The [T] / [R] ratio is decreased and the oxygen affinity is increased.The [T] / [R] ratio is decreased and the oxygen affinity is unchanged.The [T] / [R] ratio is increased and the oxygen affinity is decreased.The [T] / [R]] ratio is increased and the oxygen affinity is increased.The [T] / [R] ratio is decreased and the oxygen affinity is decreased. The Hb Yakima variation is caused by the mutation D99H, which results in ashift in oxygen binding as shown in the graph. What effect does the D99Hvariation have on the [T] / [R] ratio of Hb Yakima protein conformationcompared to Hb Normal and how does D99H alter oxygen affinity?Fractional saturation1.02Hb Yakima20Hb Normal4060po, (kPa)80100

Hemoglobin binds to oxygen and it appears in two state. The T state (tensed) and R state (relaxed).…

Answered: The Hb Yakima variation is caused by…

Biochemistry

9th Edition

ISBN:9781319114671

Author:Lubert Stryer, Jeremy M. Berg, John L. Tymoczko, Gregory J. Gatto Jr.

Publisher:Lubert Stryer, Jeremy M. Berg, John L. Tymoczko, Gregory J. Gatto Jr.

Chapter1: Biochemistry: An Evolving Science

Section: Chapter Questions

Problem 1P

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Similar questions

A family of proteins known as cupredoxins contain a single redoxactive Cu ion coordinated by a Cys, a Met, and two His residues. The reduction potentials of cupredoxins range from about 0.15 V to 0.68 V. What does this information reveal about the role of the protein component of the cupredoxins?
Consider the phenolic hydroxyl group of a particular Tyr residue in a protein. Suppose the hydroxyl group in the unfolded protein in aqueous solution, where the group is exposed to H2O, has a pKa of 10.0. If that group is found in a hydrophobic environment in the interior of the protein when the protein is folded into its native tertiary structure, would you expect the pKa of the phenolic hydroxyl to be higher or lower in the folded protein interior than in H2O? Explain your reasoning.
The last residue of the protein (tail) is Tryptophan, and the first residue (head) is labeled with IAEDANS acceptor. Estimate the length of the protein (head to tail) if the efficiency of RET measures at 0.01 Please show step by step and how to get the 34.014 number
A schematic diagram of the helical structure of cytochrome b562 is reproduced below. Thisprotein belongs to the family of -proteins that have a four-helix bundle. Number the helices 1 – 4according to their N  C direction. Indicate relative orientations of the macrodipoles of helices 1 – 4adjacent to the diagram of cytochrome b562. Remember that according to the definition of a dipolethat the arrow points towards the positive end.
Which of the following drugs is a less potent microsomal triglyceride transfer protein (MTP) inhibitor, which reduces cholesterol levels? O Structure 1 (IC50 = 23 nM) O BMS-200150 (IC50 = 2.2 µM) O Hybrid Structure (IC50 = 12 nM) %3! O Hit compound (IC50 = 36 µM) %3D Lomitapide (IC50 = 0.5 nM)
Suppose you have a solution of a protein, which contains a specific Tyr residue that has an actual (measured) pKa of 8.8. The protein binds a ligand by several noncovalent interactions, one of which is a hydrogen bond in which the Tyr phenolic hydroxyl group must serve as a hydrogen bond donor. Calculate the percentage of the protein molecules in which that tyrosyl residue's phenolic hydroxyl group could serve as a hydrogen bond donor at pH 8.5
Based on the pK, table for amino acids (Table 1.1), draw the structures with net charge, and the equilibria representing the complete deprotonation of aspartate. Table 1.1 Name pK pk2 pKR Glycine Alanine 2.4 9.8 2.3 9.7 Valine 2.3 2.4 9.6 Leucine 9.6 Isoleucine 2.4 9.7 Methionine 2.3 9.2 Phenylalanine Proline Serine 1.8 9.1 2.0 10.6 2.1 9.2 Threonine 2.6 10.4 Cysteine Asparagine Glutamine 1.8 10.8 8.3 2.0 8.8 2.2 9.1 Tyrosine Tryptophan Aspartate Glutamate 2.2 9.1 10.9 2.4 9.4 2.0 2.2 10.0 3.9 9.7 4.3 Histidine 1.8 9.2 6.0 10.8 Lysine Arginine 2.2 9.2 1.8 9.0 12.5
Consider the positively charged amino acid lysine Lys2+ 21 COOH I H&N-C-H I pH 14 12 10 8 6 4 2 0 CH₂ I CH₂ I CH₂ I CH₂ T NH₂+ 0 Nelson p85 2.18 = 2.18 PK₁ Lys+ COO™ I H₂N-C-H H₂N-C-H ī I -----) 8.95 Lysº 8.95 pK₂ pka carboxyl = 2.19 pkaamino = 9.67 pka sidechain = 4.25 COO™ I CH₂ I CH₂ I CH₂ I CH₂ I NH₂¹ 1.0 2.0 Equivalents of OH added- COO™ I H₂N-C-H I 10.79 1 10.79 pk Isoelectric point Lys CH₂2 I CH₂ I CH₂ I CH₂ T NH₂ 3.0 +H3N NH3+ T CH₂ T CH₂ CH₂ CH₂ -COO™ H Lysine (Lys, K) Physiological pH = 7.4 < pl → Amino acid is positively charged at physiological pH 1. Consider glutamate in its fully protonated form (e.g. in a pH = 1 solution) 1) Draw all the forms of glutamate at various pH 2) Calculate the pl of this amino acid 3) Sketch a titration curve showing pH as a function of added [OH-] and locate the predominant forms of histidine in the curve STEPS: 1. Find the H atoms that can be removed on the molecule 2. Associated a pka value to each removable H. 3. Draw the Aa structure at:…

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