2004Exam

2004Exam - ; cremello X chestnut "it; ’3‘ Abl/...

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Unformatted text preview: ; cremello X chestnut "it; ’3‘ Abl/ ANSW’LLS mullet Bl206/282 — Exam 1 (LR L Page3 February 12, 2004 ATLQ LO Q T Form 2 1. The following coat colors are known to be determined by alleles at one locus in horses: Palomino = golden coat with lighter mane and tail Cremello = almost white Chestnut = brown The following table gives ratios obtained from various matings between the above varieties: Al‘a ' palomino X palomino 1/4 chestnut; 1/2 palomino; ' PM Ark 1/4‘ cremello Am; ' A, Which hypothesisE gest explains the results? (3) B. c. o E one gene with two alleles showing incomplete dominance two genes showing recessive epistasis one gene with two alleles; one being dominant, the other recessive two genes showing no epistasis one gene with two alleles; one being recessive lethal 2. One form of colorblindness results from an X-linked recessive mutation )6. Two parents withnormal vision have a orblind son. What is the probability that their next child will be colorblind? ' . A. 1/2 )i" x B. 3/8 C. 1/8 . D. o x“ i ‘ X @ 1/4 is Bl206/282 — Exam 1 Page4 Februar 12 2 04 Form 2 (3. and 4.) Genes A, B, and C are independently assorting and control production of a black pigment as indicated in the pathway below. Assume that C produces an inhibitor that prevents the formation of black pigment by destroying the activity of the enzyme produced by B. A B colorless —-—————> colorless —————-—> black T C (inhibitor) Q,’ co'lorkrSS 3. A colorless individual of genotype AA BB CQ is crossed with a colorless at 0x 7 wt individual of genotype woo individual giving a colorlessfl F1 are crossed lob together. What is the ratio of Epigrfiss to black in the F2. ' A. 1 black : 63 colorless i?‘ Mr B loll, c B. 37 black : 27 colorless W05 ’4 qu/Sf 7 L C. 3 black: 1 colorless * ,MUL it A" B“ D D. 27 black : 37 colorless Wbcc b — ,3 V ® 9! black : 55 colorless , ( “LDC MW LO 1 v ' (Wt/41”” " M 4. Based on the information given above, a different cross is done in which a colorless A_A_BB_C_C individual is crossed to a colorless AA bb cc individual giving a colorless F1. After crossing F1 together, the F2 progeny segregate 3 black to 13 colorless. What conclusion can be drawn regarding gene interaction? A b shows recessive epistasis to C or c B. c shows recessive epistasis to B or b — a C. B shows dominant epistasis to C h A A B ‘5' L’ ‘3 ® Cshows dominant epistasis to B V ,5 m a W x b Low,“ 5; E. no epistasis is seen ' 1 ‘ W6C“ Ath in a UL V doth/tell mat/LO ( Ll} mac 3 I C 34 " . 6 i 5 _ l K WEE CC‘WVQ _ T , own. BI206/282 — Exam 1 Page5 February 12I 2004 Form 2 5. In Drosophila a mutant strain has plum-colored eyes. A cross between a plum- eyed male and a plum-eyed female gives 2 plum—eyed : 1 red-eyed (wild type). A second mutant strain of Drosophila called stubble has short bristles instead. of the normal long bris/tlgs. A cross between a stubble female and a stubble male gives 2 stubble : 1 normal bristled. What will be the relative proportion of phenotypes in the adult progeny of a cross between twowplum-eyed,“stubble-bristled-flies. (Assume that the enes are located on different autosomes. g ) \A M Vt M M i>£ .m m ', \ W A 3 plum-eyed; normal bristled 7' W I V? W 2 red-eyed, stubble “‘7 . S ‘9 1 red-eyed, normal bristled 34‘ v mu» -‘ VL'VHQ , z - r d B. 1 plum-eYed, normal bristled 7 fim‘w Q he gt; 2 plum-eyed, stubble1 red-eyed, stubble 8‘9 x 1 red-eyed, normal bristled A. 6 plum-eyed, stubble C. 2 plum-eyed, stubble 1 red-eyed, normal bristled ‘ A: D. 9 plum-eyed, stubble 3 plum—eyed, normal bristled 2’ H, 7i \ He not 3 red-eyed, stubble 83 lg 53 W “Y Yul“ 5+ 1 red-eyed, normal bristled Mm V? g l y‘aka WQQWU/NM /l/ 7' -\7\ Wm @ 4 plum-eyed, stubble >aw‘130‘0”w¢' ' W‘gg W85 i3?” mm ‘ 2 plum-eyed, normal bristled ‘0 H V 2 red-eyed, stubble AA H” m i” 1 red-eyed, normal bristled AAWfi (WW/q) ; i/w, W’f‘hfi‘l‘lyé.‘ 2%; (6. 7.) The following five possible answers are to be used for questions 6. and7. Only one answer is correct for each question, however answers can be used more than once. ’ A. prophase l of mitosis B. prophase l of meiosis C. prophase ll of meiosis D. anaphase of mitosis AND anaphase ll of meiosis E. anaphase l of meiosis 6. When do homologous chromosomes pair and undergo recombination? B 7. When do sister chromatids begin to separate? D BI206/282 — Exam 1 Page6 February 12, 2004 Form 2 8. Which of the following statements is true regarding sex determination: In Drosophila, sex is determined by a balance between the number of sets of autosomes and the number of X chromosomes It. In humans, sex is det ined by a balance between the number of sets of autosomes and the numb r of Y chromosomes XXY is malz in humans and female in fruitflies A. I 2'“ B. H C. III {I2} I and III E. all of the above 9. If a genetic disease is inherited on the basis of an autosomalmdonli‘nant gene, one expect to find which of the following? " m . If/a_c_hild,has the disease, one of his or her parents and grandparents also has he disease Affected mothers never have affected sons t B. C. In all cases, if both parents are affected, all of their offspring have the disease D Affected fathers can only have affected children E. Neither of his or her parents will have the disease. 10. In Drosophila, two red-eyed, long wingedflies are bred together and produce the offspring given in the following table: ‘ phenotype of progeny Females Males red-eyed, long-winged 3/4 3/8 red-eyed, vestigial-winged 1/4 1/8 white-eyed, long-winged 3/8 white-e ed, vesti ial-win ed 1/8 What are the genotypes of the parents of these progeny? W+ V+ C. X"*YM/*W and ’XWXIl W‘W ® QWY v' v and X“ X” v“ v \E;’ X‘w-r ¥v and . xm; XE twin —o\fW'\ Bl206/282 - Exam 1 Page7 February 12, 2004 Form 2 11. Assume that a cross is made between AaBb and aabb plants and the offspring occur in the following numbers: 106 Aabb, 94 aaBb, 52 AaBb, and 48 aabb. These results are consistent with which of the following hypothesis regarding the two genes: A. sex-linked inheritance with 30% crossing over é linkage with 50% crossing over ‘ . linkage with approximately 33 map units between the two gene loci D. independent assortment with no epistasis E. fix? ““ $.31”? 100% recombination {2 Wfifl‘i‘ WK '53 file 12. A geneticist, in assessing data that fell into two phenotypic classes, counted values of 250 and 150 for the two classes. She decided to perform chi-square analysis to help distinguish between three null hypotheses: a) a 3:1 ratio; andW ratio. What‘ can she conclude about each hypothesis based on chi square testing. . 'l y L) l 'I l A. AlLthLeenuuhypothesis can be rejected by l \ 0 B. None of the three null hypothesis can be rejected C. can be rejected Wan be rejected 1:1 and 3:1 can be rejected A chi square table is given at left: degrees of 0.05 freedom prob h 00 L400 6 l ‘7 iii/ll “SM n“ ‘tct l" A é¥:\ 2 i” w m a" \ \o‘jlb flab!" , ‘33 7) BI206/282 — Exam 1 Page8 Februar 12 2004 Form2 13. Use the following two-point recombination data to map the genes listed. Pick the map that best aproximates the data. ene pair — distance in map units ene pair — distance in ma units a,b — 50 Si b,d — 13 be — 50 m a,d — 38 c,d —] 50 b,c — 50 (m 2 K 13 ’ 50 B 13 38 8 7 9/ 13 50 7 v 8 t?” 50 8 7 50 fl, 8 7 50 13 14. For the following pedigree which genetic model best explains the mode of XV inheritanc [k Aga 0‘ KM ' CI 0- IOI/O _ OOIOIIO Autosomal recessive ‘ac... . BI206/282 — Exam 1 Page 9 February 12, 2004 Form 2 (15. and 16) Female Drosophila, heterozygous for the three recessive mutations black (b) body, dumpy (dp) wings, and hooked (hk) bristles were testcrossed. Among the 1000 resulting progeny, the following phenotypes were observed: .rfi in $2.; 16. "@051? , . V "A; L} dumpy 304 yw Mag CW Wynd black, hooked 3017l ,..-,é.x2.§—;b;§¥\~;; x t; wild type 169 7 t 9.0% $12,137“ black, dumpy, hooked 172 ~ M, "I it {1+2 H % black 19 7 mass at???“ ‘ W-G dumpy, hooked 21 59 i9 talc 5 hooked 8 . V2,)“ Wt Wm, ‘ dumpy, black 6 7 ‘15") é {j b h 1+ apwfihtfi— \lfl‘i lYZ‘l Ifi-rz Which map best represents this data? 19.3.; 5.4 hk ’ b 35.5 p b 34-1 dB v 4.0 hk Wm“? t; hk 35.5 b 5.4 dp :‘fiml we“??? SW 7385‘; 35.5 b 5.4 hk M dp 34.1 ‘ b 4.0 hk W What is the value of interference in the cross? 1 ' 5r46/1000 0.73 . \_ z r- (0-3‘557Lomsq7 0‘“ 0. 19.15% Bl206/282 — Exam 1 Page10 Februar 12 2004 Form2 (17. and 18.) Neurospora crassa of the genotype age 0 are crossed with Neurospora of genotype Lb + and the following tetrads were scored. Number of tetrads scoring with a given pattern a+c abc ++c +bc ab+ a+c a + c a b c a + c a b c a b + +b+ +++ ab+ a++ ++c +b+ 137? 141: 26 25 2 3 a ’ C V 17. Which genes are linked to each other: \7-; N P 9 \3‘1 + l‘tl 77 7, ' ® a and c (79 B. b and o lo — C ,1 C. aandb \31 +3 7,77‘L1‘+1 D. all three genes are linked to each other E. each gene is on a different chromosome cl “ ‘0 fin '7"; \Hl *1 18 Which of the three genes is most clgswelymflnmkwed to its centromere? A. a l/ o .. v a. b Jail 3233. o c V D. all three genes equally close to their centromere m to r, E. no centromere linkage can be detected for any gene ” 2J3"; 0.5 O 19. Below is dihybrid yeast cell going through prophase l of meiosis. A crossover occurring between chromatids 2 and 3 is shown. To generate an NPD tetrad which two chromatids would need to be involved in second crossover? Bl206/282 — Exam 1 Page 11 February 12, 2004 Form 2 20. If the guanine + cytosine make up 54 °/o of the nucleotides in a given sample of double-stranded DNA, then adenine would make up what percent of the bases? A. 46 % 37 % M M ,i/@ 23 % ‘ D. 13 % E. 15 % 21. Which of the following statements is/are true regarding heteroduplex DNA? Hegroduplex DNA is formed when non-sister chromatids exchange strands 41‘.” Mismatches in heteroduplex DNA will prevent gene conversiorx Heteroduplex DNA forms on both chromatids involved in the croSsover A. l B. II C. III (E) l and ill ». all of the above 22. Below is shown a recombination intermediate between two non-sister chromatids. Indicate which area contains heteroduplex DNA. Which of the following mutagens typically results in deletions or insertions? alkylating agents such ethyl methane sulfonate hydroxylating agents such as hydroxylamine base analogs such as 5-bromouracil intercalating agents such as proflavin deaminating agents such as nitrous acid mccflows Bl206/282 — Exam 1 Page12 Februar 12 2004 Form2 24. The bacterial repair system that corrects mismatched bases after DNA synthesis is able to discriminate between the old and newly made DNA strands because: A. DNA polymerase is attached to the new strand B. older DNA is more likely to contain errors @ older DNA contains methyl groups at specific sites D. newer DNA contains uracil in stead of thymine E. the old strand is pink and the new strand is blue 25. Which of the following statements is/are true regarding a complementation test? . Only dominant alleles canbe tested )( If two mutations are in the same gene, they can complement if the mutations m p > 0.001 cM apart. I. A large deletion that removes two adjacent genes should fail to complement single base mutations in either gene A. l B. H g? I” ,. landlll E? all of the above 26. in Drosophila, mutants a, b, c, d, e, f, and 9, all have the same phenotype: the absence of red pigment in the eyes. In pairwise combinations in complementation tests, the follow results were produced, where "+" indicates the wild-type eye color and "-" indicates the mutant henot pe. -nnnn nun—— u-— --—— n---—— u----— -----— Mi l-l------ \Zi LEV How many different genes are identified from this analysis? ’ WAGON—L A. Cg; E. Bl206/282 — Exam 1 Page13 February 12, 2004 Form 2 27. Below is shown a map of a series of deletion mutants of the rI/A gene (or cistron) of T4 phage. The region deleted in strain is indicated by the size of the open box. For example mutant 1272 deletes all 6 regions (A1 - A6) of the r/IA gene, while mutant PT1 deletes only region A4 - A6. PK rII Acistron deletion mutants You have Isolated a single base mutation in HM and you cross it to the above deletion mutants. You find that you can generate wildztype‘recombinants from crosses of your mutant A105, Ragga 131:], fiand 1‘2_4_1,‘but not with 1272. In which region of the r]! A flgfine dOes your newly isolated mutation lie? AA. A1 B. A2 o. A4 D. A5 E. A6 Bl206/282 - Exam 1 Page14 Februar 12 2004 Form2 28. For a self-fertile organism with the genotype AaBch in which A, B, and C are unlinked genes", what proportion of the mes produced will be a b c? U 1/8 J m B. 1/6 1.3 l C. 1/4 (a l : i D. 1/2 E. 1 29. If A is dominant to a, the phenotype of AA and Aa are indistinguishable. Which of the following methbds could be used to determine the genotypic identity of an organism that shows the A phenotype. ® test cross- ll. chifiq-u’are test Gfi progeny test /2(. | /B./ ll 1 lll I . l and III E. all of the above An ‘ 4A 30. A purple-flowered plant is crossed with a white-flowered pea plant. All of the F1 plants prod’uoe pur l f 5. When the F1 plants are allowed to self-pollinate, 401 of the F2 plants have purple flowers and 131 have white flowers. What are the genotypic ratios of thei1 and 52 generations? (P is used to represent thedominant allele and p is used to represent the recessive allele.) . wapu * WW“ 9 l“ (’1? ‘A. F1 - all PP, F2 -1PP:2Pp:1pp :\ vwwzu (’3 ® F1 -all Pp; F2-1PP:2Pp:1pp \ “we F1w-1PP:1'Pp; F2 - 3PP:1 Pp «1 L101 Vwi‘oUl 2‘3 W l 1:), F1 - 1PP:1 pp; F2 - 3PP11PP ‘1 ‘1‘“ E. F1 - all Pp; F2 - 2sz1pp 3MP” w 31. Which of the following crosses is expected result in 4 phenotypic classes of progeny found equal proportions? @ AaBb x aabb B. AaBb X AABb K AaBb X AaBb /D./ Aabb X aabb )2 AABB x aabb ’ t sawtfi. is w Bl206/282 — Exam 1 Ni Y7 S ‘ {SEW Page15 5 Februar 12 2004 ‘0‘0 Form2 32. A normal Drosophila has both brown and scarlet pigment granules in the eye, which account for the wildfiypaeymolor. brown (b) is a recessive allele on g; Ci chromosome 2 that when homozygous, results in the absence scarlet so the eyes ‘9 J appear brown. scarlet (st) is recessive allele on chromosome 3 that whe’nm g _ . homozygous results in scarlet eyes because of the absence of brownfwflies ‘ 5‘9 31 make no pigment and thus have @ejgyks. A pure breeding brown-eye fly is lob x 53 crossed to a pure breeding scarlet-eyed fly to create an F1 generation and then male 4/ and female F1 flies are crossed to create an F2 generation. What is the phenotypic ‘9‘? 53" ratio predicted for the F1 and F2 ratios? F1 - all white-eyed; F2 9 white-eyed: 3 brown-eyed : 3 scarlet-eyed : 1 wild type {Biz F1 - all wild type; F2 9 wild type : 7 white-eyed C F1 - all wild type; F2 1 wild type : 1 brown-eyed : 1 scarlet-eyed : 1 white- .B.’ F1 - 1 scarlet-eyed : 1 brown-eyed F2 3 wild type : 1 white-eyed (E) F1 - all wild type; F2 9 wild type : 3 brown-eyed : 3 scarlet-eyed : 1 white- eyed ‘ we 3T 37 v» BBStst; Bio STst 33. Consider the three independently assorting gene pairs Aa, Bb and Cc each of which affects a different body characteristic. In each case the upper letter signifies the dominant allele and the lowercase letter the recessive allele. The cross Aa Bb Cc x Aa Bb Cc is performed. What is the probability of obtaining zmthat show the dominant phenotype for exactly two of the three genes? (Assume no epistasis.) fl ‘3 A. 9/4 LP/LO (3/ ’17 (’h )1 3’ 16L . 9/16 a/ a, l/ , @ 27/64 .11 J ’3 -- . ‘2/3 1 T T a. 37 E. 1/64 “i 34. In humans, the three alleles l", [3, and i constitute an allelic series that determines the ABO blood group system. Which of the following is not possible? A. an 0 child from both parents having A blood type an 0 child from both parents having B blood type an 0 child from one parent with O and one parent with AB blood type B”. an A child from both parents being having AB blood type WE. an 0 child from one parent with A and one parent with B blood type END ...
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2004Exam - ; cremello X chestnut "it; ’3‘ Abl/...

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