopsin is a fascinating protein synthesized by the salt-loving halophile, Halobacterium halobium. This membrane protein is purple due to the presence of retinal, a chromophore associated with light absorption. It forms what are known as "purple patches" within the cell membrane, which are aggregate formations of these proteins. But bacteriorhodopsin is not just for show; it performs the critical function of being a light-activated proton pump. This means it harnesses light energy to move protons across the cell membrane, thereby generating a proton gradient used to fuel various cellular processes. Bacteriorhodopsin exemplifies the intricate ways in which life utilizes environmental resources, like light, to power cellular activities.
Alpha Helix
The alpha helix is a common structural element found in proteins, including bacteriorhodopsin. In this context, alpha helices are incredibly important as they allow the protein to span the cell membrane. Each helix contains amino acid residues that add up to a length allowing them to completely cross the membrane. The structure is stabilized by hydrogen bonds between the backbone of the amino acids, making it ideal to serve as a bridge over the hydrophobic (water-repelling) core of the membrane. For bacteriorhodopsin, each alpha helix must cross the 45 Å thick membrane. By using an average rise of 1.5 Å per residue, scientists calculate that at least 30 amino acids are needed per segment to support this crossing.
Membrane Protein
Membrane proteins, like bacteriorhodopsin, play vital roles in cellular function and communication. They are embedded within the phospholipid bilayer of cell membranes, often involved in transportation, signaling, and enzymatic activities. Membrane proteins can be aligned to span the membrane, such as transmembrane proteins like bacteriorhodopsin, or they can be associated with one side of the membrane. Bacteriorhodopsin's function as a proton pump exemplifies a dynamic role, where by using structural features like alpha helices, it creates passages through the membrane's thick hydrophobic layers. These proteins are essential in maintaining the cell's environment and allowing for selective exchange of substances necessary for life.
Proton Pump
In the world of cellular biology, proton pumps are critical for energy transduction. Bacteriorhodopsin, acting as a proton pump, uses light to transport protons from the inside to the outside of the cell. This proton transport creates a difference in charge and proton concentration across the membrane, known as a proton gradient. Such gradients are a potential form of energy, utilized by cells to perform a multitude of tasks, particularly the synthesis of ATP, the cell's energy currency. By moving protons across the membrane, bacteriorhodopsin not only demonstrates a unique mechanism of energy capture and conversion but also highlights the intricate design of cellular machinery that sustains life in extreme environments. This process is essential for cellular energy production and showcases the elegance of protein function and structure.

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Most popular questions from this chapter

Which structural biology method (CD, x-ray crystallography, NMR, or cryo-EM) is best suited to each task? a. Obtaining an ultra-high resolution \((<1.5 \AA)\) structure of a drug bound to its protein target b. Obtaining a low-to-medium resolution (5-10 \AA) reconstruction of the \(11 \mathrm{MDa}(11,000,000 \mathrm{Da})\) bacterial flagellar motor c. Identifying the protonation state and \(\mathrm{p} K_{\mathrm{a}}\). of a His side chain in an enzyme active site d. Determining whether a protein is intrinsically disordered or contains secondary structure elements

Protein-Folding Therapies The Food and Drug Administration recently approved the drug lumacaftor for the treatment of cystic fibrosis in patients with the F508 \(\Delta\) CFTR mutation. This mutation is a genetically encoded deletion of amino acid F508 from the protein. About \(2 / 3\) of cystic fibrosis patients have this mutation, and lumacaftor is one of the first drugs that functions as a pharmacological chaperone to correct a defect in the protein-folding process. However, lumacaftor is not always effective in treating patients who have other CFTR mutations that result in misfolding. Why is lumacaftor able to correct the misfolding of some mutant CFTR proteins and not others?

Some natural proteins are rich in disulfide bonds, and their mechanical properties, such as tensile strength, viscosity, and hardness, correlate with the degree of disulfide bonding. a. Glutenin, a wheat protein rich in disulfide bonds, imparts the cohesive and elastic character of dough made from wheat flour. Similarly, the hard, tough nature of tortoise shell results from the extensive disulfide bonding in its \(a\) keratin. What is the molecular basis for the correlation between disulfide-bond content and mechanical properties of the protein? b. Most globular proteins denature and lose their activity when they are briefly heated to \(65^{\circ} \mathrm{C}\). However, the denaturation of globular proteins that contain multiple disulfide bonds often requires longer heat exposure at higher temperatures. One such protein is bovine pancreatic trypsin inhibitor (BPTI), which has 58 amino acid residues in a single peptide chain and contains three disulfide bonds. After a solution of denatured BPTI is cooled, the protein regains its activity. What is the molecular basis for this property of BPTI?

Properties of the Peptide Bond In x-ray studies of crystalline peptides, Linus Pauling and Robert Corey found that the \(\mathrm{C}-\mathrm{N}\) bond in the peptide link is intermediate in length (1.32 Å) between a typical \(\mathrm{C}-\mathrm{N}\) single bond \(\left(1.49 \AA^{\circ}\right)\) and \(\mathrm{a} \mathrm{C}=\mathrm{N}\) double bond \((1.27\) A). They also found that the peptide bond is planar (all four atoms attached to the C-N group are located in the same plane) and that the two \(a\)-carbon atoms attached to the \(\mathrm{C}-\mathrm{N}\) are always trans to each other (on opposite sides of the peptide bond). a. What does the length of the \(\mathrm{C}-\mathrm{N}\) bond in the peptide linkage indicate about its strength and its bond order (i.e., whether it is single, double, or triple)? b. What do Pauling and Corey's observations tell us about the ease of rotation about the \(\mathrm{C}-\mathrm{N}\) peptide bond?

Margaret Oakley Dayhoff originated the idea of protein superfamilies after noticing that proteins with diverse amino acid sequences can have similar tertiary structures. Why can protein structure be more highly conserved than individual amino acid sequences?

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