0 276222.3333 0.25 270736.3333 0.5 310845 1 223241.6667 1.5 248714 2 262044 2.5 297373.3333 3 269513.6667 3.5 158465 4 189415 4.5 2252 5 66921 5.25 91895.33333 5.5 94238 5.75 90974 6 97186 Fluorescence intensity (340nm) Fluorescence Intensity (340nm) 350000 300000 250000 200000 150000 100000 50000 350000 300000 250000 200000 150000 100000 50000 0 0 1 2 3 y 7128.9x+283002 4 5 6 7 [Gdn HCI] M y=-75905x+491949 y=5043.2x+65205 0 0 1 2 3 4 5 6 7 [Gdn HCI] M 350000 y=7128.9x+283002 300000 250000 200000 Fluorescence intensity (340nm) 150000 100000 50000 0 y=-75905x+491949 y=5043.2x+65205 0 1 2 3 4 5 6 7 [Gdn HCl] M

What is the Keq of Reaction #3? Show your work & clearly label your answer. Is Reaction #3 favorable under mammalian cellular conditions when the concentration of dGTP is 0.2 M, dCMP is 20 mM, dGMP-dCMP is 7 mM, and PPi is 10 mM?  Use △G to support your answer.

(please type answer fast).

Consider the following process: NaCl(s) water−→−−water→  Na+(aq) + Cl-(aq); ΔH = +4.2 kJ/mol Under what conditions (temperature) would you expect this process to be spontaneous?  Select a relatively high temperature and a relatively low temperature, calculate (estimate) the relative values of ΔG to support your conclusion. The entropy increases in this process Please insert an image of your work

If 287.9 umol of enzyme X has a Vmax = 47.8 mmol/sec, what is the value of kcat %3D sec-1? Please report answer with 1 decimal place. Please do not report units. Your Answer: Answer units MacBook Air 888 F5 F4 F3 F2 %23 %24 %24

3L5

The formation constants at 25°C for Fe(CN)4-6 and Fe(EDTA)2– are 1.00 x 1037 and 2.10 x 1014, respectively. Answer the questions below. 1) Calculate K under standard conditions for the reaction Fe(EDTA)2−(aq) + 6CN−(aq) ----> Fe(CN)4−6​(aq) + EDTA4−(aq) 2) Calculate ΔG° for the reaction. (kJ/mol)

A one-to-one protein (P)-ligand (L) complexation (P + L PL) has a dissociation equilibrium constant (Kd) value of 100 nM at 25°C, and the Kd remains the same at 37°C. 1) What is AS of binding at 25°C? Assume ACp of the binding is 0 over the temperature range. AS = 1.34E2 kJ/(mol*K) (note the unit!!) (sig. fig =3) 2) What is the concentration of the PL complex formed at equilibrium when you mix 0.20 uM (microM) of Protein and 1.0 uM of Ligand together at 37°C? PL at equilibrium = 8.1E-1 uM (note the unit!!) (sig. fig =2)

In the table below, name each functional group in anserine that is ionizable and record an estimated pKa

Glucose-1-phosphate has a ΔG°′ value of −20.9 kJ/mol, whereas that for glucose-6-phosphate is −12.5 kJ/mol. After reviewing the molecular structures of these compounds, explain why there is such a difference in these values.