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24 - 43 Let us-c'all the haplotype in which E1 is pic'sent...

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Unformatted text preview: 43. Let us-c'all the haplotype in which E1 is pic'sent and E2 is not B, What bands will be seen when DNA from a BB homozygote is cut with Eco'RI and probed? a) 2hand 5 lib band-s '-_'b_) 5 and 6 kb 5) 7‘, 5. i1 and 2 kh d} 4 kl: only e15 kh only f} lland 4 kb 9'9, 7 and 21:13 h]. 12 and 6 kb i)some other away- '44. Let us call the haplotype in which E2 is present and El is not C. What bands will be seen when DNAfrom a CC 'hmnozygote is cut with EcoRI and probed? a) 4 and 5 khbands b) 5 and 6 kb c.) '7, 5, 4 and 2 kh . d)'4 kb only e) 5 lib only i)" 12 and 4 kb g] 9, 7 and 2 kb h) 12 and 6 kh ijsorne other array '45. Let us call the h‘aplotype in which'both El and E2 are absent-D. What bands will be seen when DNA firom a. DD homozygote is-eut with EcoRI and probed“? -a14 and 5 kh hands _ b): 5. and 6 id: (:17, 5, 4 and 2 95b £114 kb only e15 his only f1 12. and 4 kb g) 94 7 audit kb 11112 and 6 kb ijs-orne other array d6. Whittbancls would be seen in EcoR-l restricted DNA from an AC heteroz-ygote? a} '7, 6, 5 and '4 lib b) 1'2, 6 and 4.l~':b i:-1_ 1'2. 5_ and 4k]: d1 12, 7 and 5 kb e1 7 and 6 kh 1) some other array '47 ll; in a given random-mating population, 60% of the chromosomes have the E31 site and 30% have the E2 site. what wouldsbc the 'fiequcncy ofthe A"haplotypcs".’ a1 ..18 b1 .12 c1 .42 d] .28 e) .25 {1 some other number 48'. What would be the frequency of'the AAge‘notyp'e‘? a) .1-8 h) .0324 c1 .424 d) .25 'e) .1764 f) some other number "49 (2 pts]. Whatproportion ofpeoplewould be BD heterozygotes‘? a} .36 b) .l 18- {91.235 (11 .18 e) some -'0 that-number " 5.0. What would be theleast frequent genotype inth‘is papulation? alAA bj'BB c1021. ('11 DD e1AB one g1AD 1013c 0131) jtco kjc'an‘ttell 51 (2 pm). What would be the expected freqt‘lency of AA x BC matings in this-populatio n‘? a) .0016 h) .0033 c) .0065 d) .5 e1 .0625 11 some other frequency Two normal pnrems produced a girl with hem-Opii'iii'a (FY—linked. recessive). Assuming n'ompa‘temi‘nl is ruled-out. there are rife-w possible atpia-nntionsfor how she-could have-been born with hemophilia. '52. A possibility is a)'inorn was hetcroZygo'u's (1111'), and there was a new 11 mutation on the X' dad. gave the girl . 'b') mom was HIT-and dad was hY c1dad‘s-Xhad a paracentrie. inveision d) dad’s 15; had aperitentn'c iii-Version e one of these - 53. Another possibility is-I/{cllthe girl 'is 45,X because of nondisjunction in mom Rb) she lsflix because of nondisjuncti'on in dad (:1 she is=47.XX‘X and has Hhh 6) .she lost both X phi-amosomes during. embryonic developmenhso has no copy of'H e) none of these 54. Yet another possibility is @'-_).-.onc of her X5 experienced an inversion during embrylanic development '0) her two X5. underwent crossing over within the h locus c) she isHh, butby chance the X carrying the H allele was inactivated in the majority ol'her cells d) she is H11, but in her 11 is epistatic to H is) none of'thes'e ' 55. Which of‘the explanations you Selected in the above 3 questions would you pick ifyou were told the girl was exceptionally short? The one from question 'a) 52 h) 53 E1 54 (31) none e] can‘t tell '. , . . 56._Gi\-'en that the girl is exceptionally short= she is probably autozygous atthc hemophilia locus. T F 517.. Givcnyou-r answer to #55. how'likely is she tohave hemophiliac children? 61‘ half her kids Will behemophiliac- b1 half her'sons will he hern'ophil'iac c) it d‘i'qae‘ndsdn her mate’s genotype d) 'she'can‘thave hemophiliac kids e) can‘t-toll A geneticist isolated DNA from a strain off. calf. that was hisfidino+ impiapimn-l" meihionine+fiflaié+ and able to use lac-rose as a carbon source finch). She used this DNA to transform a strain amorrophiefor his. trp, met. foi. om! iambic to use lactose. S-hethgn plated equal sized sorirpies ofonnsfqrrwd cells onto satiric-MS media, with the results shown. ...
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