{"id":25589,"date":"2025-06-19T05:21:12","date_gmt":"2025-06-19T05:21:12","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=25589"},"modified":"2025-06-19T05:21:14","modified_gmt":"2025-06-19T05:21:14","slug":"aspirin-is-an-irreversible-inhibitor-of-pgh2-synthase-reacting-with-a-ser-residue-in-the-active-site","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/aspirin-is-an-irreversible-inhibitor-of-pgh2-synthase-reacting-with-a-ser-residue-in-the-active-site\/","title":{"rendered":"Aspirin is an irreversible inhibitor of PGH2 synthase, reacting with a Ser residue in the active site"},"content":{"rendered":"\n<p>Aspirin is an irreversible inhibitor of PGH2 synthase, reacting with a Ser residue in the active site. You have incubated 5 \u00c2\u00b5M enzyme with 1 mM aspirin and have found that 85% of the enzyme was inactivated in 30 minutes. A) Suppose you wanted to use 5 mM aspirin and inactivate 99% of the enzyme. For how long should you allow the reaction to proceed? B) Suppose that you incubate 1 \u00c2\u00b5M enzyme with 1 mM aspirin for 30 minutes. Would you expect more or less than 85% of the enzyme to be inactivated? Explain your reasoning. C) Suppose that you incubate 1 mM enzyme with 1 mM aspirin for 30 minutes. Would you expect more or less than 85% of the enzyme to be inactivated? Explain your reasoning.<\/p>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-vivid-green-cyan-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p>We can use <strong>pseudo-first-order kinetics<\/strong> for irreversible inhibition when the inhibitor (aspirin) is in large excess over the enzyme.<\/p>\n\n\n\n<p>The equation for first-order decay is:[E][E0]=e\u2212kt\\frac{[E]}{[E_0]} = e^{-kt}[E0\u200b][E]\u200b=e\u2212kt<\/p>\n\n\n\n<p>Given:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>85% inactivated \u2192 15% active enzyme remains<\/li>\n\n\n\n<li>[E]\/[E0]=0.15[E]\/[E_0] = 0.15[E]\/[E0\u200b]=0.15<\/li>\n\n\n\n<li>[I]=1[I] = 1[I]=1 mM<\/li>\n\n\n\n<li>[E0]=5[E_0] = 5[E0\u200b]=5 \u00b5M<\/li>\n\n\n\n<li>Time = 30 min<\/li>\n<\/ul>\n\n\n\n<p>First, solve for <strong>k<\/strong>:0.15=e\u2212k\u22c530\u21d2ln\u2061(0.15)=\u221230k\u21d2k=\u2212ln\u2061(0.15)30\u22481.897130\u22480.06324&nbsp;min\u221210.15 = e^{-k \\cdot 30} \\Rightarrow \\ln(0.15) = -30k \\Rightarrow k = \\frac{-\\ln(0.15)}{30} \\approx \\frac{1.8971}{30} \\approx 0.06324\\ \\text{min}^{-1}0.15=e\u2212k\u22c530\u21d2ln(0.15)=\u221230k\u21d2k=30\u2212ln(0.15)\u200b\u2248301.8971\u200b\u22480.06324&nbsp;min\u22121<\/p>\n\n\n\n<p>Now, you want 99% inactivation \u2192 1% enzyme remains active:0.01=e\u2212kt0.01 = e^{-kt}0.01=e\u2212kt<\/p>\n\n\n\n<p>Solve for <strong>t<\/strong> with [I]=5[I] = 5[I]=5 mM (5\u00d7 more inhibitor):<\/p>\n\n\n\n<p>Rate increases proportionally:knew=5\u00d70.06324=0.3162&nbsp;min\u22121k_{\\text{new}} = 5 \\times 0.06324 = 0.3162\\ \\text{min}^{-1}knew\u200b=5\u00d70.06324=0.3162&nbsp;min\u221210.01=e\u22120.3162t\u21d2ln\u2061(0.01)=\u22120.3162t\u21d2t=4.60520.3162\u224814.57&nbsp;min0.01 = e^{-0.3162t} \\Rightarrow \\ln(0.01) = -0.3162t \\Rightarrow t = \\frac{4.6052}{0.3162} \\approx 14.57\\ \\text{min}0.01=e\u22120.3162t\u21d2ln(0.01)=\u22120.3162t\u21d2t=0.31624.6052\u200b\u224814.57&nbsp;min<\/p>\n\n\n\n<p><strong>Answer A:<\/strong> Approximately <strong>14.6 minutes<\/strong> are needed to inactivate 99% of the enzyme with 5 mM aspirin.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>B)<\/strong><\/h3>\n\n\n\n<p>If you incubate <strong>1 \u00b5M enzyme<\/strong> with <strong>1 mM aspirin<\/strong> (same inhibitor concentration), you now have a <strong>larger excess of inhibitor<\/strong> (1000-fold excess vs. 200-fold before).<\/p>\n\n\n\n<p>This increases the likelihood of enzyme-inhibitor encounters. Because pseudo-first-order kinetics still applies and aspirin is in even greater excess, the inactivation will proceed <strong>faster<\/strong>.<\/p>\n\n\n\n<p><strong>Answer B:<\/strong> You would expect <strong>more than 85%<\/strong> of the enzyme to be inactivated in 30 minutes.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>C)<\/strong><\/h3>\n\n\n\n<p>Now you incubate <strong>1 mM enzyme<\/strong> with <strong>1 mM aspirin<\/strong> \u2014 the molar ratio is <strong>1:1<\/strong>.<\/p>\n\n\n\n<p>In this case, there is <strong>no excess of aspirin<\/strong>, so not all enzyme molecules can be inactivated. At most, only 1 mM of enzyme can be inactivated if all aspirin molecules react. However, this is under ideal conditions.<\/p>\n\n\n\n<p>Also, with a 1:1 ratio, the reaction no longer follows pseudo-first-order kinetics \u2014 it&#8217;s a <strong>second-order reaction<\/strong>, and competition between enzyme molecules can slow the rate.<\/p>\n\n\n\n<p><strong>Answer C:<\/strong> You would expect <strong>less than 85%<\/strong> of the enzyme to be inactivated in 30 minutes.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"852\" height=\"1024\" src=\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-63.jpeg\" alt=\"\" class=\"wp-image-25590\" srcset=\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-63.jpeg 852w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-63-250x300.jpeg 250w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-63-768x923.jpeg 768w\" sizes=\"auto, (max-width: 852px) 100vw, 852px\" \/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>Aspirin is an irreversible inhibitor of PGH2 synthase, reacting with a Ser residue in the active site. You have incubated 5 \u00c2\u00b5M enzyme with 1 mM aspirin and have found that 85% of the enzyme was inactivated in 30 minutes. A) Suppose you wanted to use 5 mM aspirin and inactivate 99% of the enzyme. [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[],"class_list":["post-25589","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/25589","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/comments?post=25589"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/25589\/revisions"}],"predecessor-version":[{"id":25591,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/25589\/revisions\/25591"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=25589"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=25589"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=25589"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}