{"id":220133,"date":"2025-05-27T14:52:53","date_gmt":"2025-05-27T14:52:53","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=220133"},"modified":"2025-05-27T14:56:37","modified_gmt":"2025-05-27T14:56:37","slug":"draw-histidine-at-ph-5-ph-8-and-ph-10","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/05\/27\/draw-histidine-at-ph-5-ph-8-and-ph-10\/","title":{"rendered":"Draw histidine at pH 5, pH 8, and pH 10"},"content":{"rendered":"\n<p>Draw histidine at pH 5, pH 8, and pH 10<\/p>\n\n\n\n<p>-please show the net charge on each of the structures<\/p>\n\n\n\n<p>-calculate the pI value of histidine<\/p>\n\n\n\n<p>the pKa values for the amino acid histidine are:<\/p>\n\n\n\n<p>pKa=1.82 for the carboxyl group<\/p>\n\n\n\n<p>pKa=6.0 for its imidazole group<\/p>\n\n\n\n<p>pKa=9.17 for its amino group<\/p>\n\n\n\n<p>please show your work, and thank you for your help!<\/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>Here is the structure of <strong>histidine<\/strong> at <strong>pH 5, pH 8, and pH 10<\/strong>, the <strong>net charges<\/strong>, and the calculation of the <strong>isoelectric point (pI)<\/strong> based on the provided pKa values:<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>1. Histidine Structures and Net Charges<\/strong><\/h3>\n\n\n\n<h4 class=\"wp-block-heading\"><strong>pKa values:<\/strong><\/h4>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Carboxyl group (\u2013COOH): pKa = 1.82<\/li>\n\n\n\n<li>Imidazole side chain: pKa = 6.00<\/li>\n\n\n\n<li>Amino group (\u2013NH3\u207a): pKa = 9.17<\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>a. pH 5<\/strong><\/h3>\n\n\n\n<p>At pH 5:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Carboxyl group<\/strong>: deprotonated (\u2013COO\u207b), because pH > 1.82<\/li>\n\n\n\n<li><strong>Amino group<\/strong>: still protonated (\u2013NH3\u207a), because pH &lt; 9.17<\/li>\n\n\n\n<li><strong>Imidazole group<\/strong>: <strong>protonated<\/strong>, because pH &lt; 6.0 \u2192 <strong>positively charged<\/strong><\/li>\n<\/ul>\n\n\n\n<p><strong>Net charge = +1 (NH3\u207a) \u22121 (COO\u207b) +1 (protonated imidazole) = +1<\/strong><\/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. pH 8<\/strong><\/h3>\n\n\n\n<p>At pH 8:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Carboxyl group<\/strong>: deprotonated (\u2013COO\u207b)<\/li>\n\n\n\n<li><strong>Amino group<\/strong>: protonated (\u2013NH3\u207a)<\/li>\n\n\n\n<li><strong>Imidazole group<\/strong>: <strong>deprotonated<\/strong>, because pH > 6.0 \u2192 <strong>neutral<\/strong><\/li>\n<\/ul>\n\n\n\n<p><strong>Net charge = +1 (NH3\u207a) \u22121 (COO\u207b) +0 (imidazole) = 0<\/strong><\/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. pH 10<\/strong><\/h3>\n\n\n\n<p>At pH 10:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Carboxyl group<\/strong>: deprotonated (\u2013COO\u207b)<\/li>\n\n\n\n<li><strong>Amino group<\/strong>: <strong>deprotonated<\/strong> to \u2013NH2 \u2192 <strong>neutral<\/strong><\/li>\n\n\n\n<li><strong>Imidazole group<\/strong>: deprotonated \u2192 neutral<\/li>\n<\/ul>\n\n\n\n<p><strong>Net charge = 0 (NH2) \u22121 (COO\u207b) +0 (imidazole) = \u22121<\/strong><\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>2. Isoelectric Point (pI) Calculation<\/strong><\/h3>\n\n\n\n<p>The pI is the pH where net charge = 0. For histidine, this occurs between the states where:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>One form has <strong>+1 charge<\/strong> (pH 5)<\/li>\n\n\n\n<li>One form has <strong>0 charge<\/strong> (pH 8)<\/li>\n<\/ul>\n\n\n\n<p>The neutral form is between:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Protonated <strong>imidazole group<\/strong> (pKa = 6.0)<\/li>\n\n\n\n<li>Protonated <strong>amino group<\/strong> (pKa = 9.17)<\/li>\n<\/ul>\n\n\n\n<p>So, take average of <strong>pKa for imidazole and amino<\/strong>: pI=6.00+9.172=15.172=7.59\\text{pI} = \\frac{6.00 + 9.17}{2} = \\frac{15.17}{2} = \\boxed{7.59}<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Summary Table<\/strong><\/h3>\n\n\n\n<figure class=\"wp-block-table\"><table class=\"has-fixed-layout\"><thead><tr><th><strong>pH<\/strong><\/th><th>Carboxyl<\/th><th>Amino<\/th><th>Imidazole<\/th><th>Net Charge<\/th><\/tr><\/thead><tbody><tr><td>5<\/td><td>\u22121<\/td><td>+1<\/td><td>+1<\/td><td><strong>+1<\/strong><\/td><\/tr><tr><td>8<\/td><td>\u22121<\/td><td>+1<\/td><td>0<\/td><td><strong>0<\/strong><\/td><\/tr><tr><td>10<\/td><td>\u22121<\/td><td>0<\/td><td>0<\/td><td><strong>\u22121<\/strong><\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Explanation (\u2248300 words)<\/strong><\/h3>\n\n\n\n<p>Histidine is a polar amino acid with a unique imidazole side chain that makes its acid\u2013base behavior distinctive. Understanding how histidine behaves at different pH values is crucial for interpreting protein structure, enzyme function, and buffering capacity.<\/p>\n\n\n\n<p>At <strong>low pH (pH 5)<\/strong>, the environment is acidic, so all ionizable groups that can hold protons do so. The carboxyl group loses its proton first (pKa 1.82), so it exists as \u2013COO\u207b. The amino group remains protonated (\u2013NH3\u207a), and the imidazole side chain, with a pKa of 6.0, is still protonated and positively charged. Thus, the net charge on histidine is <strong>+1<\/strong>.<\/p>\n\n\n\n<p>At <strong>neutral to slightly basic pH (pH 8)<\/strong>, the imidazole group loses its proton (since pH &gt; pKa 6.0), becoming neutral. The carboxyl is deprotonated (\u2013COO\u207b), and the amino group remains protonated (\u2013NH3\u207a) since the pH is still below its pKa (9.17). The charges balance out, giving histidine a <strong>net charge of 0<\/strong>. This state is called the <strong>zwitterion<\/strong>, where the molecule is electrically neutral but has charged groups.<\/p>\n\n\n\n<p>At <strong>pH 10<\/strong>, the environment is more basic than all three pKa values. Thus, the carboxyl is deprotonated (\u2013COO\u207b), the amino group is also deprotonated to \u2013NH2 (neutral), and the imidazole remains neutral. The molecule carries a <strong>net charge of \u22121<\/strong>.<\/p>\n\n\n\n<p>The <strong>isoelectric point (pI)<\/strong> of histidine lies between the pKa values where the molecule transitions between +1 and 0, and 0 and \u20131. For histidine, this is between the <strong>imidazole (6.0)<\/strong> and <strong>amino group (9.17)<\/strong>. Averaging these gives <strong>pI = 7.59<\/strong>, meaning that at this pH, histidine exists predominantly in its zwitterionic, uncharged form.<\/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-banner8-42.jpeg\" alt=\"\" class=\"wp-image-220134\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>Draw histidine at pH 5, pH 8, and pH 10 -please show the net charge on each of the structures -calculate the pI value of histidine the pKa values for the amino acid histidine are: pKa=1.82 for the carboxyl group pKa=6.0 for its imidazole group pKa=9.17 for its amino group please show your work, and [&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-220133","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/220133","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=220133"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/220133\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=220133"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=220133"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=220133"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}