Fa'04 Final

Fa'04 Final - "a) 0 b) .0625 Class): EEC-'13 {Co \ 1...

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Unformatted text preview: "a) 0 b) .0625 Class): EEC-'13 {Co \ 1 Tie, (no t Fall Chico 100 — Fall Quarter. 2004 — Final Examination, Version B Both Siberian and l limalayan snow leopards are white. Siherians have the genotype aamm, and Hintalayans are AAillM. AF leopards do not have melanoqt-‘tes. while on do. M_ leopards can produce pigment in their melanorwtes. while nun cannot. A Siberian with “shovel "shaped ears was crossed to a Himalayan with "dish " shaped ears. The Fl leopards were all white with “leaf” shaped ears. [Melanoeytes are the pigment cells in skin andfurj 1. What preportion ofthe F2 would you expect to be pigmented? a) 3M 13} 9l16 c) ll4 3l16 c) llltS I) ll2 g) [2'8 h) some other number 2. What proportion ofthe F2 would you expect to have leaf shaped ears? (Hint: lllfi of the F2 have “ruffle” shaped ears). a) 3l4 b] 9l16 c) ll4 d] 3l16 e) ll16 f) ll2 g) ll8 h) some other number 3. What is the probability that_an F2 leopard taken at random would be white with shovel ears? c).081 d).25 e) .0625 "@3152 g) Some othernumber 4 (2 pts), What is the probability that the genotype ofa randomly selected white F2 would be AaMM? h) some other number ..._—-- b)9l16 c)ll4 d).154-'c);06 25 01/2 at Its a) .532 b) .035 a) 3n 5. What is the probability that the first offspring ofa mating between a white F2 and a Himalayan would be pigmented? c).25 a) .188 e} .562 n .75 g) some other number - \l 6. What is the probability that the first offspring ofa mating between a white F2 and a Siberian would be pigmented? I ' a) BM b) 9llb __;E)',ll4 d} .154 e] .0625 f) V: g) ll8 h) some other number 7 [2 pts). What is the probability that the first offspring of a mating between two white F2 leopards would be pigmented? a)9l16 b).154 c)d_._095 d).;0625 e) ll2 f)1l16 g} some othernumber 8 (2 pts). What-His the probability that the first offspring ofa mating between two white, leaf eared F2 leopards would be pigmented and ruffle eared? d) answer to 7 times .01235 e) answer to 7 plus .198 fllfinswer to 2 plus 4l9 b) answer to 7 times 4l9 "xej'ia'nswer to 7 plus .01235 f] answer to "l times .198 g) some other number 9 (2 pts). What is the probability that the first offspring of the mating in #8 would be either pigmented and ruffle eared or pigmented and shovel cared? a) answer to #8 plus 2l3 fb} afiswer to #8 times 2l3 JYanswer to 8 times 4l9 d) answer to #8 plus 4l9 e) answer to #8 plus 9l16 /,..f) answer to #7 times ll4 g) some other number 10. Ifthe mating between an F1 leopard and a ruffle eared Siberian produced a litter of? kittens, what is the probability that 2 of the "i would be pigmented and ruffle eared? f) .8125 g) some other number .0594 b).00283 c)21 (1)0426 e).1875 ll. The inheritance of pigmentation described in the intro to this section is an example of a) variable expressivin b) failure of penetrance e)cornplementation djnepistasis e) meristic trait f) linkage g) something else Geneticists on separate continents discovered several recessive temperature sensitive mutants of a tiny diploid worm called C. elegans. All these mutants changed the shape ofthe warm when it was grown at high temperature (35°C). Normally C. elegans exists in graceful curv become straight as a ruler at 35°. Naturally these geneticists wanted es at normal (22°C) and elevated temperatures. but each ofthese mutants to ltnow whether they had all discovered the same mutant, so they mailed each other mutant worms to experiment on. T he results of several experiments are shown below. Allphenotgoic data are obtained at 35”. Cross China 1 1: China 1 x Brazil at __ France it Sweden x Brazil at ,: Sweden 2: Alaska 1: China 2 France _‘ I. Alaska California California Califorrfltii Brazil China 2 Fl all straight all curved ' all curiied ‘ all straight all strai ht all curved” all stflgbt all straight i F2 all straight 50% curved 50% curved I 1% Curved all straight 50% Curved all straight .l% curved l 12 [2 pts). The F1 ofthe France x California cross are all straight because the France and California mutants a) complement b) fail to complement e) recombine d) fail to recombine \3.‘ e) can‘t tell from the information given l3. The F2 ofthc Sweden x California cross are all straight because the Sweden and California mutants a) complement _. b) fail to complement c) recombine d) fail to recombine e) can‘t tell from the information given 14. All ofthese mutants are in the same cistron. T 5/1: 15 (2 pts). How many cistrons are represented by these mutants? 5)} b) 4 fi) 2 d) 0 e) l i) can’t tell U - _..-\ 16 (2 pts). Which ofthe mutants are alleles of China 1? 3.) Alaska in) Brazil c) California (1) China 2 e) France f) more than one g) none 11) can‘t tell 1? (2 pts). Which ofthe mutants are homoallelesgf China 1? a) Alaska b) Brazil e) California China 2 e) France _ f)_more than one g) none .r h) can’t tell \_. 18 (2 pts). Which of the mutants are heteroallelcs of China 1'? a) Alaska b) Brazil e) California d) China 2 e) France f) Sweden g) more than one it) none i) can’t tell 19 (2 pts). Which of the mutants are heteroalleles of Brazil/ifi-afiklaska b) Brazil c) C/alifoi'nia cl) China 2 e) France? f) more than one g) none h) can’t tell / / ‘ » ' “ 20 (2 pts). Fewer curved F2 arise from the Alaska x China-2 cross than from the France x California cross because a) France and California complement while Alaska and China 2 do not la) France and California recombine, while Alaska and China 2 do not c) France and California are in different genes, while AlaSka and China 2 are in the same gene {ED France and California are farther apart than Alaska and China 2 e) can’t tell 21 (2 pts). Fewer curved F2 arise from the France x California cross than from the Brazil x California cross because a) Brazil and California complement while France and California do not b) Brazil and California recombine, while France and California do not c) Brazil and California are in the sme gene, while France and California are in different genes d) Brazil and California are farther apart than France and California e) can’t tell 22 (2 pts). Which of the mutants is most likely to be non-reyertible? a) Alaska b) Brazil . c) California {' l. Pl. d) China 1 e) China 2 1‘) France g) Sweden |._-'h)_'_inore than one i) none 23 (2 pts). The scientist who found China '1 was able to discover which protein was abnormal in this mutant. He called it l _ _ _ . I ' sinuin, and discovered that the 16ttl amino acid was changed from the normal trp to cys. Which ofthe other mutants will have the normal tip at sinuin aa #16? a) all except China-2 b) all except China-2 and Alaska c) all except China-2, Alaska and Sweden d) all ofthcm e) none ofthern f) can‘t tell 24 (2 pts). China-2 have an amino' acid other than tip or cys at position #16 of sinuin. T F 25 (2 pts). The scientist who discovered-the California inutant also was able to isolate an abnormal protein from his mutant. He named the protein linein..w'Li_nein"and"sinuin may be the same polypeptide T \--- _ .—- " ' ' inst _- ‘— 26. The scientist who discovered theBrazil mutant found an abnormal protein and called it rioin. This may be the same polypeptide as linein or sinuin ' F /" 27 (2 pts) The Alaska mutant is defective in a) sinuin h) linein c) rioin d) sinuin and linein e) sinuin and rioin t) linein and rioin g) all three h) none of them i) can‘t tell I 2-8. The France mutant is defective in a) sinuin b)-linein c) rioin d) sinuin and linein e) sinuin and rioin f) linein and rioin g) all three h) none ofthem i) can‘t tell 29 (2 pts). The Sweden mutant is defective in and rioin f) linein and rioin g) all three a) sinuin b) linein c) rioin d) sinuin and linein e) sinuin h) none of them i) can’t tell \ , , I. j)'can’ttell ‘- ' l‘ .- rill ,‘3' i .I _. a /_ A naturalist captured lizardsfron: an area in the desert. She noticed some trait diflerences among individuals. and found that the allele 8 determines presence (and the recessive .6 causes absence.) ofa blue spot on the side, L determines a long tail, and its recessive allele 1 determines short. and C determines absence, and c determines presence of a crest on the head. A short tailed crested male with no blue spot was mated to a long tailedfemale with a lilue spot and '- no crest (heterozygous at the l), l and c loci). T he progeny obtained are shown below. 1/ Wh en spots and crests are not mentioned. it means they are absent.) 24 long-tailed spotted females short-tailed spotted females; . 52 shortifaiiied crested females -_ - _. 23 long-tailed crested spotted females '- 51 short-tailed females " ‘ "7125 short-tailed crested spotted females -— — 1 short-tailed crested spotted male -—— 1 long-tailed female ', 330 long-tailed crested males 7:346 long-tailed spotted males 26 short tailed crested males {'422 longwtailed males _ __ _ ‘__"’42 long tailed crested SpQfiCd males 2 short-tailed spotted males 24 short-tailed males 2000 total 30 (2 pts). How many loci are segregating? a) 2 b) 3 e) 4 d) 5 e) can’t tell " 31. A phenotypic class is niiSsing from the progeny. T F 32. How many independently assorting linkage groups are there? a) l b) 2 c) 3 d) 4 8) can’t tell —" 33. How is sex inherited in these lizards? a) XX females and KY males b) XX males and KY females c) single locus with the male allele dominant \ :1) single locus with the female allele dominant e) some other way-5'" 0 can’t tell ' 34 (2 pts). The order ofloci linked to b is c, sex independent d) l— b, c — sex locus g) some other order h) can’t tell a) l — sex locus — b, c independent 6) b — sex locus — c, 1 independent c)l—b— b) c -— l — b sex locus f) none ofthem are linked to b 35. The order ofloci linked to c is independent d) l — c, b — sex locus' g) some other order h) can‘t tell a) l — sex locus — c. b independent in) b - l r c — sex locus c) l — c — b, sex e) c — sex locus — b, 1 independent f) none ofthem are linked to c 36. The map distance between b and c is a) 0 b) 5 c) 10 d) 15 e) 20 t) 25 (ii/{50 h) some other number i) can‘t tell " 37. The map distance between b and l is a) 0 b) 5 c) 10 d)'15 e) 20 f) 25 g) 50 h) some other number i) can‘t tell " 38. The map distance between c and l is a) 0 b) 5 c) 10 d) 15 e) 20 f) 25 g) 50 h) some other number i) can’t tell 39 (2 pts). The distances between sex and b. c and l are, respectively a) 0, 0 and 50 b) 10, 50 and 5 d) 50, 50, and 50 e) 5, 10, and 50 f) 25. 15 and 5 g) 15, 0 and 50 @Sbme other number c) 5‘ 50 and 10 i) can‘ttell 40. The genotype ofthe female in the original cross was (using F and ffor sex alleles) b) LFBlll fb Cllc C)CLFBllclfb d)FlBlthb Qltc 3) IF BllL fb Cllc e) some other arrangement 41.Howmuchinterferenceisthere? :1) none b) .2 c).4 (dip e).8 f)1 g) some other number A particular cloned anonymous DNA sequence (lets call it "probe ") is 5 kl; long. it is homologous to a sequence located entirely between two variable EcoRl sites. El and E2, which are 9 id) apart. Lets say E l is on the left and E2 on the right. An invariant EcoRl site is 7 kl) to the left ofEl. a second invariant site is 4 lcb to the left ofE2. in the. region homologous to probe and a third invariant EcoRl site is 2 lab to the right ofE2. 42. Let us call the haplotype in which E] and E2 are both present A. What bands will be seen when DNA From an AA homozygote is cut with EcoRI and probed (with “‘probe“)? a) 4 and 5 kb bands b) 5 and 6 kb c) 7, 5, 4 and 2 kb cl} 4 kb only c) 5 kb only f) 4 and 12 kb g) 9. 7 and 2 kb h) 12 and 6 kb i) some other array ‘ noi- j?" 43. Let us call the haplotype in which E1 is present and E2 is not B. What bands will be seen when DNA from a BB hornozygote is cut with EcoRI and probed? a) 4 and 5 kb bands 5 and 6 kb c) 7, 5, 4 and 2 kb d) 4 kb only e) 5 kb only f) 12 and 4 kb g) 9, 7 and 2 kb h) 12 and 6 kb i)some other array 44. Let us call the haplotype in which E2 is present and El is not C. What bands will be seen when DNA from a CC homozygote is cut with EcoRl and probed? a) 4 and 5 kb bands b) 5 and 6 kb c) 7, 5, 4 and 2 kb v d) 4 kb only e) 5 kb only 0 l2 and 4 kb g) 9, 7 and 2 kb h) 12 and 6 kb i)some other array 45. Let us call the haplotype in which both El and E2 are absent D. What bands will be seen when DNA from a DD homozygote is cut with EcoRI and probed“? a) 4 and 5 kb hands b) 5 and 6 kb c) "i, 5, 4 and 2 kb d) 4 kb only e) 5 kb only f) 12 and 4 kb g) 9, 7 and 2 kb 312 and 6 kb i)sorne other array 46. What bands would be seen in EcoRl restricted DNA from an AC heterozygote? a) 7, 6, 5 and 4 kb b) 12, 6 and 4 kl) l2, 5 and 4 kb d) 12, 7 and 5 kb e) "i and 6 kh f) some other array 47 If, in a given random-mating population, 60% of the chromosomes have the E1 site and 30% have the E2 site, what would be the frequency ofthe A haplotypes? a) .18 b) .12 c) .42 d) .28 e) .25 0 some other number 48. What would be the frequency ofthe AA genotype? a) .18 b) .0324 c) .424 d) .25 e) .1764 f) some other number 49 (2 pts). What proportion ofpeople would be BD heterozygotes? a) .36 b) .1 18 {c} .235 d) .18 e) some other number "- 50. What would be the least frequent genotype in this population? a) AA b) BB c) CC d) DD e) AB 1') AC g) AD h) BC i) ED j) CD k) can”! tell 51 {2 pts). What would be the expected frequency of AA x BC matings in this population? a) .0016 b) .0033 c) .0065 d) .5 e) .0625 1‘) some other frequency Two normal parents produced a girl with hemophilia {X—iinked recessive). Assuming non-pateritinl is ruled out. there are afewpossibie cxpianationsfor how she could have been born with hemophilia. 52. A possibility is a) mom was heterozygous (11h), and there was a new h mutation on the X dad gave the girl 13) inom was Hh and dad was hY c) dad‘s X had a paracentric inversion d) dad's X had a pericentn'c inversion e one of these 53. Another possibility is-//a')’the girl is 45,X because of nondisjunction in morn ..b) she is 45,): because of nondisjunction in dad c) she is 47.}(XX and has Hhh d) she lost both X chromosomes during embryonic development, so has no copy of H e) none ofthese 54. Yet another possibility is @)-.one of her Xs experienced an inversion during embryonic development 13) her two Xs underwent crossing over within the h locus c) she is Hh, but by chance the X carrying the H allele was inactivated in the majority ofher cells (:1) she is Hh, but in her h is epistatic to H c) none of these ' 55. Which ofthe explanations you selected in the above 3 questions would you pick if you were told the girl was exceptionally short? The one from question a) 52 b) 53 I e) 54 d) none e) can’t tell 56. Given that the girl is exceptionally short. she is probably autozygous at the hemophilia locus. T F 57. Given your answer to #55, how likely is she to have hemophiliac children? {3- half her kids will be hemophiliac b) half her sons will be hemophiliac c) it depends on her mate’s genotype d)"she can‘t have hemophiliac kids e) can’t tell A geneticist isolated DNA from a strain ofE. coii that was histidine+ tryptophan+ methionine+ foiate+ and chic to use lactose as a carbon source {lac+). She used this DNA to transform a strain auxotrophicfor his. trp, met, fat. and tmahie to use lactose. She then piated equal sized sampies oftransfot'med ceiis onto various media, with the results ShO‘H'ti. / Media in which Inc is indicated have lactose as the mic carbon source. {fine is not indicated, giucose is the carbon source. medium contains Fcolonies fol lae met tin 1 SS. Which locus is furthest from the lac locus? a) fol b) his I fol his lac tfll c) met cl} tip 6) can't tell his lac mm fol his lac met 59. Which locus is furthest from the fol locus“? a) lac h).his c) met (1) trp e} can’t tell 60 (2 pts). What is the order of loci? fa)--fo‘l:his—lae—n1et-trp ,bfl'ac- his-mct~ttp~fol ' c) his-lae-met—fol-trp d) met—lae—fol-his-trp - - - - i" ' -. '\ l C‘ e) mct~fol—lac—rrp-hts i) can’t tell |i\_u't.\i' at, '5 -. i- " \ i _ . _ _ a. r" i: w r The ceHs represented at rt gin were observed in a tissue sotnpt’e from u prev-tonst unknown species ofduetr. YA - E-‘S‘L‘ I I :I I It 61. Cell I) in the figure at right is in the first ofthe stages represented. What is the rder in which the other cells would oeeflurf.’ ,. aficEFGH b) EFCBGAH ,efrBAGCHE. ."‘d) BQFEGAH e)OEEBACGH i) can't tell I / 62. The process being represented by these cells is limited to germ cells F 63. These cells were derived from a a) male b) female e) can’t tell The X chromosome in this species is a) acrocentric b) metacentric e) submetaeeutric (1) can’t tell 65. How many pairs of autosomes are present in the karyotype of this species? a] 3 b) 2 c) 1 9 d) none e) can‘t ten \1 . \_.I.\ . -. ' . ‘ . . - x I- r' I --../-. - (if I " ,."‘,-._J-,:.' i . | 556, Which cells are haploid? a) all ofthem b} H only c) D, E, F and H _‘.d)/A, C, G and H e) E, F, G and It) A and H only g) E and H only h) none i) can‘t tell '— r 67. Which cells contain single stranded DNA? 1" i '5') all of them b) H only c) D, E, F and H d) A, C, G and H e) E, F, G and H f) A and H only g] E and H only h) none i) can’t tel] 68 (2 pts). If this duck was heterozygous for a paraeentrie inversion in an autosome, and crossing over occurred in the loop, which cell would show a bridge? (Write letter corresponding to cell on answer sheet) 69 (2 pts). If this duck was heterozygous for a reciprocal translocation between two autosomes, which cell would show a quadrivalent? (write letter corresponding to cell on answer sheet} The N terminoiportion oft: normttiprotein is inet—ser—voi—{ys-pro-iie-iy.s~vnt..._ (The dots show the protein continues) A tnutttnt form ofthe protein has the complete sequence met—cyS-ser-in-thr-osp. 70 (2 pts). What type of mutation is this?I a) missense b) nonsense e) silent d) deletion frame shift e) insertion frame shift f) can’t tell 71 (2 pts). Counting the A of the initial AUG as position #1, at what position in the mRNA is the abnormality located? a)4 c)6 d)? e)8 f)9 g)lO h)ll i)l2 "£2. The abnormality in the mRNA involves a a} A b) T e) C d} G e) U i) can’t tell ...
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This note was uploaded on 09/18/2010 for the course BICD 100 taught by Professor Nehring during the Fall '08 term at UCSD.

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Fa'04 Final - "a) 0 b) .0625 Class): EEC-'13 {Co \ 1...

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