{"id":235,"date":"2016-05-02T13:46:18","date_gmt":"2016-05-02T17:46:18","guid":{"rendered":"https:\/\/pressbooks.bccampus.ca\/conceptsofbiologygunness\/?post_type=chapter&#038;p=235"},"modified":"2016-05-02T13:46:18","modified_gmt":"2016-05-02T17:46:18","slug":"8-2-laws-of-inheritance","status":"publish","type":"chapter","link":"https:\/\/pressbooks.bccampus.ca\/conceptsofbiologygunness\/chapter\/8-2-laws-of-inheritance\/","title":{"raw":"8.2 Laws of Inheritance","rendered":"8.2 Laws of Inheritance"},"content":{"raw":"<div>\n<div class=\"bcc-box bcc-highlight\">\n<h3>Learning Objectives<\/h3>\nBy the end of this section, you will be able to:\n<ul><li>Explain the relationship between genotypes and phenotypes in dominant and recessive gene systems<\/li>\n\t<li>Use a Punnett square to calculate the expected proportions of genotypes and phenotypes in a monohybrid cross<\/li>\n\t<li>Explain Mendel\u2019s law of segregation and independent assortment in terms of genetics and the events of meiosis<\/li>\n\t<li>Explain the purpose and methods of a test cross<\/li>\n<\/ul><\/div>\n<\/div>\n<p id=\"fs-idp8170416\">The seven characteristics that Mendel evaluated in his pea plants were each expressed as one of two versions, or traits. Mendel deduced from his results that each individual had two discrete copies of the characteristic that are passed individually to offspring. We now call those two copies <strong>genes<\/strong>, which are carried on chromosomes. The reason we have two copies of each gene is that we inherit one from each parent. In fact, it is the chromosomes we inherit and the two copies of each gene are located on paired chromosomes. Recall that in meiosis these chromosomes are separated out into haploid gametes. This <strong>separation, or segregation<\/strong>, of the homologous chromosomes means also that only one of the copies of the gene gets moved into a gamete. The offspring are formed when that gamete unites with one from another parent and the two copies of each gene (and chromosome) are restored.<\/p>\n<p id=\"fs-idp8399888\">For cases in which a single gene controls a single characteristic, a diploid organism has two genetic copies that may or may not encode the same version of that characteristic. For example, one individual may carry a gene that determines white flower color and a gene that determines violet flower color. Gene variants that arise by mutation and exist at the same relative locations on homologous chromosomes are called<strong> alleles<\/strong>. Mendel examined the inheritance of genes with just two allele forms, but it is common to encounter more than two alleles for any given gene in a natural population.<\/p>\n\n<section id=\"fs-idp91312\"><h1>Phenotypes and Genotypes<\/h1>\n<p id=\"fs-idp353872\">Two alleles for a given gene in a diploid organism are expressed and interact to produce physical characteristics. The observable traits expressed by an organism are referred to as its <strong>phenotype<\/strong>. An organism\u2019s underlying genetic makeup, consisting of both the physically visible and the non-expressed alleles, is called its <strong>genotype<\/strong>. Mendel\u2019s hybridization experiments demonstrate the difference between phenotype and genotype. For example, the phenotypes that Mendel observed in his crosses between pea plants with differing traits are connected to the diploid genotypes of the plants in the P, F<sub>1<\/sub>, and F<sub>2<\/sub> generations. We will use a second trait that Mendel investigated, seed color, as an example. Seed color is governed by a single gene with two alleles. The yellow-seed allele is dominant and the green-seed allele is recessive. When true-breeding plants were cross-fertilized, in which one parent had yellow seeds and one had green seeds, all of the F<sub>1<\/sub> hybrid offspring had yellow seeds. That is, the hybrid offspring were phenotypically identical to the true-breeding parent with yellow seeds. However, we know that the allele donated by the parent with green seeds was not simply lost because it reappeared in some of the F<sub>2<\/sub> offspring (<a href=\"#figure8.5\" class=\"autogenerated-content\">Figure 8.5<\/a>). Therefore, the F<sub>1<\/sub> plants must have been genotypically different from the parent with yellow seeds.<\/p>\n<p id=\"fs-idp100656\">The P plants that Mendel used in his experiments were each <strong>homozygous<\/strong> for the trait he was studying. Diploid organisms that are <span>homozygous<\/span> for a gene have two identical alleles, one on each of their homologous chromosomes. The genotype is often written as <em>YY<\/em> or <em>yy<\/em>, for which each letter represents one of the two alleles in the genotype. The dominant allele is capitalized and the recessive allele is lower case. The letter used for the gene (seed color in this case) is usually related to the dominant trait (yellow allele, in this case, or \u201c<em>Y<\/em>\u201d). Mendel\u2019s parental pea plants always bred true because both produced gametes carried the same allele. When P plants with contrasting traits were cross-fertilized, all of the offspring were<strong> heterozygous<\/strong> for the contrasting trait, meaning their genotype had different alleles for the gene being examined. For example, the F<sub>1<\/sub> yellow plants that received a <em>Y<\/em> allele from their yellow parent and a <em>y<\/em> allele from their green parent had the genotype <em>Yy<\/em>.<\/p>\n\n<figure id=\"figure8.5\">\n\n[caption id=\"attachment_212\" align=\"aligncenter\" width=\"500\"]<a href=\"http:\/\/opentextbc.ca\/biology\/wp-content\/uploads\/sites\/96\/2015\/02\/Figure_08_02_01.jpg\"><img src=\"https:\/\/pressbooks.bccampus.ca\/conceptsofbiologygunness\/wp-content\/uploads\/sites\/34\/2016\/05\/Figure_08_02_01.jpg\" alt=\"By the end of this section, you will be able to:      Explain the relationship between genotypes and phenotypes in dominant and recessive gene systems     Use a Punnett square to calculate the expected proportions of genotypes and phenotypes in a monohybrid cross     Explain Mendel&#x2019;s law of segregation and independent assortment in terms of genetics and the events of meiosis     Explain the purpose and methods of a test cross\" class=\"wp-image-212\" height=\"342\" width=\"500\"\/><\/a> Figure 8.5 Phenotypes are physical expressions of traits that are transmitted by alleles. Capital letters represent dominant alleles and lowercase letters represent recessive alleles. The phenotypic ratios are the ratios of visible characteristics. The genotypic ratios are the ratios of gene combinations in the offspring, and these are not always distinguishable in the phenotypes.[\/caption]\n\n<span id=\"fs-idp172960\">\n<\/span>\n\n<\/figure><section id=\"fs-idp9702720\"><h2>Law of Dominance<\/h2>\n<p id=\"fs-idp8507088\">Our discussion of homozygous and heterozygous organisms brings us to why the F<sub>1<\/sub> heterozygous offspring were identical to one of the parents, rather than expressing both alleles. In all seven pea-plant characteristics, one of the two contrasting alleles was dominant, and the other was recessive. Mendel called the dominant allele the expressed unit factor; the recessive allele was referred to as the latent unit factor. We now know that these so-called unit factors are actually genes on homologous chromosomes. For a gene that is expressed in a dominant and recessive pattern, homozygous dominant and heterozygous organisms will look identical (that is, they will have different genotypes but the same phenotype), and the recessive allele will only be observed in homozygous recessive individuals.<\/p>\n\n<table id=\"tab-ch08_02_01\" style=\"border-color: #000000\" summary=\"\"><thead><tr><th colspan=\"4\">Correspondence between Genotype and Phenotype for a Dominant-Recessive Characteristic.<\/th>\n<\/tr><tr><th>\n<\/th><th>Homozygous<\/th>\n<th>Heterozygous<\/th>\n<th>Homozygous<\/th>\n<\/tr><\/thead><tbody><tr><td>Genotype<\/td>\n<td><em>YY<\/em><\/td>\n<td><em>Yy<\/em><\/td>\n<td><em>yy<\/em><\/td>\n<\/tr><tr><td>Phenotype<\/td>\n<td>yellow<\/td>\n<td>yellow<\/td>\n<td>green<\/td>\n<\/tr><\/tbody><\/table><p id=\"fs-idp6144\">Mendel\u2019s <span>law of dominance<\/span> states that in a heterozygote, one trait will conceal the presence of another trait for the same characteristic. For example, when crossing true-breeding violet-flowered plants with true-breeding white-flowered plants, all of the offspring were violet-flowered, even though they all had one allele for violet and one allele for white. Rather than both alleles contributing to a phenotype, the dominant allele will be expressed exclusively. The recessive allele will remain latent, but will be transmitted to offspring in the same manner as that by which the dominant allele is transmitted. The recessive trait will only be expressed by offspring that have two copies of this allele (<a href=\"#figure8.6\" class=\"autogenerated-content\">Figure 8.6<\/a>), and these offspring will breed true when self-crossed.<\/p>\n\n<figure id=\"figure8.6\">\n\n[caption id=\"attachment_213\" align=\"aligncenter\" width=\"400\"]<a href=\"http:\/\/opentextbc.ca\/biology\/wp-content\/uploads\/sites\/96\/2015\/02\/Figure_08_02_02.jpg\"><img src=\"https:\/\/pressbooks.bccampus.ca\/conceptsofbiologygunness\/wp-content\/uploads\/sites\/34\/2016\/05\/Figure_08_02_02.jpg\" alt=\"Photo shows a mother with an albino child.\" class=\"wp-image-213\" height=\"500\" width=\"400\"\/><\/a> Figure 8.6 The allele for albinism, expressed here in humans, is recessive. Both of this child\u2019s parents carried the recessive allele.[\/caption]\n\n<span id=\"fs-idp20960\">\n<\/span>\n\n<\/figure><\/section><\/section><section id=\"fs-idp9652112\"><h1>Monohybrid Cross and the Punnett Square<\/h1>\n<p id=\"fs-idp15696\">When fertilization occurs between two true-breeding parents that differ by only the characteristic being studied, the process is called a <span>monohybrid<\/span> cross, and the resulting offspring are called monohybrids. Mendel performed seven types of monohybrid crosses, each involving contrasting traits for different characteristics. Out of these crosses, all of the F<sub>1<\/sub> offspring had the phenotype of one parent, and the F<sub>2<\/sub> offspring had a 3:1 phenotypic ratio. On the basis of these results, Mendel postulated that each parent in the monohybrid cross contributed one of two paired unit factors to each offspring, and every possible combination of unit factors was equally likely.<\/p>\n<p id=\"fs-idp7191648\">The results of Mendel\u2019s research can be explained in terms of probabilities, which are mathematical measures of likelihood. The probability of an event is calculated by the number of times the event occurs divided by the total number of opportunities for the event to occur. A probability of one (100 percent) for some event indicates that it is guaranteed to occur, whereas a probability of zero (0 percent) indicates that it is guaranteed to not occur, and a probability of 0.5 (50 percent) means it has an equal chance of occurring or not occurring.<\/p>\n<p id=\"fs-idp7192032\">To demonstrate this with a monohybrid cross, consider the case of true-breeding pea plants with yellow versus green seeds. The dominant seed color is yellow; therefore, the parental genotypes were <em>YY <\/em>for the plants with yellow seeds and <em>yy<\/em> for the plants with green seeds. A <span>Punnett square<\/span>, devised by the British geneticist Reginald Punnett, is useful for determining probabilities because it is drawn to predict all possible outcomes of all possible random fertilization events and their expected frequencies. <a href=\"#figure8.9\" class=\"autogenerated-content\">Figure 8.9<\/a> shows a Punnett square for a cross between a plant with yellow peas and one with green peas. To prepare a Punnett square, all possible combinations of the parental alleles (the genotypes of the gametes) are listed along the top (for one parent) and side (for the other parent) of a grid. The combinations of egg and sperm gametes are then made in the boxes in the table on the basis of which alleles are combining. Each box then represents the diploid genotype of a zygote, or fertilized egg. Because each possibility is equally likely, genotypic ratios can be determined from a Punnett square. If the pattern of inheritance (dominant and recessive) is known, the phenotypic ratios can be inferred as well. For a monohybrid cross of two true-breeding parents, each parent contributes one type of allele. In this case, only one genotype is possible in the F<sub>1<\/sub> offspring. All offspring are <em>Yy<\/em> and have yellow seeds.<\/p>\n<p id=\"fs-idp90112\">When the F<sub>1 <\/sub>offspring are crossed with each other, each has an equal probability of contributing either a <em>Y<\/em> or a <em>y<\/em> to the F<sub>2<\/sub> offspring. The result is a 1 in 4 (25 percent) probability of both parents contributing a <em>Y<\/em>, resulting in an offspring with a yellow phenotype; a 25 percent probability of parent A contributing a <em>Y<\/em> and parent B a <em>y<\/em>, resulting in offspring with a yellow phenotype; a 25 percent probability of parent A contributing a <em>y<\/em> and parent B a <em>Y<\/em>, also resulting in a yellow phenotype; and a (25 percent) probability of both parents contributing a <em>y<\/em>, resulting in a green phenotype. When counting all four possible outcomes, there is a 3 in 4 probability of offspring having the yellow phenotype and a 1 in 4 probability of offspring having the green phenotype. This explains why the results of Mendel\u2019s F<sub>2<\/sub> generation occurred in a 3:1 phenotypic ratio. Using large numbers of crosses, Mendel was able to calculate probabilities, found that they fit the model of inheritance, and use these to predict the outcomes of other crosses.<\/p>\n\n<\/section><section id=\"fs-idp8393424\"><h1>Law of Segregation<\/h1>\n<p id=\"fs-idp315728\">Observing that true-breeding pea plants with contrasting traits gave rise to F<sub>1<\/sub> generations that all expressed the dominant trait and F<sub>2<\/sub> generations that expressed the dominant and recessive traits in a 3:1 ratio, Mendel proposed the <span>law of segregation<\/span>. This law states that <strong>paired unit factors (genes) must segregate equally into gametes<\/strong> such that offspring have an equal likelihood of inheriting either factor. For the F<sub>2<\/sub> generation of a monohybrid cross, the following three possible combinations of genotypes result: homozygous dominant, heterozygous, or homozygous recessive. Because heterozygotes could arise from two different pathways (receiving one dominant and one recessive allele from either parent), and because heterozygotes and homozygous dominant individuals are phenotypically identical, the law supports Mendel\u2019s observed 3:1 phenotypic ratio. The equal segregation of alleles is the reason we can apply the Punnett square to accurately predict the offspring of parents with known genotypes. The physical basis of Mendel\u2019s law of segregation is the first division of meiosis in which the homologous chromosomes with their different versions of each gene are segregated into daughter nuclei. This process was not understood by the scientific community during Mendel\u2019s lifetime (<a href=\"#figure8.7\" class=\"autogenerated-content\">Figure 8.7<\/a>).<\/p>\n\n<figure id=\"figure8.7\">\n\n[caption id=\"attachment_214\" align=\"aligncenter\" width=\"600\"]<a href=\"http:\/\/opentextbc.ca\/biology\/wp-content\/uploads\/sites\/96\/2015\/02\/Figure_08_02_03.jpg\"><img src=\"https:\/\/pressbooks.bccampus.ca\/conceptsofbiologygunness\/wp-content\/uploads\/sites\/34\/2016\/05\/Figure_08_02_03.jpg\" alt=\"Homologous pairs of chromosomes line up at the metaphase plate during metaphase I of meiosis. The homologous chromosomes with their different versions of each gene are segregated into daughter nuclei.\" class=\"wp-image-214\" height=\"287\" width=\"600\"\/><\/a> Figure 8.7 The first division in meiosis is shown.[\/caption]\n\n<span id=\"fs-idp8324816\">\n<\/span>\n\n<\/figure><section id=\"fs-idp320048\"><h2>Test Cross<\/h2>\n<p id=\"fs-idp8164432\">Beyond predicting the offspring of a cross between known homozygous or heterozygous parents, Mendel also developed a way to determine whether an organism that expressed a dominant trait was a heterozygote or a homozygote. Called the <span>test cross<\/span>, this technique is still used by plant and animal breeders. In a test cross, the <strong>dominant-expressing organism is crossed with an organism that is homozygous recessive<\/strong> for the same characteristic. If the dominant-expressing organism is a homozygote, then all F<sub>1<\/sub> offspring will be heterozygotes expressing the dominant trait (<a href=\"#figure8.8\" class=\"autogenerated-content\">Figure 8.8<\/a>). Alternatively, if the dominant-expressing organism is a heterozygote, the F<sub>1<\/sub> offspring will exhibit a 1:1 ratio of heterozygotes and recessive homozygotes (<a href=\"#figure8.9\" class=\"autogenerated-content\">Figure 8.9<\/a>). The test cross further validates Mendel\u2019s postulate that pairs of unit factors segregate equally.<\/p>\n\n<figure id=\"figure8.8\">\n\n[caption id=\"attachment_215\" align=\"aligncenter\" width=\"400\"]<a href=\"http:\/\/opentextbc.ca\/biology\/wp-content\/uploads\/sites\/96\/2015\/02\/Figure_08_02_04.jpg\"><img src=\"https:\/\/pressbooks.bccampus.ca\/conceptsofbiologygunness\/wp-content\/uploads\/sites\/34\/2016\/05\/Figure_08_02_04.jpg\" alt=\"In a test cross, a parent with a dominant phenotype but unknown genotype is crossed with a recessive parent. If the parent with the unknown phenotype is homozygous dominant, all the resulting offspring will have at least one dominant allele. If the parent with the unknown phenotype is heterozygous, 50 percent of the offspring will inherit a recessive allele from both parents and will have the recessive phenotype.\" class=\"wp-image-215\" height=\"496\" width=\"400\"\/><\/a> Figure 8.8 A test cross can be performed to determine whether an organism expressing a dominant trait is a homozygote or a heterozygote.[\/caption]\n\n<span id=\"fs-idp5481664\">\n<\/span>\n\n<\/figure><figure id=\"figure8.9\">\n\n[caption id=\"attachment_216\" align=\"aligncenter\" width=\"400\"]<a href=\"http:\/\/opentextbc.ca\/biology\/wp-content\/uploads\/sites\/96\/2015\/02\/Figure_08_02_05.png\"><img src=\"https:\/\/pressbooks.bccampus.ca\/conceptsofbiologygunness\/wp-content\/uploads\/sites\/34\/2016\/05\/Figure_08_02_05.png\" alt=\"A test cross can be performed to determine whether an organism expressing a dominant trait is a homozygote or a heterozygote.\" class=\"wp-image-216\" height=\"597\" width=\"400\"\/><\/a> Figure 8.9 This Punnett square shows the cross between plants with yellow seeds and green seeds. The cross between the true-breeding P plants produces F1 heterozygotes that can be self-fertilized. The self-cross of the F1 generation can be analyzed with a Punnett square to predict the genotypes of the F2 generation. Given an inheritance pattern of dominant\u2013recessive, the genotypic and phenotypic ratios can then be determined.[\/caption]\n\n<span id=\"fs-idp348496\">\n<\/span>\n\n<\/figure><p id=\"fs-idp308144\">In pea plants, round peas (<em>R<\/em>) are dominant to wrinkled peas (<em>r<\/em>). You do a test cross between a pea plant with wrinkled peas (genotype <em>rr<\/em>) and a plant of unknown genotype that has round peas. You end up with three plants, all which have round peas. From this data, can you tell if the parent plant is homozygous dominant or heterozygous?<\/p>\nYou cannot be sure if the plant is homozygous or heterozygous as the data set is too small: by random chance, all three plants might have acquired only the dominant gene even if the recessive one is present.\n\n<\/section><\/section><section id=\"fs-idp8300992\"><h1>Law of Independent Assortment<\/h1>\n<p id=\"fs-idm22880\">Mendel\u2019s <span>law of independent assortment<\/span> states that <strong>genes do not influence each other with regard to the sorting of alleles into gametes<\/strong>, and every possible combination of alleles for every gene is equally likely to occur. Independent assortment of genes can be illustrated by the <span>dihybrid<\/span> cross, a cross between two true-breeding parents that express different traits for two characteristics. Consider the characteristics of seed color and seed texture for two pea plants, one that has wrinkled, green seeds (<em>rryy<\/em>) and another that has round, yellow seeds (<em>RRYY<\/em>). Because each parent is homozygous, the law of segregation indicates that the gametes for the wrinkled\u2013green plant all are <em>ry<\/em>, and the gametes for the round\u2013yellow plant are all <em>RY<\/em>. Therefore, the F<sub>1<\/sub> generation of offspring all are <em>RrYy <\/em>(<a href=\"#figure8.10\" class=\"autogenerated-content\">Figure 8.10<\/a>).<\/p>\n\n<div id=\"fs-idp107968\" class=\"art-connection\"><figure id=\"figure8.10\">\n\n[caption id=\"attachment_217\" align=\"aligncenter\" width=\"500\"]<a href=\"http:\/\/opentextbc.ca\/biology\/wp-content\/uploads\/sites\/96\/2015\/02\/Figure_08_02_06.png\"><img src=\"https:\/\/pressbooks.bccampus.ca\/conceptsofbiologygunness\/wp-content\/uploads\/sites\/34\/2016\/05\/Figure_08_02_06.png\" alt=\"This illustration shows a dihybrid cross between pea plants. In the P generation, a plant that has the homozygous dominant phenotype of yellow, round peas is crossed with a plant with the homozygous recessive phenotype of green, wrinkled peas. The resulting F_{1} offspring have a heterozygous genotype and yellow, round peas. Self-pollination of the F_{1} generation results in F_{2} offspring with a phenotypic ratio of 9:3:3:1 for round&#x2013;yellow, round&#x2013;green, wrinkled&#x2013;yellow, and wrinkled&#x2013;green peas, respectively.\" class=\"wp-image-217\" height=\"440\" width=\"500\"\/><\/a> Figure 8.10 A dihybrid cross in pea plants involves the genes for seed color and texture. The P cross produces F1 offspring that are all heterozygous for both characteristics. The resulting 9:3:3:1 F2 phenotypic ratio is obtained using a Punnett square.[\/caption]\n\n<span id=\"fs-idp8499696\">\n<\/span>\n\n<\/figure><p id=\"fs-idp232864\">In pea plants, purple flowers (<em>P<\/em>) are dominant to white (<em>p<\/em>), and yellow peas (<em>Y<\/em>) are dominant to green (<em>y<\/em>). What are the possible genotypes and phenotypes for a cross between <em>PpYY<\/em> and <em>ppYy<\/em> pea plants? How many squares would you need to complete a Punnett square analysis of this cross?<\/p>\nThe possible genotypes are PpYY, PpYy, ppYY, and ppYy. The former two genotypes would result in plants with purple flowers and yellow peas, while the latter two genotypes would result in plants with white flowers with yellow peas, for a 1:1 ratio of each phenotype. You only need a 2 \u00d7 2 Punnett square (four squares total) to do this analysis because two of the alleles are homozygous.\n\n<\/div>\n<p id=\"fs-idp7195168\">The gametes produced by the F<sub>1<\/sub> individuals must have one allele from each of the two genes. For example, a gamete could get an <em>R<\/em> allele for the seed shape gene and either a <em>Y<\/em> or a <em>y<\/em> allele for the seed color gene. It cannot get both an <em>R<\/em> and an <em>r<\/em> allele; each gamete can have only one allele per gene. The law of independent assortment states that a gamete into which an <em>r<\/em> allele is sorted would be equally likely to contain either a <em>Y<\/em> or a <em>y<\/em> allele. Thus, there are four equally likely gametes that can be formed when the <em>RrYy<\/em> heterozygote is self-crossed, as follows: <em>RY<\/em>, <em>rY<\/em>, <em>Ry<\/em>, and <em>ry<\/em>. Arranging these gametes along the top and left of a 4 \u00d7 4 Punnett square gives us 16 equally likely genotypic combinations. From these genotypes, we find a phenotypic ratio of 9 round\u2013yellow:3 round\u2013green:3 wrinkled\u2013yellow:1 wrinkled\u2013green. These are the offspring ratios we would expect, assuming we performed the crosses with a large enough sample size.<\/p>\n<p id=\"fs-idp8161072\">The physical basis for the law of independent assortment also lies in meiosis I, in which the different homologous pairs line up in random orientations. Each gamete can contain any combination of paternal and maternal chromosomes (and therefore the genes on them) because the orientation of tetrads on the metaphase plane is random (<a href=\"#figure8.11\" class=\"autogenerated-content\">Figure 8.11<\/a>).<\/p>\n\n<figure id=\"figure8.11\">\n\n[caption id=\"attachment_218\" align=\"aligncenter\" width=\"500\"]<a href=\"http:\/\/opentextbc.ca\/biology\/wp-content\/uploads\/sites\/96\/2015\/02\/Figure_08_02_07.jpg\"><img src=\"https:\/\/pressbooks.bccampus.ca\/conceptsofbiologygunness\/wp-content\/uploads\/sites\/34\/2016\/05\/Figure_08_02_07-1024x904.jpg\" alt=\"Homologous pairs of chromosomes line up at the metaphase plate during metaphase I of meiosis. The homologous chromosomes, with their different versions of each gene, are randomly segregated into daughter nuclei, resulting in a variety of possible genetic arrangements.\" class=\"wp-image-218\" height=\"441\" width=\"500\"\/><\/a> Figure 8.11 The random segregation into daughter nuclei that happens during the first division in meiosis can lead to a variety of possible genetic arrangements.[\/caption]\n\n\u00a0\n\n<\/figure><\/section><section id=\"fs-idm14496\" class=\"summary\"><h1>Probability Basics<\/h1>\n<p id=\"fs-id1771773\">Probabilities are mathematical measures of likelihood. The empirical probability of an event is calculated by dividing the number of times the event occurs by the total number of opportunities for the event to occur. It is also possible to calculate theoretical probabilities by dividing the number of times that an event is expected to occur by the number of times that it could occur. Empirical probabilities come from observations, like those of Mendel. Theoretical probabilities come from knowing how the events are produced and assuming that the probabilities of individual outcomes are equal. A probability of one for some event indicates that it is guaranteed to occur, whereas a probability of zero indicates that it is guaranteed not to occur. An example of a genetic event is a round seed produced by a pea plant. In his experiment, Mendel demonstrated that the probability of the event \u201cround seed\u201d occurring was one in the F<sub>1<\/sub> offspring of true-breeding parents, one of which has round seeds and one of which has wrinkled seeds. When the F<sub>1<\/sub> plants were subsequently self-crossed, the probability of any given F<sub>2<\/sub> offspring having round seeds was now three out of four. In other words, in a large population of F<sub>2<\/sub> offspring chosen at random, 75 percent were expected to have round seeds, whereas 25 percent were expected to have wrinkled seeds. Using large numbers of crosses, Mendel was able to calculate probabilities and use these to predict the outcomes of other crosses.<\/p>\n\n<section id=\"fs-id1805576\"><h2>The Product Rule and Sum Rule<\/h2>\n<p id=\"fs-id1798638\">Mendel demonstrated that the pea-plant characteristics he studied were transmitted as discrete units from parent to offspring. As will be discussed, Mendel also determined that different characteristics, like seed color and seed texture, were transmitted independently of one another and could be considered in separate probability analyses. For instance, performing a cross between a plant with green, wrinkled seeds and a plant with yellow, round seeds still produced offspring that had a 3:1 ratio of green:yellow seeds (ignoring seed texture) and a 3:1 ratio of round:wrinkled seeds (ignoring seed color). The characteristics of color and texture did not influence each other.<\/p>\n<p id=\"fs-id1780876\">The <span>product rule<\/span> of probability can be applied to this phenomenon of the independent transmission of characteristics. The product rule states that the probability of two independent events occurring together can be calculated by multiplying the individual probabilities of each event occurring alone. To demonstrate the product rule, imagine that you are rolling a six-sided die (D) and flipping a penny (P) at the same time. The die may roll any number from 1\u20136 (D<sub>#<\/sub>), whereas the penny may turn up heads (P<sub>H<\/sub>) or tails (P<sub>T<\/sub>). The outcome of rolling the die has no effect on the outcome of flipping the penny and vice versa. There are 12 possible outcomes of this action, and each event is expected to occur with equal probability.<\/p>\n\n<table id=\"tab-ch12-01-02\" summary=\"\"><thead><tr><th colspan=\"2\">Twelve Equally Likely Outcomes of Rolling a Die and Flipping a Penny<\/th>\n<\/tr><tr><th>Rolling Die<\/th>\n<th>Flipping Penny<\/th>\n<\/tr><\/thead><tbody><tr><td>D<sub>1<\/sub><\/td>\n<td>P<sub>H<\/sub><\/td>\n<\/tr><tr><td>D<sub>1<\/sub><\/td>\n<td>P<sub>T<\/sub><\/td>\n<\/tr><tr><td>D<sub>2<\/sub><\/td>\n<td>P<sub>H<\/sub><\/td>\n<\/tr><tr><td>D<sub>2<\/sub><\/td>\n<td>P<sub>T<\/sub><\/td>\n<\/tr><tr><td>D<sub>3<\/sub><\/td>\n<td>P<sub>H<\/sub><\/td>\n<\/tr><tr><td>D<sub>3<\/sub><\/td>\n<td>P<sub>T<\/sub><\/td>\n<\/tr><tr><td>D<sub>4<\/sub><\/td>\n<td>P<sub>H<\/sub><\/td>\n<\/tr><tr><td>D<sub>4<\/sub><\/td>\n<td>P<sub>T<\/sub><\/td>\n<\/tr><tr><td>D<sub>5<\/sub><\/td>\n<td>P<sub>H<\/sub><\/td>\n<\/tr><tr><td>D<sub>5<\/sub><\/td>\n<td>P<sub>T<\/sub><\/td>\n<\/tr><tr><td>D<sub>6<\/sub><\/td>\n<td>P<sub>H<\/sub><\/td>\n<\/tr><tr><td>D<sub>6<\/sub><\/td>\n<td>P<sub>T<\/sub><\/td>\n<\/tr><\/tbody><\/table><p id=\"fs-id2316750\">Of the 12 possible outcomes, the die has a 2\/12 (or 1\/6) probability of rolling a two, and the penny has a 6\/12 (or 1\/2) probability of coming up heads. By the product rule, the probability that you will obtain the combined outcome 2 and heads is: (D<sub>2<\/sub>) x (P<sub>H<\/sub>) = (1\/6) x (1\/2) or 1\/12. Notice the word \u201cand\u201d in the description of the probability. The \u201cand\u201d is a signal to apply the product rule. For example, consider how the product rule is applied to the dihybrid cross: the probability of having both dominant traits in the F<sub>2<\/sub> progeny is the product of the probabilities of having the dominant trait for each characteristic, as shown here:<\/p>\n\n<div id=\"eip-772\">\n\n\u00a0\n<div style=\"text-align: center\" class=\"MathJax_Display\"><span id=\"MathJax-Element-1-Frame\" class=\"MathJax\"><span id=\"MathJax-Span-1\" class=\"math\"><span><span><span id=\"MathJax-Span-2\" class=\"mrow\"><span id=\"MathJax-Span-3\" class=\"semantics\"><span id=\"MathJax-Span-4\" class=\"mrow\"><span id=\"MathJax-Span-5\" class=\"mrow\"><span id=\"MathJax-Span-6\" class=\"mfrac\"><span><span><span id=\"MathJax-Span-7\" class=\"mn\">3<\/span><span\/><span><span id=\"MathJax-Span-8\" class=\"mn\"><span\/><span><span>\/4<\/span><span\/><\/span><\/span><span id=\"MathJax-Span-9\" class=\"mtext\">\u00a0<\/span><span id=\"MathJax-Span-10\" class=\"mo\">\u00d7<\/span><span id=\"MathJax-Span-11\" class=\"mtext\">\u00a0<\/span><span id=\"MathJax-Span-12\" class=\"mfrac\"><span><span><span id=\"MathJax-Span-13\" class=\"mn\">3<\/span><span\/><span><span id=\"MathJax-Span-14\" class=\"mn\"><span\/><span><span>\/4<\/span><span\/><\/span><\/span><span id=\"MathJax-Span-15\" class=\"mtext\">\u00a0<\/span><span id=\"MathJax-Span-16\" class=\"mo\">=<\/span><span id=\"MathJax-Span-17\" class=\"mtext\">\u00a0<\/span><span id=\"MathJax-Span-18\" class=\"mtext\"><span>\u200b<\/span><\/span><span id=\"MathJax-Span-19\" class=\"mfrac\"><span><span><span id=\"MathJax-Span-20\" class=\"mn\">9\/16<\/span><\/span><span><span><span\/><\/span><\/span><\/span><\/span><\/span><\/span><span\/><\/span><span\/><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/div>\n<\/div>\n<p id=\"fs-id1512071\">On the other hand, the <span>sum rule<\/span> of probability is applied when considering two mutually exclusive outcomes that can come about by more than one pathway. The sum rule states that the probability of the occurrence of one event or the other event, of two mutually exclusive events, is the sum of their individual probabilities. Notice the word \u201cor\u201d in the description of the probability. The \u201cor\u201d indicates that you should apply the sum rule. In this case, let\u2019s imagine you are flipping a penny (P) and a quarter (Q). What is the probability of one coin coming up heads and one coin coming up tails? This outcome can be achieved by two cases: the penny may be heads (P<sub>H<\/sub>) and the quarter may be tails (Q<sub>T<\/sub>), or the quarter may be heads (Q<sub>H<\/sub>) and the penny may be tails (P<sub>T<\/sub>). Either case fulfills the outcome. By the sum rule, we calculate the probability of obtaining one head and one tail as [(P<sub>H<\/sub>) \u00d7 (Q<sub>T<\/sub>)] + [(Q<sub>H<\/sub>) \u00d7 (P<sub>T<\/sub>)] = [(1\/2) \u00d7 (1\/2)] + [(1\/2) \u00d7 (1\/2)] = 1\/2. You should also notice that we used the product rule to calculate the probability of P<sub>H<\/sub> and Q<sub>T<\/sub>, and also the probability of P<sub>T<\/sub> and Q<sub>H<\/sub>, before we summed them. Again, the sum rule can be applied to show the probability of having just one dominant trait in the F<sub>2<\/sub> generation of a dihybrid cross:<\/p>\n\n<div id=\"fs-id1760400\">\n\n\u00a0\n<div style=\"text-align: center\" class=\"MathJax_Display\"><span id=\"MathJax-Element-2-Frame\" class=\"MathJax\"><span id=\"MathJax-Span-23\" class=\"math\"><span><span><span id=\"MathJax-Span-24\" class=\"mrow\"><span id=\"MathJax-Span-25\" class=\"semantics\"><span id=\"MathJax-Span-26\" class=\"mrow\"><span id=\"MathJax-Span-27\" class=\"mrow\"><span id=\"MathJax-Span-28\" class=\"mfrac\"><span><span><span id=\"MathJax-Span-29\" class=\"mn\">3\/16 + 3\/4 = 15\/16<\/span><\/span><\/span><\/span><span id=\"MathJax-Span-42\" class=\"mfrac\"><span><span><span>\n<\/span><span\/><\/span><\/span><\/span><\/span><\/span><\/span><span\/><\/span><span\/><\/span><\/span><\/span><\/span><\/div>\n<div class=\"MathJax_Display\">\n<\/div>\n<table id=\"tab-ch12-01-03\" summary=\"\"><thead><tr><th colspan=\"2\">The Product Rule and Sum Rule<\/th>\n<\/tr><tr><th>Product Rule<\/th>\n<th>Sum Rule<\/th>\n<\/tr><\/thead><tbody><tr><td>For independent events A and B, the probability (P) of them both occurring (A <em>and<\/em> B) is (P<sub>A<\/sub> \u00d7 P<sub>B<\/sub>)<\/td>\n<td>For mutually exclusive events A and B, the probability (P) that at least one occurs (A <em>or<\/em> B) is (P<sub>A<\/sub> + P<sub>B<\/sub>)<\/td>\n<\/tr><\/tbody><\/table><p id=\"fs-id1570444\">To use probability laws in practice, it is necessary to work with large sample sizes because small sample sizes are prone to deviations caused by chance. The large quantities of pea plants that Mendel examined allowed him calculate the probabilities of the traits appearing in his F<sub>2<\/sub> generation. As you will learn, this discovery meant that when parental traits were known, the offspring\u2019s traits could be predicted accurately even before fertilization.<\/p>\n\n\n[caption id=\"attachment_847\" align=\"aligncenter\" width=\"300\"]<a href=\"http:\/\/opentextbc.ca\/biology\/wp-content\/uploads\/sites\/96\/2015\/03\/Figure_12_02_11.jpg\"><img src=\"https:\/\/pressbooks.bccampus.ca\/conceptsofbiologygunness\/wp-content\/uploads\/sites\/34\/2016\/05\/Figure_12_02_11-300x264.jpg\" alt=\"This is a pedigree of a family that carries the recessive disorder alkaptonuria. In the second generation, an unaffected mother and an affected father have three children. One child has the disorder, so the genotype of the mother must be Aa and the genotype of the father is aa. One unaffected child goes on to have two children, one affected and one unaffected. Because her husband was not affected, she and her husband must both be heterozygous. The genotype of their unaffected child is unknown, and is designated A?. In the third generation, the other unaffected child had no offspring, and his genotype is therefore also unknown. The affected third-generation child goes on to have one child with the disorder. Her husband is unaffected and is labeled &#x201C;3.&#x201D; The first generation father is affected and is labeled &#x201C;1.&#x201D; The first generation mother is unaffected and is labeled &#x201C;2.&#x201D;  The Art Connection question asks the genotype of the three numbered individuals.\" class=\"wp-image-847 size-medium\" height=\"264\" width=\"300\"\/><\/a> Figure 8.12[\/caption]\n\nAlkaptonuria is a recessive genetic disorder in which two amino acids, phenylalanine and tyrosine, are not properly metabolized. Affected individuals may have darkened skin and brown urine, and may suffer joint damage and other complications. In this pedigree, individuals with the disorder are indicated in blue and have the genotype <em>aa<\/em>. Unaffected individuals are indicated in yellow and have the genotype <em>AA<\/em> or <em>Aa<\/em>. Note that it is often possible to determine a person\u2019s genotype from the genotype of their offspring. For example, if neither parent has the disorder but their child does, they must be heterozygous. Two individuals on the pedigree have an unaffected phenotype but unknown genotype. Because they do not have the disorder, they must have at least one normal allele, so their genotype gets the \u201c<em>A?<\/em>\u201d designation.\n\nWhat are the genotypes of the individuals labeled 1, 2 and 3?\n<h1>Section Summary<\/h1>\n<p id=\"fs-idp8291920\">When true-breeding, or homozygous, individuals that differ for a certain trait are crossed, all of the offspring will be heterozygous for that trait. If the traits are inherited as dominant and recessive, the F<sub>1<\/sub> offspring will all exhibit the same phenotype as the parent homozygous for the dominant trait. If these heterozygous offspring are self-crossed, the resulting F<sub>2<\/sub> offspring will be equally likely to inherit gametes carrying the dominant or recessive trait, giving rise to offspring of which one quarter are homozygous dominant, half are heterozygous, and one quarter are homozygous recessive. Because homozygous dominant and heterozygous individuals are phenotypically identical, the observed traits in the F<sub>2<\/sub> offspring will exhibit a ratio of three dominant to one recessive.<\/p>\n<p id=\"fs-idp8297328\">Mendel postulated that genes (characteristics) are inherited as pairs of alleles (traits) that behave in a dominant and recessive pattern. Alleles segregate into gametes such that each gamete is equally likely to receive either one of the two alleles present in a diploid individual. In addition, genes are assorted into gametes independently of one another. That is, in general, alleles are not more likely to segregate into a gamete with a particular allele of another gene.<\/p>\n\n<\/div><\/section><section id=\"fs-idp8298336\" class=\"art-exercise\"><div class=\"bcc-box bcc-info\">\n<h3>Exercises<\/h3>\n<section class=\"art-exercise\"><div id=\"fs-idp384944\">\n<div id=\"fs-idp385200\">\n<p id=\"fs-idp385456\">\u00a0In pea plants, round peas (<em>R<\/em>) are dominant to wrinkled peas (<em>r<\/em>). You do a test cross between a pea plant with wrinkled peas (genotype <em>rr<\/em>) and a plant of unknown genotype that has round peas. You end up with three plants, all which have round peas. From this data, can you tell if the parent plant is homozygous dominant or heterozygous?<\/p>\n\n<\/div>\n<div id=\"fs-idp210784\">\n<p id=\"fs-idp356992\">\u00a0Answer: You cannot be sure if the plant is homozygous or heterozygous as the data set is too small: by random chance, all three plants might have acquired only the dominant gene even if the recessive one is present.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"fs-idp358352\">\n<div id=\"fs-idp358608\">\n<p id=\"fs-idp8155328\">In pea plants, purple flowers (<em>P<\/em>) are dominant to white (<em>p<\/em>), and yellow peas (<em>Y<\/em>) are dominant to green (<em>y<\/em>). What are the possible genotypes and phenotypes for a cross between <em>PpYY<\/em> and <em>ppYy<\/em> pea plants? How many squares would you need to complete a Punnett square analysis of this cross?<\/p>\n\n<\/div>\n<div id=\"fs-idp8356544\">\n<p id=\"fs-idp8356928\">Answer: The possible genotypes are <em>PpYY<\/em>, <em>PpYy<\/em>, <em>ppYY<\/em>, and <em>ppYy<\/em>. The former two genotypes would result in plants with purple flowers and yellow peas, while the latter two genotypes would result in plants with white flowers with yellow peas, for a 1:1 ratio of each phenotype. You only need a 2 \u00d7 2 Punnett square (four squares total) to do this analysis because two of the alleles are homozygous.<\/p>\n\n<\/div>\n<\/div>\n<\/section><section id=\"fs-idp8335648\" class=\"multiple-choice\"><h1>Multiple Choice<\/h1>\n<div id=\"fs-idp8336528\">\n<div id=\"fs-idp8336784\">\n<p id=\"fs-idp41328\">The observable traits expressed by an organism are described as its ________.<\/p>\nA) phenotype\n\nB) genotype\n\nC) alleles\n\nD) zygote\n\n<\/div>\n<div id=\"fs-idp129328\">\n<p id=\"fs-idp129840\">A<\/p>\n\n<\/div>\n<\/div>\n<div id=\"fs-idp130480\">\n<div id=\"fs-idp130736\">\n<p id=\"fs-idp130864\">A recessive trait will be observed in individuals that are ________ for that trait.<\/p>\nA) heterozygous\n\nB) homozygous or heterozygous\n\nC) homozygous\n\nD) diploid\n\n<\/div>\n<div id=\"fs-idp7427056\">\n<p id=\"fs-idp7427568\">C<\/p>\n\n<\/div>\n<\/div>\n<div id=\"eip-idp98973408\">\n<div id=\"eip-idp98422272\">\n<p id=\"eip-idp111921824\">What are the types of gametes that can be produced by an individual with the genotype <em>AaBb<\/em>?<\/p>\n<em>A) Aa<\/em>, <em>Bb<\/em>\n\n<em>B) AA<\/em>, <em>aa<\/em>, <em>BB<\/em>, <em>bb<\/em>\n\n<em>C) AB<\/em>, <em>Ab<\/em>, <em>aB<\/em>, <em>ab<\/em>\n\n<em>D) AB<\/em>, <em>ab<\/em>\n\n<\/div>\n<div id=\"eip-idp31308480\">\n<p id=\"eip-idp145405808\">C<\/p>\n\n<\/div>\n<\/div>\n<div id=\"eip-idp176464\">\n<div id=\"eip-idp57297296\">\n<p id=\"eip-idp12363232\">What is the reason for doing a test cross?<\/p>\nA) to identify heterozygous individuals with the dominant phenotype\n\nB) to determine which allele is dominant and which is recessive\n\nC) to identify homozygous recessive individuals in the F<sub>2<\/sub>\n\nD) to determine if two genes assort independently\n\n<\/div>\n<div id=\"eip-idp53378432\">\n<p id=\"eip-idm19345264\">A<\/p>\n\n<\/div>\n<\/div>\n<\/section><section id=\"fs-idp7428336\" class=\"free-response\"><h1>Free Response<\/h1>\n<div id=\"fs-idp338672\">\n<div id=\"fs-idp338928\">\n<p id=\"fs-idp339056\">Use a Punnett square to predict the offspring in a cross between a dwarf pea plant (homozygous recessive) and a tall pea plant (heterozygous). What is the phenotypic ratio of the offspring?<\/p>\n\n<\/div>\n<div id=\"fs-idp339648\">\n<p id=\"fs-idp340032\">The Punnett square would be 2 \u00d7 2 and will have <em>T<\/em> and <em>T<\/em> along the top and <em>T<\/em> and <em>t<\/em> along the left side. Clockwise from the top left, the genotypes listed within the boxes will be <em>Tt<\/em>, <em>Tt<\/em>, <em>tt<\/em>, and <em>tt<\/em>. The phenotypic ratio will be 1 tall:1 dwarf.<\/p>\n\n<\/div>\n<\/div>\n<div id=\"eip-idp91747856\">\n<div id=\"eip-idp111475264\">\n<p id=\"eip-idp143481520\">Use a Punnett square to predict the offspring in a cross between a tall pea plant (heterozygous) and a tall pea plant (heterozygous). What is the genotypic ratio of the offspring?<\/p>\n\n<\/div>\n<div id=\"eip-idp27640512\">\n<p id=\"eip-idp17733792\">The Punnett square will be 2 \u00d7 2 and will have <em>T<\/em> and <em>t<\/em> along the top and <em>T<\/em> and <em>t<\/em> along the left side. Clockwise from the top left, the genotypes listed within the boxes will be <em>TT<\/em>, <em>Tt<\/em>, <em>Tt<\/em>, and <em>tt<\/em>. The genotypic ratio will be 1<em>TT<\/em>:2<em>Tt<\/em>:1<em>tt<\/em>.<\/p>\n\n<\/div>\n<\/div>\n<\/section><\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Glossary<\/h3>\n<strong>allele: <\/strong>one of two or more variants of a gene that determines a particular trait for a characteristic\n\n<strong>dihybrid: <\/strong>the result of a cross between two true-breeding parents that express different traits for two characteristics\n\n<strong>genotype: <\/strong>the underlying genetic makeup, consisting of both physically visible and non-expressed alleles, of an organism\n\n<strong>heterozygous: <\/strong>having two different alleles for a given gene on the homologous chromosomes\n\n<strong>homozygous: <\/strong>having two identical alleles for a given gene on the homologous chromosomes\n\n<strong>law of dominance: <\/strong>in a heterozygote, one trait will conceal the presence of another trait for the same characteristic\n\n<strong>law of independent assortment: <\/strong>genes do not influence each other with regard to sorting of alleles into gametes; every\u00a0possible combination of alleles is equally likely to occur\n\n<strong>law of segregation: <\/strong>paired unit factors (i.e., genes) segregate equally into gametes such that offspring have an equal likelihood of inheriting any combination of factors\n\n<strong>monohybrid: <\/strong>the result of a cross between two true-breeding parents that express different traits for only one characteristic\n\n<strong>phenotype: <\/strong>the observable traits expressed by an organism\n\n<strong>Punnett square: <\/strong>a visual representation of a cross between two individuals in which the gametes of each individual are denoted along the top and side of a grid, respectively, and the possible zygotic genotypes are recombined at each box in the grid\n\n<strong>test cross: <\/strong>a cross between a dominant expressing individual with an unknown genotype and a homozygous recessive individual; the offspring phenotypes indicate whether the unknown parent is heterozygous or homozygous for the dominant trait\n\n<\/div>\n\u00a0\n\n<\/section><div>\n<\/div><\/section>","rendered":"<div>\n<div class=\"bcc-box bcc-highlight\">\n<h3>Learning Objectives<\/h3>\n<p>By the end of this section, you will be able to:<\/p>\n<ul>\n<li>Explain the relationship between genotypes and phenotypes in dominant and recessive gene systems<\/li>\n<li>Use a Punnett square to calculate the expected proportions of genotypes and phenotypes in a monohybrid cross<\/li>\n<li>Explain Mendel\u2019s law of segregation and independent assortment in terms of genetics and the events of meiosis<\/li>\n<li>Explain the purpose and methods of a test cross<\/li>\n<\/ul>\n<\/div>\n<\/div>\n<p id=\"fs-idp8170416\">The seven characteristics that Mendel evaluated in his pea plants were each expressed as one of two versions, or traits. Mendel deduced from his results that each individual had two discrete copies of the characteristic that are passed individually to offspring. We now call those two copies <strong>genes<\/strong>, which are carried on chromosomes. The reason we have two copies of each gene is that we inherit one from each parent. In fact, it is the chromosomes we inherit and the two copies of each gene are located on paired chromosomes. Recall that in meiosis these chromosomes are separated out into haploid gametes. This <strong>separation, or segregation<\/strong>, of the homologous chromosomes means also that only one of the copies of the gene gets moved into a gamete. The offspring are formed when that gamete unites with one from another parent and the two copies of each gene (and chromosome) are restored.<\/p>\n<p id=\"fs-idp8399888\">For cases in which a single gene controls a single characteristic, a diploid organism has two genetic copies that may or may not encode the same version of that characteristic. For example, one individual may carry a gene that determines white flower color and a gene that determines violet flower color. Gene variants that arise by mutation and exist at the same relative locations on homologous chromosomes are called<strong> alleles<\/strong>. Mendel examined the inheritance of genes with just two allele forms, but it is common to encounter more than two alleles for any given gene in a natural population.<\/p>\n<section id=\"fs-idp91312\">\n<h1>Phenotypes and Genotypes<\/h1>\n<p id=\"fs-idp353872\">Two alleles for a given gene in a diploid organism are expressed and interact to produce physical characteristics. The observable traits expressed by an organism are referred to as its <strong>phenotype<\/strong>. An organism\u2019s underlying genetic makeup, consisting of both the physically visible and the non-expressed alleles, is called its <strong>genotype<\/strong>. Mendel\u2019s hybridization experiments demonstrate the difference between phenotype and genotype. For example, the phenotypes that Mendel observed in his crosses between pea plants with differing traits are connected to the diploid genotypes of the plants in the P, F<sub>1<\/sub>, and F<sub>2<\/sub> generations. We will use a second trait that Mendel investigated, seed color, as an example. Seed color is governed by a single gene with two alleles. The yellow-seed allele is dominant and the green-seed allele is recessive. When true-breeding plants were cross-fertilized, in which one parent had yellow seeds and one had green seeds, all of the F<sub>1<\/sub> hybrid offspring had yellow seeds. That is, the hybrid offspring were phenotypically identical to the true-breeding parent with yellow seeds. However, we know that the allele donated by the parent with green seeds was not simply lost because it reappeared in some of the F<sub>2<\/sub> offspring (<a href=\"#figure8.5\" class=\"autogenerated-content\">Figure 8.5<\/a>). Therefore, the F<sub>1<\/sub> plants must have been genotypically different from the parent with yellow seeds.<\/p>\n<p id=\"fs-idp100656\">The P plants that Mendel used in his experiments were each <strong>homozygous<\/strong> for the trait he was studying. Diploid organisms that are <span>homozygous<\/span> for a gene have two identical alleles, one on each of their homologous chromosomes. The genotype is often written as <em>YY<\/em> or <em>yy<\/em>, for which each letter represents one of the two alleles in the genotype. The dominant allele is capitalized and the recessive allele is lower case. The letter used for the gene (seed color in this case) is usually related to the dominant trait (yellow allele, in this case, or \u201c<em>Y<\/em>\u201d). Mendel\u2019s parental pea plants always bred true because both produced gametes carried the same allele. When P plants with contrasting traits were cross-fertilized, all of the offspring were<strong> heterozygous<\/strong> for the contrasting trait, meaning their genotype had different alleles for the gene being examined. For example, the F<sub>1<\/sub> yellow plants that received a <em>Y<\/em> allele from their yellow parent and a <em>y<\/em> allele from their green parent had the genotype <em>Yy<\/em>.<\/p>\n<figure id=\"figure8.5\">\n<figure id=\"attachment_212\" aria-describedby=\"caption-attachment-212\" style=\"width: 500px\" class=\"wp-caption aligncenter\"><a href=\"http:\/\/opentextbc.ca\/biology\/wp-content\/uploads\/sites\/96\/2015\/02\/Figure_08_02_01.jpg\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/conceptsofbiologygunness\/wp-content\/uploads\/sites\/34\/2016\/05\/Figure_08_02_01.jpg\" alt=\"By the end of this section, you will be able to:      Explain the relationship between genotypes and phenotypes in dominant and recessive gene systems     Use a Punnett square to calculate the expected proportions of genotypes and phenotypes in a monohybrid cross     Explain Mendel&#x2019;s law of segregation and independent assortment in terms of genetics and the events of meiosis     Explain the purpose and methods of a test cross\" class=\"wp-image-212\" height=\"342\" width=\"500\" \/><\/a><figcaption id=\"caption-attachment-212\" class=\"wp-caption-text\">Figure 8.5 Phenotypes are physical expressions of traits that are transmitted by alleles. Capital letters represent dominant alleles and lowercase letters represent recessive alleles. The phenotypic ratios are the ratios of visible characteristics. The genotypic ratios are the ratios of gene combinations in the offspring, and these are not always distinguishable in the phenotypes.<\/figcaption><\/figure>\n<p><span id=\"fs-idp172960\"><br \/>\n<\/span><\/p>\n<\/figure>\n<section id=\"fs-idp9702720\">\n<h2>Law of Dominance<\/h2>\n<p id=\"fs-idp8507088\">Our discussion of homozygous and heterozygous organisms brings us to why the F<sub>1<\/sub> heterozygous offspring were identical to one of the parents, rather than expressing both alleles. In all seven pea-plant characteristics, one of the two contrasting alleles was dominant, and the other was recessive. Mendel called the dominant allele the expressed unit factor; the recessive allele was referred to as the latent unit factor. We now know that these so-called unit factors are actually genes on homologous chromosomes. For a gene that is expressed in a dominant and recessive pattern, homozygous dominant and heterozygous organisms will look identical (that is, they will have different genotypes but the same phenotype), and the recessive allele will only be observed in homozygous recessive individuals.<\/p>\n<table id=\"tab-ch08_02_01\" style=\"border-color: #000000\" summary=\"\">\n<thead>\n<tr>\n<th colspan=\"4\">Correspondence between Genotype and Phenotype for a Dominant-Recessive Characteristic.<\/th>\n<\/tr>\n<tr>\n<th>\n<\/th>\n<th>Homozygous<\/th>\n<th>Heterozygous<\/th>\n<th>Homozygous<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>Genotype<\/td>\n<td><em>YY<\/em><\/td>\n<td><em>Yy<\/em><\/td>\n<td><em>yy<\/em><\/td>\n<\/tr>\n<tr>\n<td>Phenotype<\/td>\n<td>yellow<\/td>\n<td>yellow<\/td>\n<td>green<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p id=\"fs-idp6144\">Mendel\u2019s <span>law of dominance<\/span> states that in a heterozygote, one trait will conceal the presence of another trait for the same characteristic. For example, when crossing true-breeding violet-flowered plants with true-breeding white-flowered plants, all of the offspring were violet-flowered, even though they all had one allele for violet and one allele for white. Rather than both alleles contributing to a phenotype, the dominant allele will be expressed exclusively. The recessive allele will remain latent, but will be transmitted to offspring in the same manner as that by which the dominant allele is transmitted. The recessive trait will only be expressed by offspring that have two copies of this allele (<a href=\"#figure8.6\" class=\"autogenerated-content\">Figure 8.6<\/a>), and these offspring will breed true when self-crossed.<\/p>\n<figure id=\"figure8.6\">\n<figure id=\"attachment_213\" aria-describedby=\"caption-attachment-213\" style=\"width: 400px\" class=\"wp-caption aligncenter\"><a href=\"http:\/\/opentextbc.ca\/biology\/wp-content\/uploads\/sites\/96\/2015\/02\/Figure_08_02_02.jpg\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/conceptsofbiologygunness\/wp-content\/uploads\/sites\/34\/2016\/05\/Figure_08_02_02.jpg\" alt=\"Photo shows a mother with an albino child.\" class=\"wp-image-213\" height=\"500\" width=\"400\" \/><\/a><figcaption id=\"caption-attachment-213\" class=\"wp-caption-text\">Figure 8.6 The allele for albinism, expressed here in humans, is recessive. Both of this child\u2019s parents carried the recessive allele.<\/figcaption><\/figure>\n<p><span id=\"fs-idp20960\"><br \/>\n<\/span><\/p>\n<\/figure>\n<\/section>\n<\/section>\n<section id=\"fs-idp9652112\">\n<h1>Monohybrid Cross and the Punnett Square<\/h1>\n<p id=\"fs-idp15696\">When fertilization occurs between two true-breeding parents that differ by only the characteristic being studied, the process is called a <span>monohybrid<\/span> cross, and the resulting offspring are called monohybrids. Mendel performed seven types of monohybrid crosses, each involving contrasting traits for different characteristics. Out of these crosses, all of the F<sub>1<\/sub> offspring had the phenotype of one parent, and the F<sub>2<\/sub> offspring had a 3:1 phenotypic ratio. On the basis of these results, Mendel postulated that each parent in the monohybrid cross contributed one of two paired unit factors to each offspring, and every possible combination of unit factors was equally likely.<\/p>\n<p id=\"fs-idp7191648\">The results of Mendel\u2019s research can be explained in terms of probabilities, which are mathematical measures of likelihood. The probability of an event is calculated by the number of times the event occurs divided by the total number of opportunities for the event to occur. A probability of one (100 percent) for some event indicates that it is guaranteed to occur, whereas a probability of zero (0 percent) indicates that it is guaranteed to not occur, and a probability of 0.5 (50 percent) means it has an equal chance of occurring or not occurring.<\/p>\n<p id=\"fs-idp7192032\">To demonstrate this with a monohybrid cross, consider the case of true-breeding pea plants with yellow versus green seeds. The dominant seed color is yellow; therefore, the parental genotypes were <em>YY <\/em>for the plants with yellow seeds and <em>yy<\/em> for the plants with green seeds. A <span>Punnett square<\/span>, devised by the British geneticist Reginald Punnett, is useful for determining probabilities because it is drawn to predict all possible outcomes of all possible random fertilization events and their expected frequencies. <a href=\"#figure8.9\" class=\"autogenerated-content\">Figure 8.9<\/a> shows a Punnett square for a cross between a plant with yellow peas and one with green peas. To prepare a Punnett square, all possible combinations of the parental alleles (the genotypes of the gametes) are listed along the top (for one parent) and side (for the other parent) of a grid. The combinations of egg and sperm gametes are then made in the boxes in the table on the basis of which alleles are combining. Each box then represents the diploid genotype of a zygote, or fertilized egg. Because each possibility is equally likely, genotypic ratios can be determined from a Punnett square. If the pattern of inheritance (dominant and recessive) is known, the phenotypic ratios can be inferred as well. For a monohybrid cross of two true-breeding parents, each parent contributes one type of allele. In this case, only one genotype is possible in the F<sub>1<\/sub> offspring. All offspring are <em>Yy<\/em> and have yellow seeds.<\/p>\n<p id=\"fs-idp90112\">When the F<sub>1 <\/sub>offspring are crossed with each other, each has an equal probability of contributing either a <em>Y<\/em> or a <em>y<\/em> to the F<sub>2<\/sub> offspring. The result is a 1 in 4 (25 percent) probability of both parents contributing a <em>Y<\/em>, resulting in an offspring with a yellow phenotype; a 25 percent probability of parent A contributing a <em>Y<\/em> and parent B a <em>y<\/em>, resulting in offspring with a yellow phenotype; a 25 percent probability of parent A contributing a <em>y<\/em> and parent B a <em>Y<\/em>, also resulting in a yellow phenotype; and a (25 percent) probability of both parents contributing a <em>y<\/em>, resulting in a green phenotype. When counting all four possible outcomes, there is a 3 in 4 probability of offspring having the yellow phenotype and a 1 in 4 probability of offspring having the green phenotype. This explains why the results of Mendel\u2019s F<sub>2<\/sub> generation occurred in a 3:1 phenotypic ratio. Using large numbers of crosses, Mendel was able to calculate probabilities, found that they fit the model of inheritance, and use these to predict the outcomes of other crosses.<\/p>\n<\/section>\n<section id=\"fs-idp8393424\">\n<h1>Law of Segregation<\/h1>\n<p id=\"fs-idp315728\">Observing that true-breeding pea plants with contrasting traits gave rise to F<sub>1<\/sub> generations that all expressed the dominant trait and F<sub>2<\/sub> generations that expressed the dominant and recessive traits in a 3:1 ratio, Mendel proposed the <span>law of segregation<\/span>. This law states that <strong>paired unit factors (genes) must segregate equally into gametes<\/strong> such that offspring have an equal likelihood of inheriting either factor. For the F<sub>2<\/sub> generation of a monohybrid cross, the following three possible combinations of genotypes result: homozygous dominant, heterozygous, or homozygous recessive. Because heterozygotes could arise from two different pathways (receiving one dominant and one recessive allele from either parent), and because heterozygotes and homozygous dominant individuals are phenotypically identical, the law supports Mendel\u2019s observed 3:1 phenotypic ratio. The equal segregation of alleles is the reason we can apply the Punnett square to accurately predict the offspring of parents with known genotypes. The physical basis of Mendel\u2019s law of segregation is the first division of meiosis in which the homologous chromosomes with their different versions of each gene are segregated into daughter nuclei. This process was not understood by the scientific community during Mendel\u2019s lifetime (<a href=\"#figure8.7\" class=\"autogenerated-content\">Figure 8.7<\/a>).<\/p>\n<figure id=\"figure8.7\">\n<figure id=\"attachment_214\" aria-describedby=\"caption-attachment-214\" style=\"width: 600px\" class=\"wp-caption aligncenter\"><a href=\"http:\/\/opentextbc.ca\/biology\/wp-content\/uploads\/sites\/96\/2015\/02\/Figure_08_02_03.jpg\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/conceptsofbiologygunness\/wp-content\/uploads\/sites\/34\/2016\/05\/Figure_08_02_03.jpg\" alt=\"Homologous pairs of chromosomes line up at the metaphase plate during metaphase I of meiosis. The homologous chromosomes with their different versions of each gene are segregated into daughter nuclei.\" class=\"wp-image-214\" height=\"287\" width=\"600\" \/><\/a><figcaption id=\"caption-attachment-214\" class=\"wp-caption-text\">Figure 8.7 The first division in meiosis is shown.<\/figcaption><\/figure>\n<p><span id=\"fs-idp8324816\"><br \/>\n<\/span><\/p>\n<\/figure>\n<section id=\"fs-idp320048\">\n<h2>Test Cross<\/h2>\n<p id=\"fs-idp8164432\">Beyond predicting the offspring of a cross between known homozygous or heterozygous parents, Mendel also developed a way to determine whether an organism that expressed a dominant trait was a heterozygote or a homozygote. Called the <span>test cross<\/span>, this technique is still used by plant and animal breeders. In a test cross, the <strong>dominant-expressing organism is crossed with an organism that is homozygous recessive<\/strong> for the same characteristic. If the dominant-expressing organism is a homozygote, then all F<sub>1<\/sub> offspring will be heterozygotes expressing the dominant trait (<a href=\"#figure8.8\" class=\"autogenerated-content\">Figure 8.8<\/a>). Alternatively, if the dominant-expressing organism is a heterozygote, the F<sub>1<\/sub> offspring will exhibit a 1:1 ratio of heterozygotes and recessive homozygotes (<a href=\"#figure8.9\" class=\"autogenerated-content\">Figure 8.9<\/a>). The test cross further validates Mendel\u2019s postulate that pairs of unit factors segregate equally.<\/p>\n<figure id=\"figure8.8\">\n<figure id=\"attachment_215\" aria-describedby=\"caption-attachment-215\" style=\"width: 400px\" class=\"wp-caption aligncenter\"><a href=\"http:\/\/opentextbc.ca\/biology\/wp-content\/uploads\/sites\/96\/2015\/02\/Figure_08_02_04.jpg\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/conceptsofbiologygunness\/wp-content\/uploads\/sites\/34\/2016\/05\/Figure_08_02_04.jpg\" alt=\"In a test cross, a parent with a dominant phenotype but unknown genotype is crossed with a recessive parent. If the parent with the unknown phenotype is homozygous dominant, all the resulting offspring will have at least one dominant allele. If the parent with the unknown phenotype is heterozygous, 50 percent of the offspring will inherit a recessive allele from both parents and will have the recessive phenotype.\" class=\"wp-image-215\" height=\"496\" width=\"400\" \/><\/a><figcaption id=\"caption-attachment-215\" class=\"wp-caption-text\">Figure 8.8 A test cross can be performed to determine whether an organism expressing a dominant trait is a homozygote or a heterozygote.<\/figcaption><\/figure>\n<p><span id=\"fs-idp5481664\"><br \/>\n<\/span><\/p>\n<\/figure>\n<figure id=\"figure8.9\">\n<figure id=\"attachment_216\" aria-describedby=\"caption-attachment-216\" style=\"width: 400px\" class=\"wp-caption aligncenter\"><a href=\"http:\/\/opentextbc.ca\/biology\/wp-content\/uploads\/sites\/96\/2015\/02\/Figure_08_02_05.png\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/conceptsofbiologygunness\/wp-content\/uploads\/sites\/34\/2016\/05\/Figure_08_02_05.png\" alt=\"A test cross can be performed to determine whether an organism expressing a dominant trait is a homozygote or a heterozygote.\" class=\"wp-image-216\" height=\"597\" width=\"400\" \/><\/a><figcaption id=\"caption-attachment-216\" class=\"wp-caption-text\">Figure 8.9 This Punnett square shows the cross between plants with yellow seeds and green seeds. The cross between the true-breeding P plants produces F1 heterozygotes that can be self-fertilized. The self-cross of the F1 generation can be analyzed with a Punnett square to predict the genotypes of the F2 generation. Given an inheritance pattern of dominant\u2013recessive, the genotypic and phenotypic ratios can then be determined.<\/figcaption><\/figure>\n<p><span id=\"fs-idp348496\"><br \/>\n<\/span><\/p>\n<\/figure>\n<p id=\"fs-idp308144\">In pea plants, round peas (<em>R<\/em>) are dominant to wrinkled peas (<em>r<\/em>). You do a test cross between a pea plant with wrinkled peas (genotype <em>rr<\/em>) and a plant of unknown genotype that has round peas. You end up with three plants, all which have round peas. From this data, can you tell if the parent plant is homozygous dominant or heterozygous?<\/p>\n<p>You cannot be sure if the plant is homozygous or heterozygous as the data set is too small: by random chance, all three plants might have acquired only the dominant gene even if the recessive one is present.<\/p>\n<\/section>\n<\/section>\n<section id=\"fs-idp8300992\">\n<h1>Law of Independent Assortment<\/h1>\n<p id=\"fs-idm22880\">Mendel\u2019s <span>law of independent assortment<\/span> states that <strong>genes do not influence each other with regard to the sorting of alleles into gametes<\/strong>, and every possible combination of alleles for every gene is equally likely to occur. Independent assortment of genes can be illustrated by the <span>dihybrid<\/span> cross, a cross between two true-breeding parents that express different traits for two characteristics. Consider the characteristics of seed color and seed texture for two pea plants, one that has wrinkled, green seeds (<em>rryy<\/em>) and another that has round, yellow seeds (<em>RRYY<\/em>). Because each parent is homozygous, the law of segregation indicates that the gametes for the wrinkled\u2013green plant all are <em>ry<\/em>, and the gametes for the round\u2013yellow plant are all <em>RY<\/em>. Therefore, the F<sub>1<\/sub> generation of offspring all are <em>RrYy <\/em>(<a href=\"#figure8.10\" class=\"autogenerated-content\">Figure 8.10<\/a>).<\/p>\n<div id=\"fs-idp107968\" class=\"art-connection\">\n<figure id=\"figure8.10\">\n<figure id=\"attachment_217\" aria-describedby=\"caption-attachment-217\" style=\"width: 500px\" class=\"wp-caption aligncenter\"><a href=\"http:\/\/opentextbc.ca\/biology\/wp-content\/uploads\/sites\/96\/2015\/02\/Figure_08_02_06.png\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/conceptsofbiologygunness\/wp-content\/uploads\/sites\/34\/2016\/05\/Figure_08_02_06.png\" alt=\"This illustration shows a dihybrid cross between pea plants. In the P generation, a plant that has the homozygous dominant phenotype of yellow, round peas is crossed with a plant with the homozygous recessive phenotype of green, wrinkled peas. The resulting F_{1} offspring have a heterozygous genotype and yellow, round peas. Self-pollination of the F_{1} generation results in F_{2} offspring with a phenotypic ratio of 9:3:3:1 for round&#x2013;yellow, round&#x2013;green, wrinkled&#x2013;yellow, and wrinkled&#x2013;green peas, respectively.\" class=\"wp-image-217\" height=\"440\" width=\"500\" \/><\/a><figcaption id=\"caption-attachment-217\" class=\"wp-caption-text\">Figure 8.10 A dihybrid cross in pea plants involves the genes for seed color and texture. The P cross produces F1 offspring that are all heterozygous for both characteristics. The resulting 9:3:3:1 F2 phenotypic ratio is obtained using a Punnett square.<\/figcaption><\/figure>\n<p><span id=\"fs-idp8499696\"><br \/>\n<\/span><\/p>\n<\/figure>\n<p id=\"fs-idp232864\">In pea plants, purple flowers (<em>P<\/em>) are dominant to white (<em>p<\/em>), and yellow peas (<em>Y<\/em>) are dominant to green (<em>y<\/em>). What are the possible genotypes and phenotypes for a cross between <em>PpYY<\/em> and <em>ppYy<\/em> pea plants? How many squares would you need to complete a Punnett square analysis of this cross?<\/p>\n<p>The possible genotypes are PpYY, PpYy, ppYY, and ppYy. The former two genotypes would result in plants with purple flowers and yellow peas, while the latter two genotypes would result in plants with white flowers with yellow peas, for a 1:1 ratio of each phenotype. You only need a 2 \u00d7 2 Punnett square (four squares total) to do this analysis because two of the alleles are homozygous.<\/p>\n<\/div>\n<p id=\"fs-idp7195168\">The gametes produced by the F<sub>1<\/sub> individuals must have one allele from each of the two genes. For example, a gamete could get an <em>R<\/em> allele for the seed shape gene and either a <em>Y<\/em> or a <em>y<\/em> allele for the seed color gene. It cannot get both an <em>R<\/em> and an <em>r<\/em> allele; each gamete can have only one allele per gene. The law of independent assortment states that a gamete into which an <em>r<\/em> allele is sorted would be equally likely to contain either a <em>Y<\/em> or a <em>y<\/em> allele. Thus, there are four equally likely gametes that can be formed when the <em>RrYy<\/em> heterozygote is self-crossed, as follows: <em>RY<\/em>, <em>rY<\/em>, <em>Ry<\/em>, and <em>ry<\/em>. Arranging these gametes along the top and left of a 4 \u00d7 4 Punnett square gives us 16 equally likely genotypic combinations. From these genotypes, we find a phenotypic ratio of 9 round\u2013yellow:3 round\u2013green:3 wrinkled\u2013yellow:1 wrinkled\u2013green. These are the offspring ratios we would expect, assuming we performed the crosses with a large enough sample size.<\/p>\n<p id=\"fs-idp8161072\">The physical basis for the law of independent assortment also lies in meiosis I, in which the different homologous pairs line up in random orientations. Each gamete can contain any combination of paternal and maternal chromosomes (and therefore the genes on them) because the orientation of tetrads on the metaphase plane is random (<a href=\"#figure8.11\" class=\"autogenerated-content\">Figure 8.11<\/a>).<\/p>\n<figure id=\"figure8.11\">\n<figure id=\"attachment_218\" aria-describedby=\"caption-attachment-218\" style=\"width: 500px\" class=\"wp-caption aligncenter\"><a href=\"http:\/\/opentextbc.ca\/biology\/wp-content\/uploads\/sites\/96\/2015\/02\/Figure_08_02_07.jpg\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/conceptsofbiologygunness\/wp-content\/uploads\/sites\/34\/2016\/05\/Figure_08_02_07-1024x904.jpg\" alt=\"Homologous pairs of chromosomes line up at the metaphase plate during metaphase I of meiosis. The homologous chromosomes, with their different versions of each gene, are randomly segregated into daughter nuclei, resulting in a variety of possible genetic arrangements.\" class=\"wp-image-218\" height=\"441\" width=\"500\" \/><\/a><figcaption id=\"caption-attachment-218\" class=\"wp-caption-text\">Figure 8.11 The random segregation into daughter nuclei that happens during the first division in meiosis can lead to a variety of possible genetic arrangements.<\/figcaption><\/figure>\n<p>\u00a0<\/p>\n<\/figure>\n<\/section>\n<section id=\"fs-idm14496\" class=\"summary\">\n<h1>Probability Basics<\/h1>\n<p id=\"fs-id1771773\">Probabilities are mathematical measures of likelihood. The empirical probability of an event is calculated by dividing the number of times the event occurs by the total number of opportunities for the event to occur. It is also possible to calculate theoretical probabilities by dividing the number of times that an event is expected to occur by the number of times that it could occur. Empirical probabilities come from observations, like those of Mendel. Theoretical probabilities come from knowing how the events are produced and assuming that the probabilities of individual outcomes are equal. A probability of one for some event indicates that it is guaranteed to occur, whereas a probability of zero indicates that it is guaranteed not to occur. An example of a genetic event is a round seed produced by a pea plant. In his experiment, Mendel demonstrated that the probability of the event \u201cround seed\u201d occurring was one in the F<sub>1<\/sub> offspring of true-breeding parents, one of which has round seeds and one of which has wrinkled seeds. When the F<sub>1<\/sub> plants were subsequently self-crossed, the probability of any given F<sub>2<\/sub> offspring having round seeds was now three out of four. In other words, in a large population of F<sub>2<\/sub> offspring chosen at random, 75 percent were expected to have round seeds, whereas 25 percent were expected to have wrinkled seeds. Using large numbers of crosses, Mendel was able to calculate probabilities and use these to predict the outcomes of other crosses.<\/p>\n<section id=\"fs-id1805576\">\n<h2>The Product Rule and Sum Rule<\/h2>\n<p id=\"fs-id1798638\">Mendel demonstrated that the pea-plant characteristics he studied were transmitted as discrete units from parent to offspring. As will be discussed, Mendel also determined that different characteristics, like seed color and seed texture, were transmitted independently of one another and could be considered in separate probability analyses. For instance, performing a cross between a plant with green, wrinkled seeds and a plant with yellow, round seeds still produced offspring that had a 3:1 ratio of green:yellow seeds (ignoring seed texture) and a 3:1 ratio of round:wrinkled seeds (ignoring seed color). The characteristics of color and texture did not influence each other.<\/p>\n<p id=\"fs-id1780876\">The <span>product rule<\/span> of probability can be applied to this phenomenon of the independent transmission of characteristics. The product rule states that the probability of two independent events occurring together can be calculated by multiplying the individual probabilities of each event occurring alone. To demonstrate the product rule, imagine that you are rolling a six-sided die (D) and flipping a penny (P) at the same time. The die may roll any number from 1\u20136 (D<sub>#<\/sub>), whereas the penny may turn up heads (P<sub>H<\/sub>) or tails (P<sub>T<\/sub>). The outcome of rolling the die has no effect on the outcome of flipping the penny and vice versa. There are 12 possible outcomes of this action, and each event is expected to occur with equal probability.<\/p>\n<table id=\"tab-ch12-01-02\" summary=\"\">\n<thead>\n<tr>\n<th colspan=\"2\">Twelve Equally Likely Outcomes of Rolling a Die and Flipping a Penny<\/th>\n<\/tr>\n<tr>\n<th>Rolling Die<\/th>\n<th>Flipping Penny<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>D<sub>1<\/sub><\/td>\n<td>P<sub>H<\/sub><\/td>\n<\/tr>\n<tr>\n<td>D<sub>1<\/sub><\/td>\n<td>P<sub>T<\/sub><\/td>\n<\/tr>\n<tr>\n<td>D<sub>2<\/sub><\/td>\n<td>P<sub>H<\/sub><\/td>\n<\/tr>\n<tr>\n<td>D<sub>2<\/sub><\/td>\n<td>P<sub>T<\/sub><\/td>\n<\/tr>\n<tr>\n<td>D<sub>3<\/sub><\/td>\n<td>P<sub>H<\/sub><\/td>\n<\/tr>\n<tr>\n<td>D<sub>3<\/sub><\/td>\n<td>P<sub>T<\/sub><\/td>\n<\/tr>\n<tr>\n<td>D<sub>4<\/sub><\/td>\n<td>P<sub>H<\/sub><\/td>\n<\/tr>\n<tr>\n<td>D<sub>4<\/sub><\/td>\n<td>P<sub>T<\/sub><\/td>\n<\/tr>\n<tr>\n<td>D<sub>5<\/sub><\/td>\n<td>P<sub>H<\/sub><\/td>\n<\/tr>\n<tr>\n<td>D<sub>5<\/sub><\/td>\n<td>P<sub>T<\/sub><\/td>\n<\/tr>\n<tr>\n<td>D<sub>6<\/sub><\/td>\n<td>P<sub>H<\/sub><\/td>\n<\/tr>\n<tr>\n<td>D<sub>6<\/sub><\/td>\n<td>P<sub>T<\/sub><\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p id=\"fs-id2316750\">Of the 12 possible outcomes, the die has a 2\/12 (or 1\/6) probability of rolling a two, and the penny has a 6\/12 (or 1\/2) probability of coming up heads. By the product rule, the probability that you will obtain the combined outcome 2 and heads is: (D<sub>2<\/sub>) x (P<sub>H<\/sub>) = (1\/6) x (1\/2) or 1\/12. Notice the word \u201cand\u201d in the description of the probability. The \u201cand\u201d is a signal to apply the product rule. For example, consider how the product rule is applied to the dihybrid cross: the probability of having both dominant traits in the F<sub>2<\/sub> progeny is the product of the probabilities of having the dominant trait for each characteristic, as shown here:<\/p>\n<div id=\"eip-772\">\n<p>\u00a0<\/p>\n<div style=\"text-align: center\" class=\"MathJax_Display\"><span id=\"MathJax-Element-1-Frame\" class=\"MathJax\"><span id=\"MathJax-Span-1\" class=\"math\"><span><span><span id=\"MathJax-Span-2\" class=\"mrow\"><span id=\"MathJax-Span-3\" class=\"semantics\"><span id=\"MathJax-Span-4\" class=\"mrow\"><span id=\"MathJax-Span-5\" class=\"mrow\"><span id=\"MathJax-Span-6\" class=\"mfrac\"><span><span><span id=\"MathJax-Span-7\" class=\"mn\">3<\/span><span><span><span id=\"MathJax-Span-8\" class=\"mn\"><span><span><span>\/4<\/span><span><\/span><\/span><span id=\"MathJax-Span-9\" class=\"mtext\">\u00a0<\/span><span id=\"MathJax-Span-10\" class=\"mo\">\u00d7<\/span><span id=\"MathJax-Span-11\" class=\"mtext\">\u00a0<\/span><span id=\"MathJax-Span-12\" class=\"mfrac\"><span><span><span id=\"MathJax-Span-13\" class=\"mn\">3<\/span><span><span><span id=\"MathJax-Span-14\" class=\"mn\"><span><span><span>\/4<\/span><span><\/span><\/span><span id=\"MathJax-Span-15\" class=\"mtext\">\u00a0<\/span><span id=\"MathJax-Span-16\" class=\"mo\">=<\/span><span id=\"MathJax-Span-17\" class=\"mtext\">\u00a0<\/span><span id=\"MathJax-Span-18\" class=\"mtext\"><span>\u200b<\/span><\/span><span id=\"MathJax-Span-19\" class=\"mfrac\"><span><span><span id=\"MathJax-Span-20\" class=\"mn\">9\/16<\/span><\/span><span><span><span><\/span><\/span><\/span><\/span><\/span><\/span><span><\/span><span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/div>\n<\/div>\n<p id=\"fs-id1512071\">On the other hand, the <span>sum rule<\/span> of probability is applied when considering two mutually exclusive outcomes that can come about by more than one pathway. The sum rule states that the probability of the occurrence of one event or the other event, of two mutually exclusive events, is the sum of their individual probabilities. Notice the word \u201cor\u201d in the description of the probability. The \u201cor\u201d indicates that you should apply the sum rule. In this case, let\u2019s imagine you are flipping a penny (P) and a quarter (Q). What is the probability of one coin coming up heads and one coin coming up tails? This outcome can be achieved by two cases: the penny may be heads (P<sub>H<\/sub>) and the quarter may be tails (Q<sub>T<\/sub>), or the quarter may be heads (Q<sub>H<\/sub>) and the penny may be tails (P<sub>T<\/sub>). Either case fulfills the outcome. By the sum rule, we calculate the probability of obtaining one head and one tail as [(P<sub>H<\/sub>) \u00d7 (Q<sub>T<\/sub>)] + [(Q<sub>H<\/sub>) \u00d7 (P<sub>T<\/sub>)] = [(1\/2) \u00d7 (1\/2)] + [(1\/2) \u00d7 (1\/2)] = 1\/2. You should also notice that we used the product rule to calculate the probability of P<sub>H<\/sub> and Q<sub>T<\/sub>, and also the probability of P<sub>T<\/sub> and Q<sub>H<\/sub>, before we summed them. Again, the sum rule can be applied to show the probability of having just one dominant trait in the F<sub>2<\/sub> generation of a dihybrid cross:<\/p>\n<div id=\"fs-id1760400\">\n<p>\u00a0<\/p>\n<div style=\"text-align: center\" class=\"MathJax_Display\"><span id=\"MathJax-Element-2-Frame\" class=\"MathJax\"><span id=\"MathJax-Span-23\" class=\"math\"><span><span><span id=\"MathJax-Span-24\" class=\"mrow\"><span id=\"MathJax-Span-25\" class=\"semantics\"><span id=\"MathJax-Span-26\" class=\"mrow\"><span id=\"MathJax-Span-27\" class=\"mrow\"><span id=\"MathJax-Span-28\" class=\"mfrac\"><span><span><span id=\"MathJax-Span-29\" class=\"mn\">3\/16 + 3\/4 = 15\/16<\/span><\/span><\/span><\/span><span id=\"MathJax-Span-42\" class=\"mfrac\"><span><span><span><br \/>\n<\/span><span><\/span><\/span><\/span><\/span><\/span><\/span><span><\/span><span><\/span><\/span><\/span><\/span><\/span><\/span><\/span><\/div>\n<div class=\"MathJax_Display\">\n<\/div>\n<table id=\"tab-ch12-01-03\" summary=\"\">\n<thead>\n<tr>\n<th colspan=\"2\">The Product Rule and Sum Rule<\/th>\n<\/tr>\n<tr>\n<th>Product Rule<\/th>\n<th>Sum Rule<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>For independent events A and B, the probability (P) of them both occurring (A <em>and<\/em> B) is (P<sub>A<\/sub> \u00d7 P<sub>B<\/sub>)<\/td>\n<td>For mutually exclusive events A and B, the probability (P) that at least one occurs (A <em>or<\/em> B) is (P<sub>A<\/sub> + P<sub>B<\/sub>)<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p id=\"fs-id1570444\">To use probability laws in practice, it is necessary to work with large sample sizes because small sample sizes are prone to deviations caused by chance. The large quantities of pea plants that Mendel examined allowed him calculate the probabilities of the traits appearing in his F<sub>2<\/sub> generation. As you will learn, this discovery meant that when parental traits were known, the offspring\u2019s traits could be predicted accurately even before fertilization.<\/p>\n<figure id=\"attachment_847\" aria-describedby=\"caption-attachment-847\" style=\"width: 300px\" class=\"wp-caption aligncenter\"><a href=\"http:\/\/opentextbc.ca\/biology\/wp-content\/uploads\/sites\/96\/2015\/03\/Figure_12_02_11.jpg\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/pressbooks.bccampus.ca\/conceptsofbiologygunness\/wp-content\/uploads\/sites\/34\/2016\/05\/Figure_12_02_11-300x264.jpg\" alt=\"This is a pedigree of a family that carries the recessive disorder alkaptonuria. In the second generation, an unaffected mother and an affected father have three children. One child has the disorder, so the genotype of the mother must be Aa and the genotype of the father is aa. One unaffected child goes on to have two children, one affected and one unaffected. Because her husband was not affected, she and her husband must both be heterozygous. The genotype of their unaffected child is unknown, and is designated A?. In the third generation, the other unaffected child had no offspring, and his genotype is therefore also unknown. The affected third-generation child goes on to have one child with the disorder. Her husband is unaffected and is labeled &#x201c;3.&#x201d; The first generation father is affected and is labeled &#x201c;1.&#x201d; The first generation mother is unaffected and is labeled &#x201c;2.&#x201d;  The Art Connection question asks the genotype of the three numbered individuals.\" class=\"wp-image-847 size-medium\" height=\"264\" width=\"300\" \/><\/a><figcaption id=\"caption-attachment-847\" class=\"wp-caption-text\">Figure 8.12<\/figcaption><\/figure>\n<p>Alkaptonuria is a recessive genetic disorder in which two amino acids, phenylalanine and tyrosine, are not properly metabolized. Affected individuals may have darkened skin and brown urine, and may suffer joint damage and other complications. In this pedigree, individuals with the disorder are indicated in blue and have the genotype <em>aa<\/em>. Unaffected individuals are indicated in yellow and have the genotype <em>AA<\/em> or <em>Aa<\/em>. Note that it is often possible to determine a person\u2019s genotype from the genotype of their offspring. For example, if neither parent has the disorder but their child does, they must be heterozygous. Two individuals on the pedigree have an unaffected phenotype but unknown genotype. Because they do not have the disorder, they must have at least one normal allele, so their genotype gets the \u201c<em>A?<\/em>\u201d designation.<\/p>\n<p>What are the genotypes of the individuals labeled 1, 2 and 3?<\/p>\n<h1>Section Summary<\/h1>\n<p id=\"fs-idp8291920\">When true-breeding, or homozygous, individuals that differ for a certain trait are crossed, all of the offspring will be heterozygous for that trait. If the traits are inherited as dominant and recessive, the F<sub>1<\/sub> offspring will all exhibit the same phenotype as the parent homozygous for the dominant trait. If these heterozygous offspring are self-crossed, the resulting F<sub>2<\/sub> offspring will be equally likely to inherit gametes carrying the dominant or recessive trait, giving rise to offspring of which one quarter are homozygous dominant, half are heterozygous, and one quarter are homozygous recessive. Because homozygous dominant and heterozygous individuals are phenotypically identical, the observed traits in the F<sub>2<\/sub> offspring will exhibit a ratio of three dominant to one recessive.<\/p>\n<p id=\"fs-idp8297328\">Mendel postulated that genes (characteristics) are inherited as pairs of alleles (traits) that behave in a dominant and recessive pattern. Alleles segregate into gametes such that each gamete is equally likely to receive either one of the two alleles present in a diploid individual. In addition, genes are assorted into gametes independently of one another. That is, in general, alleles are not more likely to segregate into a gamete with a particular allele of another gene.<\/p>\n<\/div>\n<\/section>\n<section id=\"fs-idp8298336\" class=\"art-exercise\">\n<div class=\"bcc-box bcc-info\">\n<h3>Exercises<\/h3>\n<section class=\"art-exercise\">\n<div id=\"fs-idp384944\">\n<div id=\"fs-idp385200\">\n<p id=\"fs-idp385456\">\u00a0In pea plants, round peas (<em>R<\/em>) are dominant to wrinkled peas (<em>r<\/em>). You do a test cross between a pea plant with wrinkled peas (genotype <em>rr<\/em>) and a plant of unknown genotype that has round peas. You end up with three plants, all which have round peas. From this data, can you tell if the parent plant is homozygous dominant or heterozygous?<\/p>\n<\/div>\n<div id=\"fs-idp210784\">\n<p id=\"fs-idp356992\">\u00a0Answer: You cannot be sure if the plant is homozygous or heterozygous as the data set is too small: by random chance, all three plants might have acquired only the dominant gene even if the recessive one is present.<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-idp358352\">\n<div id=\"fs-idp358608\">\n<p id=\"fs-idp8155328\">In pea plants, purple flowers (<em>P<\/em>) are dominant to white (<em>p<\/em>), and yellow peas (<em>Y<\/em>) are dominant to green (<em>y<\/em>). What are the possible genotypes and phenotypes for a cross between <em>PpYY<\/em> and <em>ppYy<\/em> pea plants? How many squares would you need to complete a Punnett square analysis of this cross?<\/p>\n<\/div>\n<div id=\"fs-idp8356544\">\n<p id=\"fs-idp8356928\">Answer: The possible genotypes are <em>PpYY<\/em>, <em>PpYy<\/em>, <em>ppYY<\/em>, and <em>ppYy<\/em>. The former two genotypes would result in plants with purple flowers and yellow peas, while the latter two genotypes would result in plants with white flowers with yellow peas, for a 1:1 ratio of each phenotype. You only need a 2 \u00d7 2 Punnett square (four squares total) to do this analysis because two of the alleles are homozygous.<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section id=\"fs-idp8335648\" class=\"multiple-choice\">\n<h1>Multiple Choice<\/h1>\n<div id=\"fs-idp8336528\">\n<div id=\"fs-idp8336784\">\n<p id=\"fs-idp41328\">The observable traits expressed by an organism are described as its ________.<\/p>\n<p>A) phenotype<\/p>\n<p>B) genotype<\/p>\n<p>C) alleles<\/p>\n<p>D) zygote<\/p>\n<\/div>\n<div id=\"fs-idp129328\">\n<p id=\"fs-idp129840\">A<\/p>\n<\/div>\n<\/div>\n<div id=\"fs-idp130480\">\n<div id=\"fs-idp130736\">\n<p id=\"fs-idp130864\">A recessive trait will be observed in individuals that are ________ for that trait.<\/p>\n<p>A) heterozygous<\/p>\n<p>B) homozygous or heterozygous<\/p>\n<p>C) homozygous<\/p>\n<p>D) diploid<\/p>\n<\/div>\n<div id=\"fs-idp7427056\">\n<p id=\"fs-idp7427568\">C<\/p>\n<\/div>\n<\/div>\n<div id=\"eip-idp98973408\">\n<div id=\"eip-idp98422272\">\n<p id=\"eip-idp111921824\">What are the types of gametes that can be produced by an individual with the genotype <em>AaBb<\/em>?<\/p>\n<p><em>A) Aa<\/em>, <em>Bb<\/em><\/p>\n<p><em>B) AA<\/em>, <em>aa<\/em>, <em>BB<\/em>, <em>bb<\/em><\/p>\n<p><em>C) AB<\/em>, <em>Ab<\/em>, <em>aB<\/em>, <em>ab<\/em><\/p>\n<p><em>D) AB<\/em>, <em>ab<\/em><\/p>\n<\/div>\n<div id=\"eip-idp31308480\">\n<p id=\"eip-idp145405808\">C<\/p>\n<\/div>\n<\/div>\n<div id=\"eip-idp176464\">\n<div id=\"eip-idp57297296\">\n<p id=\"eip-idp12363232\">What is the reason for doing a test cross?<\/p>\n<p>A) to identify heterozygous individuals with the dominant phenotype<\/p>\n<p>B) to determine which allele is dominant and which is recessive<\/p>\n<p>C) to identify homozygous recessive individuals in the F<sub>2<\/sub><\/p>\n<p>D) to determine if two genes assort independently<\/p>\n<\/div>\n<div id=\"eip-idp53378432\">\n<p id=\"eip-idm19345264\">A<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section id=\"fs-idp7428336\" class=\"free-response\">\n<h1>Free Response<\/h1>\n<div id=\"fs-idp338672\">\n<div id=\"fs-idp338928\">\n<p id=\"fs-idp339056\">Use a Punnett square to predict the offspring in a cross between a dwarf pea plant (homozygous recessive) and a tall pea plant (heterozygous). What is the phenotypic ratio of the offspring?<\/p>\n<\/div>\n<div id=\"fs-idp339648\">\n<p id=\"fs-idp340032\">The Punnett square would be 2 \u00d7 2 and will have <em>T<\/em> and <em>T<\/em> along the top and <em>T<\/em> and <em>t<\/em> along the left side. Clockwise from the top left, the genotypes listed within the boxes will be <em>Tt<\/em>, <em>Tt<\/em>, <em>tt<\/em>, and <em>tt<\/em>. The phenotypic ratio will be 1 tall:1 dwarf.<\/p>\n<\/div>\n<\/div>\n<div id=\"eip-idp91747856\">\n<div id=\"eip-idp111475264\">\n<p id=\"eip-idp143481520\">Use a Punnett square to predict the offspring in a cross between a tall pea plant (heterozygous) and a tall pea plant (heterozygous). What is the genotypic ratio of the offspring?<\/p>\n<\/div>\n<div id=\"eip-idp27640512\">\n<p id=\"eip-idp17733792\">The Punnett square will be 2 \u00d7 2 and will have <em>T<\/em> and <em>t<\/em> along the top and <em>T<\/em> and <em>t<\/em> along the left side. Clockwise from the top left, the genotypes listed within the boxes will be <em>TT<\/em>, <em>Tt<\/em>, <em>Tt<\/em>, and <em>tt<\/em>. The genotypic ratio will be 1<em>TT<\/em>:2<em>Tt<\/em>:1<em>tt<\/em>.<\/p>\n<\/div>\n<\/div>\n<\/section>\n<\/div>\n<div class=\"bcc-box bcc-success\">\n<h3>Glossary<\/h3>\n<p><strong>allele: <\/strong>one of two or more variants of a gene that determines a particular trait for a characteristic<\/p>\n<p><strong>dihybrid: <\/strong>the result of a cross between two true-breeding parents that express different traits for two characteristics<\/p>\n<p><strong>genotype: <\/strong>the underlying genetic makeup, consisting of both physically visible and non-expressed alleles, of an organism<\/p>\n<p><strong>heterozygous: <\/strong>having two different alleles for a given gene on the homologous chromosomes<\/p>\n<p><strong>homozygous: <\/strong>having two identical alleles for a given gene on the homologous chromosomes<\/p>\n<p><strong>law of dominance: <\/strong>in a heterozygote, one trait will conceal the presence of another trait for the same characteristic<\/p>\n<p><strong>law of independent assortment: <\/strong>genes do not influence each other with regard to sorting of alleles into gametes; every\u00a0possible combination of alleles is equally likely to occur<\/p>\n<p><strong>law of segregation: <\/strong>paired unit factors (i.e., genes) segregate equally into gametes such that offspring have an equal likelihood of inheriting any combination of factors<\/p>\n<p><strong>monohybrid: <\/strong>the result of a cross between two true-breeding parents that express different traits for only one characteristic<\/p>\n<p><strong>phenotype: <\/strong>the observable traits expressed by an organism<\/p>\n<p><strong>Punnett square: <\/strong>a visual representation of a cross between two individuals in which the gametes of each individual are denoted along the top and side of a grid, respectively, and the possible zygotic genotypes are recombined at each box in the grid<\/p>\n<p><strong>test cross: <\/strong>a cross between a dominant expressing individual with an unknown genotype and a homozygous recessive individual; the offspring phenotypes indicate whether the unknown parent is heterozygous or homozygous for the dominant trait<\/p>\n<\/div>\n<p>\u00a0<\/p>\n<\/section>\n<div>\n<\/div>\n<\/section>\n","protected":false},"author":16,"menu_order":14,"template":"","meta":{"pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-235","chapter","type-chapter","status-publish","hentry"],"part":181,"_links":{"self":[{"href":"https:\/\/pressbooks.bccampus.ca\/conceptsofbiologygunness\/wp-json\/pressbooks\/v2\/chapters\/235","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/pressbooks.bccampus.ca\/conceptsofbiologygunness\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/pressbooks.bccampus.ca\/conceptsofbiologygunness\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/conceptsofbiologygunness\/wp-json\/wp\/v2\/users\/16"}],"version-history":[{"count":1,"href":"https:\/\/pressbooks.bccampus.ca\/conceptsofbiologygunness\/wp-json\/pressbooks\/v2\/chapters\/235\/revisions"}],"predecessor-version":[{"id":851,"href":"https:\/\/pressbooks.bccampus.ca\/conceptsofbiologygunness\/wp-json\/pressbooks\/v2\/chapters\/235\/revisions\/851"}],"part":[{"href":"https:\/\/pressbooks.bccampus.ca\/conceptsofbiologygunness\/wp-json\/pressbooks\/v2\/parts\/181"}],"metadata":[{"href":"https:\/\/pressbooks.bccampus.ca\/conceptsofbiologygunness\/wp-json\/pressbooks\/v2\/chapters\/235\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/pressbooks.bccampus.ca\/conceptsofbiologygunness\/wp-json\/wp\/v2\/media?parent=235"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/conceptsofbiologygunness\/wp-json\/pressbooks\/v2\/chapter-type?post=235"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/conceptsofbiologygunness\/wp-json\/wp\/v2\/contributor?post=235"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/pressbooks.bccampus.ca\/conceptsofbiologygunness\/wp-json\/wp\/v2\/license?post=235"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}