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The folding and unfolding rate constants for a myoglobin mutant have been determined. The unfolding rate constant
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Biochemistry: Concepts and Connections (2nd Edition)
- Given a tripeptide Cys-His-Lys, Cys: Pk1 = 1.71; Pk2 = 10.78; PkR = 8.33 His: Pk1 = 1.82; Pk2 = 9.17; PkR = 6.0 Lys: Pk1 = 2.18; Pk2 = 8.95; PkR = 10.53 draw the protonic equilibria for the tripeptide what is the IpH?arrow_forwardGiven a tripeptide Cys-His-Lys, Cys: Pk1 = 1.71; Pk2 = 10.78; PkR = 8.33 His: Pk1 = 1.82; Pk2 = 9.17; PkR = 6.0 Lys: Pk1 = 2.18; Pk2 = 8.95; PkR = 10.53 draw the protonic equilibria for the tripeptide what is the IpH? What is the dominant structure at pH 2.0? What is the first buffering region of the tripeptide?arrow_forwardProtein concentration can readily be determined using the Beer-Lambert law: A = e l c where A = absorbance e = molar absorption coefficient (M-1cm-1) l = light path length (cm) c = concentration (M) If the molar absorption coefficient at 280 nm for yeast ADH is 48860 M-1cm-1 and a 10 mL solution of the protein has an absorbance at 280 nm of 0.4 (as measured by a spectrometer with pathlength 1 cm), then what is the concentration of the protein solution (in μM)? i.e. concentration = ______ μM If the molecular weight of the protein is 36849, what is its concentration in mg/mL? i.e. concentration = _______ mg/mL For each part of the question, show your calculations to arrive at your answers.arrow_forward
- Finally, discuss the consequences of 9.2 uM for actin polymerization in a cell, considering critical concentrations values of 0.1 uM for barbed (plus) end addition and 0.6 uM for pointed (minus) end addition of actin. Recall that at G-actin concentrations above the critical concentration, actin filaments elongate. When the G-actin concentration is greater than the Cc for the barbed end (C (T)) but less than the Cc for the pointed end (Cc (D)), the filament maintains a steady state length due to treadmilling, where the rate of ATP actin addition at the barbed end equals the rate of ADP-actin at the pointed end.arrow_forwardGiven a tripeptide Cys-His-Lys, Cys: Pk1 = 1.5; Pk2 = 10.8; PkR = 8.5 His: Pk1 = 1.6; Pk2 = 9.0; PkR = 7.0 Lys: Pk1 = 2.2; Pk2 = 8.5; PkR = 9.8 a.draw the protonic equilibria for the tripeptide. b.what is the IpH? c.What is the dominant structure at pH 3.0? d.What is the first buffering region of the tripeptide?arrow_forwardThe proximal histidine residues have been replaced by glycine residues by mutation of the cloned genes for both the α and β subunits of hemoglobin. With the tetrameric mutant hemoglobin (all subunits being mutant, α H F8 G, β H F8 G), it was found that the “proximal” coordination bonds to hemes in the mutant protein could be replaced by having the small molecule imidazole in the buffers. Oxygen binding curves for the tetrameric mutant hemoglobin were measured. A. The degree of cooperativity in oxygen binding for the mutant hemoglobin (with imidazole present) would be expected to 1) increase 2) decrease 3) not be affected) compared with the normal protein. B. Justify your answer to part A in terms of what you know about the structural basis of cooperativity in hemoglobin. C. How would the Hill coefficient for the mutant be expected to change compared with nH for normal hemoglobin, which is ~3?arrow_forward
- 10) The saturation curves for two myoglobin mutants are shown below. a) Estimate the K, values for both mutants. (Refer to your notes for the interpretation of K with respect to binding 100 90 80 curves.) 70 b) Based on the K, values, which mutant binds oxygen more tightly? Briefle explain why. 60 50 40 30 -% saturaton (mutant 1) 20 +* saturaton (mutant 2) 10 20 40 60 80 100 pO2 (torr) % saturationarrow_forwardSickle cell anemia is caused by a point mutation in the β-globin chain of hemoglobin. Glutamic acid is replaced by Valine. HBB sequence in normal adult hemoglobin (Hb A): Leu-Thr-Pro-Glu-Glu-Lys-Ser HBB sequence in mutant adult hemoglobin (Hb S): Leu-Thr-Pro-Val-Glu-Lys-Ser What effect does this mutation have on the structure and function of the protein? Predict what would happen to the RBC if the glutamic acid was replaced with asparagine instead of valine.arrow_forwardThe enzyme lysozyme kills certain bacteria by attacking a sugar called N-acetylglucosamine (NAG) in their cell walls. At an enzyme concentration of 2 × 10-6 M, the maximum rate for substrate (NAG) reaction, found at high substrate concentration, is 1 × 10-6 mol L¯'s1. The rate is reduced by a factor of 2 when the substrate concentration is reduced to 6 x 10-6 M. Determine the Michaelis-Menten constants Km and k, for lysozyme.arrow_forward
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