{"id":253106,"date":"2025-07-11T21:37:06","date_gmt":"2025-07-11T21:37:06","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=253106"},"modified":"2025-07-11T21:37:08","modified_gmt":"2025-07-11T21:37:08","slug":"determine-the-fraction-of-vmax-that-would-be-obtained-at-the-following-substrate-concentrations","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/07\/11\/determine-the-fraction-of-vmax-that-would-be-obtained-at-the-following-substrate-concentrations\/","title":{"rendered":"Determine the fraction of Vmax that would be obtained at the following substrate concentrations."},"content":{"rendered":"\n<p>Determine the fraction of Vmax that would be obtained at the following substrate concentrations. SHOW ALL WORK: a.) [S] = 0.5Km b.) [S] = 2Km c.) [S] = 10Km<\/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 determine the fraction of Vmax at different substrate concentrations, we use the Michaelis-Menten equation. The Michaelis-Menten equation relates the rate of reaction (v) to the substrate concentration [S] and the maximum velocity (Vmax) of the enzyme: v=Vmax[S]Km+[S]v = \\frac{V_{\\text{max}} [S]}{K_m + [S]}v=Km\u200b+[S]Vmax\u200b[S]\u200b<\/p>\n\n\n\n<p>Where:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>vvv is the reaction velocity at substrate concentration [S][S][S],<\/li>\n\n\n\n<li>VmaxV_{\\text{max}}Vmax\u200b is the maximum reaction velocity,<\/li>\n\n\n\n<li>KmK_mKm\u200b is the Michaelis constant (the substrate concentration at which the reaction rate is half of VmaxV_{\\text{max}}Vmax\u200b).<\/li>\n<\/ul>\n\n\n\n<p>The fraction of VmaxV_{\\text{max}}Vmax\u200b is found by calculating: vVmax=[S]Km+[S]\\frac{v}{V_{\\text{max}}} = \\frac{[S]}{K_m + [S]}Vmax\u200bv\u200b=Km\u200b+[S][S]\u200b<\/p>\n\n\n\n<p>Now, let\u2019s calculate this fraction for the given substrate concentrations:<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">a.) [S]=0.5Km[S] = 0.5 K_m[S]=0.5Km\u200b<\/h3>\n\n\n\n<p>Substitute [S]=0.5Km[S] = 0.5 K_m[S]=0.5Km\u200b into the formula: vVmax=0.5KmKm+0.5Km=0.5Km1.5Km=13\\frac{v}{V_{\\text{max}}} = \\frac{0.5 K_m}{K_m + 0.5 K_m} = \\frac{0.5 K_m}{1.5 K_m} = \\frac{1}{3}Vmax\u200bv\u200b=Km\u200b+0.5Km\u200b0.5Km\u200b\u200b=1.5Km\u200b0.5Km\u200b\u200b=31\u200b<\/p>\n\n\n\n<p>Thus, the fraction of VmaxV_{\\text{max}}Vmax\u200b at [S]=0.5Km[S] = 0.5 K_m[S]=0.5Km\u200b is <strong>1\/3<\/strong> or approximately 33.33%.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">b.) [S]=2Km[S] = 2 K_m[S]=2Km\u200b<\/h3>\n\n\n\n<p>Substitute [S]=2Km[S] = 2 K_m[S]=2Km\u200b into the formula: vVmax=2KmKm+2Km=2Km3Km=23\\frac{v}{V_{\\text{max}}} = \\frac{2 K_m}{K_m + 2 K_m} = \\frac{2 K_m}{3 K_m} = \\frac{2}{3}Vmax\u200bv\u200b=Km\u200b+2Km\u200b2Km\u200b\u200b=3Km\u200b2Km\u200b\u200b=32\u200b<\/p>\n\n\n\n<p>Thus, the fraction of VmaxV_{\\text{max}}Vmax\u200b at [S]=2Km[S] = 2 K_m[S]=2Km\u200b is <strong>2\/3<\/strong> or approximately 66.67%.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">c.) [S]=10Km[S] = 10 K_m[S]=10Km\u200b<\/h3>\n\n\n\n<p>Substitute [S]=10Km[S] = 10 K_m[S]=10Km\u200b into the formula: vVmax=10KmKm+10Km=10Km11Km=1011\\frac{v}{V_{\\text{max}}} = \\frac{10 K_m}{K_m + 10 K_m} = \\frac{10 K_m}{11 K_m} = \\frac{10}{11}Vmax\u200bv\u200b=Km\u200b+10Km\u200b10Km\u200b\u200b=11Km\u200b10Km\u200b\u200b=1110\u200b<\/p>\n\n\n\n<p>Thus, the fraction of VmaxV_{\\text{max}}Vmax\u200b at [S]=10Km[S] = 10 K_m[S]=10Km\u200b is <strong>10\/11<\/strong> or approximately 90.91%.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Summary:<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li>At [S]=0.5Km[S] = 0.5 K_m[S]=0.5Km\u200b, the fraction of VmaxV_{\\text{max}}Vmax\u200b is 1\/3 or 33.33%.<\/li>\n\n\n\n<li>At [S]=2Km[S] = 2 K_m[S]=2Km\u200b, the fraction of VmaxV_{\\text{max}}Vmax\u200b is 2\/3 or 66.67%.<\/li>\n\n\n\n<li>At [S]=10Km[S] = 10 K_m[S]=10Km\u200b, the fraction of VmaxV_{\\text{max}}Vmax\u200b is 10\/11 or 90.91%.<\/li>\n<\/ul>\n\n\n\n<p>These calculations show that as the substrate concentration increases relative to the KmK_mKm\u200b, the reaction velocity approaches VmaxV_{\\text{max}}Vmax<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/07\/learnexams-banner9-74.jpeg\" alt=\"\" class=\"wp-image-253107\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>Determine the fraction of Vmax that would be obtained at the following substrate concentrations. SHOW ALL WORK: a.) [S] = 0.5Km b.) [S] = 2Km c.) [S] = 10Km The Correct Answer and Explanation is: To determine the fraction of Vmax at different substrate concentrations, we use the Michaelis-Menten equation. The Michaelis-Menten equation relates the [&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-253106","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/253106","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=253106"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/253106\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=253106"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=253106"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=253106"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}