2003Exam

2003Exam - Bl206/282 - Exam 1 Page 2 February ‘13, 2003 l...

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Unformatted text preview: Bl206/282 - Exam 1 Page 2 February ‘13, 2003 l ,. Form 0 1. "For a self-fertile:rga‘nismwith:thegenotypeéeéléBbCofinxwhich A, B, and C are _ @7 1/64 1 x, 4 X M I ‘ B. 1/16 rt ”‘ ”* (qr/«1X ma ‘lw C. 1/4 " D. 1/2 E. 3/4 2 Among the crosses below, which WI” give a ratio of genotypes? A AA-BBX aabb 441% B AaBb x AaBb 11-1—5“ c. AaBb X. aabb A 0» it 9' AW" 3 «Q. _ AaBBX AaBB A A El» 595 5w“; ,E AAbb X aaBB Pm i5“? qa (Mo Goa. loin 3. Anallele is: A. one of the bases found in DNA . @2 one of the variant forms gene found within a species 8‘. one of two sister chromatids D. the term for the same gene found in different species B. the dominant form of a gene 4. AlbiniSm, the lack of pigmentationjn humans, results from an autosomal rmene (a). Two parents with normal pigmentation have an albino girl. What is tlfiprobability that their neLtchilg will be a normally pigmented girl? é) 3/8 A“ M M‘ B 1/4 40mm":; A w 3- 2;: r «it 1 . . fl i E 3’4 07100513 '"" 6 Bl206/282 - Exam 1 Page 3 February 13, 2003 _ , , Form 0 5. Drosophilarthewdominaneallele“ Cy. (Cur/y) resultsrin curlyr wings. The cross Cy Cy+ x CyCy+ results in a ratio of2 curly : 1 wildtype F1 progeny. The cross between the curly F1 also gives a ratio of 2 curly ; ] wildtype F2 progeny. Which hypothesis(es) is (are) consistent with‘th‘is observation? /l. the curly wing phenotype is controlled by one gene “ll. 25% of the F1 and F2 generation are absentV V Ill. the Cy allele is dominant for curly wings and recessive forieihallly t/ A. l B C D. l and ill @ all of the above \ may W54 lab M r» l; a a bi ‘ 1 Assume that a cross is made betweenIAaBd and/gabbtplants and the offspring i r in the following numbers: 106 Aabb, 94 aaBb, ~1§ ’2 Aa‘lifi} andfiaggpfi? These suits are consistent with the followmg circumstance: L 2 l A. sex-linked inheritance with 30% crossing over me)» “why B. ,linkage with 50% crossing over ((3 linkage with approximately 33 map units between the two gene loci D: independent assortment I E. 100% recombination (109*,4 Q egg-g = ‘30?) ‘ 3% ’0‘“? ‘M v? Ho 7', 7. What types of ratios are likely to occur in crosses dealing with a sigg gene pair A 9:3:3z1. 9:3:4 ‘ B. 1:1:1:1,1:4:6:4:1 @. 3:1,1:1,1:2:1é-‘ D 9:7,13:3 E 15:1,12:3:1' Bl206/282 — Exam 1 Page 4 February'13’, 2003 Form 0 8. The following coat colors are known to bedetermined:by-r'alleles at one locus in horses ' ’Palamino = golden coat with lighter mane andtail Ciérnsllg almqstlwhite Chestnut: brown The following table gives ratios obtainedmfrgm. vanefles: “various matings between the above Ratio of - all oremello all alamino M .___.. I'O en Mr AA oremello X oremello AA chestnut X chestnut o. a oremello X ohestnu palamino t not palamino ><l % Lama it’j Mum”; b f3 “*6 rite-aw «Max 3 one gene with two alleles showing incomplete dominance Bf two genes showing reCessive epistasis a. one gene with two alleles; one being dominant, the other recessive E5. two genes showing no epistasis E. one genewith' two alleles; one begin recessive lethal 9. all—white Crosses? L \i“ W -~ (P60) A. - {LA A'b' /B' 1 golden: 1 white A ’lola @sz X“ 1/ C. V 6‘ (k D. A & B 88w " om k ‘ pad“); 21* 'B&C Ld‘l‘é‘bu éxf/lla aflw (513 [BB *4!» a at; ax CL: ~55 AA (5b '5“ ma, lob aaeov >‘ [A A J , \ldw’ A l lk LQ~~MVWZIYW. ,,l,...., f7" 5.1 WWW WWW flwwx WWW (. y}m% tgjliwor \ - Bl206/282 —‘ Exam 1 Page 5 v Februar 112332003 Form 0 ' ’10. ialn‘f’nosphilafftzforlcefdibristles are'caused‘by an. and X-linked recessive mutation and brown eyes are caused by an autosomal recessive mutation. A female fly that is homozygous for normal bristles and red eyes is crossed to a male fl with fork ‘ ’and brown eyes. The Flare‘crossed to produce an F2 generation. Which of the following predictions are likely regarding the F2 population: \(l. 100 % of the females will be normal bristled \/ll. brown eyes will be found equally between males a females /lll. disregarding the sex of the flies, a 923:3:1 ratio will be seen It. B. H ’ mgr-«w! thflwr X EMF 0 M. C I” W {MK 6M 1“ E, g l and I” w) at all of the above ‘9 2 .XP 3 WEE >.x/va c ,. Be 3% X mo 3 >4 % E (11. - 12.) The following five possible answers are to be used for questions 11. and 12. Only one answer is correct for each question, however answers can be used more than once. A. prophase l of meiosis B. metaphase | of meiosis C. metaphase of mitosis AND metaphase ll of meiosis D. anaphase l of meiosis E. anaphase of mitosis AND anaphase ll of meiosis 11. When do homologous replicated chromosomes separate and move toward opposite poles? D 12. When do sister chromatids begin to separate? 8 \ aBl206/282 — Exam 1 Page 6 “(February‘13f2003 . . , , Form 0 13. For the followingapedlg’reeywhat istthe 'mostéa-ilikelyr-moet'of inheritance. Assume the trait is rare in the population: 125% 4% fl ‘fléfl fag, M *" A?" a Autosomal dominant X-linked dominant None of the above Vi” In Drosphila, sex IS determined by a balance between the number of sets of autosomes and the number of X chromosomes » x ll. inhumane, segg is determined by a balance between the number of sets of autosomes and the number of Y chromosomes will. MY is, male in Mans and female in fruitflies z I is" B: II E. ill (I) l and m E. all of the above ‘Blzflwflfizvfifixam 1 ‘ Page 7 Februar "£2.13 112003 v . ~ Form 0 . r(15. andétliSE-XA"ho~mozygous2strain of yellow corniscrossedwitha homozygous strain of purple corn. The F1 are crossed and produce 119 purple kernels and 89 yellow kernels. 15. Use the chi square test to determine if the F2 ratio fits better to a 3:1 ratio or to a 9:7 ratio. In addition, based on your answer, predict how many genes are segregating in the cross? A. A 3:1 ratio fits; one gene is involved B. A 3:1 ratio fits; two genes are involved C. A 9:7 ratio fits; one gene is involved CD) A 9:7 ratio fits; two genes are involved E. Neither a 3:1 nor a 9:7 ratio fits and thus the number of genes involved cannot be determined. g 4. ( ' ‘ 205 degrees of 0.05 . W134 ‘0 T) A chisquare table is given at left (\n— 55W y, (EELEW as l NOT 7» l W m, ‘3; 0|}? ’ L a 7 1 (V‘llfi'lflj'+ (gal/fl) __‘ [01$ 7w “'7 q I Mfg? 16. Regardless of your answer to the previous question, assume that you find X additional evidence supporting a the 3:1 hypothesis. Which phenomenon could help explain the data. P x A interference W W A: CB} reduced penetrance ‘, 9 E". unequal crossing over M l Q, E. incomplete-eleminauae W 36 E. epistaeis no r 17. Wheffifollowing drawingsrepresents the X‘chromosomes of a particular female fruitflyyouare working with. One chromosome contains a recessive mutant allele of the carnation gene (can and dominant mutant allele of Bar-eyes (Bar). This chromosome also‘has a constriction that is microscopically visible. The other X chromosome has wild-type alleles of both carnation (can) and Bar-eyes (Beth) and the chromosome has piece of the Y chromosome attached to one end. 3 Bar W After mating this female to wild-type males, a portion of the male offspring have wild— type carnation eyes and mutant Bar-eyes. What is the structure of the X chromosome , in these male flies? A. 02$ Bl206/282 —» Exam 1 Page 9 February ’13,‘ 2003 Form 0 (18, 19‘ and”20)i"Female“Drosophila;"heterozygous for the three recessive mutations spineless (sp), hairless (h), and claret (cl) were testcrossed. Among the 1000 progeny from a trihybrid F1 female that was testcrossed, the following phenotypes were observed: spineless 318 74? his}; §me 2» claret, hairless ‘3tgfigfl.h~d” claret, spineless 130 sf m C‘ > WWW“ Ct hairless 14Qfir’f l" hairless, claret, spineless gm???» Azwa 3r wild type 38 set" M ‘ hairless, spineless 12 g9 V.“ “J‘ waste. “WVL " L‘ claret ~ 18 9‘91” if U 18 Which gene is in the middle? A spineless B. claret LC). hairless There is no answer D or E for this question 19. What is the value of interference in the cross? wily/113:1” 7 50cm alrlgtgflf ., ,0 CM mm W“ A 1 a aln "Wigs ‘ B 0.74 (9 mm C. 0.67 s :2 "’“M 3 ,05 D. 0.34 amt" \ ' :5: Oh “m . t E 0 xi 2 \, 4,; » o Llofifl‘b , If two tthybLids of the type used above are crossed, what is the probability of ining phenotypically W198? Remember that male fruitflies do not undergo combination. hawks; ct dani- A. 0.10 ' “CV/ shd k 6, ht’C‘r/g r’nu e@ 0.175 Si 5 l’ l9 c. 0.35 D. 0.3855 E. 0.7 BI206/282 — Exam 1 Page 11 Eebruar "313 2003 Form 0 21. You cross a'"‘ll/IA'Ta trp4 yeast strain tora MATa h‘is3 strain to create a diploid which you then induce to sporulate. You dissect 100 tetrads and obtain 3,2.P.D, 35 T and 3WD. Which of the following statements is/are correct conclusions: “I. trp4 and hiss ar® *ll. trp4 and his? areJinked no recombination has occurred between‘Lrg4 Offlfifi and a centre mere @ I B. ll C. III D. I and lll E. all of the above 22. If adenine makes up g; "/0 of the nucleotides in a given sample of double- stranded DNA, then guanine would make up what percent of the bases? A. 22 °/o ’l a B. 3 % 7ft: fa c 40 % u” 3t @ 27% W 9:14;) E. 15% 23. DNA is replicated through a process that is: A conservative B liberal C. dispersive @ semi-conservative E compassionate 24. When resolution of a Holliday structure (a recombination intermediate) occurs by making the second set of nicks on different DNA strands than the first pair of nicks, this would result in: ' recombination between two outside, markers no observable recombination between outside markers no heteroduplex formation no gene conversion no meiosis WPOWQ Bl206/282 — Exam 1 - Page 12 February 13, 2003 Form 0 25. Below is shown a recomination intermediate between two non—sister chromatids. Using the letters on the figure, inidicate which area contains heteroduplex DNA. 26. in your organic chemistry lab you have developed a new low-calorie sweetener. Being a trained molecular biologist (as well as an excellent organic chemist) you decide to test the mutagenicity of your new sweetener before exclusively licensing the sweetener to Starbuck's. You decide to use the Ame's test which tests for the ability of your chemical to increase the reversion rate of a his‘ mutation in Salmonella. N W; aded to his‘ b distilled water distilled water + rat liver enz mes sweetener sweetener+ rat lier enz mes Which conclusion is most consistent with this data A. The sweetener and’its conversion products are not mutagenic, so start learning how to say 'thank you' in Swedish 8. The sweetener and its conversion products are equally mutagenic 8. Rat liver enzymes are highly mutagenic AD. The sweetener is more mutagenic than its conversion products ® The sweetener is not mutagenic, but its conversion products are mutagenic Bl206/282 — Exam 1 Page 13 Februar 13 2003 Form 0 (27. and 28.) You have isolated ten new tryptophan—requiring mutants of yeast. in order to place them into complementation groups, you make diploids for each combination. A "+" indicates growth on unsupplemented medium. indicates absence of growth on unsupplemented medium. 10 A. B. 5 © 3 D. 1 E. o 28. Mutant 10 fails to complement mutants that represent more than one complementation group. Which hypothesis(es) is (are) consistent with this observation? V I. vmutant 10 contains mutations in two different genes )4”. mutant 10 is a member of the same complementation group that 3 is V Ill. “mutant 10 could be a deletion which removes at least two genes A. | B. ll C. lll @ l and m E. all of the above ;.:~.B‘l206/2'82 éiExam 1 Page 14 \ . ~7‘Febwar “1.3;:2003 » Form 0 29. LaMVhich rof the following statements-vislaretrue regarding “a complementation test? /l. VOnly recessive alleles can be tested >< ll. if two mutations are in the same gene, they can complement if the mutations/are at different ends of the gene /lll. A large deletion that removes two adjacentgenes should fail to complement single base mutations in either gene A. l B. ll C. III ©. l and III E. all of the above ‘30. Below is shown a map of a series of deletion mutants of the rI/A gene (or cistron) of T4 phage. The region deleted is indicated by the size of the box. For example mutant 1272 deletes all 6 regions (A1 - A6) of the rl/A gene, while mutant PT7 deletes only region A4 - A6. g“ r/lACiStron x! \L Q} l\) deletion mutants z: ¢w You have isolated a single base mutation in r/IA and you cross it to the above deletion mutants. You find that you can only generate wild-tyge recombinants with the mutant is crossed with strain A 105. In which region of the rll A gene does your mutation lie? \ A. A1 B. A2 C. A4 - {9. A5 E. A6 » feBlZfiGZZBZ’t-afixam 1 ' Page 1.5 ‘Eebnuar. “231.1,3’42003 ‘ F0rm£0 i:31. 'iwl'n'i'N‘eurospbra, compounds A, B, C,‘ D, and E are-“chemical‘components of a biosynthetic pathway. Four mutations were found that affected this pathway and they were tested for their ability to be rescued by the various compounds. The table below showsr’the results "of “the exeriment. “+” indicates growth; “—“ indicates no growth. media su lement W A 1 B. 2 r9. 3 we}? WM D 4 E none of the above 32. Several different antigens can be detected in blood tests. The following four traits were tested for each individual shown below. ABO type IA and /B codominant, irecessive Rh type Rh+ dominant to Fih’ MN type M and N codominant X9“) type Xgm” dominant to Xgl‘?) All of the blood types are autosomal except for Xgla) which is X linked. Mother Which, if any, of the alleged fathers could be the real father? A. Alleged father 1 B. Alleged father 2 Alleged father 3 D. Alleged father 4 E. None of the alleged fathers are possible é-iv‘ziBl206/282 '— Exam 1 f‘aFebizuar 2003 H__ : FormO W 33. *er'th'e fruitfly the yellowbody(y)~mutation andt'thercarnation eye (car) mutation map 65_cM_apart on the X Chromosome according to published reports. A friend of yours decides to determine the distance herself and scores the male progeny from cross between a female heterozygous for both mutations and a wild-type male. What map distance will your friend obtain for the distance between the two mutations? A. 0 CM 25 CM «(V/1 50 cM D. 62.5 CM E 125 CM co " i=- The components of of a Wde: ( base, sugar, and phosphate 3 tryptophan and leucine base,- amino acids mRNA, tRNA and rRNA phosphorus and sulfur ;/ WSW. END ...
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2003Exam - Bl206/282 - Exam 1 Page 2 February ‘13, 2003 l...

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