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disorders). Punnett squares are useful tools that apply the rules of probability and meiosis to predict the possible outcomes of genetic crosses. Test crosses are done to determine whether or not an individual is homozygous or heterozygous by crossing the individual with a homozygous recessive. Information presented and the examples highlighted in the section support concepts outlined in Big Idea 3 of the AP® Biology Curriculum Framework. The learning objectives (LO) listed in the Curriculum Framework provide a transparent foundation for the AP® Biology course, an inquiry-based laboratory experience, instructional activities, and AP® exam questions. A learning objective merges required content with one or more of the seven science practices (SP). Big Idea 3 Enduring Understanding 3.A Living systems store, retrieve, transmit and respond to information essential to life processes. Heritable information provides for continuity of life. 478 Chapter 12 | Mendel's Experiments and Heredity Essential Knowledge 3.A.3 The chromosomal basis of inheritance proposed by Mendel provides an understanding of the pattern of passage of genes from parent to offspring. Science Practice Science Practice Learning Objective 1.1 The student can create representations and models of natural or man-made phenomena and systems in the domain. 7.2 The student can connect concepts in and across domain(s) to generalize or extrapolate in and/or across enduring understandings and/or big ideas. 3.12 The student is able to construct a representation (e.g., Punnett square) that connects the process of meiosis to the passage of traits from parent to offspring. Essential Knowledge 3.A.3 The chromosomal basis of inheritance proposed by Mendel provides an understanding of the pattern of passage of genes from parent to offspring. Science Practice Learning Objective 3.1 The student can pose scientific questions. 3.13 The student is able to pose questions about ethical, social or medical issues surrounding human genetic disorders. Essential Knowledge 3.A.3 The chromosomal basis of inheritance proposed by Mendel provides an understanding of the pattern of passage of genes from parent to offspring. Science Practice Learning Objective 2.2 The student can apply mathematical routines to quantities that describe natural phenomena. 3.14 The student is able to apply mathematical routines to determine Mendelian patterns of inheritance provided by data sets. Essential Knowledge 3.A.4 The inheritance patterns of many traits cannot be explained by simple Mendelian genetics. Science Practice Learning Objective 6.5 The student can evaluate alternative scientific explanations. 3.15 The |
student is able to explain deviations from Mendel’s model of the inheritance of traits. Essential Knowledge 3.A.4 The inheritance patterns of many traits cannot be explained by simple Mendelian genetics. Science Practice Learning Objective 6.3 The student can articulate the reasons that scientific explanations and theories are refined or replaced. 3.16 The student is able to explain how the inheritance pattern of many traits cannot be accounted for by Mendelian genetics. Essential Knowledge 3.A.4 The inheritance patterns of many traits cannot be explained by simple Mendelian genetics. Science Practice Learning Objective 1.2 The student can describe representations and models of natural or man-made phenomena and systems in the domain. 3.17 The student is able to describe representations of an appropriate example of inheritance patterns that cannot be explained by Mendel’s model of the inheritance of traits. The Science Practice Challenge Questions contain additional test questions for this section that will help you prepare for the AP exam. These questions address the following standards: [APLO 3.12][APLO 3.14][APLO 3.16][APLO 3.11][APLO 3.13][APLO 3.17] 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 the expression of genes 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 linear order of genes. In other words, peas are diploid organisms in that they have two copies of each chromosome. The same is true for many other plants and for virtually all animals. Diploid organisms utilize meiosis to produce haploid gametes, which contain one copy of each homologous chromosome that unite This OpenStax book is available for free at http://cnx.org/content/col12078/1.6 Chapter 12 | Mendel's Experiments and Heredity 479 at fertilization to create a diploid zygote. 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. Gene variants that arise by mutation and exist at the same relative locations on homologous chromosomes are called alleles. 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. Phenotypes and Genotypes 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 phenotype. An organism’s underlying genetic makeup, consisting of both physically visible and non-expressed alleles, is called its genotype. Mendel’s hybridization experiments demonstrate the difference between phenotype and genotype. When true-breeding plants in which one parent had yellow pods and one had green pods were cross-fertilized, all of the F1 hybrid offspring had yellow pods. That is, the hybrid offspring were phenotypically identical to the true-breeding parent with yellow pods. However, we know that the allele donated by the parent with green pods was not simply lost because it reappeared in some of the F2 offspring. Therefore, the F1 plants must have been genotypically different from the parent with yellow pods. The P1 plants that Mendel used in his experiments were each homozygous for the trait he was studying. Diploid organisms that are homozygous at a given gene, or locus, have two identical alleles for that gene on their homologous chromosomes. Mendel’s parental pea plants always bred true because both of the gametes produced carried the same trait. When P1 plants with contrasting traits were cross-fertilized, all of the offspring were heterozygous for the contrasting trait, meaning that their genotype reflected that they had different alleles for the gene being examined. Dominant and Recessive Alleles Our discussion of homozygous and heterozygous organisms brings us to why the F1 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 chromosome pairs. 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). The recessive allele will only be observed in homozygous recessive individuals (Table 12. |
4). Human Inheritance in Dominant and Recessive Patterns Dominant Traits Recessive Traits Achondroplasia Brachydactyly Albinism Cystic fibrosis Huntington’s disease Duchenne muscular dystrophy Marfan syndrome Galactosemia Neurofibromatosis Phenylketonuria Widow’s peak Wooly hair Sickle-cell anemia Tay-Sachs disease Table 12.4 This is a table of human inheritance traits and categorizes dominate versus recessive patterns. Several conventions exist for referring to genes and alleles. For the purposes of this chapter, we will abbreviate genes using the first letter of the gene’s corresponding dominant trait. For example, violet is the dominant trait for a pea plant’s flower color, so the flower-color gene would be abbreviated as V (note that it is customary to italicize gene designations). Furthermore, 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 VV, a homozygous recessive pea plant with white flowers as vv, and a heterozygous pea plant with violet flowers as Vv. The Punnett Square Approach for a Monohybrid Cross When fertilization occurs between two true-breeding parents that differ in only one characteristic, the process is called a 480 Chapter 12 | Mendel's Experiments and Heredity monohybrid cross, and the resulting offspring are monohybrids. Mendel performed seven monohybrid crosses involving contrasting traits for each characteristic. On the basis of his results in F1 and F2 generations, 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. To demonstrate a monohybrid cross, consider the case of true-breeding pea plants with yellow versus green pea seeds. The dominant seed color is yellow; therefore, the parental genotypes were YY for the plants with yellow seeds and yy for the plants with green seeds, respectively. A Punnett square, devised by the British geneticist Reginald Punnett, can be drawn that applies the rules of probability to predict the possible outcomes of a genetic cross or mating and their expected frequencies. To prepare a Punn |
ett 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, representing their meiotic segregation into haploid gametes. Then the combinations of egg and sperm are made in the boxes in the table to show which alleles are combining. Each box then 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 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. All offspring are Yy and have yellow seeds (Figure 12.4). Figure 12.4 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 F1 heterozygotes with a yellow phenotype. Punnett square analysis can be used to predict the genotypes of the F2 generation. A self-cross of one of the Yy heterozygous offspring can be represented in a 2 × 2 Punnett square because each parent can donate one of two different alleles. Therefore, the offspring can potentially have one of four allele combinations: YY, Yy, yY, or yy (Figure 12.4). Notice that there are two ways to obtain the Yy genotype: a Y from the egg and a y from the sperm, or a y from the egg and a Y from the sperm. Both of these possibilities must be counted. Recall that Mendel’s peaplant characteristics behaved in the same way in reciprocal crosses. Therefore, the two possible heterozygous combinations produce offspring that are genotypically and phenotypically identical despite their dominant and recessive alleles deriving This OpenStax book is available for free at http://cnx.org/content/col12078/1.6 Chapter 12 | Mendel's Experiments and Heredity 481 from different parents. They are grouped together. Because fertilization is a random event, we expect each combination to be equally likely and for the offspring to exhibit a ratio of YY:Yy:yy genotypes of 1:2 |
:1 (Figure 12.4). Furthermore, because the YY and Yy 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. Indeed, working with large sample sizes, Mendel observed approximately this ratio in every F2 generation resulting from crosses for individual traits. Mendel validated these results by performing an F3 cross in which he self-crossed the dominant- and recessive-expressing F2 plants. When he self-crossed the plants expressing green seeds, all of the offspring had green seeds, confirming that all green seeds had homozygous genotypes of yy. When he self-crossed the F2 plants expressing yellow seeds, he found that one-third of the plants bred true, and two-thirds of the plants segregated at a 3:1 ratio of yellow:green seeds. In this case, the true-breeding plants had homozygous (YY) genotypes, whereas the segregating plants corresponded to the heterozygous (Yy) genotype. When these plants self-fertilized, the outcome was just like the F1 self-fertilizing cross. The Test Cross Distinguishes the Dominant Phenotype 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 test cross, this technique is still used 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. If the dominant-expressing organism is a homozygote, then all F1 offspring will be heterozygotes expressing the dominant trait (Figure 12.5). Alternatively, if the dominant expressing organism is a heterozygote, the F1 offspring will exhibit a 1:1 ratio of heterozygotes and recessive homozygotes (Figure 12.5). The test cross further validates Mendel’s postulate that pairs of unit factors segregate equally. 482 Chapter 12 | Mendel's Experiments and Heredity Figure 12.5 A test cross can be performed to determine whether an organism expressing a dominant trait is a homozygote or a heterozygote. In pea plants, round peas (R) are dominant to wrinkled peas |
(r).You do a test cross between a pea plant with wrinkled peas (genotype rr) 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? a. The data set is too small to predict the genotype of the round pea plant. Assuming that the unknown parent is heterozygous, the probability of having only round pea plants from a random sample of 3 progeny will be 1 8. b. The genotype of the unknown round pea plant is Rr. Assuming that the unknown parent is heterozygous, then the probability of having only round pea plants from a random sample of 3 progeny will be 1 4. c. The genotype of the unknown round pea plant is Rr. Assuming that the unknown parent is heterozygous, the probability of having only round pea plants from a random sample of 3 progeny will be 1 2. d. The data set is too small to predict the genotype of the round pea plant. The probability of having only round pea plants from a random sample of 3 progeny will be 1 6. Many human diseases are genetically inherited. A healthy person in a family in which some members suffer from a recessive genetic disorder may want to know if he or she has the disease-causing gene and what risk exists of passing the disorder on to his or her offspring. Of course, doing a test cross in humans is unethical and impractical. Instead, geneticists use pedigree This OpenStax book is available for free at http://cnx.org/content/col12078/1.6 Chapter 12 | Mendel's Experiments and Heredity 483 analysis to study the inheritance pattern of human genetic diseases (Figure 12.6). 484 Chapter 12 | Mendel's Experiments and Heredity Figure 12.6 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 |
aa. Unaffected individuals are indicated in yellow and have the genotype AA or Aa. Note that it is often possible to determine a person’s 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 “A?” designation. This OpenStax book is available for free at http://cnx.org/content/col12078/1.6 Chapter 12 | Mendel's Experiments and Heredity 485 Using the pedigree above, what are the genotypes of the individuals labeled 1, 2 and 3? a. 1- aa, 2-AA, 3- AA b. 1-aa, 2-Aa, 3- Aa c. 1-aa, 2-Aa, 3-AA d. 1-Aa, 2-Aa, 3-Aa Alternatives to Dominance and Recessiveness Mendel’s experiments with pea plants suggested that: (1) two “units” or alleles exist for every gene; (2) alleles maintain their integrity in each generation (no blending); and (3) in the presence of the dominant allele, the recessive allele is hidden and makes no contribution to the phenotype. Therefore, recessive alleles can be “carried” and not expressed by individuals. Such heterozygous individuals are sometimes referred to as “carriers.” Further genetic studies in other plants and animals have shown that much more complexity exists, but that the fundamental principles of Mendelian genetics still hold true. In the sections to follow, we consider some of the extensions of Mendelism. If Mendel had chosen an experimental system that exhibited these genetic complexities, it’s possible that he would not have understood what his results meant. Incomplete Dominance Mendel’s results, that traits are inherited as dominant and recessive pairs, contradicted the view at that time that offspring exhibited a blend of their parents’ traits. However, the heterozygote phenotype occasionally does appear to be intermediate between the two parents. For example, in the snapdragon, Antirrhinum majus (Figure 12.7), a cross between a homozygous parent with |
white flowers (CWCW) and a homozygous parent with red flowers (CRCR) will produce offspring with pink flowers (CRCW). (Note that different genotypic abbreviations are used for Mendelian extensions to distinguish these patterns from simple dominance and recessiveness.) This pattern of inheritance is described as incomplete dominance, denoting 486 Chapter 12 | Mendel's Experiments and Heredity the expression of two contrasting alleles such that the individual displays an intermediate phenotype. The allele for red flowers is incompletely dominant over the allele for white flowers. However, the results of a heterozygote self-cross can still be predicted, just as with Mendelian dominant and recessive crosses. In this case, the genotypic ratio would be 1 CRCR:2 CRCW:1 CWCW, and the phenotypic ratio would be 1:2:1 for red:pink:white. Figure 12.7 These pink flowers of a heterozygote snapdragon result “storebukkebruse”/Flickr) from incomplete dominance. (credit: Codominance A variation on incomplete dominance is codominance, in which both alleles for the same characteristic are simultaneously expressed in the heterozygote. An example of codominance is the MN blood groups of humans. The M and N alleles are expressed in the form of an M or N antigen present on the surface of red blood cells. Homozygotes (LMLM and LNLN) express either the M or the N allele, and heterozygotes (LMLN) express both alleles equally. In a self-cross between heterozygotes expressing a codominant trait, the three possible offspring genotypes are phenotypically distinct. However, the 1:2:1 genotypic ratio characteristic of a Mendelian monohybrid cross still applies. Multiple Alleles Mendel implied that only two alleles, one dominant and one recessive, could exist for a given gene. We now know that this is an oversimplification. Although individual humans (and all diploid organisms) can only have two alleles for a given gene, multiple alleles may exist at the population level such that many combinations of two alleles are observed. Note that when many alleles exist for the same gene, the convention is to denote the most common phenotype or genotype among wild animals as the wild type (often abbreviated “+”); this is considered |
the standard or norm. All other phenotypes or genotypes are considered variants of this standard, meaning that they deviate from the wild type. The variant may be recessive or dominant to the wild-type allele. An example of multiple alleles is coat color in rabbits (Figure 12.8). Here, four alleles exist for the c gene. The wild-type version, C+C+, is expressed as brown fur. The chinchilla phenotype, cchcch, is expressed as black-tipped white fur. The Himalayan phenotype, chch, has black fur on the extremities and white fur elsewhere. Finally, the albino, or “colorless” phenotype, cc, is expressed as white fur. In cases of multiple alleles, dominance hierarchies can exist. In this case, the wildtype allele is dominant over all the others, chinchilla is incompletely dominant over Himalayan and albino, and Himalayan is dominant over albino. This hierarchy, or allelic series, was revealed by observing the phenotypes of each possible heterozygote offspring. This OpenStax book is available for free at http://cnx.org/content/col12078/1.6 Chapter 12 | Mendel's Experiments and Heredity 487 Figure 12.8 Four different alleles exist for the rabbit coat color (C) gene. The complete dominance of a wild-type phenotype over all other mutants often occurs as an effect of “dosage” of a specific gene product, such that the wild-type allele supplies the correct amount of gene product whereas the mutant alleles cannot. For the allelic series in rabbits, the wild-type allele may supply a given dosage of fur pigment, whereas the mutants supply a lesser dosage or none at all. Interestingly, the Himalayan phenotype is the result of an allele that produces a temperaturesensitive gene product that only produces pigment in the cooler extremities of the rabbit’s body. Alternatively, one mutant allele can be dominant over all other phenotypes, including the wild type. This may occur when the mutant allele somehow interferes with the genetic message so that even a heterozygote with one wild-type allele copy expresses the mutant phenotype. One way in which the mutant allele can interfere is by enhancing the function of the wildtype gene product or changing its distribution in the body. One example of this is the Antennapedia mutation in Drosophila (Figure 12 |
.9). In this case, the mutant allele expands the distribution of the gene product, and as a result, the Antennapedia heterozygote develops legs on its head where its antennae should be. Figure 12.9 As seen in comparing the wild-type Drosophila (left) and the Antennapedia mutant (right), Antennapedia mutant has legs on its head in place of antennae. the 488 Chapter 12 | Mendel's Experiments and Heredity Multiple Alleles Confer Drug Resistance in the Malaria Parasite Malaria is a parasitic disease in humans that is transmitted by infected female mosquitoes, including Anopheles gambiae (Figure 12.10a), and is characterized by cyclic high fevers, chills, flu-like symptoms, and severe anemia. Plasmodium falciparum and P. vivax are the most common causative agents of malaria, and P. falciparum is the most deadly (Figure 12.10b). When promptly and correctly treated, P. falciparum 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. (a) (b) Figure 12.10 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) In Southeast Asia, Africa, and South America, P. falciparum has developed resistance to the anti-malarial drugs chloroquine, mefloquine, and sulfadoxine-pyrimethamine. P. falciparum, which is haploid during the life stage in which it is infectious to humans, has evolved multiple drug-resistant mutant alleles of the dhps gene. Varying degrees of sulfadoxine resistance are associated with each of these alleles. Being haploid, P. falciparum needs only one drug-resistant allele to express this trait. In Southeast Asia, different sulfadoxine-resistant |
alleles of the dhps gene are localized to different geographic regions. This is a common evolutionary phenomenon that occurs because drug-resistant mutants arise in a population and interbreed with other P. falciparum isolates in close proximity. Sulfadoxineresistant parasites cause considerable human hardship in regions where this drug is widely used as an over-the-counter malaria remedy. As is common with pathogens that multiply to large numbers within an infection cycle, P. falciparum evolves relatively rapidly (over a decade or so) in response to the selective pressure of commonly used anti-malarial drugs. For this reason, scientists must constantly work to develop new drugs or drug combinations to combat the worldwide malaria burden. [2] According to this passage, why does P. falciparum only need one drug-resistant dhps allele to express the drug resistance trait? a. The drug-resistant dhps allele is co-dominant with the wild type allele. b. Only one dhps allele is present during all stages of the P. falciparum life cycle. c. Only one dhps allele is present when P. falciparum is infectious. d. The drug-resistant dhps allele prevents the wild type allele from being expressed. 2. Sumiti Vinayak, et al., “Origin and Evolution of Sulfadoxine Resistant Plasmodium falciparum,” Public Library of Science Pathogens 6, no. 3 (2010): e1000830, doi:10.1371/journal.ppat.1000830. This OpenStax book is available for free at http://cnx.org/content/col12078/1.6 Chapter 12 | Mendel's Experiments and Heredity 489 X-Linked Traits In humans, as well as in many other animals and some plants, the sex of the individual is determined by sex chromosomes. The sex chromosomes are one pair of non-homologous chromosomes. Until now, we have only considered inheritance patterns among non-sex chromosomes, or autosomes. In addition to 22 homologous pairs of autosomes, human females have a homologous pair of X chromosomes, whereas human males have an XY chromosome pair. Although the Y chromosome contains a small region of similarity to the X chromosome so that they can pair during meiosis, the Y chromosome is much shorter and contains many fewer genes. When a gene being examined is |
present on the X chromosome, but not on the Y chromosome, it is said to be X-linked. Eye color in Drosophila was one of the first X-linked traits to be identified. Thomas Hunt Morgan mapped this trait to the X chromosome in 1910. Like humans, Drosophila males have an XY chromosome pair, and females are XX. In flies, the wild-type eye color is red (XW) and it is dominant to white eye color (Xw) (Figure 12.11). Because of the location of the eye-color gene, reciprocal crosses do not produce the same offspring ratios. Males are said to be hemizygous, because they have only one allele for any X-linked characteristic. Hemizygosity makes the descriptions of dominance and recessiveness irrelevant for XY males. Drosophila males lack a second allele copy on the Y chromosome; that is, their genotype can only be XWY or XwY. In contrast, females have two allele copies of this gene and can be XWXW, XWXw, or XwXw. Figure 12.11 In Drosophila, several genes determine eye color. The genes for white and vermilion eye colors are located on the X chromosome. Others are located on the autosomes. 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. In an X-linked cross, the genotypes of F1 and F2 offspring depend on whether the recessive trait was expressed by the male or the female in the P1 generation. With regard to Drosophila eye color, when the P1 male expresses the white-eye phenotype and the female is homozygous red-eyed, all members of the F1 generation exhibit red eyes (Figure 12.12). The F1 females are heterozygous (XWXw), and the males are all XWY, having received their X chromosome from the homozygous dominant P1 female and their Y chromosome from the P1 male. A subsequent cross between the XWXw female and the XWY male would produce only red-eyed females (with XWXW or XWXw genotypes) and both red- and white-eyed males (with XWY or XwY genotypes). Now, consider a cross between a homozyg |
ous white-eyed female and a male with red eyes. The F1 generation would exhibit only heterozygous red-eyed females (XWXw) and only white-eyed males (XwY). Half of the F2 females would be red-eyed (XWXw) and half would be white-eyed (XwXw). Similarly, half of the F2 males would be redeyed (XWY) and half would be white-eyed (XwY). 490 Chapter 12 | Mendel's Experiments and Heredity Figure 12.12 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. What ratio of offspring would result from a cross between a white-eyed male fruit fly and a female fruit fly that is heterozygous for red eye color? a. b. c. d. 25% of the offspring are males and hemizygous dominant with red eyes and 25% are male and hemizygous recessive with white eyes. 25% are female and heterozygous with red eyes and 25% are females and homozygous recessive with white eyes. 50% of the offspring are male and hemizygous dominant with red eyes and 50% are male hemizygous recessive with white eyes. 50% are female and heterozygous with red eyes and 50% are female and homozygous recessive with white eyes. 25% of the males are hemizygous dominant with red eyes and 50% of the male are hemizygous recessive with white eyes. 25% females are heterozygous with red eyes and 50% of the females are homozygous with white eyes. 50% of the males are hemizygous dominant with red eyes and 25% of the male are hemizygous recessive with white eyes. 50% females are heterozygous with red eyes and 25% of the females are homozygous recessive with white eyes Discoveries in fruit fly genetics can be applied to human genetics. When a female parent is homozygous for a recessive Xlinked trait, she will pass the trait on to 100 percent of her offspring. Her male offspring are, therefore, destined to express the trait, as they will inherit their father's Y chromosome. In humans, the alleles for certain conditions (some forms of color blindness, hemophilia, and muscular dystrophy) are X |
-linked. Females who are heterozygous for these diseases 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 This OpenStax book is available for free at http://cnx.org/content/col12078/1.6 Chapter 12 | Mendel's Experiments and Heredity 491 carrier status to half of their daughters; therefore, recessive X-linked traits appear more frequently in males than females. In some groups of organisms with sex chromosomes, the sex with the non-homologous sex chromosomes is the female rather than the male. This is the case for all birds. In this case, sex-linked traits will be more likely to appear in the female, in which they are hemizygous. Human Sex-linked Disorders Sex-linkage studies in Morgan’s laboratory provided the fundamentals for understanding X-linked recessive disorders in humans, which include red-green color blindness, and Types A and B hemophilia. Because human males need to inherit only one recessive mutant X allele to be affected, X-linked disorders are disproportionately observed in males. Females must inherit recessive X-linked alleles from both of their parents in order to express the trait. When they inherit one recessive X-linked mutant allele and one dominant X-linked wild-type allele, they are carriers of the trait and are typically unaffected. Carrier females can manifest mild forms of the trait due to the inactivation of the dominant allele located on one of the X chromosomes. However, female carriers can contribute the trait to their sons, resulting in the son exhibiting the trait, or they can contribute the recessive allele to their daughters, resulting in the daughters being carriers of the trait (Figure 12.13). Although some Y-linked recessive disorders exist, typically they are associated with infertility in males and are therefore not transmitted to subsequent generations. Figure 12.13 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 she will have a 50 percent chance of being a carrier like her mother. 492 Chapter 12 | Mendel's Experiments and Heredity Watch this video (http://openstaxcollege.org/l/sex-linked_trts) to learn more about sex-linked traits. The most common form of hemophilia affects 1 out of every 5 |
,000 male births worldwide, but the condition is much rarer in females. Explain why this is the case. a. Females need two mutated X chromosomes to be hemophilic. b. Females need one mutated X chromosome to be hemophilic. c. Females do not inherit mutated X chromosomes. d. Females need two mutated X chromosomes to not be hemophilic. Lethality A large proportion of genes in an individual’s genome are essential for survival. Occasionally, a nonfunctional allele for an essential gene can arise by mutation and be transmitted in a population as long as individuals with this allele also have a wild-type, functional copy. The wild-type allele functions at a capacity sufficient to sustain life and is therefore considered to be dominant over the nonfunctional allele. However, consider two heterozygous parents that have a genotype of wildtype/nonfunctional mutant for a hypothetical essential gene. In one quarter of their offspring, we would expect to observe individuals that are homozygous recessive for the nonfunctional allele. Because the gene is essential, these individuals might fail to develop past fertilization, die in utero, or die later in life, depending on what life stage requires this gene. An inheritance pattern in which an allele is only lethal in the homozygous form and in which the heterozygote may be normal or have some altered non-lethal phenotype is referred to as recessive lethal. For crosses between heterozygous individuals with a recessive lethal allele that causes death before birth when homozygous, only wild-type homozygotes and heterozygotes would be observed. The genotypic ratio would therefore be 2:1. In other instances, the recessive lethal allele might also exhibit a dominant (but not lethal) phenotype in the heterozygote. For instance, the recessive lethal Curly allele in Drosophila affects wing shape in the heterozygote form but is lethal in the homozygote. A single copy of the wild-type allele is not always sufficient for normal functioning or even survival. The dominant lethal inheritance pattern is one in which an allele is lethal both in the homozygote and the heterozygote; this allele can only be 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 because, as you might expect, the allele only lasts one generation and is not transmitted. However, just as the |
recessive lethal allele might not immediately manifest the phenotype of death, dominant lethal alleles also 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. An example of this in humans is Huntington’s disease, in which the nervous system gradually wastes away (Figure 12.14). People who are heterozygous for the dominant Huntington allele (Hh) will inevitably develop the fatal disease. However, the onset of Huntington’s disease may not occur until age 40, at which point the afflicted persons may have already passed the allele to 50 percent of their offspring. This OpenStax book is available for free at http://cnx.org/content/col12078/1.6 Chapter 12 | Mendel's Experiments and Heredity 493 Figure 12.14 The neuron in the center of this micrograph (yellow) has nuclear inclusions characteristic of Huntington’s disease (orange area in the center of the neuron). Huntington’s disease occurs when an abnormal dominant allele for the Huntington gene is present. (credit: Dr. Steven Finkbeiner, Gladstone Institute of Neurological Disease, The TaubeKoret Center for Huntington's Disease Research, and the University of California San Francisco/Wikimedia) Activity This section includes descriptions of genetically-inherited human diseases, such as sickle cell anemia, alkaptonuria, hemophilia, color blindness and Huntington’s disease. One issue surrounding genetic issue disorders is the right to privacy. Can you think of other examples of ethical, social, or medical surrounding human genetic disorders? Lab Investigation Investigate inheritance patterns in an organism of choice, such as Wisconsin Fast Plants or Drosophila melanogaster, by performing several genetic crosses and comparing expected and observed phenotypic ratios. Virtual labs exploring Mendelian inheritance patterns are also available online. Think About It • In pea plants, round peas (R) are dominant to wrinkles peas (r) (Figure 12.5). You do a test cross between a pea plant with wrinkled peas (genotype rr) and a plant of unknown genotype that has round peas (genotype either RR or Rr). You end up with three offspring plants, all which have round peas. Based on the phenotype of the offspring plants, can you deduce the genotype of the round pea parent plant? If the round pea parent plant is |
heterozygous, calculate the probability that a random sample of 3 progeny peas will all be round. • Can a human male be a carrier of red-green color blindness? Justify your answer. • In pea plants, violet flowers (V) are dominant to white flowers (v). What are the possible genotypes and phenotypes for a cross between Vv and vv pea plants? Use a Punnett square to show all work. 494 Chapter 12 | Mendel's Experiments and Heredity 12.3 | Laws of Inheritance In this section, you will explore the following questions: • What is the relationship between Mendel’s law of segregation and independent assortment in terms of genetics and the events of meiosis? • How can the forked-lined method and probability rules be used to calculate the probability of genotypes and phenotypes from multiple gene crosses? • How do linkage, cross-over, epistasis, and recombination violate Mendel’s laws of inheritance? Connection for AP® Courses As was described previously, Mendel proposed that genes are inherited as pairs of alleles that behave in a dominant and recessive pattern. During meiosis, alleles segregate, or separate, such that each gamete is equally likely to receive either one of the two alleles present in the diploid individual. Mendel called this phenomenon the law of segregation, which can be demonstrated in a monohybrid cross. In addition, genes carried on different chromosomes sort into gametes independently of one another. This is Mendel’s law of independent assortment. This law can be demonstrated in a dihybrid cross involving two different traits located on different chromosomes. Punnett squares can be used to predict genotypes and phenotypes of offspring involving one or two genes. Although chromosomes sort independently into gametes during meiosis, Mendel’s law of independent assortment refers to genes, not chromosomes. In humans, single chromosomes may carry more than 1,000 genes. Genes located close together on the same chromosome are said to be linked genes. When genes are located in close proximity on the same chromosome, their alleles tend to be inherited together unless recombination occurs. This results in offspring ratios that violate Mendel’s law of independent assortment. Genes that are located far apart on the same chromosome are likely to assort independently. The rules of probability can help to sort this out (pun intended). The law states that alleles of |
different genes assort independently of one another during gamete formation. Information presented and the examples highlighted in the section support concepts outlined in Big Idea 3 of the AP® Biology Curriculum Framework. The Learning Objectives listed in the Curriculum Framework provide a transparent foundation for the AP® Biology course, an inquiry-based laboratory experience, instructional activities, and AP® exam questions. A learning objective merges required content with one or more of the seven Science Practices. Big Idea 3 Enduring Understanding 3.A Living systems store, retrieve, transmit and respond to information essential to life processes. Heritable information provides for continuity of life. Essential Knowledge 3.A.3 The chromosomal basis of inheritance provides an understanding of the pattern of passage (transmission) of genes from parent to offspring. Science Practice Learning Objective 2.2 The student can apply mathematical routines to quantities that describe natural phenomena. 3.14 The student is able to apply mathematical routines to determine Mendelian patterns of inheritance provided by data. Essential Knowledge 3.A.4 The inheritance pattern of many traits cannot be explained by simple Mendelian genetics. Science Practice Learning Objective 6.5 The student can evaluate alternative scientific explanations. 3.15 The student is able to explain deviations from Mendel’s model of the inheritance of traits. This OpenStax book is available for free at http://cnx.org/content/col12078/1.6 Chapter 12 | Mendel's Experiments and Heredity 495 Essential Knowledge 3.A.4 The inheritance pattern of many traits cannot be explained by simple Mendelian genetics. Science Practice Learning Objective 6.3 The student can articulate the reasons that scientific explanations and theories are refined or replaced. 3.16 The student is able to explain how the inheritance patterns of many traits cannot be accounted for by Mendelian genetics. Essential Knowledge 3.A.4 The inheritance pattern of many traits cannot be explained by simple Mendelian genetics. Science Practice Learning Objective 1.2 The student can describe representations and models of natural or man-made phenomena and systems in the domain. 3.17 The student is able to describe representations of an appropriate example of inheritance patterns that cannot be explained by Mendel’s model of the inheritance of traits. The Science Practice Challenge Questions contain additional test questions for this section that will help you prepare for the AP exam. These questions address the following standards: [APLO 3.11][APLO 3.15][APLO 3.14][APLO |
3.17][APLO 3.12] Mendel generalized the results of his pea-plant experiments into four postulates, some of which are sometimes called “laws,” that describe the basis of dominant and recessive inheritance in diploid organisms. As you have learned, more complex extensions of Mendelism exist that do not exhibit the same F2 phenotypic ratios (3:1). Nevertheless, these laws summarize the basics of classical genetics. Pairs of Unit Factors, or Genes Mendel proposed first that paired unit factors of heredity were transmitted faithfully from generation to generation by the dissociation and reassociation of paired factors during gametogenesis and fertilization, respectively. After he crossed peas with contrasting traits and found that the recessive trait resurfaced in the F2 generation, Mendel deduced that hereditary factors must be inherited as discrete units. This finding contradicted the belief at that time that parental traits were blended in the offspring. Alleles Can Be Dominant or Recessive Mendel’s law of dominance states that in a heterozygote, one trait will conceal the presence of another trait for the same characteristic. 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 by the same manner in which the dominant allele is transmitted. The recessive trait will only be expressed by offspring that have two copies of this allele (Figure 12.15), and these offspring will breed true when self-crossed. Since Mendel’s experiments with pea plants, other researchers have found that the law of dominance does not always hold true. Instead, several different patterns of inheritance have been found to exist. 496 Chapter 12 | Mendel's Experiments and Heredity Figure 12.15 The child in the photo expresses albinism, a recessive trait. Equal Segregation of Alleles Observing that true-breeding pea plants with contrasting traits gave rise to F1 generations that all expressed the dominant trait and F2 generations that expressed the dominant and recessive traits in a 3:1 ratio, Mendel proposed the law of segregation. This law states that paired unit factors (genes) must segregate equally into gametes such that offspring have an equal likelihood of inheriting either factor. For the F2 generation of a monohybrid cross, the following three possible combinations of genotypes could result: homozyg |
ous 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’s 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’s 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. The role of the meiotic segregation of chromosomes in sexual reproduction was not understood by the scientific community during Mendel’s lifetime. Independent Assortment Mendel’s law of independent assortment states that genes do not influence each other with regard to the sorting of alleles into gametes, and every possible combination of alleles for every gene is equally likely to occur. The independent assortment of genes can be illustrated by the dihybrid 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 green, wrinkled seeds (yyrr) and another that has yellow, round seeds (YYRR). Because each parent is homozygous, the law of segregation indicates that the gametes for the green/wrinkled plant all are yr, and the gametes for the yellow/round plant are all YR. Therefore, the F1 generation of offspring all are YyRr (Figure 12.16). This OpenStax book is available for free at http://cnx.org/content/col12078/1.6 Chapter 12 | Mendel's Experiments and Heredity 497 Figure 12.16 This dihybrid cross of pea plants involves the genes for seed color and texture. In pea plants, purple flowers (P) are dominant to white flowers (p) and yellow peas (Y) are dominant to green peas (y). What are the possible genotypes and phenotypes for a cross between PpYY and ppYy pea plants? What is the minimum number of squares that you need to do a Punnett square analysis of this cross? a. ppYY, Ppyy, ppYY, ppy |
y yielding white flowers with yellow peas, purple flowers with yellow peas, and white flowers with green peas. You can find this with a 3×3 Punnett square. b. PPYY, PpYy, ppYY, ppyy yielding purple flowers with yellow peas, white flowers with yellow peas, and white flowers with green peas. You can find this with a 2×2 Punnett square. c. Ppyy, PpYy, ppYY, ppyy yielding purple flowers with green peas, purple flowers with yellow peas, white flowers with yellow peas, and white flowers with green peas. You can find this with a 3×3 Punnett square. d. PpYY, PpYy, ppYY, ppYy yielding purple flowers with yellow peas, and white flowers with yellow peas. You can find this with a 2×2 Punnett square. For the F2 generation, the law of segregation requires that each gamete receive either an R allele or an r allele along with either a Y allele or a y allele. The law of independent assortment states that a gamete into which an r allele sorted would be equally likely to contain either a Y allele or a y allele. Thus, there are four equally likely gametes that can be formed when the YyRr heterozygote is self-crossed, as follows: YR, Yr, yR, and yr. Arranging these gametes along the top and left of a 4 × 4 Punnett square (Figure 12.16) gives us 16 equally likely genotypic combinations. From these genotypes, we infer a phenotypic ratio of 9 round/yellow:3 round/green:3 wrinkled/yellow:1 wrinkled/green (Figure 12.16). These are the offspring ratios we would expect, assuming we performed the crosses with a large enough sample size. Because of independent assortment and dominance, the 9:3:3:1 dihybrid phenotypic ratio can be collapsed into two 3:1 ratios, characteristic of any monohybrid cross that follows a dominant and recessive pattern. Ignoring seed color and considering only seed texture in the above dihybrid cross, we would expect that three quarters of the F2 generation offspring would be round, and one quarter would be wrinkled. Similarly, isolating only seed color, we would assume that three quarters of the F2 offspring would be yellow and one quarter |
would be green. The sorting of alleles for texture and color are independent events, so we can apply the product rule. Therefore, the proportion of round and yellow F2 offspring is 498 Chapter 12 | Mendel's Experiments and Heredity expected to be (3/4) × (3/4) = 9/16, and the proportion of wrinkled and green offspring is expected to be (1/4) × (1/4) = 1/16. These proportions are identical to those obtained using a Punnett square. Round, green and wrinkled, yellow offspring can also be calculated using the product rule, as each of these genotypes includes one dominant and one recessive phenotype. Therefore, the proportion of each is calculated as (3/4) × (1/4) = 3/16. The law of independent assortment also indicates that a cross between yellow, wrinkled (YYrr) and green, round (yyRR) parents would yield the same F1 and F2 offspring as in the YYRR x yyrr cross. 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. Forked-Line Method When more than two genes are being considered, the Punnett-square method becomes unwieldy. For instance, examining a cross involving four genes would require a 16 × 16 grid containing 256 boxes. It would be extremely cumbersome to manually enter each genotype. For more complex crosses, the forked-line and probability methods are preferred. To prepare a forked-line diagram for a cross between F1 heterozygotes resulting from a cross between AABBCC and aabbcc parents, we first create rows equal to the number of genes being considered, and then segregate the alleles in each row on forked lines according to the probabilities for individual monohybrid crosses (Figure 12.17). We then multiply the values along each forked path to obtain the F2 offspring probabilities. Note that this process is a diagrammatic version of the product rule. The values along each forked pathway can be multiplied because each gene assorts independently. For a trihybrid cross, the F2 phenotypic ratio is 27:9:9:9:3: |
3:3:1. Figure 12.17 The forked-line method can be used to analyze a trihybrid cross. Here, the probability for color in the F2 generation occupies the top row (3 yellow:1 green). The probability for shape occupies the second row (3 round:1 wrinked), and the probability for height occupies the third row (3 tall:1 dwarf). The probability for each possible combination of traits is calculated by multiplying the probability for each individual trait. Thus, the probability of F2 offspring having yellow, round, and tall traits is 3 × 3 × 3, or 27. Probability Method While the forked-line method is a diagrammatic approach to keeping track of probabilities in a cross, the probability method gives the proportions of offspring expected to exhibit each phenotype (or genotype) without the added visual assistance. Both methods make use of the product rule and consider the alleles for each gene separately. Earlier, we examined the phenotypic proportions for a trihybrid cross using the forked-line method; now we will use the probability method to examine the genotypic proportions for a cross with even more genes. For a trihybrid cross, writing out the forked-line method is tedious, albeit not as tedious as using the Punnett-square method. To fully demonstrate the power of the probability method, however, we can consider specific genetic calculations. For instance, for a tetrahybrid cross between individuals that are heterozygotes for all four genes, and in which all four genes are sorting independently and in a dominant and recessive pattern, what proportion of the offspring will be expected to be homozygous recessive for all four alleles? Rather than writing out every possible genotype, we can use the probability method. We know that for each gene, the fraction of homozygous recessive offspring will be 1/4. Therefore, multiplying this fraction for each of the four genes, (1/4) × (1/4) × (1/4) × (1/4), we determine that 1/256 of the offspring will be quadruply homozygous recessive. For the same tetrahybrid cross, what is the expected proportion of offspring that have the dominant phenotype at all four loci? We can answer this question using phenotypic proportions, but let’s do it the hard way—using genotypic proportions. The question asks for the proportion of offspring that are 1) homozyg |
ous dominant at A or heterozygous at A, and 2) homozygous at B or heterozygous at B, and so on. Noting the “or” and “and” in each circumstance makes clear where to This OpenStax book is available for free at http://cnx.org/content/col12078/1.6 Chapter 12 | Mendel's Experiments and Heredity 499 apply the sum and product rules. The probability of a homozygous dominant at A is 1/4 and the probability of a heterozygote at A is 1/2. The probability of the homozygote or the heterozygote is 1/4 + 1/2 = 3/4 using the sum rule. The same probability can be obtained in the same way for each of the other genes, so that the probability of a dominant phenotype at A and B and C and D is, using the product rule, equal to 3/4 × 3/4 × 3/4 × 3/4, or 27/64. If you are ever unsure about how to combine probabilities, returning to the forked-line method should make it clear. Rules for Multihybrid Fertilization Predicting the genotypes and phenotypes of offspring from given crosses is the best way to test your knowledge of Mendelian genetics. Given a multihybrid cross that obeys independent assortment and follows a dominant and recessive pattern, several generalized rules exist; you can use these rules to check your results as you work through genetics calculations (Table 12.5). To apply these rules, first you must determine n, the number of heterozygous gene pairs (the number of genes segregating two alleles each). For example, a cross between AaBb and AaBb heterozygotes has an n of 2. In contrast, a cross between AABb and AABb has an n of 1 because A is not heterozygous. General Rules for Multihybrid Crosses General Rule Number of Heterozygous Gene Pairs Number of different F1 gametes Number of different F2 genotypes Given dominant and recessive inheritance, the number of different F2 phenotypes Table 12.5 2n 3n 2n Linked Genes Violate the Law of Independent Assortment Although all of Mendel’s pea characteristics behaved according to the law of independent assortment, we now know that some allele combinations are not inherited independently |
of each other. Genes that are located on separate nonhomologous chromosomes will always sort independently. However, each chromosome contains hundreds or thousands of genes, organized linearly on chromosomes like beads on a string. The segregation of alleles into gametes can be influenced by linkage, in which genes that are located physically close to each other on the same chromosome are more likely to be inherited as a pair. However, because of the process of recombination, or “crossover,” it is possible for two genes on the same chromosome to behave independently, or as if they are not linked. To understand this, let’s consider the biological basis of gene linkage and recombination. Homologous chromosomes possess the same genes in the same linear order. The alleles may differ on homologous chromosome pairs, but the genes to which they correspond do not. In preparation for the first division of meiosis, homologous chromosomes replicate and synapse. Like genes on the homologs align with each other. At this stage, segments of homologous chromosomes exchange linear segments of genetic material (Figure 12.18). This process is called recombination, or crossover, and it is a common genetic process. Because the genes are aligned during recombination, the gene order is not altered. Instead, the result of recombination is that maternal and paternal alleles are combined onto the same chromosome. Across a given chromosome, several recombination events may occur, causing extensive shuffling of alleles. 500 Chapter 12 | Mendel's Experiments and Heredity Figure 12.18 The process of crossover, or recombination, occurs when two homologous chromosomes align during meiosis and exchange a segment of genetic material. Here, the alleles for gene C were exchanged. The result is two recombinant and two non-recombinant chromosomes. When two genes are located in close proximity on the same chromosome, they are considered linked, and their alleles tend to be transmitted through meiosis together. To exemplify this, imagine a dihybrid cross involving flower color and plant height in which the genes are next to each other on the chromosome. If one homologous chromosome has alleles for tall plants and red flowers, and the other chromosome has genes for short plants and yellow flowers, then when the gametes are formed, the tall and red alleles will go together into a gamete and the short and yellow alleles will go into other gametes. These are called the parental gen |
otypes because they have been inherited intact from the parents of the individual producing gametes. But unlike if the genes were on different chromosomes, there will be no gametes with tall and yellow alleles and no gametes with short and red alleles. If you create the Punnett square with these gametes, you will see that the classical Mendelian prediction of a 9:3:3:1 outcome of a dihybrid cross would not apply. As the distance between two genes increases, the probability of one or more crossovers between them increases, and the genes behave more like they are on separate chromosomes. Geneticists have used the proportion of recombinant gametes (the ones not like the parents) as a measure of how far apart genes are on a chromosome. Using this information, they have constructed elaborate maps of genes on chromosomes for well-studied organisms, including humans. Mendel’s seminal publication makes no mention of linkage, and many researchers have questioned whether he encountered linkage but chose not to publish those crosses out of concern that they would invalidate his independent assortment postulate. The garden pea has seven chromosomes, and some have suggested that his choice of seven characteristics was not a coincidence. However, even if the genes he examined were not located on separate chromosomes, it is possible that he simply did not observe linkage because of the extensive shuffling effects of recombination. Testing the Hypothesis of Independent Assortment To better appreciate the amount of labor and ingenuity that went into Mendel’s experiments, proceed through one of Mendel’s dihybrid crosses. Question: What will be the offspring of a dihybrid cross? Background: Consider that pea plants mature in one growing season, and you have access to a large garden in which you can cultivate thousands of pea plants. There are several true-breeding plants with the following pairs of This OpenStax book is available for free at http://cnx.org/content/col12078/1.6 Chapter 12 | Mendel's Experiments and Heredity 501 traits: tall plants with inflated pods, and dwarf plants with constricted pods. Before the plants have matured, you remove the pollen-producing organs from the tall/inflated plants in your crosses to prevent self-fertilization. Upon plant maturation, the plants are manually crossed by transferring pollen from the dwarf/constricted plants to the stigmata of the tall/inflated plants. |
Hypothesis: Both trait pairs will sort independently according to Mendelian laws. When the true-breeding parents are crossed, all of the F1 offspring are tall and have inflated pods, which indicates that the tall and inflated traits are dominant over the dwarf and constricted traits, respectively. A self-cross of the F1 heterozygotes results in 2,000 F2 progeny. Test the hypothesis: Because each trait pair sorts independently, the ratios of tall:dwarf and inflated:constricted are each expected to be 3:1. The tall/dwarf trait pair is called T/t, and the inflated/constricted trait pair is designated I/ i. Each member of the F1 generation therefore has a genotype of TtIi. Construct a grid analogous to Figure 12.16, in which you cross two TtIi individuals. Each individual can donate four combinations of two traits: TI, Ti, tI, or ti, meaning that there are 16 possibilities of offspring genotypes. Because the T and I alleles are dominant, any individual having one or two of those alleles will express the tall or inflated phenotypes, respectively, regardless if they also have a t or i allele. Only individuals that are tt or ii will express the dwarf and constricted alleles, respectively. As shown in Figure 12.19, you predict that you will observe the following offspring proportions: tall/inflated:tall/constricted:dwarf/ inflated:dwarf/constricted in a 9:3:3:1 ratio. Notice from the grid that when considering the tall/dwarf and inflated/ constricted trait pairs in isolation, they are each inherited in 3:1 ratios. Figure 12.19 This figure shows all possible combinations of offspring resulting from a dihybrid cross of pea plants that are heterozygous for the tall/dwarf and inflated/constricted alleles. Test the hypothesis: You cross the dwarf and tall plants and then self-cross the offspring. For best results, this is repeated with hundreds or even thousands of pea plants. What special precautions should be taken in the crosses and in growing the plants? Analyze your data: You observe the following plant phenotypes in the F2 generation: 2706 tall/inflated, 930 tall/ constricted, 888 dwarf/inflated, and 300 dwarf/constricted. Reduce these findings to a ratio and determine if they |
are consistent with Mendelian laws. Form a conclusion: Were the results close to the expected 9:3:3:1 phenotypic ratio? Do the results support the prediction? What might be observed if far fewer plants were used, given that alleles segregate randomly into gametes? Try to imagine growing that many pea plants, and consider the potential for experimental error. For instance, what would happen if it was extremely windy one day? 502 Chapter 12 | Mendel's Experiments and Heredity Think About It In the shepherd’s-purse plant (Capsella bursa-pastoris), seed shape is controlled by two genes, A and B. When both the A and B loci are homozygous recessive (aabb), the seeds are ovoid. However, if the dominant allele for either or both of these genes is present, the seeds are triangular. Based on this information, what are the expected phenotypic ratios for a cross between plants that are heterozygous for both traits? What is the expected ratio of phenotypes from a dihybrid cross? How do you explain the difference between the expected dihybrid cross ratio and ratio observed in the shepherd’s-purse plant? Epistasis Mendel’s studies in pea plants implied that the sum of an individual’s phenotype was controlled by genes (or as he called them, unit factors), such that every characteristic was distinctly and completely controlled by a single gene. In fact, single observable characteristics are almost always under the influence of multiple genes (each with two or more alleles) acting in unison. For example, at least eight genes contribute to eye color in humans. Eye color in humans is determined by multiple genes. Use the Eye Color Calculator (http://openstaxcollege.org/l/ eye_color_calc) to predict the eye color of children from parental eye color. A couple produces a green-eyed child. Both of the parents have brown eyes. Explain how this is genetically possible. a. Both parents are homozygous for the dominant trait of brown eyes. b. Both parents are heterozygous, having the green trait on the green-blue eye gene. c. Both parents are heterozygous with the recessive trait of brown eyes. d. Both parents are homozygous having the green trait on the green-blue eye gene. In some cases, several genes can contribute to aspects of a common phenotype without |
their gene products ever directly interacting. In the case of organ development, for instance, genes may be expressed sequentially, with each gene adding to the complexity and specificity of the organ. Genes may function in complementary or synergistic fashions, such that two or more genes need to be expressed simultaneously to affect a phenotype. Genes may also oppose each other, with one gene modifying the expression of another. In epistasis, the interaction between genes is antagonistic, such that one gene masks or interferes with the expression of another. “Epistasis” is a word composed of Greek roots that mean “standing upon.” The alleles that are being masked or silenced are said to be hypostatic to the epistatic alleles that are doing the masking. Often the biochemical basis of epistasis is a gene pathway in which the expression of one gene is dependent on the function of a gene that precedes or follows it in the pathway. An example of epistasis is pigmentation in mice. The wild-type coat color, agouti (AA), is dominant to solid-colored fur (aa). However, a separate gene (C) is necessary for pigment production. A mouse with a recessive c allele at this locus is unable to produce pigment and is albino regardless of the allele present at locus A (Figure 12.20). Therefore, the genotypes AAcc, Aacc, and aacc all produce the same albino phenotype. A cross between heterozygotes for both genes (AaCc x AaCc) This OpenStax book is available for free at http://cnx.org/content/col12078/1.6 Chapter 12 | Mendel's Experiments and Heredity 503 would generate offspring with a phenotypic ratio of 9 agouti:3 solid color:4 albino (Figure 12.20). In this case, the C gene is epistatic to the A gene. Figure 12.20 In mice, the mottled agouti coat color (A) is dominant to a solid coloration, such as black or gray. A gene at a separate locus (C) is responsible for pigment production. The recessive c allele does not produce pigment, and a mouse with the homozygous recessive cc genotype is albino regardless of the allele present at the A locus. Thus, the C gene is epistatic to the A gene. Ep |
istasis can also occur when a dominant allele masks expression at a separate gene. Fruit color in summer squash is expressed in this way. Homozygous recessive expression of the W gene (ww) coupled with homozygous dominant or heterozygous expression of the Y gene (YY or Yy) generates yellow fruit, and the wwyy genotype produces green fruit. However, if a dominant copy of the W gene is present in the homozygous or heterozygous form, the summer squash will produce white fruit regardless of the Y alleles. A cross between white heterozygotes for both genes (WwYy × WwYy) would produce offspring with a phenotypic ratio of 12 white:3 yellow:1 green. Finally, epistasis can be reciprocal such that either gene, when present in the dominant (or recessive) form, expresses the same phenotype. In the shepherd’s purse plant (Capsella bursa-pastoris), the characteristic of seed shape is controlled by two genes in a dominant epistatic relationship. When the genes A and B are both homozygous recessive (aabb), the seeds are ovoid. If the dominant allele for either of these genes is present, the result is triangular seeds. That is, every possible genotype other than aabb results in triangular seeds, and a cross between heterozygotes for both genes (AaBb x AaBb) would yield offspring with a phenotypic ratio of 15 triangular:1 ovoid. As you work through genetics problems, keep in mind that any single characteristic that results in a phenotypic ratio that totals 16 is typical of a two-gene interaction. Recall the phenotypic inheritance pattern for Mendel’s dihybrid cross, which considered two non-interacting genes—9:3:3:1. Similarly, we would expect interacting gene pairs to also exhibit ratios expressed as 16 parts. Note that we are assuming the interacting genes are not linked; they are still assorting independently into gametes. 504 Chapter 12 | Mendel's Experiments and Heredity For an excellent review of Mendel’s experiments and to perform your own crosses and identify patterns of inheritance, visit the Mendel’s Peas (http://openstaxcollege.org/l/mendels_peas) web lab. Explain how Mendel’s experiments help modern-day farmers breed crops that exhibit preferred traits, like |
tall height or large fruit size. a. by providing information about the variety of pea plants b. by providing information about soil condition c. by providing information about organic feritilizers d. by providing information about the inheritance of traits and the concept of dominance This OpenStax book is available for free at http://cnx.org/content/col12078/1.6 Chapter 12 | Mendel's Experiments and Heredity 505 KEY TERMS allele gene variations that arise by mutation and exist at the same relative locations on homologous chromosomes autosomes any of the non-sex chromosomes blending theory of inheritance hypothetical inheritance pattern in which parental traits are blended together in the offspring to produce an intermediate physical appearance codominance in a heterozygote, complete and simultaneous expression of both alleles for the same characteristic continuous variation inheritance pattern in which a character shows a range of trait values with small gradations rather than large gaps between them dihybrid result of a cross between two true-breeding parents that express different traits for two characteristics discontinuous variation inheritance pattern in which traits are distinct and are transmitted independently of one another dominant trait which confers the same physical appearance whether an individual has two copies of the trait or one copy of the dominant trait and one copy of the recessive trait dominant lethal inheritance pattern in which an allele is lethal both in the homozygote and the heterozygote; this allele can only be transmitted if the lethality phenotype occurs after reproductive age epistasis antagonistic interaction between genes such that one gene masks or interferes with the expression of another F1 F2 first filial generation in a cross; the offspring of the parental generation second filial generation produced when F1 individuals are self-crossed or fertilized with each other genotype underlying genetic makeup, consisting of both physically visible and non-expressed alleles, of an organism hemizygous presence of only one allele for a characteristic, as in X-linkage; hemizygosity makes descriptions of dominance and recessiveness irrelevant heterozygous having two different alleles for a given gene on the homologous chromosome homozygous having two identical alleles for a given gene on the homologous chromosome hybridization offspring process of mating two individuals that differ with the goal of achieving a certain characteristic in their incomplete dominance in a heterozygote, expression of two contrasting alleles such that the individual displays an intermediate phenotype law of dominance in a heterozygote, one trait will conceal the presence of another trait for the same |
ressed traits are described as recessive. When the offspring in Mendel’s experiment were self-crossed, the F2 offspring exhibited the dominant trait or the recessive trait in a 3:1 ratio, confirming that the recessive trait had been transmitted faithfully from the original P0 parent. Reciprocal crosses generated identical F1 and F2 offspring ratios. By examining sample sizes, Mendel showed that his crosses behaved reproducibly according to the laws of probability, and that the traits were inherited as independent events. Two rules in probability can be used to find the expected proportions of offspring of different traits from different crosses. To find the probability of two or more independent events occurring together, apply the product rule and multiply the probabilities of the individual events. The use of the word “and” suggests the appropriate application of the product rule. To find the probability of two or more events occurring in combination, apply the sum rule and add their individual probabilities together. The use of the word “or” suggests the appropriate application of the sum rule. 12.2 Characteristics and Traits When true-breeding or homozygous individuals that differ for a certain trait are crossed, all of the offspring will be heterozygotes for that trait. If the traits are inherited as dominant and recessive, the F1 offspring will all exhibit the same phenotype as the parent homozygous for the dominant trait. If these heterozygous offspring are self-crossed, the resulting F2 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 F2 offspring will exhibit a ratio of three dominant to one recessive. 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 This OpenStax book is available for free at http://cnx.org/content/col12078/1.6 Chapter 12 | Mendel's Experiments and Heredity 507 for any given gene, it is common for more than two alleles |
of a gene to exist in a population. In humans, as in many animals and some plants, 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. Finally, some alleles can be lethal. Recessive lethal alleles are only lethal in homozygotes, but dominant lethal alleles are fatal in heterozygotes as well. 12.3 Laws of Inheritance 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, alleles are generally not more likely to segregate into a gamete with a particular allele of another gene. A dihybrid cross demonstrates independent assortment when the genes in question are on different chromosomes or distant from each other on the same chromosome. For crosses involving more than two genes, use the forked line or probability methods to predict offspring genotypes and phenotypes rather than a Punnett square. Although chromosomes sort independently into gametes during meiosis, Mendel’s law of independent assortment refers to genes, not chromosomes, and a single chromosome may carry more than 1,000 genes. When genes are located in close proximity on the same chromosome, their alleles tend to be inherited together. This results in offspring ratios that violate Mendel's law of independent assortment. However, recombination serves to exchange genetic material on homologous chromosomes such that maternal and paternal alleles may be recombined on the same chromosome. This is why alleles on a given chromosome are not always inherited together. Recombination is a random event occurring anywhere on a chromosome. Therefore, genes that are far apart on the same chromosome are likely to still assort independently because of recombination events that occurred in the intervening chromosomal space. Whether or not they are sorting independently, genes may interact at the level of gene products such that the expression of an allele for one gene masks or modifies the expression of an allele for a different gene. This is called epistasis. REVIEW QUESTIONS 1. Mendel performed hybridizations by transferring pollen to the female |
ova from what part of the male plant? a. anther b. pistil c. d. stigma seed 2. Which is one of the seven characteristics that Mendel observed in pea plants? a. b. c. d. flower size leaf shape seed texture stem color 3. Imagine you are performing a cross involving garden offspring would you expect if you pea plants. What F1 cross true-breeding parents with green seeds and yellow seeds? Yellow seed color is dominant over green. a. 100 percent yellow-green seeds b. 100 percent yellow seeds c. 50 percent yellow, 50 percent green seeds d. 25 percent green, 75 percent yellow seeds 4. Consider a cross to investigate the pea pod texture trait, involving constricted or inflated pods. Mendel found that the traits behave according to a dominant/recessive pattern in which inflated pods were dominant. If you performed this cross and obtained 650 inflated-pod plants in the F2 generation bred from true-breading stock, approximately how many constricted-pod plants would you expect to have? a. 600 b. 165 c. 217 d. 468 5. The observable traits expressed by an organism are described as its a. alleles b. genotype c. phenotype d. zygote 6. A recessive trait will be observed in individuals that are what for that trait? a. diploid b. heterozygous c. homozygous or heterozygous d. homozygous 7. If black and white true-breeding mice are mated and the result is all gray offspring, what inheritance pattern would 508 Chapter 12 | Mendel's Experiments and Heredity this be indicative of? a. codominance b. dominance c. incomplete dominance d. multiple alleles 8. The ABO blood groups in humans are controlled by the IA, IB, and I alleles. The IA allele encodes the A blood group antigen, IB encodes B, and I encodes O. Both A and B are dominant to O. If a heterozygous blood type A parent (iAi) and a heterozygous blood type B parent (iBi) mate, one quarter of their offspring will have AB blood type (IAIB) in which both antigens are expressed equally. Therefore, the ABO blood groups are an example of _______. a. codominance and incomplete dominance b. incomplete dominance only c. multiple alleles and incomplete dominance d. multiple alleles and codomin |
ance 9. In a mating between two individuals that are heterozygous for a recessive lethal allele that is expressed in utero, what genotypic ratio (homozygous dominant : heterozygous : homozygous recessive) would you expect to observe in the off-spring? a. b. c. d 10. The forked line and probability methods make use of what probability rule? a. monohybrid rule b. product rule c. d. sum rule test cross 11. This OpenStax book is available for free at http://cnx.org/content/col12078/1.6 In pea plants, smooth seeds (S) are dominant to wrinkled seeds (s). In a genetic cross of two plants that are heterozygous for the seed shape trait, the Punnett square is shown. What is the missing genotype? a. SS b. Ss c. d. sS ss 12. If the inheritance of two traits fully obeys Mendelian laws of inheritance, where may you assume that the genes are located? a. on any autosomal chromosome or chromosomes b. on Y chromosomes c. on the same chromosome d. on separate chromosomes 13. How many different offspring genotypes are expected in a trihybrid cross between parents heterozygous for all three traits? How many phenotypes are expected if the traits behave in a dominant and recessive pattern? a. 64 genotypes; 16 phenotypes b. 16 genotypes; 64 phenotypes c. 8 genotypes; 27 phenotypes d. 27 genotypes; 8 phenotypes 14. Four-o’ clock flowers may be red, pink or white. In the crossing of a true-breeding red and true-breeding white plants, all the offspring are pink. Use a Punnett square to determine the correct genotype of the offspring if the red parent has genotype RR and the white parent has genotype rr. a. RR and Rr b. Rr and rr c. Rr only d. RR only Chapter 12 | Mendel's Experiments and Heredity 509 15. Which cellular process underlies Mendel’s law of independent assortment? a. Chromosomes align randomly during meiosis. b. Chromosomes can exchange genetic material during crossover. c. Gametes contain half the number of chromosomes of somatic cells. d. Daughter cells are genetically identical to parent cells after mitosis. 16 |
. While studying meiosis, you observe that gametes receive one copy of each pair of homologous chromosomes and one copy of the sex chromosomes. This observation is the physical explanation of Mendel’s law of ______. a. dominance b. c. d. independent assortment random distribution of traits segregation 17. In some primroses, the petal color blue is dominant. A cross between a true-breed blue primrose and a white primrose yields progeny with white petals. A second gene at another locus prevented the expression of the dominant coat color. This is an example of ______. a. codominance b. hemizygosity c. incomplete dominance d. epistasis 18. CRITICAL THINKING QUESTIONS 20. Describe one reason why the garden pea was an excellent model system for studying inheritance. a. The garden pea has flowers that close tightly to promote cross-fertilization. b. The garden pea has flowers that close tightly to prevent cross-fertilization. c. The garden pea does not mature in one season and is a perennial plant. d. Male and female reproductive parts attain maturity at different times, promoting selffertilization. 21. How would you perform a reciprocal cross to test stem height in the garden pea? Purple flowers (P) are dominant over red flowers (p) and long pollen grains are dominant over round pollen grains. When purple flowers and long pollen grain plants were crossed with plants with white flowers and round pollen grains, all the F1 pollen grains. The F1 are in the table. What conclusions about the physical relationship between the traits can be drawn from the experiment? plants showed purple flowers and long plants were crossed and the results a. The traits are probably linked. b. The traits follow the law of independent assortment c. The traits are located on different chromosomes d. There was epistasis. 19. When the expression of one gene pair masks or modifies the expression of another, the genes show _______. a. codominance b. epistasis c. incomplete dominance d. partial linkage 510 Chapter 12 | Mendel's Experiments and Heredity a. First cross is performed by transferring the a. Yes, males can be the carriers of red-green color blindness, as color blindness is autosomal dominant. b. No, males cannot be the carriers of red-green color blindness, as color blindness is X-linked. c. No, males cannot |
be the carriers of red-green color blindness, as color blindness is Y-linked. d. Yes, males can be the carriers of red-green color blindness, as color blindness is autosomal recessive. 25. Use the probability method to calculate the genotypes and genotypic proportions of a cross between AABBCc and Aabbcc parents. a. Possible genotypes are AABBcc, AaBbCc, AaBbcc and the ratio 1 : 2 : 1. b. Possible genotypes are AABbcc, AaBbCc, AaBbcc and the ratio 1 : 3 : 1. c. Possible genotypes are AABbCc, AABbcc, AaBbCc, AaBbcc and the ratio 1 : 1 : 1 : 1. d. Possible genotypes are AABbcc, AaBbCC, AaBbcc and the ratio 1 : 1 : 1. 26. How does the segregation of traits result in different combinations of gametes at the end of meiosis? pollen of a heterozygous tall plant to the stigma of a true breeding dwarf plant. Second cross is performed by transferring the pollen of a heterozygous dwarf plant to the stigma of a true breeding tall plant. b. First cross is performed by transferring the pollen of a true breeding tall plant to the stigma of a true breeding dwarf plant. Second cross is performed by transferring the pollen of a true breeding dwarf plant to the stigma of a true breeding tall plant. c. First cross is performed by transferring the pollen of a true breeding tall plant to the stigma of a heterozygous dwarf plant. Second cross is performed by transferring the pollen of a heterozygous dwarf plant to the stigma of a true breeding tall plant. d. First cross is performed by transferring the pollen of a heterozygous tall plant to the stigma of a heterozygous dwarf plant. Second cross is performed by transferring the pollen of a heterozygous tall plant to the stigma of a heterozygous dwarf plant. 22. Flower position in pea plants is determined by a gene with axial and terminal alleles. Given that axial is dominant to terminal, list all of the possible F1 and F2 genotypes and phenotypes from a cross involving parents that are homozygous for each trait. Express genotypes with conventional genetic abbreviations. a. F1: All AA-axial; F2 |
: AA-Axial and aa-terminal. b. F1: All aa-terminal; F2: AA-Axial and Aa- terminal. c. F1: AA-axial and Aa-terminal; F2: All AA-axial. d. F1: All Aa-axial; F2: AA-Axial, Aa-Axial, and aa-terminal. 23. 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? a. 1 Tall : 1 dwarf b. 1 tall : 2 dwarf c. 3 tall : 1 dwarf d. 1 dwarf : 4 tall 24. Can a human male be a carrier of red-green color blindness? This OpenStax book is available for free at http://cnx.org/content/col12078/1.6 Chapter 12 | Mendel's Experiments and Heredity 511 27. In Section 12.3, “Laws of Inheritance,” an example of epistasis was given for summer squash. Cross white WwYy heterozygotes to demonstrate the phenotypic ratio of 12 white : 3 yellow : 1 green that was given in the text. a. 12 offspring are white, as the W gene is epistatic to the Y gene. Three offspring are yellow, because w is not epistatic. Green offspring is obtained when the recessive form of both genes (wwyy) are present. b. 12 offspring are white as W gene is hypostatic to Y gene. Three offspring are yellow because Y is epistatic to w. Green offspring is obtained when the dominant form of both the genes (WWYY) is present. c. 12 offspring are white as W gene is dominant. Three offspring are yellow because Y is dominant and w is recessive. Green offspring is obtained when the recessive form of both the genes (wwyy) is present, showing codominance. d. 12 offspring are white as W is epistatic to Y gene. Three offspring are yellow because Y is hypostatic to w. Green offspring is obtained when the recessive form of both the genes (wwyy) are present, showing codominance. a. The chromosomes randomly align during metaphase I at the equator, and |
separation of homologous chromosomes occurs during anaphase I. Similarly separation of sister chromatids occurs at anaphase II of meiosis II. At the end of meiosis II, four different gametic combinations are produced, each containing a haploid set of chromosomes. b. The chromosomes randomly align during anaphase I at the equator. Separation of bivalent chromosomes occur during metaphase I of meiosis I. Similarly, separation of sister chromatids occurs at metaphase II of meiosis II. At the end of meiosis II, four different gametic combinations are produced, each containing a haploid set of chromosomes. c. The chromosomes randomly align during prophase I at the equator, and separation of sister chromatids occurs during metaphase I of meiosis I. Similarly separation of bivalent chromosomes occur at metaphase II of meiosis II. At the end of meiosis II, four different gametic combinations are produced, each containing a diploid set of chromosomes. d. The chromosomes randomly align during prophase I at the equator, and separation of bivalent chromosomes occur during anaphase I of meiosis I. Similarly, separation of homologous chromosomes occurs at metaphase II of meiosis II. At the end of meiosis II, four different gametic combinations are produced, each containing a diploid set of chromosomes. TEST PREP FOR AP® COURSES 28. The trait for widow’s peak can be considered a monoallelic dominant trait in humans. If a man with a widow’s peak and a woman with a straight hairline have a child together, what is the probability that the child will inherit the widow’s peak if you know that the father’s mother had a straight hairline? a. 0.25 b. 0.5 c. 0.75 d. 1 29. Don’t like Brussels sprouts? Blame your genes. The chemical PTC (phenylthiocarbamide), which is nearly identical to a compound found in the cabbage family, tastes very bitter for some people. Others cannot detect a taste. The ability to taste PTC is incompletely dominant and is controlled by a gene on chromosome 7. A woman who finds Brussel sprouts mildly distasteful (in other words, who can taste PTC weakly) has a child with a man who hates Brussel sprouts (in other words, who can taste PTC strongly). What is |
the probability that their son likes Brussel sprouts (in other words, cannot taste PTC)? a. 0 b. 0.25 c. 0.5 d. 1 30. Tay-Sachs disease is an autosomal recessive disorder that causes severe problems in neurons. Children who receive two copies of the gene rarely live beyond the age of five. There is no cure for the disease. During a genetic screening, a couple is told that both partners carry the recessive gene. What kind of issue must the couple confront? a. b. scientific financial c. ethical d. educational 31. A couple has three daughters. What is the probability that the next child they have will be a daughter? 512 Chapter 12 | Mendel's Experiments and Heredity a. b. c. d. 0% 25% 50% 100% 32. What is the probability that a couple will have three daughters? a. Gray body and cinnabar eyes are dominant. b. Eye color is sex-linked. c. Body color is sex-linked. d. Gray body and red eyes are dominant. 36. b. c. d. 33. Petunias can be blue, red, or violet. When a blue flower is crossed with a red flower, all the resulting flowers are violet. When a violet flower is crossed with a red flower, about half of the flowers are violet and half are red. How do you characterize the color trait? a. complete dominance b. codominance c. d. incomplete dominance sex-linked 34. Petunias can be blue, red, or violet. When a blue flower is crossed with a red flower, all the resulting flowers are violet. Two violet petunias are crossed. Which is the most probable result of the cross? a. b. 75% of the flowers are blue and 25% of the flowers are red. 50% of the flowers are blue and 50% of the flowers are red. c. 75% of the flowers are red and 25% are blue. d. 25% of the flowers are blue, 50% of the flowers are violet, and 25% of the flowers are red. 35. Fruit flies (Drosophila melanogaster) with a wild-type phenotype have gray bodies and red eyes. Certain mutations can cause changes to these traits. Mutant flies may have a black body and/or cinnabar eyes. To study the genetics of these traits, a researcher crossed a truebreeding |
wild-typed male fly with a true-breeding female fly with a black body and cinnabar eyes. All of the F1 progeny displayed a wild type phenotype. Which of the following is correct about the traits observed? generation were crossed with Female flies from the F1 true-breeding male flies with black bodies and cinnabar eyes. The table represents the predicted outcome and the data obtained from the cross. What was the assumption that lead to the predicted numbers? a. The traits assort independently. b. The traits are located on the X chromosome. c. The traits are on the same chromosome. d. The female flies were homozygous for wild type alleles. 37. Cats can be black, yellow, or calico (black and yellow patches). Coat color is carried on the X chromosome. What type of inheritance is color coat in cats? 1. Complete dominance 2. Codominance 3. Incomplete dominance 4. Sex-linked a. 2 b. 3 c. 2,4 d. 3,4 38. Cats can be black, yellow or calico (black and yellow patches). Coat color is carried on the X chromosome. A yellow cat is crossed with a black cat. Assume that the offspring are both male and female. What are the phenotypes of the offspring and in what proportions? a. All the cats are yellow. b. All the cats are black. c. All the cats are calico. d. There is not enough information to answer the question. SCIENCE PRACTICE CHALLENGE QUESTIONS 39. The gene SLC24A5 encodes an antiporter membrane protein that exchanges sodium for calcium (R. Ginger et This OpenStax book is available for free at http://cnx.org/content/col12078/1.6 Chapter 12 | Mendel's Experiments and Heredity 513 al., JBC, 2007). This process has a role in the synthesis of the melanosomes that cause skin pigmentation. A mutation in this gene affecting a single amino acid occurs in humans. The homozygous mutant gene is found in 99% of humans with European origins. Both the wild type and mutant display codominance. A. Representing the wild-type form of the gene as +/+ and the mutant form of the gene as m/m for two homozygous parents, construct a Punnett square for this cross using the first grid below. Annotate your representation to |
identify the phenotypes with high (H), intermediate (I), and low (L) melanosome production. Use the second grid to represent an F2 generation from the offspring of the first cross. Use annotation to show the phenotype. SCLO24A5 are codominant. The definition of the statistic 2 = ∑ (Oi − Ei)2 Χc Ei where X is the chi-square test statistic, c is the significant level of the test (we will use 0.05), O is the observed value for variable i, and E is the expected value for variable i. The Chi-square statistic table is provided in the AP Biology Exam. Degrees of Freedom.05 3.84 5.99 7.82 9.49 11.07 12.59 14.07 15.51 Table 12.9 F1 m m + + Table 12.6 F2 Table 12.7 B. Draw sister chromatids at anaphase II for both parents in the F1 generation and annotate your drawing to identify each genotype of the gametes using the cells of the Punnett square. C. Explain which of Mendel’s laws is violated by codominance. D. Suppose that these data were available to evaluate the claim that the wild-type and mutant forms of SLC24A5 are codominant: F2 Phenotype Observed Expected H I L Table 12.8 1206 2238 1124 Complete the table. Explain the values expected in terms of the genotype of the offspring. E. Using a c2 statistic at the 95% confidence level, evaluate the claim that the wild-type and mutant forms of 40. Adrenoleukodystrophy (ALD) is a genetic disorder in which lipids with very high molecular weights are not metabolized and accumulate within cells. Accumulation of these fats in the brain damages the myelin that surrounds nerves. This progressive disease has two causes: an autosomal recessive allele, which causes neonatal ALD, and a mutation in the ABCD1 gene located on the X chromosome. A controversial treatment is the use of Lorenzo’s oil, which is expensive; despite this treatment, neurological degradation persists in many patients. Gene therapy as a potential treatment is currently in trials but is also very costly. An infant patient exhibits symptoms of neonatal ALD, which are difficult to distinguish from the X-linked form of the disease. The infant’s physician |
consults electronic health records to construct a pedigree showing family members who also presented symptoms similar to ALD. The pedigree is shown in this diagram. The infant patient is circled. Symbols for males (o) and females (m) are filled when symptoms are present. Figure 12.21 A. Using the pedigree, explain which form of ALD (neonatal or X-linked) is present in the infant. B. Sharing of digital records among health providers is one method proposed to improve the quality and reduce the cost of health care in the U.S. The privacy of electronic 514 Chapter 12 | Mendel's Experiments and Heredity health records is a concern. Pose three questions that must be addressed in developing policies that balance the costs of treatments and diagnoses, patient quality of life, and risks to individual privacy. 41. Two genes, A and B, are located adjacent to each other (linked) on the same chromosome. In the original cross (P0), one parent is homozygous dominant for both traits (AB), whereas the other parent is recessive (ab). Characteristic Alleles Chromosome Seed color yellow (I) / green (i) Seed coat & flowers colored (A) / white (a) Mature pods smooth (V) / wrinkled (v) from leaf axils (Fa) / umbellate at top of plant (fa) > 1 m (Le) / ~0.5 m (le) green (Gp) / yellow (gp) smooth (R) / wrinkled (r) Flower stalk Height Unripe pods Mature seeds Table 12.10 1 1 4 4 4 5 7 A. Describe the distribution of genotypes and phenotypes in F1. B. Describe the distribution of genotypes and phenotypes when F1 is crossed with the ab parent. C. Describe the distribution of genotypes and phenotypes when F1 is crossed with the AB parent. D. Explain the observed non-Mendelian results in terms of the violation of the laws governing Mendelian genetics. 42. Gregor Mendel’s 1865 paper described experiments on the inheritance of seven characteristics of Pisum sativum shown in the first column in the table below. Many years later, based on his reported outcomes and analysis of the inheritance of a single characteristic, Mendel developed the concepts of genes, their alleles, and dominance. These concepts are defined in the second column of the table using conventional symbols |
for the dominant allele for each characteristic. Even later, the location of each of these genes on one of the seven chromosomes in P. sativum were determined, as shown in the third column. A. Before the acceptance of what Mendel called “factors” This OpenStax book is available for free at http://cnx.org/content/col12078/1.6 as the discrete units of inheritance, the accepted model was that the traits of progeny were “blended” traits of the parents. Evaluate the evidence provided by Mendel’s experiments in disproving the blending theory of inheritance. B. Mendel published experimental data and analysis for two experiments involving the inheritance of more than a single characteristic. He examined two-character inheritance of seed shape and seed color. He also reported three-character inheritance of seed shape, seed color, and flower color. Evaluate the evidence provided by the multiple-character experiments. Identify which of the following laws of inheritance depend upon these multiplecharacter experiments for support: a. During gamete formation, the alleles for each gene segregate from each other so that each gamete carries only one allele for each gene. b. Genes for different traits can segregate independently during the formation of gametes. c. Some alleles are dominant, whereas others are recessive. An organism with at least one dominant allele will display the effect of the dominant allele. d. All three laws can be inferred from the single- character experiments. C. As shown in the table above, some chromosomes contain the gene for more than one of the seven characteristics Mendel studied, for example, seed color and flowers. The table below shows, with filled cells above the dashed diagonal line, the combinations of characteristics for which Mendel reported results. In the cells below the dotted diagonal line, identify with an X each cell where deviations from the law or laws identified in part B might be expected. Figure 12.22 D. Explain the reasons for the expected deviations for those combinations of characteristics identified in part C. E. In one of the experiments reported by Mendel, deviations from the law identified in part B might be expected. Explain how the outcomes of this experiment were consistent with Mendel’s laws. 43. A dihybrid cross involves two traits. A cross of parental types AaBb and AaBb can be represented with a Punnett square: Chapter 12 | Mendel's Experiments and Hered |
ity 515 representation is to calculate the number of each type of genotype in the offspring directly by counting the unique permutations of the letters representing the alleles. For example, the probability of the cross AaBbCc × AaBbCc is 3 (AA, Aa, aA) × 3 (BB, Bb, bB) × 3 (CC, Cc, cC)/64 = 27/64. B. Using the probability method, calculate the likelihood of these phenotypes from each trihybrid cross: • • recessive in all traits from the cross AaBbCc × aabbcc recessive in the gene with alleles C and c and dominant in the other two traits from the cross AaBbCc × AaBbCc • dominant in the gene with alleles A and a and recessive in the other two traits from the cross AaBbcc × AaBbCc C. The probability method is an easy way to calculate the likelihood of each particular phenotype, but it doesn’t simultaneously display the probability of all possible phenotypes. The forked line representation described in the text allows the entire phenotypic distribution to be displayed. Using the forked line method, calculate the probabilities in a cross between AABBCc and Aabbcc parents: • all traits are recessive: aabbcc • • • traits are dominant at each loci, A?B?C? traits are dominant at two genes and recessive at the third traits are dominant at one gene and recessive at the other two 44. Construct a representation showing the connection between the process of meiosis and the transmission of six possible phenotypes from parents to F2 offspring. The phenotypes are labeled A, a, B, b and C, c. Expression of each phenotype is controlled by a separate Mendelian gene. Your representation should show the proportion of every possible combination of phenotypes (e.g., ABC, AbC, etc.) that will be present in the F2 offspring. Figure 12.23 This representation clearly organizes all of the possible genotypes and reveals the 9:3:3:1 distribution of phenotypes and a 4×4 grid of 16 cells. Expressed as a fraction of the 16 possible genotypes of the offspring, the phenotypic ratio describes the probability of each phenotype among the offspring: 3 (AA, Aa, aA) × 3 (BB |
, bB, Bb)/16 = 9/16; 3 (AA, Aa, aA) × 1 (bb) /16 = 3/16; 1 (aa) × 3 (BB, bB, Bb) = 3/16; and 1 (aa) × 1 (bb) = 1/16. A. Using the probability method, calculate the likelihood of these phenotypes from each dihybrid cross: • recessive in the gene with alleles A and a from the cross AaBb × aabb • dominant in both genes from the cross AaBb × aabb • • recessive in both genes from the cross AaBb × aabb recessive in either gene from the cross AaBb × aabb A Punnett square representation of a trihybrid cross, such as the self-cross of AaBbCc, is more cumbersome because there are eight columns and rows (2×2×2 ways to choose parental genotypes) and 64 cells. A less tedious 516 Chapter 12 | Mendel's Experiments and Heredity This OpenStax book is available for free at http://cnx.org/content/col12078/1.6 Chapter 13 | Modern Understandings of Inheritance 517 13 | MODERN UNDERSTANDINGS OF INHERITANCE Figure 13.1 Chromosomes are threadlike nuclear structures consisting of DNA and proteins that serve as the repositories for genetic information. The chromosomes depicted here were isolated from a fruit fly’s salivary gland, stained with dye, and visualized under a microscope. Akin to miniature bar codes, chromosomes absorb different dyes to produce characteristic banding patterns, which allows for their routine identification. (credit: modification of work by “LPLT”/Wikimedia Commons; scale-bar data from Matt Russell) Chapter Outline 13.1: Chromosomal Theory and Genetic Linkages 13.2: Chromosomal Basis of Inherited Disorders Introduction According to the United Nations Office on Drugs and Crime, approximately 95% of those who commit homicide are men. While behavior is shaped by the environment one grows up and lives in, genetics also play a role. For example, scientists have discovered genes that appear to increase one’s tendency to exhibit aggressive behavior. One of the genes, called MAOA, is located on the X chromosome. In one recent study involving a group of male prisoners in Finland, scientists |
found that the prisoners who inherited a variant of this gene were between 5% and 10% more likely to have committed a violent crime. Men only have one copy of the gene, since men only have one X chromosome. Women, however, have two copies of the X chromosome and therefore two copies of the gene. Therefore, women who inherit the variant allele will most likely also have a normal allele to counteract its effects. It is important to note that many men inherit the variant copy of MAOA and only some commit violent crimes. The environment seems to play a much more critical role. You can read more about nature/nurture roles in crime in this article (http://openstaxcollege.org/l/32whysomany). 518 Chapter 13 | Modern Understandings of Inheritance 13.1 | Chromosomal Theory and Genetic Linkages In this section, you will explore the following question: • What is the relationship among genetic linkage, crossing over, and genetic variation? Connection for AP® Courses Proposed independently by Sutton and Boveri in the early 1900s, the Chromosomal Theory of Inheritance states that chromosomes are vehicles of genetic heredity. As we have discovered, patterns of inheritance are more complex than Mendel could have imagined. Mendel was investigating the behavior of genes. He was fortunate in choosing traits coded by genes that happened to be on different chromosomes or far apart on the same chromosome. When genes are linked or near each other on the same chromosome, patterns of segregation and independent assortment change. In 1913, Sturtevant devised a method to assess recombination frequency and infer the relative positions and distances of linked genes on a chromosome based on the average number of crossovers between them during meiosis. The content presented in this section supports the Learning Objectives outlined in Big Idea 3 of the AP® Biology Curriculum Framework. The AP® Learning Objectives merge essential knowledge content with one or more of the seven Science Practices. These objectives provide a transparent foundation for the AP® Biology course, along with inquiry-based laboratory experiences, instructional activities, and AP® exam questions. Big Idea 3 Enduring Understanding 3.A Living systems store, retrieve, transmit and respond to information essential to life processes. Heritable information provides for continuity of life. Essential Knowledge 3.A.2 In eukaryotes, heritable information is passed to the next generation via processes that include the cell cycle and mitosis or meiosis plus fertilization. Science Practice Learning Objective 7. |
1 The student can connect phenomena and models across spatial and temporal scales. 3.10 The student is able to represent the connection between meiosis and increased genetic diversity necessary for evolution. Essential Knowledge 3.A.3 The chromosomal basis of inheritance provides an understanding of the pattern of passage (transmission) of genes from parent to offspring. Science Practice Science Practice Learning Objective 1.1 The student can create representations and models of natural or man-made phenomena and systems in the domain. 7.2 The student can connect concepts in and across domain(s) to generalize or extrapolate in and/or across enduring understandings and/or big ideas. 3.12 The student is able to construct a representation that connects the process of meiosis to the passage of traits from parent to offspring. The Science Practice Challenge Questions contain additional test questions for this section that will help you prepare for the AP exam. These questions address the following standards: [APLO 3.2][APLO 3.11][APLO 3.14][APLO 3.15][APLO 3.28][APLO 3.26][APLO 3.17][APLO 4.22] Long before chromosomes were visualized under a microscope, the father of modern genetics, Gregor Mendel, began studying heredity in 1843. With the improvement of microscopic techniques during the late 1800s, cell biologists could stain and visualize subcellular structures with dyes and observe their actions during cell division and meiosis. With each mitotic division, chromosomes replicated, condensed from an amorphous (no constant shape) nuclear mass into distinct X-shaped bodies (pairs of identical sister chromatids), and migrated to separate cellular poles. This OpenStax book is available for free at http://cnx.org/content/col12078/1.6 Chapter 13 | Modern Understandings of Inheritance 519 Chromosomal Theory of Inheritance The speculation that chromosomes might be the key to understanding heredity led several scientists to examine Mendel’s publications and re-evaluate his model in terms of the behavior of chromosomes during mitosis and meiosis. In 1902, Theodor Boveri observed that proper embryonic development of sea urchins does not occur unless chromosomes are present. That same year, Walter Sutton observed the separation of chromosomes into daughter cells during meiosis (Figure 13.2). Together, these observations led to the development of the Chromosomal Theory of Inheritance, which identified chromosomes as the genetic |
material responsible for Mendelian inheritance. Figure 13.2 (a) Walter Sutton and (b) Theodor Boveri are credited with developing the Chromosomal Theory of Inheritance, which states that chromosomes carry the unit of heredity (genes). The Chromosomal Theory of Inheritance was consistent with Mendel’s laws and was supported by the following observations: • During meiosis, homologous chromosome pairs migrate as discrete structures that are independent of other chromosome pairs. • The sorting of chromosomes from each homologous pair into pre-gametes appears to be random. • Each parent synthesizes gametes that contain only half of their chromosomal complement. • Even though male and female gametes (sperm and egg) differ in size and morphology, they have the same number of chromosomes, suggesting equal genetic contributions from each parent. • The gametic chromosomes combine during fertilization to produce offspring with the same chromoso<|endoftext|>me number as their parents. Despite compelling correlations between the behavior of chromosomes during meiosis and Mendel’s abstract laws, the Chromosomal Theory of Inheritance was proposed long before there was any direct evidence that traits were carried on chromosomes. Critics pointed out that individuals had far more independently segregating traits than they had chromosomes. It was only after several years of carrying out crosses with the fruit fly, Drosophila melanogaster, that Thomas Hunt Morgan provided experimental evidence to support the Chromosomal Theory of Inheritance. Genetic Linkage and Distances Mendel’s work suggested that traits are inherited independently of each other. Morgan identified a 1:1 correspondence between a segregating trait and the X chromosome, suggesting that the random segregation of chromosomes was the physical basis of Mendel’s model. This also demonstrated that linked genes disrupt Mendel’s predicted outcomes. The fact that each chromosome can carry many linked genes explains how individuals can have many more traits than they have chromosomes. However, observations by researchers in Morgan’s laboratory suggested that alleles positioned on the same chromosome were not always inherited together. During meiosis, linked genes somehow became unlinked. Homologous Recombination In 1909, Frans Janssen observed chiasmata—the point at which chromatids are in contact with each other and may exchange segments—prior to the first division of meiosis. He suggested that alleles become unlinked and chromosomes physically exchange segments. As chromosomes condensed and paired with |
their homologs, they appeared to interact at distinct points. Janssen suggested that these points corresponded to regions in which chromosome segments were exchanged. It is now known that the pairing and interaction between homologous chromosomes, known as synapsis, does more than 520 Chapter 13 | Modern Understandings of Inheritance simply organize the homologs for migration to separate daughter cells. When synapsed, homologous chromosomes undergo reciprocal physical exchanges at their arms in a process called homologous recombination, or more simply, “crossing over.” To better understand the type of experimental results that researchers were obtaining at this time, consider a heterozygous individual that inherited dominant maternal alleles for two genes on the same chromosome (such as AB) and two recessive paternal alleles for those same genes (such as ab). If the genes are linked, one would expect this individual to produce gametes that are either AB or ab with a 1:1 ratio. If the genes are unlinked, the individual should produce AB, Ab, aB, and ab gametes with equal frequencies, according to the Mendelian concept of independent assortment. Because they correspond to new allele combinations, the genotypes Ab and aB are nonparental types that result from homologous recombination during meiosis. Parental types are progeny that exhibit the same allelic combination as their parents. Morgan and his colleagues, however, found that when such heterozygous individuals were test crossed to a homozygous recessive parent (AaBb × aabb), both parental and nonparental cases occurred. For example, 950 offspring might be recovered that were either AaBb or aabb, but 50 offspring would also be obtained that were either Aabb or aaBb. These results suggested that linkage occurred most often, but a significant minority of offspring were the products of recombination. This OpenStax book is available for free at http://cnx.org/content/col12078/1.6 Chapter 13 | Modern Understandings of Inheritance 521 Figure 13.3 Inheritance patterns of unlinked and linked genes are shown. In (a), two genes are located on different chromosomes so independent assortment occurs during meiosis. The offspring have an equal chance of being the parental type (inheriting the same combination of traits as the parents) or a nonparental type (inheriting a different combination of traits than the parents). In (b), two |
genes are very close together on the same chromosome so that no crossing over occurs between them. The genes are therefore always inherited together and all of the offspring are the parental type. In (c), two genes are far apart on the chromosome such that crossing over occurs during every meiotic event. The recombination frequency will be the same as if the genes were on separate chromosomes. (d) The actual recombination frequency of fruit fly wing length and body color that Thomas Morgan observed in 1912 was 17 percent. A crossover frequency between 0 percent and 50 percent indicates that the genes are on the same chromosome and crossover occurs some of the time. In a test cross for two characteristics such as the one shown here, can the predicted frequency of recombinant offspring be 60%? Why or why not? a. Yes, the predicted offspring frequencies range from 0% to 100% b. No, the predicted offspring frequencies cannot be higher than 30%. c. Yes, the predicted offspring frequencies range from 0% to 60%. d. No, the predicted offspring frequencies range from 0% to 50%. 522 Chapter 13 | Modern Understandings of Inheritance Think About It A test cross involving F1 dihybrid flies produces more parental-type offspring than recombinant-type offspring. How can you explain these observed results? Genetic Markers for Cancers Scientists have used genetic linkage to discover the location in the human genome of many genes that cause disease. They locate disease genes by tracking inheritance of traits through generations of families and creating linkage maps that measure recombination among groups of genetic “markers.” The two BRCA genes, mutations which can lead to breast and ovarian cancers, were some of the first genes discovered by genetic mapping. Women who have family histories of these cancers can now be screened to determine if one or both of these genes carry a mutation. If so, they can opt to have their breasts and ovaries surgically removed. This decreases their chances of getting cancer later in life. The actress Angelia Jolie brought this to the public’s attention when she opted for surgery in 2014 and again in 2015 after doctors found she carried a mutated BRCA1 gene. Which of the following statements most accurately describes domestication syndrome? a. Genes responsible for temperament are on the same chromosome as genes responsible for certain facial features. b. A single gene codes for both temperament and certain facial features, such as jaw size. c. Genes responsible for mild temperament are only expressed when genes encoding a cute face |
are also present. d. The products of genes encoding temperament interact with the products of genes encoding facial features. Genetic Maps Janssen did not have the technology to demonstrate crossing over so it remained an abstract idea that was not widely accepted. Scientists thought chiasmata were a variation on synapsis and could not understand how chromosomes could break and rejoin. Yet, the data were clear that linkage did not always occur. Ultimately, it took a young undergraduate student and an “all-nighter” to mathematically elucidate the problem of linkage and recombination. In 1913, Alfred Sturtevant, a student in Morgan’s laboratory, gathered results from researchers in the laboratory, and took them home one night to mull them over. By the next morning, he had created the first “chromosome map,” a linear representation of gene order and relative distance on a chromosome (Figure 13.4). This OpenStax book is available for free at http://cnx.org/content/col12078/1.6 Chapter 13 | Modern Understandings of Inheritance 523 Figure 13.4 This genetic map orders Drosophila genes on the basis of recombination frequency. Which of the following statements is true? a. Recombination of the red/brown eye and long/short aristae alleles will occur more frequently than recombination of the alleles for wing length and body color. b. Recombination of the body color and red/cinnabar eye alleles will occur more frequently than recombination of the alleles for wing length and aristae length. c. Recombination of the body color and aristae length alleles will occur more frequently than recombination of red/brown eye alleles and the aristae length alleles. d. Recombination of the gray/black body color and long/short aristae alleles will not occur. As shown in Figure 13.4, by using recombination frequency to predict genetic distance, the relative order of genes on chromosome 2 could be inferred. The values shown represent map distances in centimorgans (cM), which correspond to recombination frequencies (in percent). Therefore, the genes for body color and wing size were 65.5 − 48.5 = 17 cM apart, indicating that the maternal and paternal alleles for these genes recombine in 17 percent of offspring, on average. To construct a chromosome map, Sturtevant assumed that genes were ordered serially on thread |
like chromosomes. He also assumed that the incidence of recombination between two homologous chromosomes could occur with equal likelihood anywhere along the length of the chromosome. Operating under these assumptions, Sturtevant postulated that alleles that were far apart on a chromosome were more likely to dissociate during meiosis simply because there was a larger region over which recombination could occur. Conversely, alleles that were close to each other on the chromosome were likely to be inherited together. The average number of crossovers between two alleles—that is, their recombination frequency—correlated with their genetic distance from each other, relative to the locations of other genes on that chromosome. Considering the example cross between AaBb and aabb above, the frequency of recombination could be calculated as 50/1000 = 0.05. That is, the likelihood of a crossover between genes A/a and B/b was 0.05, or 5 percent. Such a result would indicate that the genes were definitively linked, but that they were far enough apart for crossovers to occasionally occur. Sturtevant divided his genetic map into map units, or centimorgans (cM), in which a recombination frequency of 0.01 corresponds to 1 cM. By representing alleles in a linear map, Sturtevant suggested that genes can range from being perfectly linked (recombination frequency = 0) to being perfectly unlinked (recombination frequency = 0.5) when genes are on different chromosomes or genes are separated very far apart on the same chromosome. Perfectly unlinked genes correspond to the frequencies predicted by Mendel to assort independently in a dihybrid cross. A recombination frequency of 0.5 indicates that 50 percent of offspring are recombinants and the other 50 percent are parental types. That is, every type of allele combination is represented with equal frequency. This representation allowed Sturtevant to additively calculate distances between several genes on the same chromosome. However, as the genetic distances approached 0.50, his predictions became less accurate because it was not clear whether the genes were very far apart on the same chromosome or on different chromosomes. In 1931, Barbara McClintock and Harriet Creighton demonstrated the crossover of homologous chromosomes in corn plants. Weeks later, homologous recombination in Drosophila was demonstrated microscopically by Curt Stern. Stern 524 Chapter 13 | Modern Understandings of Inheritance observed several X-linked phenotypes that were associated with a structurally |
unusual and dissimilar X chromosome pair in which one X was missing a small terminal segment, and the other X was fused to a piece of the Y chromosome. By crossing flies, observing their offspring, and then visualizing the offspring’s chromosomes, Stern demonstrated that every time the offspring allele combination deviated from either of the parental combinations, there was a corresponding exchange of an X chromosome segment. Using mutant flies with structurally distinct X chromosomes was the key to observing the products of recombination because DNA sequencing and other molecular tools were not yet available. It is now known that homologous chromosomes regularly exchange segments in meiosis by reciprocally breaking and rejoining their DNA at precise locations. Review Sturtevant’s process (http://openstaxcollege.org/l/gene_crossover). to create a genetic map on the basis of recombination frequencies here Genetic diversity is the total number of genetic characteristics in a species. Explain how chromosomal crossover contributes to genetic diversity. a. Chromosomal crossover is a specific, non-random process during which chromosomes are linked together and exchange DNA, contributing to the genetic diversity. b. Chromosomal crossover occurs during meiosis when chromosome pairs are linked and exchange DNA. Thus, crossover increases the variance of genetic combinations in the haploid gamete cell. c. Chromosomal crossover results in the inheritance of genetic material by offspring and the recombination event is not variable in frequency or location. d. Chromosomal crossover occurs during the mitotic process when chromosomes are linked together and recombination takes place, increasing the variance of genetic combinations in the haploid mitotic cells formed from mitosis. Mendel’s Mapped Traits Homologous recombination is a common genetic process, yet Mendel never observed it. Had he investigated both linked and unlinked genes, it would have been much more difficult for him to create a unified model of his data on the basis of probabilistic calculations. Researchers who have since mapped the seven traits investigated by Mendel onto the seven chromosomes of the pea plant genome have confirmed that all of the genes he examined are either on separate chromosomes or are sufficiently far apart as to be statistically unlinked. Some have suggested that Mendel was enormously lucky to select only unlinked genes, whereas others question whether Mendel discarded any data suggesting linkage. In any case, Mendel consistently observed independent assortment because he examined genes that were effectively unlinked. 13.2 | Chromosomal Basis of Inherited Disorders In this |
section, you will explore the following question: • What are the genetic consequences that result from nondisjunction and errors in chromosome structure through inversions and translocations? Connection for AP® Courses The number, size, shape, and banding patterns of chromosomes make them easily identifiable in a karyogram and allows for the assessment of many chromosomal abnormalities. Although the cell cycle, mitosis, and meiosis are highly regulated to This OpenStax book is available for free at http://cnx.org/content/col12078/1.6 Chapter 13 | Modern Understandings of Inheritance 525 prevent errors, the processes are not perfect. One example is the failure of homologous chromosomes or sister chromatids to separate properly during meiosis I or meiosis II (a phenomenon referred to as nondisjunction). This results in gametes with too many or too few chromosomes. Disorders in chromosome number (aneuploidy) are typically lethal to the embryo, although a few trisomic genotypes are viable (e.g., Down syndrome). Because of X inactivation, aberrations in sex chromosomes typically have milder phenotypic effects (e.g., Turner syndrome) than aneuploidy. Sometimes segments of chromosome are duplicated, deleted, or rearranged by inversion or translocation. These aberrations can result in problematic phenotypic effects. Diagnostic testing can detect many of these chromosomal disorders in individuals well before birth, resulting in medical, ethical, and civic issues, such as the right to privacy. A condition in which an organism has more than the normal number of chromosome sets (two for diploid species) is called polyploidy. Polyploidy resulting in odd numbers of chromosomes is rare because it results in sterile organisms. One set of chromosomes has no pair so meiosis cannot proceed normally. In contrast, polyploidy resulting in even chromosome numbers is very common in the plant kingdom. Polyploid plants tend to be larger and more robust than individuals with the normal number of chromosomes. Information presented and the examples highlighted in the section support concepts outlined in Big Idea 3 of the AP® Biology Curriculum Framework. The Learning Objectives listed in the Curriculum Framework provide a transparent foundation for the AP® Biology course, an inquiry-based laboratory experience, instructional activities, and AP® exam questions. A Learning Objective merges required content with one or more of the seven Science Practices. Big Idea 3 End |
uring Understanding 3.A Living systems store, retrieve, transmit and respond to information essential to life processes. Heritable information provides for continuity of life. Essential Knowledge 3.A.2 In eukaryotes, heritable information is passed to the next generation via processes that include the cell cycle and mitosis or meiosis plus fertilization. Science Practice Learning Objective 6.2 The student can construct explanations of phenomena based on evidence produced through scientific practices. 3.9 The student is able to construct an explanation, using visual representations or narratives, as to how DNA in chromosomes is transmitted to the next generation via mitosis, or meiosis followed by fertilization. Essential Knowledge 3.A.3 The chromosomal basis of inheritance provides an understanding of the pattern of passage (transmission) of genes from parent to offspring. Science Practice Learning Objective Big Idea 3 Enduring Understanding 3.C Essential Knowledge Science Practice Science Practice Learning Objective 3.1 The student can pose scientific questions. 3.13 The student is able to pose questions about ethical, social or medical issues surrounding human genetic disorders. Living systems store, retrieve, transmit and respond to information essential to life processes. The processing of genetic information is imperfect and is a source of genetic variation. 3.A.3 Changes in genotype can result in changes in phenotype. 6.4 The student can make claims and predictions about natural phenomena based on scientific theories and models. 7.2 The student can connect concepts in and across domain(s) to generalize or extrapolate in and/or across enduring understandings and/or big ideas. 3.24 The student is able to predict how a change in genotype, when expressed as a phenotype, provides a variation that can be subject to natural selection. 526 Chapter 13 | Modern Understandings of Inheritance Inherited disorders can arise when chromosomes behave abnormally during meiosis. Chromosome disorders can be divided into two categories: abnormalities in chromosome number and chromosomal structural rearrangements. Because even small segments of chromosomes can span many genes, chromosomal disorders are characteristically dramatic and often fatal. Identification of Chromosomes The isolation and microscopic observation of chromosomes forms the basis of cytogenetics and is the primary method by which clinicians detect chromosomal abnormalities in humans. A karyotype is the number and appearance of chromosomes, and includes their length, banding pattern, and centromere position. To obtain a view of an individual’s karyotype, cytologists photograph the chromosomes and then cut and paste each chromosome into |
a chart, or karyogram, also known as an ideogram (Figure 13.5). Figure 13.5 This karyotype is of a female human. Notice that homologous chromosomes are the same size, and have the same centromere positions and banding patterns. A human male would have an XY chromosome pair instead of the XX pair shown. (credit: Andreas Blozer et al) In a given species, chromosomes can be identified by their number, size, centromere position, and banding pattern. In a human karyotype, autosomes or “body chromosomes” (all of the non–sex chromosomes) are generally organized in order of size from largest (chromosome 1) to smallest (chromosome 22). (The X and Y chromosomes, the 23rd pair, are not autosomes.) However, chromosome 21 is actually shorter than chromosome 22. This was discovered after the naming of Down syndrome as trisomy 21, reflecting how this disease results from possessing one extra chromosome 21 (three total). Not wanting to change the name of this disease, scientists retained the original numbering system. The chromosome “arms” projecting from either end of the centromere may be designated as short or long, depending on their relative lengths. The short arm is abbreviated p (for “petite”), whereas the long arm is abbreviated q (because it follows “p” alphabetically). Each arm is further subdivided and denoted by a number. For example, locus 3 on the short arm of chromosome 21 is denoted 21p3. Using this naming system, locations on chromosomes can be described consistently in the scientific literature. This OpenStax book is available for free at http://cnx.org/content/col12078/1.6 Chapter 13 | Modern Understandings of Inheritance 527 Geneticists Use Karyograms to Identify Chromosomal Aberrations is referred to as the “father of modern genetics,” he performed his experiments with Although Mendel none of the tools that the geneticists of today routinely employ. One such powerful cytological technique is karyotyping, a method in which traits characterized by chromosomal abnormalities can be identified from a single cell. To observe an individual’s karyotype, a person’s cells (like white blood cells) are first collected from a blood sample or other tissue. In the laboratory, the isolated cells are stimulated to begin |
actively dividing. A chemical called colchicine is then applied to cells to arrest condensed chromosomes in metaphase. Cells are then made to swell using a hypotonic solution so the chromosomes spread apart. Finally, the sample is preserved in a fixative and applied to a slide. The geneticist then stains chromosomes with one of several dyes to better visualize the distinct and reproducible banding patterns of each chromosome pair. Following staining, the chromosomes are viewed using bright-field microscopy. A common stain choice is the Giemsa stain. Giemsa staining results in approximately 400–800 bands (of tightly coiled DNA and condensed proteins) arranged along all of the 23 chromosome pairs; an experienced geneticist can identify each band. In addition to the banding patterns, chromosomes are further identified on the basis of size and centromere location. To obtain the classic depiction of the karyotype in which homologous pairs of chromosomes are aligned in numerical order from longest to shortest, the geneticist obtains a digital image, identifies each chromosome, and manually arranges the chromosomes into this pattern (Figure 13.5). At its most basic, the karyogram may reveal genetic abnormalities in which an individual has too many or too few chromosomes per cell. Examples of this are Down Syndrome, which is identified by a third copy of chromosome 21, and Turner Syndrome, which is characterized by the presence of only one X chromosome in women instead of the normal two. Geneticists can also identify large deletions or insertions of DNA. For instance, Jacobsen Syndrome—which involves distinctive facial features as well as heart and bleeding defects—is identified by a deletion on chromosome 11. Finally, the karyotype can pinpoint translocations, which occur when a segment of genetic material breaks from one chromosome and reattaches to another chromosome or to a different part of the same chromosome. During Mendel’s lifetime, inheritance was an abstract concept that could only be inferred by performing crosses and observing the traits expressed by offspring. By observing a karyogram, today’s geneticists can actually visualize the chromosomal composition of an individual to confirm or predict genetic abnormalities in offspring, even before birth. Disorders in Chromosome Number Of all of the chromosomal disorders, abnormalities in chromosome number are the most obviously identifiable from a karyogram. Disorders of chromosome number include the duplication or loss of entire chromosomes, as well as changes in the number of complete sets of chromosomes. They are caused by nondisjunction, which occurs |
when pairs of homologous chromosomes or sister chromatids fail to separate during meiosis. Misaligned or incomplete synapsis, or a dysfunction of the spindle apparatus that facilitates chromosome migration, can cause nondisjunction. The risk of nondisjunction occurring increases with the age of the parents. Nondisjunction can occur during either meiosis I or II, with differing results (Figure 13.6). If homologous chromosomes fail to separate during meiosis I, the result is two gametes that lack that particular chromosome and two gametes with two copies of the chromosome. If sister chromatids fail to separate during meiosis II, the result is one gamete that lacks that chromosome, two normal gametes with one copy of the chromosome, and one gamete with two copies of the chromosome. 528 Chapter 13 | Modern Understandings of Inheritance Figure 13.6 Nondisjunction occurs when homologous chromosomes or sister chromatids fail to separate during meiosis, resulting in an abnormal chromosome number. Nondisjunction may occur during meiosis I or meiosis II. Which of the following statements about nondisjunction is true? a. Nondisjunction only results in gametes with n+1 or n-1 chromosomes. b. Nondisjunction occurring during meiosis II results in 50% normal gametes. c. Nondisjunction during meiosis I results in 50% normal gametes. d. Nondisjunction always results in four different kinds of gametes. Aneuploidy An individual with the appropriate number of chromosomes for their species is called euploid; in humans, euploidy corresponds to 22 pairs of autosomes and one pair of sex chromosomes. An individual with an error in chromosome number is described as aneuploid, a term that includes monosomy (loss of one chromosome) or trisomy (gain of an extraneous chromosome). Monosomic human zygotes missing any one copy of an autosome invariably fail to develop to birth because they lack essential genes. This underscores the importance of “gene dosage” in humans. Most autosomal trisomies also fail to develop to birth; however, duplications of some of the smaller chromosomes (13, 15, 18, 21, or 22) can result in offspring that survive for several weeks to many years. Trisomic individuals suffer from |
a different type of genetic imbalance: an excess in gene dose. Individuals with an extra chromosome may synthesize an abundance of the gene products encoded by that chromosome. This extra dose (150 percent) of specific genes can lead to a number of functional challenges and often precludes development. The most common trisomy among viable births is that of chromosome 21, which corresponds to Down Syndrome. Individuals with this inherited disorder are characterized by short stature and stunted digits, facial distinctions that include a broad skull and large tongue, and significant developmental delays. The incidence of Down syndrome is correlated with maternal age; older women are more likely to become pregnant with fetuses carrying the trisomy 21 genotype (Figure 13.7). This OpenStax book is available for free at http://cnx.org/content/col12078/1.6 Chapter 13 | Modern Understandings of Inheritance 529 Figure 13.7 The incidence of having a fetus with trisomy 21 increases dramatically with maternal age. Visualize chromosome (http://openstaxcollege.org/l/down_syndrome). addition the of a that leads to Down syndrome in this video simulation With increasing age, women are at increased risk of having a baby with a chromosomal abnormality such as Down syndrome. Why is age a risk factor? a. Cells are more likely to make mistakes as we age due to an increase in the nondisjunction of cells during cell division, which is the reason for the occurrence of Down syndrome. b. The chance of this disorder increases with age due to increased mistakes in the mitotic cells with age. c. There are increased risks of translocation mutations with age, even though other mutation rates are constant, and this increases the risk. d. The risk of having a child with Down syndrome is associated primarily with lifestyle factors that change with age. Polyploidy An individual with more than the correct number of chromosome sets (two for diploid species) is called polyploid. For instance, fertilization of an abnormal diploid egg with a normal haploid sperm would yield a triploid zygote. Polyploid animals are extremely rare, with only a few examples among the flatworms, crustaceans, amphibians, fish, and lizards. Polyploid animals are sterile because meiosis cannot proceed normally and instead produces mostly aneuploid daughter cells that cannot yield viable zygotes. Rarely, polyploid animals can reproduce asexually by haplodi |
ploidy, in which an unfertilized egg divides mitotically to produce offspring. In contrast, polyploidy is very common in the plant kingdom, and polyploid plants tend to be larger and more robust than euploids of their species (Figure 13.8). 530 Chapter 13 | Modern Understandings of Inheritance Figure 13.8 As with many polyploid plants, this triploid orange daylily (Hemerocallis fulva) is particularly large and robust, and grows flowers with triple the number of petals of its diploid counterparts. (credit: Steve Karg) Sex Chromosome Nondisjunction in Humans trisomies and monosomies. Therefore, Humans display dramatic deleterious effects with autosomal it may seem counterintuitive that human females and males can function normally, despite carrying different numbers of the X chromosome. Rather than a gain or loss of autosomes, variations in the number of sex chromosomes are associated with relatively mild effects. In part, this occurs because of a molecular process called X inactivation. Early in development, when female mammalian embryos consist of just a few thousand cells (relative to trillions in the newborn), one X chromosome in each cell inactivates by tightly condensing into a quiescent (dormant) structure called a Barr body. The chance that an X chromosome (maternally or paternally derived) is inactivated in each cell is random, but once the inactivation occurs, all cells derived from that one will have the same inactive X chromosome or Barr body. By this process, females compensate for their double genetic dose of X chromosome. In so-called “tortoiseshell” cats, embryonic X inactivation is observed as color variegation (Figure 13.9). Females that are heterozygous for an X-linked coat color gene will express one of two different coat colors over different regions of their body, corresponding to whichever X chromosome is inactivated in the embryonic cell progenitor of that region. Figure 13.9 In cats, the gene for coat color is located on the X chromosome. In the embryonic development of female cats, one of the two X chromosomes is randomly inactivated in each cell, resulting in a tortoiseshell pattern if the cat has two different alleles for coat color. Male cats, having only one X chromosome, never exhibit a tortoiseshell coat color. (credit: Michael Bodega |
) An individual carrying an abnormal number of X chromosomes will inactivate all but one X chromosome in each of her cells. However, even inactivated X chromosomes continue to express a few genes, and X chromosomes must reactivate for the proper maturation of female ovaries. As a result, X-chromosomal abnormalities are typically associated with mild mental and physical defects, as well as sterility. If the X chromosome is absent altogether, the individual will not develop in This OpenStax book is available for free at http://cnx.org/content/col12078/1.6 Chapter 13 | Modern Understandings of Inheritance 531 utero. Several errors in sex chromosome number have been characterized. Individuals with three X chromosomes, called triploX, are phenotypically female but express developmental delays and reduced fertility. The XXY genotype, corresponding to one type of Klinefelter syndrome, corresponds to phenotypically male individuals with small testes, enlarged breasts, and reduced body hair. More complex types of Klinefelter syndrome exist in which the individual has as many as five X chromosomes. In all types, every X chromosome except one undergoes inactivation to compensate for the excess genetic dosage. This can be seen as several Barr bodies in each cell nucleus. Turner syndrome, characterized as an X0 genotype (i.e., only a single sex chromosome), corresponds to a phenotypically female individual with short stature, webbed skin in the neck region, hearing and cardiac impairments, and sterility. Duplications and Deletions In addition to the loss or gain of an entire chromosome, a chromosomal segment may be duplicated or lost. Duplications and deletions often produce offspring that survive but exhibit physical and mental abnormalities. Duplicated chromosomal segments may fuse to existing chromosomes or may be free in the nucleus. Cri-du-chat (from the French for “cry of the cat”) is a syndrome associated with nervous system abnormalities and identifiable physical features that result from a deletion of most of 5p (the small arm of chromosome 5) (Figure 13.10). Infants with this genotype emit a characteristic high-pitched cry on which the disorder’s name is based. Figure 13.10 This individual with cri-du-chat syndrome is shown at two, four, nine, and 12 years of age. (credit: Paola Cerruti Mainardi) Chromosomal Structural Rearrangements Cy |
tologists have characterized numerous structural rearrangements in chromosomes, but chromosome inversions and translocations are the most common. Both are identified during meiosis by the adaptive pairing of rearranged chromosomes with their former homologs to maintain appropriate gene alignment. If the genes carried on two homologs are not oriented correctly, a recombination event could result in the loss of genes from one chromosome and the gain of genes on the other. This would produce aneuploid gametes. Chromosome Inversions A chromosome inversion is the detachment, 180° rotation, and reinsertion of part of a chromosome. Inversions may occur 532 Chapter 13 | Modern Understandings of Inheritance in nature as a result of mechanical shear, or from the action of transposable elements (special DNA sequences capable of facilitating the rearrangement of chromosome segments with the help of enzymes that cut and paste DNA sequences). Unless they disrupt a gene sequence, inversions only change the orientation of genes and are likely to have more mild effects than aneuploid errors. However, altered gene orientation can result in functional changes because regulators of gene expression could be moved out of position with respect to their targets, causing aberrant levels of gene products. An inversion can be pericentric and include the centromere, or paracentric and occur outside of the centromere (Figure 13.11). A pericentric inversion that is asymmetric about the centromere can change the relative lengths of the chromosome arms, making these inversions easily identifiable. Figure 13.11 Pericentric inversions include the centromere, and paracentric inversions do not. A pericentric inversion can change the relative lengths of the chromosome arms; a paracentric inversion cannot. When one homologous chromosome undergoes an inversion but the other does not, the individual is described as an inversion heterozygote. To maintain point-for-point synapsis during meiosis, one homolog must form a loop, and the other homolog must mold around it. Although this topology can ensure that the genes are correctly aligned, it also forces the homologs to stretch and can be associated with regions of imprecise synapsis (Figure 13.12). Figure 13.12 When one chromosome undergoes an inversion but the other does not, one chromosome must form an inverted loop to retain point-for-point interaction during synapsis. This inversion pairing is essential to maintaining |
gene alignment during meiosis and to allow for recombination. This OpenStax book is available for free at http://cnx.org/content/col12078/1.6 Chapter 13 | Modern Understandings of Inheritance 533 The Chromosome 18 Inversion Not all structural rearrangements of chromosomes produce nonviable, impaired, or infertile individuals. In rare instances, such a change can result in the evolution of a new species. In fact, a pericentric inversion in chromosome 18 appears to have contributed to the evolution of humans. This inversion is not present in our closest genetic relatives, the chimpanzees. Humans and chimpanzees differ cytogenetically by pericentric inversions on several chromosomes and by the fusion of two separate chromosomes in chimpanzees that correspond to chromosome two in humans. The pericentric chromosome 18 inversion is believed to have occurred in early humans following their divergence from a common ancestor with chimpanzees approximately five million years ago. Researchers characterizing this inversion have suggested that approximately 19,000 nucleotide bases were duplicated on 18p, and the duplicated region inverted and reinserted on chromosome 18 of an ancestral human. A comparison of human and chimpanzee genes in the region of two genes—ROCK1 and USP14—that are adjacent on chimpanzee chromosome 17 (which corresponds to human chromosome 18) are more distantly positioned on human chromosome 18. This suggests that one of the inversion breakpoints occurred between these two genes. Interestingly, humans and chimpanzees express USP14 at distinct levels in specific cell types, including cortical cells and fibroblasts. Perhaps the chromosome 18 inversion in an ancestral human repositioned specific genes and reset their expression levels in a useful way. Because both ROCK1 and USP14 encode cellular enzymes, a change in their expression could alter cellular function. It is not known how this inversion contributed to hominid evolution, but it appears to be a significant factor in the divergence of humans from other primates. this inversion indicates that [1] According to the passage, which of the following events are believed to have occurred after humans diverged from their common ancestor with chimpanzees? a. Paracentric inversions occurred at several chromosomes, including human chromosome 18. b. Two separate chromosomes underwent a pericentric inversion, then fused to form chromosome 2 in humans. c. 19, 000 nucleotide bases were duplicated, inverted, and reinserted at human chromosome 18. d. The ROCK1 and US |
P14 genes were duplicated in early humans, which increased expression of these genes. Translocations A translocation occurs when a segment of a chromosome dissociates and reattaches to a different, nonhomologous chromosome. Translocations can be benign or have devastating effects depending on how the positions of genes are altered with respect to regulatory sequences. Reciprocal translocations result from the exchange of chromosome segments between two nonhomologous chromosomes such that there is no gain or loss of genetic information (Figure 13.13). 1. Violaine Goidts et al., “Segmental duplication associated with the human-specific inversion of chromosome 18: a further example of the impact of segmental duplications on karyotype and genome evolution in primates,” Human Genetics. 115 (2004):116-122 534 Chapter 13 | Modern Understandings of Inheritance Figure 13.13 A reciprocal translocation occurs when a segment of DNA is transferred from one chromosome to another, nonhomologous chromosome. (credit: modification of work by National Human Genome Research/USA) Activity A Day in the Life. Compose a short story, PowerPoint presentation, video, poem, or significant piece of art to describe a day in the life of a teenager afflicted with a single gene disorder or chromosomal abnormality. You need to include the causes and effects of the disorder and pose a question about a social, medical, or ethical issue(s) associated with human genetic disorders. Think About It Create a series of representations to show how nondisjunction can result in a trisomic zygote from a cell with 2n = 4. This OpenStax book is available for free at http://cnx.org/content/col12078/1.6 Chapter 13 | Modern Understandings of Inheritance 535 KEY TERMS aneuploid individual with an error in chromosome number; includes deletions and duplications of chromosome segments autosome any of the non-sex chromosomes centimorgan (cM) (also, map unit) relative distance that corresponds to a recombination frequency of 0.01 Chromosomal Theory of Inheritance theory proposing that chromosomes are the vehicles of genes and that their behavior during meiosis is the physical basis of the inheritance patterns that Mendel observed chromosome inversion detachment, 180° rotation, and reinsertion of a chromosome arm euploid individual with the appropriate number of chromosomes for their species homologous recombination process by which homolog |
ous chromosomes undergo reciprocal physical exchanges at their arms, also known as crossing over karyogram photographic image of a karyotype karyotype number and appearance of an individuals chromosomes; includes the size, banding patterns, and centromere position monosomy otherwise diploid genotype in which one chromosome is missing nondisjunction failure of synapsed homologs to completely separate and migrate to separate poles during the first cell division of meiosis nonparental (recombinant) type progeny resulting from homologous recombination that exhibits a different allele combination compared with its parents paracentric inversion that occurs outside of the centromere parental types progeny that exhibits the same allelic combination as its parents pericentric inversion that involves the centromere polyploid individual with an incorrect number of chromosome sets recombination frequency average number of crossovers between two alleles; observed as the number of nonparental types in a population of progeny translocation process by which one segment of a chromosome dissociates and reattaches to a different, nonhomologous chromosome trisomy otherwise diploid genotype in which one entire chromosome is duplicated X inactivation condensation of X chromosomes into Barr bodies during embryonic development in females to compensate for the double genetic dose CHAPTER SUMMARY 13.1 Chromosomal Theory and Genetic Linkages The Chromosomal Theory of inheritance, proposed by Sutton and Boveri, states that chromosomes are the vehicles of genetic heredity. Neither Mendelian genetics nor gene linkage is perfectly accurate; instead, chromosome behavior involves segregation, independent assortment, and occasionally, linkage. Sturtevant devised a method to assess recombination frequency and infer the relative positions and distances of linked genes on a chromosome on the basis of the average number of crossovers in the intervening region between the genes. Sturtevant correctly presumed that genes are arranged in serial order on chromosomes and that recombination between homologs can occur anywhere on a chromosome with equal likelihood. Whereas linkage causes alleles on the same chromosome to be inherited together, homologous recombination biases alleles toward an inheritance pattern of independent assortment. 536 Chapter 13 | Modern Understandings of Inheritance 13.2 Chromosomal Basis of Inherited Disorders The number, size, shape, and banding pattern of chromosomes make them easily identifiable in a karyogram and allows for the assessment of many chromosomal abnormalities. Disorders in chromosome number, or aneuploidies, are typically lethal |
to the embryo, although a few trisomic genotypes are viable. Because of X inactivation, aberrations in sex chromosomes typically have milder phenotypic effects. Aneuploidies also include instances in which segments of a chromosome are duplicated or deleted. Chromosome structures may also be rearranged, for example by inversion or translocation. Both of these aberrations can result in problematic phenotypic effects. Because they force chromosomes to assume unnatural topologies during meiosis, inversions and translocations are often associated with reduced fertility because of the likelihood of nondisjunction. REVIEW QUESTIONS 1. When comparing humans (or in Drosophila), are Xlinked recessive traits observed more frequently in males, in similar numbers between males and females, more frequently in females, or is the frequency different depending on the trait? Why? a. b. c. d. in more males than females in more females than males in males and females equally in different distributions depending on the trait 2. Which recombination frequency corresponds to perfect linkage and violates the law of independent assortment? a. b. c. d. 0 0.25 0.5 0.75 3. Which recombination frequency corresponds to independent assortment and the absence of linkage? a. b. c. d. 0 0.25 0.5 0.75 4. This OpenStax book is available for free at http://cnx.org/content/col12078/1.6 Based on the diagram, which of the following statements is true? a. Recombination of the body color and red/ cinnabar eye alleles will occur more frequently than recombination of the alleles for wing length and aristae length. b. Recombination of the body color and aristae length alleles will occur more frequently than recombination of red/ brown eye alleles and the aristae length alleles. c. Recombination of the gray/black body color and long/short aristae alleles will not occur. d. Recombination of the red/brown eye and long/ short aristae alleles will occur more frequently than recombination of the alleles for wing length and body color. 5. Which of the following codes describes position 12 on the long arm of chromosome 13? Chapter 13 | Modern Understandings of Inheritance 537 a. 13p12 b. 13q12 c. 12p13 d. 12q13 6. Assume a |
pericentric inversion occurred in one of two homologs prior to meiosis. The other homolog remains normal. During meiosis, what structure, if any, would CRITICAL THINKING QUESTIONS 7. Which best describes the Chromosomal Theory of Inheritance? a. The theory was proposed by Charles Darwin. It describes the units of inheritance between parents and offspring as well as the processes by which those units control offspring development. b. The theory was proposed by Boveri-Sutton. It describes linkage, recombination, and crossing over and states that Mendelian genes have specific loci on chromosomes, which undergo segregation and independent assortment. c. The theory was proposed by Charles Darwin. It states the Mendelian genes have two alternate forms and undergo independent assortment. It helped increase understanding of linkage and recombination. d. The theory was proposed by Boveri-Sutton. It describes the units of inheritance between parents and offspring as well as the processes by which those units control development in offspring. 8. In a test cross for two characteristics (dihybrid cross), can the predicted frequency of recombinant offspring be 60%? Why or why not? a. No. The predicted frequency of recombinant offspring ranges from 0% (for linked traits) to 50% (for unlinked traits) because of both parental and nonparental cases. b. Yes. The predicted frequency of recombinant offspring can be 60% if genes are located very far from each other. c. Yes. The predicted frequency can be 60% if crossing over occurs during every meiotic event. d. No. The predicted frequency can never be 60% due to the presence of mutations such as deletions. 9. Choose the statement that best describes how nondisjunction (see Figure 13.6) can result in an aneuploid zygote. these homologs assume in order to pair accurately along their lengths? a. V formation b. cruciform c. a loop d. pairing would not be possible a. Nondisjunction only occurs when homologous chromosomes do not separate during meiosis I, resulting in the formation of gametes containing n+1 and n-1 chromosomes. b. Nondisjunction only occurs when sister chromatids do not separate in meiosis II, resulting in the formation of gametes containing n+1 and n-1 chromosomes. c. Nondisjunction is the failure of homologous chromosomes |
to separate during meiosis I or the failure of sister chromatids to separate during meiosis II, leading to the formation of n+1/n-1/n chromosomes. d. Nondisjunction occurs when the sister chromatids fail to separate during mitosis II, resulting in the formation of gametes containing n+1 and n-1/n chromosomes. 10. Select the answer that correctly identifies the various chromosomal aberrations and their respective genetic consequence. a. nondisjunction - aneuploid gametes; duplication - physical and mental abnormalities; deletion lethal to a diploid organism; inversion chromosomal breaks in gene; translocations effects depend on how positions of genes are altered b. nondisjunction - physical and mental abnormalities; inversion - genetic imbalance; duplication - aneuploid gametes; translocations chromosomal breaks in the gene; deletion effects depend on how positions of genes are altered c. deletion -aneuploid gametes; translocations - physical and mental abnormalities; duplication effects depend on positions of genes; nondisjunction - causes genetic imbalance lethal to a diploid organism; aneuploidy - leads to various syndromes d. nondisjunction - chromosomal breaks in gene; duplication - physical and mental abnormalities; deletion - genetic imbalance lethal to a diploid organism; inversion - aneuploid gametes; translocations - effects depend on positions of genes 538 Chapter 13 | Modern Understandings of Inheritance TEST PREP FOR AP® COURSES 11. The figure represents a Drosophila linkage map for genes A-E. The numbers between the gene loci are the relative map units between each gene. Based on the linkage map, which two genes are most likely to segregate together? 14. a. Number Observed: AaBb (46), Aabb (4), aaBb (4), Aabb (46) b. Number Observed: AaBb (4), Aabb (46), aaBb (46), Aabb (4) c. Number Observed: AaBb (25), Aabb (25), aaBb (25), Aabb (25) d. Number Observed: AaBb (50), Aabb (0), aaBb (0), Aabb (50) a. A and B b. B and C c. |
C and D d. D and E 12. A test cross was made between true-breeding EEWW flies and eeww flies. The resulting F1 generation was then crossed with eeww flies. 100 offspring in the F2 generation were examined, and it was discovered that the E and W genes were not linked. Which is the correct genotype of the F2 if the genes were not linked? offspring if the genes were linked and a. Linked: 50% EeWw and 50% eeww; not linked: 25% EeWw, 25%, Eeww 25% eeWw, and 25% eeww b. Linked: 25% Eeww, 50% eeWw; not linked: parental genotypes EeWw and eeww. c. Linked genotypes (EeWw and eeww) and recombinant genotypes (Eeww and eeWw) in the F2 generation are nearly the same irrespective of their linkage. d. Linked: mostly with parental genotypes, Eeww and eeWw; unlinked: 25% EeWw and eeww with 75% Eeww and eeWw. 13. A cross was made with true-breeding AABB flies and true-breeding aabb flies. The resulting F1 then crossed with true-breeding aabb flies. Based on the linkage map, which of the following F2 genotype ratios is most likely to be observed? generation generation was This OpenStax book is available for free at http://cnx.org/content/col12078/1.6 Which of the following symptoms is most likely associated with the disorder shown in the karyotype? a. b. lethality infertility c. heart and bleeding defects d. short stature and stunted growth 15. Choose the correct option amongst the following that describes the disorder shown in the karyotype and the social, ethical or medical issue related to the disorder. Chapter 13 | Modern Understandings of Inheritance 539 a. Down syndrome (47 XY +21) causes intellectual disability, vision problems, congenital heart disease, and susceptibility to cancer. Healthcare providers often do not discuss the positive aspects of raising a child with Down syndrome and often provide out of date information. b. Klinefelter syndrome (47 XY +21) causes intellectual disability, vision problems, congenital heart disease, and susceptibility |
to cancer. Arguments are often made against abortion of an affected fetus. c. Klinefelter syndrome (47 XXY) causes sterility and reduced testosterone production. Arguments are often made against informing insurance companies about a diagnosis of this disease. d. Down syndrome (47 XXY) causes sterility and lower testosterone production. Arguments are often made against informing insurance companies about a diagnosis of this disease. 16. Which of the following gene orders is the most likely outcome of an inversion mutation in the chromosome shown? a. RSTUV b. RRSTUV c. RSUV d. RTSUV 17. With the help the diagram given, choose the most appropriate statement describing nondisjunction and its genetic consequences. a. Nondisjunction occurs when a homologous pair is unable to separate during meiosis I, resulting in the formation of gametes containing n+1 and n-1 chromosomes. This is called aneuploidy. b. Nondisjunction occurs due to the inability of sister chromatids to separate during meiosis II, resulting in the formation of gametes containing n+1 and n-1 chromosomes. This results in heart and bleeding defects. c. Nondisjunction is the failure of homologous chromosomes to separate during meiosis I or failure of sister chromatids to separate during meiosis II. This results in aneuploid gametes. d. Nondisjunction occurs when a pair of homologous chromosomes fails to segregate during meiosis II resulting in the formation of gametes containing n+1, n-1, or n numbers of chromosomes. This results in abnormal growth patterns. 18. If the effects of Klinefelter syndrome are compared to the effects of Down Syndrome, this disorder is ______. a. more severe than Down syndrome, due to gene deletions in this syndrome b. more severe than Down syndrome, due to trisomy in Klinefelter syndrome c. d. less severe than Down syndrome, due to monosomy in Down Syndrome less severe than Down syndrome, due to Xinactivation in this disorder 540 Chapter 13 | Modern Understandings of Inheritance SCIENCE PRACTICE CHALLENGE QUESTIONS 19. Drosophila that are true breeding for the traits straight wings (S) and red eyes (R) are crossed with flies that are true breeding for curved wings (s) and brown eyes (r). A |
test cross is then made between the offspring and the truebreeding ssrr flies. A. Use the symbols S, s, R, and r to construct a representation of the parental genotypes in the test cross. B. If these genes are located on different chromosomes, use a Punnett square to construct a representation of the offspring of the test cross. C. Predict the distribution of genotypes and phenotypes resulting from the test cross. D. As it happens, these genes are both on chromosome II as shown below. Use the symbols S, s, R, and r to construct a representation of the parental and recombinant genotypes in the test cross. E. Suppose that 500 flies are produced in the test cross. Apply mathematical methods to calculate the expected number of recombinant offspring using the linear map units (LMU) shown in the diagram below. Figure 13.14 20. Studies like the one described in question AP12.1 were carried out by Morgan and Sturtevant beginning in 1911. The discovery of linkage was made by Bateson and Punnett in 1900. They crossed a true-breeding purple (P) plant with long seeds (L) with a true-breeding red (r) plant with round seeds (l). They then performed a self-cross between the F1 generations. They obtained the F2 data shown below. This OpenStax book is available for free at http://cnx.org/content/col12078/1.6 Phenotype Genotype(s) Observed Expected purple, long purple, round red, long red, round total Table 13.1 4831 390 393 1338 6952 A. Use the symbols P, p, L, and l to construct a representation of the F2 genotypes and complete the second column in the table. B. Complete the fourth column of the table above by recording the values of the predicted numbers of plants with each genotype. C. Apply a c2 test at the 95% confidence level to evaluate the claim that these data confirm linkage. The definition of the statistic χ 2 = ∑ (o − e)2 e provided in the AP Biology Exam. and this table are D. At first, Bateson and Punnett did not see that these genes are located on the same chromosome and proceeded to measure the linkage distance between them, taking the first step toward creating a gene map. Justify the selection of data and the procedure from which data could be collected that would have provided the |
necessary evidence to confirm linkage and recombination. 21. Review the observations that provided researchers with evidence in support of the Chromosomal Theory of Inheritance. A. Evaluate the dependence of these observations on improvements in a critical technology during the period from 1850 to 1940. Identify this technology and describe how this technology allowed scientists to make the connection between chromosomes and genes. (As a hint, the name “chromosome” is taken from the Greek word chroma, which means colored or stained.) B. Mendel’s laws of inheritance are explained by the chromosomal theory. Use these observations to justify: • • the law of segregation the law of independent assortment 22. Errors in the transmission of genetic information to future generations are essential. Otherwise, organisms could not evolve over time. Some errors in the synthesis of Chapter 13 | Modern Understandings of Inheritance 541 new DNA during S phase in either meiosis or mitosis are not repaired. These errors usually involve single nucleotides. Errors that occur during prophase I of meiosis that are not corrected can involve the exchange of sequences between homologous chromosomes (duplications) or even nonhomologous chromosomes (translocations). Duplications are usually retained, and the organism remains viable without a change in phenotype. Translocations are usually lethal or significantly alter phenotype. In eukaryotes, duplications and the shuffling of parental genes through recombination are important sources of variation. Construct an explanation of the role of duplication as a source of raw material for future mutations and selection and contrast this type of variation with recombination. 23. Bacteria and Archaea reproduce asexually, and genetic material is in a closed loop. In both domains, genetic material is transferred horizontally, and polyploidy is common. Polypoidy is common in plants and occurs in invertebrate animals but is less common in vertebrates. In all domains, multiple copies of genes (gene duplication) are common. Based on this information, compare and contrast the mechanisms that provide genetic variation in the three domains: Bacteria, Archaea, and Eukarya. 542 Chapter 13 | Modern Understandings of Inheritance This OpenStax book is available for free at http://cnx.org/content/col12078/1.6 Chapter 14 | DNA Structure and Function 543 14 | DNA STRUCTURE AND FUNCTION Figure 14.1 Michael Morton went to jail in 1986 for the murder of this wife. Twenty |
-five years later, in 2011, he was exonerated of her murder by DNA evidence. (credit: Lauren Gerson) Chapter Outline 14.1: Historical Basis of Modern Understanding 14.2: DNA Structure and Sequencing 14.3: Basics of DNA Replication 14.4: DNA Replication in Prokaryotes 14.5: DNA Replication in Eukaryotes 14.6: DNA Repair Introduction Each person’s DNA is unique, and it is possible to detect differences among individuals within a species on the basis of these unique features. DNA analysis has many practical applications, including identifying criminals (forensics), determining paternity, tracing genealogy, identifying pathogens, researching archeological finds, tracing disease outbreaks, and studying human migration patterns. In the medical field, DNA is used in diagnostics, new vaccine development, and cancer therapy. It is often possible to determine predisposition to diseases by sequencing genes. Sometimes an innocent person is erroneously convicted of a crime and sent to jail. Between 2000 and 2015, evidence from DNA was used to exonerate over 250 innocent people. Twenty of those people were on death row after being convicted of a murder they didn’t commit. To learn more about the intense scientific and legal processes used to exonerate those wrongfully convicted, go to The Innocence Project website here (http://www.openstaxcollege.org/l/32innocence). 544 Chapter 14 | DNA Structure and Function 14.1 | Historical Basis of Modern Understanding In this section, you will explore the following questions: • What is transformation of DNA? How do Griffith’s experiments in 1928 relate to our modern understanding of DNA and how it works? • What are key historic experiments that helped identify DNA as the genetic material? • What are Chargaff’s rules of nitrogenous base pairing? Connection for AP® Courses Today the three letters “DNA” have become synonymous with crime solving, paternity testing, human identification, and genetic testing. All of these procedures are possible because of the discovery, in the middle of the twentieth century, that DNA is the genetic material. The results of several classic experiments set the stage for an explosion of our knowledge about DNA and how it stores and transmits genetic information. DNA was first isolated from white blood cells by Miescher in the 1860s. Over fifty years later, Griffith’s work transforming strains of the bacterium Streptococcus pneumoniae provided the first clue that DNA |
and not protein (as others argued) is the universal molecule of heredity. Griffith’s conclusions were later supported by Avery, MacLeod, and McCarty. Subsequent experiments by Hershey and Chase using the bacteriophage T2 proved decisively that DNA is the genetic material. Shortly thereafter, Chargaff determined the ratios of adenine, thymine, cytosine, and guanine in DNA, suggesting paired relationships (A = T and C = G). He also found that the percentages of A, T, C, and G are different for different species. All of these historically important experiments shaped our current understanding of DNA. The content presented in this section supports the learning objectives outlined in Big Ideas 3 and 4 of the AP® Biology Curriculum Framework. The AP® learning objectives merge essential knowledge content with one or more of the seven science practices. These objectives provide a transparent foundation for the AP® Biology course, along with inquiry-based laboratory experiences, instructional activities, and AP® exam questions. Big Idea 3 Enduring Understanding 3.A Essential Knowledge Science Practice Learning Objective Essential Knowledge Science Practice Learning Objective Big Idea 4 Living systems store, retrieve, transmit and respond to information essential to life processes. Heritable information provides for continuity of life. 3.A.1 DNA, and in some cases RNA, is the primary source of heritable information. 6.5 The student can evaluate alternative scientific explanations. 3.1 The student is able to construct scientific explanations that use the structures and mechanisms of DNA to support the claim that DNA is the primary source of heritable information. 3.A.1 DNA, and in some cases RNA, is the primary source of heritable information. 4.1 The student can justify the selection of the kind of data needed to answer a particular scientific question. 3.2 The student is able to justify the selection of data from historical investigations that support the claim that DNA is the source of heritable information. Biological systems interact, and these systems and their interactions possess complex properties. This OpenStax book is available for free at http://cnx.org/content/col12078/1.6 Chapter 14 | DNA Structure and Function 545 Enduring Understanding 4.A Interactions within biological systems lead to complex properties. Essential Knowledge 4.A.1 The subcomponents of biological molecules and their sequence determine the properties of that molecule. Science Practice Learning Objective 7.1 The student can connect phenomena and models across spatial and temporal scales. 4. |
1 The student is able to explain the connection between the sequence and the subcomponents of a biological polymer and its properties. The Science Practice Challenge Questions contain additional test questions for this section that will help you prepare for the AP exam. These questions address the following standards: [APLO 3.2][APLO 3.28][APLO 1.11][APLO 1.16][APLO 3.1][APLO 4.1] Modern understandings of DNA have evolved from the discovery of nucleic acid to the development of the double-helix model. In the 1860s, Friedrich Miescher (Figure 14.2), a physician by profession, was the first person to isolate phosphaterich chemicals from white blood cells or leukocytes. He named these chemicals (which would eventually be known as RNA and DNA) nuclein because they were isolated from the nuclei of the cells. Figure 14.2 Friedrich Miescher (1844–1895) discovered nucleic acids. 546 Chapter 14 | DNA Structure and Function To see Miescher conduct an experiment step-by-step, click through this review (http://openstaxcollege.org/l/ miescher_levene) of how he discovered the key role of DNA and proteins in the nucleus. Why were Phoebus Levene’s discoveries important to our current understanding of DNA? a. Phoebus Levene believed that the four nucleotides in DNA are not linked or repeated in the same pattern and that they are held together by phosphodiester bonds. b. He discovered that the nucleotides were held together by phosphodiester bonds, in which two phosphate groups bind two sugars together. This discovery led to our current understanding of DNA. c. He believed that proteins were less likely the vehicles for hereditary information. Later he discovered the four nucleotides in DNA which were linked together and repeated in a wide variety of different ways. d. He believed inaccurately that the four nucleotides in DNA repeated over in the same pattern. Also, he discovered that the nucleotides were held together by phosphodiester bonds in which the phosphate group binds two sugars together. A half century later, British bacteriologist Frederick Griffith was perhaps the first person to show that hereditary information could be transferred from one cell to another “horizontally,” rather than by descent. In 1928, he reported the first demonstration of bacterial transformation, a process in which external DNA is taken |
up by a cell, thereby changing morphology and physiology. He was working with Streptococcus pneumoniae, the bacterium that causes pneumonia. Griffith worked with two strains, rough (R) and smooth (S). The R strain is non-pathogenic (does not cause disease) and is called rough because its outer surface is a cell wall and lacks a capsule; as a result, the cell surface appears uneven under the microscope. The S strain is pathogenic (disease-causing) and has a capsule outside its cell wall. As a result, it has a smooth appearance under the microscope. Griffith injected the live R strain into mice and they survived. In another experiment, when he injected mice with the heat-killed S strain, they also survived. In a third set of experiments, a mixture of live R strain and heat-killed S strain were injected into mice, and—to his surprise—the mice died. Upon isolating the live bacteria from the dead mouse, only the S strain of bacteria was recovered. When this isolated S strain was injected into fresh mice, the mice died. Griffith concluded that something had passed from the heat-killed S strain into the live R strain and transformed it into the pathogenic S strain, and he called this the transforming principle (Figure 14.3). These experiments are now famously known as Griffith's transformation experiments. This OpenStax book is available for free at http://cnx.org/content/col12078/1.6 Chapter 14 | DNA Structure and Function 547 Figure 14.3 Two strains of S. pneumoniae were used in Griffith’s transformation experiments. The R strain is nonpathogenic. The S strain is pathogenic and causes death. When Griffith injected a mouse with the heat-killed S strain and a live R strain, the mouse died. The S strain was recovered from the dead mouse. Thus, Griffith concluded that something had passed from the heat-killed S strain to the R strain, transforming the R strain into S strain in the process. (credit "living mouse": modification of work by NIH; credit "dead mouse": modification of work by Sarah Marriage) Scientists Oswald Avery, Colin MacLeod, and Maclyn McCarty (1944) were interested in exploring this transforming principle further. They isolated the S strain from the dead mice and isolated the proteins and nucleic acids, namely RNA and DNA, as these were possible candidates for the molecule of heredity. They conducted a systematic elimination study. They used |
enzymes that specifically degraded each component and then used each mixture separately to transform the R strain. They found that when DNA was degraded, the resulting mixture was no longer able to transform the bacteria, whereas all of the other combinations were able to transform the bacteria. This led them to conclude that DNA was the transforming principle. Forensic Scientists and DNA Analysis DNA evidence was used for the first time to solve an immigration case. The story started with a teenage boy returning to London from Ghana to be with his mother. Immigration authorities at the airport were suspicious of him, thinking that he was traveling on a forged passport. After much persuasion, he was allowed to go live with his mother, but the immigration authorities did not drop the case against him. All types of evidence, including photographs, were provided to the authorities, but deportation proceedings were started nevertheless. Around the same time, Dr. Alec Jeffreys of Leicester University in the United Kingdom had invented a technique known as DNA fingerprinting. The immigration authorities approached Dr. Jeffreys for help. He took DNA samples from the mother and three of her children, plus an unrelated mother, and compared the samples with the boy’s DNA. Because the biological father was not in the picture, DNA from the three children was compared with the boy’s DNA. He found a match in the boy’s DNA for both the mother and his three siblings. He concluded that the boy was indeed the mother’s son. including documents, handwriting, Forensic scientists analyze many items, firearms, and biological samples. They analyze the DNA content of hair, semen, saliva, and blood, and compare it with a database of DNA profiles of known criminals. Analysis includes DNA isolation, sequencing, and sequence analysis; most forensic DNA analysis involves polymerase chain reaction (PCR) amplification of short tandem repeat (STR) loci and electrophoresis to determine the length of the PCR-amplified fragment. Only mitochondrial DNA is sequenced for forensics. Forensic scientists are expected to appear at court hearings to present their findings. They are usually employed in crime labs of city and state government agencies. Geneticists experimenting with DNA techniques also work for scientific and research organizations, pharmaceutical industries, and college and university labs. Students wishing to pursue a career as a forensic scientist should have at least a bachelor's degree in chemistry, biology, or physics, and preferably some experience working in a laboratory. 548 Chapter 14 | DNA Structure and Function Activity DNA Necklace. 1) Using a molecular modeling kit (or |
an online virtual kit such as jmol), create a model of each of the 4 nucleotides in DNA, based on structural diagrams found in this chapter or elsewhere online. 2) Identify where each nucleotide hydrogen-bonds with its complementary base. Add these bonds to secure the two pairs of nucleotides together. How does the hydrogen bonding differ between the two pairs of complementary bases? 3) Now look at a structural diagram of a complete DNA molecule. Based on the diagram, connect your two pairs of nucleotides together along your DNA’s sugar-phosphate backbone (depending on your model kit, you may have to first disconnect the hydrogen bonds between the complementary bases). Which atoms and molecules did you have to remove and add to create the sugar-phosphate backbone? Think About It Explain why radioactive sulfur and phosphorus were used to label T2 bacteriophages in the Hershey-Chase experiments. How did the results of these experiments contribute to the identification of DNA as the genetic material? Experiments conducted by Martha Chase and Alfred Hershey in 1952 provided confirmatory evidence that DNA was the genetic material and not proteins. Chase and Hershey were studying a bacteriophage, which is a virus that infects bacteria. Viruses typically have a simple structure: a protein coat, called the capsid, and a nucleic acid core that contains the genetic material, either DNA or RNA. The bacteriophage infects the host bacterial cell by attaching to its surface, and then it injects its nucleic acids inside the cell. The phage DNA makes multiple copies of itself using the host machinery, and eventually the host cell bursts, releasing a large number of bacteriophages. Hershey and Chase labeled one batch of phage with radioactive sulfur, 35S, to label the protein coat. Another batch of phage were labeled with radioactive phosphorus, 32P. Because phosphorous is found in DNA, but not protein, the DNA and not the protein would be tagged with radioactive phosphorus. Each batch of phage was allowed to infect the cells separately. After infection, the phage bacterial suspension was put in a blender, which caused the phage coat to be detached from the host cell. The phage and bacterial suspension was spun down in a centrifuge. The heavier bacterial cells settled down and formed a pellet, whereas the lighter phage particles stayed in the supernatant. In the tube that contained phage labeled with 35S, the supernatant contained the radio |
actively labeled phage, whereas no radioactivity was detected in the pellet. In the tube that contained the phage labeled with 32P, the radioactivity was detected in the pellet that contained the heavier bacterial cells, and no radioactivity was detected in the supernatant. Hershey and Chase concluded that it was the phage DNA that was injected into the cell and carried information to produce more phage particles, thus providing evidence that DNA was the genetic material and not proteins (Figure 14.4). This OpenStax book is available for free at http://cnx.org/content/col12078/1.6 Chapter 14 | DNA Structure and Function 549 Figure 14.4 In Hershey and Chase's experiments, bacteria were infected with phage radiolabeled with either 35S, which labels protein, or 32P, which labels DNA. Only 32P entered the bacterial cells, indicating that DNA is the genetic material. Around this same time, Austrian biochemist Erwin Chargaff examined the content of DNA in different species and found that the amounts of adenine, thymine, guanine, and cytosine were not found in equal quantities, and that it varied from species to species, but not between individuals of the same species. He found that the amount of adenine equals the amount of thymine, and the amount of cytosine equals the amount of guanine, or A = T and G = C. This is also known as Chargaff’s rules. This finding proved immensely useful when Watson and Crick were getting ready to propose their DNA double helix model. 14.2 | DNA Structure and Sequencing In this section, you will explore the following questions: • What is the molecular structure of DNA? • What is the Sanger method of DNA sequencing? What is an application of DNA sequencing? • What are the similarities and differences between eukaryotic and prokaryotic DNA? Connection for AP® Courses The currently accepted model of the structure of DNA was proposed in 1953 by Watson and Crick, who made their model after seeing a photograph of DNA that Franklin had taken using X-ray crystallography. The photo showed the molecule’s double-helix shape and dimensions. The two strands that make up the double helix are complementary and anti-parallel in nature. That is, one strand runs in the 5' to 3' direction, whereas the complementary strand runs in the 3' to 5' |
direction. (The significance of directionality will be important when we explore how DNA copies itself.) DNA is a polymer of nucleotides that consists of deoxyribose sugar, a phosphate group, and one of four nitrogenous bases—A, T, C, and G—with a purine always pairing with a pyrimidine (as Chargaff found). The genetic “language” of DNA is found in sequences of the nucleotides. During cell division each daughter cell receives a copy of DNA in a process called replication. In the years since the discovery of the structure of DNA, many technologies, including DNA sequencing, have been developed that enable us to better understand DNA and its role in our genomes. Information presented and the examples highlighted in the section support concepts outlined in Big Idea 3 of the AP® Biology Curriculum Framework. The Learning Objectives listed in the Curriculum Framework provide a transparent foundation for the AP® Biology course, an inquiry-based laboratory experience, instructional activities, and AP® exam 550 Chapter 14 | DNA Structure and Function questions. A Learning Objective merges required content with one or more of the seven science practices. Big Idea 3 Enduring Understanding 3.A Essential Knowledge Science Practice Learning Objective Essential Knowledge Science Practice Learning Objective Essential Knowledge Science Practice Learning Objective Living systems store, retrieve, transmit and respond to information essential to life processes. Heritable information provides for continuity of life. 3.A.1 DNA, and in some cases RNA, is the primary source of heritable information. 6.5 The student can evaluate alternative scientific explanations. 3.1 The student is able to construct scientific explanations that use the structures and mechanisms of DNA to support the claim that DNA is the primary source of heritable information. 3.A.1 DNA, and in some cases RNA, is the primary source of heritable information. 4.1 The student can justify the selection of the kind of data needed to answer a particular scientific question. 3.2 The student is able to justify the selection of data from historical investigations that support the claim that DNA is the source of heritable information. 3.A.1 DNA, and in some cases RNA, is the primary source of heritable information. 6.4 The student can make claims and predictions about natural phenomena based on scientific theories and models. 3.5 The student can justify the claim that humans can manipulate heritable information by identifying at least two commonly used technologies. The Science Practice Challenge Questions contain additional test questions for this section that will help you |
prepare for the AP exam. These questions address the following standards: [APLO 3.3][APLO 3.5][APLO 3.13] The building blocks of DNA are nucleotides. The important components of the nucleotide are a nitrogenous base, deoxyribose (5-carbon sugar), and a phosphate group (Figure 14.5). The nucleotide is named depending on the nitrogenous base. The nitrogenous base can be a purine such as adenine (A) and guanine (G), or a pyrimidine such as cytosine (C) and thymine (T). This OpenStax book is available for free at http://cnx.org/content/col12078/1.6 Chapter 14 | DNA Structure and Function 551 Figure 14.5 Each nucleotide is made up of a sugar, a phosphate group, and a nitrogenous base. The sugar is deoxyribose in DNA and ribose in RNA. The nucleotides combine with each other by covalent bonds known as phosphodiester bonds or linkages. The purines have a double ring structure with a six-membered ring fused to a five-membered ring. Pyrimidines are smaller in size; they have a single six-membered ring structure. The carbon atoms of the five-carbon sugar are numbered 1', 2', 3', 4', and 5' (1' is read as “one prime”). The phosphate residue is attached to the hydroxyl group of the 5' carbon of one sugar of one nucleotide and the hydroxyl group of the 3' carbon of the sugar of the next nucleotide, thereby forming a 5'-3' phosphodiester bond. In the 1950s, Francis Crick and James Watson worked together to determine the structure of DNA at the University of Cambridge, England. Other scientists like Linus Pauling and Maurice Wilkins were also actively exploring this field. Pauling had discovered the secondary structure of proteins using X-ray crystallography. In Wilkins’ lab, researcher Rosalind Franklin was using X-ray diffraction methods to understand the structure of DNA. Watson and Crick were able to piece together the puzzle of the DNA molecule on the basis of Franklin's data because Crick had also studied X-ray diffraction (Figure 14.6). In 1962, James Watson, Francis Crick, and Maurice Wilkins were awarded the Nobel Prize in |
Medicine. Unfortunately, by then Franklin had died, and Nobel prizes are not awarded posthumously. Figure 14.6 The work of pioneering scientists (a) James Watson, Francis Crick, and Maclyn McCarty led to our present day understanding of DNA. Scientist Rosalind Franklin discovered (b) the X-ray diffraction pattern of DNA, which helped to elucidate its double helix structure. (credit a: modification of work by Marjorie McCarty, Public Library of Science) Watson and Crick proposed that DNA is made up of two strands that are twisted around each other to form a right-handed helix. Base pairing takes place between a purine and pyrimidine; namely, A pairs with T and G pairs with C. Adenine and thymine are complementary base pairs, and cytosine and guanine are also complementary base pairs. The base pairs are stabilized by hydrogen bonds; adenine and thymine form two hydrogen bonds and cytosine and guanine form three hydrogen bonds. The two strands are anti-parallel in nature; that is, the 3' end of one strand faces the 5' end of the other strand. The sugar and phosphate of the nucleotides form the backbone of the structure, whereas the nitrogenous bases are 552 Chapter 14 | DNA Structure and Function stacked inside. Each base pair is separated from the other base pair by a distance of 0.34 nm, and each turn of the helix measures 3.4 nm. Therefore, ten base pairs are present per turn of the helix. The diameter of the DNA double helix is 2 nm, and it is uniform throughout. Only the pairing between a purine and pyrimidine can explain the uniform diameter. The twisting of the two strands around each other results in the formation of uniformly spaced major and minor grooves (Figure 14.7). Figure 14.7 DNA has (a) a double helix structure and (b) phosphodiester bonds. The (c) major and minor grooves are binding sites for DNA binding proteins during processes such as transcription (the copying of RNA from DNA) and replication. Activity Read Watson and Crick’s original Nature article, “Molecular Structure of Nucleic Acids: A Structure for Deoxyribose Nucleic Acid,” How did Watson and Crick’s model build on the findings of Rosalind Franklin? How did their model of DNA build on the findings of Hers |
hey and Chase, and others, showing that DNA can encode and pass information on to the next generation? Think About It Watson and Crick’s work determined the structure of DNA. However, it was still relatively unknown how DNA encoded information into genes. Select one modern form of biotechnology and research its basic methods online. Examples include gene sequencing, DNA fingerprinting, PCR (polymerase chain reaction), genetically-modified food, etc. Briefly describe your chosen technology, and what benefits it provides us. Then describe how Watson and Crick’s findings were vital to the development of your chosen technology. This OpenStax book is available for free at http://cnx.org/content/col12078/1.6 Chapter 14 | DNA Structure and Function 553 DNA Sequencing Techniques Until the 1990s, the sequencing of DNA (reading the sequence of DNA) was a relatively expensive and long process. Using radiolabeled nucleotides also compounded the problem through safety concerns. With currently available technology and automated machines, the process is cheap, safer, and can be completed in a matter of hours. Fred Sanger developed the sequencing method used for the human genome sequencing project, which is widely used today (Figure 14.8). Visit this site (http://openstaxcollege.org/l/DNA_sequencing) to watch a video explaining the DNA sequence reading technique that resulted from Sanger’s work. Describe one advantage and a possible limitation to Sanger’s method. a. Sanger’s method can be used to sequence more than one strand at a time which is less time consuming. Challenges of Sanger’s method includes its decreased accuracy to sequence DNA strands. b. Sanger’s method is a reliable and accurate way of sequencing DNA strands. However, only one strand at a time can be sequenced at a time. Also, it can look for one base only at a time which can be time consuming. c. Sanger’s method is highly inexpensive and less accurate. However, it is not readily adaptable to commercial kits. d. Sanger’s method is less time consuming and highly accurate. However, it is more expensive than other methods available for sequencing. The method is known as the dideoxy chain termination method. The sequencing method is based on the use of chain terminators, the dideoxynucleotides (ddNTPs). The dideoxynucleot |
ides, or ddNTPSs, differ from the deoxynucleotides by the lack of a free 3' OH group on the five-carbon sugar. If a ddNTP is added to a growing a DNA strand, the chain is not extended any further because the free 3' OH group needed to add another nucleotide is not available. By using a predetermined ratio of deoxyribonucleotides to dideoxynucleotides, it is possible to generate DNA fragments of different sizes. Figure 14.8 In Frederick Sanger's dideoxy chain termination method, dye-labeled dideoxynucleotides are used to generate DNA fragments that terminate at different points. The DNA is separated by capillary electrophoresis on the basis of size, and from the order of fragments formed, the DNA sequence can be read. The DNA sequence readout is shown on an electropherogram that is generated by a laser scanner. The DNA sample to be sequenced is denatured or separated into two strands by heating it to high temperatures. The DNA is divided into four tubes in which a primer, DNA polymerase, and all four nucleotides (A, T, G, and C) are added. In addition to each of the four tubes, limited quantities of one of the four dideoxynucleotides are added to each tube 554 Chapter 14 | DNA Structure and Function respectively. The tubes are labeled as A, T, G, and C according to the ddNTP added. For detection purposes, each of the four dideoxynucleotides carries a different fluorescent label. Chain elongation continues until a fluorescent dideoxy nucleotide is incorporated, after which no further elongation takes place. After the reaction is over, electrophoresis is performed. Even a difference in length of a single base can be detected. The sequence is read from a laser scanner. For his work on DNA sequencing, Sanger received a Nobel Prize in chemistry in 1980. Sanger’s genome sequencing has led to a race to sequence human genomes at a rapid speed and low cost, often referred to as the $1000 in one day sequence. Learn more by selecting the Sequencing at Speed animation here (http://openstaxcollege.org/l/DNA_and_genomes). Explain how fast DNA sequencing can change the way doctors treat disease. a. Faster genetic sequencing will help in quick analysis of the genetic makeup of bacteria that can cause |
diseases in humans for better and more efficient treatments. Also, sequencing of a cancerous cell’s DNA can provide better ways to treat or prevent cancer. b. Fast DNA sequencing can help us quickly analyze the genetic information of existing only bacteria (not new strains) only that cause disease in humans, which may lead to more efficient treatments. c. Fast DNA sequencing can help doctors to treat and diagnose diseases which are not rare in populations. d. Faster genetic sequencing can be used to treat and prevent a few types of cancers and thus increase the life expectancy of patients suffering from the diseases. Gel electrophoresis is a technique used to separate DNA fragments of different sizes. Usually the gel is made of a chemical called agarose. Agarose powder is added to a buffer and heated. After cooling, the gel solution is poured into a casting tray. Once the gel has solidified, the DNA is loaded on the gel and electric current is applied. The DNA has a net negative charge and moves from the negative electrode toward the positive electrode. The electric current is applied for sufficient time to let the DNA separate according to size; the smallest fragments will be farthest from the well (where the DNA was loaded), and the heavier molecular weight fragments will be closest to the well. Once the DNA is separated, the gel is stained with a DNA-specific dye for viewing it (Figure 14.9). This OpenStax book is available for free at http://cnx.org/content/col12078/1.6 Chapter 14 | DNA Structure and Function 555 Figure 14.9 DNA can be separated on the basis of size using gel electrophoresis. (credit: James Jacob, Tompkins Cortland Community College) 556 Chapter 14 | DNA Structure and Function [1] Neanderthal Genome: How Are We Related? The first draft sequence of the Neanderthal genome was recently published by Richard E. Green et al. in 2010. Neanderthals are the closest ancestors of present-day humans. They were known to have lived in Europe and Western Asia before they disappeared from fossil records approximately 30,000 years ago. Green’s team studied almost 40,000-year-old fossil remains that were selected from sites across the world. Extremely sophisticated means of sample preparation and DNA sequencing were employed because of the fragile nature of the bones and heavy microbial contamination. In their study, the scientists were able to sequence some four billion base pairs. The Neanderthal sequence was compared with that |
of present-day humans from across the world. After comparing the sequences, the researchers found that the Neanderthal genome had 2 to 3 percent greater similarity to people living outside Africa than to people in Africa. While current theories have suggested that all present-day humans can be traced to a small ancestral population in Africa, the data from the Neanderthal genome may contradict this view. Green and his colleagues also discovered DNA segments among people in Europe and Asia that are more similar to Neanderthal sequences than to other contemporary human sequences. Another interesting observation was that Neanderthals are as closely related to people from Papua New Guinea as to those from China or France. This is surprising because Neanderthal fossil remains have been located only in Europe and West Asia. Most likely, genetic exchange took place between Neanderthals and modern humans as modern humans emerged out of Africa, before the divergence of Europeans, East Asians, and Papua New Guineans. Several genes seem to have undergone changes from Neanderthals during the evolution of presentday humans. These genes are involved in cranial structure, metabolism, skin morphology, and cognitive development. One of the genes that is of particular interest is RUNX2, which is different in modern day humans and Neanderthals. This gene is responsible for the prominent frontal bone, bell-shaped rib cage, and dental differences seen in Neanderthals. It is speculated that an evolutionary change in RUNX2 was important in the origin of modern-day humans, and this affected the cranium and the upper body. According to the passage, which statement best describes the relationship between humans and Neanderthals? a. Early humans emerged from Africa, then spread out to populate different parts of the globe. An isolated population of these early humans interbred with Neanderthals. b. Early humans interbred with Neanderthals, emerged from Africa, then spread out to populate different parts of the globe. c. Early humans emerged from Africa, interbred with Neanderthals, then spread out to populate different parts of the globe. d. Early humans did not interbreed with Neanderthals, but we have many genetic similarities because we share a common ancestor. 1. Richard E. Green et al., “A Draft Sequence of the Neandertal Genome,” Science 328 (2010): 710-22. This OpenStax book is available for free at http://cnx.org/content/col12078/1.6 Chapter 14 | DNA Structure and Function 557 Watch Svant |
e Pääbo’s talk (http://openstaxcollege.org/l/neanderthal) explaining the Neanderthal genome research at the 2011 annual TED (Technology, Entertainment, Design) conference. Which of the statements gives the best explanation for the wider genetic variation in the human population in Africa than the rest of the world? a. It has been suggested that all humans most likely descended from Africa. This is supported by the research that genetic variance in Africa was also found in the rest of the world. b. The theory that humans descended from Africa was supported by the research that most of the human genomes tested outside of Africa had close ties to the genomes of people in Africa but a genetic variance in Africa was not found in the rest of the world. c. Humans have most likely descended from Africa. This research is supported by the fact that all the human genomes tested outside of Africa had close ties to the genomes of people in Africa. Also, there is a genetic variance in Africa that was not found in the rest of the world. d. The transition to modern humans occurred within Africa which was sudden. Thus, human genomes tested outside of Africa had close ties to the genomes of people in Africa. DNA Packaging in Cells When comparing prokaryotic cells to eukaryotic cells, prokaryotes are much simpler than eukaryotes in many of their features (Figure 14.10). Most prokaryotes contain a single, circular chromosome that is found in an area of the cytoplasm called the nucleoid. 558 Chapter 14 | DNA Structure and Function Figure 14.10 A eukaryote contains a well-defined nucleus, whereas in prokaryotes, the chromosome lies in the cytoplasm in an area called the nucleoid. In eukaryotic cells, DNA and RNA synthesis occur in a separate compartment from protein synthesis. In prokaryotic cells, both processes occur together. What advantages might there be to separating the processes? What advantages might there be to having them occur together? a. Compartmentalization in eukaryotic cells enables the building of more complex proteins and RNA products. In prokaryotes, the advantage is that RNA and protein synthesis occurs much more quickly because it occurs in a single compartment. b. Compartmentalization in prokaryotic cells enables the building of more complex proteins and RNA products. In eukaryotes, the advantage is that RNA and protein synthesis occurs much more quickly because they |
occur in a single compartment. c. Compartmentalization in eukaryotic cells enables the building of simpler proteins and RNA products. In prokaryotes, the advantage is only simpler proteins and RNA products because complex ones are not needed. d. Compartmentalization in eukaryotic cells enables the building of more complex proteins and RNA products. In prokaryotes, the advantage is that RNA and protein synthesis takes more time because it occurs in a single compartment. The size of the genome in one of the most well-studied prokaryotes, E.coli, is 4.6 million base pairs (approximately 1.1 mm, if cut and stretched out). So how does this fit inside a small bacterial cell? The DNA is twisted by what is known as supercoiling. Supercoiling means that DNA is either under-wound (less than one turn of the helix per 10 base pairs) or over-wound (more than 1 turn per 10 base pairs) from its normal relaxed state. Some proteins are known to be involved in the supercoiling; other proteins and enzymes such as DNA gyrase help in maintaining the supercoiled structure. Eukaryotes, whose chromosomes each consist of a linear DNA molecule, employ a different type of packing strategy to fit their DNA inside the nucleus (Figure 14.11). At the most basic level, DNA is wrapped around proteins known as histones to form structures called nucleosomes. The histones are evolutionarily conserved proteins that are rich in basic amino acids and form an octamer. The DNA (which is negatively charged because of the phosphate groups) is wrapped tightly around the histone core. This nucleosome is linked to the next one with the help of a linker DNA. This is also known as the “beads on a string” structure. This is further compacted into a 30 nm fiber, which is the diameter of the structure. At the metaphase stage, the chromosomes are at their most compact, are approximately 700 nm in width, and are found in association with scaffold proteins. In interphase, eukaryotic chromosomes have two distinct regions that can be distinguished by staining. The tightly packaged region is known as heterochromatin, and the less dense region is known as euchromatin. Heterochromatin usually contains genes that are not expressed, and is found in the regions of the centromere and telomeres. The e |
uchromatin usually contains genes that are transcribed, with DNA packaged around nucleosomes but not further compacted. This OpenStax book is available for free at http://cnx.org/content/col12078/1.6 Chapter 14 | DNA Structure and Function 559 Figure 14.11 These figures illustrate the compaction of the eukaryotic chromosome. 14.3 | Basics of DNA Replication In this section, you will explore the following questions: • How does the structure of DNA provide for the process of replication? • How did the Meselson and Stahl experiments support the semi-conservative nature of replication? Connection for AP® Courses The Watson and Crick model suggested a way in which DNA could be replicated during cell division. Basically, the two strands unwind and separate where the hydrogen bonds connect the nucleotides. Each parental strand then serves as a template for a new, complementary daughter strand. Replication is said to be semi-conservative because the original information encoded in each parental strand is conserved (kept) in the daughter molecules. Thus, a newly replicated molecule of DNA consists of one “old” strand and one “new” strand. Meselson and Stahl used density differences in nitrogen isotopes to investigate replication, and their experiments supported the semi-conservative model. However, the process of replication is more complex than their model’s simple description. Information presented and the examples highlighted in the section support concepts outlined in Big Idea 3 and Big Idea 4 of the AP® Biology Curriculum Framework. The Learning Objectives listed in the Curriculum Framework provide a transparent foundation for the AP® Biology course, an inquiry-based laboratory experience, instructional activities, and AP® exam questions. A Learning Objective merges required content with one or more of the seven Science Practices. 560 Chapter 14 | DNA Structure and Function Big Idea 3 Enduring Understanding 3.A Essential Knowledge Living systems store, retrieve, transmit and respond to information essential to life processes. Heritable information provides for continuity of life. 3.A.1 DNA, and in some cases RNA, is the primary source of heritable information. Science Practice 1.2 The student can describe representations and models of natural or man-made phenomena and systems in the domain. Learning Objective 3.3 The student is able to describe representations and models that illustrate how genetic information is copied for transmission between generations. The Science Practice Challenge Questions contain additional test questions for this section that will help you |
prepare for the AP exam. These questions address the following standards: [APLO 2.34][APLO 3.3][APLO 4.1] The elucidation of the structure of the double helix provided a hint as to how DNA divides and makes copies of itself. This model suggests that the two strands of the double helix separate during replication, and each strand serves as a template from which the new complementary strand is copied. What was not clear was how the replication took place. There were three models suggested (Figure 14.12): conservative, semi-conservative, and dispersive. Figure 14.12 The three suggested models of DNA replication. Grey indicates the original DNA strands, and blue indicates newly synthesized DNA. In conservative replication, the parental DNA remains together, and the newly formed daughter strands are together. The semi-conservative method suggests that each of the two parental DNA strands act as a template for new DNA to be synthesized; after replication, each double-stranded DNA includes one parental or “old” strand and one “new” strand. In the dispersive model, both copies of DNA have double-stranded segments of parental DNA and newly synthesized DNA interspersed. Meselson and Stahl were interested in understanding how DNA replicates. They grew E. coli for several generations in a medium containing a “heavy” isotope of nitrogen (15N) that gets incorporated into nitrogenous bases, and eventually into the DNA (Figure 14.13). This OpenStax book is available for free at http://cnx.org/content/col12078/1.6 Chapter 14 | DNA Structure and Function 561 Figure 14.13 Meselson and Stahl experimented with E. coli grown first in heavy nitrogen (15N) then in 14N. DNA grown in 15N (red band) is heavier than DNA grown in 14N (orange band), and sediments to a lower level in cesium chloride solution in an ultracentrifuge. When DNA grown in 15N is switched to media containing 14N, after one round of cell division the DNA sediments halfway between the 15N and 14N levels, indicating that it now contains fifty percent 14N. In subsequent cell divisions, an increasing amount of DNA contains 14N only. This data supports the semi-conservative replication model. (credit: modification of work by Mariana Ruiz Villareal) The E. coli culture was then shifted into medium containing 14N |
and allowed to grow for one generation. The cells were harvested and the DNA was isolated. The DNA was centrifuged at high speeds in an ultracentrifuge. Some cells were allowed to grow for one more life cycle in 14N and spun again. During the density gradient centrifugation, the DNA is loaded into a gradient (typically a salt such as cesium chloride or sucrose) and spun at high speeds of 50,000 to 60,000 rpm. Under these circumstances, the DNA will form a band according to its density in the gradient. DNA grown in 15N will band at a higher density position than that grown in 14N. Meselson and Stahl noted that after one generation of growth in 14N after they had been shifted from 15N, the single band observed was intermediate in position in between DNA of cells grown exclusively in 15N and 14N. This suggested either a semi-conservative or dispersive mode of replication. The DNA harvested from cells grown for two generations in 14N formed two bands: one DNA band was at the intermediate position between 15N and 14N, and the other corresponded to the band of 14N DNA. These results could only be explained if DNA replicates in a semi-conservative manner. Therefore, the other two modes were ruled out. During DNA replication, each of the two strands that make up the double helix serves as a template from which new strands are copied. The new strand will be complementary to the parental or “old” strand. When two daughter DNA copies are formed, they have the same sequence and are divided equally into the two daughter cells. 562 Chapter 14 | DNA Structure and Function Click through this tutorial (http://openstaxcollege.org/l/DNA_replicatio2) on DNA replication. One theory of aging is that the body’s ability to fix mistakes in its DNA decreases as we age. How can this affect DNA replication? a. Aging causes accumulation of DNA mutations and DNA damage of only the nuclear DNA and the mistakes will be passed down to new cells causing age related diseases. b. Aging results in ineffective DNA repair mechanism so that the mistakes in the DNA will be passed down to new cells. This could lead to the development of age-related diseases. c. Aging causes DNA polymerase to function abnormally. This is the sole reason which causes defects in DNA replication. d. DNA replication of only fast growing cells is affected by aging. Activity Design (but do not implement) |
an experiment to test the three models of DNA replication. Summarize the results you would expect if each of the three models of DNA replication were correct. Assume you have access in a laboratory to the following: an experimental organism such as E. coli, an unlimited variety of isotopes, test tube and centrifuge, and organic growth media. 14.4 | DNA Replication in Prokaryotes In this section, you will explore the following questions: • How is DNA replicated in prokaryotes, and what are the roles of the leading and lagging strands and Okazaki fragments in the process? • What is the role of DNA polymerase and other enzymes and proteins in supporting replication? Connection for AP® Courses As was stated previously, DNA replication is more complex than simply unzipping the double helix and making new complementary strands. Replication in prokaryotes starts from a sequence of nucleotides on the chromosome called the origin of replication—the point at which the DNA opens up or unzips. The enzyme helicase opens up the DNA at the point where hydrogen bonds connect the strands, resulting in the formation of a Y-shaped replication fork. Single-strand binding proteins keep the fork open. The enzyme primase synthesizes RNA primers to initiate DNA synthesis by DNA polymerase, which can add nucleotides only in the 5' to 3' direction. DNA polymerase recognizes the 3'-OH end as its landing site; thus, polymerase “reads” the template strand in the 3' to 5' direction and builds the complementary DNA polymer in the 5' to 3' direction. One strand—called the leading strand—is synthesized continuously in the direction of the replication fork (the direction in which helicase is separating the two strands), with polymerase adding new nucleotides one-by-one. However, This OpenStax book is available for free at http://cnx.org/content/col12078/1.6 Chapter 14 | DNA Structure and Function 563 replication of the other strand—called the lagging strand—occurs in a direction away from the replication fork, in short stretches of DNA known as Okazaki fragments. (Think of the activities on the lagging strand as analogous to trying to walk on a moving sidewalk that is moving in the opposite direction.) The RNA primers are replaced by DNA nucleotides, and ligase seals the DNA, creating phosphodiester bonds between the 3'-OH of one |
end and the 5'-phosphate of the other strand. The replicated DNA molecules now consist of one original template strand and one newly synthesized strand. Information presented and the examples highlighted in the section support concepts outlined in Big Idea 3 of the AP® Biology Curriculum Framework. The Learning Objectives listed in the Curriculum Framework provide a transparent foundation for the AP® Biology course, an inquiry-based laboratory experience, instructional activities, and AP® exam questions. A Learning Objective merges required content with one or more of the seven Science Practices. Big Idea 3 Enduring Understanding 3.A Essential Knowledge Living systems store, retrieve, transmit and respond to information essential to life processes. Heritable information provides for continuity of life. 3.A.1 DNA, and in some cases RNA, is the primary source of heritable information. Science Practice 1.2 The student can describe representations and models of natural or man-made phenomena and systems in the domain. Learning Objective 3.3 The student is able to describe representations and models that illustrate how genetic information is copied for transmission between generations. The Science Practice Challenge Questions contain additional test questions for this section that will help you prepare for the AP exam. These questions address the following standards: [APLO 4.3][APLO 1.18][APLO 1.21] DNA replication has been extremely well studied in prokaryotes primarily because of the small size of the genome and the mutants that are available. E. coli has 4.6 million base pairs in a single circular chromosome and all of it gets replicated in approximately 42 minutes, starting from a single origin of replication and proceeding around the circle in both directions. This means that approximately 1000 nucleotides are added per second. The process is quite rapid and occurs without many mistakes. DNA replication employs a large number of proteins and enzymes, each of which plays a critical role during the process. One of the key players is the enzyme DNA polymerase, also known as DNA pol, which adds nucleotides one by one to the growing DNA chain that are complementary to the template strand. The addition of nucleotides requires energy; this energy is obtained from the nucleotides that have three phosphates attached to them, similar to ATP which has three phosphate groups attached. When the bond between the phosphates is broken, the energy released is used to form the phosphodiester bond between the incoming nucleotide and the growing chain. In prokaryotes, three main types of polymerases are known: DNA pol I |
, DNA pol II, and DNA pol III. It is now known that DNA pol III is the enzyme required for DNA synthesis; DNA pol I and DNA pol II are primarily required for repair. How does the replication machinery know where to begin? It turns out that there are specific nucleotide sequences called origins of replication where replication begins. In E. coli, which has a single origin of replication on its one chromosome (as do most prokaryotes), it is approximately 245 base pairs long and is rich in AT sequences. The origin of replication is recognized by certain proteins that bind to this site. An enzyme called helicase unwinds the DNA by breaking the hydrogen bonds between the nitrogenous base pairs. ATP hydrolysis is required for this process. As the DNA opens up, Y-shaped structures called replication forks are formed. Two replication forks are formed at the origin of replication and these get extended bi- directionally as replication proceeds. Single-strand binding proteins coat the single strands of DNA near the replication fork to prevent the single-stranded DNA from winding back into a double helix. DNA polymerase is able to add nucleotides only in the 5' to 3' direction (a new DNA strand can be only extended in this direction). It also requires a free 3'-OH group to which it can add nucleotides by forming a phosphodiester bond between the 3'-OH end and the 5' phosphate of the next nucleotide. This essentially means that it cannot add nucleotides if a free 3'-OH group is not available. Then how does it add the first nucleotide? The problem is solved with the help of a primer that provides the free 3'-OH end. Another enzyme, RNA primase, synthesizes an RNA primer that is about five to ten nucleotides long and complementary to the DNA. Because this sequence primes the DNA synthesis, it is appropriately called the primer. DNA polymerase can now extend this RNA primer, adding nucleotides one by one that are complementary to the template strand (Figure 14.14). 564 Chapter 14 | DNA Structure and Function Figure 14.14 A replication fork is formed when helicase separates the DNA strands at the origin of replication. The DNA tends to become more highly coiled ahead of the replication fork. Topoisomerase breaks and reforms DNA’s phosphate backbone ahead of the replication fork, thereby relieving the pressure that results from this supercoiling. Single-strand binding proteins bind to the single- |
stranded DNA to prevent the helix from re-forming. Primase synthesizes an RNA primer. DNA polymerase III uses this primer to synthesize the daughter DNA strand. On the leading strand, DNA is synthesized continuously, whereas on the lagging strand, DNA is synthesized in short stretches called Okazaki fragments. DNA polymerase I replaces the RNA primer with DNA. DNA ligase seals the gaps between the Okazaki fragments, joining the fragments into a single DNA molecule. (credit: modification of work by Mariana Ruiz Villareal) You isolate a cell strain in which the joining together of Okazaki fragments is impaired and suspect that a mutation has occurred in an enzyme found at the replication fork. Which enzyme is most likely to be mutated? a. DNA ligase b. DNA polymerase III c. helicase d. topoisomerase The replication fork moves at the rate of 1000 nucleotides per second. DNA polymerase can only extend in the 5' to 3' direction, which poses a slight problem at the replication fork. As we know, the DNA double helix is anti-parallel; that is, one strand is in the 5' to 3' direction and the other is oriented in the 3' to 5' direction. One strand, which is complementary to the 3' to 5' parental DNA strand, is synthesized continuously towards the replication fork because the polymerase can add nucleotides in this direction. This continuously synthesized strand is known as the leading strand. The other strand, complementary to the 5' to 3' parental DNA, is extended away from the replication fork, in small fragments known as Okazaki fragments, each requiring a primer to start the synthesis. Okazaki fragments are named after the Japanese scientist who first discovered them. The strand with the Okazaki fragments is known as the lagging strand. The leading strand can be extended by one primer alone, whereas the lagging strand needs a new primer for each of the short Okazaki fragments. The overall direction of the lagging strand will be 3' to 5', and that of the leading strand 5' to 3'. A protein called the sliding clamp holds the DNA polymerase in place as it continues to add nucleotides. The sliding clamp is a ring-shaped protein that binds to the DNA and holds the polymerase in place. Topoisomerase prevents the over-winding of the DNA double helix ahead of the replication fork as the DNA is opening up; it does so by causing temporary |
nicks in the DNA helix and then resealing it. As synthesis proceeds, the RNA primers are replaced by DNA. The primers are removed by the exonuclease activity of DNA pol I, and the gaps are filled in by deoxyribonucleotides. The nicks that remain between the newly synthesized DNA (that replaced the RNA primer) and the previously synthesized DNA are sealed by the enzyme DNA ligase that catalyzes the formation of phosphodiester linkage between the 3'-OH end of one nucleotide and the 5' phosphate end of the other fragment. Once the chromosome has been completely replicated, the two DNA copies move into two different cells during cell division. The process of DNA replication can be summarized as follows: 1. DNA unwinds at the origin of replication. This OpenStax book is available for free at http://cnx.org/content/col12078/1.6 Chapter 14 | DNA Structure and Function 565 2. Helicase opens up the DNA-forming replication forks; these are extended bidirectionally. 3. Single-strand binding proteins coat the DNA around the replication fork to prevent rewinding of the DNA. 4. Topoisomerase binds at the region ahead of the replication fork to prevent supercoiling. 5. Primase synthesizes RNA primers complementary to the DNA strand. 6. DNA polymerase starts adding nucleotides to the 3'-OH end of the primer. 7. Elongation of both the lagging and the leading strand continues. 8. RNA primers are removed by exonuclease activity. 9. Gaps are filled by DNA pol by adding dNTPs. 10. The gap between the two DNA fragments is sealed by DNA ligase, which helps in the formation of phosphodiester bonds. Table 14.1 summarizes the enzymes involved in prokaryotic DNA replication and the functions of each. Prokaryotic DNA Replication: Enzymes and Their Functions Enzyme/protein Specific Function DNA pol I DNA pol II DNA pol III Helicase Ligase Primase Exonuclease activity removes RNA primer and replaces with newly synthesized DNA Repair function Main enzyme that adds nucleotides in the 5'-3' direction Opens the DNA helix by breaking hydrogen bonds between the nitrogenous bases Seals the gaps between the Okazaki fragments to create one continuous DNA strand Synthesizes RNA primers needed to start replication Sliding Clamp Helps to hold the |
DNA polymerase in place when nucleotides are being added Topoisomerase Single-strand binding proteins (SSB) Table 14.1 Helps relieve the stress on DNA when unwinding by causing breaks and then resealing the DNA Binds to single-stranded DNA to avoid DNA rewinding back. 566 Chapter 14 | DNA Structure and Function Review the full process of DNA replication here (http://openstaxcollege.org/l/replication_DNA). Explain why errors in DNA replication are rare events in cells. a. Errors in DNA replication are rare events in a cell due to the presence of DNA ligase enzyme which fixes mistakes in the copying process. b. Polymerase I is solely responsible for proofreading and fixing mistakes in the copying process, which explains why so few mistakes are made. c. Polymerase I and II are responsible for proofreading and fixing mistakes in the copying process which explains why errors in DNA replication are rare. d. Errors in DNA replication are rare events in cells due to the action of DNA helicase. Activity Use the model of DNA you constructed in Section 14.2 to demonstrate the process of replication in prokaryotes, showing how the activities differ on the leading and lagging strands. You need to add to your model by including enzymes and other proteins involved in the replication process. Think About It You isolate a DNA strand in which the joining together of Okazaki fragments is impaired and suspect that a mutation has occurred in an enzyme found at the replication fork. Which enzyme is most likely mutated? 14.5 | DNA Replication in Eukaryotes In this section, you will explore the following questions: • What are the similarities and differences between DNA replication in eukaryotes and prokaryotes? • What is the role of telomerase in DNA replication? Connection for AP® Courses Concepts and examples described in this section are not in scope for AP. However, the roles of telomeres and telomerase in aging and cancer are informative and build on your knowledge of DNA replication in prokaryotes. Eukaryotic genomes are much more complex and larger in size than prokaryotic genomes. The human genome has three billion base pairs per haploid set of chromosomes, and 6 billion base pairs are replicated during the S phase of the cell cycle. There are multiple origins of replication on the eukaryotic chromosome; humans can have up to 100,000 origins of replication. The rate of replication is approximately |
100 nucleotides per second, much slower than prokaryotic replication. This OpenStax book is available for free at http://cnx.org/content/col12078/1.6 Chapter 14 | DNA Structure and Function 567 In yeast, which is a eukaryote, special sequences known as Autonomously Replicating Sequences (ARS) are found on the chromosomes. These are equivalent to the origin of replication in E. coli. The number of DNA polymerases in eukaryotes is much more than prokaryotes: 14 are known, of which five are known to have major roles during replication and have been well studied. They are known as pol α, pol β, pol γ, pol δ, and pol ε. The essential steps of replication are the same as in prokaryotes. Before replication can start, the DNA has to be made available as template. Eukaryotic DNA is bound to basic proteins known as histones to form structures called nucleosomes. The chromatin (the complex between DNA and proteins) may undergo some chemical modifications, so that the DNA may be able to slide off the proteins or be accessible to the enzymes of the DNA replication machinery. At the origin of replication, a pre-replication complex is made with other initiator proteins. Other proteins are then recruited to start the replication process (Table 14.2). A helicase using the energy from ATP hydrolysis opens up the DNA helix. Replication forks are formed at each replication origin as the DNA unwinds. The opening of the double helix causes over-winding, or supercoiling, in the DNA ahead of the replication fork. These are resolved with the action of topoisomerases. Primers are formed by the enzyme primase, and using the primer, DNA pol can start synthesis. While the leading strand is continuously synthesized by the enzyme pol δ, the lagging strand is synthesized by pol ε. A sliding clamp protein known as PCNA (Proliferating Cell Nuclear Antigen) holds the DNA pol in place so that it does not slide off the DNA. RNase H removes the RNA primer, which is then replaced with DNA nucleotides. The Okazaki fragments in the lagging strand are joined together after the replacement of the RNA primers with DNA. The gaps that remain are sealed by DNA ligase, which forms the phosphodiester bond. Telomere Replication |
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