{"id":241,"date":"2014-11-12T22:02:18","date_gmt":"2014-11-12T22:02:18","guid":{"rendered":"http:\/\/courses.candelalearning.com\/novabiology\/?post_type=chapter&#038;p=241"},"modified":"2018-07-13T19:49:46","modified_gmt":"2018-07-13T19:49:46","slug":"characteristics-and-traits","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/nemcc-biology1v2\/chapter\/characteristics-and-traits\/","title":{"raw":"Characteristics and Traits","rendered":"Characteristics and Traits"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Objectives<\/h3>\r\nBy the end of this section, you will be able to:\r\n<ul>\r\n \t<li>Define basic genetic terms<\/li>\r\n \t<li>Develop a Punnett square and calculate the expected proportions in a monohybrid cross<\/li>\r\n \t<li>Explain the purpose and methods of a test cross<\/li>\r\n \t<li>Identify several non-Mendelian inheritance patterns<\/li>\r\n<\/ul>\r\n<\/div>\r\nThe seven characteristics that Mendel evaluated in his pea plants were each expressed as one of two versions, or traits. The physical expression of characteristics is accomplished through gene expression carried on chromosomes. The genetic makeup of peas consists of two similar, or homologous, copies of each chromosome, one from each parent. Each pair of homologous chromosomes has the same order of genes.\u00a0 Diploid organisms utilize meiosis to produce haploid gametes, which contain one copy of each homologous chromosome that unite at fertilization creating a diploid zygote.\r\n\r\nFor 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 a characteristic. Alternate forms of a gene are known as <span style=\"text-decoration: underline\">alleles<\/span>.\u00a0 An allele will have the same genes in same corresponding area(loci) on a pair of homologous chromosomes.\u00a0 Although Mendel only examined the inheritance of genes with just two allele forms, it is common to encounter more than two alleles for any given gene in a natural population.\r\n<h2>Phenotypes and Genotypes<\/h2>\r\nTwo 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 <b>phenotype<\/b>. An organism\u2019s underlying genetic makeup, both physically visible and non-expressed alleles, is called its <b>genotype<\/b>. Mendel\u2019s hybridization experiments demonstrate the difference between phenotype and genotype. When true-breeding plants, one with yellow pods and one with green pods, were cross-fertilized, all of the F<sub>1<\/sub> hybrid offspring had yellow pods. The hybrid offspring were phenotypically identical to the true-breeding parent with yellow pods. We know that the allele donated by the parent with green pods was not lost because it reappeared in some of the F<sub>2<\/sub> offspring. The F<sub>1<\/sub> plants must have been genotypically different from the parent with yellow pods.\r\n\r\nThe P<sub>1<\/sub> plants used in Mendel's experiments were identical for the trait being studied. Diploid organisms that are <b>homozygous<\/b> at a given gene, or locus, have two identical alleles for that gene on their homologous chromosomes. Mendel\u2019s parental pea plants always bred true because both of the gametes produced carried the same trait. When P<sub>1<\/sub> plants with contrasting traits were cross-fertilized, all of the offspring were <b>heterozygous<\/b> for the contrasting trait.\u00a0 Their genotype reflected different alleles for the gene being examined.\r\n<h3>Dominant and Recessive Alleles<\/h3>\r\nIn Mendel's experiments, why were the F<sub>1 <\/sub>offspring identical to one of the parents? \u00a0 In all seven pea-plant characteristics, one of the two contrasting alleles was dominant, and the other was recessive. We know they are actually genes on homologous chromosome pairs. For a gene expression, homozygous dominant and heterozygous organisms will have the same phenotype but differ in genotype.\u00a0 The recessive allele will only be observed in homozygous recessive individuals, or when the dominant allele is absent.\u00a0 Table 1 shows a few of the disorders inherited by these dominant and recessive alleles.\r\n<table>\r\n<thead>\r\n<tr>\r\n<th colspan=\"2\">Table 1. Human Inheritance in Dominant and Recessive Patterns<\/th>\r\n<\/tr>\r\n<tr>\r\n<th>Dominant Traits<\/th>\r\n<th>Recessive Traits<\/th>\r\n<\/tr>\r\n<\/thead>\r\n<tbody>\r\n<tr>\r\n<td>Achondroplasia<\/td>\r\n<td>Albinism<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Brachydactyly<\/td>\r\n<td>Cystic fibrosis<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Huntington\u2019s disease<\/td>\r\n<td>Duchenne muscular dystrophy<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Marfan syndrome<\/td>\r\n<td>Galactosemia<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Neurofibromatosis<\/td>\r\n<td>Phenylketonuria<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Widow\u2019s peak<\/td>\r\n<td>Sickle-cell anemia<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Wooly hair<\/td>\r\n<td>Tay-Sachs disease<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\nSeveral methods exist for referring to genes and alleles. For our purposes, we will abbreviate genes using the first letter of the gene\u2019s corresponding dominant trait. For example, violet is the dominant trait for a pea plant\u2019s flower color, so the flower-color gene would be abbreviated as <em>V<\/em>.\u00a0 We will use uppercase and lowercase letters to represent dominant and recessive alleles, respectively. Therefore, we would refer to the genotype of a homozygous dominant pea plant with violet flowers as <em>VV<\/em>, a heterozygous pea plant with violet flowers as Vv, and a homozygous recessive pea plant with white flowers as <em>vv.<\/em>\r\n<h2>The Punnett Square Approach for a Monohybrid Cross<\/h2>\r\nWhen fertilization between two true-breeding parents differing in only one characteristic occurs, the process is known as a<b>monohybrid<\/b> cross with the resulting offspring being monohybrids. Mendel performed seven monohybrid crosses for each plant characteristic. Based on his results, Mendel believed that each parent contributed one of two paired factors to each offspring with every possible combination of factors equally likely.\r\n\r\nConsider the case of true-breeding pea plants with yellow versus green pea seeds. The dominant seed color is yellow, so the parental genotypes were <em>YY<\/em> for the plants with yellow seeds and <em>yy<\/em> for the plants with green seeds, respectively. A <b>Punnett square<\/b>, devised by the British geneticist Reginald Punnett, can be drawn applying the rules of probability to predict the possible outcomes of a genetic cross and their expected frequencies. To prepare a Punnett square, all possible combinations of the parental alleles are listed along the top (for one parent) and side (for the other parent) of a grid.\u00a0 This arrangement represents their meiotic segregation into haploid gametes. The combinations of egg and sperm are placed in the boxes to show which alleles are combining. Each box represents the diploid genotype of a zygote, or fertilized egg, that could result from this mating. Because each possibility is equally likely, genotypic ratios can be determined from a Punnett square. If the pattern of inheritance (dominant or recessive) is known, the phenotypic ratios can understood. In a monohybrid cross of two true-breeding parents, each parent contributes one type of allele. In our example, only one genotype is possible. All offspring are <em>Yy(genotype)<\/em> and have yellow seeds(phenotype) (Figure\u00a01).\r\n\r\n[caption id=\"attachment_1445\" align=\"aligncenter\" width=\"544\"]<img class=\"size-full wp-image-1445\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/198\/2016\/11\/28184030\/Figure_12_02_01.jpg\" alt=\"This illustration shows a monohybrid cross. In the P generation, one parent has a dominant yellow phenotype and the genotype YY, and the other parent has the recessive green phenotype and the genotype yy. Each parent produces one kind of gamete, resulting in an F_{1} generation with a dominant yellow phenotype and the genotype Yy. Self-pollination of the F_{1} generation results in an F_{2} generation with a 3 to 1 ratio of yellow to green peas. One out of three of the yellow pea plants has a dominant genotype of YY, and 2 out of 3 have the heterozygous phenotype Yy. The homozygous recessive plant has the green phenotype and the genotype yy.\" width=\"544\" height=\"784\" \/> Figure\u00a01. In the P generation, pea plants that are true-breeding for the dominant yellow phenotype are crossed with plants with the recessive green phenotype. This cross produces F<sub>1<\/sub> heterozygotes with a yellow phenotype. Punnett square analysis can be used to predict the genotypes of the F<sub>2<\/sub> generation.[\/caption]\r\n\r\nWhat result would occur is we self-pollinated the<sub> \u00a0<\/sub>F<sub>1<\/sub> offspring?\u00a0 Each parent can donate one of two different alleles. Therefore, the offspring can potentially have one of four allele combinations: <em>YY<\/em>, <em>Yy<\/em>, <em>yY<\/em>, or <em>yy<\/em> . There are two ways to obtain the <em>Yy<\/em> genotype: a <em>Y<\/em> from the egg and a <em>y<\/em> from the sperm, or a <em>y<\/em> from the egg and a <em>Y<\/em> from the sperm. Both possibilities must be counted. The two possible heterozygous combinations produce offspring that are genotypically and phenotypically identical, despite the origin of the alleles. \u00a0 Because fertilization is rando, we expect each combination to be equally likely and for the offspring to exhibit a ratio of <em>YY<\/em>:<em>Yy<\/em>:<em>yy<\/em> genotypes of 1:2:1 (Figure\u00a01). Because the <em>YY<\/em> and <em>Yy<\/em> offspring have yellow seeds and are phenotypically identical, applying the sum rule of probability, we expect the offspring to exhibit a phenotypic ratio of 3 yellow:1 green.\u00a0 Working with large sample sizes, Mendel observed this ratio in every F<sub>2<\/sub> generation resulting from crosses for individual traits.\r\n<h3>Test Cross<\/h3>\r\nMendel developed a way to determine whether an organism that expressed a dominant trait was homozygous or heterozygous. Called a <b>test cross<\/b>, this technique is still used today by plant and animal breeders. In a test cross, the dominant-expressing organism is crossed with an organism that is homozygous recessive for the same characteristic. \u00a0 If the dominant-expressing organism is a homozygote, then all F<sub>1<\/sub> offspring will be heterozygotes expressing the dominant trait (Figure\u00a02).\u00a0 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 (Figure\u00a02). The test cross further validates Mendel\u2019s belief that pairs of factors segregate equally.\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Art Connection<\/h3>\r\n[caption id=\"attachment_1447\" align=\"aligncenter\" width=\"544\"]<img class=\"size-full wp-image-1447\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/198\/2016\/11\/28184115\/Figure_12_02_02.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 of the resulting offspring will have at least one dominant allele. If the parent with the unknown phenotype is heterozygous, fifty percent of the offspring will inherit a recessive allele from both parents and will have the recessive phenotype.\" width=\"544\" height=\"675\" \/> Figure\u00a02.\u00a0 A test cross can be performed to determine whether an organism expressing a dominant trait is a homozygote or a heterozygote.[\/caption]\r\n\r\nIn 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 round pea parent plant is homozygous dominant or heterozygous? If the round pea parent plant is heterozygous, what is the probability that a random sample of 3 progeny peas will all be round?\r\n\r\n<\/div>\r\n<h2><strong>PEDIGREE CHARTS<\/strong><\/h2>\r\nMany human diseases are inherited through our genetics. \u00a0 What are my chances of inheriting a particular disease? \u00a0 Can I pass defective genes to my offspring?\u00a0 Geneticists use pedigree charts to study inheritance patterns and help answer these types of questions.\u00a0 Figure\u00a0 3 shows a pedigree chart for alkaptonuria.\u00a0 Individuals with this disorder have dark skin, brown urine, and suffer joint damage.\u00a0 For our purposes, we will discuss three inheritance patterns in pedigree analysis.\r\n\r\n(1)\u00a0 autosomal recessive - affected child may have neither parent affected(carriers); if both parents affected then all children affected\r\n\r\n(2)\u00a0 autosomal dominant - unaffected child has affected parents; affected child has at least one affected parent; all heterozygotes affected\r\n\r\n(3)\u00a0 X-linked recessive - affected son can have normal parents; if mom affected then all sons affected; more males than females affected\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Art Connection<\/h3>\r\n[caption id=\"attachment_1448\" align=\"aligncenter\" width=\"544\"]<img class=\"size-full wp-image-1448\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/198\/2016\/11\/28184146\/Figure_12_02_03.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 \u201c3.\u201d The first generation father is affected and is labeled \u201c1.\u201d The first generation mother is unaffected and is labeled \u201c2.\u201d The Art Connection question asks the genotype of the three numbered individuals. \" width=\"544\" height=\"478\" \/> Figure\u00a03. In this pedigree analysis for alkaptonuria, individuals with the disorder are indicated in blue and have the genotype aa. Unaffected individuals are indicated in yellow and have the genotype AA or Aa.[\/caption]\r\n\r\nAlkaptonuria is a recessive genetic disorder in which two amino acids are not properly metabolized. Affected individuals may have darkened skin, brown urine,\u00a0 and may suffer joint damage and other complications. Looking at Figure\u00a03, we can see that it is often possible to determine a person\u2019s genotype from the genotype of their offspring. 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.\r\n\r\nWhat are the genotypes of the individuals labeled 1, 2 and 3?\r\n\r\n<\/div>\r\n<h2>BEYOND MENDEL<\/h2>\r\n<span style=\"text-decoration: underline\">Incomplete Dominance<\/span>\r\n\r\n[caption id=\"attachment_1449\" align=\"alignright\" width=\"350\"]<img class=\" wp-image-1449\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/198\/2016\/11\/28184218\/Figure_12_02_04.jpg\" alt=\"Photo is of a snapdragon with a pink flower.\" width=\"350\" height=\"466\" \/> Figure\u00a04. These pink flowers of a heterozygote snapdragon result from incomplete dominance. (credit: \u201cstorebukkebruse\u201d\/Flickr)[\/caption]\r\n\r\nMendel\u2019s results contradicted the view that offspring exhibited a blend of their parents\u2019 traits. But sometimes blending does exist.\u00a0 In the snapdragon(Figure 4), a cross between a homozygous parent with white flowers (<em>C<sup class=\"sup\">W<\/sup>C<sup class=\"sup\">W<\/sup><\/em>) and a homozygous parent with red flowers (<em>C<sup class=\"sup\">R<\/sup>C<sup class=\"sup\">R<\/sup><\/em>) will produce offspring with pink flowers (<em>C<sup class=\"sup\">R<\/sup>C<sup class=\"sup\">W<\/sup><\/em>).\u00a0 Note that different genotypic abbreviations are used to distinguish these patterns from basic Mendelian genetics. This pattern of inheritance is described as <b>incomplete dominance<\/b>, where the heterozygote is an intermediate between the dominant and recessive traits.\u00a0 The allele for red flowers is incompletely dominant over the allele for white flowers.\u00a0 The results of a heterozygote self-cross can still be predicted.\u00a0 In this case, the genotypic ratio would be 1 <em>C<sup class=\"sup\">R<\/sup>C<sup class=\"sup\">R<\/sup><\/em>:2 <em>C<sup class=\"sup\">R<\/sup>C<sup class=\"sup\">W<\/sup><\/em>:1 <em>C<sup class=\"sup\">W<\/sup>C<sup class=\"sup\">W<\/sup><\/em>, and the phenotypic ratio would be 1:2:1 for red:pink:white.\r\n\r\n<span style=\"text-decoration: underline\">Codominance<\/span>\r\n\r\nA variation on incomplete dominance is <b>codominance<\/b>.\u00a0 Codominance allows both alleles for the same characteristic to be expressed in the heterozygote. An example of codominance is familial hypercholesterolemia.\u00a0 Individuals that inherit this as a heterozygote may exhibit cardiovascular disease early in life(age 30) due to high cholesterol deposits.\u00a0 Other health issues may arise as the cholesterol also deposits in the skin, tendons, and corneas. Most of us inherit as a homozygous dominant case. Recessive inheritance of this disorder results in early death, generally as an infant.\r\n\r\n<span style=\"text-decoration: underline\">Multiple Alleles\u00a0<\/span>\r\n\r\nMendel thought that only two alleles could exist for a given gene.\u00a0 Although individuals can only have two alleles for a given gene, multiple alleles exist and are observed. An example of multiple alleles is in the coat color of rabbits (Figure\u00a05).\u00a0 Four alleles exist for the <em>c<\/em> gene. The wild-type version, <em>C<sup class=\"sup\">+<\/sup>C<sup class=\"sup\">+<\/sup><\/em>, is expressed as brown fur. The chinchilla phenotype, <em>c<sup class=\"sup\">ch<\/sup>c<sup class=\"sup\">ch<\/sup><\/em>, is shown as black-tipped white fur. The Himalayan phenotype, <em>c<sup class=\"sup\">h<\/sup>c<sup class=\"sup\">h<\/sup><\/em>, has white fur with black furred extremities . Finally, the albino, or \u201ccolorless\u201d phenotype, <em>cc<\/em>, is expressed as white fur.\u00a0 The wild-type allele is dominant over all the others, chinchilla is incompletely dominant over Himalayan and albino, and Himalayan is dominant over albino. By observing the phenotypes of each possible heterozygote offspring, it is easy to understand this dominance hierarchy.\r\n\r\n[caption id=\"attachment_1450\" align=\"aligncenter\" width=\"800\"]<img class=\"size-full wp-image-1450\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/198\/2016\/11\/28184300\/Figure_12_02_05.jpg\" alt=\"This illustration shows the four different variants for coat color in rabbits at the c allele. The genotype CC produces the wild type phenotype, which is brown. The genotype c^{ch}c^{ch} produces the chinchilla phenotype, which is black-tipped white fur. The genotype c^{h}c^{h} produces the Himalayan phenotype, which is white on the body and black on the extremities. The genotype cc produces the recessive phenotype, which is white\" width=\"800\" height=\"574\" \/> Figure\u00a05. Four different alleles exist for the rabbit coat color (C) gene.[\/caption]\r\n\r\n[caption id=\"attachment_1451\" align=\"alignright\" width=\"350\"]<img class=\" wp-image-1451\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/198\/2016\/11\/28184330\/Figure_12_02_06.jpg\" alt=\"This photo shows Drosophila that has normal antennae on its head, and a mutant that has legs on its head.\" width=\"350\" height=\"345\" \/> Figure\u00a06. As seen in comparing the wild-type Drosophila (left) and the Antennapedia mutant (right), the Antennapedia mutant has legs on its head in place of antennae.[\/caption]\r\n\r\nThe complete dominance of one phenotype over all other often occurs as an effect of \u201cdosage\u201d of a specific gene product.\u00a0 In rabbits, one allele may supply a given dosage of fur pigment, while the others receive a lesser dosage or none at all. Interestingly, the Himalayan phenotype is the result of an allele producing a temperature-sensitive gene product that only produces pigment in the cooler extremities of the rabbit\u2019s body.\r\n\r\nAlternatively, one mutant allele can be dominant over all other phenotypes.\u00a0 This may occur when the mutant allele somehow interferes with the genetic message so that even a heterozygote expresses the mutant phenotype. The mutant allele can interfere by enhancing the function of a gene product or changing its distribution in the body. One example is in <i>Drosophila<\/i>, the fruit fly. In this case, the mutant allele expands the distribution of the gene product and the heterozygote develops legs on its head instead of antennae(Figure 6).\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Evolution Connection<\/h3>\r\n<h4>Multiple Alleles Confer Drug Resistance in the Malaria Parasite<\/h4>\r\nMalaria is a parasitic disease in humans that is transmitted by infected female mosquitoes, including <em>Anopheles gambiae<\/em> (Figure7a), and is characterized by\u00a0 high fevers, chills, flu-like symptoms, and severe anemia. <em>Plasmodium falciparum<\/em> and <em>P. vivax<\/em> are the most common causative agents of malaria, and <em>P. falciparum<\/em> is the most deadly (Figure\u00a07b)<em>.<\/em> When promptly and correctly treated, <em>P. falciparum<\/em> malaria has a mortality rate of 0.1 percent. However, in some parts of the world, the parasite has evolved resistance to commonly used malaria treatments, so the most effective malarial treatments can vary by geographic region.\r\n\r\n[caption id=\"attachment_1452\" align=\"aligncenter\" width=\"707\"]<img class=\"size-full wp-image-1452\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/198\/2016\/11\/28184426\/Figure_12_02_07.png\" alt=\"Photo a shows the Anopheles gambiae mosquito, which carries malaria. Photo b shows a micrograph of sickle-shaped Plasmodium falciparum, the parasite that causes malaria. The Plasmodium is about 0.75 microns across.\" width=\"707\" height=\"326\" \/> Figure\u00a07. The (a) Anopheles gambiae, or African malaria mosquito, acts as a vector in the transmission to humans of the malaria-causing parasite (b) Plasmodium falciparum, here visualized using false-color transmission electron microscopy. (credit a: James D. Gathany; credit b: Ute Frevert; false color by Margaret Shear; scale-bar data from Matt Russell)[\/caption]\r\n\r\n<\/div>\r\n<span style=\"text-decoration: underline\">X-linked Traits<\/span>\r\n\r\nIn many life forms, the sex of the individual is determined by sex chromosomes. The sex chromosomes are one pair of non-homologous chromosomes. In our previous discussions,\u00a0 we have considered inheritance of the autosomes.\u00a0 Autosomes are all chromosomes of inheritance except sex. In addition to 22 homologous pairs of autosomes, human females have a homologous pair of X chromosomes.\u00a0 Human males have an XY chromosome pair. Although the Y chromosome contains a small region of similarity to the X chromosome allowing them to pair during meiosis, the Y chromosome is much shorter and contains fewer genes. When a gene being examined is present on the X chromosome, but not on the Y chromosome, it is said to be <b>X-linked<\/b>.\r\n\r\n[caption id=\"attachment_1453\" align=\"alignright\" width=\"350\"]<img class=\"wp-image-1453\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/198\/2016\/11\/28184500\/Figure_12_02_08.jpeg\" alt=\"Photo shows six fruit flies, each with a different eye color.\" width=\"350\" height=\"439\" \/> Figure\u00a08. In Drosophila, the gene for eye color is located on the X chromosome. Clockwise from top left are brown, cinnabar, sepia, vermilion, white, and red. Red eye color is wild-type and is dominant to white eye color.[\/caption]\r\n\r\nEye color in <em>Drosophila<\/em> was one of the first X-linked traits to be identified. \u00a0 In 1910, Thomas Hunt Morgan mapped this trait to the X chromosome.\u00a0 Like humans, <em>Drosophila<\/em> males have an XY chromosome pair, and females are XX. In flies, the wild-type eye color is red (X<sup class=\"sup\"><em>W<\/em><\/sup>) and is dominant to white eye color (X<sup class=\"sup\"><em>w<\/em><\/sup>) (Figure\u00a08). Because of the location of the eye-color gene, reciprocal crosses do not produce the same offspring ratios. Males are said to be <b>hemizygous<\/b>. They have only one allele for any X-linked characteristic. Hemizygosity makes the descriptions of dominance and recessiveness irrelevant for XY males. <em>Drosophila<\/em> males lack a second allele copy on the Y chromosome so their genotype can only be X<sup class=\"sup\"><em>W<\/em><\/sup>Y or X<sup class=\"sup\"><em>w<\/em><\/sup>Y. \u00a0 Females have two allele copies and can be X<sup class=\"sup\"><em>W<\/em><\/sup>X<sup class=\"sup\"><em>W<\/em><\/sup>, X<sup class=\"sup\"><em>W<\/em><\/sup>X<sup class=\"sup\"><em>w<\/em><\/sup>, or X<sup class=\"sup\"><em>w<\/em><\/sup>X<sup class=\"sup\"><em>w<\/em><\/sup>.\r\n\r\nIn an X-linked cross, the genotypes of F<sub>1<\/sub> and F<sub>2<\/sub> offspring depended on whether the recessive trait was expressed by the male or the female in the P<sub>1<\/sub> generation. With regard to <em>Drosophila<\/em> eye color, when the P<sub>1<\/sub> male expresses the white-eye phenotype and the female is homozygous red-eyed, all members of the F<sub>1<\/sub> generation exhibit red eyes. The F<sub>1<\/sub> females are heterozygous (X<em><sup class=\"sup\">W<\/sup><\/em>X<i><sup>w)<\/sup><\/i>) and the males are all X<em><sup class=\"sup\">W<\/sup><\/em>Y. A subsequent cross between the X<em><sup class=\"sup\">W<\/sup><\/em>X<em><sup class=\"sup\">w<\/sup><\/em> female and the X<em><sup class=\"sup\">W<\/sup><\/em>Y male would produce only red-eyed females (X<em><sup class=\"sup\">W<\/sup><\/em>X<em><sup class=\"sup\">W<\/sup><\/em> or X<em><sup class=\"sup\">W<\/sup><\/em>X<em><sup class=\"sup\">w<\/sup><\/em> genotypes) and both red- and white-eyed males (X<em><sup class=\"sup\">W<\/sup><\/em>Y or X<em><sup class=\"sup\">w<\/sup><\/em>Y genotypes).\u00a0 Consider a cross between a homozygous white-eyed female and a male with red eyes(Figure 9). The F<sub>1<\/sub> generation would exhibit only heterozygous red-eyed females (X<em><sup class=\"sup\">W<\/sup><\/em>X<em><sup class=\"sup\">w<\/sup><\/em>) and only white-eyed males (X<em><sup class=\"sup\">w<\/sup><\/em>Y). Half of the F<sub>2<\/sub> females would be red-eyed (X<em><sup class=\"sup\">W<\/sup><\/em>X<em><sup class=\"sup\">w<\/sup><\/em>) and half would be white-eyed (X<em><sup class=\"sup\">w<\/sup><\/em>X<em><sup class=\"sup\">w<\/sup><\/em>).\u00a0 Similarly, half of the F<sub>2<\/sub> males would be red-eyed (X<em><sup class=\"sup\">W<\/sup><\/em>Y) and half would be white-eyed (X<em><sup class=\"sup\">w<\/sup><\/em>Y).\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Art Connection<\/h3>\r\n[caption id=\"attachment_1454\" align=\"aligncenter\" width=\"725\"]<img class=\"size-full wp-image-1454\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/198\/2016\/11\/28184601\/Figure_12_02_09.jpeg\" alt=\"This illustration shows a Punnett square analysis of fruit fly eye color, which is a sex-linked trait. A red-eyed male fruit fly with the genotype X^{w}Y is crossed with a white-eyed female fruit fly with the genotype X^{w}X^{w}. All of the female offspring acquire a dominant W allele from the father and a recessive w allele from the mother, and are therefore heterozygous dominant with red eye color. All of the male offspring acquire a recessive w allele from the mother and a Y chromosome from the father and are therefore hemizygous recessive with white eye color.\" width=\"725\" height=\"729\" \/> Figure\u00a09. Punnett square analysis is used to determine the ratio of offspring from a cross between a red-eyed male fruit fly and a white-eyed female fruit fly.[\/caption]\r\n\r\nWhat ratio of offspring would result from a cross between a white-eyed male and a female that is heterozygous for red eye color?\r\n\r\n<\/div>\r\nWe can apply discoveries in fruit fly genetics to human genetics. When a female parent is homozygous for a recessive X-linked trait, she will pass the trait on to 100 percent of her offspring. Her male offspring are destined to express the trait, as they will inherit their father's Y chromosome. In humans, the alleles for certain conditions, such as some forms of color blindness, hemophilia, and muscular dystrophy),are X-linked. Heterozygous females of a disease are said to be carriers and may not exhibit any phenotypic effects. These females will pass the disease to half of their sons and will pass carrier status to half of their daughters.\u00a0 For this reason, recessive X-linked traits appear more frequently in males than females.\r\n<h2>Human Sex-linked Disorders<\/h2>\r\nSex-linkage studies provided the fundamentals for understanding X-linked recessive disorders in humans,\u00a0 including red-green color blindness, and Types A and B hemophilia. Because human males need to inherit only one recessive X allele to be affected, X-linked disorders are disproportionately observed in males. Females must inherit recessive X-linked alleles from both\u00a0 parents in order to express the trait. When they inherit one recessive X-linked allele and one dominant X-linked allele, they are carriers of the trait and are typically unaffected.\u00a0 However, female carriers can contribute the trait to their sons, resulting in the son exhibiting the trait.\u00a0 They can contribute the recessive allele to their daughters, resulting in the daughters being carriers of the trait (Figure\u00a010). Although some Y-linked recessive disorders exist, typically they are associated with male infertility and are not transmitted to subsequent generations.\r\n\r\n[caption id=\"attachment_1455\" align=\"aligncenter\" width=\"655\"]<img class=\"size-full wp-image-1455\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/198\/2016\/11\/28184635\/Figure_12_02_10.jpeg\" alt=\"A diagram shows an unaffected father with a dominant allele and an unaffected carrier mother with an x-linked recessive allele. Four figures of offspring are shown representing the various resulting genetic combinations: unaffected son, unaffected daughter, affected son, and unaffected carrier daughter.\" width=\"655\" height=\"719\" \/> Figure\u00a010. The son of a woman who is a carrier of a recessive X-linked disorder will have a 50 percent chance of being affected. A daughter will not be affected, but will have a 50 percent chance of being a carrier like her mother.[\/caption]\r\n\r\n<div class=\"textbox shaded\">\r\n<h3>Link to Learning<\/h3>\r\nWatch this video to learn more about sex-linked traits.\r\n\r\nhttps:\/\/youtu.be\/-ROhfKyxgCo\r\n\r\n<\/div>\r\n<h2>Lethality<\/h2>\r\nMany genes in an individual\u2019s genome are essential for survival. Occasionally, a nonfunctional allele for an essential gene can arise by mutation and be transmitted as long as individuals also have a functional copy. The functional allele works at a capacity sufficient to sustain life and is considered to be dominant over the nonfunctional allele.\u00a0 An inheritance pattern in which an allele is only lethal in the homozygous form is a recessive lethal.\u00a0 In this case, the heterozygote may be normal or have some altered non-lethal phenotype.\r\n\r\n[caption id=\"attachment_1456\" align=\"alignright\" width=\"300\"]<img class=\"wp-image-1456\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/198\/2016\/11\/28184712\/Figure_12_02_11.jpeg\" alt=\"Micrograph shows a neuron with nuclear inclusions characteristic of Huntington\u2019s disease.\" width=\"300\" height=\"389\" \/> Figure\u00a011. The neuron in the center of this micrograph (yellow) has nuclear inclusions characteristic of Huntington\u2019s disease (orange area in the center of the neuron). Huntington\u2019s disease occurs when an abnormal dominant allele for the Huntington gene is present. (credit: Dr. Steven Finkbeiner, Gladstone Institute of Neurological Disease, The Taube-Koret Center for Huntington's Disease Research, and the University of California San Francisco\/Wikimedia)[\/caption]\r\n\r\nA single copy of the wild-type allele is not always sufficient for normal functioning or survival. The <b>dominant lethal<\/b> inheritance pattern is when an allele is lethal in the homozygote and the heterozygote\u00a0 This allele is transmitted if the lethality phenotype occurs after reproductive age. Individuals with mutations that result in dominant lethal alleles fail to survive, even in the heterozygote form. Dominant lethal alleles are very rare.\u00a0 As you might expect, the allele only lasts one generation and therefore not transmitted. However, just as the recessive lethal allele might not immediately manifest the phenotype of death, dominant lethal alleles might not be expressed until adulthood. Once the individual reaches reproductive age, the allele may be unknowingly passed on, resulting in a delayed death in both generations.\u00a0 In humans, Huntington\u2019s disease, where the nervous system gradually wastes away, is an example of this dominant lethal allele(Figure\u00a011). People, heterozygous for Huntington allele (<em>Hh<\/em>), will inevitably develop the fatal disease. The onset of Huntington\u2019s disease may not occur until age 40, at which point the afflicted person may have already passed the allele to 50 percent of their offspring.\r\n<h2>Section Summary<\/h2>\r\nWhen true-breeding and homozygous recessive individuals that differ for a certain trait are crossed, all of the offspring will be heterozygotes for that trait.\u00a0 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.\u00a0 Using the product and sum rules, genetic outcomes can be approximated to help in our further understanding.\r\n\r\nAlleles do not always behave in dominant and recessive patterns. Incomplete dominance describes situations in which the heterozygote exhibits a phenotype that is intermediate between the homozygous phenotypes. Codominance describes the simultaneous expression of both of the alleles in the heterozygote. Although diploid organisms can only have two alleles for any given gene, it is common for more than two alleles of a gene to exist.\r\n\r\nIn many life forms, females have two X chromosomes and males have one X and one Y chromosome. Genes that are present on the X but not the Y chromosome are said to be X-linked, such that males only inherit one allele for the gene, and females inherit two. Some alleles can be lethal. Recessive lethal alleles are only lethal in homozygotes, but dominant lethal alleles are fatal in both homozygotes and heterozygotes.\r\n<div class=\"textbox exercises\">\r\n<h3>Additional Self Check Questions<\/h3>\r\n<ol>\r\n \t<li>In pea plants, round peas (<em>R<\/em>) are dominant to wrinkled peas (<em>r<\/em>).\u00a0 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. The cross gives you all round peas.\u00a0 Can you tell if the round pea parent plant is homozygous dominant or heterozygous?<\/li>\r\n \t<li>What is the tile given to Mendel?<\/li>\r\n \t<li>What ratio of offspring would result from a cross between a white-eyed male and a female that is heterozygous for red eye color?<\/li>\r\n \t<li>The gene for flower position in pea plants exists as axial or terminal alleles. Given that axial is dominant to terminal, list all of the possible F<sub>1<\/sub> and F<sub>2<\/sub> genotypes and phenotypes from a cross involving parents that are homozygous for each trait. Express genotypes with conventional genetic abbreviations.<\/li>\r\n \t<li>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?<\/li>\r\n \t<li>Can a human male be a carrier of red-green color blindness?<\/li>\r\n<\/ol>\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Answers<\/h3>\r\n<ol>\r\n \t<li>Based on your data, the round pea plant would be RR.<\/li>\r\n \t<li>Mendel is known as the \"father of genetics\".<\/li>\r\n \t<li>Half of the female offspring would be heterozygous (X<sup class=\"sup\"><em>W<\/em><\/sup>X<sup class=\"sup\"><em>w<\/em><\/sup>) with red eyes, and half would be homozygous recessive (X<sup class=\"sup\"><em>w<\/em><\/sup>X<sup class=\"sup\"><em>w<\/em><\/sup>) with white eyes. Half of the male offspring would be hemizygous dominant (X<sup class=\"sup\"><em>W<\/em><\/sup>Y) withe red eyes, and half would be hemizygous recessive (X<sup class=\"sup\"><em>w<\/em><\/sup>Y) with white eyes.<\/li>\r\n \t<li>Because axial is dominant, the gene would be designated as <em>A<\/em>. F<sub>1<\/sub> would be all heterozygous <em>Aa<\/em> with axial phenotype. F<sub>2<\/sub> would have possible genotypes of <em>AA<\/em>, <em>Aa<\/em>, and <em>aa<\/em>; these would correspond to axial, axial, and terminal phenotypes, respectively.<\/li>\r\n \t<li>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.<\/li>\r\n \t<li>No, males can only express color blindness. They cannot carry it because an individual needs two X chromosomes to be a carrier.<\/li>\r\n<\/ol>\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Objectives<\/h3>\n<p>By the end of this section, you will be able to:<\/p>\n<ul>\n<li>Define basic genetic terms<\/li>\n<li>Develop a Punnett square and calculate the expected proportions in a monohybrid cross<\/li>\n<li>Explain the purpose and methods of a test cross<\/li>\n<li>Identify several non-Mendelian inheritance patterns<\/li>\n<\/ul>\n<\/div>\n<p>The seven characteristics that Mendel evaluated in his pea plants were each expressed as one of two versions, or traits. The physical expression of characteristics is accomplished through gene expression carried on chromosomes. The genetic makeup of peas consists of two similar, or homologous, copies of each chromosome, one from each parent. Each pair of homologous chromosomes has the same order of genes.\u00a0 Diploid organisms utilize meiosis to produce haploid gametes, which contain one copy of each homologous chromosome that unite at fertilization creating a diploid zygote.<\/p>\n<p>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 a characteristic. Alternate forms of a gene are known as <span style=\"text-decoration: underline\">alleles<\/span>.\u00a0 An allele will have the same genes in same corresponding area(loci) on a pair of homologous chromosomes.\u00a0 Although Mendel only examined the inheritance of genes with just two allele forms, it is common to encounter more than two alleles for any given gene in a natural population.<\/p>\n<h2>Phenotypes and Genotypes<\/h2>\n<p>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 <b>phenotype<\/b>. An organism\u2019s underlying genetic makeup, both physically visible and non-expressed alleles, is called its <b>genotype<\/b>. Mendel\u2019s hybridization experiments demonstrate the difference between phenotype and genotype. When true-breeding plants, one with yellow pods and one with green pods, were cross-fertilized, all of the F<sub>1<\/sub> hybrid offspring had yellow pods. The hybrid offspring were phenotypically identical to the true-breeding parent with yellow pods. We know that the allele donated by the parent with green pods was not lost because it reappeared in some of the F<sub>2<\/sub> offspring. The F<sub>1<\/sub> plants must have been genotypically different from the parent with yellow pods.<\/p>\n<p>The P<sub>1<\/sub> plants used in Mendel&#8217;s experiments were identical for the trait being studied. Diploid organisms that are <b>homozygous<\/b> at a given gene, or locus, have two identical alleles for that gene on their homologous chromosomes. Mendel\u2019s parental pea plants always bred true because both of the gametes produced carried the same trait. When P<sub>1<\/sub> plants with contrasting traits were cross-fertilized, all of the offspring were <b>heterozygous<\/b> for the contrasting trait.\u00a0 Their genotype reflected different alleles for the gene being examined.<\/p>\n<h3>Dominant and Recessive Alleles<\/h3>\n<p>In Mendel&#8217;s experiments, why were the F<sub>1 <\/sub>offspring identical to one of the parents? \u00a0 In all seven pea-plant characteristics, one of the two contrasting alleles was dominant, and the other was recessive. We know they are actually genes on homologous chromosome pairs. For a gene expression, homozygous dominant and heterozygous organisms will have the same phenotype but differ in genotype.\u00a0 The recessive allele will only be observed in homozygous recessive individuals, or when the dominant allele is absent.\u00a0 Table 1 shows a few of the disorders inherited by these dominant and recessive alleles.<\/p>\n<table>\n<thead>\n<tr>\n<th colspan=\"2\">Table 1. Human Inheritance in Dominant and Recessive Patterns<\/th>\n<\/tr>\n<tr>\n<th>Dominant Traits<\/th>\n<th>Recessive Traits<\/th>\n<\/tr>\n<\/thead>\n<tbody>\n<tr>\n<td>Achondroplasia<\/td>\n<td>Albinism<\/td>\n<\/tr>\n<tr>\n<td>Brachydactyly<\/td>\n<td>Cystic fibrosis<\/td>\n<\/tr>\n<tr>\n<td>Huntington\u2019s disease<\/td>\n<td>Duchenne muscular dystrophy<\/td>\n<\/tr>\n<tr>\n<td>Marfan syndrome<\/td>\n<td>Galactosemia<\/td>\n<\/tr>\n<tr>\n<td>Neurofibromatosis<\/td>\n<td>Phenylketonuria<\/td>\n<\/tr>\n<tr>\n<td>Widow\u2019s peak<\/td>\n<td>Sickle-cell anemia<\/td>\n<\/tr>\n<tr>\n<td>Wooly hair<\/td>\n<td>Tay-Sachs disease<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<p>Several methods exist for referring to genes and alleles. For our purposes, we will abbreviate genes using the first letter of the gene\u2019s corresponding dominant trait. For example, violet is the dominant trait for a pea plant\u2019s flower color, so the flower-color gene would be abbreviated as <em>V<\/em>.\u00a0 We will use uppercase and lowercase letters to represent dominant and recessive alleles, respectively. Therefore, we would refer to the genotype of a homozygous dominant pea plant with violet flowers as <em>VV<\/em>, a heterozygous pea plant with violet flowers as Vv, and a homozygous recessive pea plant with white flowers as <em>vv.<\/em><\/p>\n<h2>The Punnett Square Approach for a Monohybrid Cross<\/h2>\n<p>When fertilization between two true-breeding parents differing in only one characteristic occurs, the process is known as a<b>monohybrid<\/b> cross with the resulting offspring being monohybrids. Mendel performed seven monohybrid crosses for each plant characteristic. Based on his results, Mendel believed that each parent contributed one of two paired factors to each offspring with every possible combination of factors equally likely.<\/p>\n<p>Consider the case of true-breeding pea plants with yellow versus green pea seeds. The dominant seed color is yellow, so the parental genotypes were <em>YY<\/em> for the plants with yellow seeds and <em>yy<\/em> for the plants with green seeds, respectively. A <b>Punnett square<\/b>, devised by the British geneticist Reginald Punnett, can be drawn applying the rules of probability to predict the possible outcomes of a genetic cross and their expected frequencies. To prepare a Punnett square, all possible combinations of the parental alleles are listed along the top (for one parent) and side (for the other parent) of a grid.\u00a0 This arrangement represents their meiotic segregation into haploid gametes. The combinations of egg and sperm are placed in the boxes to show which alleles are combining. Each box represents the diploid genotype of a zygote, or fertilized egg, that could result from this mating. Because each possibility is equally likely, genotypic ratios can be determined from a Punnett square. If the pattern of inheritance (dominant or recessive) is known, the phenotypic ratios can understood. In a monohybrid cross of two true-breeding parents, each parent contributes one type of allele. In our example, only one genotype is possible. All offspring are <em>Yy(genotype)<\/em> and have yellow seeds(phenotype) (Figure\u00a01).<\/p>\n<div id=\"attachment_1445\" style=\"width: 554px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-1445\" class=\"size-full wp-image-1445\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/198\/2016\/11\/28184030\/Figure_12_02_01.jpg\" alt=\"This illustration shows a monohybrid cross. In the P generation, one parent has a dominant yellow phenotype and the genotype YY, and the other parent has the recessive green phenotype and the genotype yy. Each parent produces one kind of gamete, resulting in an F_{1} generation with a dominant yellow phenotype and the genotype Yy. Self-pollination of the F_{1} generation results in an F_{2} generation with a 3 to 1 ratio of yellow to green peas. One out of three of the yellow pea plants has a dominant genotype of YY, and 2 out of 3 have the heterozygous phenotype Yy. The homozygous recessive plant has the green phenotype and the genotype yy.\" width=\"544\" height=\"784\" \/><\/p>\n<p id=\"caption-attachment-1445\" class=\"wp-caption-text\">Figure\u00a01. In the P generation, pea plants that are true-breeding for the dominant yellow phenotype are crossed with plants with the recessive green phenotype. This cross produces F<sub>1<\/sub> heterozygotes with a yellow phenotype. Punnett square analysis can be used to predict the genotypes of the F<sub>2<\/sub> generation.<\/p>\n<\/div>\n<p>What result would occur is we self-pollinated the<sub> \u00a0<\/sub>F<sub>1<\/sub> offspring?\u00a0 Each parent can donate one of two different alleles. Therefore, the offspring can potentially have one of four allele combinations: <em>YY<\/em>, <em>Yy<\/em>, <em>yY<\/em>, or <em>yy<\/em> . There are two ways to obtain the <em>Yy<\/em> genotype: a <em>Y<\/em> from the egg and a <em>y<\/em> from the sperm, or a <em>y<\/em> from the egg and a <em>Y<\/em> from the sperm. Both possibilities must be counted. The two possible heterozygous combinations produce offspring that are genotypically and phenotypically identical, despite the origin of the alleles. \u00a0 Because fertilization is rando, we expect each combination to be equally likely and for the offspring to exhibit a ratio of <em>YY<\/em>:<em>Yy<\/em>:<em>yy<\/em> genotypes of 1:2:1 (Figure\u00a01). Because the <em>YY<\/em> and <em>Yy<\/em> offspring have yellow seeds and are phenotypically identical, applying the sum rule of probability, we expect the offspring to exhibit a phenotypic ratio of 3 yellow:1 green.\u00a0 Working with large sample sizes, Mendel observed this ratio in every F<sub>2<\/sub> generation resulting from crosses for individual traits.<\/p>\n<h3>Test Cross<\/h3>\n<p>Mendel developed a way to determine whether an organism that expressed a dominant trait was homozygous or heterozygous. Called a <b>test cross<\/b>, this technique is still used today by plant and animal breeders. In a test cross, the dominant-expressing organism is crossed with an organism that is homozygous recessive for the same characteristic. \u00a0 If the dominant-expressing organism is a homozygote, then all F<sub>1<\/sub> offspring will be heterozygotes expressing the dominant trait (Figure\u00a02).\u00a0 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 (Figure\u00a02). The test cross further validates Mendel\u2019s belief that pairs of factors segregate equally.<\/p>\n<div class=\"textbox key-takeaways\">\n<h3>Art Connection<\/h3>\n<div id=\"attachment_1447\" style=\"width: 554px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-1447\" class=\"size-full wp-image-1447\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/198\/2016\/11\/28184115\/Figure_12_02_02.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 of the resulting offspring will have at least one dominant allele. If the parent with the unknown phenotype is heterozygous, fifty percent of the offspring will inherit a recessive allele from both parents and will have the recessive phenotype.\" width=\"544\" height=\"675\" \/><\/p>\n<p id=\"caption-attachment-1447\" class=\"wp-caption-text\">Figure\u00a02.\u00a0 A test cross can be performed to determine whether an organism expressing a dominant trait is a homozygote or a heterozygote.<\/p>\n<\/div>\n<p>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 round pea parent plant is homozygous dominant or heterozygous? If the round pea parent plant is heterozygous, what is the probability that a random sample of 3 progeny peas will all be round?<\/p>\n<\/div>\n<h2><strong>PEDIGREE CHARTS<\/strong><\/h2>\n<p>Many human diseases are inherited through our genetics. \u00a0 What are my chances of inheriting a particular disease? \u00a0 Can I pass defective genes to my offspring?\u00a0 Geneticists use pedigree charts to study inheritance patterns and help answer these types of questions.\u00a0 Figure\u00a0 3 shows a pedigree chart for alkaptonuria.\u00a0 Individuals with this disorder have dark skin, brown urine, and suffer joint damage.\u00a0 For our purposes, we will discuss three inheritance patterns in pedigree analysis.<\/p>\n<p>(1)\u00a0 autosomal recessive &#8211; affected child may have neither parent affected(carriers); if both parents affected then all children affected<\/p>\n<p>(2)\u00a0 autosomal dominant &#8211; unaffected child has affected parents; affected child has at least one affected parent; all heterozygotes affected<\/p>\n<p>(3)\u00a0 X-linked recessive &#8211; affected son can have normal parents; if mom affected then all sons affected; more males than females affected<\/p>\n<div class=\"textbox key-takeaways\">\n<h3>Art Connection<\/h3>\n<div id=\"attachment_1448\" style=\"width: 554px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-1448\" class=\"size-full wp-image-1448\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/198\/2016\/11\/28184146\/Figure_12_02_03.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 \u201c3.\u201d The first generation father is affected and is labeled \u201c1.\u201d The first generation mother is unaffected and is labeled \u201c2.\u201d The Art Connection question asks the genotype of the three numbered individuals.\" width=\"544\" height=\"478\" \/><\/p>\n<p id=\"caption-attachment-1448\" class=\"wp-caption-text\">Figure\u00a03. In this pedigree analysis for alkaptonuria, individuals with the disorder are indicated in blue and have the genotype aa. Unaffected individuals are indicated in yellow and have the genotype AA or Aa.<\/p>\n<\/div>\n<p>Alkaptonuria is a recessive genetic disorder in which two amino acids are not properly metabolized. Affected individuals may have darkened skin, brown urine,\u00a0 and may suffer joint damage and other complications. Looking at Figure\u00a03, we can see that it is often possible to determine a person\u2019s genotype from the genotype of their offspring. 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<\/div>\n<h2>BEYOND MENDEL<\/h2>\n<p><span style=\"text-decoration: underline\">Incomplete Dominance<\/span><\/p>\n<div id=\"attachment_1449\" style=\"width: 360px\" class=\"wp-caption alignright\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-1449\" class=\"wp-image-1449\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/198\/2016\/11\/28184218\/Figure_12_02_04.jpg\" alt=\"Photo is of a snapdragon with a pink flower.\" width=\"350\" height=\"466\" \/><\/p>\n<p id=\"caption-attachment-1449\" class=\"wp-caption-text\">Figure\u00a04. These pink flowers of a heterozygote snapdragon result from incomplete dominance. (credit: \u201cstorebukkebruse\u201d\/Flickr)<\/p>\n<\/div>\n<p>Mendel\u2019s results contradicted the view that offspring exhibited a blend of their parents\u2019 traits. But sometimes blending does exist.\u00a0 In the snapdragon(Figure 4), a cross between a homozygous parent with white flowers (<em>C<sup class=\"sup\">W<\/sup>C<sup class=\"sup\">W<\/sup><\/em>) and a homozygous parent with red flowers (<em>C<sup class=\"sup\">R<\/sup>C<sup class=\"sup\">R<\/sup><\/em>) will produce offspring with pink flowers (<em>C<sup class=\"sup\">R<\/sup>C<sup class=\"sup\">W<\/sup><\/em>).\u00a0 Note that different genotypic abbreviations are used to distinguish these patterns from basic Mendelian genetics. This pattern of inheritance is described as <b>incomplete dominance<\/b>, where the heterozygote is an intermediate between the dominant and recessive traits.\u00a0 The allele for red flowers is incompletely dominant over the allele for white flowers.\u00a0 The results of a heterozygote self-cross can still be predicted.\u00a0 In this case, the genotypic ratio would be 1 <em>C<sup class=\"sup\">R<\/sup>C<sup class=\"sup\">R<\/sup><\/em>:2 <em>C<sup class=\"sup\">R<\/sup>C<sup class=\"sup\">W<\/sup><\/em>:1 <em>C<sup class=\"sup\">W<\/sup>C<sup class=\"sup\">W<\/sup><\/em>, and the phenotypic ratio would be 1:2:1 for red:pink:white.<\/p>\n<p><span style=\"text-decoration: underline\">Codominance<\/span><\/p>\n<p>A variation on incomplete dominance is <b>codominance<\/b>.\u00a0 Codominance allows both alleles for the same characteristic to be expressed in the heterozygote. An example of codominance is familial hypercholesterolemia.\u00a0 Individuals that inherit this as a heterozygote may exhibit cardiovascular disease early in life(age 30) due to high cholesterol deposits.\u00a0 Other health issues may arise as the cholesterol also deposits in the skin, tendons, and corneas. Most of us inherit as a homozygous dominant case. Recessive inheritance of this disorder results in early death, generally as an infant.<\/p>\n<p><span style=\"text-decoration: underline\">Multiple Alleles\u00a0<\/span><\/p>\n<p>Mendel thought that only two alleles could exist for a given gene.\u00a0 Although individuals can only have two alleles for a given gene, multiple alleles exist and are observed. An example of multiple alleles is in the coat color of rabbits (Figure\u00a05).\u00a0 Four alleles exist for the <em>c<\/em> gene. The wild-type version, <em>C<sup class=\"sup\">+<\/sup>C<sup class=\"sup\">+<\/sup><\/em>, is expressed as brown fur. The chinchilla phenotype, <em>c<sup class=\"sup\">ch<\/sup>c<sup class=\"sup\">ch<\/sup><\/em>, is shown as black-tipped white fur. The Himalayan phenotype, <em>c<sup class=\"sup\">h<\/sup>c<sup class=\"sup\">h<\/sup><\/em>, has white fur with black furred extremities . Finally, the albino, or \u201ccolorless\u201d phenotype, <em>cc<\/em>, is expressed as white fur.\u00a0 The wild-type allele is dominant over all the others, chinchilla is incompletely dominant over Himalayan and albino, and Himalayan is dominant over albino. By observing the phenotypes of each possible heterozygote offspring, it is easy to understand this dominance hierarchy.<\/p>\n<div id=\"attachment_1450\" style=\"width: 810px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-1450\" class=\"size-full wp-image-1450\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/198\/2016\/11\/28184300\/Figure_12_02_05.jpg\" alt=\"This illustration shows the four different variants for coat color in rabbits at the c allele. The genotype CC produces the wild type phenotype, which is brown. The genotype c^{ch}c^{ch} produces the chinchilla phenotype, which is black-tipped white fur. The genotype c^{h}c^{h} produces the Himalayan phenotype, which is white on the body and black on the extremities. The genotype cc produces the recessive phenotype, which is white\" width=\"800\" height=\"574\" \/><\/p>\n<p id=\"caption-attachment-1450\" class=\"wp-caption-text\">Figure\u00a05. Four different alleles exist for the rabbit coat color (C) gene.<\/p>\n<\/div>\n<div id=\"attachment_1451\" style=\"width: 360px\" class=\"wp-caption alignright\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-1451\" class=\"wp-image-1451\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/198\/2016\/11\/28184330\/Figure_12_02_06.jpg\" alt=\"This photo shows Drosophila that has normal antennae on its head, and a mutant that has legs on its head.\" width=\"350\" height=\"345\" \/><\/p>\n<p id=\"caption-attachment-1451\" class=\"wp-caption-text\">Figure\u00a06. As seen in comparing the wild-type Drosophila (left) and the Antennapedia mutant (right), the Antennapedia mutant has legs on its head in place of antennae.<\/p>\n<\/div>\n<p>The complete dominance of one phenotype over all other often occurs as an effect of \u201cdosage\u201d of a specific gene product.\u00a0 In rabbits, one allele may supply a given dosage of fur pigment, while the others receive a lesser dosage or none at all. Interestingly, the Himalayan phenotype is the result of an allele producing a temperature-sensitive gene product that only produces pigment in the cooler extremities of the rabbit\u2019s body.<\/p>\n<p>Alternatively, one mutant allele can be dominant over all other phenotypes.\u00a0 This may occur when the mutant allele somehow interferes with the genetic message so that even a heterozygote expresses the mutant phenotype. The mutant allele can interfere by enhancing the function of a gene product or changing its distribution in the body. One example is in <i>Drosophila<\/i>, the fruit fly. In this case, the mutant allele expands the distribution of the gene product and the heterozygote develops legs on its head instead of antennae(Figure 6).<\/p>\n<div class=\"textbox key-takeaways\">\n<h3>Evolution Connection<\/h3>\n<h4>Multiple Alleles Confer Drug Resistance in the Malaria Parasite<\/h4>\n<p>Malaria is a parasitic disease in humans that is transmitted by infected female mosquitoes, including <em>Anopheles gambiae<\/em> (Figure7a), and is characterized by\u00a0 high fevers, chills, flu-like symptoms, and severe anemia. <em>Plasmodium falciparum<\/em> and <em>P. vivax<\/em> are the most common causative agents of malaria, and <em>P. falciparum<\/em> is the most deadly (Figure\u00a07b)<em>.<\/em> When promptly and correctly treated, <em>P. falciparum<\/em> malaria has a mortality rate of 0.1 percent. However, in some parts of the world, the parasite has evolved resistance to commonly used malaria treatments, so the most effective malarial treatments can vary by geographic region.<\/p>\n<div id=\"attachment_1452\" style=\"width: 717px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-1452\" class=\"size-full wp-image-1452\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/198\/2016\/11\/28184426\/Figure_12_02_07.png\" alt=\"Photo a shows the Anopheles gambiae mosquito, which carries malaria. Photo b shows a micrograph of sickle-shaped Plasmodium falciparum, the parasite that causes malaria. The Plasmodium is about 0.75 microns across.\" width=\"707\" height=\"326\" \/><\/p>\n<p id=\"caption-attachment-1452\" class=\"wp-caption-text\">Figure\u00a07. The (a) Anopheles gambiae, or African malaria mosquito, acts as a vector in the transmission to humans of the malaria-causing parasite (b) Plasmodium falciparum, here visualized using false-color transmission electron microscopy. (credit a: James D. Gathany; credit b: Ute Frevert; false color by Margaret Shear; scale-bar data from Matt Russell)<\/p>\n<\/div>\n<\/div>\n<p><span style=\"text-decoration: underline\">X-linked Traits<\/span><\/p>\n<p>In many life forms, the sex of the individual is determined by sex chromosomes. The sex chromosomes are one pair of non-homologous chromosomes. In our previous discussions,\u00a0 we have considered inheritance of the autosomes.\u00a0 Autosomes are all chromosomes of inheritance except sex. In addition to 22 homologous pairs of autosomes, human females have a homologous pair of X chromosomes.\u00a0 Human males have an XY chromosome pair. Although the Y chromosome contains a small region of similarity to the X chromosome allowing them to pair during meiosis, the Y chromosome is much shorter and contains fewer genes. When a gene being examined is present on the X chromosome, but not on the Y chromosome, it is said to be <b>X-linked<\/b>.<\/p>\n<div id=\"attachment_1453\" style=\"width: 360px\" class=\"wp-caption alignright\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-1453\" class=\"wp-image-1453\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/198\/2016\/11\/28184500\/Figure_12_02_08.jpeg\" alt=\"Photo shows six fruit flies, each with a different eye color.\" width=\"350\" height=\"439\" \/><\/p>\n<p id=\"caption-attachment-1453\" class=\"wp-caption-text\">Figure\u00a08. In Drosophila, the gene for eye color is located on the X chromosome. Clockwise from top left are brown, cinnabar, sepia, vermilion, white, and red. Red eye color is wild-type and is dominant to white eye color.<\/p>\n<\/div>\n<p>Eye color in <em>Drosophila<\/em> was one of the first X-linked traits to be identified. \u00a0 In 1910, Thomas Hunt Morgan mapped this trait to the X chromosome.\u00a0 Like humans, <em>Drosophila<\/em> males have an XY chromosome pair, and females are XX. In flies, the wild-type eye color is red (X<sup class=\"sup\"><em>W<\/em><\/sup>) and is dominant to white eye color (X<sup class=\"sup\"><em>w<\/em><\/sup>) (Figure\u00a08). Because of the location of the eye-color gene, reciprocal crosses do not produce the same offspring ratios. Males are said to be <b>hemizygous<\/b>. They have only one allele for any X-linked characteristic. Hemizygosity makes the descriptions of dominance and recessiveness irrelevant for XY males. <em>Drosophila<\/em> males lack a second allele copy on the Y chromosome so their genotype can only be X<sup class=\"sup\"><em>W<\/em><\/sup>Y or X<sup class=\"sup\"><em>w<\/em><\/sup>Y. \u00a0 Females have two allele copies and can be X<sup class=\"sup\"><em>W<\/em><\/sup>X<sup class=\"sup\"><em>W<\/em><\/sup>, X<sup class=\"sup\"><em>W<\/em><\/sup>X<sup class=\"sup\"><em>w<\/em><\/sup>, or X<sup class=\"sup\"><em>w<\/em><\/sup>X<sup class=\"sup\"><em>w<\/em><\/sup>.<\/p>\n<p>In an X-linked cross, the genotypes of F<sub>1<\/sub> and F<sub>2<\/sub> offspring depended on whether the recessive trait was expressed by the male or the female in the P<sub>1<\/sub> generation. With regard to <em>Drosophila<\/em> eye color, when the P<sub>1<\/sub> male expresses the white-eye phenotype and the female is homozygous red-eyed, all members of the F<sub>1<\/sub> generation exhibit red eyes. The F<sub>1<\/sub> females are heterozygous (X<em><sup class=\"sup\">W<\/sup><\/em>X<i><sup>w)<\/sup><\/i>) and the males are all X<em><sup class=\"sup\">W<\/sup><\/em>Y. A subsequent cross between the X<em><sup class=\"sup\">W<\/sup><\/em>X<em><sup class=\"sup\">w<\/sup><\/em> female and the X<em><sup class=\"sup\">W<\/sup><\/em>Y male would produce only red-eyed females (X<em><sup class=\"sup\">W<\/sup><\/em>X<em><sup class=\"sup\">W<\/sup><\/em> or X<em><sup class=\"sup\">W<\/sup><\/em>X<em><sup class=\"sup\">w<\/sup><\/em> genotypes) and both red- and white-eyed males (X<em><sup class=\"sup\">W<\/sup><\/em>Y or X<em><sup class=\"sup\">w<\/sup><\/em>Y genotypes).\u00a0 Consider a cross between a homozygous white-eyed female and a male with red eyes(Figure 9). The F<sub>1<\/sub> generation would exhibit only heterozygous red-eyed females (X<em><sup class=\"sup\">W<\/sup><\/em>X<em><sup class=\"sup\">w<\/sup><\/em>) and only white-eyed males (X<em><sup class=\"sup\">w<\/sup><\/em>Y). Half of the F<sub>2<\/sub> females would be red-eyed (X<em><sup class=\"sup\">W<\/sup><\/em>X<em><sup class=\"sup\">w<\/sup><\/em>) and half would be white-eyed (X<em><sup class=\"sup\">w<\/sup><\/em>X<em><sup class=\"sup\">w<\/sup><\/em>).\u00a0 Similarly, half of the F<sub>2<\/sub> males would be red-eyed (X<em><sup class=\"sup\">W<\/sup><\/em>Y) and half would be white-eyed (X<em><sup class=\"sup\">w<\/sup><\/em>Y).<\/p>\n<div class=\"textbox key-takeaways\">\n<h3>Art Connection<\/h3>\n<div id=\"attachment_1454\" style=\"width: 735px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-1454\" class=\"size-full wp-image-1454\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/198\/2016\/11\/28184601\/Figure_12_02_09.jpeg\" alt=\"This illustration shows a Punnett square analysis of fruit fly eye color, which is a sex-linked trait. A red-eyed male fruit fly with the genotype X^{w}Y is crossed with a white-eyed female fruit fly with the genotype X^{w}X^{w}. All of the female offspring acquire a dominant W allele from the father and a recessive w allele from the mother, and are therefore heterozygous dominant with red eye color. All of the male offspring acquire a recessive w allele from the mother and a Y chromosome from the father and are therefore hemizygous recessive with white eye color.\" width=\"725\" height=\"729\" \/><\/p>\n<p id=\"caption-attachment-1454\" class=\"wp-caption-text\">Figure\u00a09. Punnett square analysis is used to determine the ratio of offspring from a cross between a red-eyed male fruit fly and a white-eyed female fruit fly.<\/p>\n<\/div>\n<p>What ratio of offspring would result from a cross between a white-eyed male and a female that is heterozygous for red eye color?<\/p>\n<\/div>\n<p>We can apply discoveries in fruit fly genetics to human genetics. When a female parent is homozygous for a recessive X-linked trait, she will pass the trait on to 100 percent of her offspring. Her male offspring are destined to express the trait, as they will inherit their father&#8217;s Y chromosome. In humans, the alleles for certain conditions, such as some forms of color blindness, hemophilia, and muscular dystrophy),are X-linked. Heterozygous females of a disease are said to be carriers and may not exhibit any phenotypic effects. These females will pass the disease to half of their sons and will pass carrier status to half of their daughters.\u00a0 For this reason, recessive X-linked traits appear more frequently in males than females.<\/p>\n<h2>Human Sex-linked Disorders<\/h2>\n<p>Sex-linkage studies provided the fundamentals for understanding X-linked recessive disorders in humans,\u00a0 including red-green color blindness, and Types A and B hemophilia. Because human males need to inherit only one recessive X allele to be affected, X-linked disorders are disproportionately observed in males. Females must inherit recessive X-linked alleles from both\u00a0 parents in order to express the trait. When they inherit one recessive X-linked allele and one dominant X-linked allele, they are carriers of the trait and are typically unaffected.\u00a0 However, female carriers can contribute the trait to their sons, resulting in the son exhibiting the trait.\u00a0 They can contribute the recessive allele to their daughters, resulting in the daughters being carriers of the trait (Figure\u00a010). Although some Y-linked recessive disorders exist, typically they are associated with male infertility and are not transmitted to subsequent generations.<\/p>\n<div id=\"attachment_1455\" style=\"width: 665px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-1455\" class=\"size-full wp-image-1455\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/198\/2016\/11\/28184635\/Figure_12_02_10.jpeg\" alt=\"A diagram shows an unaffected father with a dominant allele and an unaffected carrier mother with an x-linked recessive allele. Four figures of offspring are shown representing the various resulting genetic combinations: unaffected son, unaffected daughter, affected son, and unaffected carrier daughter.\" width=\"655\" height=\"719\" \/><\/p>\n<p id=\"caption-attachment-1455\" class=\"wp-caption-text\">Figure\u00a010. The son of a woman who is a carrier of a recessive X-linked disorder will have a 50 percent chance of being affected. A daughter will not be affected, but will have a 50 percent chance of being a carrier like her mother.<\/p>\n<\/div>\n<div class=\"textbox shaded\">\n<h3>Link to Learning<\/h3>\n<p>Watch this video to learn more about sex-linked traits.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-1\" title=\"Sex-linked traits | Biomolecules | MCAT | Khan Academy\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/-ROhfKyxgCo?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<\/div>\n<h2>Lethality<\/h2>\n<p>Many genes in an individual\u2019s genome are essential for survival. Occasionally, a nonfunctional allele for an essential gene can arise by mutation and be transmitted as long as individuals also have a functional copy. The functional allele works at a capacity sufficient to sustain life and is considered to be dominant over the nonfunctional allele.\u00a0 An inheritance pattern in which an allele is only lethal in the homozygous form is a recessive lethal.\u00a0 In this case, the heterozygote may be normal or have some altered non-lethal phenotype.<\/p>\n<div id=\"attachment_1456\" style=\"width: 310px\" class=\"wp-caption alignright\"><img loading=\"lazy\" decoding=\"async\" aria-describedby=\"caption-attachment-1456\" class=\"wp-image-1456\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/198\/2016\/11\/28184712\/Figure_12_02_11.jpeg\" alt=\"Micrograph shows a neuron with nuclear inclusions characteristic of Huntington\u2019s disease.\" width=\"300\" height=\"389\" \/><\/p>\n<p id=\"caption-attachment-1456\" class=\"wp-caption-text\">Figure\u00a011. The neuron in the center of this micrograph (yellow) has nuclear inclusions characteristic of Huntington\u2019s disease (orange area in the center of the neuron). Huntington\u2019s disease occurs when an abnormal dominant allele for the Huntington gene is present. (credit: Dr. Steven Finkbeiner, Gladstone Institute of Neurological Disease, The Taube-Koret Center for Huntington&#8217;s Disease Research, and the University of California San Francisco\/Wikimedia)<\/p>\n<\/div>\n<p>A single copy of the wild-type allele is not always sufficient for normal functioning or survival. The <b>dominant lethal<\/b> inheritance pattern is when an allele is lethal in the homozygote and the heterozygote\u00a0 This allele is transmitted if the lethality phenotype occurs after reproductive age. Individuals with mutations that result in dominant lethal alleles fail to survive, even in the heterozygote form. Dominant lethal alleles are very rare.\u00a0 As you might expect, the allele only lasts one generation and therefore not transmitted. However, just as the recessive lethal allele might not immediately manifest the phenotype of death, dominant lethal alleles might not be expressed until adulthood. Once the individual reaches reproductive age, the allele may be unknowingly passed on, resulting in a delayed death in both generations.\u00a0 In humans, Huntington\u2019s disease, where the nervous system gradually wastes away, is an example of this dominant lethal allele(Figure\u00a011). People, heterozygous for Huntington allele (<em>Hh<\/em>), will inevitably develop the fatal disease. The onset of Huntington\u2019s disease may not occur until age 40, at which point the afflicted person may have already passed the allele to 50 percent of their offspring.<\/p>\n<h2>Section Summary<\/h2>\n<p>When true-breeding and homozygous recessive individuals that differ for a certain trait are crossed, all of the offspring will be heterozygotes for that trait.\u00a0 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.\u00a0 Using the product and sum rules, genetic outcomes can be approximated to help in our further understanding.<\/p>\n<p>Alleles do not always behave in dominant and recessive patterns. Incomplete dominance describes situations in which the heterozygote exhibits a phenotype that is intermediate between the homozygous phenotypes. Codominance describes the simultaneous expression of both of the alleles in the heterozygote. Although diploid organisms can only have two alleles for any given gene, it is common for more than two alleles of a gene to exist.<\/p>\n<p>In many life forms, females have two X chromosomes and males have one X and one Y chromosome. Genes that are present on the X but not the Y chromosome are said to be X-linked, such that males only inherit one allele for the gene, and females inherit two. Some alleles can be lethal. Recessive lethal alleles are only lethal in homozygotes, but dominant lethal alleles are fatal in both homozygotes and heterozygotes.<\/p>\n<div class=\"textbox exercises\">\n<h3>Additional Self Check Questions<\/h3>\n<ol>\n<li>In pea plants, round peas (<em>R<\/em>) are dominant to wrinkled peas (<em>r<\/em>).\u00a0 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. The cross gives you all round peas.\u00a0 Can you tell if the round pea parent plant is homozygous dominant or heterozygous?<\/li>\n<li>What is the tile given to Mendel?<\/li>\n<li>What ratio of offspring would result from a cross between a white-eyed male and a female that is heterozygous for red eye color?<\/li>\n<li>The gene for flower position in pea plants exists as axial or terminal alleles. Given that axial is dominant to terminal, list all of the possible F<sub>1<\/sub> and F<sub>2<\/sub> genotypes and phenotypes from a cross involving parents that are homozygous for each trait. Express genotypes with conventional genetic abbreviations.<\/li>\n<li>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?<\/li>\n<li>Can a human male be a carrier of red-green color blindness?<\/li>\n<\/ol>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Answers<\/h3>\n<ol>\n<li>Based on your data, the round pea plant would be RR.<\/li>\n<li>Mendel is known as the &#8220;father of genetics&#8221;.<\/li>\n<li>Half of the female offspring would be heterozygous (X<sup class=\"sup\"><em>W<\/em><\/sup>X<sup class=\"sup\"><em>w<\/em><\/sup>) with red eyes, and half would be homozygous recessive (X<sup class=\"sup\"><em>w<\/em><\/sup>X<sup class=\"sup\"><em>w<\/em><\/sup>) with white eyes. Half of the male offspring would be hemizygous dominant (X<sup class=\"sup\"><em>W<\/em><\/sup>Y) withe red eyes, and half would be hemizygous recessive (X<sup class=\"sup\"><em>w<\/em><\/sup>Y) with white eyes.<\/li>\n<li>Because axial is dominant, the gene would be designated as <em>A<\/em>. F<sub>1<\/sub> would be all heterozygous <em>Aa<\/em> with axial phenotype. F<sub>2<\/sub> would have possible genotypes of <em>AA<\/em>, <em>Aa<\/em>, and <em>aa<\/em>; these would correspond to axial, axial, and terminal phenotypes, respectively.<\/li>\n<li>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.<\/li>\n<li>No, males can only express color blindness. They cannot carry it because an individual needs two X chromosomes to be a carrier.<\/li>\n<\/ol>\n<\/div>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-241\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Biology. <strong>Authored by<\/strong>: Open Stax. <strong>Project<\/strong>: http:\/\/cnx.org\/contents\/185cbf87-c72e-48f5-b51e-f14f21b5eabd@9.17:1\/Biology. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">All rights reserved content<\/div><ul class=\"citation-list\"><li>Sex Linked Traits. <strong>Provided by<\/strong>: Khan Academy. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/youtu.be\/-ROhfKyxgCo\">http:\/\/youtu.be\/-ROhfKyxgCo<\/a>. <strong>License<\/strong>: <em>Other<\/em>. <strong>License Terms<\/strong>: Standard YouTube license<\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":18,"menu_order":5,"template":"","meta":{"_candela_citation":"[{\"type\":\"copyrighted_video\",\"description\":\"Sex Linked Traits\",\"author\":\"\",\"organization\":\"Khan Academy\",\"url\":\"http:\/\/youtu.be\/-ROhfKyxgCo\",\"project\":\"\",\"license\":\"other\",\"license_terms\":\"Standard YouTube license\"},{\"type\":\"cc\",\"description\":\"Biology\",\"author\":\"Open Stax\",\"organization\":\"\",\"url\":\"\",\"project\":\"http:\/\/cnx.org\/contents\/185cbf87-c72e-48f5-b51e-f14f21b5eabd@9.17:1\/Biology\",\"license\":\"cc-by\",\"license_terms\":\"\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-241","chapter","type-chapter","status-publish","hentry"],"part":231,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/nemcc-biology1v2\/wp-json\/pressbooks\/v2\/chapters\/241","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/nemcc-biology1v2\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/nemcc-biology1v2\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/nemcc-biology1v2\/wp-json\/wp\/v2\/users\/18"}],"version-history":[{"count":26,"href":"https:\/\/courses.lumenlearning.com\/nemcc-biology1v2\/wp-json\/pressbooks\/v2\/chapters\/241\/revisions"}],"predecessor-version":[{"id":1678,"href":"https:\/\/courses.lumenlearning.com\/nemcc-biology1v2\/wp-json\/pressbooks\/v2\/chapters\/241\/revisions\/1678"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/nemcc-biology1v2\/wp-json\/pressbooks\/v2\/parts\/231"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/nemcc-biology1v2\/wp-json\/pressbooks\/v2\/chapters\/241\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/nemcc-biology1v2\/wp-json\/wp\/v2\/media?parent=241"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/nemcc-biology1v2\/wp-json\/pressbooks\/v2\/chapter-type?post=241"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/nemcc-biology1v2\/wp-json\/wp\/v2\/contributor?post=241"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/nemcc-biology1v2\/wp-json\/wp\/v2\/license?post=241"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}