BIL 250 2007 Midterm II Form 2

BIL 250 2007 Midterm II Form 2 - Spring 2007 Midterm II...

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Unformatted text preview: Spring 2007' Midterm II, Form 2 Dr Alex Wilson NAME: COMPREHENSION QUESTIONS (1 point each) 1. Mobile genetic elements are found only in prokaryotes. a. True ® False (2 Human X and Y chromosomes synapse during meiosis. a. True 0 CB: False 3. Polyploid individuals have two or more incomplete sets of chromosomes. a. True Cb) False ‘3 . 4. Bacterial mutants that require additional nutrients in their growth media are called auxotrophs. C#: G: True 2 False O\ 5. Antibiotic resistance can be transferred from one bacterial cell to another by conjugation. True 0\ b. False 6. Plasmids do not have to integrate into the host cell chromosome in order to be r licated. True r x) b. False" a ‘ 7. The fly, Drosophila melanogaster, cannot be transformed using plasmids derived from transposable elements. a. True ® False b 8. The physical distance between gene loci on a chromosome doesn‘t affect crossover interference. a. True (BI. False b 9. Linked genes always exhibit recombination frequencies of less than 50%. True Q . False 10. In many sequenced genomes, a significant percentage of genes have no known function. @ True 0\ M g f 0' 5P b. False “i 0 x J /“. '- ‘e" \‘ /11.£ sequence database of ESTs (expressed sequence tags) contains only the 3' QMOK g r anscribed sequences of a genome. S x” -"\ é? “ s, @ True a a £82,? b 0\ U ‘L . ,Q‘fi-‘J‘ 12. Loci that are far apart have a higher recombination rate than loci that are Q69 close together. True 0\ b. False 13.Transformati0n is the process of converting one bacterial species into xv. another. ‘ yuan; model. at Craft“ 0“ 3" a. True in “Ha \ mi- DNA tit‘“ X‘Wfis‘fi ® False b 7 t m MP b can cl” . . t 33‘ 14. Nonsense mutations do not alter DNA sequence information. d “ { ' (ii Yr)- - v‘ / at .\ I a. True @3615? 1d“ Spl' "\iv @ False b h t.‘ XDNA glycosylases recognize and cleave phosphodiester bonds in DNA. qoé L \O \‘3 v ’ u... . 4‘ ® True i r 7 .. m \o‘” L e I. 0a 7‘“ \L toe . b. False cc- f C ‘ mm?“ Wmi'flxi'k \n 2 A“ I it. . Midterm 11, Form 2 Dr Alex Wilson Spring 2007 NAME: car: MULTIPLE CHOICE QUESTIONS - SELECT 91!: OF THE FOLLOWING ANSWERS (1 point each) 16. FISH stands for 6'9. fluorescent in situ hybridization b. first induced strand hybrid F1 insertion segment homolog flanking insertion sequence hybrid fluorescent insertion segment hybrid 0\ one We type of mutation in which a pyrimidine is substituted for a purine is g Silent Transition Q Transversion C, d. Transitory e. Nonsense 18. Hybridization of single-stranded wild-type DNA with DNA from mutations caused by IS elements characteristically shows (through electron microscopy) 3. Chi structures C e51. ix" b. Unpaired tails (C3 A single-stranded loop representing the IS element d. A single-stranded loop representing wild-type DNA e. Theta structures 19. Trinucleotlde repeat disorders are the result of a. A high rate of mutation throughout the genome B (E) Extensive duplication of a single codon c. Deviation from the genetic code in human mitochondria d. tRNAs failing to recognize specific codons e. tranversions of the DNA bases in the coding sequence \ g 1 \ 05¢ 1’ I ' F ‘ - l '-®During mutagenic treatment with nitrous acid, an adenine deaminates to form hypoxanthine, which bonds like guanine. The mutational event would be C. "' 1’ U a. ATto CG AT a (A b. ATto GC b c. ATtoTA ‘. . E g d. GCtoAT ,1” '3 " _ H e. GCtoTA ‘Cji * A L_.':' In a mammal how many inactivated X chromosomes (Barr bodies) would be present in cells of individuals who are XXX? 21. 2 aeéewm Ln-thMNI-‘D . Retrgttallsposons move via an intermediate that is _,.--“ ‘ a. A double-stranded lollipop .\\ b. A retrovirus \ c. double—stranded RNA \J {553‘ d single—stranded DNA ‘\ 4“" \“L " single-stranded RNA \L Sui? Blob 9% KW“ 4 3 sol 6“ Spring 2007 Midterm 11, Form 2 Dr Alex Wilson NAME: C#: 23.1n E. coil, mutation arising during repair is mostly by a. Thymine dimmer splitting b. Excision repair SOS repair Q d. Mismatch repair e. RecombinatiOn repair 24. Fra ile X syndromelis caused by Trinucleotide repeats; (at b. Free radicals c. Microdeletions d. S-Bromouracil a e. Depurination 25. A specific pair pf primers is used to amplify and sequence a pair of . * microsatellite alleles p and q in a woman (q migrated further than p). The 34/ same primers revealed a different pair of bands, r and s, in her husband (p a“: fix migrated father than rand q migrated farther than 5). Which microsatelllte 3' a a genotypes are possible in the children of this couple (assuming all alleles are (0 Q" ‘ unlinked). I r Qt C" a. Onlypq orrs Ci )4 i’ Q a L - g“ b._ Only'pr or qs A r { I (a; _, c. Only ps or rq \3 v (4| . @ pr or p5 or qr or qs e. The children would not have any of the bands from the parents transgenic Arabidopsis plant contains two copies of transgene T, one on chromosome 1 and one on chromosome 2. What percentage of the gametes ' 7 \‘PJL/ wrll NOT contain the transgene. ‘0 a. 2% (it ’1 ._._--:.L—'—-a~ " b. 10% Ar“ OS c. 25% 4/: (an ii— I —i— d. 50% C, ,_ W e. 10096 = 27. A virulent phage is always (3) Lytic b. Temperate Q C. Capable of lysogeny d. Capable of producing a prophage e. Capable of zygotic induction 23. A linear DNA. molecular has 3 target sites for restriction enzyme EcoRI. What is the maximum number of fragments that can be produced if a sample of the molecule is only partly digested? a 3 b. 4 c. 7 (l e w 29. A circular DNA molecular has n target sites for restriction enzyme EcoRI. How many fragments will be produced after complete digestioLf— a.n-1 ___._.____-——-— @n M c.n+1 lD J # ._. —— d. 2n-1 Spring 2007 Midterm II, Form 2 Dr Alex Wilson NAME: C#: 30. Four clones (A, B, C & D) of human genomic DNA are tested for sequence- tagged sites 1 through 5. A shows 2 & 3; B shows 2 & 5; C shows 1 El 5; D shows 3 8: 4. What is the order of the clones in their contlg? 2' <1 S I. Q 3 4 S 4 l © *CjBAo A . _. _x_.,. I: _l r Q d r3 )c y r) 1 .I//_/ (r) y y C .IJ.‘ . 31.The mutation causing the recessive disease for sickle-cell anemia removes one MstII restriction site from the globin gene so that a probe, instead of hybridizing to two fragments of 1.1 kb and 0.2 kb, hybridizes to one 1.3 kb fragment. For two parents to have a 25% chance of a child with sickle—cell anemia a. They must both show only fragments of 1.1 and 0.2 kb d b. They must both show fragments of 1.3 and 0.2 kb c. One must show a 1.3 and the other a 1.1 and 0.2 kb fragment They must both show all three types of fragment {I I. he F, G and H loci are linked in the order written. There are 30 map units between F and G, and 30 map units between G and H. If a plant FG H/fg h is testcrossed, what proportion of progeny plants will be fgh/fgh if there is no __interference? 4 a. 0.590 W t Cdi'i/‘Efi‘fl. “ W“ *8 ,2 ‘\_- a :l, %The maize gene sh and bz are linked, 40 map units apart.’ ’0 (i ’v ‘ ' If a pjant so: bz/sh of is selfed, what proportion of the progeny will be w hbztshbz?‘ r c wl Q W“, {"‘5. 0.40 ' I”? r F 3‘er is 1 / S in i? i ‘ Cr 035 b. 0.20 J 1 C- 0-50 w— ..... _.___.-+- h 3i- __ a "Ifl‘ O '1 'MO {am e b2 _. 1 . H - d H . . 3“ r; '1’ 0'04 ‘ ._ ‘0 .1 _ e "fizflflif %.A B \ST search is done to ‘/ 4‘ ‘ ‘r‘ 4i " a find homologous gene or protein sequences 3 .1 Q __ r“ .i ._ 9“" find the chromosomal location of a sequence ‘ L‘ c. predict the three-dimensional structure of a protein from its amino 0 . \Owt") [\an acid sequence 091m cl. find restriction sites and SNPs in a sequence. “Spy-Aiuii e. determine the conditions under which a gene is expressed. 35. What is the function of dideoxynucleotides in SWDQ? a. They act as primers for DNA polymerase. I b. They act as primers for reverse transcriptase. L"? ék+ _ has} c. They cut the sequenced DNA at specific sites. SC , 0F cl. They allow only the specific sequencing of the RNAs of a genome. ' (E) They stop synthesis at a specific site, so the base at that site can be ' DN“ m e ‘” determined schéwnex. A Spring 2007 NAME: Midterm II, Form 2 Dr Alex Wilson C#: Use the following diagram for question 36 ddCTP rea ion ddATP reaction si ddTTP reaction :1de reaction fragment of DNA is cloned into a plasmid with a sequencing primer binding te. After dideoxy sequencing, the gel pattern shown above is obtained. What was the sequence of the DNA strand that acted as the template in the sequencing reaction? yaw; ’ a. 5' GCTAGCA 3' :f i b. 5' ACGATCG 3' C; ("17w m c1 T- . - . S'TGCTAGC 3' i . 5' com-cm 3' section of a genome is cut with three enzymes: A, B, and C. Cutting with A A find B yields a 10-kb fragment. Cutting with B and C yields a 2-kb fragment. What is the expected result from a digest with A and C, if the C site lies in between the A and B sites? i9 38. Old and new strands of DNA can be recognized by )AThe terms cis and trans describe 0\ 4§<Linked genes a. a 12-kb fragment b. an 8-kb fragment 5) an 8-kb and a Z-kb fragment d. a 10—kb and a Z-kb fragment e. a lO-kb, an B-kb, and a 2-kb fragment DNA glycosylases methylation patterns . 3' -) 5’ exonuclease activity d. altered DNA structure e. AP endonucleases a. lo a. the configuration of alleles on homologous chromosomes \b‘ the relative position of different loci on the same chromosome M che relative position of different loci on different chromosomes . the ability of a deletion to uncover a recessive allele e.\ the ability to use deletions for mapping genes assort randomly X can crossover and recombinm/ are allelic K co-segregate will segregate independently X a. G) c. d. e. Spring 2007 Midterm II, Form 2 Dr Alex Wilson NAME: (213': SHORT ANSWER QUESTIONS 41.If the genes assort Independently, what will be the progeny genotypes and their proportions in a cross lit/a ; gfb {z A/a ; b/b ? (6 points) P1: Wimp/4 M», '/4 a81'l4ab m '- "lq M a "a as X : V8; A A Eb ‘, H5 A 1w":- ~. 7/35 /‘-. LLB-‘0 1, l lag; A 0 "rs til-‘1‘) }\\ C\.\:‘J1‘CJ --| I :6 IL" In) ,' 'K-' r.' r; I] 0 In I ‘- \. I l ‘ ‘ I I {/8 é; Tm We Amie ‘i'eAA'n‘r; '/4 Ame "4 A “~70 i ’g M“: m 'l ‘ I I 42.The protein encoded by the alkaptonuria gene is 445 amino acids long, yet the gene spans 60 kb. a. Roughly how many times larger is this gene than it needs to be? (2 mm“) 445. 4%” a a x g —_ “L QJCJ 06L M r _ (A I. int; "4' I‘H‘qm l“; b. How is it possible that this gene is larger than it needs to be? (2 points) )4 .. . O r-( .; ' ' 9 film/K ,-'-\ (re? WNC7 fie!) DU) n J -. _ J J .; ' Mn; \L 5 M . (Hm 9N If) . o W 3% Inwns' 1'0 b" 5/31. (£14 5 i 43. Five YAC clones of human DNA (YAC A through YAC E) were tested for sequence-tagged sites STSl though STSS. The results are shown in the i” [3, ,. following table, in which a plus sign (+) shows that the YAC contains that / - ' ' ' ' ) STS: It?” -J(jlj)[ YAC 2 3 4 5 6 t. i m“ A + + + - - - , B +, — - + - - PM: 9255‘.» a c — + + - - + ~ ' D - — ' - + + — C E — + - - - + a. What is the order of the STSs in this genome? (2 points) i gay-asks - _ 1 . b. Align the YACs in o a contig (2 points) "pink '4 C l \\\\ w D l L l A 't l r u i Q» \J $51 M. fig 7 Spring 2007 Midterm 11, Form 2 Dr Alex Wilson NAME: C#: 44.After Drosophila DNA has been treated with a restriction enzyme, the fragments are attached to plasmids and selected as clones in E. coli. With the use of this “shotgun” technique, every DNA sequence of Drosophila in a library can be recovered. How would you identify a clone that contains DNA coding for the protein actin, whose amino acid sequence is known? (2 points) I] (4' r' \i f3 1—7.! (“a 2 L; (J r L F; .“I J f w: p. (j "y f‘a’ -_l- , LAJ’ ¢(lr if: /?rfi51 yin» twat/1”" H" spill}! (J/CJ» m , my» . . I: Q " Eb" )1 'r'" '1 :‘3‘ x _’ j/l'r'v’ g’lfli‘ 'I fix/71': 45. Why are frames’hift mutations more likely than missense mutations to result in proteins that lack normal function? (4 points) (HINT: Be sure to define what a frameshift mutation is and what a missense mutation is as well as answering the question) , U. {W it, A} F F. _' f-‘n' - /’Y‘ o 1'0: Fir id" '17 0/‘, 3:4,.._ IL. ,I f. ,'.agn=./_rr~»/-i f/ Q’Qi/f/‘é'fl f: m» %/9-’ //F/ijJ/‘/J” n" hr n J, an ’ fif’bvr’ J _..-- fiof— ncreflm'l}. .f( d; J / {if/LN 6/? , C/ I pp Ll. ( In: I. /‘:, if {fit/WEE!" 3")er a I f {/2 .’ v j . / pain if 8 Kurd—.1 Spring 2007 Midterm II, Form 2 Dr Alex Wilson NAME: C#: WORKED PROBLEM (5 points plus 5 bonus points = 10 points) 46. In chickens there are 12 map units separating the genes (F) frizzled and Crest (Cr) and 8 map units separating Cr and Stocky (S). The genes are ordered F- Cr-S. In the table below predict the number of progeny you expect to result from crossing a heterozygous female whose genes are in cis and a recessive homozygous male. There are a total of 10,000 progeny. Hint: use the symbol + to designate the wild-type allele. (4 points) what is the genotype of the mother? (1 point) ' I; Q fry-(War ;3* 3L}; a What is the genotype of the father? (1 point) :8) f¥,.c.r.cr,~3 g In the “Type” column below, indicate whether the progeny phenotype is the result of a double crossover, a single crossover or if it is a parental phenotype (4 points) Progeny Phenotype Number of Progeny A ...
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BIL 250 2007 Midterm II Form 2 - Spring 2007 Midterm II...

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