1. The human hemoglobin molecule, like all mammalian he-moglobins, is comprised of two a-chains and two ß-chains con- ₂taining 141 and 146 amino acid residues, respectively. Be-cause the molecule possesses two-fold symmetry, there area1-a2, B1-B2, and a1-B2 interfaces formed by amino acidsidechains through which structural changes are transmittedunderlying ligand binding. The most important of these is thea1-B2 interface that is illustrated in the diagram on the right.All of the sidechain interactions across the a1-B2 interface arehydrophobic 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. Itsapproximate location in the Hb molecule is represented by thered double arrow in the diagram on the right.His FG4 97Asp G1 (99)Tyr C7(42)Hb mutant(a) T State (deoxy)95The C6 41(a)The diagram below on the right-hand side illustrates the polar Asp(a94). . . Asn (102)interaction that stabilizes the R (oxy) conformation of human hemoglobin (HbA). This sidechain inter-action underlies the allosteric, structural transition of hemoglobin and helps to maintain its tetramerstructure. In the table below the properties of several mutants of hemoglobin involving the a94 positionare compared at pH ~ 7.4.38POTCDA 44HbAHb BassettHb RoanneHb Sunshine SethoxygenationAmino acid changeAsp(a94)AlaAsp(a94) GluAsp(a94) HisAsp(a94)Asn(b)The diagram on the right compares O2-dissoci-ation curves for human hemoglobins characterized by dif-ferent p50 values. If Ko2 represents the equilibrium disso-ciation constant for oxygen binding, e.g.,Hb(02)nHb + nO₂show the relationship between Ko2 and p50.Assign one of the curves (a, b, . . . , e) that best approxi-mates the Ko2 of each of the mutant hemoglobins in thetable above including HbA where Ko2 represents the equi-librium dissociation constant for oxygen binding. If no curveis present for any of the mutant hemogobins, state the rea-son why the mutant is not represented. Indicate on thegraph where its p50 would lie.a C terminusHis FG4 97HbAPERCENT SATURATIONB₂OBC100-JOINTnн2.81.42.11.11.030-0 20BFGFGAsn G4 (102)Hb TitusvilleIn the style shown on the right for HbA, make a schematicdrawing of the potential (a94)... (B102) hydrogen bond-ing relationship for each mutant hemoglobin wherein thedots connect the donor hydrogen (D-H) and acceptor hy-drogen (A) atoms, assuming that it could be formed in theabsence of structural distortion of the protein. The changein amino acid residue is represented, e.g., for Hb Bassettas Asp(a94)Ala, meaning that residue Asp at position a94 has been replaced by Ala.094WHHỆ CHÍSWITCH(b) R State (oxy)a/b/c/d/eAsp G1 (94)01B₂C terminusThe CG 41p50(mm Hg)26409303338CDP 44BID2PARTIAL PRESSURE OF OXYGENveins musclelungsC=O8C-CH₂CHNHα₁40 60 80 100 120 140OXYGEN PRESSURE(millimeters of mercury)

A table summarizing the characteristics of many hemoglobin mutants affecting the 94 position is…

(a) The diagram below on the right-hand side illustrates the polar Asp(a94)... Asn (102) interaction that stabilizes the R (oxy) conformation of human hemoglobin (HbA). This sidechain inter- action underlies the allosteric, structural transition of hemoglobin and helps to maintain its tetramer structure. In the table below the properties of several mutants of hemoglobin involving the a94 position are compared at pH ~ 7.4.

(a) The diagram below on the right-hand side illustrates the polar Asp(a94)... Asn (102) interaction that stabilizes the R (oxy) conformation of human hemoglobin (HbA). This sidechain inter- action underlies the allosteric, structural transition of hemoglobin and helps to maintain its tetramer structure. In the table below the properties of several mutants of hemoglobin involving the a94 position are compared at pH ~ 7.4.

Biology: The Unity and Diversity of Life (MindTap Course List)

15th Edition

ISBN:9781337408332

Author:Cecie Starr, Ralph Taggart, Christine Evers, Lisa Starr

Publisher:Cecie Starr, Ralph Taggart, Christine Evers, Lisa Starr

Chapter9: From Dna To Protein

Section: Chapter Questions

Problem 5SQ: The main function of a DNA molecule is to ________ . a. store heritable information b. carry a...

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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?
4.3 Draw a fractional binding curve (θ v.s [L]) for a protein binding to one molecule of L with a Kd of 10 mM.
A protein-ligand binding reaction is run. At equilibrium, half the protein is ligand bound, the unboundligand concentration is 0.657 nM. Calculate the koff value for this reaction. Assume the kon value is typical ofprotein-ligand interactions.
1. How many different tripeptides are possible? How many carboxyl terminals of polypeptide chains are present in a molecule of hemoglobin? 2. You have isolated a pentapeptide composed of four glycine residues and one lysine residue that resides at the C-terminus of the peptide. Using the information provided in the legend of Figure 2.27, if the pK of the side chain of lysine is 10 and the pK of the terminal carboxyl group is 4, what is the structure of the peptide at pH 7? At pH 12? 3. The side chains of glutamic acid (pK 4.3) and arginine (pK 12.5) can form an ionic bond under certain conditions. Draw the relevant portions of the side chains and indicate whether or not an ionic bond could form at the following: (a) pH 4; (b) pH 7; (c) pH 12; (d) pH 13.
Consider a globular protein that contains an alpha-helix within it: a) Suppose there is a ligand binding pocket in this alpha-helix that contains residues of Leu (2), Phe (5), Gly (6), Ala (18), Tyr (22), Tyr (28), Gly (30), Cys (45), His (48) and Asp (49). What does this tell us about the likely nature or characteristics of a potential binding ligand? I said tha because 6/10 of the amino acids in this binding pocket are non-polar, that the ligand is also non-polar, but I think the question needs more description about the ligand. b) Briefly describe the driving forces and energetics behind folding of globular proteins (use terms like ΔH, ΔS, and ΔG) I said that H = -, S = +, G = - Folding driving force would likely be entropy change from non-polar residues interacting with a solvent (ex: water). Hence, non-polar residues would be buried in the globular protein structure from folding, causing entropy to increase for any liberated solvent molecules. Not 100% sure if this is correct,…
In a molecular disease of hemoglobin, Hemoglobin Rainier, Tyr 145β is replaced by Cys, which forms a disulfide bond with another Cysresidue in the same subunit. This prevents the formation of ion pairs that normally stabilize the T state. How does hemoglobin Rainier differ from normal hemoglobin with respect to the Hillcoefficient? Explain your answers.
What impact do mutations in the alpha-chain usually have, if any, on hemoglobin's function? A. T-to-R-state transition B. dimer-dimer stabilization C. Allosteric Regulation D. A & B E. A & C F. A,B, & C
What determines when an amino acid ionized? i.e. Why are both Lysine (pKa 10.53) and Argenine (pKa 12.48) ionized at 11 pH?
Vasopressin: CYFQNCPRG Oxytocin: CYIQNCPLG Both of these peptides will react with reducing agents such as beta mercaptoethnol (BME). What side group reacts with this reducing agent? What does the fact that these peptides both react with BME tell you about the side groups and therefore this peptide?
Pathogenic E. coil have recently been shown to degrade tight junction proteins during infection. How would this provide an advantage to the bacteria?
The main function of a DNA molecule is to ________ . a. store heritable information b. carry a translatable message c. form peptide bonds between amino acids.
In a molecular disease of hemoglobin, Hemoglobin Rainier, Tyr 145β is replaced by Cys, which forms a disulfide bond with another Cysresidue in the same subunit. This prevents the formation of ion pairs that normally stabilize the T state. How does hemoglobin Rainier differ from normal hemoglobin with respect to the Hill coefficient