protein A on-rate += 25 °C binding Determinat and C э Kon = 1.6 x 106 M²ts² 8-1 free energy in complex ас -35k 1/mol мм μM → affinity → off-rate for dissociation
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- Acetic acid is the principal ingredient in vinegar as shown; that's why it tastes sour. At equilibrium, a solution contains [CH3CO2H] = 0.0787 M and [H3 O+] = [CH3 CO2−] = 0.00118 M. What is the value of Ka for acetic acid?During movement, muscle cells require large amounts of ATP to fuel their contractile apparatus. These cells contain high levels of creatine phosphate (Figure Q13–11), which has a standard free-energy change (ΔG°) for hydrolysis of its phosphate bond of –10.3 kcal/mole. why is this a useful compound to store energy?In a 0.1000 M acetic acid solution at 25 degrees celsius , the acid ionizes to the extent of about 1.34 %. Since each molecule of acetic acid which ionizes produces 1 H+ ion and 1 C2H3O2- ion, the concentration in the solution are: HC2H3O2 < -----------> H+ + C2H3O2-
- 4. Order: 40,000 units heparin in 1L of 0.9% NaCl to infuse at 25 mL/h. Calculate the hourly heparin dosage in mL/hour.?A 70-kg adult human (154 lb) could meet his orher entire energy needs for one day by eating 3 moles ofglucose (540 g). (We do not recommend this.) Each mol-ecule of glucose generates 30 molecules of ATP when it isoxidized to CO2. The concentration of ATP is maintained incells at about 2 mM, and a 70-kg adult has about 25 litersof intracellular fluid. Given that the ATP concentrationremains constant in cells, calculate how many times perday, on average, each ATP molecule in the body is hydro-lyzed and resynthesized.Calculate the concentration of an EDTA solution of which 24.22 mL were needed to titrate the Ca2+ions present in a 50.00 mL solution containing 246.7 mg CaCO3. Express your answer in terms of (a) molar concentration of EDTA (b) Ca Titer (mg Ca per mL of EDTA) MM Ca= 40.078 solve it in a way that it is in one equation already and cancelling all the values just like given in th epicture
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- Estimate the TKN associated with a sample having 50 mg/L of cell tissue and 10 mg/L of ammonia. Assume cell tissue has a molecular composition of C5H7O2N.An antibody binds to another protein with anequilibrium constant, K, of 5 × 109 M–1. When it binds toa second, related protein, it forms three fewer hydrogenbonds, reducing its binding affinity by 11.9 kJ/mole. Whatis the K for its binding to the second protein? (Free-energychange is related to the equilibrium constant by the equa-tion ΔG° = –2.3 RT log K, where R is 8.3 × 10–3 kJ/(mole K)and T is 310 K.)Acetocholinesterase is an enzyme possessing a single active site that metabolizesacetylcholine with a turn over number of 1.4 x 10^4s-1. How many grams of acetylcholine(molecular formula C7NO2H16+) will 2.16 x 10^-6 g acetocholinesterase metabolize in 60minutes? (The enzyme’s molecular mass is 4.2 x 10^4 g/mol).