In Figure 1-6, the students have 1 of 15 different heights,plus there are two height classes (4′11″ and 5′ 0″) forwhich there are no observed students. That is a total of17 height classes. If a single Mendelian gene can account for only two classes of a trait (such as purple orwhite flowers), how many Mendelian genes would beminimally required to explain the observation of 17height classes? Continuous variation for heightFIGURE 1-6 Students atthe Connecticut AgricultureCollege in 1914 show arange of heights. RonaldFisher proposed thatcontinuously variable traitslike human height arecontrolled by multipleMendelian genes. [A. F.Blakeslee, "Corn and Men,"Journal of Heredity 5, 11,4:104:115:05:15:25:35:45:55:65:75:85:95:105:116:06:16:21914, 511-518.]

Name: In Figure 1-6, the students have 1 of 15 different heights,plus there
Uploaded: 2024-03-20T06:43:04.000Z
Channel: Bartleby
Description: In Figure 1-6, the students have 1 of 15 different heights,plus there are two height classes (4′11″ and 5′ 0″) forwhich there are no observed students. That is a total of17 height classes. If a single Mendelian gene can account for only two classes o

This is example of polygenic inheritance. in this there are large number of genes involved for…

Science

Biology

Continuous variation for height FIGURE 1-6 Students at the Connecticut Agriculture College in 1914 show a range of heights. Ronald Fisher proposed that continuously variable traits like human height are controlled by multiple Mendelian genes. [A. F. Blakeslee, "Corn and Men," Journal of Heredity 5, 11, 4:10 4:11 5:0 5:1 5:2 5:3 5:4 5:5 5:6 5:7 5:8 5:9 5:10 5:11 6:0 6:1 6:2 1914, 511-518.]

Continuous variation for height FIGURE 1-6 Students at the Connecticut Agriculture College in 1914 show a range of heights. Ronald Fisher proposed that continuously variable traits like human height are controlled by multiple Mendelian genes. [A. F. Blakeslee, "Corn and Men," Journal of Heredity 5, 11, 4:10 4:11 5:0 5:1 5:2 5:3 5:4 5:5 5:6 5:7 5:8 5:9 5:10 5:11 6:0 6:1 6:2 1914, 511-518.]

Human Heredity: Principles and Issues (MindTap Course List)

11th Edition

ISBN:9781305251052

Author:Michael Cummings

Publisher:Michael Cummings

Chapter19: Population Genetics And Human Evolution

Section: Chapter Questions

Problem 10QP

See similar textbooks

Similar questions

From a biological viewpoint, speculate as to why many traits seemto fit a normal distribution. Students with a strong background inmath and statistics may want to explain how a normal distributionis generated, and what it means. Can you think of biological examplesthat do not fit a normal distribution?
Alkaptonuria is a recessive autosomal genetic disorder associated with darkening of the urine. In the United States, approximately 1 out of every 250,000people has alkaptonuria.a. Assuming Hardy-Weinberg equilibrium, estimatethe frequency of the allele responsible for this trait.b. What proportion of people in the U.S. populationare carriers for this trait? In this population, whatis the ratio of carriers to individuals affected byalkaptonuria?c. If a woman without alkaptonuria had a child withthis trait with one husband then remarried, what isthe chance that a child produced by her secondmarriage would have alkaptonuria?d. Alkaptonuria is a relatively benign condition, sothere is little selective advantage to individualswith any genotype; as a result, your assumptionof Hardy-Weinberg equilibrium in part (a) is reasonable. Could you also use the assumption ofHardy-Weinberg equilibrium to estimate the allelefrequencies and carrier frequencies of more severerecessive autosomal conditions…
The allele for a window's peak (hairline) is dominant over the allele for the straight hairline. In a population of 500 individuals, 25% show the recessive phenotype. How many individuals would you expect to be homozygous dominant and heterozygous for the trait?
Small population size causes genetic drift because ofchance sampling of different alleles from one generation to the next. We can predict how much geneticdrift occurs for a given population size using binomialsampling statistics. With a population of size N, wecan estimate that 95% of the time the allele frequency(p) in the next generation will be withinthe confidence interval of p ± 1.96 (√p(1 − p)2N ),where p(1 − p)2Nis an estimate of the statistical variancein allele frequencies from one generation to the nextwith random sampling of 2N alleles each generation. a. What is the confidence interval for p = 0.5 whenN = 100,000?b. What is the confidence interval for p = 0.5 whenN = 10?c. How are the results in parts (a) and (b) related tothe consequences of a population bottleneck?
Two traits with similar phenotypic variance exist in apopulation. One trait has two major genes and sixminor loci that influence the phenotypic value, andthe second trait has 12 minor loci and no major genesaffecting the phenotypic value. Does this informationtell you which trait you should expect to respondmost consistently to selection? Explain why it doesor does not.
A genetics researcher determines that the broad-sense heritability of height among Southwestern University undergraduate students is 0.90. Which of the following conclusions would be reasonable? Explain your answer. a. Sally is a Southwestern University undergraduate student, so 10% of her height is determined by nongenetic factors. b. Ninety percent of variation in height among all undergraduate students in the United States is due to genetic differences. c. Ninety percent of the height of Southwestern University undergraduate students is determined by genes. d. Ten percent of the variation in height among Southwestern University undergraduate students is determined by variation in nongenetic factors. e. Because the heritability of height among Southwestern University students is so high, any change in the students’ environment will have minimal effect on their height.
6. You have identified an autosomal gene that contributesto tail size in male guppies, with a dominant allele B forlarge tails and a recessive allele b for small tails. Femaleguppies of all genotypes have similar tail sizes. Youknow that female guppies usually mate with males withthe largest tails, but the effects of population densityand the ratio of the sexes on this preference have notbeen studied. You therefore place an equal number ofmales in three tanks. In tank 1, the number of femalesis twice the number of males. In tank 2, the numbers ofmales and females are equal. In tank 3, half as manyfemales as males are present. After mating, you findthe following proportions of small-tailed males amongthe progeny: tank 1, 16%; tank 2, 25%; tank 3, 30%.a. In your original population (before the animals wereplaced in the three tanks), 25% of the males havesmall tails. Assuming that the allele frequencies inmales and females are the same, calculate the frequencies of B and b in your…
Let’s suppose that weight in a species of mammal is polygenic, andeach gene exists as a heavy and light allele. If the allele frequenciesin the population are equal for both types of alleles (i.e., 50%heavy alleles and 50% light alleles), what percentage of individualswill be homozygous for the light alleles in all of the genesaffecting this trait, if the trait was determined by the followingnumber of genes?A. TwoB. ThreeC. Four
a) Suppose the frequency of the recessive allele a for the recessive condition albinism is 0.2 Assume the Hardy-Weinberg distribution. ai) What is the frequency of albinism in the general population? aii) What fraction of the population are carriers? b) In a large population of mice, 0.58 have the recessive genetic condition rsiancer (fictional). Assume the Hardy-Weinberg distribution. What fraction of mice are carriers? Giver your answer with two digits parts the decimal point. c) In a large population of squirrels, 0.14 have the recessive genetic condition nlourianfilindr (fictional). Assume the Hardy-Weinberg distribution. What is the frequency of the dominant allele in this population? Give your answer with two digits parts the decimal point. d) In a large population of chimpanzees, there is a recessive genetic condition telactalase (fictional). Assume the Hardy-Weinberg distribution. If the frequency of dominant allele is 0.77 then what fraction of the population has…
. A chicken breeder is working with a population in whichthe mean number of eggs laid per hen in one month is 28and the variance is 5 eggs2. The narrow-sense heritabilityis known to be 0.8. Given this information, can the breeder expect that the population will respond to selectionfor an increase in the number of eggs per hen in the nextgeneration?a. No, applying selection is always risky and a breedernever knows what to expect.b. No, a breeder needs to know the broad-senseheritability to know what to expect.c. Yes, since the narrow-sense heritability is close to1 (0.8), then we would expect selective breeding couldlead to increased egg production in the next generation.d. Yes, since the variance is greater than 0.e. Both c and d are correct.
Sixteen percent (16%) of the population is homozygous recessive for a trait. What are the frequencies of the dominant and recessive alleles of that trait?
Cystic fibrosis is a homozygous recessive condition that affects 1 in 10,000 of the Hispanic populationin the United States. Calculate the frequency of the dominant allele, the frequency of the recessiveallele, and the percentage of heterozygous individuals (carriers) in the Hispanic population.Frequency of thedominant alleleFrequency of therecessive allele% homozygous dominant% homozygous recessive% heterozygous