Ch 17 TB_Final
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Ch 17 TB_Final

Course Number: BIO 269, 2012

College/University: Cleveland State

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Chapter 17 Population and Evolutionary Genetics True/False 1. All genetic drift arises from sampling errors and chance. (T) 2. A population may be in Hardy-Weinberg equilibrium for one locus but not for others. (T) 3. Inbreeding results in an increase in the frequency of heterozygotes compared to the results of random mating. (F) 4. Genetic variation must exist within a population before evolution can take...

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17 Population Chapter and Evolutionary Genetics True/False 1. All genetic drift arises from sampling errors and chance. (T) 2. A population may be in Hardy-Weinberg equilibrium for one locus but not for others. (T) 3. Inbreeding results in an increase in the frequency of heterozygotes compared to the results of random mating. (F) 4. Genetic variation must exist within a population before evolution can take place. (T) 5. Genetic differences in large, randomly breeding populations remain constant if migration occurs. (F) 6. Evolution cannot be directly observed. (F) 7. Evolution occurs in individuals, as well as in groups. (F) 8. Sympatric speciation is the most common source of the origin of species. (F) 9. For most protein-encoding genes, the synonymous rate of change is considerably higher than the nonsynonymous rate. (T) 10. Different parts of genes evolve at different rates. (T) Multiple Choice 11-12. Rapid changes in allelic frequencies by e take place in populations that are d . a. mutation b. evolving c. migrating *d. small *e. genetic drift 13-14. Migration tends to c genetic variation between subpopulations and a genetic variation within each subpopulation. *a. increase. b. not change. Chapter 17 *c. reduce. d. standardize. e. invert. 15. Which of the following evolutionary forces does NOT change allele frequencies? *a. nonrandom mating b. mutation c. selection d. drift e. migration 16. Which agent of evolution would tend to increase the frequency of all recessive phenotypes in a population relative to what is expected from HardyWeinberg equilibrium? a. Genetic drift *b. Inbreeding c. Natural selection d. Positive assortative mating e. Negative assortative mating 17. DNA typing is used to compare evidence DNA (E) left at a crime scene to two suspects (S1 and S2). Suspect 1 is excluded by the evidence, but suspect 2 remains included. What is the expected frequency of suspect 2s genotype if the allele frequencies in the population are f(A1) = 0.1, f(A2) = 0.2, and f(A3) = 0.7 and the population is at Hardy-Weinberg equilibrium? E S1 S1 S2 a. 0.01 *b. 0.04 A1 c. 0.14 A2 d. 0.28 e. 0.49 A 3 18. Five male and 5 female astronauts set off on a voyage that after several hundred years will establish a new colony of humans in another galaxy. What will likely be true of the colony that is established? a. The genetic structure of the colony will be about the same as that of the population of humans on Earth. b. The colony will contain the same amount of genetic diversity as is found among humans on Earth. c. The colony will contain more genetic diversity than is found among humans on Earth. *d. The colony will contain much less genetic diversity than is found among humans on Earth. e. None of the above 19. The gene pool in the population described in question 18 will reflect which agent of evolution? *a. Founder effect Population and Evolutionary Genetics b. Bottleneck effect c. Natural selection d. Gene flow e. Assortative mating 20. If tall humans choose tall mates and short humans choose short mates, what will be the most likely result? a. An increase in heterozygosity at all gene pairs b. An increase in homozygosity at all gene pairs c. An increase in heterozygosity at the gene pairs that control height *d. An increase in homozygosity at the gene pairs that control height e. An increase in the frequency of tall alleles in the population 21. In watchamakallits black fur is dominant to white. In a certain population of watchamakallits, white fur occurs in a frequency of 1/100. What is the frequency of the recessive allele in the population, assuming Hardy-Weinberg equilibrium? a. 0.01 *b. 0.1 c. 0.5 d. 0.9 e. 0.99 22. What is the frequency of heterozygous watchamakallits in the population described in question 21, assuming Hardy-Weinberg equilibrium? a. 0.01 b. 0.02 *c. 0.18 d. 0.82 e. 0.99 23-24. c speciation arises in the absence of any geographic barrier to gene flow, whereas b speciation is initiated by a geographic barrier to gene flow. (a) Autocratic. *(b) Allopatric. *(c) Sympatric. (d) Parapatric. (e) Peripatric. 25. ______ are sets of genes that are similar in sequence but encode different products. (a) Codons (b) Genomes (c) Exons (d) Pseudogenes *(e) Multigene families 26. Which region of a gene should have the highest rates of substitution? (a) first position of a codon Chapter 17 (b) second position of a codon *(c) third position of a codon (d) exons (e) untranslated regions 27. ______ characteristics evolved from the same character in a common ancestor. (a) Similar (b) Paralogous *(c) Homologous (d) Phylogenetic (e) Parsimonious 28. Any change in a population from Hardy-Weinberg equilibrium can be defined as ______. a. mutation b. genetic equilibrium c. migration *d. evolution e. natural selection Matching For each number, match the reproductive isolating mechanism best associated with the phenomenon described. 29. trying to mate a Saint Bernard with a Chihuahua (a) 30. a male fiddler crab waving its claw to attract a mate (a) 31. a sheep adapted to mountains mating with a sheep adapted to desert producing offspring maladapted to either habitat (b) 32. fireflies flashing their light-emitting organs (a) 33. the hybrid offspring of two species dies before birth (b) 34. the hybrid offspring of two species dies after birth (b) 35. two plant species flowering at different times (a) a. prezygotic b. postzygotic Short Answer Discussion 36. Because real-life populations are, of course, not infinitely large, why is the Hardy-Weinberg condition of an infinitely large population usually met for natural populations? Population and Evolutionary Genetics It is usually met because most natural populations are large enough that chance factors (i.e., random genetic drift) have negligible effects on allele frequencies. 37. What effect does mutation have on genetic variation? It increases or creates it. 38. What is the first step in the formation of two new species from a previously existing species? Give an example of how this might occur. genetic isolation of the subpopulations, usually by a vicariance event (some geographical barrier like a river or mountain range rising) 39. What two factors are thought to play a critical role in sympatric speciation? (strong) natural selection and assortative mating 40. How might gene duplication provide a mechanism for the addition of new genes with novel functions? After a gene duplicates, there are two copies of the sequence, one of which is free to change and potentially take on a new function. 41. How has molecular data helped resolve the evolutionary relationships among distantly related taxa? Trying to assess the evolutionary history of distantly related organisms is often difficult because the organisms have few characteristics in common. All organisms have certain molecular traits in common, such as ribosomal RNA sequences and some fundamental proteins. These molecules offer a valid basis for comparisons among all organisms. 42. The Hardy-Weinberg law (equation) is a mathematical model in which allele frequencies in populations remain constant from generation to generation. Given all the conditions that must be met for the Hardy-Weinberg equation to be valid, why is this equation useful for studying population genetics? By specifying the ideal conditions that must be met for allele frequencies to remain constant in populations, the H-W law can identify evolutionary forces that can cause changes in allele frequencies in the real world. Remember, the H-W law is a model and thus, by definition provides a simplistic view of reality, which serves as a starting point and basis for comparison for further examination and description of real populations. 43. Huntington disease is caused by a single dominant gene and results in progressive mental and neurological damage. The disease is usually symptomatic when a person is between 30 and 50 years old and the patient Chapter 17 usually dies within 15 years of diagnosis. Approximately 1 in 25,000 Caucasians have this disease. Huntington disease has not been associated with any other disease, now or in the past. Why might natural selection not have eliminated such a deleterious gene from the population? Natural selection acts through reproduction, and most individuals with Huntington disease reproduce before discovering they have the disease (although genetic testing may have an impact on this). 44. Evolutionary biologists usually define population size using the effective population size rather than the census number. What is the effective population size and what factors can affect it? Population size is usually defined as the number of individuals in a group. But, the evolution of a gene pool depends only on those individuals who contribute genes to the next generation. Population geneticists usually define population size as the equivalent number of breeding adults, the effective population size (Ne). Several factors determine the equivalent number of breeding adults, including sex ratio, variation between individuals in reproductive success, fluctuations in population size, age structure of the population, and whether mating is random. 45. Why is sympatric speciation thought to be so rare in nature? Specifically, which evolutionary forces act against sympatric speciation and what evolutionary forces might act to allow sympatric speciation? In sympatry, gene flow will overwhelm the forces of natural selection, drift, and assortative mating. If natural selection and assortative mating are strong enough, you could get sympatric speciation. Extended Answer Discussion 46. Distinguish between overdominance and underdominance and their consequences for selection. Underdominance occurs when the heterozygote has lower fitness than both homozygotes (W11 > W12 < W22). Underdominance leads to an unstable equilibrium; here, allelic frequencies will not change as long as they are at equilibrium but, if they are disturbed from the equilibrium point by some other evolutionary force, they will move away from equilibrium until one allele eventually becomes fixed. Overdominance occurs when the heterozygote has higher fitness than the fitnesses of the two homozygotes (W11 < W12 > W22). With overdominance, both alleles are favored in the heterozygote, and neither allele is eliminated from the population. The allelic frequencies change with Population and Evolutionary Genetics overdominant selection until a stable equilibrium is reached, at which point there is no further change. 47. Discuss the role of the availability of whole-genome sequences in understanding the evolutionary process. Whole-genome sequences are providing new information about how genomes evolve and about the processes that shape the size, complexity, and organization of genomes. Whole-genome sequences make apparent the duplication of individual genes, chromosomal segments, and even entire genomes. Genome sequences from different organisms also might provide evidence for horizontal transfer of genes. Problems and Calculations 48. A kind new of tulip is produced that develops only purple or pink flowers. Assume that flower color is controlled by a single-gene locus, and that the purple allele (C) is dominant to the pink allele (c). A random sample of 1000 tulips from a large cultivated field yields 847 purple flowers, and 153 pink flowers. a. Determine the frequency of the purple and pink alleles in this field population. b. Estimate the proportion of all purple flowering plants that are heterozygotes and homozygotes. a. Assuming the population is in H-W equilibrium, let p = the dominant purple allele (C), and q = the recessive allele (c). Applying the H-W equation, the frequency of the pink allele = q2 = 153/1000 = 0.153; and q = (0.153)1/2 = 0.39. Because p + q = 1, p = 1 q = 0.61. b. If the population is in H-W equilibrium, p2 + 2pq + q2 = 1. For the heterozygotes, 2pq = 2(0.61)(0.39) = 0.476. Therefore, 47.6%, or (0.476)(1000) = 476 of all the purple flowered plants in this population are expected to be heterozygotes (Cc), and consequently, (p2)(1000) = (0.61)2 (1000) = 372 plants are expected to be homozygous dominants (CC). 49. You are studying a very large population of crocodiles on the Nile River in Africa and have identified a newly arisen (by mutation) allele at the C locus. The initial allele frequency of the mutant allele c is 0.01. You have also determined that the allele acts additively. On a moonless night, you genotype nesting females and count the number of eggs they lay. You find that, on average, "CC" females produce 98 eggs, "Cc" produce 99 eggs, and "cc" produce 100 eggs. In this species of crocodile, all eggs hatch and survive to maturity. a. Will the "c" allele increase, decrease, or stay the same in the next generation? b. Will you likely be able to observe the change in the allele frequency? Why? c. In the long run, what will happen to the frequencies of the "C" and "c" alleles in this population? Chapter 17 a. The cc genotype has the highest fitness and its relative fitness indicates that c will increase in the next generation. b. Because the absolute differences in fitness are just one egg, the change in allele or genotype frequencies is likely to be very slow and, therefore, hard to observe. c. If the directional selection is consistent, even though the differences in fitness are very small, the C allele will be eventually (many generations) eliminated from the population and the c allele will be fixed (frequency of one). Of course, we assume that the allele is not lost from the population through random drift (which should not be common, given the stated large population size). 50. A woman was murdered and the female suspect, her half sister, was apprehended. Skin from underneath the victims fingernails was removed and DNA fingerprinted. Blood collected from the suspect and the victim was also fingerprinted and compared. Several different loci, with known allele frequencies (indicated below) in the womens ethnic group, were analyzed in the samples. Five loci (A-E) in the fingernail evidence and the suspects blood matched. Calculate the probability that the skin underneath the victims fingernails is from someone other than the suspect. Fingernail evidence Victims blood Suspects blood A B C D E Known allele frequencies for victims ethnic group: Locus A (heterozygous): Frequency of allele 1 = 0.65; frequency of allele 2 = 0.35 Locus B (homozygous): Allele frequency = 0.25 Locus C (homozygous): Allele frequency = 0.4 Locus D (heterozygous): Frequency of allele 1 = 0.15; frequency of allele 2 = 0.85 Locus E (homozygous): Allele frequency = 0.22 Population and Evolutionary Genetics For locus A: 2pq = 2(0.65)(0.35) = 0.455 For locus B: p2 = (0.25)2 = 0.0625 For locus C: p2 = (0.4)2 = 0.16 For locus D: 2pq = 2(0.15)(0.85) = 0.255 For locus E: p2 = (0.22)2 = 0.0484. Therefore, using the multiplicative rule, the overall probability that the skin underneath the suspects fingernails is from someone other than the victim = (0.455)(0.0625)(0.16)(0.255)(0.0484) = 0.0000561561 = approximately 1/18,000. Therefore, there is about one chance in 18,000 that the skin underneath the victims fingernail is not the suspects. 51. You are studying cannibals in Borneo and want to determine if a specific village population is in H-W equilibrium with respect to the single-gene locus segregating the two co-dominant M and N blood type alleles (LM, LN). The following are sample data from two generations of the cannibals: Generation 1 (G1) 2 (G2) Phenotypes M MN N 324 709 177 287 665 123 Total # genotypes 1210 1075 a. Calculate the genotype frequencies for both generations. b. Calculate frequencies for the LM and LN alleles for each generation. c. Determine whether the G2 of this population of Borneo cannibals is in H-W equilibrium. a. Genotype frequencies: Generation 1 Generation 2 LM LM = 324/1210 = 0.268 287/1075 = 0.267 665/1075 = 0.619 LM LN = 709/1210 = 0.586 LN LN = 177/1210 = 0.146 123/1075 = 0.114 Note the genotype frequencies for both generations are virtually identical for LM LM, but not as close for LM LN and LN LN. b. Allele frequencies: For the LM allele in G1: 2(324) + (709) = 1357. Total # of all alleles in G1 = 1,357 + 709 + 2(177) = 2420. Therefore, the frequency of the LM allele in G1 = 1357/2420 = 0.561. For the LN allele in G1: (709) + 2(177) = 1063. Therefore, the frequency of the LN allele in G1 = 1063/2420 = 0.439. For the LM allele in G2: 2(287) + (665) = 1239. Total # of all alleles in G2 = 1239 + 665 + 2(123) = 2150. Therefore, the frequency of the LM allele in G2 = 1239/2150 = 0.576. For the LN allele in G2: (665) +2(123) = 911. Therefore, the frequency of the LN allele in G2 = 911/2,150 = 0.424. Chapter 17 c. Let p = frequency of LM in G2 (0.576), and q = frequency of LN (0.424). At equilibrium, p2 + 2pq + q2 = 1. The observed number of LMLM individuals in G2 = 287; the expected number is p2 (1075) = (0.576)2 (1075) = 357. The observed number of LNLN individuals in G2 = 123; the expected number = q2 = (0.424)2(1075) = 193. The observed number of LMLN individuals in G2 = 665; the expected number = 2pq (1075) = 2(0.576)(0.424)(1075) = 525. Perform a chi-square test to test for H-W equilibrium: (287 357)2/(357) + (123 193)2/(193) + (665 525)2/(525) = 13.7 + 25.4 + 37.3 = 76.4. Degrees of freedom = 3 1 1 = 1. Chi-square critical value = 3.814, which is much less than 76.4. Therefore, this is a highly significant chisquare statistic, and the hypothesis must therefore be rejected. We conclude that the M-N allele frequencies are not in equilibrium in this population. 52. Like the ABO blood type antigens, the co-dominant M-N antigens are also present in human red blood cells. A sample of 5631 individuals in a population was examined for M-N antigens, and the results follow. a. Calculate the frequency of each allele in the population. b. Are the M-N genotypes in H-W equilibrium? Blood type M (LMLM): 1245 individuals Blood type MN (LMLN): 3421 individuals Blood type N (LNLN): 965 individuals a. The total number of alleles sampled = 2(1245) + 2(3421) + 2(965) = 2(5631) = 11,262. The total number of LM alleles in the sample = p = 2(1245) + (3421) = 5911; and the frequency of the LM allele = 5911/11,262 = 0.525. The total number of LN alleles = q = 2(965) + (3421) = 5351; and the frequency of the LN allele = 5351/11,262 = 0.475. Note that p + q = 1. b. p2 (LMLM ) + 2pq (LMLN ) + q2 (LNLN ) = 1. For (LMLM): p2 = (0.525)2 = 0.276. Therefore, (5631)(0.276) = 1554. For (LMLN): 2pq = 2(0.525)(0.475) = 0.49875. Therefore, (5631)(0.498) = 2808. For (LNLN): q2 = (0.475)2 = 0.225625. Therefore, (5631)(0.225) = 1270. Test for equilibrium using the chi-square test: (1245 1554)2/(1554) + (3421 2808)2/(2808) + (965 1270)2/(1270) = 61.4 + 133.8 + 73.2 = 268.4 The chi-square statistic is clearly so large that our hypothesis must be rejected, and thus we assume that the population is not in H-W equilibrium with respect to the M-N blood type antigens. 53. You discover a certain species of weed growing in soil contaminated with toxic PCBs, and later determine that the PCB resistance is due to a single dominant allele. a. If 45% of the seeds from a randomly mating population of resistant weeds will germinate in contaminated soil, what is the frequency of the PCB-resistance allele? Population and Evolutionary Genetics b. Among all the plants that germinate, what proportion will be heterozygous? c. What proportion will be homozygous dominant? a. Because 45% of the seeds germinate, the 55% that do not germinate must be homozygous recessives (q2). The frequency of the recessive allele (q) is therefore (0.55)1/2 = 0.742; and the frequency of the dominant resistance allele = p = 1 q = 1 0.742 = 0.258. b. The expected frequency of heterozygotes = 2pq = 2(0.258)(0.742) = 0.383, and the overall frequency of surviving genotypes that are heterozygotes = (0.383)/(0.45) = 0.85 (85%). c. The expected frequency of homozygous dominants = p2 = (0.258)2 = 0.067, and the overall frequency of surviving genotypes that are homozygote dominant = (0.067)/(0.45) = 0.15 (15%). 54. The frequencies of two co-dominant alleles (X and Y) for a single gene locus (I) are the same in three different populations that you are studying. Interestingly, however, you find that the following genotypic frequencies are not the same. a. What are the frequencies of each allele in the three populations? b. Which population most closely approximates Hardy-Weinberg equilibrium? Explain your Register to View AnswerThe genotype distribution of which population could have been produced from inbreeding? Explain your answer. I XI X Population 1 2 3 I XI Y IYIY 0.50 0.60 0.42 0.40 0.20 0.56 0.10 0.20 0.02 (a) Note that the problem states that the allele frequencies are the same in the three populations, so each frequency needs to be calculated only once. For population 1, the frequency of the IX allele = p = 0.50 + 1/2(0.40) = 0.70, and the frequency of the IY allele = q = 0.10 + 1/2 (0.40) = 0.30. Applying the same formula to populations 2 and 3 will produce the same allele frequencies. (b) The equilibrium distribution would be p2 = 0.49 for XX, 2pq = 0.42 for XY, and q2 = 0.09 for YY. Population 1 most closely approximates the equilibrium distribution. (c) Inbreeding would produce an excess of homozygous individuals and a dearth of heterozygotes compared to the H-W equilibrium distribution. Therefore, population 2s distribution could be produced from inbreeding. 55. The diagram below represents the ranges of three species (A, B, and C), which are separated by a mountain range and river. As an expert geologist, you hypothesize that the area was originally occupied by a single species. Its range Chapter 17 was split first by the mountain range. Later, the river formed, separating populations on the eastern side of the mountains. If your hypothesis is correct, what should the phylogenetic relationships be among the three species? Draw a phylogeny as your answer. N B A S C Register to View Answer A B

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LSU - CE - 3400
LSU - CE - 3400
LSU - CE - 3400
LSU - CE - 3400
LSU - CE - 3400
LSU - CE - 3400
LSU - CE - 3400
LSU - CE - 3400
LSU - EE - 2950
EE 2950 Test 1 Spring 2012 DO NOT MARK ANSWERS ON TEST. USE Scantron! March 5, 2012 1. The power absorbed by a resistance can be expressed as: a.* !&quot; !&quot;!&quot; !&quot; b. !&quot; !&quot;!&quot; !&quot; c. !&quot; !&quot;!&quot; !&quot;
LSU - EE - 2950
EE 2950 Test 2, Spring 2012 DO NOT MARK ANSWERS ON TEST. USE Scantron!April 2, 2012 () = + [(0) ] + = I. = 0II. = 02III. =Use the circuit below for questions 1-4:C5+ () -3 = 75IV. = () = = = 0 = +-725V1 () = 1. Classify the ci
LSU - EE - 2950
./.'.EE 2950 Test 1 Fall 2011 DO NOT MARK ANSWERS ON TEST. USE Scantron!September 27, 2011 v21. It is your first day on the job and you are asked to determine what is wrong with the circuit that ispowering a motor that is malfunctioning on a piece
LSU - EE - 2950
/EE 2950 Test 1 Fall 20 I0 DO NOT MARK ANSWERS ON TEST.Thursday, September 23,20101. The power absorbed by a resistance can be expressed as:a.dq dwdt dqb . .'!:!ic. ~dwdw (itd.dwdw dtdw ~dq dw150Vson resistor.SOD._~o:r:- 0c-.:I&quot;i~O-1-
LSU - EE - 2950
LSU - EE - 2950
LSU - ECON - 2030
Week316:26Giventheproductionpossibilitiesfrontiershown,pointjis attainableforAmanda.(graphwithJat(0,5)oncurve) Productionpossibilityfrontier:curveorlinewithmaxpointsonitGiventheproductionpossibilitiesfrontiershown,pointPis inefficientforAmanda.(gra
LSU - ECON - 2030
TODAYS MENU: Monday 10 September 2012I. BUSINESSA. Practice Problems1. Chapter 4: 1, 3-7, 10-20, 22B. First Exam: Two weeks from today (24 September)II. SUBSTANCEA. Buyers side of market: DemandRational Buyer: Benefit&gt;costReservation Price: what y
LSU - ECON - 2030
TODAYS MENU: Wednesday 12 September 2012I. BUSINESSA. Practice Problems1. Chapter 4: 1, 3-7, 10-20, 222. Chapter 5: 1-5, 9, 11, 14B. First Exam: One week from Monday (24 September) Chapters: (1,2,4,5,7)II. SUBSTANCEA. Sellers side of market: Supply
LSU - ECON - 2030
TODAYS MENU: Monday 27 August 2012I. BUSINESSA. Practice Problems1. Chapter 2: 1, 2, 4-8, 10II. SUBSTANCEA. Note on graphing: Appendix to Chapter 2B. Opportunity Cost Application: International Trade1. Questions to Answera. If workers in the Unite
LSU - CE - 2700
Sarah L. AbramsCE 2700 Project 1September 30, 2011The Boston Big DigThe Central Artery/Tunnel Project, known unofficially as the Big Dig since completion,was a megaproject in Boston. The project routed the Central Artery (Interstate 93), the mainhig
LSU - CE - 2700
Sarah Abrams, Zach Jordan, Reid PlaucheCE 2700 Project 2November 18, 2011The Nicholson-Brightside/Lee Turn Lane ProposalWhat comes to mind when one says the word intersection? Bad traffic and wrecks aretwo things that are often instantly thought of.