2-011 Exam 3 KEY F07

2-011 Exam 3 KEY F07 - 2:011 Fall 07 Exam 3 110 pts/2hr KEY...

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Unformatted text preview: 2:011 Fall 07 Exam 3 110 pts/2hr KEY Page 1 If you have any comments on keyed answers (e.g. alternative correct keys) you must submit them to the appropriate instructor (Dr Frankel or Dr Klahn) immediately if these comments are to be taken into account in grading. Part I Genetics Questions 1a. Draw below a diagram of a tetrad in prophase of the first meiotic division. You need not show crossovers but you should be sure to include the centromere(s) (2 points). a a A A (b, c, and d count one point each) b. How many chromatids are there in a tetrad? four c. How many chromosomes are there in a tetrad? two d. How many DNA double helices are there in a tetrad? four e. Now assume that an Aa heterozygote is undergoing meiosis. In your diagram (above), label the A and a alleles, placing them correctly relative to each other on the components of the tetrad as it might appear prior to crossing-over (2 points). --------------------------------------------------------------------------------------------------------------2. The genes for long aristae (L) and brown eyes (B) appear to be unlinked. That is, when appropriate crosses (e.g. l+l+b+b+ llbb) are performed, 50% of the testcross progeny are derived from recombinant gametes produced by the l+lb+b doubly heterozygous F1. Yet the L and B genes are both located on the same (second) chromosome of Drosophila melanogaster. Indicate the procedure that geneticists working with Drosophila use in order to arrive at that conclusion. You need only describe the principle behind the procedure, not any detailed results. It is helpful to draw a labeled diagram to assist in your explanation. (3 points) They performed a series of crosses involving genes that were situated between L and B on the same chromosome, e.g. L X, X Y, Y Z, and Z B. L x Y z B l X y Z b 2:011 Fall 07 Exam 3 110 pts/2hr KEY Page 2 You will need the genetic code table reprinted below in its RNA version(from Campbell & Reece, Biology, 7th ed. p. 314) to answer question #3 3. The base triplet CGA on the coding strand of DNA corresponds to the CGA triplet of mRNA which codes for the amino acid arginine. a. Give below one example of a single base-substitution in each category. Use the DNA version (T, not U) of the genetic code. (1 point each):: A synonymous substitution: CGA CGT, CGC, or CGG or AGA A missense substitution: CGA GGA or CTA or CCA or CAA A nonsense (chain-terminating) substitution: CGA TGA b. Which of these base-substitutions would be likely to have the most severe phenotypic effect? Nonsense (chain terminating) (1 point) Why? (2 points) It would result in premature termination of a polypeptide chain, with only a fragment of the normal protein being produced. 2:011 Fall 07 Exam 3 110 pts/2hr KEY Page 3 4. Match the regions of the cloning vector shown below to their use in the generation of recombinant DNA: The regions: (A) Ori: origin of replication (B) MCS: multiple cloning site (C) ampicillin-resistance gene D) LacZ: gene from lac operon that encodes beta-galactosidase (also includes an i gene and p and o sites). In the following four items, enter the letter of the correct answer from the above list in the blanks (this is a "matching quiz": each of the above regions is the correct answer for just one of the following four items). (1 point each; 4 total) 1. The region with multiple restriction sites for insertion of foreign DNA B 2. The gene that allows selection of plasmid vectors that have succeeded in infecting an E. coli cell C. 3. The binding site for E. coli DNA polymerase. A. 4. The gene that allows us to test for whether or not foreign DNA has been plasmid vector D . inserted into the 2:011 Fall 07 Exam 3 110 pts/2hr KEY Page 4 Section II. Evolution Multiple Choice. 2 pts each. Fill in the dot on your Scantron form corresponding to the best option for each question below with the letter by that option. For security, also circle the letter on this form. 1. Which of the following is not a necessary requirement for evolution through natural selection? a a. Individuals can change phenotypes during their lifetime to adapt to changes in the environment. b. Parents that differ from the average phenotype produce progeny that share those differences. c. Differences in phenotype can cause differences in reproductive success. d. More individuals are born than the environment can support. e. All of the above are required for evolution through natural selection. 2. Which is the smallest unit within which evolution is possible? A single a. cell. b. individual. c. population. d. species. e. community. 3. The Uniformitarian Principle states that a. alleles of a single locus occur in equal frequencies in natural populations. b. the lengths of all geological periods are equal. X c. physical and chemical laws have operated in the same way over the earth's history. d. the fitnesses of all alleles at a locus are equal. e. linear macromolecules at the origin of life contained equal proportions of left and right-handed enantiomers. 4. One inference we can make from the Uniformitarian Principle is that a. all populations in a species have equal potential for evolution. h b. because geological process act slowly now, the earth is very old. c. the duration of all geological periods will be the same. d. the rate of evolution through natural selection will be the same at all loci in a species. e. all species in a taxon have evolved equally from their shared most recent common ancestor. 5. Which of the following statements is most correct? Genetic variation is maintained in populations by a. directional selection and mutation but lost through random drift. b. directional selection but lost through random drift and mutation. c. migration and random drift but lost through directional selection. : d. migration and mutation but lost through random drift. e. mutation and directional selection but lost through random drift and migration. 6. Under which of the following circumstances would the `biological species' concept not be useful to determine if individuals in different populations are members of different species? The two populations a. show only small phenotypic differences. b. do not have a fossil history. c. are sympatric. d. both have high levels of genetic variation. e. only reproduce asexually. you can't use an interbreeding criterion if there's no breeding 2:011 Fall 07 Exam 3 110 pts/2hr KEY Page 5 7.. The most likely result of prolonged decrease in population size of any population is a. increased adaptation to the local environment. b. increased migration rates to other environments. p c. loss of rare alleles from populations. thru random genetic drift d. loss of common alleles from populations. e. reduced rates of mutation per locus. 8. Which of the following characteristics is not an advantage for the use of molecular sequence data in phylogenetic constructions? a. Any sequence includes a large number of characters (positions). b. Different sequence positions are not equivalent in their probability of change. c. Some sequences are found in all groups of living things. d. A given sequence tends to evolve at the same rate in different species. e. All sequences are subject to mutation. 9. A variety of experiments have been conducted in which the environmental circumstances of the early (pre-life) Earth are simulated. The important result from these experiments is the observation that, under these circumstances, a. small organic molecules are readily formed even in the absence of living cells. b. atmospheric oxygen was essential for the origin of living organisms. c. life can evolve only if free energy spontaneously increases in chemical systems. d. enzymatic catalysis is not necessary for replication of DNA. e. enzymatic catalysis is necessary for replication of DNA. 10. Phospholipid bilayers are suggested as components of precellular living systems because such bilayers a. increase the rates of catabolic reactions. b. increase the rates of anabolic reactions L c. assemble spontaneously in water. d. actively transport compounds from one side of the bilayer to the other. e. occur in nature only in left-handed enantiomers. 11. Layered clays may have been a site of catalysis in the origin of precellular life. Catalysis in such clays occurs because the clays a. increase the rate of movement of molecules in solution : b. restrict the movement of molecules to two dimensions. c. generate heat through abiotic reactions. d. exclude oxygen. e. move precursor molecules to positions of highest reactivity. 12. In the RNA world hypothesis for the origin of precellular life, it is argued that living systems first evolved through forms in which RNA was used for information storage and catalysis. It is argued that DNA then `took over' information storage from RNA. Such a takeover would be selected for because DNA a. can fold into 3-dimensional tertiary structures while RNA cannot. b. is more efficient at catalysis than is RNA. c. has more kinds of bases than RNA and therefore can encode more information. a d. readily forms double helices that are less likely to be broken down than RNA. e. can form complexes with polypeptides more readily than RNA. 2:011 Fall 07 Exam 3 110 pts/2hr KEY Page 6 13. The significance of organelles in the evolution of eukaryotes is that organelles a. are essential for sexual reproduction. b. produced the oxygen that was required for eukaryotic life. D c. increase rates of movement of metabolites within the cell. d. provide resistance to dehydration. e. increase the number of chromosomal copies available to the cell. 14. Which of the following sets of taxa correctly summarizes the "5-Kingdom" system of classification of life? a. Viruses, archaebacteria, eubacteria, plants, animals b. Monera, eubacteria, plants, animals, fungi p c. Monera, protista, plants, animals, fungi d. Archaebacteria, eubacteria, protista, plants, animals e. Archaebacteria, eubacteria, plants, animals, fungi 15. Which of the following species, practicing sexual reproduction, could not produce heritable genetic variation in its progeny? a. a triploid (3n) species b. a tetraploid (4n) species c. a diploid species in which there is only one chromosomal type (n =1) d. a species that is a hybrid of two different parental species. e. a species in which there is no crossing-over between homologues. NO KEY!! This question was misworded and was therefore excluded Use the diagram of a 'standardized' plant life cycle below as needed to answer the next 3 questions. A is a mature sporophyte; the order of events in the life cycle follows the arrows. The number of individuals shown in each stage of the life cycle is not intended to communicate useful information! A E B D C 16. Which stages are diploid? a. A only b. A and B c. C only d. B, C, and D e. E and A 17. Which of the following characteristics is not a correct difference between algae and land plants in the expression of this life cycle? In algae, a. photosynthesis occurs in both stages A and C, in land plants in only A or C. b. cells at stage B all have the same morphology, in land plants they can be different. X c. cells in stage D have different morphologies, in land plants they are all the same. d. cells at stage D can disperse, but in land plants they are mainly enclosed in stage C. e. All of the options above state a correct difference between algae and land plants. 2:011 Fall 07 Exam 3 110 pts/2hr KEY Page 7 18. In seed plants, ecological dispersal is undertaken by stage(s) a. B, D, E b. B only c. D only a d. E only i.e. the fertilized embryo, or `seed' e. A only 19. The evolution of life stages in the land plants has increased enclosure of some life stages inside parental structures. The main selective pressure for this evolutionary trend is a a. dehydration of gametes b. dehydration of adults c. predation on gametes d. predation on adults e. competition for resources by all life stages. 20. The evolution of ribs and vertebrae in the chordates was necessary for a. herbivory a b. sinusoidal swimming c. filter feeding d. burrowing lifestyle e. respiration using gills 21. Which of the phylogenies a through e is congruent with phylogeny `X'? ( A, B, C, D, E are species). A B C D E E D C B A X D E C A B a b E C D A B A B C D E E D C B A c d e 22. If parasite populations are restricted in range to the bodies of their hosts, the phylogeny of parasitic species will be congruent with the phylogeny of their host species. This prediction follows from the argument that a a. speciation of the host prevents gene flow between parasite populations. b. the host and parasite populations are subject to the same mutations. c. the parasite population will decrease the fitness of the host population. d. parasitic species are strongly selected to overcome their ecological dependence on the host species. e. the host and parasite have shared a common ancestor and therefore share limits of evolutionary response inherited from that ancestor. 2:011 Fall 07 Exam 3 110 pts/2hr KEY Page 8 2:011 Fall 07 Exam 3 110 pts/2hr KEY Page 9 Section III. Evolution Short Answer. (1) You may continue answers on the backs of sheets if you identify those answers by number and show us where to find them. (2) Show Your Work on any problems. That is, insert the correct numbers in the correct places in the correct formulae. If you show your work, you do not need to calculate the final result unless it is needed in a further calculation. Show your complete line of reasoning on all "explanation" questions (3) Answers to these questions are not intended to be long or complex. Usually an adequate answer will require one good sentence for each point the question is worth. 1. (4 pt) Word matching. Insert the best word from the list below beside each allantois dorsal nerve cord jaws sympatry amnion notochord vertebrae anisogamy coelom heterospory isomorphy pharyngeal gills phoresy yolk sac chorion lungs plasmid lungs_or chorion__________ analogous to stomates of plants note: as I pointed out in lecture, lungs, chorion, and stomates are all enclosed gas exchange structures (thus resist dehydration) but pharyngeal gills are not enclosed. Since this question has two correct answers already, I don't feel I should accept a third. amnion____________ hydraulic cushion for embryos heterospory___________ results in gametophytes of different genders phoresy_______ transport by animals 2:011 Fall 07 Exam 3 110 pts/2hr KEY Page 10 2. (5 pt) There are three genes (A, B, C) in diploid fruitflies, each with a dominant allele (respectively, alleles A, B, or C) and a deleterious recessive allele (respectively, alleles a, b, or c). The three genes are located on different chromosomes, as follows: -- the A gene is autosomal. -- the B gene is located on the portion of the Y chromosome not homologous to the X chromosome. (in flies, females are XX, males are XY) -- the C gene is located on the portion of the X-chromosome not homologous to the Y chromosome. You set up a population in which each allele at each locus has the frequency 0.50. You now expose this population to selection against the deleterious recessive alleles for five generations. Assume the fitness of the recessive homozygotes is the same across all 3 genes. a. (2 pt) Which of the following options best predicts the expected decline in frequency of these 3 recessive alleles in this 5 generations of selection, from greatest to least decline in frequency? (circle letter(s) by best option) i. a, b, c ii. b, a, c iii. c, b, a **** iv. b, c, a. b. (3 pt) Justify your answer to part (a.) above by completing the section below for the b and c alleles. (don't do a vs b or a vs c, because the same ideas apply in all three comparisons). Allele __b____ will decline in frequency faster than allele __c____ because: Because the B gene is always haploid, the `b' allele can never occur in heterozygous form. It is always expressed in the haploid form and therefore is always exposed to selection. Because the C gene is diploid in females, the c allele can occur in heterozygous form in females in at least some generations therefore will not be selected against in those cases. not needed: [the A gene is diploid all the time so an `a' allele has the highest probability of all 3 alleles of occurring in heterozygotes, therefore the least chance of being selected out.] 2:011 Fall 07 Exam 3 110 pts/2hr KEY Page 11 3. (7 pt) The autosomal T locus in cats has two alleles, T and t. It controls tail morphology, as shown by the table below. Fitness values of these phenotypes are also listed. genotype phenotype W(fitness) TT straight (normal) tail 0.80 Tt bent (kinked) tail 1.00 tt stub tail 0.50 In a population of 150 cats (100 female, 50 male) you find the following tail genotypes as a function of gender. Allele frequencies for females (p = freq T, q = freq t) are also shown. Female Male TT 60 5 Tt 10 10 tt 30 35 p 0.65 q 0.35 0.20 0.80 e.g. freq(T) = p in males = ((2x5)+ (1x10))/100 = 0.20 a. (2 pt) what are the frequencies of the T and t alleles in males? Enter them in the table above. Show your work! (if you cannot get those values, check here ______ and use p, q = 0.50 in all further questions) b. (3 pt) Assuming these cats mate at random, predict the relative frequencies of the following phenotypes in their progeny (show work): b. kink tail (0.65)(0.80) + (0.20)(0.35) this will be equal to the sum of the frequencies of the two ways to get a heterozygote; that is, prop of kinks = (freq(T) in males x freq (t) in females) + (freq(t) in males x freq (T) in females) stub tail (0.35)(0.80) [= freq (t) in females x freq (t) in males] (note: taking average allele frequencies across genders will not work because males and females are not present in the same numbers; average allele frequencies would be biased towards females) c. (2 pt) Calculate the fitness of a `T' allele in the females of the population above. (hint: with what gamete must a `T' allele in a female fuse to form a zygote?) = 0.20(0.80) + 0.80(1) in homozygotes in heterozygotes --use male allele frequencies because an egg has to fuse with a sperm, not another egg, to make a zygote 2:011 Fall 07 Exam 3 110 pts/2hr KEY Page 12 4. (10 pt) Farmer Jack has two insect crop pests, aphids and cutworms. Aphids are small (0.01 gm) and have a generation time of two weeks; cutworms are larger (10 gm) and reproduce once a year. Aphid density is usually about 1000 times that of cutworms. Both feed on the same crop, and there is food for both all the time. Adults of both species disperse for short distances (within fields) by flying. Before 1990 Jack never used chemical pesticides, but in 1990-95 he treated twenty 160 acre fields with pesticide `X'. All individuals of both pest species were equally exposed to X. Treatments were repeated several times at the same dosage from 1990 to 1995. X degrades rapidly (within 1-2 weeks) in nature, so do not worry about it accumulating in the population. Initially the populations of both pests exposed to X dropped drastically. Aphids went extinct in 3 fields and cutworms in 15 fields. However, by 1995 the aphid populations had returned to their pre-1990 densities in the 17 fields in which they survived, but the cutworms remained very scarce in the 5 fields they still occupied. a a. Explain (two (2) reasons) the difference in population response to agent X between aphids and cutworms. Use only the 'facts' given above. Treat the fields as being isolated from each other and from all other outside effects. 1) because aphids have much larger population sizes than cutworms,(1 pt) there was a much greater chance that each aphid population would include a few individuals that had a few alleles for X-resistance(2 pt) Therefore aphids survived in most fields, cutworms didn't 2) because aphids have a shorter generation time (1 pt), therefore higher reproductive rate(1 pt) , the population of X-resistant aphids can return to high densities in a short time interval.(1 pt) b (4 pt). Jack treated all of each field with agent X (see fig 1). Farmer Jill applied X to fields of the same size, following the same application schedule, but in each field she sprayed only the center, leaving a 'surround' of unsprayed land in each field (fig 2). The average time course of aphid densities (aphids/acre) in the sprayed portion of each field in which they survive is shown in fig 3. fig 1 farmer Jack farmer Jill fig. 2 fig. 3 X X not Xed . aphid density * Jack's fields Jill's fields ** * * * * * * * * 1990 1995 Using only the facts above, explain how Jill's sprayed fields suffered lower aphid infestation in 1995 than Jack's did. Assume each field is totally isolated. By not spraying the surround, Jill maintained a population of non-resistant aphids. (1 pt) Aphids from these populations continually dispersed into the sprayed region. (1 pt) Because pesticide resistance (almost any phenotype) carries a metabolic 'cost' (1 pt) nonresistance would tend to spread in the resistant population, only to be selected against when X is applied. (1 pt) .The balance of these two forces maintains some nonresistance in Jill's sprayed regions. (1 pt) (4 pt max) This question was inspired by current Federal guidelines on genetically engineered crops. Jill is obeying the law, Jack is not--look where Jack ends up! 2:011 Fall 07 Exam 3 110 pts/2hr KEY Page 13 5. (6 pt) The vicariance model of speciation is diagrammed below. In this model, a large parent population P is split into two 'daughter populations' (A, B) by a geographic feature such as an enlarging lake. A and B are each about half the size of the parent population P. Over time A and B evolve into two different species, A* and B*. A A P Lake A* BB time, and population movement B* B* a. (2 pt) Individuals of any population cannot disperse across the lake. Why is the absence of dispersal important for speciation in this model? absence of gene flow between populations means that unique features (changes in allele freqs due to mutation, selection, or random drift) that evolve in each cannot be transmitted to the other population; therefore, differences accumulate between them. b. (2 pt) There are different environments on the different shores of the lake. What important effect does environmental differences between lake shores have in this model? adaptation to different environments is a source of differences between A and B lines. c. (2 pt) Imagine that, as shown, the ranges of A* and B* overlap, so individuals of these two related species may mate. The reinforcement hypothesis argues that there is selection for intrinsic barriers to interspecific mating, such as species-specific mating signals, at this point. What is the selective force for such barriers? Individuals that mate with heterospecifics (i.e. A with B, B with A) produce hybrid progeny that are generally subviable or subfertile; individuals that don't make such mistakes have higher fitness because their progeny can survive and reproduce. 2:011 Fall 07 Exam 3 110 pts/2hr KEY Page 14 6. (10 pt) Protistans are single-celled eukaryotic organisms. Two species of protistans (A and B) have fluorescent 2-membraned organelles (FA, FB respectively) that contain a small chromosome. A scientist suggests that the fluorescent organelles (FA, FB) evolved as prokaryotic endosymbionts of the two protistan species from a free living fluorescent prokaryote species (R). To test this hypothesis, she measures divergence between A, B, FA, FB, and R. Her data set is 100 shared bases in an intron of a protein-coding gene found in the genome of all these taxa. Her results, as number of differences between each pair of taxa, are shown in the difference matrix below. R FB FA B A 52 52 52 32 B 52 52 52 FA 24 12 FB 24 a. (4 pt) Use the difference matrix above to build a rooted phylogeny for these 5 taxa. Draw it below. Do not include branch lengths. Use only these data to build the phylogeny. Cluster A and B, on one branch. Cluster R , then FA and FB on the other b. (2 pt) what assumption did you have to make to determine the position of the common ancestor of these 5 groups? Assume this condition is true for all remaining questions. The rate of evolution in the intron is the same across all branches of the phylogeny. c. (2 pt) Does the phylogeny support the hypothesis that the fluorescent organelles (F) are prokaryotic endosymbionts of the two protistan species? circle one yes no can't tell from these data justify your answer: yes: the organelles originate on the same branch as `R' d. (2 pt) The scientist further hypothesizes that species A and B evolved fluorescent organelles independently of each other. That is, fluorescent organelles are a convergence in A and B. The data support that hypothesis. Explain how they support it. According to the D values, which are directly proportional to time, R arose more recently than did either of A, B, therefore F (which evolved from R--see part c) cannot have been present in the mrca of A and B. Rather, each had to acquire F later, and therefore independently of each other. 2:011 Fall 07 Exam 3 110 pts/2hr KEY Page 15 7. (6 pts) New (extra) copies of genes in a gene family are generated through three processes: polyploidization, unequal crossing over, and replicative transposition. a. Each of these three processes predicts a different pattern of positions of the gene (family) copies within the genome. Briefly describe the predictions for any two of the processes. (3 pt) Note: You can answer verbally and/or use clearly labeled diagram(s) --whatever is easiest and clearest. You don't have to describe the mechanisms, just the predictions. unequal crossing-over all gene copies will be found in a row on one chromosome polyploidization copies of the gene family will be on different chromosomes, but the homologous genes within the family will be in the same order replicative transposition --new copies will be scattered through the genome at random positions b. Describe two(2) points of functional significance of new extra gene copies--that is, what extra function can they potentially confer? (3 pt) (1) increases transcription rates of the mRNA coded by the gene; (2) if one copy is inactivated (e.g. by mutation) other copies will still maintain the function (3) extra copies can evolve totally new functions ...
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This note was uploaded on 03/28/2008 for the course CHEM 2-011 taught by Professor Larsen during the Fall '07 term at University of Iowa.

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