{"id":219580,"date":"2025-05-26T11:19:06","date_gmt":"2025-05-26T11:19:06","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=219580"},"modified":"2025-05-26T11:19:08","modified_gmt":"2025-05-26T11:19:08","slug":"many-drugs-function-by-acting-as-inhibitors-of-particular-enzyme-reactions","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/05\/26\/many-drugs-function-by-acting-as-inhibitors-of-particular-enzyme-reactions\/","title":{"rendered":"Many drugs function by acting as inhibitors of particular enzyme reactions."},"content":{"rendered":"\n<p>Many drugs function by acting as inhibitors of particular enzyme reactions. If an enzyme\u2019s Vmax is 15 units\/min\/mg protein, with a Km of 1.25 \u00b5M in the absence of inhibitor, but in the presence of 5 \u00b5M inhibitor the Vmax is 6 units\/min\/mg protein, with the same Km, what is the velocity of the reaction in the presence of 5 \u00b5M inhibitor at a substrate concentration of 2.50 \u00b5M? (A) 2 units\/min\/mg protein (B) 4 units\/min\/mg protein (C) 6 units\/min\/mg protein (D) 8 units\/min\/mg protein (E) 10 units\/min\/mg protein<\/p>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-1-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p>To solve this problem, we use the <strong>Michaelis-Menten equation<\/strong>: v=Vmax\u2061[S]Km+[S]v = \\frac{V_{\\max} [S]}{K_m + [S]}<\/p>\n\n\n\n<p>Where:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>vv is the reaction velocity,<\/li>\n\n\n\n<li>Vmax\u2061V_{\\max} is the maximum velocity,<\/li>\n\n\n\n<li>[S][S] is the substrate concentration,<\/li>\n\n\n\n<li>KmK_m is the Michaelis constant.<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\">Given:<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li>In the <strong>presence<\/strong> of inhibitor:\n<ul class=\"wp-block-list\">\n<li>Vmax\u2061=6V_{\\max} = 6 units\/min\/mg protein<\/li>\n\n\n\n<li>Km=1.25\u2009\u03bcMK_m = 1.25 \\, \\mu M<\/li>\n<\/ul>\n<\/li>\n\n\n\n<li>Substrate concentration, [S]=2.50\u2009\u03bcM[S] = 2.50 \\, \\mu M<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\">Step-by-step Calculation:<\/h3>\n\n\n\n<p>v=6\u00d72.501.25+2.50=153.75=4\u2009units\/min\/mg&nbsp;proteinv = \\frac{6 \\times 2.50}{1.25 + 2.50} = \\frac{15}{3.75} = 4 \\, \\text{units\/min\/mg protein}<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Correct Answer:<\/h3>\n\n\n\n<p><strong>(B) 4 units\/min\/mg protein<\/strong><\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Explanation <\/h3>\n\n\n\n<p>Many drugs inhibit enzymes to regulate biochemical pathways. These inhibitors can be classified by their effects on enzyme kinetics, especially on two key parameters: <strong>Vmax\u2061V_{\\max}<\/strong> and <strong>KmK_m<\/strong>.<\/p>\n\n\n\n<p>The Michaelis-Menten equation models how reaction rate depends on substrate concentration. In this problem, the enzyme exhibits a Vmax\u2061V_{\\max} of 15 units\/min\/mg protein and a KmK_m of 1.25 \u00b5M under normal conditions. When 5 \u00b5M of an inhibitor is added, the Vmax\u2061V_{\\max} drops to 6 units\/min\/mg protein, but KmK_m remains unchanged.<\/p>\n\n\n\n<p>A constant KmK_m indicates that the <strong>affinity<\/strong> of the enzyme for the substrate is unaffected. This suggests the inhibitor is likely <strong>non-competitive<\/strong>, binding to an allosteric site rather than the active site. Non-competitive inhibitors reduce the overall number of functional enzymes, lowering Vmax\u2061V_{\\max}, but not altering substrate binding (KmK_m).<\/p>\n\n\n\n<p>To calculate the reaction rate at a substrate concentration of 2.50 \u00b5M, we apply the Michaelis-Menten equation using the inhibited Vmax\u2061V_{\\max} (6 units\/min\/mg protein) and the unchanged KmK_m (1.25 \u00b5M). The result: v=6\u00d72.51.25+2.5=153.75=4\u2009units\/min\/mg&nbsp;proteinv = \\frac{6 \\times 2.5}{1.25 + 2.5} = \\frac{15}{3.75} = 4 \\, \\text{units\/min\/mg protein}<\/p>\n\n\n\n<p>Thus, the reaction velocity under these inhibited conditions is <strong>4 units\/min\/mg protein<\/strong>, matching choice <strong>(B)<\/strong>.<\/p>\n\n\n\n<p>This analysis highlights how enzyme inhibitors affect reaction kinetics and how mathematical modeling helps understand drug actions at the molecular level.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/05\/learnexams-banner8-22.jpeg\" alt=\"\" class=\"wp-image-219581\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>Many drugs function by acting as inhibitors of particular enzyme reactions. If an enzyme\u2019s Vmax is 15 units\/min\/mg protein, with a Km of 1.25 \u00b5M in the absence of inhibitor, but in the presence of 5 \u00b5M inhibitor the Vmax is 6 units\/min\/mg protein, with the same Km, what is the velocity of the reaction [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-219580","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/219580","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=219580"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/219580\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=219580"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=219580"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=219580"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}