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Fall 2009 Bio. Gen. Genetics Test 1

Fall 2009 Bio. Gen. Genetics Test 1 - F091 WW 7 2 1...

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Unformatted text preview: F091 WW 7% 2 1. Multiple Choice Questions " . \The localized DNA sequences of two individuals are shown below: Individual 1: AAGGCTCAT TTCCGAGTA .T 2% 5’ Such differences in the localized sequences between individuals can be used as /—...________, milestones in mapping anal sis, also known as fingle DNA markers Single DNA sequence markers c. Single nucleotide polymorphismsu/ d. Simple sequence length polymorphisms e. Restriction fragment length polymorphisms (RFLP) IV. The localized DNA sequences of two individuals are shown below: 1 z 3 *1 Individual 1: CGATTCAATAATAATAfiTiATCGTTACG I: )0ng 5014/ ? :—,____.__._.. GCATTG'I'I'ATTATTATTATTAGCAATGC ’7 c , “Lanai Individual 2: CGAWCAATAATAATMTAA AATAATCGTTACG GCATTGTTATTA‘ITATTATTATTATTAGCAATG‘C I Such differences in the localized sequences between individuals can be used as milestones in mapping analysis, also known as 3/ Single DNA markers fl Single DNA sequence markers 0. Single nucleotide polymorphisms Simple sequence length polymorphisms . ‘ Restriction fragment length polymorphisms (RFLP) 3. The following figure shows an example of using DNA fingerprint technology to establish a linkage between DNA markers and a disease allele: Dominant phenotype (Pl 3 a: :- m a Im'fll'l'lunlflb Each marker (A, B, C, D, E, F, G, H, and I) is represented by a band that contains a minisateilite region flanked by restriction cuts. a. Marker B and C both are possibly linked with the diseased allele. b. Marker C and D both are possibly linked with the diseased allele. 0. Marker C and H both are possibly linked with the diseased allele. @ Marker c and I both are possibly linked with the diseased allele. e. Marker H and I both are possibly linked with the diseased allele. 4. The following figure shows a microsatellite locus linking to a disease gene for a family of six children. The banding pattern shown in the figure is generated by PCR (Polymerase chain reaction). This pattern is illustrated with the use of four different sized microsatellite “alleles”, M’ through M’m. Parental genotypes We, III-II Illflfiflfix p M“ KEY '* 4- PCR Primers P Seminal-It disease allele {KEITH Mittens-Elma tapeats M' — M"” Maiecular markers shun I—lfu MM“ M “an“ m mu.- -.: rum w H. I rmmn ma 6...)me Banding patterns of parents and children Mn” M" M' PCR products Which one of the markers is probably in cis configuration to the diseased allele p? a. M” e . C. M97! d. M9999 e. All of the above. 5. bacteria can grow on minimal medium while require supplementation ofone or more cellular building blocks or nutrients. a. Heterotrophic, auxotrophic b. Auxotrophic, prototrophic ,. Prototrophic, auxotrOphie g Phototrophic, heterotrophic 6. Which ofthe following describes the relationship between Hfr strains and F' strains? a. An Hfr strain donates DNA to an F" strain, thus converting it into an F' strain. ‘/ b. After DNA is transferred from an Hfr to an F', the Hfr strain becomes an F’ strain.’ . . When an Hfr strain is used for transduction, it is called an F’ strain. - flit} strains can be converted into F' strains through the liberation f the F factor and | ""3 some flanking bacterial DNA from the bacterial chromosome. /0 "JINResistance to antibiotics, once obtained by a bacterium, s rea bacterial population. The most common mechanism of ansfer of resistance enes is: J”— )(transformation. ~ furs oer bfzygotic induction. c. general transduction. "Elj‘jspecialized transduction. re. conjugation. 8. -Many Hfr strains can be derived from the same F)r strain shown in this map. / \. Y R :/ .._-'— Which one of the following gene orders could not be obtained through’hirglktleguency recornbinflon? _ aQATEZ b. P C D B S cZYRSB r$AQPC e. All can be obtained 9. For recessive mutations that cause disease only when both copies are present, heterozygous carriers are disease free due to haplosufficiency. This means is; g. Jr ____.——-—-_._.. the dominant allele makes enough protein product to offset the effect of one bad J recessive allele. the dominant allele does not interact with the product of the recessive allele and therefore no disease is produced. ' the recessive makes a protein that is only half of what is required for normal function. that both dominant and recessive alleles need the effect ofanother gene(s) in order for the recessive effect to be overcomed. 10. Human blood group phenotypes are controlled by a. é e. incompletely dominant alleles completely dominant alleles (so-dominant alleles pleiotropy none of the above I 1. In a I945 trial, a woman accused Charlie Chaplin of fathering her child. The ABO blood types were as follows: Woman type A ie A. .. Charlie Chaplin type 0 Child type B What conclusion would you make—as ajury member—about this trial? Charlie Chaplin is definitely the father of the child. Charlie Chaplin could not be the father of the child. c. Charlie Chaplin has 50% possibility of being the father ofthe child. (1. Not enough information to make a decision. 12. In Mendelian genetics, genes may interact to produce different dihybrid ratio. Which of the following is most likely due to interactions between genes? a. l: b. l: @ 9: cl. 9: l :1: :l :4 :3: L.L-.Ji..n-ll\t-3'—l l 13. The type of dominance is determined by the molecular functions of the alleles ofa gene and by the investigative level of analysis. The following electrophoresis diagram shows the protein ofboth alleles. Positions to which hemo- QIOblfls have Hemoglobin types l Phenotype Genotype m 9'3““! present Sickle-cell HbSI-Hbg S and A trait Sickte-cell anemia ”b5! “b5 S Norma! HM! Hb‘" A Migration "gm-.154 rum-m...- m fl‘ullk Anal: an. mu lam-m a 200mm li-i'nwuvull and Conny-III! This diagram illustrates a cgdominantrelationship between sickle-cell trait and normal trait at ' T ' 3. Cellular level b. Organismal level [a] Molecular level .- LCK All ofthe above. 14. A snapdragon that bred true for white petals was crossed to a plant that bred true for yellow petals. All the F] had white petals. The F] individuals were then crossed, giving F2 plants with the following phenotypes: 'Tr' L. J' + \l L \l /'7 t White 245 (l w ‘l £1 L x . s. . Yellow (ii in 51‘. 3; f Spotted 19 j 5: ._ ‘ {; Cflt‘puul Ira"; Propose a genetic explanation of these results: /g./ it is an one gene showing dominantfrecessive relationship between yellow and white color alleles. b. it is an one gene showing c-odominant relationship between yellow and white color alleles. c. it is an one gene showing incomplete dominant relationship between yellow and white color alleles. (d; it is a two gene effect showing some interactions between two genes. ,x‘ 15/ Avery, McLeod, and McCarty’s experiment showed that: a. the transforming principle could not be carried out in vitro. fithe conversion of Type 1113 to Type HR was due to mutation. c. the process of heat-killing cells was not entirely effective. d. DNAse treatment eliminates all transforming activity in heat-killed 1113 cells. 16. Which of the following statements is incorrect about DNA structure? [eff he two strands of a double helix are complementary and anti-parallel. . The structure was first predicted by Watson and Crick in 1953 to be arranged as a double-stranded helix. .Q'he final structure disproved the work leading to Chargaff‘s rule. . All of the above are correct. 17. The functions associated with telomeres are to: a. prevent ribonucleases from degrading the ends ofiinear RNA primer molecules. flaliow the fusion of broken chromosomal ends. Q facilitate replication of chromosomes without the loss oftermini.|/ Leia/ensure the appropriate segregation of chromosomes. ,9? provide chromosomal anchorage to spindle-fibers. ' T:232.A;ttte;2‘ 18. An RNA molecule has the follo_y_v_in centages of bases: A = 23%, U = 42%, C = 21%, and G = 14%. Is this RNA 151$ or double stranded? What would be the percentages of bases in the template strand of the DNA that contains the gene for this RNA? at. single stranded; A = 23%, U = 42%, C = 21%, and G = 14% b. single stranded; A = 23%, T = 42%, C = 2.1%, and G = 14% ‘ @single stranded; A z 42%, T = 23%, (3 = 21%, and C = 14% / [dfdouble stranded; A = 42%, T = 23%, G = 21%, and C = 14% //‘6' double stranded; A = 23%, T = 42%, c = 21%, and G = 14% 01M Refer to the following figures for the following two questions. | II,N—C—l’l mil-il‘LrliH II (x; ‘ H j “ o o o "f "I: ”I I O—ELO—slr—meL-omcn, -- 4. . i a o 0 H a L H H CH :..._C.H{ H a b. $00” "H,N—C--H ,l-lE e. 19. Which diagram shows a nucleotide with allurine base? --.-——’--'_ [Ir/"ZN . . ‘igure a D b. Figure b ’49: Figure 0 . Figure d "\ ”Figure e ' ‘ | 20 Which diagram shows a nucleotide that would be used to make RNA? fligurea N ANGLE) Mgl-M— QN A. . Figure b \NLAW (b. one \i.-"Figure d iFigure e 21. You learn that a Mars lander has retrieved a bacterial sample from the polar ice caps. You obtain a sample ofthis bacteria and perform the Meselson and Stahl’s experiment to determine how the Mars bacteria replicates its DNA. Based on the following equilibrium centrifugation results, what type of replication would you propose for this new bacteria? band position before afterist after2nd after 3rd N replication repiication replication 14 in N15 in N14 in N14 in N14 a. conservative replication b. semi conservative replication @dispersive replication ()5. discontinuous replication /e.’ continuous replication 22. Which enzyme used in DNA replication most closely resembles RNApolymerase: a. DNA polymerase I. b. DNA polymerase II]. c. primase. d. telomerase. e. helicase. 23. An in vim: transcription system that contains a bacterial gene initiates transcription, but from —'—'-—-_. random points on the DNA. Which of the following proteins most likely is missing from the real—CTN— '- a. sigma factor ‘/ b. rho factor c. RNA polymerase ll (1. TATA-binding protein 24. The fact that some organisms contain much larger amounts of DNA than apparently "needed'I and that some relatively closely related organisms may have vastly different amounts of DNA is more typical in a. viruses than in bacteria. _ b. RNA viruses than in DNA viruses __=T:@ eukaryotes than in prokaryotes d. the family "alphoid" rather than the diphloid family e. prokaryotes than in eukaryotes 25. Some mutants from different organisms exhibit the same recessive phenotype. In order to determine whether such mutations belongs to he same gene or not, which test would you _fl perform? a. test cross b. epistasis test @complementation test d. allelic series test e. biochemical test 10 II. True and False: 26.Solving the replication problem at chromosome ends is not relevant for the leading DNA strand. A. True \ -| B./False 27. Te'lomerase uses an inherent RNA template to synthesize new DNA. “A: j/True B. False 28. All RNAs are translated. A. True lfialse 29. Initiation of transcription does not require a primer. A. True B. False 30. Dr. Frederick Griffith demonstrated that a transforming chemical from dead bacteria could change the genetic information of living bacteria. A. True B. False 31. Transduction requires direct contact between bacterial cells. “1W3...- ._ . 3.! 32. Plasmids do not have to integrate into the host cell chromosome in order to be replicated. A. True 8. False 11 33. Specialized transduction is used to map distances between phage genes. A. True B. False 34. A complementation group is a complete set of the genes controlling a phenotypic trait. A. True B. False 35. Epistasis involves intra-allelic gene interaction. A. True B. False 12 [11. Worked Problems and Short Answer Question 1. The following table contains a list of statements that apply to replication, transcription, both, or neither. ln each,empty box, put a check mark if that statement applies to replication or transcription. If it doesn’t apply, leave it blank. Replication Transcription The new strand is made 5’ to 3’. The new strand is made 3’ to 5”. The new strand is identical to the template strand, with - the exception ofU’s replacing T’s. The new strand is complementary to the template strand. J The template strand is RNA. - The product is DNA. J The product is RNA. - _ An RNA primer is required to initiate synthesis. -- Synthesis of the new strand is initiated at a promoter. The process is done only during the S—phase ofthe cell cycle. The process is done only during the G l-phase ofthe cell cycle. 13 w Create a 2. Several mutants are isolated, all of which require ompound G for growth. The compounds (1 to V) in the biosynthetic pathway to G are known, but their order in the pathway is not known. Each compound is “fie—Sted'for its ability to support the growth of each mutant (1 to 5). In the following table, a plus sign indicates growth and a minus sign indicates no growth. Su - - lementar com ounds tested a. What is the order of compounds from l to V in the pathway? W "7 It? IE- 7 I 790% ./ b. Would a heterojaqon composed of double mutants L2 and 2&4 and 1,4 grow on a minimal medium? ' Ij‘L ll (Draw the pathway forWin a singleheterokarypon IQEPSWEF this {question} __,.._____ __...-_..._,. r r. Ll : Wen fl? ' mun “0+ 3mm -' / 9H: lU—bflfiL—Jflfl 7’ I‘ll: 331 411331 14 ' P: L.“ r PIX ifiop u (i) What is the name of the structure shown above? / (ii) What is its function in transcription? . 673‘" ‘iflf‘fli—(AL (iii) What is the probable base composition of the region marked by the “A” arrow? U (iv) What is the probabie base composition of the region marked by the “B" arrow? C (v) If the region marked with the “B” arrow was mutated to: GGCCGGTGC, what effect would this have and why? - -’ -‘” '- " 3 m herp‘m hop 7/ [In li‘; agrarian I" “C9 ip'W'ir‘J‘L ”ironcéumtv—i an .2 7_ i M Meaty C s, as ‘ ILW: Mega-(3;? A} ”the Com? Pogsiifi--- hat—L or game! berm—‘- so r:dwrs«c*°“C _, , o ,5 obi-T borfy'S behemo- L F} f} ”Limb rat/é: {2W7 v4 “gym ' 50%,) GA. OHM“ i -~ I v’ - . r"r I'u' ‘ b’hfifig Git")- QS fiifl'l‘a SJ Sir? I .- ¥;_l‘j"( '- ,3“ W55- Mp 41L. LZFPLL . . , at f} _ g (- “P t3 0MP [OF'€u~o- v" “Vi W e: - -- -- , {1 4C r m -"‘C'-" 5/ ‘7/' 15 ...
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