N. . 2160 bedoctant) and th and the electron accepts Problem ?

Hi! I’m a little confused with this problem and was hoping to be guided through!

N. . 2160bedoctant) and thand the electron acceptsProblem ?Thermolysan enzyme thatt greatly iny increases the ratete of hydrolysis of the amids band in ebacteria produce thermalysin to break downa defensebolyypeptides.somespecies of pathonproduced byy the immune systsystem. Therettherefore, small molecules that Inhibsnhibit thermalysin may bections thatpepimber of eor treating infecomprises three "binding pactCompound 1interacts withth thermolysinat a specific sitevaried the18 51, 51′, and 52denoted as 51, sd 52′, To design effectlyeffective thermolysin inn Inhibitors, chemists systematicallyof the substituent (R)uent (R) near the 52′ pocket and then measmeasured the thermadynamics ates of binding alIdentity of thethe different Inent inhibitors to thermolysin.Val180Phe 114AsnHIPLeutaaAsn116Glut4aVal 198PhetaoLOUPOP16 180Tyr 157Tyr19aHis142His140Glu160(a) The dissociation constant (K;) for compound 1 was determined to be 1.81 jM, and the entropychange of binding (AS") was determined to be 11.1 cal/mol-K. Assuming that the bindingexperiment was performed under standard biochemical conditions, calculate the enthalpy ofbinding (AH"’) for compound 1. Assume T = 25 "C.(b) When compound 2 was studied, it was found to bind 11 times more tightly to thermolysin thancompound 1. The enthalpy of binding (AH") of 2 to thermolysin was determined to be -7.7kcal/mol. Is the entropy change for binding compound 2 more or less favorable than for compound1? Express your answer quantitatively by reporting the appropriate value in units of cal/mol . K.

0 replies

Leave a Reply

Want to join the discussion?
Feel free to contribute!

Leave a Reply

Your email address will not be published. Required fields are marked *