{"id":209024,"date":"2025-04-28T09:36:35","date_gmt":"2025-04-28T09:36:35","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=209024"},"modified":"2025-04-28T09:36:36","modified_gmt":"2025-04-28T09:36:36","slug":"what-is-the-probability-that-their-offspring-will-have-mid-digital-hair","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/04\/28\/what-is-the-probability-that-their-offspring-will-have-mid-digital-hair\/","title":{"rendered":"What is the probability that their offspring will have mid-digital hair"},"content":{"rendered":"\n<p>What is the probability that their offspring will have mid-digital hair? What is the probability that two offspring in a row will have no freckles? What is the probability of two offspring in a row having mid-digital hair and freckles?<br>Mid-Digital Hair and Freckles<br>Finger mid-digital hair (M) is dominant over the recessive trait of no mid-digital hair (m). Freckles (F) is dominant over the recessive trait of no freckles (f). (It is important to note that this example of mid-digital hair and freckles cannot be explained with simple genetics, but for the purpose of this question, assume that it can.)<\/p>\n\n\n\n<p>A woman that is heterozygous for freckles and has no mid-digital hair is married to a man with no freckles that is heterozygous for mid-digital hair. The couple is expecting a child.<\/p>\n\n\n\n<p>Draw a Punnett square of this cross then answer the following questions:<\/p>\n\n\n\n<p>What is the probability that their offspring will have mid-digital hair?<br>What is the probability that two offspring in a row will have no freckles?<br>What is the probability of two offspring in a row having mid-digital hair and freckles?<\/p>\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>To solve this problem, we need to consider the genetics of both traits, mid-digital hair and freckles. Let&#8217;s break down the information provided:<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Key Information:<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Mid-Digital Hair (M)<\/strong>: Dominant trait for the presence of mid-digital hair, with the recessive trait for no mid-digital hair being (m).<\/li>\n\n\n\n<li><strong>Freckles (F)<\/strong>: Dominant trait for the presence of freckles, with the recessive trait for no freckles being (f).<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\">Parental Genotypes:<\/h3>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Mother<\/strong>: Heterozygous for freckles (Ff) and has no mid-digital hair (mm).<\/li>\n\n\n\n<li><strong>Father<\/strong>: No freckles (ff) and heterozygous for mid-digital hair (Mm).<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\">Punnett Square Setup:<\/h3>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Mother&#8217;s genotype<\/strong>: Ff (freckles), mm (no mid-digital hair).<\/li>\n\n\n\n<li><strong>Father&#8217;s genotype<\/strong>: ff (no freckles), Mm (mid-digital hair).<\/li>\n<\/ol>\n\n\n\n<h3 class=\"wp-block-heading\">Punnett Square for Mid-Digital Hair:<\/h3>\n\n\n\n<p>To determine the probability of the offspring inheriting mid-digital hair, we focus on the possible combinations for the mid-digital hair allele:<\/p>\n\n\n\n<figure class=\"wp-block-table\"><table class=\"has-fixed-layout\"><thead><tr><th><\/th><th>M<\/th><th>m<\/th><\/tr><\/thead><tbody><tr><td>m<\/td><td>Mm<\/td><td>mm<\/td><\/tr><tr><td>m<\/td><td>Mm<\/td><td>mm<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n<p>The offspring could either inherit:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Mm<\/strong> (mid-digital hair) or<\/li>\n\n\n\n<li><strong>mm<\/strong> (no mid-digital hair).<\/li>\n<\/ul>\n\n\n\n<p>Therefore, <strong>50%<\/strong> of the offspring will have mid-digital hair (Mm), and the other <strong>50%<\/strong> will not (mm).<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Punnett Square for Freckles:<\/h3>\n\n\n\n<p>Now, let&#8217;s calculate the probability of the offspring inheriting freckles, focusing on the alleles for freckles:<\/p>\n\n\n\n<figure class=\"wp-block-table\"><table class=\"has-fixed-layout\"><thead><tr><th><\/th><th>F<\/th><th>f<\/th><\/tr><\/thead><tbody><tr><td>f<\/td><td>Ff<\/td><td>ff<\/td><\/tr><tr><td>f<\/td><td>Ff<\/td><td>ff<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n<p>The offspring could either inherit:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>Ff<\/strong> (freckles) or<\/li>\n\n\n\n<li><strong>ff<\/strong> (no freckles).<\/li>\n<\/ul>\n\n\n\n<p>Therefore, <strong>50%<\/strong> of the offspring will have freckles (Ff), and the other <strong>50%<\/strong> will not (ff).<\/p>\n\n\n\n<h3 class=\"wp-block-heading\">Probability of Two Offspring in a Row Having No Freckles:<\/h3>\n\n\n\n<p>To find the probability of two offspring in a row having no freckles, we multiply the probability of a single offspring having no freckles by itself:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Probability of no freckles (ff) = <strong>50%<\/strong> or 0.5.<\/li>\n\n\n\n<li>Probability of two offspring in a row having no freckles = 0.5 \u00d7 0.5 = <strong>0.25<\/strong> or <strong>25%<\/strong>.<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\">Probability of Two Offspring in a Row Having Mid-Digital Hair and Freckles:<\/h3>\n\n\n\n<p>The probability of an offspring having both mid-digital hair and freckles is the product of the two individual probabilities:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Probability of mid-digital hair (Mm) = <strong>50%<\/strong> or 0.5.<\/li>\n\n\n\n<li>Probability of freckles (Ff) = <strong>50%<\/strong> or 0.5.<\/li>\n\n\n\n<li>Probability of both traits = 0.5 \u00d7 0.5 = <strong>0.25<\/strong> or <strong>25%<\/strong>.<\/li>\n<\/ul>\n\n\n\n<p>Now, to find the probability of two offspring having both mid-digital hair and freckles, we multiply the probability of one offspring having both traits by itself:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Probability of two offspring having both traits = 0.25 \u00d7 0.25 = <strong>0.0625<\/strong> or <strong>6.25%<\/strong>.<\/li>\n<\/ul>\n\n\n\n<h3 class=\"wp-block-heading\">Final Answers:<\/h3>\n\n\n\n<ol class=\"wp-block-list\">\n<li><strong>Probability of offspring having mid-digital hair<\/strong>: <strong>50%<\/strong>.<\/li>\n\n\n\n<li><strong>Probability of two offspring in a row having no freckles<\/strong>: <strong>25%<\/strong>.<\/li>\n\n\n\n<li><strong>Probability of two offspring in a row having mid-digital hair and freckles<\/strong>: <strong>6.25%<\/strong>.<\/li>\n<\/ol>\n\n\n\n<h3 class=\"wp-block-heading\">Explanation:<\/h3>\n\n\n\n<p>The probability calculations are based on Mendelian genetics, where dominant traits are expressed when at least one dominant allele is present. In this case, the presence of mid-digital hair and freckles is determined by dominant alleles (M and F). The use of Punnett squares helps to visualize all possible combinations of alleles and determine the likelihood of various phenotypes in the offspring. The product rule is applied to calculate the probability of multiple events happening in sequence (e.g., two offspring inheriting specific traits).<\/p>\n","protected":false},"excerpt":{"rendered":"<p>What is the probability that their offspring will have mid-digital hair? What is the probability that two offspring in a row will have no freckles? What is the probability of two offspring in a row having mid-digital hair and freckles?Mid-Digital Hair and FrecklesFinger mid-digital hair (M) is dominant over the recessive trait of no mid-digital [&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-209024","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/209024","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=209024"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/209024\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=209024"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=209024"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=209024"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}