5. calcium (ca") binding to the protein calmodulin (cam) is given by the mechanism ca2+ + cam k=1.0x108 m~15-1 ca/cam at 298 k (a) calculate the initial rate of formation of ca/cam if (ca/cam]o = 0 and [ca²+]o = [cam]o = 1.0 x 10m at 298 k. do the same if [ca²+] = [camo = 1.0 x 10^ m. (b) write a general expression to calculate the concentration of ca/cam as a function of time (ca/cam](t)) by assuming that [ca´"]o = [cam]o and [ca/cam]o = 0. (c) how much product (ca/cam) will form in 1.0 ms, if [ca2+]o = [cam]o = 1.0 mm and [ca/cam]o = 0. (d) how long will it take in seconds) to form 5.0 x 10-8 m of ca/cam under physiological conditions ([ca2+] = [cam]o = 1.0 x 10-7 m and [ca/cam]o = 0)? (e) the dissociation of ca²+ from ca/cam can be described as: k =? ca/cam shoemi ca2+ + cam ca2+ + cam k = 108 m-15-1 a temperature jump experiment measured a relaxation time (t = = to 0.10 s at 298 k under physiological conditions ([ca²+] = [cm]o = 1.0 x 10-7 m). what is ka([ca2+1, +[caml.)+k) equal the value of k., and dissociation constant (ky = )?