Unformatted text preview: 33. Assume Elvis hasone copy ofthe normalallele. What is the probability his other copy is the polydactyly allele? a) pb)
p/(l .4p) e) .4pt(1.6p) d) .913 e) .lp f) zero '3) some other number i1) can’t tell All theeopies ofthis gene'ean be divided into three types. ﬂﬂﬂp are penetrant P and .411 are non—peaetrant P.
Elvis‘s other'eopy is not penetrant P, but could be either ofthe other possibilities. h" The'tabie shows iheﬁequeneies ofMN bioad groupiogoes in sompfes, ofpeopleﬁom n few different populations. PopulationL If}; typeM type M? . L type N
A L . I 45 210 L 245
B ul— LITCL 90 a t , , 420 —»—. L 93 _
c: #er .252 336 L 112 r“:
1'.) '1" .490 ”300 I L— 96
E j r if: 250 350 L 259 1' 34. Which of the populations are random mating? a) A and C b) A only e) 151,13 and Ed) B, Cand D e} I) only 1‘} B and D. g) some other combination A and C are the only ones in which the genotype frequencies fit. p2 MM, 2pq MN and q1 NN. 0 3'5. PopulationA an) a possible case of inbreedin g be‘cause 'ﬂty'pe' MN) is less than 2pq b) a possible case of inbreedingi beoause ﬁtype MN} is more than 2pc] e] random mating becausefﬂtypelﬂ) 5:331? Square
because f(ty_pe N) is the square root of KN] e] none of the preceding. ' v) 17/3}; Population B is a} a possible case of inbreeding, beeauseﬁtype MN
0 because f(type MN) is more than 2pq‘$) random mating because f[type‘l‘i) is thesqnare of EN) :1) random mating because fag/pom is thesquare root of f(N) e) none of the preceding There is an excess of helerozygotes 37. Population C is a) a possible case ofinbreeding,
because f(typ_e MN) is more than 2pq c)_' random mating : n lor Lad’6 because fityp‘e N) is thesquare root of KN) e}'none of the pree'eding “i in an" t“? l‘léjsi’
\j/ in 6,. 113:1?“ r11} of RN) d) random mating
SchA k is?“
Kb 3} is less than qu h) a'possib'le case ofinbreeding, because ﬁfyp‘e MN)is 1655 than 2pc] b) a possible ease of inbreeding,
because f(typ"e N) is the square of RN} cl) random mating _ . Population D is a) a possible ease of inbreeding. because f(typ_e MN) is less ghan qu'b) a possible case of : breeding, because fitype MN) is more than lpq a) random rn mating because [(type Ellis the square root of f(N) e) none of the preoeding Q9. Population E is a) a possible ease of inbreeding,
inbreeding, betauSe' fIt'ype MN) is more than que} ran mating because f(typ'e N) is the squareroot of HM) e) none. of the preceding ating‘beeauseﬁtype N) is the squareot"f(_N) 'd') random because f(type MN) is less than 2pq b) a possible case of
dom mating because KtypeN) isthe square of RN) d) random Q43. In population A, if the average inbreeding coefﬁcient was .2, the "number of type MN people'wouid have been
a) mm b.) 1.68 c121i d) 1'99 e) 415 f) 260 g) imijossible to 'tell (11430qu ...
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 Fall '08
 Nehring

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