Question

# The antibiotic paromomycin (P) forms a complex with a particular 27-nucleotide RNA construct (R), (see Anal....

The antibiotic paromomycin (P) forms a complex with a particular 27-nucleotide RNA construct (R), (see Anal. Biochem. 2000, 280(2), 264-71). Dissociation of this complex can be described schematically (in aqueous solution) as: RP↔R+P KD =0.520* at25oC where the equilibrium constant KD is commonly referred to as the dissociation constant. Assuming one begins with 0.0780M of RP complex (only):

(a) Determine the equilibrium amounts of [RP] and [P] at 25 oC.

(b) Suppose that the value of KD changes to 1.47* when the temperature is raised to 37 oC. Does the equilibrium of part (a) shift to the right (R); shift to the left (L); or remained unchanged (U).

(c) Is this process (reaction) exothermic or endothermic or is ∆H = zero at equilibrium (put correct answer/word in the blank)?

(d) Starting from the final equilibrium amounts determined in part (a) at 25 oC, determine the equilibrium amounts of [RP] and [P] if the temperature is raised to 37 oC. Hints: quadratics should use the positive roots and keep three significant digits. If pressed for time, set up rxn table/quadratics for partial credit. (25 points)

These are the answers I just need to know how to get them: [RP] = 0.009 and [P] = 0.069 (somewhat rounded); R; endothermic; [RP] = 0.0038 and [P] = 0.074.

Solution:

a) The equilibrium constant,

RP <---> R + P

Initial 0.0780 0 0

change -x x x

Equilibrium 0.0780-x x x

by quadratic equation solving we get,

x = 0.0688 , -0.5888 by taking positive value x=0.0688 M

[R]=[P]=0.0688 M

[RP]=0.0780-0.0688=0.0092 M

b) once the value changes to 1.47 the reaction scheme shifts to left side as [RP] concentration increases

c) The reaction is endothermic as temperature increament only shifts the equilibrium condition hence antibiotic gets activated in the body where the body temperature favours the reaction kinetics of the antibiotic dose.

d) in increased temerature reaction shifts left side hence the reaction scheme will be,

[R] + [P] <------> [RP]

.

similarly by plugging values kd=1.47 and computing for quadratic equation as done in part (a) we get,

[RP]=0.0038 M and [R]=[P]=0.074 M

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