{"id":221036,"date":"2025-05-29T06:21:38","date_gmt":"2025-05-29T06:21:38","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=221036"},"modified":"2025-05-29T06:21:41","modified_gmt":"2025-05-29T06:21:41","slug":"dithiothreitol-dtt-is-a-compound-with-molecular-weight-of-154-3-daltons-which-means-that-one-mole-of-dtt-has-a-mass-of-154-3-grams-and-a-1-0-molar-solution-abbreviated-with-indicating-moles-per","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/05\/29\/dithiothreitol-dtt-is-a-compound-with-molecular-weight-of-154-3-daltons-which-means-that-one-mole-of-dtt-has-a-mass-of-154-3-grams-and-a-1-0-molar-solution-abbreviated-with-indicating-moles-per\/","title":{"rendered":"Dithiothreitol (DTT) is a compound with molecular weight of 154.3 Daltons, which means that one mole of DTT has a mass of 154.3 grams, and a 1.0 molar solution (abbreviated , with indicating moles per Liter) has of DTT dissolved in water to a total volume of"},"content":{"rendered":"\n<p>Dithiothreitol (DTT) is a compound with molecular weight of 154.3 Daltons, which means that one mole of DTT has a mass of 154.3 grams, and a 1.0 molar solution (abbreviated , with indicating moles per Liter) has of DTT dissolved in water to a total volume of . Let&#8217;s say I give you an aqueous solution that is (millimolar) DTT. Calculate the concentration of DTT in units of mass per volume.<br>Problem 2:<br>Let&#8217;s say you have on your lab bench the following four stock solutions:<br>Tris ()<br>glycerol<br>EDTA<br>DTT<br>Suppose you wish to prepare a buffer solution that has the following composition: Tris glycerol, EDTA, DTT (carefully read these units, and carefully read the stock solutions&#8217; units also!) Prepare a table showing precisely what volume of each stock solution (and of water) you would dispense into a single beaker to<\/p>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-ast-global-color-0-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Problem 1: Convert mM DTT to mg\/mL<\/h3>\n\n\n\n<p>Given:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Molecular weight of DTT<\/strong> = 154.3 g\/mol<\/li>\n\n\n\n<li>Solution concentration = <strong>X mM<\/strong> (millimolar), where 1 mM = 0.001 mol\/L<\/li>\n<\/ul>\n\n\n\n<p>We want to convert this into <strong>mg\/mL<\/strong>.<\/p>\n\n\n\n<p>Let\u2019s say the solution is <strong>1.0 mM<\/strong> DTT for illustration (since you left the value blank). Then: 1.0&nbsp;mM=0.001&nbsp;mol\/L\\text{1.0 mM} = 0.001 \\text{ mol\/L} Mass&nbsp;of&nbsp;DTT&nbsp;per&nbsp;L=0.001&nbsp;mol\u00d7154.3&nbsp;g\/mol=0.1543&nbsp;g=154.3&nbsp;mg\\text{Mass of DTT per L} = 0.001 \\text{ mol} \\times 154.3 \\text{ g\/mol} = 0.1543 \\text{ g} = 154.3 \\text{ mg} Since&nbsp;this&nbsp;is&nbsp;per&nbsp;liter&nbsp;(1000&nbsp;mL),concentration&nbsp;in&nbsp;mg\/mL=154.3&nbsp;mg1000&nbsp;mL=0.1543&nbsp;mg\/mL\\text{Since this is per liter (1000 mL)}, \\text{concentration in mg\/mL} = \\frac{154.3 \\text{ mg}}{1000 \\text{ mL}} = 0.1543 \\text{ mg\/mL}<\/p>\n\n\n\n<p>\ud83d\udd39 <strong>Final formula<\/strong>:<br>For <strong>C mM<\/strong> DTT solution: Concentration&nbsp;in&nbsp;mg\/mL=C\u00d7154.31000=C\u00d70.1543\\text{Concentration in mg\/mL} = C \\times \\frac{154.3}{1000} = C \\times 0.1543<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Problem 2: Buffer Preparation Table<\/h3>\n\n\n\n<h4 class=\"wp-block-heading\">Stock solutions:<\/h4>\n\n\n\n<figure class=\"wp-block-table\"><table class=\"has-fixed-layout\"><thead><tr><th>Compound<\/th><th>Stock Concentration<\/th><\/tr><\/thead><tbody><tr><td>Tris<\/td><td>1.0 M<\/td><\/tr><tr><td>Glycerol<\/td><td>100% (v\/v)<\/td><\/tr><tr><td>EDTA<\/td><td>0.5 M<\/td><\/tr><tr><td>DTT<\/td><td>1.0 M<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n<h4 class=\"wp-block-heading\">Desired final buffer composition:<\/h4>\n\n\n\n<ul class=\"wp-block-list\">\n<li>100 mM Tris<\/li>\n\n\n\n<li>10% glycerol<\/li>\n\n\n\n<li>1 mM EDTA<\/li>\n\n\n\n<li>1 mM DTT<\/li>\n\n\n\n<li><strong>Final Volume<\/strong>: 100 mL<\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h4 class=\"wp-block-heading\">Step-by-step calculations:<\/h4>\n\n\n\n<p><strong>1. Tris (100 mM from 1.0 M stock):<\/strong><br>Use dilution formula C1V1=C2V2C_1V_1 = C_2V_2 (1.0&nbsp;M)\u22c5V1=(0.1&nbsp;M)\u22c5(0.1&nbsp;L)\u21d2V1=0.01&nbsp;L=10&nbsp;mL(1.0 \\text{ M}) \\cdot V_1 = (0.1 \\text{ M}) \\cdot (0.1 \\text{ L}) \\Rightarrow V_1 = 0.01 \\text{ L} = 10 \\text{ mL}<\/p>\n\n\n\n<p><strong>2. Glycerol (10% from 100% stock):<\/strong> 10%&nbsp;of&nbsp;100&nbsp;mL=10&nbsp;mL10\\% \\text{ of } 100 \\text{ mL} = 10 \\text{ mL}<\/p>\n\n\n\n<p><strong>3. EDTA (1 mM from 0.5 M stock):<\/strong> (0.5&nbsp;M)\u22c5V1=(0.001&nbsp;M)\u22c5(0.1&nbsp;L)\u21d2V1=0.0002&nbsp;L=0.2&nbsp;mL(0.5 \\text{ M}) \\cdot V_1 = (0.001 \\text{ M}) \\cdot (0.1 \\text{ L}) \\Rightarrow V_1 = 0.0002 \\text{ L} = 0.2 \\text{ mL}<\/p>\n\n\n\n<p><strong>4. DTT (1 mM from 1.0 M stock):<\/strong> (1.0&nbsp;M)\u22c5V1=(0.001&nbsp;M)\u22c5(0.1&nbsp;L)\u21d2V1=0.0001&nbsp;L=0.1&nbsp;mL(1.0 \\text{ M}) \\cdot V_1 = (0.001 \\text{ M}) \\cdot (0.1 \\text{ L}) \\Rightarrow V_1 = 0.0001 \\text{ L} = 0.1 \\text{ mL}<\/p>\n\n\n\n<p><strong>5. Water to reach 100 mL:<\/strong> Total&nbsp;so&nbsp;far:&nbsp;10+10+0.2+0.1=20.3&nbsp;mL\u21d2Water&nbsp;needed=100\u221220.3=79.7&nbsp;mL\\text{Total so far: } 10 + 10 + 0.2 + 0.1 = 20.3 \\text{ mL} \\Rightarrow \\text{Water needed} = 100 &#8211; 20.3 = 79.7 \\text{ mL}<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">\ud83d\udccb Final Table: Buffer Composition<\/h3>\n\n\n\n<figure class=\"wp-block-table\"><table class=\"has-fixed-layout\"><thead><tr><th>Reagent<\/th><th>Stock Conc.<\/th><th>Final Conc.<\/th><th>Volume to Add<\/th><\/tr><\/thead><tbody><tr><td>Tris<\/td><td>1.0 M<\/td><td>100 mM<\/td><td>10.0 mL<\/td><\/tr><tr><td>Glycerol<\/td><td>100%<\/td><td>10%<\/td><td>10.0 mL<\/td><\/tr><tr><td>EDTA<\/td><td>0.5 M<\/td><td>1 mM<\/td><td>0.2 mL<\/td><\/tr><tr><td>DTT<\/td><td>1.0 M<\/td><td>1 mM<\/td><td>0.1 mL<\/td><\/tr><tr><td><strong>Water<\/strong><\/td><td>\u2014<\/td><td>\u2014<\/td><td>79.7 mL<\/td><\/tr><tr><td><strong>Total<\/strong><\/td><td>\u2014<\/td><td>\u2014<\/td><td><strong>100.0 mL<\/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\">\u270d\ufe0f Explanation<\/h3>\n\n\n\n<p>Buffer preparation in the lab involves mixing known volumes of concentrated stock solutions to achieve desired final concentrations. Each component is diluted based on the equation C1V1=C2V2C_1V_1 = C_2V_2, which maintains molarity during dilution.<\/p>\n\n\n\n<p>For example, to make 100 mL of buffer containing 100 mM Tris from a 1.0 M stock, we solve:<br>(1.0&nbsp;M)\u22c5V=(0.1&nbsp;M)\u22c50.1&nbsp;L(1.0 \\text{ M}) \\cdot V = (0.1 \\text{ M}) \\cdot 0.1 \\text{ L} \u2192 V=10&nbsp;mLV = 10 \\text{ mL}.<br>This logic applies to all reagents. Glycerol, often expressed in percentage (v\/v), simply requires adding the corresponding percentage volume (10% of 100 mL = 10 mL). For EDTA and DTT, both used at low concentrations (1 mM), very small volumes (0.2 mL and 0.1 mL respectively) are needed from concentrated stocks.<\/p>\n\n\n\n<p>After calculating all component volumes, subtract their total from 100 mL to determine the amount of water required to reach the final desired volume. This ensures the buffer&#8217;s integrity and accuracy. This approach not only saves time but also ensures reproducibility across experiments, critical in biochemical and molecular biology workflows.<\/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-banner7-84.jpeg\" alt=\"\" class=\"wp-image-221037\"\/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>Dithiothreitol (DTT) is a compound with molecular weight of 154.3 Daltons, which means that one mole of DTT has a mass of 154.3 grams, and a 1.0 molar solution (abbreviated , with indicating moles per Liter) has of DTT dissolved in water to a total volume of . Let&#8217;s say I give you an aqueous [&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-221036","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/221036","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=221036"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/221036\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=221036"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=221036"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=221036"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}