{"id":165058,"date":"2024-11-11T07:10:17","date_gmt":"2024-11-11T07:10:17","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=165058"},"modified":"2024-11-11T07:10:19","modified_gmt":"2024-11-11T07:10:19","slug":"given-these-parameters-what-is-the-%ce%b4g-for-k-ion-transport-into-the-frog-muscle-cell-in-the-presence-of-ach","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2024\/11\/11\/given-these-parameters-what-is-the-%ce%b4g-for-k-ion-transport-into-the-frog-muscle-cell-in-the-presence-of-ach\/","title":{"rendered":"Given these parameters, what is the \u0394G for K+ ion transport into the frog muscle cell in the presence of Ach"},"content":{"rendered":"\n<p>Given these parameters, what is the \u0394G for K+ ion transport into the frog muscle cell in the presence of Ach? \u0394G = <em>_<\/em> kJ\/mol<br>Dr. Knowsalot smiles and nods. &#8220;Good,&#8221; she says. &#8220;Now let&#8217;s see whether K+ ions pass through the open Ach receptors into the cell. As you might remember, these channels are permeable to Na, K+, and Ca\u00b2+. For a typical frog muscle cell, the concentration of K+ is roughly 124 mM inside the cell and 2.30 mM outside the cell.&#8221; Remember, to calculate \u0394G for an ion, you simply add the equations for the chemical and electrostatic components together. \u0394G = RT in (C2\/C1) + ZFVm. As mentioned earlier, a typical animal cell resting membrane potential is \u201360 mV, and this process occurs at body temperature, which is 310 K.<\/p>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-6-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p>To calculate the change in Gibbs free energy (\u0394G) for the transport of K\u207a ions into the frog muscle cell in the presence of acetylcholine (Ach), we will use the formula:<\/p>\n\n\n\n<p>[<br>\\Delta G = RT \\ln\\left(\\frac{C_2}{C_1}\\right) + ZFV_m<br>]<\/p>\n\n\n\n<p>Where:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>( R ) is the gas constant, which is 8.314 J\/mol\u00b7K.<\/li>\n\n\n\n<li>( T ) is the temperature in Kelvin (310 K, as given).<\/li>\n\n\n\n<li>( C_1 ) is the concentration of K\u207a outside the cell (2.30 mM).<\/li>\n\n\n\n<li>( C_2 ) is the concentration of K\u207a inside the cell (124 mM).<\/li>\n\n\n\n<li>( Z ) is the charge of the ion, which for K\u207a is +1.<\/li>\n\n\n\n<li>( F ) is the Faraday constant, which is 96,485 C\/mol.<\/li>\n\n\n\n<li>( V_m ) is the membrane potential, which is given as -60 mV (-0.060 V in volts).<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\">Step 1: Calculate the chemical component (( \\Delta G_{\\text{chem}} ))<\/h3>\n\n\n\n<p>This part of the equation involves the concentration gradient and is calculated using the formula:<\/p>\n\n\n\n<p>[<br>\\Delta G_{\\text{chem}} = RT \\ln\\left(\\frac{C_2}{C_1}\\right)<br>]<\/p>\n\n\n\n<p>Substitute the known values:<\/p>\n\n\n\n<p>[<br>\\Delta G_{\\text{chem}} = (8.314 \\, \\text{J\/mol\u00b7K}) \\times (310 \\, \\text{K}) \\times \\ln\\left(\\frac{124 \\, \\text{mM}}{2.30 \\, \\text{mM}}\\right)<br>]<\/p>\n\n\n\n<p>[<br>\\Delta G_{\\text{chem}} = (8.314 \\times 310) \\times \\ln(53.91)<br>]<\/p>\n\n\n\n<p>[<br>\\Delta G_{\\text{chem}} = 2577.34 \\times 3.986<br>]<\/p>\n\n\n\n<p>[<br>\\Delta G_{\\text{chem}} = 10244.96 \\, \\text{J\/mol} = 10.245 \\, \\text{kJ\/mol}<br>]<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 2: Calculate the electrostatic component (( \\Delta G_{\\text{electro}} ))<\/h3>\n\n\n\n<p>This part involves the membrane potential and is calculated using:<\/p>\n\n\n\n<p>[<br>\\Delta G_{\\text{electro}} = ZFV_m<br>]<\/p>\n\n\n\n<p>Substitute the known values:<\/p>\n\n\n\n<p>[<br>\\Delta G_{\\text{electro}} = (1) \\times (96,485 \\, \\text{C\/mol}) \\times (-0.060 \\, \\text{V})<br>]<\/p>\n\n\n\n<p>[<br>\\Delta G_{\\text{electro}} = -5,789.1 \\, \\text{J\/mol} = -5.789 \\, \\text{kJ\/mol}<br>]<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Step 3: Calculate the total ( \\Delta G )<\/h3>\n\n\n\n<p>Now, we add the two components:<\/p>\n\n\n\n<p>[<br>\\Delta G = \\Delta G_{\\text{chem}} + \\Delta G_{\\text{electro}} = 10.245 \\, \\text{kJ\/mol} + (-5.789 \\, \\text{kJ\/mol})<br>]<\/p>\n\n\n\n<p>[<br>\\Delta G = 4.456 \\, \\text{kJ\/mol}<br>]<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Conclusion:<\/h3>\n\n\n\n<p>The \u0394G for K\u207a ion transport into the frog muscle cell in the presence of Ach is <strong>4.46 kJ\/mol<\/strong>.<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Explanation:<\/h3>\n\n\n\n<p>The \u0394G represents the free energy change for K\u207a transport, accounting for both the concentration gradient (chemical component) and the electrostatic influence (electrostatic component). The concentration gradient favors the movement of K\u207a into the cell (positive \u0394G from chemical gradient), but the membrane potential works against this movement (negative \u0394G from electrostatic gradient). Since the total \u0394G is positive, the transport of K\u207a into the cell is not spontaneous under the given conditions.<\/p>\n","protected":false},"excerpt":{"rendered":"<p>Given these parameters, what is the \u0394G for K+ ion transport into the frog muscle cell in the presence of Ach? \u0394G = _ kJ\/molDr. Knowsalot smiles and nods. &#8220;Good,&#8221; she says. &#8220;Now let&#8217;s see whether K+ ions pass through the open Ach receptors into the cell. As you might remember, these channels are permeable [&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-165058","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/165058","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=165058"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/165058\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=165058"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=165058"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=165058"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}