Question

2) (14 pts) Shown on the next page is a schematic drawing of the mechanism for an RNase. Residues that are important for catalysis are also shown schematically. (a) Refer to the scheme and describe the roles of His 12, Lys 41, and His 119 in the catalytic mechanism. List all of the following that apply: general acid/base catalysis (GABC), covalent catalysis, electrostatic stabilization of transition state. For a GABC, indicate in which step(s) of the mechanism the residue acts as a GAC or GBC (e.g., "residue xyz acts as a GAC in the 2nd step, and as a GBC in 3rd step") (b) At pH 6.0 the value of the rate constant kcat for the catalyzed reaction is 1.2 x 103 s at 25 C for wild-type RNase, whereas the rate constant of the noncatalyzed reaction is 3.0 x 109 s1 under the same conditions. Calculate the rate acceleration for RNase at 25 C and pH 6.0. Show your work (c) Compared to the wild-type RNase, the Lys41Ala (K41A) mutant of RNase reduces transition state stabilization by 29.5 kl/mol at 25 "C and pH 6.0. Calculate keat at 25 C and pH 6.0 for the K14A mutant of RNase. Show your work and state any assumptions you make in solving this problem (d) The RNase described above is maximally active at pH 6.0. The plot of enzyme activity vs. pH shows two ionizable groups: one with a pk, 5.4 the other with a pK, 6.6. One of these corresponds to His 12 and the other to His 119. What do you predict the pK, of His 12 to be? What is the pK, of His 119? Explain your reasoning. (e) Given the information in part (d), calculate the fraction of RNase molecules that are predicted to be catalytically active at pH 6.40. In other words, what fraction of the RNase in solution at pH 6.40 will have both His residues in the protonation states that correspond to the catalytically active form of the enzyme? Show your work

Answer 1

(a) Role of His 12 : General acid/base catalysis in First step. ( GBC : as it uses lone pair on N to accept proton of -OH group in 1st step).

   Lys 41 : Electrostatic stabilization of transition state : (-NH₃⁺ of lysine stabilizes -ve charge (^-O-P-) on pentavalent phosphate intermediate (formed in step 1) by electrostatic attraction )

Role of His 119 : General acid/base catalysis in second step. ( act as GAC in 2nd step , it loses proton on imidazole ring N to 5'- OCH₂- of ribose).

(b) kcat at pH (6.0) and 25 C = 1.2*10³ s-1

k (uncatalyzed ) at pH (6.0) and 25 C = 3.0*10^-9 s-1

so rate acceleration for RNase = k(cat.) / k(uncat.) = 1.2*10³ s-1 / 3.0*10^-9 s-1 = 4.0*10¹¹

(c) we assume cataysis follows transition state theory ;

Being other thing similar except transition state energy : (at pH (6.0) and 25 C)

kcat(mut.) / kcat = ( e−(ΔG + 29.5 kJ/mol)/RT ) /  e−ΔG/RT =  e−( 29.5 kJ/mol)/RT = 6.74*10^-6

kcat = 1.2*10³ s-1

so, kcat(mut.) = 8.1*10^-3 s-1

(d) two ionizable group pKa = 5.4 and pKa = 6.6

pK_a is the negative base10 logarithm of the acid dissociation constant (K_a) ; The lower the pK_a value, the stronger the acid. It is also impacted by pKb of conjugate base also : pKa +pKb = 14

pKa = 5.4 is for His 119 ,as its conjugate base having more pKb (as it accept proton from water which more acidic than HOCH₂-).

so,  pKa = 6.6 is for His 12 ,as its conjugate base having comparatively less pKb.