Experimental results describing a protein's amino acid composition are useful for estimating the molecular weight (MW) of theentire protein. A quantitative amino acid analysis reveals that bovine cytochrome c contains 1% tryptophan (M, 204) by weight.Calculate the approximate molecular weight of bovine cytochrome c if there is 1 tryptophan residue. Please enter your answerwith three significant figures.approximate bovine cytochrome c MW:number of threonine residues:19.420.4 x10³Bovine chymotrypsinogen has a molecular weight of 25.6 kDa. Amino acid analysis shows that this enzyme is 9% threonine(M, 119). Calculate how many threonine residues are present in a molecule of bovine chymotrypsinogen. Round your answer tothe nearest whole number.IncorrectIncorrectDa

Proteins are high molecular weight polymers of amino acid residues linked together via peptide…

Answered: Experimental results describing a…

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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The Lineweaver-Burke plot was originally developed in order to "linearize" the data obtained from enzyme kinetics experiments, in order to facilitate the determination of kinetic parameters. Why is it not considered to be an accurate method for this purpose? It is very difficult to draw a straight line on a computer. It is very difficult to calculate the variables required for the "x" and "y" axis. It is more accurate to use the standard "V versus [S]" plot to determine Vmax and KM- The plot weights the least accurate data points the most heavily. It is no longer considered to be acceptable to extrapolate from known data.
5) In the below two step transformation, the first step AG is positive i.e. 1.7 kJ/mol. Yet this two step coupled reaction takes place to form Fructose-1.6-bisphosphate. Explain why?" C-H CH-OH CH,-0 H-C-OH AG"- 1.7 KJ/mol (+0.4 kcalmol) C=0 AG"=-14.2 k/mol(-3.4 kcalmol) HO-C-H HO-C-H HO-C-H H-C-OH H-C-OH H-C-OH H-C-OH H-C-OH ATP ADP H-C-OH CH-0-P. CH,-0-P-o- CH,-o- Glucose-6-phosphate Fructose-6-phosphate Fructose-1,6-bisphosphate
A series of standard proteins and an unknown enzyme were studied by gel filtration on a Sephadex G200 (the 200 refers to the maximum pore size in kDa) column. The elution volume Vel for each protein is given in Table 1 below. (a) Plot the data in the form of log Mr versus elution volume. From the line of best fit through the points for the standards, determine the Mr of the unknown enzyme. Explain why ferritin and ovomucoid behave anomalously. Table 1 - The Vel versus Mr data Protein Mr Vel (mL) Blue dextran* Lysozyme Chymotrypsin Ovalbumin Serum albumin Aldolase Urease Ferritin# Ovomucoid# Unknown 1,000,000 14,000 25,000 45,000 68,500 150,000 500,000 700,000 28,000 ⎯ 85 200 190 170 150 125 90 92 160 139 *Blue dextran is not a protein but a high-Mr carbohydrate that has a covalently bound blue dye, and it elutes with the void volume of the column. # Do not…
At equilibrium, D-galactose exists almost exclusively in its α and β pyranose forms. Aqueous solutions are freshly prepared for the α and β forms, both at the same concentration and temperature. The solution of the α form rotatesplane-polarized light +150.7°, whereas the solution of the β form rotates the light +52.8°. Over time, both solutions have the same measured rotation of +80.2°. How much of the equilibrated solution does each form account for?
The fosfomycin disc (200/50) also contains glucose-6-phosphate (G-6-P). Which ONE of the following best describes why this is the case? Select one: A. G-6-P increases the solubility of fosfomycin B. G-6-P is used to promote active transport of fosfomycin into the bacterial cell C. G-6-P is used in the formulation to correct glucose deficiencies promoted by fosfomycin D. G-6-P stabilises fosfomycin from spontaneous hydrolysis in aqueous solution E. G-6-P is an inhibitor of FosA that inactivates fosfomycin
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
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.
An enzyme uses a lysine and a histidine as general acid-base catalytic residues. In one direction of the reaction, the lysine and histidine residues are typically protonated (~NH3*, ~Im), but in the other direction of the reaction, the two residues are "reverse protonated" (~NH2, ~Hlm*) (see below scheme). A pH profile of the reaction at various pH values yielded estimates of 7.57 and 9.30 for the macroscopic ionization constants pk,' and pK2, respectively. A separate set of experiments with site-speciffic variants of the enzyme yielded an estimate of 8.78 for the microscopic ionization constant pKp. Given this information, what is the value of the microscopic ionization constant pKA? Report your answer to the nearest hundredth. Lys-NH3* His-HIm" KB H KA H* Lys-NH2 Lys-NH3* His-HIm" His-Im Kc H* Kp Lys-NH2 His-Im
Draw Lineweaver-Burk plots for the behavior of an enzyme for which the following experimental data are available. What are the Km and Vmax values for the inhibited and uninhibited rxns? Is the inhibitor competitive or noncompetitive? [S] (M) 10,000 1 x 104 2.000-5 x 104 666673 1.5 x 10-3 400- 2.5 x 10-³ 2005x103 V, No Inhibitor (Arbitrary units) 0.026 38,461 0.092 10.87 37100 0.136 7.35 Wak 0.150 -67 0.165 6.06 L slope: 0033 yint=5057 полесте um.x=0.198 km 6.5x 10-4 0.010 0.040 25 10.086 11.63 0.120 8.33 0.142 7.04 2500 V, Inhibitor Present (Arbitrary Units) 100 ↓ $op.= .009 yint: 5.21 900x12 km: 1.73 y/o conf. V10-3 321000
In a pUC19 digest for 1 ug of pUC19 (DNA conc. 282ng/ul) using 10X Cutsmart buffer, pure water and BamHI enzyme in a total volume of 40 ul, how much of each solution is added to the total volume?
An enzyme contains an active site aspartic acid with a pKa = 5.0, whichacts as a general acid catalyst. On the accompanying template, draw thecurve of enzyme activity (reaction rate) versus pH for the enzyme (assumethat the protein is stably folded between pH 2–12 and that the active siteAsp is the only ionizable residue involved in catalysis). Briefly explain theshape of your curve.
The enzyme serine hydroxymethy ltransferase (SHMT) catalyses the conversion of serine into glycine. The fo llowing table gives the initial rates, vo, for the SHMT-catalysed reaction of the substrate serine at var ious concentratio ns of serine, lSI.[S]/(mmol dm-3) 10 20 30 40vo(μmol dm-3 s-1) 1.63 2.94 4.10 4.95Use the data to determine the values of the MichaelisMenten constant, the maximum velocity of the reaction, and the maximum turnover number of the enzyme.

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