Midterm-2-Key - 1 Evolution Name{I E I Fall 2011 Student...

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Unformatted text preview: ' ' 1) Evolution Name: {I E I Fall 2011 , Student number:-____—__ Midterm Exam #2 ' There are 15 questions totaling 42 points on this exam. Point-allocation is given next to each question Your final mark for the exam will be the percentage of points out of 42. The exam is worth 20% of your final course grade. - No notes or tethook can be used during the exam. There" IS a formula sheet at the back of the exam. No assistance between students is allowed. [fa question is unclear to you raiselyour hand and an instructcir will try to help. 7001 wl‘g “Cor wae/[YrV‘e’ \ W[‘- 5 L95“: new LWN .y W9 1.) cars. {Lorrie/1* OVW~$Q US glue/n 120$ (PM; U“ I 0,an 1 _ 7 §W&S\J\ A414, GOA—W"! “‘LM MFL R (/kO-J‘r “LI 0“} CT‘RDW 10m$/ +0 S4019 fluvlg 1A ‘L‘EVV‘V‘S O MAM 111,... W" [1] [a] Identify four processes that cause allele frequencies to change' 1n a subpopulation. (2 points] Wulnl‘iow Sa-[eiz/g-tom ' I @ For“; $0, Wubroxsfi‘ow , Wolf Corral df'flsfl' . 7 _ WSUJQ/f [b] Identify two processes that by themselves do not change allele frequencies, but change'the genetic composition of a subpopulation. (1 point) ' [2] ‘A diploid population consists of 350 mm, 1102 A1A2 and 1548 AzAz genotypes. Assess whether there is evidence for inbreeding in this population. Provide support for your analysis. [3 pOints) N: ESO+ Hozgilswxz‘EOoo \ 3‘ ‘ ‘ Pa: w 2-0130 @341: (“meaty ' 2.13000 _ _ H: "09L 2: 0‘3? '4 7: 01W, ‘94; 010.302 :01 1192 ‘ @000 F."- H I: 0'Q'Q‘FO'3? .10.?2 F303. euldeM-CP‘ {+0 0’99" '- jParerareecQ 15:“) @wo Ll [b -—e F3 (D ~13 HO [3] At two loci the following haplotype counts are observed: 140 A131, 340 Ang, 60 A132, 460 A231. What' 13 the expected» number of A132 haplotypes if the haplotypes were in linkage equilibrium? (2 points] . ; Matteo 0 ~_ 340*“605dcf CPA} ”"3 - 30' 9‘ l 1’52; Mai/O - - {000 ' . I ' ‘fFr/OO: Ng’gz— V41 p32. 2—10 w. 0.2. 0 0 80 [4] An asexual population consists of 0.2 A genotypes. and 0.8 B genotypes. The per capita growth rate of the A genotype is 1. 7 and the per capita growth rate of the B genotype is 2.3. What will the frequency of the B genotype be 1n the next generation? (3 points) _ 1}— [,Jhrg ‘Prs «- Prs ~——--~ «’3‘ w l- (Vile: cook; szh-zze . _ (/3 ,, N4 “ ET; 5‘0 EMF (5 A 50ml+§>5 Oelgo> f3 -—_ 0.2.0.24 +0.5: Jr:- 0968 l ZOrS’lfll ([le :. 0 C5 0.02% [5] In a diploid and sexually reproducing population there is a genetic locus with . two alleles. The frequency of the A1 allele is 0.45 and the frequency of the A2 allele is 0.55. The relative fitnesses of genotypes are given in the table below. I ' __—_ ———I_ [a] What type of fitness model occurs at this locus? [1 point] Helm: anSUJVCL 0900xw‘l‘ofie/ ' (D [b] What is the frequency of the A1 allele'in the next generation? (Answer to an accuracy of 3 decimal places). [3 points) I ‘ :33 T, 0.451.705? 4 .Z-o.t€§«o.95.-| 4' 03,51. one :. 0,0132 ® . ° ' z - - I (D - _ - f ‘ - .90. + 35* .‘Bbq'w: OAS“ .QlDzaasa f -0_Ci€l 0 [c] What is the expected equilibrium frequency of A1? (1 point) . A .‘ -lr 0.0L! ,, . 3 ' " ’ :_‘ 0894 l fail S'l’lz, 0.0344104 I I O l [6] [a] In words, explain what determines the marginal fitness of an allele. [2 points) ' (I) GamHVog 'H/ke/ fitter/{6, OC’Cursi?\ C93 ?r010e~.103]:‘('~[ ~H/ve, wile/[e]: s- {W 0- flCWWLTVG ® ftor O» or (2-) “"(D @ flew“ 'Lfid'F'L‘ Ci> V’CZL) Cb (pm; &‘QCU$SEU~A gig SUv’U‘idorthp J'peceuuckli/ o Ladle/(‘6‘ [b] If the marginal fitness of an allele is greater than the mean fitnessof a population will the frequency of the allele increase or decrease? [1 point] "Morma CD ._ ‘ [7] What processes cause a geographic dine in allele frequency? [2 points) . [' Wlbm-Lri U-A -" sew/(erc/me . [0% OKV‘C/e; f~ 7‘ o-_ "0) [8] Describe a type of frequency dependent selection. Discuss whether this typelof frequency dependent selection can or cannot maintain genetic or phenotypic diversity in a species. [3 points) ' @10ch (as :WW‘ 9-19.95 3 O) (\Doglllr‘wcf - Ewe-7 veer/7* 0- JeVavchG/w L I «Crave/troy éoleb/I.€v0)~9-W'l seine, ‘0“ 'l* S Jag/+30% [9‘ [AU US 2,. [9] The effective population size of humans is about 10,000 to 20,000, yet our current census size is about 7,000,000,000. Indicate and explain a likely reason for why the effective size is so much lower than our current census size. [2 points) “Lc‘he'xltf, r“ €16"sz is. huwoxwg [5049 at SWWll T7O?U(CN-CU‘A. @ lb!" (Big-Cru'fi'ns‘fb'fi ' 05:, \Aus‘l‘orlcwlt‘f . 5W” {laifiufexfl‘lc/M 979’ I? r ”wafiUW—l 992+ \ftw'lF-t) . c”?(>~/~Li‘o~.l olafeu—elifi 0m WEE-g ¥0r och/ex WSWV$I 1 [10] [n lions, about 25% of adults are reproductive males and 75% are reproductive females. If a subpopulation of lions consisted of 32 adults what IS its effective population size? (2 points) NM ’ 0.7/7'31: (g /U4;‘;,0‘7'§.3Zgrch. [11] Is the change in allele frequency of a codominant and deleterious allele with ' selection coefficient 5: -0. 001 affected more by random genetic drift or selection, on ' average, in a population with an effective population size equal to 10,000? Explain why. [3 points] ‘ ’de‘ OJAS we F ‘9 . mtg at A LTargi‘rl {FU'Q % E) ——-————--—_-——-—————— : zxue yaw-tr . (”but A Wye/(cohw :_ ‘ . a ll U" 3.01 «5:9 P) loorlfiCJ/Tf) ax . a: ' :f. 0.) Sme who own age, am. magi“, is» A b‘l 99/[Cc/7UHA {5 A ', splat/Irww l/Wg Oksfga‘ief e/WSr (l) idem“?! 1W3 gaLflciLIUv" COIWZC/‘H‘fi ' - ® CQIFeE/‘R' «@Lwhw [12] Below is the gene tree for a sample of 7 alleles at algenetic locus. Hash marks in the tree indicate mutations. ' What type Of seleCtionmay be occufring at this locus? Supportyour answer with an analysis. (4 points) ‘ ' - Since [‘6‘ h/co-ro 40/ 50>?e’gjrg @- IQV~LO‘V‘C/-l“") 969,0le ‘ 5 O CCU fv’t W5 [13] What type of data can be used to detect a recent selective sweep of an allele in a sexually reproducing population undergoing recombination? Given this type of data, what would be indicative of a recent. selective sweep? (2 points) ' Lew/(5 09V. llhk0>6c9nsefiull£lwium ' O) Mame, L0. as u Wane, 01:00 (gfilflczlwcéwa-e? I . ' be 6 some, I/‘Hb o, bookie) use. ganmrce-g Q? «l, [14] [a] Identify four sources of variation that contribute to phenotypic variation in a population. [2 points] ' ' Agcdfiku e, @&£¢ol’$ gfiwfipa 7c 3M ‘1}me MAI/Vii aOW‘l'AC’MC’e . ' _ ”Z harem-l“ mixer/l OWwfii/ 59"5lrcn9l8 ' - gamut , _ EIWUIUHUWWRI I (Pontifical works £0? W‘l-cwltom/vrusrouliam) 841:. [b] What source of phenotypi’c variation contributes to narrow-sense ‘ heritability? [1 point] ado): Srfiu @ . , [15] Describe two approaches that can be used to estimate narrow- sense heritability. [2 points) ‘ Cg) AW” ‘ F d‘vSé’zbl u—[%uQ\ &ri'p°l\<’—K€V“+*O~I I I. I minim C§§ M60511“); . ® CTZB . gt)": Wonk: viii/l5» W s ...
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