{"id":186161,"date":"2025-01-24T08:14:14","date_gmt":"2025-01-24T08:14:14","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=186161"},"modified":"2025-01-24T08:14:16","modified_gmt":"2025-01-24T08:14:16","slug":"the-net-reaction-catalyzed-by-malate-dehydrogenase-is-malate-nad-oxaloacetate-nadh-h","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/01\/24\/the-net-reaction-catalyzed-by-malate-dehydrogenase-is-malate-nad-oxaloacetate-nadh-h\/","title":{"rendered":"The net reaction catalyzed by malate dehydrogenase is malate + NAD+ oxaloacetate + NADH + H +"},"content":{"rendered":"\n<p>The net reaction catalyzed by malate dehydrogenase is malate + NAD+ oxaloacetate + NADH + H + The standard reduction potentials for the half-reactions are given in the table: Calculate Delta E0&#8242; for the reaction as shown: Delta E0&#8242; = V Calculate Delta G0&#8242;: Delta G0 = kcal \/ mol<br><\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img decoding=\"async\" src=\"https:\/\/learnexams.com\/blog\/wp-content\/uploads\/2025\/01\/image-438.png\" alt=\"\" class=\"wp-image-186162\"\/><\/figure>\n\n\n\n<p><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-6-color\"><strong>The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n<p>The reaction catalyzed by malate dehydrogenase is:<\/p>\n\n\n\n<p><strong>Malate + NAD\u207a \u21cc Oxaloacetate + NADH + H\u207a<\/strong><\/p>\n\n\n\n<p>To calculate the standard change in Gibbs free energy (\u0394G\u00b0&#8217;) and the standard cell potential (\u0394E\u00b0&#8217;) for this reaction, we can use the standard reduction potentials provided for the half-reactions:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Oxaloacetate + 2H\u207a + 2e\u207b \u21cc Malate<\/strong>: E\u00b0&#8217; = -0.17 V<\/li>\n\n\n\n<li><strong>NAD\u207a + H\u207a + 2e\u207b \u21cc NADH<\/strong>: E\u00b0&#8217; = -0.32 V<\/li>\n<\/ul>\n\n\n\n<p><strong>Calculating \u0394E\u00b0&#8217;:<\/strong><\/p>\n\n\n\n<p>The standard cell potential (\u0394E\u00b0&#8217;) is calculated by subtracting the reduction potential of the oxidation half-reaction from that of the reduction half-reaction:<\/p>\n\n\n\n<p>\u0394E\u00b0&#8217; = E\u00b0&#8217; (reduction) &#8211; E\u00b0&#8217; (oxidation)<\/p>\n\n\n\n<p>In this reaction, malate is oxidized to oxaloacetate, and NAD\u207a is reduced to NADH. Therefore, the oxidation half-reaction is:<\/p>\n\n\n\n<p><strong>Malate \u21cc Oxaloacetate + 2H\u207a + 2e\u207b<\/strong><\/p>\n\n\n\n<p>The reduction half-reaction is:<\/p>\n\n\n\n<p><strong>NAD\u207a + H\u207a + 2e\u207b \u21cc NADH<\/strong><\/p>\n\n\n\n<p>Thus, \u0394E\u00b0&#8217; = (-0.32 V) &#8211; (-0.17 V) = -0.15 V<\/p>\n\n\n\n<p><strong>Calculating \u0394G\u00b0&#8217;:<\/strong><\/p>\n\n\n\n<p>The standard change in Gibbs free energy (\u0394G\u00b0&#8217;) is related to the standard cell potential (\u0394E\u00b0&#8217;) by the equation:<\/p>\n\n\n\n<p>\u0394G\u00b0&#8217; = -nF\u0394E\u00b0&#8217;<\/p>\n\n\n\n<p>Where:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>n<\/strong> = number of moles of electrons transferred (2 electrons in this reaction)<\/li>\n\n\n\n<li><strong>F<\/strong> = Faraday&#8217;s constant (96,485 C\/mol)<\/li>\n\n\n\n<li><strong>\u0394E\u00b0&#8217;<\/strong> = standard cell potential (in volts)<\/li>\n<\/ul>\n\n\n\n<p>Substituting the known values:<\/p>\n\n\n\n<p>\u0394G\u00b0&#8217; = -2 \u00d7 96,485 C\/mol \u00d7 (-0.15 V)<\/p>\n\n\n\n<p>\u0394G\u00b0&#8217; = 28,946 J\/mol<\/p>\n\n\n\n<p>Since 1 kcal = 4,184 J, \u0394G\u00b0&#8217; = 28,946 J\/mol \u00f7 4,184 J\/kcal \u2248 6.9 kcal\/mol<\/p>\n\n\n\n<p><strong>Conclusion:<\/strong><\/p>\n\n\n\n<p>The standard cell potential (\u0394E\u00b0&#8217;) for the malate dehydrogenase reaction is -0.15 V, and the standard change in Gibbs free energy (\u0394G\u00b0&#8217;) is approximately 6.9 kcal\/mol.<\/p>\n\n\n\n<p>A negative \u0394E\u00b0&#8217; indicates that the reaction is non-spontaneous under standard conditions. The positive \u0394G\u00b0&#8217; value further confirms that the reaction requires energy input to proceed. In biological systems, this reaction is coupled with other reactions to drive the overall metabolic processes.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>The net reaction catalyzed by malate dehydrogenase is malate + NAD+ oxaloacetate + NADH + H + The standard reduction potentials for the half-reactions are given in the table: Calculate Delta E0&#8242; for the reaction as shown: Delta E0&#8242; = V Calculate Delta G0&#8242;: Delta G0 = kcal \/ mol The Correct Answer and Explanation [&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-186161","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/186161","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=186161"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/186161\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=186161"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=186161"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=186161"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}