Please help with part D and H If cant do all, then as much as you can answer. Thank you ! 11D The T-state structure of hemoglobin is an example of awhereas its alpha helices are an example of a11H The glyoxylate pathway produces one molecule oftype of structure,from two molecules oftype of structure.

Introduction: Hemoglobin is a respiratory pigment found in red blood corpuscles. It is made up of…

Answered: 11D The T-state structure of hemoglobin…

Biochemistry

6th Edition

ISBN:9781305577206

Author:Reginald H. Garrett, Charles M. Grisham

Publisher:Reginald H. Garrett, Charles M. Grisham

Chapter26: Synthesis And Degradation Of Nucleotides

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1. The human hemoglobin molecule, like all mammalian he- moglobins, is comprised of two a-chains and two ß-chains con- ₂ taining 141 and 146 amino acid residues, respectively. Be- cause the molecule possesses two-fold symmetry, there are a1-a2, B1-B2, and a1-B2 interfaces formed by amino acid sidechains through which structural changes are transmitted underlying ligand binding. The most important of these is the a1-B2 interface that is illustrated in the diagram on the right. All of the sidechain interactions across the a1-B2 interface are hydrophobic except for that between Asp(a94) and Asn- (B102). This is the only polar interaction across the α1-³2 in- terface and it helps to stabilize the oxy- or R-conformation. Its approximate location in the Hb molecule is represented by the red double arrow in the diagram on the right. His FG4 97 Asp G1 (99) Tyr C7 (42) Hb mutant (a) T State (deoxy) 95 The C6 41 (a) The diagram below on the right-hand side illustrates the polar Asp(a94). . . Asn…
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:…
Match the following Class of Proteins with their corresponding Prosthetic Group Components. nucleic acids 1. hemoproteins metal ion 2. lipoproteins phosphate group 3. flavoproteins pigment 4. glycoproteins heme unit 5. phosphoproteins 6. nucleoproteins lipid 5. metalloproteins flavin nucleotides -8 chromoprotein carbohydrate
Protease enzymes cleave proteins by hydrolyzing peptide bonds. The strategy for each type of metalloprotease begins with generating a nucleophile that attacks the peptide bond that attacks the peptide carbonyl group. O Macmillan Learning On the basis of the information provided in the figure, show the next step in the mechanism for peptide-bond cleavage by a metalloprotease. Metalloproteases H R₁ HN Zn Enz 2+ R₂ Draw curved arrows on the pre-drawn structures to show the metalloprotease mechanism. If you need to reset the structures, click More followed by Reset Drawing. Select Draw Templates Groups More B - H Enz H H с R1 | : HN O | Zn 2+ B R2 N Zn Erase
In oxygenated hemoglobin, pKa 5 6.6 forthe histidines at position 146 on the -chain. In deoxygenated hemoglobin, the pKa of these residues is 8.2. How can this piece ofinformation be correlated with the Bohr effect?
In some metalloproteins such as haemoglobin and some metalloenzymes, the Fe2+ metal is not bound to the amino acid side-chains of the protien,but to a prophrin ring called a heme prosthetic group. How does the heme coordinate to the iron?
8. The chemical structure of Coenzyme A contains the following except— a phosphoanhydride moiety. a β-mercaptoethylamine residue. a lipoic acid residue. a pantothenic acid residue. an adenosine-3’- phosphate.
. Separating Glycated Hb From Normal Hb (Integrates with Chapters 5and 6.) Human hemoglobin can react with sugars in the blood(usually glucose) to form covalent adducts. The a-amino groups ofN-terminal valine in the Hb b-subunits react with the C-1 (aldehyde)carbons of monosaccharides to form aldimine adducts, whichrearrange to form very stable ketoamine products. Quantitation ofthis “glycated hemoglobin” is important clinically, especially fordiabetic individuals. Suggest at least three methods by which glycatedHb (also referred to as HbA1c) could be separated from normal Hband quantitated
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2B. S. aureus hemolysin B attacks the RBC cell membrane by hydrolyzing the sphingomyelin headgroup: ОН HN .R hemolysin B cuts this bond i) Draw a plausible mechanism of hydrolysis for this lipid headgroup. Let B- and BH be general base and general acid. 00-P-O LOR2 OR, ii) Why is this damaging to the overall membrane architecture of the RBC?
22. Describe the Induced fit of Hemoglobin, Hb (enzyme-like protein) which has quarternary structure and exhibits allosteric activation with cooperativity.
The figure shows interaction of various amino acids (Phe, Arg, and Asn) present in Hb with 2,3-DPG. The side chains of these amino acids and 2,3-DPG interactions are circled in red. HO Но Phe HN Asn H2N NH Arg You identify a mutant version of the globin chain where the Arg in circle 2 is replaced with aspartic acid in a patient. Would the Hb protein in this patient be able to transport oxygen even in the presence of 2,3-DPG? ASPARTIC ACID (asp) No, because the mutant Hb version will bind 2,3 DPG, and not oxygen Yes, because the mutant Hb version will bind 2,3 DPG, and not oxygen Yes, because the mutant Hb version will bind oxygen and not 2,3 DPG

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11D The T-state structure of hemoglobin is an example of a whereas its alpha helices are an example of a 11H The glyoxylate pathway produces one molecule of type of structure, from two molecules of type of structure.