Question

biochemistry please answer 7 and 8 since they are related and show work. Thank you

6. Penicillin is hydrolyzed and thereby rendered inactive by penicillinase, an enzyme present in some resistant bacteria. The amount of penicillin hydrolyzed in 1 minute by a solution containing purified penicillinase was measured as a function of the concentration of penicillin. Assume that the concentrations of penicillin and penicillinase do not change appreciably during the assay. See attached graph. What is the value of Kn? What is the value of Vmax? Show your work. Amount hydrolyzed (mol/min) DATA [Penicillin] (M) 0.1 x 10" 0.3 x 10 0.5 x 10 1.0 x 10 3.0 x 10 5.0 x 105 YE 7.625 x+ 1.41:24 whenxo x = lumax = 1.466 24 Vmax 1.4624 -0.68 0.11 x 10° 0.25 x 10° 0.34 x 10 0.45 x 10 0.58 x 10° 0.61 x 109 Y: 7.625 x + 1046 24 yo x= 1/km = 7.lezs = 5.21 1.4624 7. After first incubating penicillinase with Ampicillin, it is found that the affinity of the enzyme for penicillin had decreased. From this piece of information, what type of inhibition would one propose is occurring? What additional piece (or pieces) of information (if any) would be useful in further proving the type of inhibition that is occurring? 8. Based on your answer to 7, draw a double reciprocal plot (Lineweaver-Burk) for the predicted type of inhibition with correctly labeled axes, and identify the uninhibited and inhibited lines.

Answer 1

Answer-

According to the given question-

Question 7-

Penicillin is a antibiotic which are obtaind from Penicillium mould and used to treat the infections caused by the bacteria.

penicillinase- also called β-lactamase is a enzyme that are used for the disruption of penicillin internal structure and antimicrobial action of the drug is reduced.

According to the given information in the question 7 -

Penicillinase is first incubnated with ampicillin followed by incubation with penicillin and the author found that the affinity  of Penicillinase for binding with penicillin was decreased because the ampicillin molecule have affinity with the binding site of Penicillinase enzyme when incubated with ampicillin. the ampicillin also compete with the penicillin for the binding at the active site of Penicillinase so this is a example of competitive inhibition.

Competitive inhibition- It is a type of reversible reaction and the effect of inhibitor can be reversed in under some circumstances. Here the inhibitor competes with substrate for binding with the active site of enzyme because they posses some structural simmilarity. here first inhibitor binds with the enzyme and form enzyme inhibitor complex and prevent the binding of substrate with the enzyme due to which product does not form.

Michaelis menten equation for this type of inhibition can be written as -

Product + Enzyme (E)Konditor) E+sYe ES 129_ 4-1 Complet I(Inhibitor) El+S No product formation K; - [E]Cty (E11 EL, (EJO, Km[es] (u ki [s] Ki So Tatal Enzyme concentrate (E) [E], - [E]+[81]+[ES] Chry) (Blu-te si ' * 3)} [ES] , [4],[s] we know that Enzyme velocity- V.k[ES] Y, CE], [s] ki but d = {1* (1) * 4, 61, 3 Umat Sobting the Ve Valve weget. Umar (3) a km + [s]

Vrare - -- st alormal - with Inhibitor here Umax is 1. equal but km is different y ! km km (Inhibitor) (normal) Michaelis menten Curve km (Inhibitor) shift towards Right side which means Km contributor) > km (neormal)

Question 8-

Double Reciprocal blot Line weaver-berk plat 11+1 ICS] Umar within hibitor - 1 normal (km 1 1+1 Umox CSJ Umar Alumax '[s] → slope km Umar and Interupt = Umox

here by increasing in the valve of $\alpha$ we can reverse the inhibition because on increasing the $\alpha$ value the degree of inhibition in stronger due to increase in the valve of Km affinity of substrate towards the enzyme is decreases. so we increase the concentration of substrate to counter the binding of activwe site of enzyme than the inhibitor.