BIO152H-2008 - BIOISZHSF Page 1 of 20 NAME(PRINT Last/Surname First lGiven Name STUDENT NO SIGNATURE UNIVERSITY OF TORONTO MISSISSAUGA DECEMBER

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Unformatted text preview: BIOISZHSF, Page 1 of 20 NAME (PRINT): Last/Surname First lGiven Name STUDENT NO: SIGNATURE: UNIVERSITY OF TORONTO MISSISSAUGA DECEMBER 2008 FINAL EXAMINATION I BtOtSZl—ISF Introduction to Evolution and Evolutionary Genetics Anne Cordon Duration - 3 hours Aids: None You may be charged with an academic offence for possessing the following items during the writing of an exam unless otherwise specified: any unauthorized aids, including but not limited to calculators, cell phones, pagers, wristwatch calculators, personal digital assistants (PDAs), iPods, MP3 players, or any other device. if any of these items are in your possession in the area of your desk, please turn them off and put them with your belongings at the front of the room before the examination begins. A penalty may be imposed if any of these items are kept with you during the writing of your exam. Please note, students are NOT allowed to petition to RE-WRITE a final examination. EXAM QUESTIONS & Instructions 1. This exam paper has 5; questions 211 pages (including this cover page). 2. Total points = 52 (worth 35% of the final mark). All questions are of equal weight. 3. Choose the BEST answer for each question. There is 9E and ONLY ONE correct answer for each question. 4. No marks are deducted for incorrect answers, so answer all questions. _ 5. Transfer all of your answers to the Scantron computer sheet within the time allowed. No answers on the exam paper will be marked. 6. You must hand in both your Scantron computer sheet AND your exam paper. 7. Do not ask the invigilators to interpret or explain questions, words, or figures on the exam—ONLY ask for help if you think there is a question missing or a typographical error which affects the meaning of the question. SCANTRON Computer Sheet 1. Use PENCIL and erase any changes completely. 2. Write you student number in the 9 boxes and then bubble in your numbers in the correct columns. 3. Write your name, date, and course in the upper right in the spaces provided. 4. Do NOT write anything along the top or side of the Scantron sheet. 5. For each question, bubble'your answer during the time allowed. D0 m open this exam until the invigilator tells you to start. 6. On SCANTRON bubble the FORM letter indicated on the last page of the exam. Continued on page 2 BIOISZHSF, Page 2 of 20 1. How does a scientific theOry differ from a scientific hypothesis? a) There is no difference, the terms are interchangeable. b) A theory is a wen—tested explanation for more general phenomena; hypotheses treat more specific observations. c) Theories define scientific laws; hypotheses are used to set up experiments. d) Theories are not necessarily testable; hypotheses are testable. 2. Which of the following is the best example of a heritable variation? 3) skin cancer b) green eyes c) amputation d) love for classical music e) a and b i) a, b, c, and d 3. Which of the following statements about meiosis is CORRECT? a) The number of chromosomes is doubled during the DNA synthesis stage of the cell cycle. b) The second meiotic division results in the halving of the number of chromosomes in the nucleus. c) The first meiotic division results in the halving the amount of DNA in the nucleus. (1) b and c e) a, b, and c 4. Which of the following statements about mitosis is INCORRECT? a) The number of chromosomes per nucleus remains the same before and after mitosis. b) The number of chromosomes is doubled during DNA synthesis. 0) Mitosis can occur in either haploid or diploid nuclei. d) a and c e) b and c i) a, b, and c 5. Which of the following describes the most likely order of events in Speciation? a) genetic isolation, genetic drift and/or natural selection, divergence to two species b) genetic isolation, divergence to two species, genetic drift and/or natural selection c) divergence to two species, genetic drift and/or natural selection, genetic isolation d) genetic drift and/or natural selection, divergence to two species, genetic isolation Continued on page 3 BIOlSZHSF, Page 3 of 20 6. Male frogs give calls that attract female frogs to approach and mate. Researchers examined mating calls of pairs of closely related tree frogs in South America. If reinforcement of prezygotic isolation is occurring, what would you expect if you compare the calls of the two species in zones of sympatry and allopatry? a) calls would be more different in areas of allopatry b) calls would be more different in areas of sympatry 0) calls would be about the same in both areas 7. The two key factors responsible for speciation among populations are: a) postzygotio isolation and morphological change b) mutation and genetic drift 0) genetic isolation and genetic divergence (1) lack of both gene flow and mutation 8. A species has four pairs of chromosomes. How many molecules of DNA do the nuclei have during G2 phase? a) 4 b) 8 c) 16 d) 32 9. Which of a man's relatives could not be the source of any of the genes on his Y— chromosome? a) father's mother b) mother‘s father 0) mother's mother cl) father‘s father, mother’s father, father’s brother e) mother's mother, mother's father, father's mother 10. If the parents were heterozygous for three traits, what is the probability that an offspring is heterozygous for all three traits? a) 1/64 b) 2/64 o) 4f64 d) 8164 o) 32/64 11. If the parents were heterozygous for three traits, what is the probability that an offspring is heterozygous for one trait and homozygous recessive for one trait and homozygous dominant for One trait? a) U64 b) 2/64 0) 4/64 d) 8/64 e) 32/64 Continued on page 4 BIOISZHSF, Page 4 of 20 12. Which of the following pairs of genes would most likely demonstrate the lowest incidence of crossing over? a. gene a and gene b b. gene 0 and gene d 0. gene d and gene e (1. gene a and gene f 13. Which of the following individuals could produce the most variation in the possible different types of gametes? a) nesttuuwwyyzz b) RRSSTTUUWWYYZZ c) SsTtUuWWZz d) QQRRSSTTUuWWYYZZ 14. Suppose 64% of a remote mountain village can taste phenylthiocarbamide (PTC). If this population conforms to Hardy-Weinberg expectations for this gene, what percentage of the population must be homozygous tasters for this trait? a) 16% b) 32% e) 40% d) 48% e) 60% Continued on page 5 BIOISEHSF, Page 5 of 20 The next six (6) questions refer to Figures A, B, C, D The figures below represent different stages of nuclear division for the same ceil. Continued on page 6 BIOlSZI-ISF, Page 6 of 20 15. Figure B refers to what stage? a) Metaphase mitosis b) Metaphase I meiosis c) Metaphase II meiosis d) Anaphase mitosis e) Anaphase I meiosis 16. Figure C refers to What stage? a) Metaphase mitosis b) Metaphase I meiosis c) Metaphase II meiosis d) Anaphase mitosis e) Anaphase I meiosis 17. Figure D refers to what stage? a) Metaphase mitosis b) Metaphase I meiosis e) Metaphase II meiosis d) Anaphase mitosis e) Anaphase I meiosis 18. How many chromosomes are in the left “cell” depicted in Figure B? a) 2 b) 4 e) 8 d) 16 19. How many chromosomes are in the “cell” depicted in Figure C? a) 2 b) 4 c) 8 cl) 16 20. What is the diploid number in a somatic cell for this organism; what is the haploid number of a gamete for this organism? 21) 2 ; l b) 4 ; I c) 4 ; 2 d) 8 ; 2 e) 8 ; 4 Continued on page 7 BIOISZHSF, Page 7 of 20 21. At what stage of nuclear division did the error occur in Figure A, Figure B and Figure C? a) Errors represented in all three figures occurred in anaphase of mitosis. b) Figure A——-no error; Figure 13 error in anaphase I ; Figure C .error in anaphase II. c) There was no error in the distribution of chromosomes; these figures represent normal possible outcomes. d) Figure A error in anaphase I; Figure B—no error; Figure C error in anaphase II 22. If 9% of an African population is born with a severe form of sickle-cell anemia (ss), what percentage of the population will be more resistant to malaria because they are heterozygous (S's) for the sickle-cell gene? a) 42 b) 30 c) 70 d) 21 Continued on page 8 BIOISZHSF, Page 8 of 20 23. After graduation, you and 19 friends build a raft, sail to a deserted island, and start a new population, totally isolated from the world. Two of your friends are carriers of the recessive cf allele; which in ’homozygotes causes cystic fibrosis. The frequency of CF births on the mainland is .059%. How does the frequency of potential cystic fibrosis births on the island compare to the frequency of CF on the mainland? a) about the same on the island and the mainland b) about two times greater occurrence on the island than on the mainland 0) about four times greater occurrence on the island than on the mainland d) about eight times greater occurrence on the island than on the mainland e) about two times greater occurrence on the mainland than the island 24. A rather large population of Biology instructors have 360 red-sided individuals and 640 tan-sided individuals. Assume that red is totally recessive. Conditions happen to be really good this year for breeding and next year there are 1,245 young "potential" Biology instructors. Assuming that all of the Hardy-Weinberg conditions are met, how many of these would you expect to be red-sided and how many tan-sided? a) 448 red; 79'? tan b) 75 red; 1170 tan c) 747 red; 498 tan d) 500 red; 775 tan 25. If a plant species with 211 m 14 forms an allopolyploid with a plant species with 2n = 18, what would be the likely diploid (212) number of the allopolyploid? a) 16 b) 28 c) 32 d) 36 26. A storm brings two formerly separated populations of beetles together. Under the biological species concept, which of the following would show that the two populations are different species? a) One population breeds in spring, the other in fall. b) Males of the two populations have different flight patterns in courtship. c) When individuals from the two populations mate with each other in the laboratory, the eggs fail to hatch. d) They look very different in size and colour. e) a, and c are correct i) a, b, c are correct g) a, b> c, d are correct Continued on page 9 BIOiSZI-ISF, Page 9 of 20 The next five (5) questions refer to the information in the CASE: Ability to drink milk Virtually all humans are born with the ability to digest the milk sugar lactose, which allows us to drink mother's milk until we are weaned, but most of us lose this ability by the time we are 12 or 13 years old. After that, even a modest nip of the white stuff causes intestinal symptoms best described in polite company as unpleasant, which arise in part from the gas-producing fermentation of lactose by bacteria in the gut. A lucky minority of us maintains the ability to digest lactose into adulthood and can go on enjoying milk throughout life. This ability is provided by an enzyme, lactase-phlorizin hydrolase, which performs the important task of breaking lactose (a disaccharide) into monosaccharides more readily absorbed by the gut. The ability to digest lactose in adults is inherited as a dominant Mendelian trait. Lactose tolerance is also called “lactase persistence”. The ability of human adults to digest milk products is unique among mammals and a relatively new trait among humans. Being either lactose intolerant or tolerant during adulthood depends completely on your body’s ability to produce lactase, which is genetically determined. There is no scientific evidence to indicate that diet or lifestyle has an impact on your ability to produce lactase as an adult. F or example, avoiding milk completely for long periods does not lead to a change in lactase production. Milk products that are soured or otherwise treated (like yogurts and solid cheeses) contain relatively low levels of lactose and the soured products may even contain the lactase-producing Lactobacillus acidophilus. As a result, these milk products cause fewer problems for lactose intolerant people compared to other milk products. Lactose intolerance is not the same as being allergic to milk. Unlike an allergy, lactose intolerance does not involve your immune system and does not necessarily mean that you will have to completely avoid dairy products. Variation in the ability of adults to digest milk has long attracted attention as an interesting developmental phenomenon, but success in understanding the genetic underpinnings of the trait has been slow. The gene encoding lactase was first mapped in the late 1980s. Both Europeans and Africans show phenotypic variation in lactase persistence (see Figure below), but the one genetic region previously noted accounts for variance only in Europeans. African populations must harbor one or more different variants that confer similar phenotypic effects. Throughout history, one of the biggest problems humans have faced is getting enough to eat. Early populations solved this problem by being highly efficient hunter—gatherers. The advent of agriculture 10,000 years ago presented an alternative solution: keep food sources close at hand. It has long been hypothesized that in the case of dairy animals, domestication had the effect of reflecting selective pressures back at the domesticators. This is because dairy animals are useful even if you can't drink their milk, but they are much more so if you can: milk is a nutritional bonanza of fat, proteins, carbohydrates, vitamins, calcium and even water...but only if you can digest it. Thus, many have argued that whereas non-dairying populations faced little pressure to digest milk into adulthood, dairying populations were under enormous selective pressure to do so. And if they were, we should see evidence of it in their genes. Continued on page 10 BIOISZHSF, Page 10 of 20 The discovery that different genetic regions account for the same phenotype in different populations is extraordinarily interesting from an evolutionary standpoint. These findings tell us that divergent human populations have been under similar pressures in the diet—pressures involving milk——and have arrived at the same solution of prolonging lactase gene expression into adulthood. In a striking testimony to the powerful evolutionary effects culture can have on our genes, not only has the domestication of cattle driven allele frequencies in humans, but it has done so at least twice, in different regions of the world. LACTOSE PQPDLATION lNTOLERANT ADULTS U.S. European Americans 2—19 % Latinos (Hispanic Americans) 52 % African Americans 70-77 % Native Americans 95 % Asian Americans 95400 % Mexico 83 % Europe Sweden 4 9/0 Switzerland 12 % Spain 15 % Finland 18 % Estonia 28 % England 32 % Hungary 37 % Greece 88 % Jordan 79 % Africa Southern Sudan (cattle herders) 17 % Ibo and Yoruba (Nigeria) 99 % Asia Japan 90 % Thailand 99 % Australia (Aborigines) 85 % Source: Robert D. McCracken, “Lactase Deficiency: An Example of Dietary Evolution," Current Anthropology 12 (Oct-Dec. 1971, pp. 479—51?) and Norman Kretchner, “Lactose and Lactase,“ Scientific American 277 (Oct. 1972, pp. 71—78) Continued on page 11 BIOISZHSF, Page 11 of 20 27. Which continent has the greatest frequency of native people with Iactase persistence? a) Africa b) South America 0) Asia (1) Europe 6) North America t) Australia 28. What is the allele frequency of lactose intolerance in the Southern Sudan compared to Nigeria? a) Southern Sudan = 0.04 ; Nigeria = 0.99 b) Southern Sudan = 0.2 ; Nigeria = 0.99 0) Southern Sudan = 0.8 ; Nigeria = 0.01 d) Both Southern Sudan and Nigeria have the same allele frequencies e) Cannot determine the allele frequencies from the data provided 29. What is the most likely reason for the distribution of variation in Iactase persistence? a) Mutation followed by natural selection. b) Mutation followed by genetic drift (likely a founder effect). c) Mutation followed by gene flow. d) No evolutionary process was likely happening, but rather environmental influences affected variation. 30. What is the most likely reason that a relatively large proportion of people in both Switzerland and Southern Sudan have Iactase persistence? a) Gene flow (migration) b) Genetic drift (likely a founder effect) 0) Genetic drift (likely a bottleneck effect) d) Convergent evolution via natural selection following separate and distinct mutations e) No evolutionary process was likely happening, but rather environmental influences affected variation in these two areas in the same way 31. Why do the authors feel that finding different regions with the same phenotype of Iactase persistence in different pepulations is extraordinarily interesting from an evolutionary perspective? a) This example demonstrates the powerful effects that culture can have on our genes. b) Dairying populations were under enormous selective pressure to be able to digest lactose. c) Both Europeans and Africans show phenotypic variation in lactase persistence. d) Nondairying populations were not submitted to the selective pressure to be able to digest lactose. Continued on page 12 B10152H5F, Page 12 of 20 The next three (3) questions refer to the CASE: The Blue People Ruth had never been as astonished as she was the day she encountered the first of the “blue people” from Troublesome Creek. The blue woman simply walked into the rural health clinic where Ruth was a nurse. Ruth suspected the woman was having a heart attack, but the woman wasn’t concerned at all. “I’m one of the blue Fugates,” she explained to Ruth, as if it was all perfectly logical. As their conversation continued, Ruth learned from her patient that there were, in fact, many blue people living in the isolated community around Troublesome Creek. The Fugate clan in the Troublesome Creek region could be traced back to the arrival of Martin Fugate, an orphan from France, in 1820. Legend has it that Martin may have been blue; for this case, we’ll assume that Martin is blue. Maltin settled in the area and married the pale, red-headed Elizabeth Smith. Over the years, they had at least seven children. Four of them reportedly were blue (2 girls and 2 boys). The mutation believed to be responsible for the “blue people” is located in the gene that codes for the enzyme called NADH diaphorase (or NADH dependent methemoglobin reductase). It is found in large concentration in red blood cells, where the enzyme functions to return hemoglobin to a normal oxygen binding state after it has been oxidized to methemoglobin (metI-lb). MetHb cannot bind oxygen or carbon dioxide (because iron, the oxygen binding part of the heme group, is in the ferric state and binds water instead of oxygen), and gives the blood a blue tint. This oxidation process is slow, but requires enzyme mediated reduction to return to hemoglobin. The graph below shows enzyme activity over time in people that are blue, people that are not blue but may have blue children, and people that are not blue and never have blue children. we MgWTMMW W blue people of blue people mettlb unreduced (an) ‘6‘ . Normal control V 7 e m Time Red blood cell extracts from three different groups assayed For ability to reduce metHb to Hb. 32. What is the most likely inheritance pattern of the blue phenotype? a) Autosomal dominant b) Autosomal recessive c) X-linked dominant d) X-linked recessive e) Co—dominant f) Incomplete dominant Continued on page 13 BIOISZHSF, Page 13 of 20 33. What is the most likely inheritance pattern of the enzyme activity? a) Autosomal dominant b) Autosomal recessive c) X-linked dominant d) X—linked recessive e) Co-dominant f) incomplete dominant 34. What is the best explanation for the prevalence of this rare condition in Troublesome Creek? a) Natural selection b) Founder effect c) Bottleneck effect d) Gene flow e) Mutation from some environmental hazard in the area The next six (6) questions refer to the CASE: Genetic Counseling for Greg and Olga Greg and Olga were both a little worried. Starting a family presented choices and responsibilities far more long-reachin g and complex than anything either of them had encountered before, and sitting here in the reception area of the genetic counselor’s office they were beginning to feel the pressure. They had met four years earlier in the hemophilia clinic where Greg was waiting for his brother Jeff to get an injection of factor VIII, a protein that helps the blood to clot. When a person’s factor VIII level is very low (less than 1% of normal), even the smallest cuts can be troublesome and uncontrolled internal bleeding is common. Complications include swelling, joint damage, and an increased likelihood of neurological complications due to intracerebral bleeding. Even simple surgical procedures such as tooth extractions become far more risky. It was in that clinic waiting room that Greg first met Olga, who was waiting for her uncle to finish his exam and receive an injection of clotting factor. Like Jeff, Olga’s uncle also suffered from hemophilia A due to factor VIII deficiency. They are now thinking about starting a family of their own, but are concerned about the risks of passing on genetic diseases to their children. They know for example that hemophilia A is an inherited disease, and several of Greg’s relatives suffer from myotonic dystrophy, a muscle weakening disease that also runs in families and is an autosomal dominant condition. From the following text, draw out Olga and Greg’s pedigree to help answer the questions about their risks for having children affected with hemophilia A (due to factor VIII deficiency) and myotonic dystrophy (MD). Continued on page 14 NAME: 0143a 1 home m brcfiherfi, one 0%? whom has {Lac-tar \i’ii dafidemq. The hrother with the disease is unw- ried to a memo who deal: not have the disco-Be. Theo have two noting, bogs, both Wmt Mo {'Q'izhm' ‘15 can (mix; mite who (1036 not eoéléer {mm ang’nh‘mg and his pavewts misc are. (3th ch'iidren who do not sot-Ger Worn on; diseases. Theq are ail etiii i'w‘m. W mtemai grown-other is healing and had a ate-tar who died first: after birth. She: mmee mo gme-ather who was one ot- waoor mit- dven. ail taboo: new: 0(- whoim were afi-ected bq am; disease. «that anqone its aware. of: Mg grahdpaw ante had *two anthrax» we; wother and mo node. HQ mute no.5 henwphii'n but me mm doesn‘t. Mo owe worried. mo aunt {who is omftected) am that; had two Lhiicts’an, wither 0% what“ showed not; sign 06 um? disease. Their boo is sat'fii single bot their aid 90%: miner 4:0 a. hermit m“ and “mi :1 son who has hemp‘niih “R. NAMEG‘rEE I have one bro-Fixer and one sits-Far) net-Hw— a? w‘aom are married. 3715 bro-Fixer suFFei-s ‘From Factor VIII deficiencfj) bui‘ no one eta-e in mg Fawifi does. M8 hfli‘iifi!’ has two brothers and One sister. One 0? m3 males and one 0? m5 omits are a-t‘tieei‘cci b3 :nfidi‘one st'i‘i'OfJiig. i753 aé‘aFceied amt married on wacFé‘mim’ man onci i'ixcg Mile a Sauna) mafiaflig wfpcei‘cd diaugiriar. W13 oi‘kcr uncle is meF'Fcei‘cciJ as is m8 motion Our primer-3 core doe-tor has said flat because both m3 mother and male are over Fifi-3 gears aid and 55m no sfinp‘f‘omsj {4163 (it: not“ how. “Hie disease. 1733 Potter is compiefeifi annual: He was adopted From an orphanage and mike?) is Kawx chant“ his Fanny. FY33 motmmi firmdmthcr was an 0:113 (fluid who aim suFFered Gem mgei’eite cigstrcpisfi. Ha— imshand (mg granapfiker) was one. a? 5.2m oividrcn (the bags and +1M'66 girls). No one in the Fania seems 'i‘o Know much about $115 With s‘i‘o‘ihs GP m3 grandth‘i‘m er his brothers. Both 0? m3 parents are «2%;ch but all 0? m3 grandparmf‘s are deceased. BIOISZHSF, Page 14 of 20 35. What is the most likely inheritance pattern of factor VIII deficiency? a) autosomalrecessivc b) autosomal dominant c) X-Iinked recessive d) X~iinked dominant e) No evidence that this condition has a genetic basis; liker due to environmental factors. Continued on page 15 BIOISZHSF, Page 15 of 20 36. What is the probability that Greg has the allele that causes myotonic dystrophy (MD)? a) 0 b) 0.25 c) 0.5 d) 0.75 e) 1.00 37. What is the probability that Greg and Olga’s children could inherit the MD allele? a) ‘0 b) 0.25 c) 0.5 d) 0.75 e) 1.00 38. What is the probability that Olga has the allele for factor VIII deficiency? a) 0 b) 0.25 c) 0.5 d) 0.75 e) 1.00 39. What is the probability that Greg and Olga’s children could exhibit factor VIII deficiency? a) boys .5 ; girls 0 13) boys 0 ; girls 0 c) boys 1; girls 0.5 (1) boys 0.5; girls 0.5 e) boys 1: girls i 40. There is a way to figure out the odds of being a carrier even without a past family history. For example, in; order for the autosomal recessive disease cystic fibrosis (CF) to show up unexpectedly in their offspring, both Olga and Greg would have to be carriers and then each would have a l in 2 chance of passing on the defective allele to their offspring. Olga is of European descent (Swedish and German); the carrier frequency of Caucasians of European descent is 1 out of 23. Greg is Asian American and within his population group the carrier frequency is 1 out of 180. What is the probability of Greg and Olga having a baby with cystic fibrosis? a) 1 out of 4 b) 1 out of 8 e) 1 out of 1000 d) 1 out of 4000 e) 1 out of 16,000 Continued on page 16 BIOISZHSF, Page 16 of 20 Next four (4) questions refer to the CASE: Baby Pierre On March 7, 1964, the baby known as Pierre was born in a remote part of Quebec. He appeared to be a healthy six-pound twelve-ounce child, except he did not eat well. Over the weeks after his birth, he became progressively more lethargic, vomiting periodically. Most peculiarly, his urine smelled of rotten cabbage, and soon the smell permeated his clothes and body. By the time he was admitted to the hospital on September 14, his muscles were weak and his ribs were showing. On November 30, baby Pierre vomited blood and died. It soon became increasingly apparent that other babies in the Chicoutimi area of Quebec Province had similar symptoms, and people recalled similar deaths in this remote area 120 miles north of Quebec City. Some families lost several children to Pierre‘s disease. In those families stricken, it soon became clear that the parents were normal, but about one quarter of their children were afflicted. Boys and girls were equally afflicted. After further investigation it was determined that, Baby Pierre and the other stricken children were victims of hereditary tyrosinemia. The children lacked the normal gene which produces a liver enzyme that breaks down the amino acid tyrosine. Without the enzyme, tyrosine builds up in the liver and kidneys leading to the cabbage—like smell of the urine. Lethal side—effects follow (as of 1997 a liver transplant was the only long-term treatment of the disease). Generations I 41. Based on the information provided about this case, what is the most likely cause? a) A pollutant is causing mutations. b) The disease is caused by an X—linked recessive condition. 0) The disease is caused by a dominant allele. d) Baby Pierre's parents are homozygous for a recessive allele causing the disease. e) The disease is an autosomal recessive condition. 42. What is the probability that female K will have a normal child if she marries a person who is a carrier for tyrosinemia? a) 0 b) 0.25 c) 0.5 d) at least 0.75 e) 1.00 Continued on page 17 BIOISZHSF, Page 17 of 20 43. What is the probability of female K having an affected child if she marries her ' cousin M? a) 0.25 (1 child in 4 will be affected) b) K is normal so must be homozygous for the normal allele; M and all of M’s siblings are normal, so M must be homozygous normal—so no chance of having an affected child c) K looks normal but may be heterozygous; M looks normal but may be heterozygous so the probability is at most 1/8 for having an affected child. d) K looks normal but may be heterozygous; M looks normal and because all of M’s siblings and parents are normal, M must be homozygous. e) All of K & M’s children will be affected. 44. What is the most likely reason for the large number of tyrosinemia cases in Chicoutimi? A) A high mutation rate. B) A selective advantage for tyrosinemia in this part of Quebec. C) Founder effect. D) Nutritional patterns in the people; large quantities of tyrosine in the diet. E) Heterozygote advantage where the allele for tyrosinemia has beneficial effects as well as harmful effects. 45. What is the best explanation for how genetic isolation and genetic divergence is occurring in organisms such as hawthorn maggot flies, even though populations occupy the same geographic area? a) Body size has changed due to disruptive selection. b) One population recently became triploid, and hybrid offspring cannot undergo meiosis correctly. c) A vicariance event occurred when nonnative host plants were introduced. d) Members of the different populations feed and mate on different types of fruit. 46. Imagine a lake containing a single population of snails. During a period of drought, the shallow middle of the lake dries up, creating two separate populations. The snails, being totally aquatic, cannot cross from one lake to the other. At some point in the future, the climate becomes wetter and the lake refills. Assuming an annual generation, how many generation must the populations have remained divided for reproductive isolation between them to evolve? a) 50 b) 5000 c) 500,000 d) In theory, Speciation could occur within a few generations, or it could take thousands of generations. There is no general answer to this question. Continued on page 18 BIOISZHSF, Page 18 of 20 4’7. Refer to the figures below. Which intermountain pairing of individuals from jumping spider populations seemed to confer a significant disadvantage? £2 £5 , me {see} 'Lr-I «I. m 4:: T: D m talents" ti M e; c: 9 i (it) number of viable efl'spring IQ 35‘ s: " nus n+5, MR I} Q: t; s S S x R G s R within a range 53 >" n: m n [8 n= 0 an it's ne‘) :1 --21 a) s XR b) G x either S or R noxs d) All matings have the same advantage/disadvantage for latency but G matings seem to have fewer viable offspring with either other G or R individuals. 48. What is the most likely reason male lions have manes? a) Male lions have manes to shield their vulnerable head and neck from the teeth and claws of other lions (as first proposed by Charles Darwin). b) The mane’s function is a signal to other male lions and serves as an indicator for overall fitness. c) The mane’s function is a signal to both other male lions and female lions and serves as an indicator for overall fitness (good genes). 49. What did Peyton West’s study of African lions find waslwere the most important feature(s) of the male lion’s mane to other lions? a) For both male and female lions, dark colour was the most important attribute of the male lion’s mane. A dark mane is an indication of good genes and fitness due to the incredible heat retention handicap for any male with a dark mane in the hot conditions where they live. b) Female choice of mates was most influenced by the colour of the male lion’s mane (the darker the better as far as female preference was concerned); for males both mane colour and length affected their status when confronted with unfamiliar males not in the same coalition! group (darker and longer manes were both signals of greater fitness to male lions, though colour was more significant among male lions in the same group). c) For both male and female lions, the length of the male lion’s mane was the most important attribute (the longer and bushier the better as a signal of better genes). A long full mane were an indication of good genes and fitness due to the incredible heat retention handicap for any male with such a mass of fur around the neck in the hot conditions where they live. d) The length and fullness of the male lion’s mane served as a protection against male-male fights and thus was the most important feature. Male-male competition in lions is particularly nasty, so males with more neck protection fare much better in these fights overpowering other males and gaining access to females. Continued on page 19 BIOISZHSF, Page 19 of 20 For the next three (3) questions refer to the CASE: Why Sex? A team of scientists at the Imperial College London tackled the problem of why sex is good (their results are published in Nature, March 25, 2005). They decided to use yeasts, which are single- celled fungi, because they can reproduce both sexually and asexually, are easy to keep in the lab, and reproduce very rapidly. Yeasts normally reproduce asexuaily, but when they are stressed (starved, high temperatures, etc.) they will reproduce sexually. The scientists did not want this switching to occur, so they genetically manipulated one asexual strain. They deleted the two genes (SPOil and SP013) required for normal meiosis, so that sexual reproduction was impossible. Now they had two pure strains—asexual and sexual. ' The Imperial College team decided to compare the reproductive rate of the asexual vs. the sexual yeasts in two different environments: harsh and benign. That is, “fitness” would be measured by comparing the growth rate relative to the non-manipulated ancestral strain. The benign enviromnent had plenty of nutrients although glucose was limited so that growth was not uncontrolled. The harsh environment had the same glucose concentration but was at a higher temperature and had more demanding osmotic conditions. 1.0 0.8 (18 o4 0.2 Natural Logarith of Relative Fitness _ O _ m O: -G Mitotic Generation Figure 1. The change in natural logarithm of fitness of asexual and sexual popuiaflens of yeast in benign and harsh environments. Points shew fitness measurements for individual pepuiafions. (x) Asexuai strains in tit-enemas environment: (n; Sexual in the benign environment: (magma: in the harsh environment; (aJ'SeXual in theihareh anvironméfit Continued on page 20 d) BIOISZHSF, Page 20 of 20 50. How was fitness measured in this experiment? Number of organisms over time in each test population compared to the number of offspring in the original population. Growth rate in each sexual population compared to each asexual population. Growth rate after specific numbers of generations. _ The size of each population as measured on agar plates and measured under the microscope. 51. Which mode of reproduction (sexual or asexual) had the greatest average fitness under benign conditions? sexuai populations asexual populations no difference in fitness between sexual and asexual reproducing populations the fitness of sexual and asexual populations was the same and showed no increase in fitness from the ancestral population 52. Which mode of reproduction (sexual or asexual) had the greatest average fitness under harsh conditions? sexual populations asexual populations no difference in fitness between sexual and asexual reproducing populations. the fitness of sexual and asexual populations was the same and showed no increase in fitness from the ancestral popuiation FORM A (code A on SCANTRON form) THE END ...
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This note was uploaded on 12/11/2011 for the course BIO 152 taught by Professor Cordon during the Spring '07 term at University of Toronto- Toronto.

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BIO152H-2008 - BIOISZHSF Page 1 of 20 NAME(PRINT Last/Surname First lGiven Name STUDENT NO SIGNATURE UNIVERSITY OF TORONTO MISSISSAUGA DECEMBER

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