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Unformatted text preview: 1 BILD3 Midterm 1:00 Peterson Hall 110 Oct 31, 2007 Name_________________ Dr. Wills 1 hour, 100 points total Section time____________ Calculators permitted, but only without cases. Read the questions carefully, and make sure that you answer all parts of each question. If diagrams are needed, make them clear and label them fully. This exam consists of 7 pages. Check to make sure you have them all. Write your name on each page. GRADER USE ONLY Question 1a ____________ 1b ____________ 1c _____________ 1d ______________ 1e ______________ 1f ______________ 2a _____________ 2b _____________ 2c _____________ 2d _____________ 3a _____________ 3b _____________ 3c _____________ 3d _____________ 4a _____________ 4b _____________ 4c _____________ 5a _____________ 5b _____________ Total ____________ 2 Name ___________________ 1. Briefly define (24 points): a) Adaptive radiation The process by which speciation occurs as a parent species occupies different ecological niches and adapts to these new environments. Over time, new species evolve from the old ones (eg Galapagos Islands) b) Exaptation A preexisting structure that carries out a particular function evolves to take up a new function c) Unequal crossing-over When during meiosis chromosomes align and separate unequally, resulting in unequal splitting up of genetic material when recombination takes place. The result is mutant chromosomes, which can lead to phenotypic effects. d) Gene pool All of the alleles/genes carried by the members of a given population e) Prezygotic isolating mechanism A mechanism that prevents two different species or groups of organisms from interbreeding by preventing the formation of a zygote. 3 f) Gene flow Exchange of genetic information between populations or subpopulations, resulting from physical migrations or gamete dispersal. Name _________________ 2. About one American in 20,000 has a condition called alkaptonuria, in which the cartilage of the nose and ears is discolored and severe spinal arthritis develops later in life. This condition is the result of homozygosity for a recessive allele. a) Assuming random mating, what is the frequency of the recessive mutant allele in the U.S. population? Show your calculations (5 points) f(aa) = q2 = 1/20,000 = 5 x 10-5 f(a) = q = q 2 = 5 x 10-5 = 7.07 x 10-3 b) What is the frequency of the heterozygotes who carry the normal and the mutant allele? (5 points) q = 0.007. p = 1 q = 0.993. 2pq is the frequency of heterozygotes, 2 x 0.007 x 0.993 0.014. c) Explain clearly why you must assume random mating in order to make these calculations. (5 points) 4 The assumption is necessary if the Hardy-Weinberg equations are to be used. If mating is non-random, the real genotype frequencies will not match the predictions. Name _________________ 2d) Alkaptonuria is about half as common in the US population as albinism, which is the result of homozygosity for a recessive mutant allele at another locus. Suggest one reason for this difference. (5 points) Several possible answers, all of which are fine: 1. There could be stronger selection against alkaptonurics, because they are less fit than albinos. 2. There could be different mutation rates at the two loci 3. The US population could be a mix of populations with different histories, so that the frequencies at these loci in the founding populations were different. 4. Albinism heterozygotes are fitter than average, so that balancing selection has increased the frequency of this mutant allele. 3. You are investigating a population of Lake Victoria cichlid fish in which the males are larger and more brightly colored than the females. This is a clear case of sexual dimorphism. The fish can be kept in an aquarium, so they are easy to study. a) You would like to test whether intersexual selection is responsible for this sexual dimorphism. Briefly outline an experiment to test this hypothesis. (10 points) 5 Design (5 points): Use a tank or a set of tanks in which females have access to all the male phenotypes, keep track of the matings. Some suggested painting males or changing the light, which were also OK answers. Results expected (3 points): If females choose some males preferentially, there is evidence for intersexual selection, which would lead to selection for sexual dimorphism. (2 points): It is necessary to show that no intrasexual selection is taking place among the males, which would influence the female choice. b) The average weight of males in your aquarium population is 100 g. You pick males that weigh 150 grams and mate them with a randomly chosen set of femalesthat have a mean weight of 70 grams. In the next generation, the mean weight of the males is 110 grams, and the mean weight of the females has risen to 71 grams. What is the heritability of male size? (6 points) (more room on next page) Name _________________ Heritability of male size = gain/reach Gain = mean of selected males mean of males in original population = 50g 10g Reach = mean of males in the next generation mean of males in original population = Heritability is 10/50 = 0.2 6 3 c) Why did the female size increase only slightly after selection? (5 points) 5 points. Most of the genes determining male size are sex-limited genes, expressed only in the male. 4 points. For mentioning sexual dimorphism. 3 points. For saying that female size was not selected for. Other answers received fewer points. d) Suggest a reason why your heritability estimate from part (b) of this question is less than 100%. (5 points) Full credit for saying that environmental effects play a role, or for saying that we do not know anything about the effect of genes in the females, since they were not selected for. 7 Name _________________ 4. You have a taxonomist friend who thinks that some birds you and she have been describing are really two species, based on the fact that some of the male birds have red patches on their throats and some do not. You, however, think that they are all the same species, because their phenotypes are otherwise so similar and because they live in the same geographic area. a) What definition of a species are you and your friend arguing about, and why? (5 points) The morphological species definition. Your friend thinks that morphological differences are sufficient to constitute different species. b) If all of the birds were phenotypically similar, is it possible that they could be more than one species? If so, what kind of species would they be? (5 points) Yes, it is possible. They would be cryptic species. c) Briefly outline an experiment that could be used to settle the argument between you over whether there is one species or two (10 points). Mate males with the two phenotypes with a number of females, and look for prezygotic isolating mechanisms such as mating preference. If they pairs do mate, look for postzygotic isolating mechanisms such as infertility or sterility of the hybrids. Carry the hybrids on to further generations, to see if there is any hybrid breakdown. Alternatively, compare DNA samples from the two types of males, to see how different they are. Use as a standard the differences among known species of similar birds. 8 Some students suggested that if the genetic differences between the males was due to allelic differences at a single locus, simple crosses could be used to show this and establish that the birds are all of one species. Name _________________ 5. There are over 1000 genes for odor receptor proteins in the mammalian genome. a) What is the most likely mechanism that produced this large number of genes? (5 points) Gene duplications, with subsequent mutational divergence. A mention of mutation only was worth one point. b) It has been found that about a third of the odor receptor genes in humans are pseudogenes (no longer functional), but that only about a tenth of the mouse receptor genes are pseudogenes. How did these genes become pseudogenes? Suggest one reason why humans have more pseudogenes than mice. (5 points) 9 The genes became pseudogenes because they accumulated mutations that made them nonfunctional. Humans might have more non-functional genes than mice because (a) mice depend more on a sense of smell than humans, so there would be strong selection against losing the genes in mice, or (b) humans have a higher mutation rate than mice at these loci. ...
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This test prep was uploaded on 04/08/2008 for the course BICD 100 taught by Professor Nehring during the Fall '08 term at UCSD.
- Fall '08