6-26. Given the following equilibria, calculate the concentration of each zinc speciesin a solution saturated with Zn(OH),(s) and containing [OH ] at a fixedconcentration of 3.2 x 10-7 M.Zn(OH),(s) Kp = 3 × 10 16Zn(OH)B1 = 1 × 104Zn(OH),(aq) B2 = 2 × 1010Zn(OH),Zn(OH);B3 = 8 × 10132-ВA — 3х 1015

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Answered: 6-26. Given the following equilibria,…

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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100 g of a pretreated lignocellulose has a mass composition of 35% lignin, 40% cellulose, 15% xylan, and 10% "other." During hydrolysis 75% of the cellulose and 80% of the xylan are converted to products. What is the wt% composition of the insoluble solids following hydrolysis? Assume that the "other" fraction is unaffected by the hydrolysis.
The net charge on the most prevalent form of bisphosphoglycerate in blood is what?
Hemoglobin is considered to be a tetrameric complex with a 64 kDa (α β)2. When attempting to purify hemoglobin, we must first purify the α and β monomers (about 16 kDa each) to prepare the tetramer. This is formed from the dimer intermediate: 2 α + 2 β -> 2 αβ -> (α β)2. The graph given represents a size-exclusion chromatogram after the refolding of the hemoglobin tetramer Using the size-exclusion chromatogram given, 1. Draw an SDS-Page Gel with a reducing agent such as BME using the three peaks listed on the graph.
Identify the following types of biomolecules and label and identify the functional groups
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
Paper electrophoresis of Asn, Ala, Asp, Lys and Ser mixtures at pH = 7 shows the fastest movement towards the anode The fully saponified 10g oil sample consumed 1.87g KOH, and the iodine value of the oil was 28.3. So how many unsaturated double bonds are there in each oil molecule on average? (lodine is known to have an atomic weight of 127) Given that 100g cellulose sample is completely hydrolyzed to obtain 78g glucose, then the percentage of cellulose in the sample is _% Given that the Km of catalase is 20mmol/L and the substrate concentration is 80mmol/L, then the percentage of catalase bound to the substrate is %
Which of the following drugs is a less potent microsomal triglyceride transfer protein (MTP) inhibitor, which reduces cholesterol levels? O Structure 1 (IC50 = 23 nM) O BMS-200150 (IC50 = 2.2 µM) O Hybrid Structure (IC50 = 12 nM) %3! O Hit compound (IC50 = 36 µM) %3D Lomitapide (IC50 = 0.5 nM)
13. 0.9% (m/v) NaCI solution and 5% (m/v) glucose solution are both isotonic to red blood cells. SHOW your work and watch sig figs & units. c. convert the concentration from M to % (m/v) for a 0.342 M NaC solution. (HINT: convert to g/ml and then multiple by 100%)

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