Question

The activity of an enzyme requires glutamic acid to display its

Answer 1

A)

the deprotonated form of COOH is COO-

we know that

pH = pKa + log [COO-/ COOH]

log[C00-/COOH]= pH - pKa

[C00-/COOH] = 10^(pH-pKa)

take the inverse

[COOH/COO-] = 10^(pKa-pH)

now at pH = 3.5

[COOH/COO-] = 10^(4.07-3.5)

[COOH/COO-] = 3.7

at pH = 4.5

[COOH/COO-] = 10^(4.07-4.05)

[COOH/COO-] = 0.37

for the activity to be high , there should be high conc of COOH

from the above date we see that

COOH conc is high at pH = 3.5

so the activity is more at pH 3.5

B)

now given

PH = 4.07

so

[COOH/COO-] = 10^(4.07-4.07)

[COOH/COO-] = 1

[COOH] = [COO-]

so half are in protonated state and the other half are in deprotnated state

so

0.5 fraction ( i.e half) of the enzymes will be active at pH = 4.07

C)

now 80% of maximum activity

max activity occurs when protonated form [COOH] is 100%

80% of max activity means

80% are in protonate form (COOH) and remaining 20% are in deprotonate form (C00-)

so

pH = pKa + log [C00-/ COOH]

pH = 4.07 + log [20/80]

pH = 3.468

so

at pH = 3.468 there will be 80% of maximum activity