{"id":183625,"date":"2025-01-16T17:59:12","date_gmt":"2025-01-16T17:59:12","guid":{"rendered":"https:\/\/learnexams.com\/blog\/?p=183625"},"modified":"2025-01-16T17:59:15","modified_gmt":"2025-01-16T17:59:15","slug":"in-garden-peas-yellow-y-seed-color-is-dominant-to-green-y","status":"publish","type":"post","link":"https:\/\/www.learnexams.com\/blog\/2025\/01\/16\/in-garden-peas-yellow-y-seed-color-is-dominant-to-green-y\/","title":{"rendered":"In garden peas, yellow (Y) seed color is dominant to green (y)"},"content":{"rendered":"\n<p>1- In garden peas, yellow (Y) seed color is dominant to green (y). Because yellow and green are two different forms of the same gene (seed color), they are alleles. Use a Punnett square to show a cross between two heterozygous pea plants.\u00c3\u201a\u00c2 What are the phenotype(s) of the offspring? What are the genotype(s) of the offspring?<\/p>\n\n\n\n<p>2- In garden peas, yellow (Y) seed color is dominant to green (y). Because yellow and green are two different forms of the same gene (seed color), they are alleles. Use a Punnett square to show a cross between two heterozygous pea plants.\u00c3\u201a\u00c2 What are the phenotype(s) of the offspring? What are the genotype(s) of the offspring?<\/p>\n\n\n\n<p>Use the information below to answer questions 3-6<\/p>\n\n\n\n<p>\u00c3\u201a\u00c2<\/p>\n\n\n\n<p>There may be a number of possible alleles for a given gene within a population. In a multiple allele system the dominance relationships between the various alleles must be considered. One of the more familiar examples of a multiple allelic system is that of the human ABO blood group. The gene involved codes for a protein located on the outside of red blood cell membranes. Three alleles (IA, IB and i) determine whether the protein is present or absent and which form of the protein (if any) is present. The A and B alleles code for the A and B forms of the protein and are co-dominant with each other. The O allele (i) codes for no protein and is recessive to both A and B alleles. This means there are four possible phenotypes (blood types: A, B, AB, and O). This also means there are 6 possible genotypes: IAIA, IBIB, IAi, IBi, IAIB and ii.\u00c3\u201a\u00c2 If you have the letter \u00c3\u00a2\u00e2\u201a\u00ac\u00cb\u0153O\u00c3\u00a2\u00e2\u201a\u00ac\u00e2\u201e\u00a2 anywhere in any of your genotypes, you are doing it wrong.<\/p>\n\n\n\n<p>3-Show the possible genotypes and phenotypes of the offspring from a cross between a homozygous male with type A blood and a homozygous female with type B blood.<\/p>\n\n\n\n<p>\u00c3\u201a\u00c2<\/p>\n\n\n\n<p>4- Show the possible genotypes and phenotypes of the offspring from a cross between a heterozygous male with type A blood and a heterozygous female with type B blood.<\/p>\n\n\n\n<p>5- If a man with blood type A, one of whose parents had blood type O, marries a woman with blood type AB, what are the putative genotypes of the offspring?<\/p>\n\n\n\n<p>6- A couple with the following blood types: the man has type AB and the woman has type B, discover their child has type O blood.\u00c3\u201a\u00c2 \u00c3\u201a\u00c2 \u00c3\u201a\u00c2 Is it possible that one of these \u00c3\u00a2\u00e2\u201a\u00ac\u00cb\u0153parents\u00c3\u00a2\u00e2\u201a\u00ac\u00e2\u201e\u00a2 \u00c3\u201a\u00c2 may not actually be the genetic parent of this child?\u00c3\u201a\u00c2 If so, which one, and how do you know?<\/p>\n\n\n\n<p>7- In horses coat color shows incomplete dominance: the alleles are chestnut color (Hc) and cremello (Hcr); heterozygous individuals have the phenotype palamino. What are the predicted phenotypes from a cross between a palamino and cremello colored horses? What are the genotypes?<\/p>\n\n\n\n<p>8- Red-green color blindness is inherited as an X-linked recessive (Xc).\u00c3\u201a\u00c2 If a color-blind man marries a woman who is heterozygous for normal vision, what would be the expected phenotypes of their children with reference to this character?\u00c3\u201a\u00c2 In your answer, specify in your phenotype descriptions the gender of the children.\u00c3\u201a\u00c2 (For example, don\u00c3\u00a2\u00e2\u201a\u00ac\u00e2\u201e\u00a2t just say 75% of the children would be colorblind \u00c3\u00a2\u00e2\u201a\u00ac\u00e2\u20ac\u0153 you would instead say 100 % of the daughters would be colorblind and 50% of the sons would be colorblind.\u00c3\u201a\u00c2 Note that this is not a correct answer; it is just to give you an idea of how to explain the correct phenotypes of the cross.)<em>_<\/em><\/p>\n\n\n\n<p>9.\u00c3\u201a\u00c2 \u00c3\u201a\u00c2 Hemophilia is another example of a X-linked disease caused when a recessive allele (Xh) is expressed. If a normal male reproduces with a heterozygous normal female, what are the expected genotypes and phenotypes? Will any of their daughters develop hemophilia?\u00c3\u201a\u00c2 As in the previous question, you must also give the gender of the child in your genotype and phenotype descriptions here<\/p>\n\n\n\n<ol start=\"10\" class=\"wp-block-list\">\n<li>In pea plants, seed shape and seed color are controlled by genes located on different chromosomes. Seeds may be round (R) or wrinkled (r), with the allele for round seeds being dominant. Alleles for seed color are yellow and green, with the green allele (y) recessive to the yellow (Y) allele. If you cross an individual that is homozygous round and yellow with an individual that is homozygous for wrinkled and green, what is the genotype of the F1 individuals? Set up a Punnett square for the dihybrid cross.<\/li>\n<\/ol>\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<h3 class=\"wp-block-heading\">Answer 1:<\/h3>\n\n\n\n<p>In garden peas, the gene for seed color has two alleles: yellow (Y) and green (y), where yellow is dominant over green. The cross between two heterozygous plants (Yy x Yy) can be represented using a Punnett square:<\/p>\n\n\n\n<p><strong>Punnett Square:<\/strong><\/p>\n\n\n\n<figure class=\"wp-block-table\"><table class=\"has-fixed-layout\"><thead><tr><th><\/th><th>Y<\/th><th>y<\/th><\/tr><\/thead><tbody><tr><td><strong>Y<\/strong><\/td><td>YY<\/td><td>Yy<\/td><\/tr><tr><td><strong>y<\/strong><\/td><td>Yy<\/td><td>yy<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n<p><strong>Possible Genotypes:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>YY<\/strong> (homozygous dominant)<\/li>\n\n\n\n<li><strong>Yy<\/strong> (heterozygous)<\/li>\n\n\n\n<li><strong>yy<\/strong> (homozygous recessive)<\/li>\n<\/ul>\n\n\n\n<p><strong>Possible Phenotypes:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Yellow (from YY or Yy genotypes)<\/li>\n\n\n\n<li>Green (from yy genotype)<\/li>\n<\/ul>\n\n\n\n<p>The phenotype ratio is 3 yellow: 1 green.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Answer 2:<\/h3>\n\n\n\n<p>This question appears to be identical to the first one. Please refer to the explanation above, which answers the same question regarding the cross between two heterozygous pea plants (Yy x Yy). The results are the same, with a phenotype ratio of 3 yellow: 1 green and the genotype ratio being 1 YY: 2 Yy: 1 yy.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Answer 3:<\/h3>\n\n\n\n<p>In this cross, the male has homozygous type A blood (IAIA), and the female has homozygous type B blood (IBIB).<\/p>\n\n\n\n<p><strong>Punnett Square:<\/strong><\/p>\n\n\n\n<figure class=\"wp-block-table\"><table class=\"has-fixed-layout\"><thead><tr><th><\/th><th>IA<\/th><th>IA<\/th><\/tr><\/thead><tbody><tr><td><strong>IB<\/strong><\/td><td>IAIB<\/td><td>IAIB<\/td><\/tr><tr><td><strong>IB<\/strong><\/td><td>IAIB<\/td><td>IAIB<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n<p><strong>Possible Genotypes:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>IAIB<\/strong> (heterozygous, type AB blood)<\/li>\n<\/ul>\n\n\n\n<p><strong>Possible Phenotypes:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>All offspring will have type AB blood (since both IA and IB are dominant and co-dominant).<\/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\">Answer 4:<\/h3>\n\n\n\n<p>In this cross, the male is heterozygous for type A blood (IAi), and the female is heterozygous for type B blood (IBi).<\/p>\n\n\n\n<p><strong>Punnett Square:<\/strong><\/p>\n\n\n\n<figure class=\"wp-block-table\"><table class=\"has-fixed-layout\"><thead><tr><th><\/th><th>IA<\/th><th>i<\/th><\/tr><\/thead><tbody><tr><td><strong>IB<\/strong><\/td><td>IAIB<\/td><td>IBi<\/td><\/tr><tr><td><strong>i<\/strong><\/td><td>IAi<\/td><td>ii<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n<p><strong>Possible Genotypes:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>IAIB<\/strong> (type AB blood)<\/li>\n\n\n\n<li><strong>IBi<\/strong> (type B blood)<\/li>\n\n\n\n<li><strong>IAi<\/strong> (type A blood)<\/li>\n\n\n\n<li><strong>ii<\/strong> (type O blood)<\/li>\n<\/ul>\n\n\n\n<p><strong>Possible Phenotypes:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Type A (IAi)<\/li>\n\n\n\n<li>Type B (IBi)<\/li>\n\n\n\n<li>Type AB (IAIB)<\/li>\n\n\n\n<li>Type O (ii)<\/li>\n<\/ul>\n\n\n\n<p>The expected ratio is 1 type AB: 1 type A: 1 type B: 1 type O.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Answer 5:<\/h3>\n\n\n\n<p>If a man with type A blood (whose parent had type O blood) marries a woman with type AB blood, the man must have the genotype <strong>IAi<\/strong>, because his parent must have passed on the O allele (i). The woman has genotype <strong>IAIB<\/strong>.<\/p>\n\n\n\n<p><strong>Punnett Square:<\/strong><\/p>\n\n\n\n<figure class=\"wp-block-table\"><table class=\"has-fixed-layout\"><thead><tr><th><\/th><th>IA<\/th><th>i<\/th><\/tr><\/thead><tbody><tr><td><strong>IA<\/strong><\/td><td>IAIA<\/td><td>IAi<\/td><\/tr><tr><td><strong>IB<\/strong><\/td><td>IAIB<\/td><td>IBi<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n<p><strong>Possible Genotypes:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>IAIA<\/strong> (type A blood)<\/li>\n\n\n\n<li><strong>IAi<\/strong> (type A blood)<\/li>\n\n\n\n<li><strong>IAIB<\/strong> (type AB blood)<\/li>\n\n\n\n<li><strong>IBi<\/strong> (type B blood)<\/li>\n<\/ul>\n\n\n\n<p><strong>Possible Phenotypes:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Type A (IAIA or IAi)<\/li>\n\n\n\n<li>Type AB (IAIB)<\/li>\n\n\n\n<li>Type B (IBi)<\/li>\n<\/ul>\n\n\n\n<p>The expected ratio is 1 type A: 1 type AB: 1 type B.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Answer 6:<\/h3>\n\n\n\n<p>If a man with blood type AB (IAIB) and a woman with blood type B (IBi) have a child with type O blood (ii), it is not genetically possible for the man to be the father of the child. To have type O blood (ii), the child must inherit an i allele from each parent. Since the man with blood type AB (IAIB) cannot pass on an i allele, he cannot be the biological father.<\/p>\n\n\n\n<p>Therefore, it is possible that the man is not the genetic parent of the child.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Answer 7:<\/h3>\n\n\n\n<p>In horses, coat color is controlled by incomplete dominance, with chestnut (Hc) being dominant to cremello (Hcr). Heterozygous individuals are palamino (HcHcr). A cross between a palamino (HcHcr) and a cremello (HcrHcr) horse would produce the following genotypes:<\/p>\n\n\n\n<p><strong>Punnett Square:<\/strong><\/p>\n\n\n\n<figure class=\"wp-block-table\"><table class=\"has-fixed-layout\"><thead><tr><th><\/th><th>Hc<\/th><th>Hcr<\/th><\/tr><\/thead><tbody><tr><td><strong>Hc<\/strong><\/td><td>HcHc<\/td><td>HcHcr<\/td><\/tr><tr><td><strong>Hcr<\/strong><\/td><td>HcHcr<\/td><td>HcrHcr<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n<p><strong>Possible Genotypes:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>HcHc<\/strong> (chestnut)<\/li>\n\n\n\n<li><strong>HcHcr<\/strong> (palamino)<\/li>\n\n\n\n<li><strong>HcrHcr<\/strong> (cremello)<\/li>\n<\/ul>\n\n\n\n<p><strong>Possible Phenotypes:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>Chestnut (HcHc)<\/li>\n\n\n\n<li>Palamino (HcHcr)<\/li>\n\n\n\n<li>Cremello (HcrHcr)<\/li>\n<\/ul>\n\n\n\n<p>The expected phenotype ratio is 1 chestnut: 2 palamino: 1 cremello.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Answer 8:<\/h3>\n\n\n\n<p>Color blindness is inherited as an X-linked recessive trait (Xc). In this case, a color-blind man (XcY) marries a woman who is heterozygous for normal vision (XcX). The possible offspring are as follows:<\/p>\n\n\n\n<p><strong>Punnett Square:<\/strong><\/p>\n\n\n\n<figure class=\"wp-block-table\"><table class=\"has-fixed-layout\"><thead><tr><th><\/th><th>Xc<\/th><th>Y<\/th><\/tr><\/thead><tbody><tr><td><strong>Xc<\/strong><\/td><td>XcXc<\/td><td>XcY<\/td><\/tr><tr><td><strong>X<\/strong><\/td><td>XcX<\/td><td>XY<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n<p><strong>Possible Genotypes:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>XcXc<\/strong> (colorblind daughter)<\/li>\n\n\n\n<li><strong>XcY<\/strong> (colorblind son)<\/li>\n\n\n\n<li><strong>XcX<\/strong> (normal vision daughter)<\/li>\n\n\n\n<li><strong>XY<\/strong> (normal vision son)<\/li>\n<\/ul>\n\n\n\n<p><strong>Possible Phenotypes:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>50% of the sons will be colorblind (XcY).<\/li>\n\n\n\n<li>50% of the sons will have normal vision (XY).<\/li>\n\n\n\n<li>50% of the daughters will be colorblind (XcXc).<\/li>\n\n\n\n<li>50% of the daughters will have normal vision (XcX).<\/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\">Answer 9:<\/h3>\n\n\n\n<p>Hemophilia is an X-linked recessive trait (Xh). In this case, a normal male (XY) marries a heterozygous normal female (XhX). The possible offspring are as follows:<\/p>\n\n\n\n<p><strong>Punnett Square:<\/strong><\/p>\n\n\n\n<figure class=\"wp-block-table\"><table class=\"has-fixed-layout\"><thead><tr><th><\/th><th>Xh<\/th><th>X<\/th><\/tr><\/thead><tbody><tr><td><strong>X<\/strong><\/td><td>XhX<\/td><td>XX<\/td><\/tr><tr><td><strong>Y<\/strong><\/td><td>XhY<\/td><td>XY<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n<p><strong>Possible Genotypes:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>XhX<\/strong> (normal female carrier)<\/li>\n\n\n\n<li><strong>XhY<\/strong> (hemophiliac male)<\/li>\n\n\n\n<li><strong>XX<\/strong> (normal female)<\/li>\n\n\n\n<li><strong>XY<\/strong> (normal male)<\/li>\n<\/ul>\n\n\n\n<p><strong>Possible Phenotypes:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>50% of the daughters will be carriers (XhX).<\/li>\n\n\n\n<li>50% of the sons will have hemophilia (XhY).<\/li>\n\n\n\n<li>50% of the daughters will have normal blood (XX).<\/li>\n\n\n\n<li>50% of the sons will have normal blood (XY).<\/li>\n<\/ul>\n\n\n\n<p>None of the daughters will develop hemophilia, as they need two copies of the Xh allele to express the trait.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Answer 10:<\/h3>\n\n\n\n<p>In this cross, seed shape (R = round, r = wrinkled) and seed color (Y = yellow, y = green) are controlled by genes on different chromosomes. The parents are homozygous round and yellow (RRYY) and homozygous wrinkled and green (rryy). The F1 generation will inherit one allele for each gene from each parent, resulting in the following genotype:<\/p>\n\n\n\n<p><strong>Punnett Square:<\/strong><\/p>\n\n\n\n<figure class=\"wp-block-table\"><table class=\"has-fixed-layout\"><thead><tr><th><\/th><th>RY<\/th><th>RY<\/th><\/tr><\/thead><tbody><tr><td><strong>ry<\/strong><\/td><td>RrYy<\/td><td>RrYy<\/td><\/tr><tr><td><strong>ry<\/strong><\/td><td>RrYy<\/td><td>RrYy<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n<p><strong>Genotype of F1:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li><strong>RrYy<\/strong> (all F1 individuals will be heterozygous for both traits).<\/li>\n<\/ul>\n\n\n\n<p><strong>Phenotype of F1:<\/strong><\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>All F1 offspring will have round yellow seeds because both the round shape and yellow color alleles are dominant.<\/li>\n<\/ul>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<p>This concludes the solutions to the questions.<\/p>\n\n\n\n<p><\/p>\n","protected":false},"excerpt":{"rendered":"<p>1- In garden peas, yellow (Y) seed color is dominant to green (y). Because yellow and green are two different forms of the same gene (seed color), they are alleles. Use a Punnett square to show a cross between two heterozygous pea plants.\u00c3\u201a\u00c2 What are the phenotype(s) of the offspring? What are the genotype(s) of [&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-183625","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/183625","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=183625"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/183625\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=183625"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=183625"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=183625"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}