{"id":184955,"date":"2025-01-21T19:37:52","date_gmt":"2025-01-21T19:37:52","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=184955"},"modified":"2025-01-21T19:37:57","modified_gmt":"2025-01-21T19:37:57","slug":"applying-hardy-weinberg-to-rock-pocket-mouse-field-data-dr-nachman-and-his-colleagues-collected-rock-pocket-mice-across-35-kilometers-of-the-arizona-sonoran-desert-which-included-both-dark","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/01\/21\/applying-hardy-weinberg-to-rock-pocket-mouse-field-data-dr-nachman-and-his-colleagues-collected-rock-pocket-mice-across-35-kilometers-of-the-arizona-sonoran-desert-which-included-both-dark\/","title":{"rendered":"APPLYING HARDY-WEINBERG TO ROCK POCKET MOUSE FIELD DATA Dr. Nachman and his colleagues collected rock pocket mice across 35 kilometers of the Arizona Sonoran Desert which included both dark"},"content":{"rendered":"\n

PART 2: APPLYING HARDY-WEINBERG TO ROCK POCKET MOUSE FIELD DATA Dr. Nachman and his colleagues collected rock pocket mice across 35 kilometers of the Arizona Sonoran Desert which included both dark, rocky lava outcrops and light, rocky, granite areas. They recorded substrate color and coat-color frequencies for each location. Each site was separated from any of the others by at least eight kilometers. The researchers trapped a total of 225 mice. Their data are summarized below. Field Data Summary Collecting Site Substrate Color Number of Mice Phenotype Light Dark Light 6 2 Light 85 80 3 Dark 7 4 Dark 5 Dark 45 3 42 6 Light 77 34 43 Source of data: Hoekstra, Hopi E., Kristen E. Drumm, and Michael W. Nachman. “Ecological Genetics of Adaptive Color Polymorphism in Pocket Mice: Geographic Variation in Selected and Neutral Genes.” Evolution 58, no. 6 (2004): 1329-1344. QUESTIONS 1. Calculate the overall frequencies of light-colored mice and dark-colored mice caught on light-colored substrates. frequency = number of mice of one color\/total number of mice Frequency of light-colored mice Frequency of dark-colored mice K 2. Calculate the overall frequencies of light-co\u00c3\u2026\u00e2\u20ac\u0161\u00c3\u0192\u00c2\u00b3red mice and dark-colored mice caught on dark-colored substrates. frequency = number of mice of one color\/total number of mice 3. Frequency of light-colored mice Frequency of dark-colored mice 7 3. Using the Hardy-Weinberg equation and data from the table above, determine the number of mice with the DD and Dd genotypes on the light, rocky, granite substrate. Frequency of mice with the dd genotype on light-colored substrate \u00c3\u00a2\u00e2\u201a\u00ac\u00c5\u00be71 q2 Frequency of mice with the DD genotype on light-colored substrate Frequency of mice with the Dd genotype on light-colored substrate _03 p2 26 “2pq” #3<\/p>\n\n\n\n

The correct answer and explanation is:<\/mark><\/strong><\/p>\n\n\n\n

To answer the questions using the Hardy-Weinberg principle, we’ll break the task into three parts as outlined in your prompt.<\/p>\n\n\n\n

\n\n\n\n

1. Frequency of light-colored and dark-colored mice on light-colored substrates<\/strong><\/h3>\n\n\n\n

The frequency of a phenotype is calculated as: Frequency=Number of mice of one phenotypeTotal number of mice on the substrate\\text{Frequency} = \\frac{\\text{Number of mice of one phenotype}}{\\text{Total number of mice on the substrate}}<\/p>\n\n\n\n

From the dataset, for the light-colored substrate:<\/p>\n\n\n\n

    \n
  • Light-colored mice<\/strong>: 85+80=16585 + 80 = 165<\/li>\n\n\n\n
  • Dark-colored mice<\/strong>: 3+7=103 + 7 = 10<\/li>\n\n\n\n
  • Total mice<\/strong>: 165+10=175165 + 10 = 175<\/li>\n<\/ul>\n\n\n\n

    Frequency of light-colored mice=165175=0.943\\text{Frequency of light-colored mice} = \\frac{165}{175} = 0.943 Frequency of dark-colored mice=10175=0.057\\text{Frequency of dark-colored mice} = \\frac{10}{175} = 0.057<\/p>\n\n\n\n

    \n\n\n\n

    2. Frequency of light-colored and dark-colored mice on dark-colored substrates<\/strong><\/h3>\n\n\n\n

    From the dataset, for the dark-colored substrate:<\/p>\n\n\n\n

      \n
    • Light-colored mice<\/strong>: 6+2=86 + 2 = 8<\/li>\n\n\n\n
    • Dark-colored mice<\/strong>: 45+42=8745 + 42 = 87<\/li>\n\n\n\n
    • Total mice<\/strong>: 8+87=958 + 87 = 95<\/li>\n<\/ul>\n\n\n\n

      Frequency of light-colored mice=895=0.084\\text{Frequency of light-colored mice} = \\frac{8}{95} = 0.084 Frequency of dark-colored mice=8795=0.916\\text{Frequency of dark-colored mice} = \\frac{87}{95} = 0.916<\/p>\n\n\n\n

      \n\n\n\n

      3. Hardy-Weinberg calculations for light-colored substrate<\/strong><\/h3>\n\n\n\n

      Assuming Hardy-Weinberg equilibrium, the genotypes are distributed as: p2+2pq+q2=1p^2 + 2pq + q^2 = 1<\/p>\n\n\n\n

      Where:<\/p>\n\n\n\n

        \n
      • p2p^2: Frequency of the DD<\/strong> genotype<\/li>\n\n\n\n
      • 2pq2pq: Frequency of the Dd<\/strong> genotype<\/li>\n\n\n\n
      • q2q^2: Frequency of the dd<\/strong> genotype<\/li>\n<\/ul>\n\n\n\n

        Step 1: Determine q2q^2<\/h4>\n\n\n\n

        The frequency of light-colored mice (q2q^2) on the light-colored substrate is 0.943: q2=0.943q^2 = 0.943<\/p>\n\n\n\n

        Step 2: Solve for qq<\/h4>\n\n\n\n

        q=q2=0.943\u22480.971q = \\sqrt{q^2} = \\sqrt{0.943} \\approx 0.971<\/p>\n\n\n\n

        Step 3: Solve for pp<\/h4>\n\n\n\n

        p=1\u2212q=1\u22120.971=0.029p = 1 – q = 1 – 0.971 = 0.029<\/p>\n\n\n\n

        Step 4: Calculate genotype frequencies<\/h4>\n\n\n\n

        Using Hardy-Weinberg equations:<\/p>\n\n\n\n

          \n
        • Frequency of DD (p2p^2)<\/strong>:<\/li>\n<\/ul>\n\n\n\n

          p2=(0.029)2\u22480.000841p^2 = (0.029)^2 \\approx 0.000841<\/p>\n\n\n\n

            \n
          • Frequency of Dd (2pq2pq)<\/strong>:<\/li>\n<\/ul>\n\n\n\n

            2pq=2(0.029)(0.971)\u22480.0562pq = 2(0.029)(0.971) \\approx 0.056<\/p>\n\n\n\n

            Step 5: Calculate number of mice for each genotype<\/h4>\n\n\n\n

            On the light-colored substrate (total 175175 mice):<\/p>\n\n\n\n

              \n
            • DD genotype<\/strong>:<\/li>\n<\/ul>\n\n\n\n

              p2\u00d7175=0.000841\u00d7175\u22480.15 mice\u22480p^2 \\times 175 = 0.000841 \\times 175 \\approx 0.15 \\text{ mice} \\approx 0<\/p>\n\n\n\n

                \n
              • Dd genotype<\/strong>:<\/li>\n<\/ul>\n\n\n\n

                2pq\u00d7175=0.056\u00d7175\u22489.8 mice2pq \\times 175 = 0.056 \\times 175 \\approx 9.8 \\text{ mice}<\/p>\n\n\n\n

                  \n
                • dd genotype<\/strong>:<\/li>\n<\/ul>\n\n\n\n

                  q2\u00d7175=0.943\u00d7175\u2248165 miceq^2 \\times 175 = 0.943 \\times 175 \\approx 165 \\text{ mice}<\/p>\n\n\n\n

                  \n\n\n\n

                  Explanation <\/strong><\/h3>\n\n\n\n

                  The Hardy-Weinberg principle provides a mathematical framework for understanding allele and genotype frequencies in a population under equilibrium. In this exercise, the distribution of light and dark coat colors in rock pocket mice on different substrates was analyzed. The principle assumes no evolution and random mating, allowing us to calculate expected genotype frequencies based on observed phenotypes.<\/p>\n\n\n\n

                  On the light-colored substrates, the high frequency of light-colored mice (94.3%94.3\\%) reflects strong selection against dark-colored mice. Using this phenotype frequency (q2=0.943q^2 = 0.943), the recessive allele frequency (q=0.971q = 0.971) and dominant allele frequency (p=0.029p = 0.029) were derived. Applying these to Hardy-Weinberg equations (p2p^2 for DD, 2pq2pq for Dd, q2q^2 for dd), the number of DD and Dd mice was determined to be 0 and approximately 10, respectively, among 175 mice.<\/p>\n\n\n\n

                  This data highlights the adaptation of rock pocket mice to their environment. The dominance of light-colored mice on light substrates suggests a selective advantage, as lighter coats provide camouflage against predators. Conversely, the near absence of the DD genotype and low frequency of Dd mice emphasize the rarity of dark alleles in this setting. This case study illustrates how environmental pressures shape genetic variation and population structure.<\/p>\n","protected":false},"excerpt":{"rendered":"

                  PART 2: APPLYING HARDY-WEINBERG TO ROCK POCKET MOUSE FIELD DATA Dr. Nachman and his colleagues collected rock pocket mice across 35 kilometers of the Arizona Sonoran Desert which included both dark, rocky lava outcrops and light, rocky, granite areas. They recorded substrate color and coat-color frequencies for each location. Each site was separated from any […]<\/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-184955","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/184955","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=184955"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/184955\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=184955"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=184955"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=184955"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}