1.1.5: Section 1 Summary
Summary
In this section, you have been introduced to the concept of information systems. We have reviewed several definitions, with a focus on the components of information systems: technology, people, data, and process. We have reviewed how the business use of information systems has evolved over the years, from the use of large mainframe computers for number crunching, through the introduction of the PC and networks, all the way to the era of mobile computing. During each of these phases, new innovations in software and technology allowed businesses to integrate technology more deeply.
As a Project Manager or System Analyst, you will have to have a great understanding of current technology trends and newer resources that will allow the organization you work for, or if you become a business owner, and stand a chance in the competitive world we live in. There are many resources freely available on the Internet and many resources that are available for purchase for you to understand how Information Technology has advanced and what skills you will need to have an understanding of.
Do not think that just because you already have experience building your own computer server, or have created some advanced software programs for your lab environments, you will succeed in a business environment. Businesses deal with many moving parts that require strong will and careful planning, As long as you are an individual who is open towards learning and is willing to take on the challenges that come with being a Project Manager or Systems Analyst, you will succeed.
Study Questions
- What are the four components that make up an information system?
- What are three examples of information system hardware?
- What is application software?
- What roles do people play in information systems?
- What is the definition of a process?
- In what year were restrictions on commercial use of the Internet first lifted? When were eBay and Amazon founded?
- What does it mean to say we are in a “post-PC world”?
- What is Carr’s main argument about information technology?
Exercises
- Suppose that you had to explain to a member of your family or one of your closest friends the concept of an information system. How would you define it? Write a one-paragraph description in your own words that you feel would best describe an information system to your friends or family.
- Of the five primary components of an information system (hardware, software, data, people, process), which do you think is the most important to the success of a business organization? Write a one-paragraph answer to this question that includes an example from your personal experience to support your answer.
- We all interact with various information systems every day: at the grocery store, at work, at school, even in our cars (at least some of us). Make a list of the different information systems you interact with every day. See if you can identify the technologies, people, and processes involved in making these systems work.
- Do you agree that we are in a post-PC stage in the evolution of information systems? Some people argue that we will always need the personal computer, but that it will not be the primary device used for manipulating information. Others think that a whole new era of mobile and biological computing is coming. Do some original research and make your prediction about what business computing will look like in the next generation.
- The Walmart case study introduced you to how that company used information systems to become the world’s leading retailer. Walmart has continued to innovate and is still looked to as a leader in the use of technology. Do some original research and write a one-page report detailing a new technology that Walmart has recently implemented or is pioneering.
- Wikipedia entry on "Information Systems ," as displayed on August 19, 2012. Wikipedia: The Free Encyclopedia . San Francisco: Wikimedia Foundation. http://en.Wikipedia.org/wiki/Informa...s_(discipline) . ↵
- Excerpted from Information Systems Today - Managing in the Digital World , fourth edition. Prentice-Hall, 2010. ↵
- Excerpted from Management Information Systems , twelfth edition, Prentice-Hall, 2012. ↵
- CERN's "The Birth of the Web." http://public.web.cern.ch/public/en/about/web-en.html ↵
- Walmart 2012 Annual Report. ↵


9.5E: Solving Trigonometric Equations (Exercises)
For the following exercises, find all exact solutions on the interval \([0,2 \pi)\).
39. \(\tan x+1=0\)
40. \(2 \sin (2 x)+\sqrt{2}=0\)
For the following exercises, find all exact solutions on the interval \([0,2 \pi)\).
41. \(2 \sin ^{2} x-\sin x=0\)
42. \(\cos ^{2} x-\cos x-1=0\)
43. \(2 \sin ^{2} x+5 \sin x+3=0\)
44. \(\cos x-5 \sin (2 x)=0\)
45. \(\frac{1}{\sec ^{2} x}+2+\sin ^{2} x+4 \cos ^{2} x=0\)
For the following exercises, simplify the equation algebraically as much as possible. Then use a calculator to find the solutions on the interval \([0,2 \pi)\). Round to four decimal places.
46. \(\sqrt{3} \cot ^{2} x+\cot x=1\)
47. \(\csc ^{2} x-3 \csc x-4=0\)
For the following exercises, graph each side of the equation to find the approximate solutions on the interval \([0,2 \pi)\).
48. \(20 \cos ^{2} x+21 \cos x+1=0\)
49. \(\sec ^{2} x-2\) sec \(x=15\)


3.4: The New South and the Problem of Race
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“There was a South of slavery and secession,” Atlanta Constitution editor Henry Grady proclaimed in an 1886 speech in New York. “That South is dead.” 12 Grady captured the sentiment of many white southern business and political leaders who imagined a New South that could turn its back to the past by embracing industrialization and diversified agriculture. He promoted the region’s economic possibilities and mutual future prosperity through an alliance of northern capital and southern labor. Grady and other New South boosters hoped to shape the region’s economy in the North’s image. They wanted industry and they wanted infrastructure. But the past could not be escaped. Economically and socially, the “New South” would still be much like the old.
A “New South” seemed an obvious need. The Confederacy’s failed insurrection wreaked havoc on the southern economy and crippled southern prestige. Property was destroyed. Lives were lost. Political power vanished. And four million enslaved Americans—representing the wealth and power of the antebellum white South—threw off their chains and walked proudly forward into freedom.
Emancipation unsettled the southern social order. When Reconstruction regimes attempted to grant freedpeople full citizenship rights, anxious whites struck back. From their fear, anger, and resentment they lashed out, not only in organized terrorist organizations such as the Ku Klux Klan but in political corruption, economic exploitation, and violent intimidation. White southerners took back control of state and local governments and used their reclaimed power to disenfranchise African Americans and pass “Jim Crow” laws segregating schools, transportation, employment, and various public and private facilities. The reestablishment of white supremacy after the “redemption” of the South from Reconstruction contradicted proclamations of a “New” South. Perhaps nothing harked so forcefully back to the barbaric southern past than the wave of lynchings—the extralegal murder of individuals by vigilantes—that washed across the South after Reconstruction. Whether for actual crimes or fabricated crimes or for no crimes at all, white mobs murdered roughly five thousand African Americans between the 1880s and the 1950s.
Lynching was not just murder, it was a ritual rich with symbolism. Victims were not simply hanged, they were mutilated, burned alive, and shot. Lynchings could become carnivals, public spectacles attended by thousands of eager spectators. Rail lines ran special cars to accommodate the rush of participants. Vendors sold goods and keepsakes. Perpetrators posed for photos and collected mementos. And it was increasingly common. One notorious example occurred in Georgia in 1899. Accused of killing his white employer and raping the man’s wife, Sam Hose was captured by a mob and taken to the town of Newnan. Word of the impending lynching quickly spread, and specially chartered passenger trains brought some four thousand visitors from Atlanta to witness the gruesome affair. Members of the mob tortured Hose for about an hour. They sliced off pieces of his body as he screamed in agony. Then they poured a can of kerosene over his body and burned him alive. 13
At the barbaric height of southern lynching, in the last years of the nineteenth century, southerners lynched two to three African Americans every week. In general, lynchings were most frequent in the Cotton Belt of the Lower South, where southern blacks were most numerous and where the majority worked as tenant farmers and field hands on the cotton farms of white landowners. The states of Mississippi and Georgia had the greatest number of recorded lynchings: from 1880 to 1930, Mississippi lynch mobs killed over five hundred African Americans; Georgia mobs murdered more than four hundred.
Throughout the late nineteenth and early twentieth centuries, a number of prominent southerners openly supported lynching, arguing that it was a necessary evil to punish black rapists and deter others. In the late 1890s, Georgia newspaper columnist and noted women’s rights activist Rebecca Latimer Felton—who would later become the first woman to serve in the U.S. Senate—endorsed such extrajudicial killings. She said, “If it takes lynching to protect women’s dearest possession from drunken, ravening beasts, then I say lynch a thousand a week.” 14 When opponents argued that lynching violated victims’ constitutional rights, South Carolina governor Coleman Blease angrily responded, “Whenever the Constitution comes between me and the virtue of the white women of South Carolina, I say to hell with the Constitution.” 15
Black activists and white allies worked to outlaw lynching. Ida B. Wells, an African American woman born in the last years of slavery and a pioneering anti-lynching advocate, lost three friends to a lynch mob in Memphis, Tennessee, in 1892. That year, Wells published Southern Horrors: Lynch Law in All Its Phases , a groundbreaking work that documented the South’s lynching culture and exposed the myth of the black rapist. 16 The Tuskegee Institute and the NAACP both compiled and publicized lists of every reported lynching in the United States. In 1918, Representative Leonidas Dyer of Missouri introduced federal anti-lynching legislation that would have made local counties where lynchings took place legally liable for such killings. Throughout the early 1920s, the Dyer Bill was the subject of heated political debate, but, fiercely opposed by southern congressmen and unable to win enough northern champions, the proposed bill was never enacted.
Lynching was only the violent worst of the southern racial world. Discrimination in employment and housing and the legal segregation of public and private life reflected the rise of a new Jim Crow South. So-called Jim Crow laws legalized what custom had long dictated. Southern states and municipalities enforced racial segregation in public places and in private lives. Separate coach laws were some of the first such laws to appear, beginning in Tennessee in the 1880s. Soon schools, stores, theaters, restaurants, bathrooms, and nearly every other part of public life were segregated. So too were social lives. The sin of racial mixing, critics said, had to be heavily guarded against. Marriage laws regulated against interracial couples, and white men, ever anxious of relationships between black men and white women, passed miscegenation laws and justified lynching as an appropriate extralegal tool to police the racial divide.
In politics, de facto limitations of black voting had suppressed black voters since Reconstruction. Whites stuffed ballot boxes and intimidated black voters with physical and economic threats. And then, from roughly 1890 to 1908, southern states implemented de jure, or legal, disfranchisement. They passed laws requiring voters to pass literacy tests (which could be judged arbitrarily) and pay poll taxes (which hit poor whites and poor blacks alike), effectively denying black men the franchise that was supposed to have been guaranteed by the Fifteenth Amendment. Those responsible for such laws posed as reformers and justified voting restrictions as for the public good, a way to clean up politics by purging corrupt African Americans from the voting rolls.
With white supremacy secured, prominent white southerners looked outward for support. New South boosters hoped to confront post-Reconstruction uncertainties by rebuilding the South’s economy and convincing the nation that the South could be more than an economically backward, race-obsessed backwater. And as they did, they began to retell the history of the recent past. A kind of civic religion known as the “Lost Cause” glorified the Confederacy and romanticized the Old South. White southerners looked forward while simultaneously harking back to an imagined past inhabited by contented and loyal slaves, benevolent and generous masters, chivalric and honorable men, and pure and faithful southern belles. Secession, they said, had little to do with the institution of slavery, and soldiers fought only for home and honor, not the continued ownership of human beings. The New South, then, would be built physically with new technologies, new investments, and new industries, but undergirded by political and social custom.
Henry Grady might have declared the Confederate South dead, but its memory pervaded the thoughts and actions of white southerners. Lost Cause champions overtook the South. Women’s groups, such as the United Daughters of the Confederacy, joined with Confederate veterans to preserve a pro-Confederate past. They built Confederate monuments and celebrated Confederate veterans on Memorial Day. Across the South, towns erected statues of General Robert E. Lee and other Confederate figures. By the turn of the twentieth century, the idealized Lost Cause past was entrenched not only in the South but across the country. In 1905, for instance, North Carolinian Thomas F. Dixon published a novel, The Clansman , which depicted the Ku Klux Klan as heroic defenders of the South against the corruption of African American and northern “carpetbag” misrule during Reconstruction. In 1915, acclaimed film director David W. Griffith adapted Dixon’s novel into the groundbreaking blockbuster film, Birth of a Nation . (The film almost singlehandedly rejuvenated the Ku Klux Klan.) The romanticized version of the antebellum South and the distorted version of Reconstruction dominated popular imagination. 17
While Lost Cause defenders mythologized their past, New South boosters struggled to wrench the South into the modern world. The railroads became their focus. The region had lagged behind the North in the railroad building boom of the midnineteenth century, and postwar expansion facilitated connections between the most rural segments of the population and the region’s rising urban areas. Boosters campaigned for the construction of new hard-surfaced roads as well, arguing that improved roads would further increase the flow of goods and people and entice northern businesses to relocate to the region. The rising popularity of the automobile after the turn of the century only increased pressure for the construction of reliable roads between cities, towns, county seats, and the vast farmlands of the South.
Along with new transportation networks, New South boosters continued to promote industrial growth. The region witnessed the rise of various manufacturing industries, predominantly textiles, tobacco, furniture, and steel. While agriculture—cotton in particular—remained the mainstay of the region’s economy, these new industries provided new wealth for owners, new investments for the region, and new opportunities for the exploding number of landless farmers to finally flee the land. Industries offered low-paying jobs but also opportunity for rural poor who could no longer sustain themselves through subsistence farming. Men, women, and children all moved into wage work. At the turn of the twentieth century, nearly one fourth of southern mill workers were children aged six to sixteen.
In most cases, as in most aspects of life in the New South, new factory jobs were racially segregated. Better-paying jobs were reserved for whites, while the most dangerous, labor-intensive, dirtiest, and lowest-paying positions were relegated to African Americans. African American women, shut out of most industries, found employment most often as domestic help for white families. As poor as white southern mill workers were, southern blacks were poorer. Some white mill workers could even afford to pay for domestic help in caring for young children, cleaning houses, doing laundry, and cooking meals. Mill villages that grew up alongside factories were whites-only, and African American families were pushed to the outer perimeter of the settlements.
That a “New South” emerged in the decades between Reconstruction and World War I is debatable. If measured by industrial output and railroad construction, the New South was a reality but if measured relative to the rest of the nation, it was a limited one. If measured in terms of racial discrimination, however, the New South looked much like the Old. Boosters such as Henry Grady said the South was done with racial questions but lynching and segregation and the institutionalization of Jim Crow exposed the South’s lingering racial obsessions. Meanwhile, most southerners still toiled in agriculture and still lived in poverty. Industrial development and expanding infrastructure, rather than re-creating the South, coexisted easily with white supremacy and an impoverished agricultural economy. The trains came, factories were built, and capital was invested, but the region remained mired in poverty and racial apartheid. Much of the “New South,” then, was anything but new.


3.1: Strunk and White Set the Standard
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- Cheryl E. Ball & Drew M. Loewe ed.
- West Virginia University via Digital Publishing Institute and West Virginia University Libraries
Author: Laura Lisabeth, St. John’s University
The Elements of Style , by William Strunk, Jr. and E. B. White, is well loved by many writers and teachers as the ultimate reference book on Standard English. Ask any writer or editor, and they are likely to have one of its four editions on their bookshelves or will remember it from some phase of their education. A recent survey of over a million college syllabi showed that The Elements of Style is the number one, most-assigned book in English-speaking countries and is also listed in the top ten assigned texts for almost every discipline. It is difficult to imagine that a book this popular with college professors and currently ranked first in sales on four different Amazon lists originated as a 43-page, self-published pamphlet a century ago.
Strunk first assigned The Elements of Style as a handbook of punctuation and usage in his advanced composition class at Cornell University in 1918. The popular writer E. B. White (who had been Strunk’s student in 1920), along with the Macmillan Publishing Company, turned Strunk’s pamphlet into a slightly longer version and published it in 1959. It included minor updates of Strunk’s original portion and a new chapter on style, introducing White’s signature essayistic voice into the book. Based largely on White’s reputation as a prolific writer of short pieces for Harper ’s and The New Yorker , The Elements of Style immediately became associated with a tradition of the best English language: grammatically correct, tasteful, clear, and organized prose.
But even in 1959, The Elements of Style was greeted with criticism by the field of college composition for being vague and misleading about the complex act of learning to compose academic writing. Its current appearance on so many course syllabi across the disciplines suggests a persistent dissatisfaction teachers have with the quality of student writing coupled with the misguided belief that one writing handbook can solve the problem. But more importantly, the kind of writing Strunk and White put forth as good writing is in fact a discourse that limits and excludes, not reflecting the valuable ways English is practiced in local and digital contexts and by a variety of writers from different language traditions. Insistence on the kind of English constructed by The Elements of Style is uninformed at best and, as I will show, unethical and racist at worst.
The genre to which The Elements of Style belongs—the writing handbook—is rooted in the mid-19thcentury handbook of conversation, a type of etiquette guide. Conversation handbooks became wildly popular when the prevailing cultural structure in pre-Civil War America was shifting power from an inherited social hierarchy to a larger self-made middle class. Guides to gentility were an aid to self-propelled social mobility and included elaborate rules for every kind of behavior imaginable, including a standard language usage that marked a speaker or writer as both apart from the socially inferior and fit for engaging with society’s elite people.
When Strunk and White’s style of English is prescribed as the dominant discourse taught in schools and used in other authorized places, it marginalizes the identities, knowledge, and being of many people who come from other literacy practices. Literacy is always attached to the deep ways of knowing embedded in language practices that are localized to different cultures. An individual’s cultural or racial identity is often closely linked to specific language practices that are not recognized in school where anything outside of Standard English is framed as error. This connection points to the fact that all literacies, Standard Academic English included, are not neutral but are ideological—always structuring social power as measured against themselves. Believing the Strunk and White style to be the best way to write suggests that Standard Academic English is a neutral transcription system for a universal reality and, therefore, a universal good. But, in fact, as literacy researchers argue, the English language cannot be understood apart from the many contexts in which it is embedded: home, school, workplaces, and social groups. It is not something fixed and unchanging, as the age and minimally revised text of The Elements of Style might suggest. Sociolinguists point to the ways English is already operating as a flexible medium, repurposed by American users to include, for example, Black and Latinx variations and the language and punctuation of social media, all of which expand the expressiveness of English and make it relevant to more users. These culturally and linguistically inflected ways of using English help people negotiate the identities and knowledges inherent in all the contexts in which people use language. Access to such uses of language can help many emerging academic writers to develop more competence and to perform better in school as they capitalize on existing meaningful ways of expressing knowledge.
Paradoxically, despite the public’s positive associations with prescriptive English as defined in The Elements of Style , the discourse has often been called into question by educational leaders who denounce its ineffectiveness in the teaching of writing as well as question the ethics of insisting on its use in a diverse and democratic society. As far back as 1932, the National Council of Teachers of English (NCTE) published a survey of American cultivated English usage showing that the well-educated group of adults who responded rarely abided by the strict usage rules of writing handbooks. At the time, NCTE president Ruth Weeks predicted that the survey’s findings revealing a democratization of language in the world outside school would prove that writing handbooks like The Elements of Style were useless and that American students were being unfairly judged by a classed and arbitrary set of standards for their writing. In 1974, the NCTE published a more monumental statement against the continued educational emphasis on Standard English usage. “Students’ Right to Their Own Language” holds that teachers of writing must respect the languages students bring with them to school, claiming “[L]anguage scholars long ago denied that the myth of a standard American dialect has any validity.”
One way to begin dismantling Strunk and White’s bad idea about writing is by understanding Standard Academic English as a historically formed, culturally specific language among many other languages. Reframe the notion of academic writing as a fixed, unchanging, and neutral discourse; think of it instead as a flexible toolkit of language practices that change with the user and the context. Many other literacies that could never conform to a Strunk and White standard continue to enrich ideas of what counts as good writing. With the unfolding of new media language practices, digital literacies are emerging to enable writers to compose discourse beyond anything Strunk and White could have imagined. These networked ways of writing, along with social-media inspired ways of thinking about punctuation, continue to explode definitions for what constitutes meaningful language and educated English.
Further Reading
For further reading on the ideologies of Standard English and the social situatedness of literacy, see Shirley Brice Heath’s Ways With Words (Cambridge University), Brian Street’s Literacy in Theory and Practice (Cambridge University), and James Paul Gee’s Social Linguistics and Literacies: Ideologies In Discourses (Routledge). Heath’s ethnography of two working class towns—one white and one African American—critiques what she calls the autonomous model of literacy , often prevalent in schools, which treats Standard English as a neutral discourse, free of ideology. Gee argues that to understand literacy we must consider the social context in which that literacy takes place.
There are several good books and articles that interrogate specific values of Standard English. Richard Ohmann’s “Use Definite Specific Concrete Language” ( College English ) and Ian Barnard’s Upsetting Composition Commonplaces (Utah State University) discuss the ideology implied in the concept of clarity, arguing that this value in writing forecloses the complex language necessary for critical inquiry of social conditions. Richard Lanham’s books Style: An Anti-Textbook (Paul Dry Books) and Revising Prose (Continuum) point to the clarity-brevity-sincerity approach to writing as limiting and problematic.
For further reading about the history of Standard English usage in education, see “Students Right To Their Own Language,” NCTE’s “Resolution On Language” (1974), and the NCTE publication Current English Usage by Sterling Andrus Leonard (1932), historical documents showing English educators’ awareness of the contingent nature of English language usage conventions.
Keywords
grammar, literacies, Standard English, style, writing handbook
Author Bio
Laura Lisabeth is an adjunct associate professor at St. John’s University. Her research is the cultural history of The Elements of Style . She enjoys exploring the intersections of college composition, publishing, and academic and public cultures that form the historical backdrop of each edition of this popular style guide. She is also interested in how new ways of thinking about academic writing might provide counternarratives to Strunk and White’s style and a more socially just language culture.


13.9.6E: Exercises
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Practice Makes Perfect
Solve Rational Equations
In the following exercises, solve.
Example \(\PageIndex{37}\)
\(\frac{1}{a}+\frac{2}{5}=\frac{1}{2}\)
- Answer
-
10
Example \(\PageIndex{38}\)
\(\frac{5}{6}+\frac{3}{b}=\frac{1}{3}\)
Example \(\PageIndex{39}\)
\(\frac{5}{2}−\frac{1}{c}=\frac{3}{4}\)
- Answer
-
\(\frac{4}{7}\)
Example \(\PageIndex{40}\)
\(\frac{6}{3}−\frac{2}{d}=\frac{4}{9}\)
Example \(\PageIndex{41}\)
\(\frac{4}{5}+\frac{1}{4}=\frac{2}{v}\)
- Answer
-
\(\frac{40}{21}\)
Example \(\PageIndex{42}\)
\(\frac{3}{7}+\frac{2}{3}=\frac{1}{w}\)
Example \(\PageIndex{43}\)
\(\frac{7}{9}+\frac{1}{x}=\frac{2}{3}\)
- Answer
-
−9
Example \(\PageIndex{44}\)
\(\frac{3}{8}+\frac{2}{y}=\frac{1}{4}\)
Example \(\PageIndex{45}\)
\(1−\frac{2}{m}=\frac{8}{m^2}\)
- Answer
-
−2, 4
Example \(\PageIndex{46}\)
\(1+\frac{4}{n}=\frac{21}{n^2}\)
Example \(\PageIndex{47}\)
\(1+\frac{9}{p}=−\frac{20}{p^2}\)
- Answer
-
−5, −4
Example \(\PageIndex{48}\)
\(1−\frac{7}{q}=−\frac{6}{q^2}\)
Example \(\PageIndex{49}\)
\(\frac{1}{r+3}=\frac{4}{2r}\)
- Answer
-
−6
Example \(\PageIndex{50}\)
\(\frac{3}{t−6}=\frac{1}{t}\)
Example \(\PageIndex{51}\)
\(\frac{5}{3v−2}=\frac{7}{4v}\)
- Answer
-
14
Example \(\PageIndex{52}\)
\(\frac{8}{2w+1}=\frac{3}{w}\)
Example \(\PageIndex{53}\)
\(\frac{3}{x+4}+\frac{7}{x−4}=\frac{8}{x^2−16}\)
- Answer
-
\(-\frac{4}{5}\)
Example \(\PageIndex{54}\)
\(\frac{5}{y−9}+\frac{1}{y+9}=\frac{18}{y^2−81}\)
Example \(\PageIndex{55}\)
\(\frac{8}{z−10}+\frac{7}{z+10}=\frac{5}{z^2−100}\)
- Answer
-
−13
Example \(\PageIndex{56}\)
\(\frac{9}{a+11}+\frac{6}{a−11}=\frac{7}{a^2−121}\)
Example \(\PageIndex{57}\)
\(\frac{1}{q+4}−\frac{7}{q−2}=1\)
- Answer
-
no solution
Example \(\PageIndex{58}\)
\(\frac{3}{r+10}−\frac{4}{r−4}=1\)
Example \(\PageIndex{59}\)
\(\frac{1}{t+7}−\frac{5}{t−5}=1\)
- Answer
-
−5, −1
Example \(\PageIndex{60}\)
\(\frac{2}{s+7}−\frac{3}{s−3}=1\)
Example \(\PageIndex{61}\)
\(\frac{v−10}{v^2−5v+4}=\frac{3}{v−1}−\frac{6}{v−4}\)
- Answer
-
no solution
Example \(\PageIndex{62}\)
\(\frac{w+8}{w^2−11w+28}=\frac{5}{w−7}+\frac{2}{w−4}\)
Example \(\PageIndex{63}\)
\(\frac{x−10}{x^2+8x+12}=\frac{3}{x+2}+\frac{4}{x+6}\)
- Answer
-
no solution
Example \(\PageIndex{64}\)
\(\frac{y−3}{y^2−4y−5}=\frac{1}{y+1}+\frac{8}{y−5}\)
Example \(\PageIndex{65}\)
\(\frac{z}{16}+\frac{z+2}{4z}=\frac{1}{2z}\)
- Answer
-
−4
Example \(\PageIndex{66}\)
\(\frac{a}{9}+\frac{a+3}{3a}=\frac{1}{a}\)
Example \(\PageIndex{67}\)
\(\frac{b+3}{3b}+\frac{b}{24}=\frac{1}{b}\)
- Answer
-
−8
Example \(\PageIndex{68}\)
\(\frac{c+3}{12c}+\frac{c}{36}=\frac{1}{4c}\)
Example \(\PageIndex{69}\)
\(\frac{d}{d+3}=\frac{18}{d^2−9}+4\)
- Answer
-
2
Example \(\PageIndex{70}\)
\(\frac{m}{m+5}=\frac{50}{m^2−25}+6\)
Example \(\PageIndex{71}\)
\(\frac{n}{n+2}=\frac{8}{n^2−4}+3\)
- Answer
-
1
Example \(\PageIndex{72}\)
\(\frac{p}{p+7}=\frac{98}{p^2−49}+8\)
Example \(\PageIndex{73}\)
\(\frac{q}{3q−9}−\frac{3}{4q+12}=\frac{7q^2+6q+63}{24q^2−216}\)
- Answer
-
no solution
Example \(\PageIndex{74}\)
\(\frac{r}{3r−15}−\frac{1}{4r+20}=\frac{3r^2+17r+40}{12r^2−300}\)
Example \(\PageIndex{75}\)
\(\frac{s}{2s+6}−\frac{2}{5s+5}=\frac{5s^2−3s−7}{10s^2+40s+30}\)
- Answer
-
no solution
Example \(\PageIndex{76}\)
\(\frac{t}{6t−12}−\frac{5}{2t+10}=\frac{t^2−23t+70}{12t^2+36t−120}\)
Solve a Rational Equation for a Specific Variable
In the following exercises, solve.
Example \(\PageIndex{77}\)
\(\frac{C}{r}=2π\) for r
- Answer
-
\(r=\frac{C}{2π}\)
Example \(\PageIndex{78}\)
\(\frac{I}{r}=P\) for r
Example \(\PageIndex{79}\)
\(\frac{V}{h}=lw\) for h
- Answer
-
\(h=\frac{v}{lw}\)
Example \(\PageIndex{80}\)
\(\frac{2A}{b}=h\) for b
Example \(\PageIndex{81}\)
\(\frac{v+3}{w−1}=\frac{1}{2}\) for w
- Answer
-
w=2v+7
Example \(\PageIndex{82}\)
\(\frac{x+5}{2−y}=\frac{4}{3}\) for y
Example \(\PageIndex{83}\)
\(a=\frac{b+3}{c−2}\) for c
- Answer
-
\(c=\frac{b+3+2a}{a}\)
Example \(\PageIndex{84}\)
\(m=\frac{n}{2−n}\) for n
Example \(\PageIndex{85}\)
\(\frac{1}{p}+\frac{2}{q}=4\) for p
- Answer
-
\(p=\frac{q}{4q−2}\)
Example \(\PageIndex{86}\)
\(\frac{3}{s}+\frac{1}{t}=2\) for s
Example \(\PageIndex{87}\)
\(\frac{2}{v}+\frac{1}{5}=\frac{1}{2}\) for w
- Answer
-
\(w=\frac{15v}{10+v}\)
Example \(\PageIndex{88}\)
\(\frac{6}{x}+\frac{2}{3}=\frac{1}{y}\) for y
Example \(\PageIndex{89}\)
\(\frac{m+3}{n−2}=\frac{4}{5}\) for n
- Answer
-
\(n=\frac{5m+23}{m}\)
Example \(\PageIndex{90}\)
\(\frac{E}{c}=m^2\) for c
Example \(\PageIndex{91}\)
\(\frac{3}{x}−\frac{5}{y}=\frac{1}{4}\) for y
- Answer
-
\(y=\frac{20x}{12−x}\)
Example \(\PageIndex{92}\)
\(\frac{R}{T}=W\) for T
Example \(\PageIndex{93}\)
\(r=\frac{s}{3−t}\) for t
- Answer
-
\(t=\frac{3r−s}{r}\)
Example \(\PageIndex{94}\)
\(c=\frac{2}{a}+\frac{b}{5}\) for a
Everyday Math
Example \(\PageIndex{95}\)
House Painting Alain can paint a house in 4 days. Spiro would take 7 days to paint the same house. Solve the equation \(\frac{1}{4}+\frac{1}{7}=\frac{1}{t}\) for t to find the number of days it would take them to paint the house if they worked together.
- Answer
-
\(2\frac{6}{11}\) days
Example \(\PageIndex{96}\)
Boating Ari can drive his boat 18 miles with the current in the same amount of time it takes to drive 10 miles against the current. If the speed of the boat is 7 knots, solve the equation \(\frac{18}{7+c}=\frac{10}{7−c}\) for c to find the speed of the current.
Writing Exercises
Example \(\PageIndex{97}\)
Why is there no solution to the equation \(\frac{3}{x−2}=\frac{5}{x−2}\)
- Answer
-
Answers will vary.
Example \(\PageIndex{98}\)
Pete thinks the equation \(\frac{y}{y+6}=\frac{72}{y^2−36}+4\) has two solutions, y=−6 and y=4. Explain why Pete is wrong.
Self Check
ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.
ⓑ After reviewing this checklist, what will you do to become confident for all objectives?


21.6.7: Plot an Ordered Pair
-
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Learning Outcomes
- Draw \(x\) and \(y\) axes.
- Plot a point in the xy-plane
We have already gone into detail about how to plot points on a number line, and that is very useful for single variable presentations. Now we will move to questions that involve comparing two variables. Working with two variables is frequently encountered in statistical studies and we would like to be able to display the results graphically. This is best done by plotting points in the xy-plane.
Example \(\PageIndex{1}\)
Plot the points: \((3,4)\),\((-2,1)\), and \((0,-1)\)
Solution
The first thing to do when plotting points is to sketch the x-axis and y-axis and decide on the tick marks. Here the numbers are all less than 5, so it is reasonable to count by 1's. Next, we plot the first point, \((3,4)\). This means to start at the origin, where the axes intersect. Then move 3 units to the right and 4 units up. After arriving there, we just draw a dot. For the next point, \((-2,1)\), we start at the origin, move 2 units to the left and 1 unit up and draw the dot. For the third point, \((0,-1)\), we don't move left or right at all since the x-coordinate is 0, but we do move 1 unit down and draw the dot. The plot is shown below.
Example \(\PageIndex{2}\)
A survey was done to look at the relationship between a person's age and their income. The first three answers are shown in the table below:
| Age | 49 | 24 | 35 |
| Income | 69,000 | 32,000 | 40,000 |
Graph the three points on the xy-plane.
Solution
Notice that the numbers are all relatively large. Therefore counting by 1's would not make sense. Instead, it makes better sense to count the Age axis, \(x\), by 10's and the Income axis, \(y\), by 1000's. The points are plotted below.
Exercise
A hotel manager was interested in seeing the relationship between the price per night, \(x\), that the hotel charged and the number of occupied rooms, \(y\). The results were (75,83), (100,60), (110,55), and (125,40). Plot these points in the xy-plane.


4.7: Resonance
Resonance occurs when the frequency of the inhomogeneous term matches the frequency of the homogeneous solution. To illustrate resonance in its simplest embodiment, we consider the second-order linear inhomogeneous ode \[\label{eq:1}\overset{..}{x}+\omega_0^2x=f\cos\omega t,\quad x(0)=x_0,\quad \overset{.}{x}(0)=u_0.\]
Our main goal is to determine what happens to the solution in the limit \(\omega\to\omega_0\).
The homogeneous equation has characteristic equation \[r^2+\omega_0^2=0,\nonumber\] so that \(r_{\pm}=\pm i\omega_0\). Therefore, \[\label{eq:2}x_h(t)=c_1\cos\omega_0t+c_2\sin\omega_0t.\]
To find a particular solution, we note the absence of a first-derivative term, and simply try \[x(t)=A\cos\omega t.\nonumber\]
Upon substitution into the ode, we obtain \[-\omega^2A+\omega_0^2A=f,\nonumber\] or \[A=\frac{f}{\omega_0^2-\omega^2}.\nonumber\]
Therefore, \[x_p(t)=\frac{f}{\omega_0^2-\omega^2}\cos\omega t.\nonumber\]
Our general solution is thus \[x(t)=c_1\cos\omega_0t+c_2\sin\omega_0t+\frac{f}{\omega_0^2-\omega^2}\cos\omega t,\nonumber\] with derivative \[\overset{.}{x}(t)=\omega_0(c_2\cos\omega_0t-c_1\sin\omega_0t)-\frac{f\omega}{\omega_0^2-\omega^2}\sin\omega t.\nonumber\]
Initial conditions are satisfied when \[\begin{aligned}x_0&=c_1+\frac{f}{\omega_0^2-\omega^2}, \\ u_0&=c_2\omega_0,\end{aligned}\] so that \[c_1=x_0-\frac{f}{\omega_0^2-\omega^2},\quad c_2=\frac{u_0}{\omega_0}.\nonumber\]
Therefore, the solution to the ode that satisfies the initial conditions is \[\begin{aligned}x(t)&=\left(x_0-\frac{f}{\omega_0^2-\omega^2}\right)\cos\omega_0t+\frac{u_0}{\omega_0}\sin\omega_0t+\frac{f}{\omega_0^2-\omega^2}\cos\omega t \\ &=x_0\cos\omega_0t+\frac{u_0}{\omega_0}\sin\omega_0t+\frac{f(\cos\omega t-\cos\omega_0t)}{\omega_0^2-\omega^2},\end{aligned}\] where we have grouped together terms proportional to the forcing amplitude \(f\).
Resonance occurs in the limit \(\omega\to\omega_0\); that is, the frequency of the inhomogeneous term (the external force) matches the frequency of the homogeneous solution (the free oscillation). By L’Hospital’s rule, the limit of the term proportional to \(f\) is found by differentiating with respect to \(\omega\):
\[\begin{align}\underset{\omega\to\omega_0}{\lim}\frac{f(\cos\omega t-\cos\omega_0t)}{\omega_0^2-\omega^2}&=\underset{\omega\to\omega_0}{\lim}\frac{-ft\sin\omega t}{-2\omega} \nonumber \\ &=\frac{ft\sin\omega_0t}{2\omega_0}\label{eq:3}.\end{align}\]
At resonance, the term proportional to the amplitude \(f\) of the inhomogeneous term increases linearly with \(t\), resulting in larger-and-larger amplitudes of oscillation for \(x(t)\). In general, if the inhomogeneous term in the differential equation is a solution of the corresponding homogeneous differential equation, then the correct ansatz for the particular solution is a constant times the inhomogeneous term times \(t\).
To illustrate this same example further, we return to the original ode, now assumed to be exactly at resonance, \[\overset{..}{x}+\omega_0^2x=f\cos\omega_0t,\nonumber\] and find a particular solution directly. The particular solution is the real part of the particular solution of \[\overset{..}{z}+\omega_0^2z=fe^{i\omega_0t}.\nonumber\]
If we try \(z_p = Ce^{i\omega_0t}\), we obtain \(0 = f\), showing that the particular solution is not of this form. Because the inhomogeneous term is a solution of the homogeneous equation, we should take as our ansatz \[z_p=Ate^{i\omega_0t}.\nonumber\]
We have \[\overset{.}{z}_p=Ae^{i\omega_0 t}(1+i\omega_0t),\quad\overset{..}{z}_p=Ae^{i\omega_0 t}\left(2i\omega_0-\omega_0^2t\right);\nonumber\] and upon substitution into the ode \[\begin{aligned}\overset{..}{z}_p+\omega_0^2z_p&=Ae^{i\omega_0t}\left(2i\omega_0-\omega_0^2t\right)+\omega_0^2Ate^{i\omega_0t} \\ &=2i\omega_0Ae^{i\omega_0 t} \\ &=fe^{i\omega_0t}.\end{aligned}\]
Therefore, \[A=\frac{f}{2i\omega_0},\nonumber\] and \[\begin{aligned}x_p&=\text{Re}\left\{\frac{ft}{2i\omega_0}e^{i\omega_0t}\right\} \\ &=\frac{ft\sin\omega_0t}{2\omega_0},\end{aligned}\] the same result as \(\eqref{eq:3}\).
Find a particular solution of \(\overset{..}{x}-3\overset{.}{x}-4x=5e^{-t}\).
Solution
If we naively try the ansatz \[x=Ae^{-t},\nonumber\] and substitute this into the inhomogeneous differential equation, we obtain \[A+3A-4A=5,\nonumber\] or \(0 = 5\), which is clearly nonsense. Our ansatz therefore fails to find a solution. The cause of this failure is that the corresponding homogeneous equation has solution \[x_h=c_1e^{4t}+c_2e^{-t},\nonumber\] so that the inhomogeneous term \(5e^{−t}\) is one of the solutions of the homogeneous equation. To find a particular solution, we should therefore take as our ansatz \[x=Ate^{-t},\nonumber\] with first- and second-derivatives given by \[\overset{.}{x}=Ae^{-t}(1-t),\quad\overset{..}{x}=Ae^{-t}(-2+t).\nonumber\]
Substitution into the differential equation yields \[Ae^{-t}(-2+t)-3Ae^{-t}(1-t)-4Ate^{-t}=5e^{-t}.\nonumber\]
The terms containing \(t\) cancel out of this equation, resulting in \(−5A = 5\), or \(A = −1\). Therefore, the particular solution is \[x_p=-te^{-t}.\nonumber\]


12.E: Thermodynamics - Homework
Spontaneity
-
Indicate whether the following processes are spontaneous or nonspontaneous.
- Liquid water freezing at a temperature below its freezing point
- Liquid water freezing at a temperature above its freezing point
- The combustion of gasoline
- A ball thrown into the air
- A raindrop falling to the ground
- Iron rusting in a moist atmosphere
- Answer
-
- spontaneous
- nonspontaneous
- spontaneous
- nonspontaneous
- spontaneous
- spontaneous
-
Many plastic materials are organic polymers that contain carbon and hydrogen. The oxidation of these plastics in air to form carbon dioxide and water is a spontaneous process; however, plastic materials tend to persist in the environment. Explain.
- Answer
-
Although the oxidation of plastics is spontaneous, the rate of oxidation is very slow. Plastics are therefore kinetically stable and do not decompose appreciably even over relatively long periods of time.
Entropy
-
Determine the entropy change for the combustion of gaseous propane, C
3
H
8
, under the standard conditions to give gaseous carbon dioxide and water.
- Answer
- 100.6 J/K
-
By calculating Δ
S
univ
at each temperature, determine if the melting of 1 mole of NaCl(
s
) is spontaneous at 500 °C and at 700 °C.
- Answer
- As Δ S univ < 0 at each of these temperatures, melting is not spontaneous at either of them. The given values for entropy and enthalpy are for NaCl at 298 K.
-
What assumptions are made about the thermodynamic information (entropy and enthalpy values) used to solve this problem?
- Answer
- It is assumed that these do not change significantly at the higher temperatures used in the problem.
Free Energy
-
A reaction has
= 100 kJ/mol and
Is the reaction spontaneous at room temperature? If not, under what temperature conditions will it become spontaneous?
- Answer
-
The reaction is nonspontaneous at room temperature.
Above 400 K, Δ G will become negative, and the reaction will become spontaneous.
-
Consider a system similar to the one in
Figure 16.8
, except that it contains six particles instead of four. What is the probability of having all the particles in only one of the two boxes in the case? Compare this with the similar probability for the system of four particles that we have derived to be equal to
What does this comparison tell us about even larger systems?
- Answer
- The probability for all the particles to be on one side is 1/32. This probability is noticeably lower than the 1/8 result for the four-particle system. The conclusion we can make is that the probability for all the particles to stay in only one part of the system will decrease rapidly as the number of particles increases, and, for instance, the probability for all molecules of gas to gather in only one side of a room at room temperature and pressure is negligible since the number of gas molecules in the room is very large.
-
Consider the system shown in
Figure 16.9
. What is the change in entropy for the process where the energy is initially associated with particles A and B, and the energy is distributed between two particles in different boxes (one in A-B, the other in C-D)?
- Answer
-
There is only one initial state. For the final state, the energy can be contained in pairs A-C, A-D, B-C, or B-D. Thus, there are four final possible states. \( \Delta \) S = 1.91 x 10 -23 J/K
-
At room temperature, the entropy of the halogens increases from I
2
to Br
2
to Cl
2
. Explain.
- Answer
- The masses of these molecules would suggest the opposite trend in their entropies. The observed trend is a result of the more significant variation of entropy with a physical state. At room temperature, I 2 is a solid, Br 2 is a liquid, and Cl 2 is a gas.
-
Indicate which substance in the given pairs has the higher entropy value. Explain your choices.
- C 2 H 5 OH( l ) or C 3 H 7 OH( l )
- C 2 H 5 OH( l ) or C 2 H 5 OH( g )
- 2H( g ) or H( g )
- Answer
-
- C 3 H 7 OH( l ) as it is a larger molecule (more complex and more massive), and so more microstates describing its motions are available at any given temperature.
- C 2 H 5 OH( g ) as it is in the gaseous state.
- 2H( g ), since entropy is an extensive property, and so two H atoms (or two moles of H atoms) possess twice as much entropy as one atom (or one mole of atoms).
-
Predict the sign of the entropy change for the following processes. Give a reason for your prediction.
- Na + ( aq ) + Cl - ( aq ) \( \rightarrow \) NaCl ( s )
- Answer
-
- Negative. The relatively ordered solid precipitating decreases the number of mobile ions in solution.
- Positive. There is a net increase of seven moles of gas from reactants to products.
-
Consider the decomposition of red mercury(II) oxide under standard state conditions.
- Is the decomposition spontaneous under standard state conditions?
- Above what temperature does the reaction become spontaneous?
- Answer
-
- The reaction is nonspontaneous;
- Above 566 °C the process is spontaneous
-
Calculate Δ
G
° for each of the following reactions from the equilibrium constant at the temperature given.
- Answer
-
- The reaction is nonspontaneous;
- Above 566 °C the process is spontaneous.
-
Calculate the equilibrium constant at 25 °C for each of the following reactions from the value of Δ
G
° given.
- Answer
-
- K = 41;
- K = 0.053
-
The evaporation of one mole of water at 298 K has a standard free energy change of 8.58 kJ.
- Is the evaporation of water under standard thermodynamic conditions spontaneous?
- Determine the equilibrium constant, K P , for this physical process.
- By calculating ∆ G , determine if the evaporation of water at 298 K is spontaneous when the partial pressure of water, is 0.011 atm.
- If the evaporation of water were always nonspontaneous at room temperature, wet laundry would never dry when placed outside. In order for laundry to dry, what must be the value of in the air?
- Answer
-
-
(a) Under standard thermodynamic conditions, the evaporation is nonspontaneous; (b)
- K p = 0.031
- The evaporation of water is spontaneous
- P H2O must always be less than K p or less than 0.031 atm. 0.031 atm represents air saturated with water vapor at 25 °C, or 100% humidity.
-
-
One of the important reactions in the biochemical pathway glycolysis is the reaction of glucose-6-phosphate (G6P) to form fructose-6-phosphate (F6P):
- Is the reaction spontaneous or nonspontaneous under standard thermodynamic conditions?
- Standard thermodynamic conditions imply the concentrations of G6P and F6P to be 1 M , however, in a typical cell, they are not even close to these values. Calculate Δ G when the concentrations of G6P and F6P are 120 μ M and 28 μ M respectively, and discuss the spontaneity of the forward reaction under these conditions. Assume the temperature is 37 °C.
- Answer
-
- Nonspontaneous as ΔG < 0
- ΔG = Δ G o + RT lnQ. ΔG=-2.1 kJ. The forward reaction to produce F6P is spontaneous under these conditions.
-
When ammonium chloride is added to water and stirred, it dissolves spontaneously and the resulting solution feels cold. Without doing any calculations, deduce the signs of Δ
G
, Δ
H
, and Δ
S
for this process, and justify your choices.
- Answer
-
Δ G is negative as the process is spontaneous. Δ H is positive as with the solution becoming cold, the dissolving must be endothermic. Δ S must be positive as this drives the process, and it is expected for the dissolution of any soluble ionic compound.
-
What happens to
(becomes more negative or more positive) for the following chemical reactions when the partial pressure of oxygen is increased?
- Answer
-
- Increasing the oxygen partial pressure will yield a decrease in Q and ΔG thus becomes more negative.
- Increasing the oxygen partial pressure will yield a decrease in Q and ΔG thus becomes more negative.
- Increasing the oxygen partial pressure will yield an increase in Q and ΔG thus becomes more positive


7: Phenotypic Variation and the Resemblance Between Relatives
The distinction between genotype and phenotype is one of the most useful ideas in biology.1 The genotype of an individual (the genome), for most purposes, is decided when the gametes fuse to form a zygote (individual). The phenotype of an individual represents any measurable aspect of an organism.
Your height, the amount of RNA transcribed from a given gene, what you ate last Tuesday: all of these are phenotypes. Nearly any phenotype we can choose to measure about an organism represents the outcome of the information encoded by their genome played out through an incredibly complicated developmental, physiological and/or behavioural processes that in turn interact with a myriad of environmental and stochastic factors. Honestly it boggles the mind how organisms work as well as they do, let alone that I managed to eat lunch last Tuesday.
There are many different ways to think about studying the path from genotype through to phenotype. The one we will take here is to think about how phenotypic variation among individuals in a population arises as a result of genetic variation in the population. One simple way to measure this genotype-phenotype relationship is to
calculate the phenotypic mean for each genotype at a locus. For example, Wang et al. (2018) explored the genetic basis of budset time in European aspen (Populus tremula); the effect of one specific SNP on that phenotype is shown in in Figure \(\PageIndex{2}\). Budset timing is a key trait underlying local adaptation to varying growing season length. The associated SNP falls in a gene (PtFT2) that is known to play a strong role in flowering time regulation in other plants.
One way for us to assess the relationship between genotype and phenotype is to find the best fitting linear line through the data, i.e. fit a linear regression of phenotypes for our individuals on their genotypes at a particular SNP (l):
\[X ∼ μ + alGl\]
In the equation above, X is a vector of the phenotypes of a set of individuals and Gl is our vector of genotypes at locus l, with Gi,l taking the value 0, 1, or 2 depending on whether our individual i is homozygote, heterozygote, or the alternate homozygote at our locus of interest. Here μ is our phenotypic mean. The slope of this regression line (al) has the interpretation of being the average effect of substituting a copy of allele 2 for a copy of allele 1. In our aspen example the slope is −13.6, i.e. swapping a single T for a G allele moves the budset forward by 13.6 days, such that the GG homozygote is predicted to set buds 27.2 days earlier than the T T homozygote.
As a measure of the significance of this genotype-phenotype re- lationship, we can calculate the p-value of our regression. To try to identify loci that are associated with our trait genome-wide, we can conduct this regression at each SNP we genotype in the genome. One common way to display the results of such an analysis (called
a genome-wide association study or GWAS for short) is to plot the minus logarithm of the p-value for each SNP along genome (a so-called Manhattan plot). Here’s one from Wang et al. (2018) for their aspen budset phenotype
The SNP with the most significant p-value is SNP in PtFT2. Note that other SNPs in the surrounding region also light up as showing a significant association with budset timing. This is because loci that are in linkage disequilibrium with a functional locus may in turn show an association, not because they directly affect the phenotype, but simply because the genotypes at the two loci are themselves non-randomly associated. Below is a zoomed in version (Figure 2 in Wang et al. (2018)) with SNPs coloured by the strength of their LD with the putatively functional SNP. Note how SNPs in strong LD with the functional allele (redder points) have more significant p-values.
Variation in some traits seems to have a relatively simple genetic basis. In our aspen example there is one clear large-effect locus, which explains 62% of the variation in budset. Note that even in this case, where we have an allele with a very strong effect on a phenotype, this is not an allele for budset, nor is PtFT2 a gene for budset. it is an allele that is associated with budset in the sampled environments and populations. In a different set of environments, this allele’s effects may be far smaller, and a different set of alleles may contribute to phenotype variation. PtFT2, the gene our focal SNP falls close to, is just one of many genes and molecular pathways involved in budset. A mutant screen for budset may uncover many genes with larger effects; this gene is just a locus that happens to be polymorphic in this particular set of genotyped individuals.
While phenotypic variation for some phenotypes has a relatively simple genetic basis, many phenotypes are likely much more genetically complex, involving the functional effect of many alleles at hundreds or thousands of polymorphic loci. For example hundreds of small effect loci affecting human height have been mapped in European populations to date. Such genetically complex traits are called polygenic traits.
In this chapter, we will use our understanding of the sharing of alleles between relatives to understand the phenotypic resemblance between relatives in quantitative phenotypes. This will allow us to understand the contribution of genetic variation to phenotypic variation. In the next chapter, we will then use these results to understand the evolutionary change in quantitative phenotypes in response to selection.
A simple additive model of a trait
Let’s imagine that the genetic component of the variation in our trait is controlled by L autosomal loci that act in an additive manner. The frequency of allele 1 at locus l is pl, with each copy of allele 1 at this locus increasing your trait value by al above the population mean. The phenotype of an individual, let’s call her i, is Xi. Her genotype at SNP l is Gi,l. Here Gi,l = 0, 1, or 2, representing the number of copies of allele 1 she has at this SNP. Her expected phenotype, given her genotype at all L SNPs, is then
\[E(Xi|Gi,1,··· ,Gi,L) = μ + XA,i = μ +Gi,lall\]
where μ is the mean phenotype in our population, and XA,i is the deviation away from the mean phenotype due to her genotype. Now in reality the phenotype is a function of the expression of those alleles in a particular environment. Therefore, we can think of this expected phenotype as being an average across a set of environments that occur in the population.
When we measure our individual’s observed phenotype we see
\[Xi =μ+XA,i +XE,i\]
where XE is the deviation from the mean phenotype due to the envi- ronment. This XE includes the systematic effects of the environment our individual finds herself in and all of the noise during development, growth, and the various random insults that life throws at our indi- vidual. If a reasonable number of loci contribute to variation in our trait then we can approximate the distribution of XA,i by a normal distribution due to the central limit theorem (see Figure 7.5). 2 Thus if we can approximate the distribution of the effect of environmental variation on our trait (XE,i) also by a normal distribution, which is reasonable as there are many small environmental effects, then the distribution of phenotypes within the population (Xi) will be normally distributed (see Figure \(\PageIndex{5}\)).
Note that as this is an additive model; we can decompose eqn. 7.3 into the effects of the two alleles at each locus and rewrite it as
\[\]
that our individual received from her mother (maternal alleles) and father (paternal alleles) respectively. This will come in handy in just a moment when we start thinking about the phenotypic covariance of relatives.
Now obviously this model seems silly at first sight as alleles don’t only act in an additive manner, as they interact with alleles at the same loci (dominance) and at different loci (epistasis). Later we’ll relax this assumption, however, we’ll find that if we are interested
in evolutionary change over short time-scales it is actually only the “additive component” of genetic variation that will (usually) concern us. We will define this more formally later on, but for the moment we can offer the intuition that parents only get to pass on a single allele at each locus on to the next generation. As such, it is the effect of these transmitted alleles, averaged over possible matings, that is an individual’s average contribution to the next generation (i.e. the additive effect of the alleles that their genotype consists of).
Additive genetic variance and heritability
As we are talking about an additive genetic model, we’ll talk about the additive genetic variance (VA), the phenotypic variance due to the additive effects of segregating genetic variation. This is a subset of the total genetic variance if we allow for non-additive effects.
The variance of our phenotype across individuals (VP ) we can write as
\[VP =Var(X)=Var(XA)+Var(XE)=VA +VE\]
In writing the phenotypic variance as a sum of the additive and envi- ronmental contributions, we are assuming that there is no covariance between XG,i and XE,i i.e. there is no covariance between genotype and environment.
Our additive genetic variance can be written as
\[VA = Var(XA) =V ar(Gi,lal)\]
where V ar(Gi,lal) is the contribution of locus l to the additive vari- ance among individuals. Assuming random mating, and that our loci are in linkage equilibrium, we can write our additive genetic variance as
\[\]
where the 2pl(1 − pl) term follows from the binomial sampling of two alleles per individual at each locus.
You have two biallelic SNPs contributing to variance in human height. At the first SNP you have an allele with an additive effect of 5cm which is found at a frequency of 1/10,000. At
the second SNP you have an allele with an additive effect of −0.5cm segregating at 50% frequency. Which SNP contributes more to the additive genetic variance? Explain the intuition of your answer.
Above in eqn(7.4) we decomposed the additive genetic component of XA,i as XM,i + XP,i the additive contributions of the maternal and paternal derived alleles in the ith individual. Similarly we can decompose the additive genetic variance VA as
\[VA = V ar(XA) = V ar(XM,i) + V ar(XP,i)\]
\[V ar(XM,i) = V ar(XM,i) = VA/2\]
assuming that our individuals are mating at random and that mater- nal and paternal alleles are equal in their effect in offspring.
An example of the additive basis of variation using polygenic scores.
Now we don’t usually get to see the individual loci contributing to highly polygenic traits. Instead, we only get to see the distribution
of the trait in the population. However, with the advent of GWAS in human genetics we can see some of the underlying genetics using the many trait-associated loci identified to date. Using the estimated ef- fect sizes at each locus, each one of which is tiny, we can calculate the weighted sum over an individual’s genotype as in equation 7.2. This weighted sum is called the individual’s polygenic score. To illustrate how polygenic scores work, we can take a set of 1700 SNPs6. The ef- fects of these SNPs are tiny; the median, absolute additive effect size is 0.07cm. Figure 7.6 shows the distribution of a thousand individuals’ polygenic scores calculated using these 1700 SNPs (simulated geno- types using the UKBB frequencies). The standard deviation of these polygenic scores ∼ 2cm. The individuals with higher polygenic scores for height are predicted to be taller than the individuals with lower polygenic scores.
The narrow sense heritability We would like a way to think about what proportion of the variation in our phenotype across individuals is due to genetic differences as opposed to environmental differences. Such a quantity will be key in helping us think about the evolution of phenotypes. For example, if variation in our phenotype had no genetic basis, then no matter how much selection changes the mean phenotype within a generation the trait will not change over generations.
We’ll call the proportion of the variance that is genetic the heritability, and denote it by h2. We can then write heritability as
\[\]
Remember that we are thinking about a trait where all of the alleles act in a perfectly additive manner. In this case our heritability h2 is referred to as the narrow sense heritability, the proportion of the variance explained by the additive effect of our loci. When we allow dominance and epistasis into our model, we’ll also have to define the broad sense heritability (the total proportion of the phenotypic vari- ance attributable to genetic variation).
The narrow sense heritability of a trait is a useful quantity; indeed we’ll see shortly that it is exactly what we need to understand the evolutionary response to selection on a quantitative phenotype. We can calculate the narrow sense heritability by using the resemblance between relatives. For example, if the phenotypic differences between individuals in our population were solely determined by environmental differences experienced by these different individuals, we should not expect relatives to resemble each other any more than random individuals drawn from the population. Now the obvious caveat here is that relatives also share an environment, so they may resemble each other due to shared environmental effects.
Note that the heritability is a property of a sample from the population in a particular set of environments at a particular time. Changes in the environment may change the phenotypic variance. Changes in the environment may also change how our genetic alleles are expressed through development and so change VA. Thus estimates of heritability are not transferable across environments or populations.
The covariance between relatives
People have long been fascinated by the resemblance between relatives, particularly twins (see Figure \(\PageIndex{7}\)). Families hold a special place in quantitative genetics, as remarkably we can use the resemblance between relatives to directly estimate the heritability and covariance of traits. To see this we can calculate the covariance in phenotype be- tween pairs of individuals (1 and 2) who have phenotypes X1 and X2 respectively.7 To think about imagine plotting the phenotypes of, say, sisters against each other. The x and y coordinates of each point will be the, say, heights of the pair of siblings. Do tall women tend to have tall sisters, do short women tend to have short sisters? How much do their phenotypes covary? If some of the variation in our phenotype is genetic we expect identical twins to resemble each other more than full siblings, who in turn will resemble each other more than half-sibs and so on out (see Figure \(\PageIndex{8}\)). Under our simple additive model of phenotypes we can write the covariance as
\[Cov(X1,X2) = Cov(X1M +X1P +X1E, X2M +X2P +X2E)\]
We can expand this out in terms of the covariance between the various components in these sums.
To make our task easier, we will make two commonly made assumptions:
- We can ignore the covariance of the environments between individuals (i.e. Cov(X1E , X2E ) = 0)
- We can ignore the covariance between the environment of one individual and the genetic variation in another individual (i.e. Cov(X1E , (X2M + X2P )) = 0).
The failure of these assumptions to hold can undermine our esti- mates of heritability, but we’ll return to that later. Moving forward with these assumptions, we can simplify our original expression above and write our phenotypic covariance between our pair of individuals as
\[Cov(X1, X2) = Cov(X1M , X2M )+Cov(X1M , X2P )+Cov(X1P , X2M )+Cov(X1P , X2P )\]
This equation is saying that, under our simple additive model, we can see the covariance in phenotypes between individuals as the covariance between the maternal and paternal allelic effects in our individuals. We can use our results about the sharing of alleles between relatives to obtain these covariance terms. But before we write down the general case, let’s quickly work through some examples.
The covariance between identical twins
Let’s first consider the case of a pair of identical twins, monzygotic (MZ) twins, from two unrelated parents. Our pair of twins share their maternal and paternal allele identical by descent (X1M = X2M and X1P = X2P ). As their maternal and paternal alleles are not correlated draws from the population, i.e. have no probability of being IBD as we’ve said the parents are unrelated, the covariance between their effects on the phenotype is zero (i.e. Cov(X1P , X2M ) = Cov(X1M , X2P ) = 0). In that case, eqn. 7.12 is
\[Cov(X1, X2) = Cov(X1M , X2M )+Cov(X1P , X2P ) = V ar(X1M )+V ar(X1P ) = VA\]
where the middle step follows from the fact the maternal (or sim- ilarly the paternal) allele in a pair of twins is the same allele so Cov(X1M , X2M ) = Cov(X1M , X1M ) = V ar(X1M ), as the covariance of random variable with itself is just its variance, and then the addi- tive variance of the maternal allele contribution is V ar(X1M ) = VA/2 following from eqn(7.9).
To calculate the narrow sense heritability we could then in principal divide the covariance of our pairs of MZ twins (MZ1 and MZ2) by the trait variance to give
\[\]
where ρMZ is the correlation of pairs of MZ twins (see Appendix eqn (A.43) for more on correlations). For example, we could estimate the heritability of a measure of body from the MZ correlation in Figure \(\PageIndex{9}\). In general, this simple estimator isn’t great as the correlation of identical twins includes the effects of the shared family environment of the twins (i.e. Cov(X1E,X2E)).
Moreover, it can be inflated by non-additive effects as identical twins don’t just share alleles, they share their entire genotypes, and thus resemble each other in phenotype also because of shared dominance effects (we’ll discuss non-additive effects in Section 7.1.1). Better twin-based estimates of heritability are commonly used based on the comparison of MZ vs twins that bypass some of these issues.
The covariance in phenotype between parent and child Children re- semble their biological parents because children inherit their genome from their parents (putting aside shared environments for the moment). If a mother and father are unrelated individuals, i.e. they are two random draws from the population, then this mother and her child share one allele IBD at each locus (i.e. r1 = 1 and r0 = r2 = 0). Let’s assume that our mother (ind 1) transmits her paternal allele to the child (ind 2), in which case XP1 = XM2, and so Cov(XP1,XM2) = V ar(XP 1) = 12 VA, and all the other covariances in eqn. 7.12 are zero. We’d also arrive at this result if instead we had thought of the mother transmitting her own maternal allele. Thus Cov(X1, X2) = 12 VA, we can leverage this form of the covariance to directly estimate h2 by regression.
We can estimate the narrow sense heritability through the regression of child’s phenotype on the parental mid-point phenotype. The parental mid-point phenotype is simply the average of the mum and dad’s phenotype. See Figure \(\PageIndex{10}\) for an example from song sparrows.
We denote the child’s phenotype by Xkid and mid-point phenotype by Xmid, so that if we take the regression Xkid ∼ Xmid this regres- sion has slope β = Cov(Xkid, Xmid)/V ar(Xmid). The covariance of Cov(Xkid, Xmid) = 21 VA, and V ar(Xmid) = 12 VP , as by taking the aver- age of the parents we have halved the variance, such that the slope of the regression is
\[\]
i.e. the regression of the child’s phenotype on the parental midpoint phenotype is an estimate of the narrow sense heritability. If much of the phenotypic variation is due to the (additive) differences in genotypes among individuals (h2 ≈ 1), then children will closely resemble their parents. Conversely if much of the variation is environmental (h2 ≈ 0), and there is no shared environment between parent and child, children will not resemble their parents.
Applying this heritability estimate to the Song sparrow sample we find h2 = 0.43 and h2 = 0.3 for beak depth and tarsus length respectively from Figure \(\PageIndex{10}\). So in Smith and Zach (1979) analysis, for example, 30% of the variance in tarsus length is atttributal to the additive effect of genetic differences among individuals. Smith and Zach (1979) also regressed the average offspring phenotype agains their fathers or mothers against their offspring, giving
a slope of βdad,avg.kid and βmum,kid. For tarsus length, for example, they found βdad,avg.kid = 0.19 and βmum,avg.kid = 0.17. Following a similar argument to that in eqn (7.15) we find that these slopes are βdad,kid = VA/2/VP = h2/2, and the same for mums. Thus the regression of offspring’s phenotype on a particular parent is an estimate of half the narrow-sense heritability, in line with the reduced slopes found by Smith and Zach (1979), this halfing of the slope is due to the fact that a single parent’s phenotype is a noisier estimate of the parental mid-point and so less informative about the child s phenotype. These parent specific estimates of heritability are particularly useful as they allow us to investigate sex-specific inheritance and sexual dimorphism (we’ll explore this in a later section).
Estimating heritability by these various parent-offspring regression have the issue of not controlling for environmental correlations between parent and offspring, which can inflate our estimates of hertability (as we will mistake environmentally mediated resemblance for genetics). Raising the organisms in the lab could remove much of the potential for shared environment between parent and offspring, but it also removes much of the environmental variation and we (as evolutionary geneticists) are usually not primarily interested in knowing the heritability in the lab bur rather in the field. In some organisms, notably plants, we can begin to sidestep these issues by raising offspring in a common set of randomized field conditions (a so called “common garden”). Another option is cross-foster animals, for example Smith and Dhondt (1980) returned to the song sparrow population and swapped eggs between parents nests. They found that the covariance between biological parents and children was still high despite these children being raised in a different nest, but that there was no significant covariance between foster parents and their non-biological children (see Figure \(\PageIndex{13}\) for beak depth). This suggests that family environment is not confounding the estimate of heritability in this song sparrow sample. However, such manipulations are often impossible in many systems, and issues of shared environmental covariance due to maternal resources from egg (or seed) are still present.
Despite its issues, this measure of heritability provides useful intuition and is directly relevant to our discussion of the response to selection in the next chapter. That’s because our regression allows us to attempt to predict the phenotype of the child given the phenotypes of the parents; how well we can do this depends on the slope. See Figure \(\PageIndex{12}\) for examples. If the slope is close to zero then the parental phenotypes hold little information about the phenotype of the child, while if the slope is close to one then the parental mid-point is a good predictor of the child’s phenotype. As we will see, natural selection will only efficiently drive evolution if children resemble their parents.
Thinking abour our prediction of child’s phenotpye more formally, the expected phenotype of the child given the parental phenotypes is
\[E(Xkid|Xmum, Xdad) = μ + βmid,kid(Xmid − μ) = μ + h2(Xmid − μ)\]
which follows from the definition of linear regression. So to find the child’s predicted phenotype, we simply take the mean phenotype and add on the difference between our parental mid-point and the population mean, multiplied by our narrow sense heritability.
The covariance between general pairs of relatives under an additive model The above examples above make clear that to understand
the covariance between phenotypes of relatives, we simply need to think about the alleles they share IBD. Consider a pair of relatives (1 and 2) with a probability r0, r1, and r2 of sharing zero, one, or two alleles IBD respectively. When they share zero alleles Cov((X1M + X1P ), (X2M + X2P )) = 0, when they share one allele Cov((X1M + X1P),(X2M +X2P)) = Var(X1M) = 12VA, and when they share two alleles Cov((X1M + X1P ), (X2M + X2P )) = VA. Therefore, the general covariance between two relatives is
\[\]
where F1,2 is our coefficient of kinship, i.e. the probability that two alleles sampled at random from our pair individuals 1 and 2 are IBD (see eqn (2.5)). So under a simple additive model of the genetic basis of a phenotype, to measure the narrow sense heritability we need to measure the covariance between pairs of relatives (assuming that we can remove the effect of shared environmental noise). From the covariance between relatives we can calculate VA, and we can then divide this by the total phenotypic variance to get h2.
A) In polygynous red-winged blackbird populations (i.e. males mate with several females), paternal half-sibs can be identified. Suppose that the covariance of tarsus lengths among half-sibs is 0.25 cm2 and that the total phenotypic variance is 4 cm2. Use these data to estimate h2 for tarsus length in this population.
B) Why might paternal half-sibs be preferable for measuring heritability than maternal half-sibs?
Estimating additive genetic variance across a variety of different relationships (The animal model). In many natural populations we may have access to individuals with a range of different relationships to each other (e.g. through monitoring of the paternity of individuals), but relatively few pairs of individuals for a specific relationship (e.g. sibs). We can try and use this information on various relatives as fully as possible in a mixed model framework. Building from equation 7.3, we can write an individual’s phenotype Xi as
\[Xi = μ + XA,i + XE,i\]
where XE,i ∼ N(0,VE) and XA,i is normally distributed across individuals with covariance matrix VAA, where the the entries for a pair of individuals i and j are Aij = 2Fi,j and Aii = 1. Given the matrix A we can estimate VA. We can also add fixed effects into this model to account for generation effects, additional mixed effects could also be included to account for shared environments between particular individuals (e.g. a shared nest). This approach is sometimes called the “animal model”, and is widely used to in modern quantitative gentics to estimate genetic variances and heritabilities.
Multiple traits
Traits often covary with each other, both due to environmentally in- duced effects (e.g. due to the effects of diet on multiple traits) and due to the expression of underlying genetic covariance between traits. Genetic covariance, in turn, can reflect pleiotropy, a mechanistic effect of an allele on multiple traits (e.g. variants that affect skin pigmenta- tion often affect hair color), the genetic linkage of loci independently affecting multiple traits, or the effects of assortative mating.
Consider two traits X1,i and X2,i in an individual i. These traits could be, say, the individual’s leg length and nose length. As before,
\[X1,i = μ1 + X1,A,i + X1,E,i
X2,i = μ2 + X2,A,i + X2,E,i\]
As before we can talk about the total phenotypic variance (V1,V2), environmental variance (V1,E and V2,E), and the additive genetic variance for trait one and two (VA,1, VA,2). But now we also have to consider the total covariance between trait one and trait two, V1,2 = Cov(X1,X2), as well as the environmentally induced covari- ance (VE,1,2 = Cov(X1,E,X2,E)) and the additive genetic covariance (VA,1,2 = Cov(X1,A,X2,A)). To better understand the covariance aris- ing due to pleiotropy, let’s think about a set of L SNPs contributing to our two traits. If the additive effect of an allele at the ith SNP is αi,1 and αi,2 on traits 1 and 2, then the additive covariance between our traits is
\[\]
assuming our loci are not in linkage disequilibrium. Thus a genetic correlation arises due to pleiotropy, because loci that tend to affect trait 1 also systematically affect trait 2. For example, alleles associated with later age at menarche (AAM), age at first menstrual cycle, in European women also tend to be positively associated with height (see Figure \(\PageIndex{15}\)), thereby creating a genetic correlation between AAM and height.
We can ‘store’ our variance and covariance values in matrices, a way of gathering these terms that will be useful when we discuss selection:
\[\]
\[\]
Here we’ve shown the matrices for two traits, but we can generalize this to an arbitrary number of traits.
We can estimate these quantities, in a similar way as before, by studying the covariance in different traits between relatives:
\[\]
An example of phenotype and genetic covariance are shown on the left and right of Figure \(\PageIndex{17}\) respectively. Gray treefrogs (Hyla versicolor) chorus to attract mates. Their call is made up of a trill, a note rapidly pulsed a number of times, that is then repeated after some period. Female frogs prefer males who make a lot of calls and where each of those calls have a large number of pulses. However, doing both is be very energetic, and so there is potentially a tradeoff between these two aspects of a male frog’s call. Indeed Welch et al. (2014) found in lab-reared male frogs that the pulse number and the time period between calls were positively correlated, left side of Figure \(\PageIndex{17}\), i.e. individuals were investing their energy in making either few highly pulsed calls or many calls with few pulses. This phenotypic covariance reflects underlying a genetic covariance between theses two frog call characteristics (right side Figure \(\PageIndex{17}\)). Fathers whose sons have calls with highly pulsed calls also have sons whose calls are more spaced apart.
One useful summary of a genetic covariance is the genetic correlation between two phenotypes
\[\]
where VA,1 and VA,2 are the additive genetic variance for trait 1 and 2 respectively. Here, rg tells us to what extent the additive genetic variance in two traits is correlated.
Another important application of genetic covariances is in the study of sexually antagonistic selection and the evolution of sexual dimorphism; here we’ll calculate the genetic covariance between male and female phenotypes. For example, below is the relationship between the forehead patch size for pied fly-catcher fathers and their sons and daughters. The phenotype has been standardized to have mean 0 and variance 1 in each group. The phenotypic covariance of the sample of fathers and sons is 0.35, while the phenotypic covariance of fathers and daughter is 0.23.
Assume we can ignore the effect of the shared environment in our pied fly-catcher example.
A)
What is the additive genetic covariance between male and female patch size?
B) What is the additive genetic correlation of male and female patch size? You can assume that the additive genetic variance is the same in males and females.
Non-additive variation.
Up to now we’ve assumed that our alleles contribute to our pheno- type in an additive fashion. However, that does not have to be the case as there may be non-additivity among the alleles present at a locus (dominance) or among alleles at different loci (epistasis). We can accommodate these complications into our models. We do this
by partitioning our total genetic variance into independent variance components. We’ll see that dominant and epistatically interacting loci can contribute to the additive genetic variance. In constructing these variance components models we’ll assume that we know the alleles contributing to variation in our trait and their effects, but in reality we rarely know these. However, as we’ll see we don’t need to know these details and we can partition our variance and estimate additive variance and other forms of non-additive variation using the resemblance between various types of relatives.
Dominance. To understand the effect of dominance, let’s consider how the allele that a parent transmits influences their offspring’s phenotype. A parent transmits one of their two alleles at a locus to their offspring. Assuming that individuals mate at random, this allele is paired with another allele drawn at random from the population. For example, assume your mother transmitted an allele 1 to you: with probability p it would be paired with another allele 1, and you would be a homozygote; and with probability q it’s paired with a 2 allele and you’re a heterozygote.
Now consider an autosomal biallelic locus l, with frequency p for allele 1, and genotypes 0, 1, and 2 corresponding to how many copies of allele 1 individuals carry. We’ll denote the mean phenotype of an individual with genotype 0, 1, and 2 as Xl,0, Xl,1, Xl,2 respectively. This mean is taking an average phenotype over all the environments and genetic backgrounds the alleles are present on. We’ll mean center (MC) these phenotypic values, setting X′l,0 = Xl,0 − μ, and likewise for the other genotypes.
We can think about the average (marginal) MC phenotype of an individual who received an allele 1 from their parent as the average of the MC phenotype for heterozgotes and 11 homozygotes, weighted by the probability that the individual has these genotypes, i.e. the probability they receive an additional allele 1 or an allele 2 from their other parent:
\[\]
Similarly, if your parent transmitted an 2 allele to you, your average MC phenotype would be
\[\]
Let’s now consider the average phenotype of an offspring by how many copies of the allele 1 they carry
i.e. the mean phenotype of each genotypes’ offspring averaged over all possible matings to other individuals in the population (assuming individuals mate at random). These are the additive MC genetic values (breeding values) of our genotypes. Here we are simply adding up the additive contributions of the alleles present in each genotype and ignoring any non-additive effects of genotype.
To illustrate this, in Figure \(\PageIndex{20}\) we plot two different cases of dominance relationships; in the top row an additive polymorphism and in the second row a fully dominant allele. The additive genetic values of the genotypes are shown as red dots. Note that the additive values of the genotypes line up with the observed MC phenotypic means in the top row, when our alleles interact in a completely additive manner. Our additive genetic values always fall along a linear line (the red line in our figure). The additive values are falling along the best fitting line of linear regression for our population, when phenotype is regressed against the additive genotype (0, 1, 2 copies of allele 1) across all individuals in our population. Note in the dominant case the additive genetic values differ from the observed phenotypic means, and are closer to the observed values for the genotypes that are most common in the population.
The difference in the additive effect of the two alleles al,2 − al,1 can be interpreted as an average effect of swapping an allele 1 for an allele 2; we’ll call this difference αl = al,2 − al,1. Our αl is also the slope of the regression of phenotype against genotype (the red line in Figure \(\PageIndex{20}\). Note that the slope of our regression of phenotype on genotype (αl) does not depend on the population allele frequency for our completely additive locus (top row of 7.20). In contrast, when there is dominance, the slope between genotype and phenotype (αl) is a function of allele frequency (bottom row of 7.20). When a dominant allele (1) is rare there is a strong slope of phenotype on genotype, bottom left Figure 7.20. This strong slope is because replacing a single copy of the 2 allele with a 1 allele in an individual has a big effect on average phenotype, as it will most likely move an individual from be- ing a 22 homozygote to being a 12 heterozygote. In contrast, when the dominant allele (1) is common in the population, replacing a 2 allele by a 1 allele in an individual on average has little phenotypic effect, leading to a weak slope (bottom right Figure \(\PageIndex{20}\)). This small effect is because as we are mainly turning heterozygotes into homozygotes (11), who have the same mean phenotype as each other.
As as an example of how dominance and population allele frequencies can change the additive effect of an allele, let’s consider the genetics of the age of sexual maturity in Atlantic Salmon. A single allele of large effect segregates in Atlantic Salmon that influences the sexual maturation rate in salmon (Ayllon et al., 2015; Barson et al., 2015), and hence the timing of their return from the sea to spawn (sea age). The allele falls close to the autosomal gene VGLL3 (Cousminer et al., 2013, variation at this gene in humans also influences the timing of puberty). The left side of Figure \(\PageIndex{22}\) shows the age at sexual maturity in males. The L allele associated with slower sexual maturity is recessive in males. While the LL homozygotes mature on average a whole year later, the additive effect of the allele is weak while the L allele is rare in the population. The right panel shows the effect of the L allele in females. Note how the allele is much more dominant in females, and has a much more pronounced additive effect. The dominance of an allele is not a fixed property of the allele but rather a statement of the relationship of genotype to phenotype, such that the dominance relationship between alleles may vary across phenotypes and contexts (e.g. sexes).
The variance in the population phenotype due to these additive breeding values at locus l, assuming HW proportions, is
\[\]
The total additive variance for the whole genotype can be found by summing the individual additive genetic variances over loci
\[\]
Having assigned the additive genetic variance to be the variance ex- plained by the additive contribution of the alleles at a locus, we define the dominance variance as the population variance among genotypes at a locus due to their deviation from additivity. We can calculate how much each genotypic mean deviates away from its additive prediction at locus l (the length of the arrows in Figure \(\PageIndex{20}\)). For example, the heterozygote deviates
\[\]
away from its additive genetic value, with similar expressions for each of the homozygotes (dl,0 and dl,2). We can then write the dominance variance at our locus as the genotype-frequency weighted sum of our squared dominance deviations
\[\]
Writing our total dominance variance as the sum across loci
\[\]
Having now partitioned all of the genetic variance into additive and dominant terms, we can write our total genetic variance as
\[\]
We can do this because by construction the covariance between our additive and dominant deviations for the genotypes is zero. We can define the narrow sense heritability as before h2 = VA/VP = VA/(VG + VE ), which is the proportion of phenotypic variance due to additive genetic variance. We can also define the total proportion of the phenotypic variance due to genetic differences among individuals, as the broad-sense heritability H2 = VG/(VG + VE).
The additive and dominance variance can be estimated by the re- semblance among relatives. When dominance is present in the loci influencing our trait (VD > 0), we need to modify our phenotype covariance among relatives to account for this non-additivity. Specifi- cally, our equation for the covariance among a general pair of relatives (eqn. 7.17 for additive variation) becomes
\[\]
where r2 is the probability that the pair of individuals share 2 alleles identical by descent, making the same assumptions (other than additivity) that we made in deriving eqn. 7.17. In table 7.1 we show the phenotypic covariance for some common pairs of relatives. Importantly, in the presence of dominance variance, the regression of offspring phenotype on parental midpoint still has a slope VA/VP , as a parent and offspring share precisely one of their autosomal alleles IBD but never their genotype IBD (assuming no inbreeding).
Full sibs and parent-offspring have the same covariance if there is no dominance variance (as they have the same kinship coefficient F1,2). However, when dominance effects are present (VD > 0), full-sibs resemble each other more than parent-offspring pairs. That’s because full-sibs can share both alleles (i.e. the full genotype at a locus) identical by descent. We can attempt to estimate VD by comparing different sets of relationships. For example, non-identical twins (full sibs born at same time) should have 1/2 the phenotypic covariance of identical twins if VD = 0. Therefore, we can attempt to estimate VD by look- ing at whether identical twins have more than twice the phenotypic covariance than non-identical twins.
The most important aspect of this discussion for thinking about evolutionary genetics is that the parent-offspring covariance is still only a function of VA. This is because our parent (e.g. the mother) transmits only a single allele, at each locus, to its offspring. The other allele the offspring receives is random (assuming random mating), as
it comes from the other unrelated parent (the father). Therefore, the average effect on the child’s phenotype of an allele the child receives from their mother is averaged over all possible random alleles the child could receive from their father (weighted by their frequency in the population). Thus we only care about the additive effect of the allele, as parents transmit only alleles (not genotypes) to their offspring. This means that the short-term response to selection, as described by the breeder’s equation, depends only on VA and the additive effect of alleles. Therefore, if we can estimate the narrow-sense heritability we can predict the short-term response.
While our VA predicts the short term response to selection, if alleles display dominance, our value of VA will change as alleles at our loci change in frequency. For, example as dominant alleles become common in the population their contribution to VA decreases, we can see this in Figure 7.20 our rare dominant allele (bottom left) contributes to the additive variance far more than when it is at high frequency (bottom right). So if selection favours higher values of our trait, the response to selection will push the dominant allele to higher frequency decreasing VA. Therefore, if there is dominance our value of VA will not be constant across generations.
Up to this point we have only considered dominance and not epistasis. However, we can include epistasis in a similar manner (for ex- ample among pairs of loci). This gets a little tricky to think about, so we will only briefly explain it. We can first estimate the additive effect of the alleles by considering the effect of the alleles averaging over their possible genetic backgrounds (including the other interact- ing alleles they are possibly paired with), just as before. We can then calculate the additive genetic variance from this. We can estimate the dominance variance, by calculating the residual variance among genotypes at a locus unexplained by the additive effect of the loci. We can then estimate the epistatic variance by estimating the residual vari- ance left unexplained among the two locus genotypes after accounting for the additive and dominant deviations calculated from each locus separately. In practice these high variance components are hard to estimate, and usually small as much of our variance is assigned to the additive effect. Again we would find that we mostly care about VA for predicting short-term evolution, but that the contribution of loci to the additive genetic variance will depend on the epistatic relationships among loci.
How could you use 1/2 sibs vs. full-sibs to estimate VD? Why might this be difficult in practice? Why are identical vs. non- identical twins better suited for this?
Summary
- A key concern of quantitative genetics is how phenotypic variation within populations is partitioned into environmental and genetic components of the variance.
- The additive genetic variance is The proportion of phenotypic variance is the narrow sense heritability h2 = VA/VP . These quantities are both measurements of the contribution of the current standing genetic variation in a particular set of environments and should not be thought of as fixed quantities of the population or trait.
- We can estimate the additive genetic variance and the heritability by using the resemblance of relatives, if we can experimentally remove or statistically partition out the effect if the shared environment among relatives.
- The genetic basis of variation in traits can genetically covary due to pleiotropy, assortative mating, and linkage. We can estimate the genetic covariance between traits by using the covariance in different traits among relatives.
- Alleles with dominance and epistatic effects can and do contribute to VA to the extent to which transmitting an additional copy of the allele to an offspring changes their expected phenotype. These alleles and combinations of alleles also contribute to higher order genetic variance components, the dominance and epistatic covariance.
- The magnitude of the additive, dominance, and epistatic genetic variance can change as allele frequency change and recombination changes the context in which alleles are expressed.
The additive genetic variance for leg length on mice is 10mm2.
What is the expected covariance of mice who are first cousins?
Can you construct a case where VA = 0 and VD > 0? You need just describe it qualitatively; you don’t need to work out the math. (tricker question).