MT2 Form A Answer Key

MT2 Form A Answer Key - BlS101WInter2010 Name fiNgé/EA Ki...

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Unformatted text preview: BlS101WInter2010 Name fiNgé/EA Ki: y S. O'Neill Last, Fir |D# Midterm ll FORM A February 22, 2010 There are 11 questions and a total of g pages in this examination. Please write your name at the top of each page. To receive credit you must show all your work! Value Score 10 1 Page N N _L O _x O 10 11 12 Total 10 10 1O _L 0 Honor Code: My signature below affirms that I wrote this exam in the spirit of the honor system of UCD. l neither received nor furnished any "help" during the exam, nor did I use any unauthorized references. Signature Student Authorization: I authorize the University to distribute my grades electronically posted to Gradebook in SmartSite, using my student ID and to return publicly this graded exam (either handed out after class, or after that left in a bin for me to pick up). All unretrieved exams will be discarded at the end of the quarter. Bswwmmmm wa Awyugfi keg s. O'Neill Last, Ei t Signature 1. (10 points) In a certain species of petunia, the petal color of the flower is normally purple. Two recessive mutations are induced, each of which produces white petals. The two pure-breeding mutant lines with white flowers are then crossed with the following results: Mutant 1 (white) x Mutant 2 (white) F1: all purple F2: 90 purple: 70 white Explain these data using clearly defined genetic symbols. You must indicate genotypes and phenotypes of mutant 1, mutant 2, F1 and F2 progeny. Show your work. man l x “Wm” "‘- wk‘nk whit-s. QQ'B’B x AAMZ— F -. A—q BL thplt , . F. ; c; A.%— Pwpk 3 ‘ Ho A Mk 3 ca 8'" I QQLL BIS101WInter2010 Name ANSME/L 165x S. O'Neill Last, First 2. (12 points) in chickens, the following genotypes give the following comb phenotypes. (Ms 3, qws-Hm =tl= 33 M (W. as; ..< _ mm ,m+ book. O —-zs- R/- ./. Assume independent assortment. ~ a. (3 points) What comb types will appear in the F1 and in the F2 and in what proportions if single-combed birds are crossed with birds of a true- breeding walnut strain? ' . _ w P ' S'msl-c (viv- PIP) x weLnurl"( R/-- P/_) F2, . a {a/h-J P/ C aLMd-b . 3 Y/Y‘J' P/-— ((2:43 F‘: R-r‘ P4 CWLLnu+3 3 RI-349/f- ((1.082) ‘ f/T'J‘ ff. [51.513ch b. (3 points) What are the genotypes of the parents in a walnut x rose mating from which progeny are 3/8 rose, 3/8 walnut, 1/8 pea, and 1/8 - single? ’Pkeuu-l-lfis . Possiblc 34~o+flfms ¥l @V°3"”‘3 “*1 3/3 voS-c. K/- 31/71.. 5/3 waLer Kl~3 P/fi ‘lsr Wk w/v‘“; Pl— \ 5‘: s It. ' c. (3 points) What are the gsenotypés of the parents in a’rv’vérndt xToég’ mating from which all the progeny are walnut? 11‘ Walnwt‘ (ft/”SNTA h +0 9.; his “sumo” 9—: 1’1»; pamrls must Lu. ”Manson‘s KJ \wj— bo’h/t ”tag ”0+ la.) M02 ‘l’kx would" pawu'l‘ MU‘F Les ho moan sous P/ P . d. (3 pomts) How many genotypes produce a walnut phenotype? Write them out. R/K 3 PIP Kid“; F/P R/K; P/vr- (U425 P/7L . 3 BIS 101 Winter 2010 Name ANSW E/& 352 S. O'Neill Last, First 3. (2 points) Consider the following observation: in Drosophi/a, 50 individuals are homozygous for the recessive trait eyeless, but only 40 of the 50 individuals have the eyeless phenotype. What term best describes this mode of inheritance. P€Nc+rb0Nce 0ft— J'NCDM plot-1. PeNC+VOUC.’ 4. (6 points) Match each of the following terms with the single best definition from the list provided (Use each letter only once). Best Definition (letter) Definitions 1. Trisomic D A. Lost one homologous pair of chromosomes (2n - 2) B. Lost a single chromosome (2n - 1) C. Has an extra set of chromosomes (3n instead of Zn) D. Has one extra chromosome (2n + 1) E . Has an extra pair of chromosomes (2n + 2) F. ls missing a set of chromosomes (1n instead of Zn) G. Has a chromosome number that is not an exact mutiple of the haploid set of chromosomes, or has chromosomes with a part or parts deleted or duplicated Has an extra Y chromosome Has an extra chromosome21 Has an extra chromosome 5 is missing part of one copy of chromosome 5 . Triploid . Aneuploid . Cri-du-chat baby lH WM“ 2 3 4. Nullisomic 5 6 . Down Syndrome baby 39’: 5 BIS 101 Winter 2010 Name 4 NS: 1062. K5 ya- S. O'Neill Last, ,irst 5. (10 points) A petunia is heterozygous for the following autosomal homologs (o=ce-ntromere): A---B---C—-o--—--D--—-E---F-—--- a----b----e----d.-----o_F-—c---.-f--- a. What type of inversion is this?‘ _ caster-(fink II i-nwasutaw b. Diagram the pairing configuration you would see at meiotic metaphase .-|. NUmber the four chromatids sequentially (1 to 4) from the top to the bottom of the page: c. Assume that a cross-over occurs bewan the C and D loci on Chromatid 2 and the c and d loci On chrOmatid 4. Diagram all of the Chromatids that would result from resolution of the cross-over event (as you would see them at the end of meiotic anaphaSe l): - Foam 24 + 5 315101 Winter2010 Name éNgk/Efl, /CE X S. O'Neill Last, First 6. In rice, three important disease resistance genes (A, B, C) are located on the same chromosome. The F1 of a cross between two true-breeding strains, when test-crossed, gave the progeny given below. Dominant alleles are symbolized by upper case letters; recessive phenotypes are symbolized by lower case letters. a. (2 points) What were the genotypes of the parents in the cross of the two true-breeding strains? (Lat. 5L CC. b. (2 points) What is the order of the loci? Fa/zm 4+6 BIS101VVInter201O Name flNs DUE/L KEY Last, First S. O'Neill c. (4 points) What are the recombination distances between each adjacent pair of loci (construct a map)? Show your calculations of Recombination Frequency that you used to support your map values. Ac. 8 4 w W -W :X: a,:::‘swa‘c am “4.. C/é, Amt-.3 5.0;. X10021?) A ’0 B Axe/5» RFA-/i+l¥4‘izz+‘rz-7-.. :1: “’M c- -M~ /5c'o_ 5WD A, C/K— A C. 3 c6 W 0.2,9 x I00 I 29%- 1M.u. L‘lMJ‘L d. (2 points) What is the Interference (I) value for these data? Show your calculation. Put your final answer in the box below. £ ‘D CD € V? N +5 1:: i»C Valarie. CSW/ # eXRC/‘i'i DCO ‘CV‘CWK I ; 1 ~— 0.493051waozill70 I: 3/90 BIS 101 VVInter 2010 Name flaw/51?; ICE Z S. O'Neill Last, First 7. (10 points) Mating is established between an Hfr strain and an F‘ with the following genotypes: Hfr ala+ met+ Iac+ his+ pro+ StrS X F' ala- met‘ laC' hiS' pr0' StrR a. What will be all the constituents of the medium used to select exconjugants that have aquired the ala+ allele? 5/148. lot-6f, Lac, Aisl/ro, 51"” b. A graphic representation of the time of entry of each donor gene is shown below. met+ 75 Iac+ Efrem pm a leles among 50 aIa-+ StrR exconj. his+ 10 0 5 1O 15 20 25 Minuteiju La¥:s )Cl‘i 0”) In the space below, construct a genetic map based upon time of entry and determine the distance (in minutes) between each loci and the origin. 5, + + + 4,, 4_ 1. t ‘ «3 ; (IS 0 a 3 3. ‘é .s “wfiimm. l... .. in...” .. wJ,” ._. 1‘; M8 5" edv‘j 5 :0 l2. n- 2.2. 5.4—4--.M_i.._._ .r...._L_u...-_J_.__d was? with Jr‘h”‘° b“"’ M Lu... Mfl only” 7" MIpuLcSCmifi-‘b , Noll: chawu. ‘huc X413: was “9+ prowl-.3 OLL§$u~¢£) Lotti Creek-if w‘sll iv— $iv~np KC m QuSweoL; («Rm-eoC ell/Wu) it! We, Ji5k~¢¢85\ a.“ ‘Cawstffeu‘l‘ wI‘T’k tuck what!— . B|S101W|nter2010 Name fi/VSWEK flex S. O‘Neill Last, First 8. (10 points) Fill—in answers. a. Complete the following sentences: Coniugation is the process of unidirectional transfer of genetic information through direct cellular contact between bacteria. Transduction is the process by which bacteriophages mediate the transfer of bacterial genetic information from one bacterium to another. Transformation is a process in which genetic information is transferred into bacteria by means of extracellular pieces of DNA. The pilus is a hairlike, bacterial cell surface component which allows the physical union of F+ and F' cells to take place. b. Complete the table below, which summarizes the results of conjugation in bacterial cells, in terms of any change in fertility status and whether any donor gene (other than fertility factor genes) transfer g exchange occurs: 1. change of state . donor gene transfer . no change of state . donor gene exchange . change of state . donor gene transfer W :0 BIS 101 Winter 2010 Name A’NSK/ FL k5 Z S. O‘Neill Last, F rst 9. (10 points) The diagrams below represent four different deletion mutants of a region of a bacteriophage chromosome. Each of the mutants is impaired in its ability to form plaques on a lawn of bacteria. Dashed lines represent regions that are present and areas between brackets represent regions that are absent. Deletion Mutant 1 ------------------- [ ] ............................ Deletion Mutant 2 [ ] ................................................... Deletion Mutant 3 ........................................ [ ] Deletion Mutant 4 [ ] ............................... 9 9+5 a. Suppose that you isolate a new, non-revertable mutation that is able to form wild-type recombinants with deletion strains 2 or 3, but not with deletion strains 1 or 4. In which region is the new mutation located? (Drawamap) E ]-._____.__—————————-———-——jMu:ui: Mu n M new MMHJ‘ Heck/4:5 Lit—Showing} l dotalci Space. 300 incl—pace» fleseu¢s 6“: “$172433 0” LY 19-.;~+$ '\ L up“ 5 how a. 'foiu'l" mutt-«Axon . 9 is b. Suppose that you isolate a new, revertable mutation that is able to form 9 wild-type recombinants with deletion strain 3, but not with deletion strains 9"» 1, 2 or 4. In which region is the new mutation located? (Draw a map) ', kwluj WW-.V.~._._.. [‘ - *_._~W,yw .2 , .. . m“ +0: V, + l vb-W-r Mai-ext C . }V~M MHWWWW”.-- ' " ' mwhn+ z , it? N}. 5f“: 2..-...” _.,. ., . S ',‘ . .7 - “ f :3 fit“*+3 1/ \DJ ‘5 x. ’MwflfiW. .—- «HIV! K a: Pong/f ‘ .fl‘ ’ - _W,L,,._;_- __-.._..,._ u + ONLY 29%. \C b.“ Show 0.. \avmokekfl. dti-c‘honve-zca’wsfnwfigoi :# (LewudabLQ, c. Suppose that you double-infect bacteria with deletion strains 2 and 3. If plaques are formed at a frequency that is similar to that when wild-type bacteriophage are used for infection, what can be said about the regions deleted in deletion strains 2 and 3? $95, {Zegxoys «(LL w Ahfleuu‘l’ germs. cl ’hAaJi' Moth-4+1 Mw‘i—tJHou is MOT I‘M-The 0.1L) 19%. )4; 106K $4.. Hahn-t '3 mutation place )aml vice ucusm . Difiem‘h glues Cam. aGCouu+ 40ft ’n’WS Obflmua‘tj‘ou 0“" GOMIolvewLep‘i—w‘koulbui‘ l‘é 717%? Wine, :9ka dléé‘euy‘i' negro»); 04W10/ Sonya 32p£/;30K «~le mecca MwmlalmaA-lou . N0 Afcombmafiou )5 Meat“? Lori. m (150% SW66 (50%;,qu BIS101 VVInter 2010 Name Ag Swag; 165%: t [rs S. O'Neill Last. 10. (10 points) Below is a tetranucleotide structure. Use this structure to answer parts a thru e. The nucleotides are numbered 1 to 4 (inside the sugar). For parts b and c, you may indicate nucleotides by those numbers. a. is the nucleic acid RNA or DNA? b. Which of the nucleotides are purines? L95 2L/ / c. Which nucleotide is at the 5' end? Nuclea-fl Je. I G 2.. 9+5 H2 / H2 eglw‘ Jiue—S'— Mauopumkd-e _ d. Using one-letter abbreviations, - (1 indicate the sequence of this ‘ tetranucleotide. H2 0 e. Using one-letter abbreviations, indicate the base sequence of a complimentary polynucleotide. 11 2 0+5. BIS 101 Winter 2010 Name V ~ S. O'Neill Last, First 11. (8 points) Please define the following terms: 1. Nucleoside: A pentose (sugar) [ribose in RNA, deoxyribose in DNA] bound to .a nitrogenous base (A, G, C, T or U). Simple answer: Sugar + base 2. Nucleotide: A monomeric molecule of RNA or DNA that consists of three distinct parts: a pentose (sugar) [ribose in RNA, deoxyribose in DNA], a nitrogenous base (A, G, C, T or U), and a phosphate group. Simple answer: Sugar + base + phosphate group 3. Ribose versus deoxyribose: Deoxyribose has a hydrogen in place of a hydroxyl (OH) group found on the 2’ carbon of ribo'se. Deoxyribose is found in DNA, whereas ribose is found in RNA. ”“443 o bl H-ocrr, o OH- »: .1 ii u u ‘4 like“ “Q:- (C. “QL‘ C Dacha pt; “DS'Q . 0“ all on l4 4. Phosphodiester bond (be specific; use a diagram if necessary): A covalent bond (in RNA and DNA) between sugars, with a phosphate group as a bridge. The are two ester bonds in this covalent linkage, hence the name “phosphodiester” bond. See diagram below: c'u, 3w. 0" g ' l 3 f?! c —— o —— P“ 0—- C . . u : i ‘5'», W : ”o-P=° ‘i rt 3‘“ 5+: Bu: OW '. I I s n. 0 Bow!) .. __ u. E kn- ' O cuw“ on. Grouse :3. 619“? 12. (2 id‘bifi‘t’é) A ceHrtain organism has a chromosomal DNA [G+C] content of 46%. What is the percent concentration of thymine in its chromosomal DNA? Show your work. [A+T]+[G+C]=100% 27% [A + T] = 100% - 46% = 54% [A] = m = 27% 12 ...
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