12.2: Characteristics and Traits
Skills to Develop
- Explain the relationship between genotypes and phenotypes in dominant and recessive gene systems
- Develop a Punnett square to calculate the expected proportions of genotypes and phenotypes in a monohybrid cross
- Explain the purpose and methods of a test cross
- Identify non-Mendelian inheritance patterns such as incomplete dominance, codominance, recessive lethals, multiple alleles, and sex linkage
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 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 F 1 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 F 2 offspring. Therefore, the F 1 plants must have been genotypically different from the parent with yellow pods.
The P 1 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 P 1 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 F 1 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 \(\PageIndex{1}\)).
| Dominant Traits | Recessive Traits |
|---|---|
| Achondroplasia | Albinism |
| Brachydactyly | Cystic fibrosis |
| Huntington’s disease | Duchenne muscular dystrophy |
| Marfan syndrome | Galactosemia |
| Neurofibromatosis | Phenylketonuria |
| Widow’s peak | Sickle-cell anemia |
| Wooly hair | Tay-Sachs disease |
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 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 F 1 and F 2 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 Punnett square, all possible combinations of the parental alleles are listed along the top (for one parent) and side (for the other parent) of a grid, 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 \(\PageIndex{1}\)).
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 \(\PageIndex{1}\)). 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 pea-plant 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 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 \(\PageIndex{1}\)). 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 F 2 generation resulting from crosses for individual traits.
Mendel validated these results by performing an F 3 cross in which he self-crossed the dominant- and recessive-expressing F 2 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 F 2 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 F 1 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 F 1 offspring will be heterozygotes expressing the dominant trait (Figure \(\PageIndex{2}\)). Alternatively, if the dominant expressing organism is a heterozygote, the F 1 offspring will exhibit a 1:1 ratio of heterozygotes and recessive homozygotes (Figure \(\PageIndex{2}\)). The test cross further validates Mendel’s postulate that pairs of unit factors segregate equally.
Exercise
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?
- Answer
-
You cannot be sure if the plant is homozygous or heterozygous as the data set is too small: by random chance, all three plants might have acquired only the dominant gene even if the recessive one is present. If the round pea parent is heterozygous, there is a one-eighth probability that a random sample of three progeny peas will all be round.
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 analysis to study the inheritance pattern of human genetic diseases (Figure \(\PageIndex{3}\)).
Exercise
What are the genotypes of the individuals labeled 1, 2 and 3?
- Answer
-
Individual 1 has the genotype aa . Individual 2 has the genotype Aa . Individual 3 has the genotype 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 \(\PageIndex{4}\)), a cross between a homozygous parent with white flowers ( C W C W ) and a homozygous parent with red flowers ( C R C R ) will produce offspring with pink flowers ( C R C W ). (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 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 C R C R :2 C R C W :1 C W C W , and the phenotypic ratio would be 1:2:1 for red:pink:white.
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 ( L M L M and L N L N ) express either the M or the N allele, and heterozygotes ( L M L N ) 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 \(\PageIndex{5}\)). Here, four alleles exist for the c gene. The wild-type version, C + C + , is expressed as brown fur. The chinchilla phenotype, c ch c ch , is expressed as black-tipped white fur. The Himalayan phenotype, c h c h , 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 wild-type 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.
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 temperature-sensitive 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 wild-type gene product or changing its distribution in the body. One example of this is the Antennapedia mutation in Drosophila (Figure \(\PageIndex{6}\)). 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.
Evolution Connection: 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 \(\PageIndex{7}\)a), 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 \(\PageIndex{7}\)b). 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.
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. Sulfadoxine-resistant 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. 1
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.  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 (X W ) and it is dominant to white eye color (X w ) (Figure \(\PageIndex{8}\)). 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 X W Y or X w Y. In contrast, females have two allele copies of this gene and can be X W X W , X W X w , or X w X w .
In an X-linked cross, the genotypes of F 1 and F 2 offspring depend on whether the recessive trait was expressed by the male or the female in the P 1 generation. With regard to Drosophila eye color, when the P 1 male expresses the white-eye phenotype and the female is homozygous red-eyed, all members of the F 1 generation exhibit red eyes (Figure \(\PageIndex{9}\)).  A subsequent cross between the X W X w female and the X W Y male would produce only red-eyed females (with X W X W or X W X w genotypes) and both red- and white-eyed males (with X W Y or X w Y genotypes).  The F 1 generation would exhibit only heterozygous red-eyed females (X W X w ) and only white-eyed males (X w Y). Half of the F 2 females would be red-eyed (X W X w ) and half would be white-eyed (X w X w ). Similarly, half of the F 2 males would be red-eyed (X W Y) and half would be white-eyed (X w Y).
Exercise
What ratio of offspring would result from a cross between a white-eyed male and a female that is heterozygous for red eye color?
- Answer
-
Half of the female offspring would be heterozygous (X W X w ) with red eyes, and half would be homozygous recessive (X w X w ) with white eyes. Half of the male offspring would be hemizygous dominant (X W Y) withe red yes, and half would be hemizygous recessive (X w Y) with white eyes.
Discoveries in fruit fly genetics can be applied to human genetics. When a female parent is homozygous for a recessive X-linked trait, she will pass the trait on to 100 percent of her offspring. Her male offspring are, 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 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 gender 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 \(\PageIndex{10}\)). Although some Y-linked recessive disorders exist, typically they are associated with infertility in males and are therefore not transmitted to subsequent generations.
Link to Learning
Watch this video to learn more about sex-linked traits.
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 wild-type/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.  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 \(\PageIndex{11}\)). 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.
Summary
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 F 1 offspring will all exhibit the same phenotype as the parent homozygous for the dominant trait. If these heterozygous offspring are self-crossed, the resulting F 2 offspring will be equally likely to inherit gametes carrying the dominant or recessive trait, giving rise to offspring of which one quarter are homozygous dominant, half are heterozygous, and one quarter are homozygous recessive. Because homozygous dominant and heterozygous individuals are phenotypically identical, the observed traits in the F 2 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 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.
Footnotes
- 1 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.
Glossary
- allele
- gene variations that arise by mutation and exist at the same relative locations on homologous chromosomes
- autosomes
- any of the non-sex chromosomes
- codominance
- in a heterozygote, complete and simultaneous expression of both alleles for the same characteristic
- 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
- 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
- incomplete dominance
- in a heterozygote, expression of two contrasting alleles such that the individual displays an intermediate phenotype
- monohybrid
- result of a cross between two true-breeding parents that express different traits for only one characteristic
- phenotype
- observable traits expressed by an organism
- Punnett square
- visual representation of a cross between two individuals in which the gametes of each individual are denoted along the top and side of a grid, respectively, and the possible zygotic genotypes are recombined at each box in the grid
- recessive lethal
- inheritance pattern in which an allele is only lethal in the homozygous form; the heterozygote may be normal or have some altered, non-lethal phenotype
- sex-linked
- any gene on a sex chromosome
- test cross
- cross between a dominant expressing individual with an unknown genotype and a homozygous recessive individual; the offspring phenotypes indicate whether the unknown parent is heterozygous or homozygous for the dominant trait
- X-linked
- gene present on the X, but not the Y chromosome


23: Postulates of Quantum Mechanics
In order to understand deeper quantum mechanics, scientists have derived a series of axioms that result in what are called postulates of quantum mechanics . These are, in fact, assumptions that we need to make to understand how the measured reality relates with the mathematics of quantum mechanics. It is important to notice that the postulates are necessary for the interpretation of the theory, but not for the mathematics behind it. Regarding of whether we interpret it or not, the mathematics is complete and consistent. In fact, as we will see in the next chapter, several controversies regarding the interpretation of the mathematics are still open, and different philosophies have been developed to rationalize the results. Recall also that there are different ways of writing the equation of quantum mechanics, all equivalent to each other (i.e., Schrödinger’s differential formulation and Heisenberg’s algebraic formulation that we saw in chapter 3). For these reasons, there is not an agreement on the number of postulates that are necessary to interpret the theory, and some philosophy and/or formulation might require more postulates than others. In this chapter, we will discuss the six postulates, as they are usually presented in chemistry and introductory physics textbooks and as they relate with a basic statistical interpretation of quantum mechanics. Regardless of the philosophical consideration on the meanings and numbers of the postulate, as well as their physical origin, these statements will make the interpretation of the theory a little easier, as we will see in the next chapter.


2.1: Forces, Fundamental and Composite
Newton's First Law
We now understand how to handle motion in all its forms, but really we haven’t done much in the way of physics, because we haven’t explained what causes these different motions. In ancient times, Aristotle made the observation that eventually all things seem to come to rest, which led him to conclude that a stationary condition was “natural” for everything (well, everything on Earth – heavenly bodies never seemed to stop moving). He stated that keeping things moving requires constant pushing or pulling, or it would eventually settle into a state of rest.
This is a very intuitive way of describing the nature of things, and most people even today see the world this way. It wasn’t until nearly 2000 years after Aristotle that a genius born on Christmas day would overturn that long-held belief. His name was Isaac Newton, and he claimed that in fact nature behaved in precisely the opposite manner. Newton claimed that it was not natural for objects to be at rest unless they were already at rest. If they were already moving, then it was natural for them to continue moving. He claimed that it was the fact that objects on Earth could not escape the slowing effects of pushes and pulls that accounted for them always coming to rest.
But Newton was more specific about this “natural state of motion.” He stated that the only type of motion that would continue indefinitely if undisturbed by pushes or pulls was constant velocity (speed and direction) motion. That is, any motion that involved changes of speed or direction requires a push or pull.
Newton's 1st Law of Motion
Objects at rest or in motion at a constant speed in a straight line will remain in that state unless acted upon by an external influence.
Conceptual Question
A stone is swung in a horizontal circle while tied to a string, which suddenly breaks. Which of the paths below represents the motion the rock will follow? (these are paths viewed from above)
- It depends upon whether the rock is speeding up or slowing down at the moment the string broke.
- Solution
-
(d) The rock starts off moving in a circle, which means it was accelerating and according to the first law, it must have had a force on it. From the description of the motion, the only force contributing to that net force had to be the tension force by the string. At the instant that the string breaks, the force vanishes, which means that the rock can no longer accelerate. Zero acceleration means constant velocity, which means that whatever speed and direction the rock had at the moment that the string broke, it must maintain. Note that the rock has no memory whatsoever of the fact that it was accelerating just a moment before, so it neither continues accelerating for a short time, nor does it compensate for the previous acceleration by accelerating the other way.
Definition of Force
What we have been calling “pushes and pulls” or “external influences” is called force in physics. Most people have an intuitive idea of what force is, and like so many other physics concepts, this intuition is very likely wrong. We'll start by saying what force is not , then move on to its definition.
Alert
Force is not a quantity stored in, or possessed by, an object. Force cannot be transferred from one object to another, nor can one claim that one object “has” more force than another. This can be a hard notion to shake.
Definition: Force
Force is an interaction between two objects, which comes in the form of a push or a pull.
This simple definition belies some very difficult conceptual ideas that people (like Aristotle, and indeed every human since) struggle with, as we will soon see. The trick will be for us to develop some tools we can rely on that will help us get past our misconceptions. We will develop these tools in the sections to come, but here we will focus on the nature of forces that we encounter in everyday life. But one thing we can conclude from this is that pushes and pulls have definite directions, which means that we can conclude that forces are vectors .
Individual Particles vs. Systems of Particles
When we consider forces on and/or by individual particles, we find a couple of things. First, all such forces act "at a distance." That is, particles never actually touch each other – it is useful to think of particles as being merely points, with no extension in space, which makes them touching each other rather problematic. So these particles are somehow aware of each other's presence, and exert pushes and/or pulls on one another. The second thing we find is that these forces only come in a limited variety of just 4 types: gravitation, electromagnetic, and two different types of nuclear forces. It is believed that while these forces all manifest very differently (the forces depend upon different particle properties, and vary differently with particle separation), they ultimately are different manifestations of a single force. Indeed, it was once thought that the electric and magnetic forces were distinct, until it was shown quite conclusively that they are two sides of the same coin, and they are now referred to as a single force. It also happens that modern theorists have shown that one of the nuclear forces (called the "weak nuclear force") is just a different manifestation of the electromagnetic force. This combination is therefore often referred to as the "electroweak" force amongs physicists. This particular unification of seemingly disparate forces is much harder to describe to those not fluent in the languages of high-energy physics and advanced mathematics, and so the simpler (older) claim that there are four such forces lingers. These four (three) action-at-a-distance-between-individual-particles forces are called the fundamental forces .
Suppose now that we have two collections of particles, each of which we categorize as a "system", or more crudely, as an "object." These two objects exert forces on one another in the following way: Every particle in system #1 exerts a fundamental force on every particle in system #2. The sum of all these forces we can now call a single "force between the two objects." Clearly due to its cumulative nature, it is not "fundamental," but it is still nevertheless a force (we will use the word composite to describe these non-fundamental forces), in that it will cause the affected objects to no longer remain at rest or in motion in a straight line at constant speed. Given that much of what we will discuss are macroscopic systems where objects comprised of trillions of trillions of particles exert forces on each other, it makes sense to categorize some of the more common examples of these composite forces. But under it all, it is important to remember that all of these flavors of forces are just macroscopic special cases of just a few fundamental forces.
Digression: Quantum Mechanics vs. Classical Mechanics
All of the discussion here (and later in this textbook) about fundamental forces and individual particles assumes that we are employing a "classical" mechanical model for describing the universe. We have known for a long time that the realm of the very small (i.e. individual particles) does not actually function in this manner. Nevertheless, physics is about using whatever model we like that describes nature in a consistent manner that has predictive value for the conditions imposed on it. This classical model will not work if we take an extremely close look at what is happening to particles, but here we are only using this model to get a more general sense for what is happening macroscopically – the world where we look at enormous systems of particles ("objects" like chairs and bicycles) – and for this purpose, this model serves us very well.
Van der Waals Force
There is clearly a wide chasm that must be bridged in order to move a discussion from a fundamental force between two point-like particles to a force between two cars in a traffic accident. The most important step in this daunting journey boils down to a simple observation of what happens when two small clusters (systems you can call molecules, if you prefer) of particles are brought into proximity with each other...
- When they are at just the right separation, the clusters do not exert a net force on each other. Every one of the individual particles of one cluster exerts a fundamental force on every particle in the other cluster, but the sum total of these forces is zero.
- When the clusters are moved closer together than the "perfect separation" described in the previous bullet, a strange thing happens – the sum total of fundamental forces between individual particles no longer comes to zero. When they get too close, the clusters repel each other.
- If the clusters are pulled slightly farther apart than the "perfect separation", then the composite force between the clusters becomes attractive.
This simple-yet-amazing property of this composite force (typically referred to as a Van der Waals force ) resulting from many electromagnetic forces is what we have to thank for our very existence. If systems of particles could only (as in the case of individual particles) only exert either attractive or repulsive forces on each other, then all matter would either collapse in on itself or explode. We will not study the mathematics of this type of composite force for several chapters, but we will refer back to its characteristic properties frequently later as we describe even more crude composite forces that we will work with in the macroscopic world.
Alert
An astute reader that looks up "Van der Waals forces" (or who perhaps studied them already in a chemistry class) will undoubtedly find that this name is generally given specifically to forces between particles in a gas. Indeed, this is the specific phenomenon that Van der Waals studied. But it turns out that the property is much more robust than only applying to gases, so we are taking some license here and referring to all forces that behave similarly with this moniker. In a later chapter, we will discuss a mathematical model for this kind of force called the "Lennard-Jones potential." The point is that we should not get too worked-up about labels we give this physical behavior – it is the behavior itself that is important.


4.6: Bar Graphs
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Another form of graph that you often want to plot is the
bar graph
. The main function that you can use in R to draw them is the
barplot()
function.
102
And to illustrate the use of the function, I’ll use the
finalists
variable that I introduced in Section 5.1.7. What I want to do is draw a bar graph that displays the number of finals that each team has played in over the time spanned by the
afl
data set. So, let’s start by creating a vector that contains this information. I’ll use the
tabulate()
function to do this (which will be discussed properly in Section
??
, since it creates a simple numeric vector:
> freq <- tabulate( afl.finalists )
> print( freq )
[1] 26 25 26 28 32 0 6 39 27 28 28 17 6 24 26 39 24
This isn’t exactly the prettiest of frequency tables, of course. I’m only doing it this way so that you can see the
barplot()
function in it’s “purest” form: when the input is just an ordinary numeric vector. That being said, I’m obviously going to need the team names to create some labels, so let’s create a variable with those. I’ll do this using the
levels()
function, which outputs the names of all the levels of a factor (see Section 4.7:
> teams <- levels( afl.finalists )
> print( teams )
[1] "Adelaide" "Brisbane" "Carlton" "Collingwood"
[5] "Essendon" "Fitzroy" "Fremantle" "Geelong"
[9] "Hawthorn" "Melbourne" "North Melbourne" "Port Adelaide"
[13] "Richmond" "St Kilda" "Sydney" "West Coast"
[17] "Western Bulldogs"
Okay, so now that we have the information we need, let’s draw our bar graph. The main argument that you need to specify for a bar graph is the
height
of the bars, which in our case correspond to the values stored in the
freq
variable:
> barplot( height = freq ) # specifying the argument name (panel a)
> barplot( freq ) # the lazier version (panel a)
Either of these two commands will produce the simple bar graph shown in Figure 6.21.
As you can see, R has drawn a pretty minimal plot. It doesn’t have any labels, obviously, because we didn’t actually tell the
barplot()
function what the labels are! To do this, we need to specify the
names.arg
argument. The
names.arg
argument needs to be a vector of character strings containing the text that needs to be used as the label for each of the items. In this case, the
teams
vector is exactly what we need, so the command we’re looking for is:
barplot( height = freq, names.arg = teams )
This is an improvement, but not much of an improvement. R has only included a few of the labels, because it can’t fit them in the plot. This is the same behaviour we saw earlier with the multiple-boxplot graph in Figure 6.16. However, in Figure 6.16 it wasn’t an issue: it’s pretty obvious from inspection that the two unlabelled plots in between 1987 and 1990 must correspond to the data from 1988 and 1989. However, the fact that
barplot()
has omitted the names of every team in between Adelaide and Fitzroy is a lot more problematic.
The simplest way to fix this is to rotate the labels, so that the text runs vertically not horizontally. To do this, we need to alter set the
las
parameter, which I discussed briefly in Section
??
. What I’ll do is tell R to rotate the text so that it’s always perpendicular to the axes (i.e., I’ll set
las = 2
). When I do that, as per the following command…
barplot(height = freq, # the frequencies
names.arg = teams, # the label
las = 2) # rotate the labels
… the result is the bar graph shown in Figure 6.23.  To fix this, we have to be a bit cleverer again. A simple fix would be to use shorter names rather than the full name of all teams, and in many situations that’s probably the right thing to do. However, at other times you really do need to create a bit more space to add your labels, so I’ll show you how to do that.
Changing global settings using
par()
Altering the margins to the plot is actually a somewhat more complicated exercise than you might think. In principle it’s a very simple thing to do: the size of the margins is governed by a graphical parameter called
mar
, so all we need to do is alter this parameter. First, let’s look at what the
mar
argument specifies. The
mar
argument is a vector containing four numbers: specifying the amount of space at the bottom, the left, the top and then the right. The units are “number of
lines'". The default value
for
mar
is
c(5.1, 4.1, 4.1, 2.1)`, meaning that R leaves 5.1”lines" empty at the bottom, 4.1 lines on the left and the bottom, and only 2.1 lines on the right. In order to make more room at the bottom, what I need to do is change the first of these numbers. A value of 10.1 should do the trick.
So far this doesn’t seem any different to the other graphical parameters that we’ve talked about. However, because of the way that the traditional graphics system in R works, you need to specify what the margins will be
before
calling your high-level plotting function. Unlike the other cases we’ve see, you can’t treat
mar
as if it were just another argument in your plotting function. Instead, you have to use the
par()
function to change the graphical parameters beforehand, and only then try to draw your figure. In other words, the first thing I would do is this:
> par( mar = c( 10.1, 4.1, 4.1, 2.1) )
There’s no visible output here, but behind the scenes R has changed the graphical parameters associated with the current device (remember, in R terminology all graphics are drawn onto a “device”). Now that this is done, we could use the exact same command as before, but this time you’d see that the labels all fit, because R now leaves twice as much room for the labels at the bottom. However, since I’ve now figured out how to get the labels to display properly, I might as well play around with some of the other options, all of which are things you’ve seen before:
barplot( height = freq,
names.arg = teams,
las=2,
ylab = "Number of Finals",
main = "Finals Played, 1987-2010",
density = 10,
angle = 20)
However, one thing to remember about the
par()
function is that it doesn’t just change the graphical parameters for the current
plot
. Rather, the changes pertain to any subsequent plot that you draw onto the same
device
. This might be exactly what you want, in which case there’s no problem. But if not, you need to reset the graphical parameters to their original settings. To do this, you can either close the device (e.g., close the window, or click the “Clear All” button in the Plots panel in Rstudio) or you can reset the graphical parameters to their original values, using a command like this:
> par( mar = c(5.1, 4.1, 4.1, 2.1) )


8.1: Apportionment
Apportionment is the problem of dividing up a fixed number of things among groups of different sizes. In politics, this takes the form of allocating a limited number of representatives amongst voters. This problem, presumably, is older than the United States, but the best-known ways to solve it have their origins in the problem of assigning each state an appropriate number of representatives in the new Congress when the country was formed. States also face this apportionment problem in defining how to draw districts for state representatives. The apportionment problem comes up in a variety of non-political areas too, though. Here is a simple question. Say your school district has 5 schools, some of which have a lot more students than others. The state budget allows the district to have 8 counselors. Assuming that each counselor can only serve one school, which school gets how many? Here, the “set” of 8 counselor positions is being divided up among the 5 schools. This sounds pretty realistic, right?
Apportionment Rules
- The things being divided up can exist only in whole numbers.
- We must use all of the things being divided up, and we cannot add any more.
- Each group must get at least one of the things being divided up.
- The number of things assigned to each group should be at least approximately proportional to the population of the group. (Exact proportionality isn’t possible because of the whole number requirement, but we should try to be close. If Group A is larger than Group B, then Group B shouldn’t get more of the things than Group A does.)
In terms of the apportionment of the United States House of Representatives, these rules imply:
- We can only have whole representatives (a state can’t have 3.4 representatives).
- We can only use the (currently) 435 representatives available. If one state gets another representative, another state has to lose one.
- Every state gets at least one representative.
- The number of representatives each state gets should be approximately proportional to the state population. This way, the number of constituents each representative has should be approximately equal.
We will look at four ways of solving the apportionment problem. Three of them (Lowndes’s method is the exception) have been used at various times to apportion the U.S. Congress although the method currently in use (the Huntington-Hill method) is significantly more complicated.
Hamilton’s Method
Alexander Hamilton proposed the method that now bears his name. His method was approved by Congress in 1791 but was vetoed by President Washington. (This, by the way, was the very first presidential veto in the United States). It was later adopted in 1852 and used through 1911. He begins by determining, to several decimal places, how many things each group should get. Since he was interested in the question of Congressional representation, we’ll use the language of states and representatives. He determines how many representatives each state should get by following these steps:
Hamilton's Method
- Determine how many people each representative should represent. Do this by dividing the total population of all the states by the total number of representatives. This answer is called the divisor.
- Divide each state’s population by the divisor to determine how many representatives it should have. Record this answer to several decimal places. This answer is called the quota.
Since we can only allocate whole representatives, Hamilton resolves the whole number problem as follows:
- Cut off all the decimal parts of all the quotas (but don’t forget what the decimals were). These are called the lower quotas. Add up the remaining whole numbers. This answer will always be less than or equal to the total number of representatives (and the “or equal to” part happens only in very specific circumstances that are incredibly unlikely to turn up).
- Assuming that the total from Step 3 was less than the total number of representatives, assign the remaining representatives, one each, to the states whose decimal parts of the quota were largest, until the desired total is reached.
Note on rounding: Today we have technological advantages that Hamilton (and the others) couldn’t even have imagined. Take advantage of them, and keep several decimal places.
Example \(\PageIndex{1}\)
The state of Delaware has three counties: Kent, New Castle, and Sussex. The Delaware state House of Representatives has 41 members. If Delaware wants to divide this representation along county lines (which is not required, but let’s pretend they do), let’s use Hamilton’s method to apportion them. The populations of the counties are as follows (from the 2010 Census):
| County | Population |
|---|---|
| Kent | 162,310 |
| New Castle | 538,479 |
| Sussex | 197,145 |
| Total | 897,934 |
Solution
- First, we determine the divisor: 897,934/41 = 21,900.82927
- Now we determine each county’s quota by dividing the county’s population by the divisor:
| County | Population | Quota |
|---|---|---|
| Kent | 162,310 | 7.4111 |
| New Castle | 538,479 | 24.5872 |
| Sussex | 197,145 | 9.0017 |
| Total | 897,934 |
- Removing the decimal parts of the quotas gives:
| County | Population | Quota | Initial |
|---|---|---|---|
| Kent | 162,310 | 7.4111 | 7 |
| New Castle | 538,479 | 24.5872 | 24 |
| Sussex | 197,145 | 9.0017 | 9 |
| Total | 897,934 |
- We need 41 representatives and this only gives 40. The remaining one goes to the county with the largest decimal part, which is New Castle:
| County | Population | Quota | Initial | Final |
|---|---|---|---|---|
| Kent | 162,310 | 7.4111 | 7 | 7 |
| New Castle | 538,479 | 24.5872 | 24 | 25 |
| Sussex | 197,145 | 9.0017 | 9 | 9 |
| Total | 897,934 | 40 |
Example \(\PageIndex{2}\)
Use Hamilton’s method to apportion the 75 seats of Rhode Island’s House of Representatives among its five counties.
| County | Population |
|---|---|
| Bristol | 49,875 |
| Kent | 166,158 |
| Newport | 82,888 |
| Providence | 626,667 |
| Washington | 126,979 |
| Total | 1,052,567 |
Solution
- The divisor is 1,052,567/75 = 14,034.22667
- Determine each county’s quota by dividing its population by the divisor:
| County | Population | Quota |
|---|---|---|
| Bristol | 49,875 | 3.5538 |
| Kent | 166,158 | 11.8395 |
| Newport | 82,888 | 5.9061 |
| Providence | 626,667 | 44.6528 |
| Washington | 126,979 | 9.0478 |
| Total | 1,052,567 |
- Remove the decimal part of each quota:
| County | Population | Quota | Initial |
|---|---|---|---|
| Bristol | 49,875 | 3.5538 | 3 |
| Kent | 166,158 | 11.8395 | 11 |
| Newport | 82,888 | 5.9061 | 5 |
| Providence | 626,667 | 44.6528 | 44 |
| Washington | 126,979 | 9.0478 | 9 |
| Total | 1,052,567 | 72 |
- We need 75 representatives and we only have 72, so we assign the remaining three, one each, to the three counties with the largest decimal parts, which are Newport, Kent, and Providence:
| County | Population | Quota | Initial | Final |
|---|---|---|---|---|
| Bristol | 49,875 | 3.5538 | 3 | 3 |
| Kent | 166,158 | 11.8395 | 11 | 12 |
| Newport | 82,888 | 5.9061 | 5 | 6 |
| Providence | 626,667 | 44.6528 | 44 | 45 |
| Washington | 126,979 | 9.0478 | 9 | 9 |
| Total | 1,052,567 | 72 | 75 |
Note that even though Bristol County’s decimal part is greater than .5, it isn’t big enough to get an additional representative because three other counties have greater decimal parts.
Hamilton’s method obeys something called the Quota Rule. The Quota Rule isn’t a law of any sort, but it is just an idea that some people, including Hamilton, think is a good one.
Quota Rule
The Quota Rule says that the final number of representatives a state gets should be within one of that state’s quota. Since we’re dealing with whole numbers for our final answers, each state should either go up to the next whole number above its quota, or down to the next whole number below its quota.
Controversy
After seeing Hamilton’s method, many people find that it makes sense. It is not that difficult to use, and they wonder why anyone would want another method. The problem is that Hamilton’s method is subject to several paradoxes. Three of them happened, on separate occasions, when Hamilton’s method was used to apportion the United States House of Representatives.
The Alabama Paradox is named for an incident that happened during the apportionment that took place after the 1880 census. (A similar incident happened ten years earlier involving the state of Rhode Island, but the paradox is named after Alabama.) The post-1880 apportionment had been completed, using Hamilton’s method applied to the new population numbers from the census. Then it was decided that because of the country’s growing population, the House of Representatives should be made larger. That meant that the apportionment would need to be done again, still using Hamilton’s method and the same 1880 census numbers, but with more representatives. The assumption was that some states would gain another representative and others would stay with the same number they already had (since there weren’t enough new representatives being added to give one more to every state). The paradox is that Alabama ended up losing a representative in the process even though no populations were changed and the total number of representatives increased.
The New States Paradox happened when Oklahoma became a state in 1907. Oklahoma had enough population to qualify for five representatives in Congress. Those five representatives would need to come from somewhere, though, so five states, presumably, would lose one representative each. That happened, but another unexpected thing also happened: Maine gained a representative (from New York)!
The Population Paradox happened between the apportionments after the census of 1900 and of 1910. In those ten years, Virginia’s population grew at an average annual rate of 1.07%, while Maine’s grew at an average annual rate of 0.67%. Virginia started with more people, grew at a faster rate, grew by more people, and ended up with more people than Maine. By itself, that doesn’t mean that Virginia should gain representatives or Maine shouldn’t, because there are lots of other states involved. But Virginia ended up losing a representative to Maine.
Jefferson’s Method
Thomas Jefferson, who lived before any of these paradoxes, proposed a different method for apportionment. After Washington vetoed Hamilton’s method, Jefferson’s method was adopted and used in Congress from 1791 through 1842. Jefferson, of course, had political reasons for wanting his method to be used rather than Hamilton’s. Primarily, his method favors larger states, and his own home state of Virginia was the largest in the country at the time. He would also argue that it’s the ratio of people to representatives that is the critical thing, and apportionment methods should be based on that. But the paradoxes we saw also provide mathematical reasons for concluding that Hamilton’s method isn’t so good. While Jefferson’s method might or might not be the best one to replace it, at least Jefferson provided an alternative.
The first steps of Jefferson’s method are the same as Hamilton’s method. He finds the same divisor and the same quota and cuts off the decimal parts in the same way, giving a total number of representatives that is less than the required total. The difference is in how Jefferson resolves that difference. He says that since we ended up with an answer that is too small, our divisor must have been too big. He changes the divisor by making it smaller, finding new quotas with the new divisor, cutting off the decimal parts, and looking at the new total, until we find a divisor that produces the required total.
Jefferson's Method
- Determine how many people each representative should represent. Do this by dividing the total population of all the states by the total number of representatives. This answer is called the standard divisor.
- Divide each state’s population by the divisor to determine how many representatives it should have. Record this answer to several decimal places. This answer is called the quota.
- Cut off all the decimal parts of all the quotas (but don’t forget what the decimals were). These are the lower quotas. Add up the remaining whole numbers. This answer will always be less than or equal to the total number of representatives.
- If the total from Step 3 was less than the total number of representatives, reduce the divisor and recalculate the quota and allocation. Continue doing this until the total in Step 3 is equal to the total number of representatives. The divisor we end up using is called the modified divisor or adjusted divisor
A careful reader should be able to spot the problem immediately. Jefferson’s method does not state exactly how to reduce that devisor. Let’s take a look at some examples.
Example \(\PageIndex{3}\)
We’ll return to Delaware and apply Jefferson’s method (Example \(\PageIndex{1}\)). We begin, as we did with Hamilton’s method, by finding the quotas with the original divisor, 21,900.82927:
| County | Population | Quota | Initial |
|---|---|---|---|
| Kent | 162,310 | 7.4111 | 7 |
| New Castle | 538,479 | 24.5872 | 24 |
| Sussex | 197,145 | 9.0017 | 9 |
| Total | 897,934 | 40 |
Solution
We need 41 representatives, and this divisor gives only 40. We must reduce the divisor until we get 41 representatives. Let’s try 21,500 as the divisor:
| County | Population | Quota | Initial |
|---|---|---|---|
| Kent | 162,310 | 7.5493 | 7 |
| New Castle | 538,479 | 25.0455 | 25 |
| Sussex | 197,145 | 9.1695 | 9 |
| Total | 897,934 | 41 |
This gives us the required 41 representatives, so we’re done. If we had fewer than 41, we’d need to reduce the divisor more. If we had more than 41, we’d need to choose a divisor less than the original but greater than the second choice.
Notice that with the new, lower divisor, the quota for New Castle County (the largest county in the state) increased by much more than those of Kent County or Sussex County.
Example \(\PageIndex{4}\)
We’ll apply Jefferson’s method for Rhode Island (Example \(\PageIndex{2}\)). The original divisor of 14,034.22667 gave these results:
| County | Population | Quota | Initial |
|---|---|---|---|
| Bristol | 49,875 | 3.5538 | 3 |
| Kent | 166,158 | 11.8395 | 11 |
| Newport | 82,888 | 5.9061 | 5 |
| Providence | 626,667 | 44.6528 | 44 |
| Washington | 126,979 | 9.0478 | 9 |
| Total | 1,052,567 | 72 |
Solution
We need 75 representatives and we only have 72, so we need to use a smaller divisor. Let’s try 13,500:
| County | Population | Quota | Initial |
|---|---|---|---|
| Bristol | 49,875 | 3.6944 | 3 |
| Kent | 166,158 | 12.3080 | 12 |
| Newport | 82,888 | 6.1399 | 6 |
| Providence | 626,667 | 46.4198 | 46 |
| Washington | 126,979 | 9.4059 | 9 |
| Total | 1,052,567 | 76 |
We’ve gone too far. We need a divisor that’s greater than 13,500 but less than 14,034.22667. Let’s try 13,700:
| County | Population | Quota | Initial |
|---|---|---|---|
| Bristol | 49,875 | 3.6405 | 3 |
| Kent | 166,158 | 12.1283 | 12 |
| Newport | 82,888 | 6.0502 | 6 |
| Providence | 626,667 | 45.7421 | 45 |
| Washington | 126,979 | 9.2685 | 9 |
| Total | 1,052,567 | 75 |
This works.
Notice, in comparison to Hamilton’s method, that although the results were the same, they came about in a different way, and the outcome was almost different. Providence County (the largest) almost went up to 46 representatives before Kent (which is much smaller) got to 12. Although that didn’t happen here, it can. Divisor-adjusting methods like Jefferson’s are not guaranteed to follow the quota rule!
Webster’s Method
Daniel Webster (1782-1852) proposed a method similar to Jefferson’s in 1832. It was adopted by Congress in 1842 but was replaced by Hamilton’s method in 1852. It was then adopted again in 1901. The difference is that Webster rounds the quotas to the nearest whole number rather than dropping the decimal parts. If that doesn’t produce the desired results at the beginning, he says, like Jefferson, to adjust the divisor until it does. (In Jefferson’s case, at least the first adjustment will always be to make the divisor smaller. That is not always the case with Webster’s method.)
Webster's Method
- Determine how many people each representative should represent. Do this by dividing the total population of all the states by the total number of representatives. This answer is called the standard divisor.
- Divide each state’s population by the divisor to determine how many representatives it should have. Record this answer to several decimal places. This answer is called the quota.
- Round (up or down) all the quotas to the nearest whole number. Add up the remaining whole numbers.
- If the total from Step 3 was less than the total number of representatives, reduce the divisor and recalculate the quota and allocation. If the total from step 3 was larger than the total number of representatives, increase the divisor and recalculate the quota and allocation. Continue doing this until the total in Step 3 is equal to the total number of representatives. The divisor we end up using is called the modified divisor or adjusted divisor
Example \(\PageIndex{5}\)
Again, Delaware (Example \(\PageIndex{1}\)), with an initial divisor of 21,900.82927:
Solution
| County | Population | Quota | Initial |
|---|---|---|---|
| Kent | 162,310 | 7.4111 | 7 |
| New Castle | 538,479 | 24.5872 | 25 (Rounded Up) |
| Sussex | 197,145 | 9.0017 | 9 |
| Total | 897,934 | 41 |
This gives the required total, so we’re done.
Example \(\PageIndex{6}\)
Again, Rhode Island (Example \(\PageIndex{2}\)), with an initial divisor of 14,034.22667:
Solution
| County | Population | Quota | Initial |
|---|---|---|---|
| Bristol | 49,875 | 3.5538 | 4 |
| Kent | 166,158 | 11.8395 | 12 |
| Newport | 82,888 | 5.9061 | 6 |
| Providence | 626,667 | 44.6528 | 45 |
| Washington | 126,979 | 9.0478 | 9 |
| Total | 1,052,567 | 76 |
This is too many, so we need to increase the divisor. Let’s try 14,100:
| County | Population | Quota | Initial |
|---|---|---|---|
| Bristol | 49,875 | 3.5372 | 4 |
| Kent | 166,158 | 11.7843 | 12 |
| Newport | 82,888 | 5.8786 | 6 |
| Providence | 626,667 | 44.4445 | 44 |
| Washington | 126,979 | 9.0056 | 9 |
| Total | 1,052,567 | 75 |
This works, so we’re done.
Like Jefferson’s method, Webster’s method carries a bias in favor of states with large populations, but rounding the quotas to the nearest whole number greatly reduces this bias. (Notice that Providence County, the largest, is the one that gets a representative trimmed because of the increased quota.) Also like Jefferson’s method, Webster’s method does not always follow the quota rule, but it follows the quota rule much more often than Jefferson’s method does. (In fact, if Webster’s method had been applied to every apportionment of Congress in all of American history, it would have followed the quota rule every single time.)
In 1980, two mathematicians, Peyton Young and Mike Balinski, proved what we now call the Balinski-Young Impossibility Theorem.
Balinski-Young Impossibility Theorem
Any apportionment method which always follows the quota rule will be subject to the possibility of paradoxes like the Alabama, New States, or Population Paradoxes. In other words, we can choose a method that avoids those paradoxes, but only if we are willing to give up the guarantee of following the quota rule.
Huntington-Hill Method
In 1920, no new apportionment was done because Congress couldn’t agree on the method to be used. They appointed a committee of mathematicians to investigate, and they recommended the Huntington-Hill Method. They continued to use Webster’s method in 1931, but after a second report recommending Huntington-Hill, it was adopted in 1941 and is the current method of apportionment used in Congress.
The Huntington-Hill Method is similar to Webster’s method, but attempts to minimize the percent differences of how many people each representative will represent.
Huntington-Hill Method
- Determine how many people each representative should represent. Do this by dividing the total population of all the states by the total number of representatives. This answer is called the standard divisor.
- Divide each state’s population by the divisor to determine how many representatives it should have. Record this answer to several decimal places. This answer is called the quota.
- Cut off the decimal part of the quota to obtain the lower quota, which we’ll call \(n\). Compute \(\sqrt{n(n+1)}\), which is the geometric mean of the lower quota and one value higher.
- If the quota is larger than the geometric mean, round up the quota; if the quota is smaller than the geometric mean, round down the quota. Add up the resulting whole numbers to get the initial allocation.
- If the total from Step 4 was less than the total number of representatives, reduce the divisor and recalculate the quota and allocation. If the total from step 4 was larger than the total number of representatives, increase the divisor and recalculate the quota and allocation. Continue doing this until the total in Step 4 is equal to the total number of representatives. The divisor we end up using is called the modified divisor or adjusted divisor.
Example \(\PageIndex{7}\)
Again, Delaware (Example \(\PageIndex{1}\)), with an initial divisor of 21,900.82927:
Solution
| County | Population | Quota | Lower Quota | Geom Mean | Initial |
|---|---|---|---|---|---|
| Kent | 162,310 | 7.4111 | 7 | 7.48 | 7 |
| New Castle | 538,479 | 24.5872 | 24 | 24.49 | 25 (Rounded Up) |
| Sussex | 197,145 | 9.0017 | 9 | 9.49 | 9 |
| Total | 897,934 | 41 |
This gives the required total, so we’re done.
Example \(\PageIndex{8}\)
Again, Rhode Island (Example \(\PageIndex{2}\)), with an initial divisor of 14,034.22667:
Solution
| County | Population | Quota | Lower Quota | Geom Mean | Initial |
|---|---|---|---|---|---|
| Bristol | 49,875 | 3.5538 | 3 | 3.46 | 4 |
| Kent | 166,158 | 11.8395 | 11 | 11.49 | 12 |
| Newport | 82,888 | 5.9061 | 5 | 5.48 | 6 |
| Providence | 626,667 | 44.6528 | 44 | 44.50 | 45 |
| Washington | 126,979 | 9.0478 | 9 | 9.49 | 9 |
| Total | 1,052,567 | 76 |
This is too many, so we need to increase the divisor. Let’s try 14,100:
| County | Population | Quota | Lower Quota | Geom Mean | Initial |
|---|---|---|---|---|---|
| Bristol | 49,875 | 3.5538 | 3 | 3.46 | 4 |
| Kent | 166,158 | 11.8395 | 11 | 11.49 | 12 |
| Newport | 82,888 | 5.9061 | 5 | 5.48 | 6 |
| Providence | 626,667 | 44.6528 | 44 | 44.50 | 44 |
| Washington | 126,979 | 9.0478 | 9 | 9.49 | 9 |
| Total | 1,052,567 | 75 |
This works, so we’re done.
In both these cases, the apportionment produced by the Huntington-Hill method was the same as those from Webster’s method.
Example \(\PageIndex{9}\)
Consider a small country with 5 states, two of which are much larger than the others. We need to apportion 70 representatives. We will apportion using both Webster’s method and the Huntington-Hill method.
| State | Population |
|---|---|
| A | 300,500 |
| B | 200,000 |
| C | 50,000 |
| D | 38,000 |
| E | 21,500 |
Solution
1. The total population is 610,000. Dividing this by the 70 representatives gives the divisor: 8714.286
2. Dividing each state’s population by the divisor gives the quotas
| State | Population | Quota |
|---|---|---|
| A | 300,500 | 34.48361 |
| B | 200,000 | 22.95082 |
| C | 50,000 | 5.737705 |
| D | 38,000 | 4.360656 |
| E | 21,500 | 2.467213 |
Webster’s Method
-
Using Webster’s method, we round each quota to the nearest whole number
| State | Population | Quota | Initial |
|---|---|---|---|
| A | 300,500 | 34.48361 | 34 |
| B | 200,000 | 22.95082 | 23 |
| C | 50,000 | 5.737705 | 6 |
| D | 38,000 | 4.360656 | 4 |
| E | 21,500 | 2.467213 | 2 |
- Adding these up, they only total 69 representatives, so we adjust the divisor down. Adjusting the divisor down to 8700 gives an updated allocation totaling 70 representatives
| State | Population | Quota | Initial |
|---|---|---|---|
| A | 300,500 | 34.54023 | 35 |
| B | 200,000 | 22.98851 | 23 |
| C | 50,000 | 5.74126 | 6 |
| D | 38,000 | 4.367816 | 4 |
| E | 21,500 | 2.471264 | 2 |
Huntington-Hill Method
- Using the Huntington-Hill method, we round down to find the lower quota, then calculate the geometric mean based on each lower quota. If the quota is less than the geometric mean, we round down; if the quota is more than the geometric mean, we round up.
| State | Population | Quota | Lower Quota | Geom Mean | Initial |
|---|---|---|---|---|---|
| A | 300,500 | 34.48361 | 34 | 34.49638 | 34 |
| B | 200,000 | 22.95082 | 22 | 22.49444 | 23 |
| C | 50,000 | 5.737705 | 5 | 5.477226 | 6 |
| D | 38,000 | 4.360656 | 4 | 4.472136 | 4 |
| E | 21,500 | 2.467213 | 2 | 2.44949 | 3 |
These allocations add up to 70, so we’re done.
Notice that this allocation is different from what was produced by Webster’s method. In this case, state E got the extra seat instead of state A.
The use of geometric mean actually favors smaller states, which was exactly what Huntington and Hill were trying to achieve.
Lowndes’ Method
William Lowndes (1782-1822) was a Congressman from South Carolina who proposed a method of apportionment that was more favorable to smaller states. Unlike the methods of Hamilton, Jefferson, and Webster, Lowndes’s method has never been used to apportion Congress.
Lowndes believed that an additional representative was much more valuable to a small state than to a large one. If a state already has 20 or 30 representatives, getting one more doesn’t matter very much. But if it only has 2 or 3, one more is a big deal, and he felt that the additional representatives should go where they could make the most difference.
Like Hamilton’s method, Lowndes’s method follows the quota rule. In fact, it arrives at the same quotas as Hamilton and the rest, and like Hamilton and Jefferson, it drops the decimal parts. But in deciding where the remaining representatives should go, we divide the decimal part of each state’s quota by the whole number part (so that the same decimal part with a smaller whole number is worth more because it matters more to that state).
Lowndes's Method
- Determine how many people each representative should represent. Do this by dividing the total population of all the states by the total number of representatives. This answer is called the divisor.
- Divide each state’s population by the divisor to determine how many representatives it should have. Record this answer to several decimal places. This answer is called the quota.
- Cut off all the decimal parts of all the quotas (but don’t forget what the decimals were). Add up the remaining whole numbers.
- Assuming that the total from Step 3 was less than the total number of representatives, divide the decimal part of each state’s quota by the whole number part. Assign the remaining representatives, one each, to the states whose ratio of decimal part to whole part were largest, until the desired total is reached.
Example \(\PageIndex{10}\)
We’ll do Delaware again. We begin in the same way as with Hamilton’s method:
| County | Population | Quota | Initial |
|---|---|---|---|
| Kent | 162,310 | 7.4111 | 7 |
| New Castle | 538,479 | 24.5872 | 24 |
| Sussex | 197,145 | 9.0017 | 9 |
| Total | 897,934 | 40 |
Solution
We need one more representative. To find out which county should get it, Lowndes says to divide each county’s decimal part by its whole number part, with the largest result getting the extra representative:
Kent: \(0.4111 / 7 \approx 0.0587 \)
New Castle: \(0.5872 / 24 \approx 0.0245\)
Sussex: \(0.0017 / 9 \approx 0.0002\)
The largest of these is Kent’s, so Kent gets the 41st representative:
| County | Population | Quota | Initial | Ratio | Final |
|---|---|---|---|---|---|
| Kent | 162,310 | 7.4111 | 7 | 0.0587 | 8 |
| New Castle | 538,479 | 24.5872 | 24 | 0.0245 | 24 |
| Sussex | 197,145 | 9.0017 | 9 | 0.0002 | 9 |
| Total | 897,934 | 40 | 41 |
Example \(\PageIndex{11}\)
Rhode Island, again beginning in the same way as Hamilton:
| County | Population | Quota | Initial |
|---|---|---|---|
| Bristol | 49,875 | 3.5538 | 3 |
| Kent | 166,158 | 11.8395 | 11 |
| Newport | 82,888 | 5.9061 | 5 |
| Providence | 626,667 | 44.6528 | 44 |
| Washington | 126,979 | 9.0478 | 9 |
| Total | 1,052,567 | 72 |
Solution
We divide each county’s quota’s decimal part by its whole number part to determine which three should get the remaining representatives:
Bristol: \(0.5538 / 3 \approx \quad 0.1846\)
Kent: \(0.8395 / 11 \approx \quad 0.0763\)
Newport: \(0.9061 / 5 \approx \quad 0.1812\)
Providence: \(0.6528 / 44 \approx \quad 0.0148\)
Washington: \(0.0478 / 9 \approx \quad 0.0053\)
The three largest of these are Bristol, Newport, and Kent, so they get the remaining three representatives:
| County | Population | Quota | Initial | Ratio | Final |
|---|---|---|---|---|---|
| Bristol | 49,875 | 3.5538 | 3 | 0.1846 | 4 |
| Kent | 166,158 | 11.8395 | 11 | 0.0763 | 12 |
| Newport | 82,888 | 5.9061 | 5 | 0.1812 | 6 |
| Providence | 626,667 | 44.6528 | 44 | 0.0148 | 44 |
| Washington | 126,979 | 9.0478 | 9 | 0.0053 | 9 |
| Total | 1,052,567 | 72 | 75 |
As you can see, there is no “right answer” when it comes to choosing a method for apportionment. Each method has its virtues and favors different sized states.
Reference
- References (19)
Contributors and Attributions
-
Saburo Matsumoto
CC-BY-4.0


21: Skin and Eye Infections
The human body is covered in skin , and like most coverings, skin is designed to protect what is underneath. One of its primary purposes is to prevent microbes in the surrounding environment from invading underlying tissues and organs. But in spite of its role as a protective covering, skin is not itself immune from infection. Certain pathogens and toxins can cause severe infections or reactions when they come in contact with the skin. Other pathogens are opportunistic, breaching the skin’s natural defenses through cuts, wounds, or a disruption of normal microbiota resulting in an infection in the surrounding skin and tissue. Still other pathogens enter the body via different routes—through the respiratory or digestive systems, for example—but cause reactions that manifest as skin rashes or lesions.
Nearly all humans experience skin infections to some degree. Many of these conditions are, as the name suggests, “skin deep,” with symptoms that are local and non-life-threatening. At some point, almost everyone must endure conditions like acne, athlete’s foot, and minor infections of cuts and abrasions, all of which result from infections of the skin. But not all skin infections are quite so innocuous. Some can become invasive, leading to systemic infection or spreading over large areas of skin, potentially becoming life-threatening.
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- 21.1: Anatomy and Normal Microbiota of the Skin and Eyes
- Human skin consists of two main layers, the epidermis and dermis, which are situated on top of the hypodermis, a layer of connective tissue. The skin is an effective physical barrier against microbial invasion. The skin’s relatively dry environment and normal microbiota discourage colonization by transient microbes. The skin’s normal microbiota varies from one region of the body to another. The conjunctiva of the eye is a frequent site for microbial infection; deeper infections are less common.
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- 21.2: Bacterial Infections of the Skin and Eyes
- Staphylococcus and Streptococcus cause many different types of skin infections, many of which occur when bacteria breach the skin barrier through a cut or wound. S. aureus are frequently associated with purulent skin infections that manifest as folliculitis, furuncles, or carbuncles. S. aureus is also a leading cause of staphylococcal scalded skin syndrome (SSSS). S. aureus is generally drug resistant and current MRSA strains are resistant to a wide range of antibiotics.
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- 21.3: Viral Infections of the Skin and Eyes
- Papillomas (warts) are caused by human papillomaviruses. Herpes simplex virus (especially HSV-1) mainly causes oral herpes, but lesions can appear on other areas of the skin and mucous membranes. Roseola and fifth disease are common viral illnesses that cause skin rashes; roseola is caused by HHV-6 and HHV-7 while fifth disease is caused by parvovirus 19. Viral conjunctivitis is often caused by adenoviruses and may be associated with the common cold. Herpes keratitis is caused by herpesviruses.
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- 21.4: Mycoses of the Skin and Eyes
- Mycoses can be cutaneous, subcutaneous, or systemic. Common cutaneous mycoses include tineas caused by dermatophytes of the genera Trichophyton, Epidermophyton, and Microsporum. Tinea corporis is called ringworm. Tineas on other parts of the body have names associated with the affected body part. Aspergillosis is a fungal disease caused by molds of the genus Aspergillus. Primary cutaneous aspergillosis enters through a break in the skin, such as the site of an injury or a surgical wound.
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- 21.5: Protozoan and Helminthic Infections of the Eyes
- The protozoan Acanthamoeba and the helminth Loa loa are two parasites that can breach the skin barrier, causing infections of the skin and eyes. Acanthamoeba keratitis is a parasitic infection of the eye that often results from improper disinfection of contact lenses or swimming while wearing contact lenses. Loiasis, or eye worm, is a disease endemic to Africa that is caused by parasitic worms that infect the subcutaneous tissue of the skin and eyes. It is transmitted by deerfly vectors.
Thumbnail: Multiple plantar warts have grown on this toe.