Biochemistry: Concepts and Connections (2nd Edition)
2nd Edition
ISBN: 9780134641621
Author: Dean R. Appling, Spencer J. Anthony-Cahill, Christopher K. Mathews
Publisher: PEARSON
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Chapter 3, Problem 7P
Assume that some protein molecule, in its folded native state, has one favored conformation. But when it is denatured, it becomes a “random coil,” with many possible conformations.
a. If we only consider the entropy for the protein, what must be the sign of
b. How will the contribution of
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The dissociation constant, Kd for a complex between protein A and protein B is 4.1 μM. If the two proteins are mixed together at initial concentrations of [A]= 0.025 μM and [B] = 4.7 μM, calculate (a) the equilibrium concentrations of A, B, and AB (the dimer formed by A and B) (b) the percentage of A bound to B
The major difference between a protein molecule in its native state
and in its denatured state lies in the number of conformations avail-
able. To a first approximation, the native, folded state can be thought
to have one conformation. The unfolded state can be estimated to
have three possible orientations about each bond between residues.
(a) For a protein of 100 residues, estimate the entropy change per
mole upon denaturation.
(b) What must be the enthalpy change accompanying denaturation
to allow the protein to be half-denatured at 50 °C?
(c) Will the fraction denatured increase or decrease with increasing
temperature?
If your initial absorbance from undiluted crude dialysate is 4.50, determine the fold
dilution needed to yield an absorbance of approximately 0.700. If you want to
prepare 1.0 ml of diluted crude, what volume (mL) of crude dialysate must you add?
Your Answer:
Answer
units
It is possible to estimate the molar extinction coefficient of a protein from
knowledge of its amino acid composition, as shown from your experiences with
EXPASY. From the molar extinction coefficient of tyrosine, tryptophan and cystine
(cysteine does not absorb appreciably at wavelengths >260 nm, while cystine does)
at a given wavelength, the extinction coefficient of the native protein in water can be
computed using the following equation:
e(Protein) = #(Tyr)*E(Tyr) + #(Trp)*E(Trp) + #(Cystine)*E(Cystine)
Where (if A280 measured in water): E (Tyr) = 1490, ɛ (Trp) = 5500, e (Cystine) = 125
Estimate the molar extinction coefficient of lysozyme using the above equation and
the amino acid composition data you found…
Chapter 3 Solutions
Biochemistry: Concepts and Connections (2nd Edition)
Ch. 3 - Prob. 1PCh. 3 - Given the following reactions and their...Ch. 3 - The decomposition of crystalline N2O5...Ch. 3 - The oxidation of glucose to CO2 and water is a...Ch. 3 - Prob. 5PCh. 3 - In another key reaction in glycolysis,...Ch. 3 - Assume that some protein molecule, in its folded...Ch. 3 - When a hydrophobic substance like a hydrocarbon is...Ch. 3 - It is observed that as temperature is increased,...Ch. 3 - Suppose a reaction has Ho and So values...
Ch. 3 - Prob. 11PCh. 3 - Prob. 12PCh. 3 - Prob. 13PCh. 3 - Undergoing moderate activity, an average person...Ch. 3 - The major difference between a protein molecule in...Ch. 3 - Prob. 16PCh. 3 - Prob. 17PCh. 3 - Consider the degradation of glucose to pyruvate by...Ch. 3 - a. Consider the malate dehydrogenase reaction,...Ch. 3 - Bovine ribonuclease folds with Ho = -280 kJmol-1...
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- You are given an equimolar (0.10 mM) mixture of Ubiquitin protein that is free/pure of any other macromolecules (e.g. nucleic acids) in a pH 7.0 phosphate buffer How can you purify Ubiquitin using Ion-Exchange chromatography (IEC)? Propose a plan using steps to purify this protein. Basic rules: Affinity chromatography (e.g. His Tag purification) is not an allowed step because these proteins are in their native state (i.e. – do not have a polyhistidine tag). Also, you are not allowed to rely solely on the color of certain proteins (e.g. cytochrome C and GFP) in your characterization/proposal. You are allowed to buffer exchange (i.e. – switch buffers) but try to keep your pH’s in a reasonable range (5.5 – 8.5) or you risk denaturing your protein. Assume proteins with the same charge (positive or negative) are not easily separated using IEC. Assume all proteins can be resolved/visualized on an SDS-PAGE gel. Ubiquitin Molecular weight (8663.02); pI (6.56); ε (M-1 cm-1) ignoring C’s…arrow_forwardA set of biomolecules listed in the table at right are in solution at pH 6.8, when they are passed through the multi-step separation process shown below. The sample passes through an ultrafilter with MWCO of 50,000 (assume this filter gives a perfect separation of molecules above and below the MWCO), and the retentate then flows through an ion exchange column filled with beads that have a positive charge under the operating conditions (pks = 7.3). Inlet Biomolecule Mixture pH 6.8 Ultrafiltration MWCO 50,000 Permeate Retentate lon Exchange Column Biomolecule Molecular weight pl A 68,000 4.2 B 92,600 9.6 C 144,000 A-3. Which biomolecules (A, B, C, etc.) exit the process in the permeate stream? K C-3. Which biomolecules are retained inside the ion exchange column? D E F G H B-3. Which biomolecules exit the process after passing through the ion exchange column? 5,800 68,000 Packed with DEAE cellulose beads, which have positively charged groups below a pH of 7.3 156,000 45,000 16,000 None…arrow_forwardthe enzyme LCK catalyzes the hydrolysis of the peptide KEY in 5.0 mM phosphate buffer at pH 7.0. The enzyme is known to obey Michaelis-Menten Kinetics. Under the conditions of this experiment: Vmax= 10 mmol/min/mg Km (KEY) = 0.20 mM Draw the complete structure of the tripeptide KEY and calculate its molecular weight Sketch the titration curve of the above tripeptide and label it.Determine the net charge of the peptide at each pk value Calculate the PI pf the peptide and draw the structure of the peptide at the isoelectric point.arrow_forward
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