6-26. Given the following equilibria, calculate the concentration of each zinc species in a solution saturated with Zn(OH),(s) and containing [OH ] at a fixed concentration of 3.2 x 10-7 M. Zn(OH),(s) Ksp = 3 × 10 16 Zn(OH)* B1 = 1 × 104 Zn(OH),(aq) B2 = 2 × 1010 Zn(OH), Zn(OH); B3 = 8 × 1013 B4 = 3 × 1015

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%)