BIO 198 - Exam 4 - 2006 Key

BIO 198 - Exam 4 - 2006 Key - NAME W5 (5 You may pick up...

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Unformatted text preview: NAME W5 (5 You may pick up your exam from Professor Sia next semester or, if you wish, we will put your exam in the hall (next semester) near Professor Sia's office (HH 3268). Please put my exam in the hall BIO 198 GENETICS ' FINAL EXAM December 18, 2006 200 Points 3 hours (~1.1 Points/Min) Instructions: 1) Write your name at the top of each page. 2) Answer questions, in the space provided, using ink. If you change an answer, clearly cross out the information you do not Wish to be considered by the graders. Do NOT use white—out 3) Please write neatly and be concise. Use pictures or diagrams if appropriate. 1) /11 7) /11 2) us 3) /1o 3) /13 9) /18 4) /27 10)_/12 5) /36 11)_/15 6) /14 12)__/15 TOTAL /200 NAME fidSWéfl 53$ 1. (11 points) Below is the pedigree ofa family wim several members exhibiting a rare genetic disorder. I W WV ll :l W x“x“ W W” m X“)? W W W“ x”; $21 an; a a. (3 points) What is the likely mode of inheritance for the disorder shown above? X‘ /fnkté(é{W (“YMMV/n b. (5 points) Assign genotypes to all ofthe individuals in the pedigree in generation II. 3‘ A 5 in bdrm V2. {Pb W e. (3 points) The individual labeled [Ila has asked you the probability that he will have a child with the disorder. Based on your answer in A, what will you tell him? 0 " 'amafiwmf/ ‘ fly’wmggfii. 2M5 WW “WK war a“ "We 3 ‘e w“ gm Mmrm was. 3“” W‘- A— 1.1 k perms, M93? 5n»: cOVNLU‘ MCW \ NAME MSW/42R Q 2. (18 points) a. Draw a replication fork. On this figure, indicate leading and lagging strands, 5’ and 3’ ends, and indicate the following proteins (in a few words, describe the function of each protein in the following list): i. DNA Polymerase III « (int/NY /’ may W ” r0 oral: ” ii, fi—clamp I flrfldégr"? e) Z ’ f4 lbw iii. ligase -gLfgfl/Cfié 07-14%”? sfwhw°( iv. DNA Polymerasel- 4mm; flfi-Mffl W5 A/fl/W “9/ pm” v. DnaB (helicase) _ flnwrnéu [ll/{)6 vi. DnaG (pn'mase) I Ia, Mn [W3 (IQ/VII) v11. Ssb v 5”“ 969M” NAME ANSI/MK Kéy 3. (13 points) You have obtained the following short DNA sequence (only one strand shown) from the middle of a bacterial gene. It is known that transcription of this sequence proceeds from right to left (i.e. the promoter is far to the right). 5 ’ -TGCTAGCTAACT—3 ’ 5’ K CQA'TCGAT T6 A 5’ a. (3 points) What is the mRNA sequence (drawn 5’ to 3’) encoded by this part of theiene? _ 5/A6, U'A CUAQAB/ tPfB‘q-Ela We 29“ “W096?! b. (6 points) What peptide sequence is present in the protein encoded by this gene? Label the amino and carboxyl ends of the peptide. C A n 8 iii 3 a 2 G t, c 1 A , G U N: c. (4 points) If you know a codon sequence, can you determine the exact sequence of the tRNA anticodon? Explain. N», M lame/g, MW 4; Mm. / 7 NAME ddfiWF/LICEF 4. (27 points) Comparison of the bacterial BER AP endonuclease sequence with he database of yeast proteins reveals the presence of a predicted protein with homology to the bacterial enzyme. a. (5 points) How would you clone the gene for expression in E. coli? (Just give the general strategy, not the details). My. MW M»ng m w Mg? M bag fig/Lana- ‘ W710; p46 Mp{ W 54% am, {ca—6b W552?» W414)? b. (8 points) Sketch a design for a linear DNA molecule you might construct to produce a knockout of the gene encoding this AP endonuclease homolog in yeast. Briefly explain how this DNA would be us‘ed to produce the knockout. ’V gob? 0/0 AW'ID/Djv) 1L” “‘9” m” g I M (NF 6 a} APO/6F @ UI NAME Mew/6K [C61 Question 4 continued. c.i. (6 points) If an AP endonuclease homolog is also found in analysis of the mouse genome, sketch the constmct you might use to generate a knockout in a mouse. Please describe why each component is included. WWW/hwhm 97 we ()3 N604 ék M Mario ’ . rig/104w? fpl’h’fl’HSW 255/ : QWXS c. ii. (5 points) If you successfully knockout this gene in mouse embryonic stem (ES) cells derived from a mouse of the genotype A/A B/B (Where A/A: brown, a/a =black, and B = the wild—type allele of the predicted AP endonuclease gene), and you introduce these ES cells into a blastocyst of the a/a B/B genotype, what is the genotype of the resulting chimeric mouse? List the different types of gametes this mouse might produce. ll/ifib hasaw ( / c. iii. (3 points) What cross would you perform with the chimeric mouse to generate mice that carry the knockout allele in all cells? What color will these mice be? NAME Aflgflt’flKEZ 5, (36 points) a. (5 points) In 2 or 3 sentences explain how the chemistry of the bases contributes to spontaneous mutagenesis? flu Was 6M- ar/sf m M/érflw/C/ flaw/awry flm/ guy w/ fA/ firm %m1/V 14 dc/JZ/fiaa. @flgéx m m;% w/ fl/k/MX 6’54. What/sz M&/ 4/17 W/yzuww flflV/flzh fimuééj/ Vm/Mzwd fl/WmVMm fiflWg/AL/M ¢p flaw—6’ away? 972; b. (6 points) In laboratory experiments with haploid yeast, a Wild—type strain may ShOW 15% killing at a particular dose of UV light while a strain with a knockout allele of the REV3 gene (an essential component of a translesion polymerase) shows 95% killing at the same dose. What happens to the frequency of mutation in these strains after UV exposure and why? m flamfi a We % We“; éCZW fix flMg/ég7hW490‘ M Maw/pend. M5722; yaw/m flp/ M/fl/Wfl/ ’57 pm #1570 42/7/ [2/04 I WZMWL/ go W 55% Wy7/ @011 M 9774% MW fl/ M flw gum/M073 ,0/7/ flo/ fiA/W flow/flaky. NAME IV ’ (fl Question 5 continued, c. (10 points) Using clearly—labeled diagrams, describe the model for prokaryotic nucleotide excision repair (NER). (AW/A 1/ UVVE 1*» Mcoam‘a [-6907, +1— m X l/ (Ax/VB 01”“)!de /le€h‘§ Re \K r— "l-Z ‘W, *6 zoo/l 9741/449st ~> d. (2 points) Give an example of a DNA lesion that is repaired by NER. 7, now e. (5 points) What is meant by “transcription—coupled’ repair? ‘ I/A/rmrazangww Vfifi yfim yzq flaw 71¢; WC/fip/ZWML flW/IAm A/E/a Almefl/l (92/5/25- NAME INSWKZ K52 Question 5 continued f. (4 points) What is the major difference between the mismatch repair pathway in bacteria and eukaryotes? W210)“ 7b g2? M/W Wfl/ZA gfiM‘Q/C" Wag. 1‘1 g. (4 points) What type of mutation (e. g. promoter, nonsense, missense, frameshift, indel) is most likely to produce a temperature sensitive mutation? (Explain) 2/ , 2/ u L " Mex/m. flx/Zafa» W:% m/ flflfzéfimfl fl /W W 25 /%W§/ Mf/“m' J" 4/77/ b1 MM§9%{ A/mswy W’V/ ’“74 may (/2; fiflnflflfdw/flé/Wa W” flmzm/fl M grew/>744. NAME flay/5g K52 6, (14 points) a. (4 points) How do RecA filaments activate the bacterial SOS response? WW SSD/Vfi éflmd W/fltéfl ; 7M [04 WWI/46745 W/ Laxfl Mfllfgmmdg’fi “145,; gwb tmwfivww/cfié . %%6 4% $§wfiffi MAJ/W W 41%”? /7L5'c:§f ‘57: l W /M «Malt/C A54 . leéoerwf 4544/4; / WK flu 27:52) 4, Mae yam/1 b04401 4&5 wfl/ fiof bl WI?” .9 flwgmfflfi‘dm dj’m/Ikfi/WZS 7" b. (4 points) What is the role of RecA in recombinational repair? fl/z/s 7%; 344“ a; a? ssflA/A Fem/I flmI/fig’fims /39/V,9 Mudflhhoa Sfim/ Wééfl/Mgc/ , c. (6 points) Draw a diagram of a Holliday junction. Label the 5’ and 3’ ends of the DNA strands and use cross-bars (as in a ladder) to represent base-pairs, with half bars to represent any unpaired bases. ~Har€ bars = ~ 1 r ? ;/ 5" i/ v r’ 9’ WULfi/{le = ’t‘ ’9 mlartttfii = + .9: 10 WWW 7. (11 points) Indicate whether the following statements are true of DNA transposons (D), LTR retrotransposons (L), non-LTR retrotransposons (N), all three (A) or none (none). List all answers that are correct. These mobile elements are examples of “selfish DNA” A The DNA copies of these elements are synthesized using a 3’ hydroxyl generated by nickng at the target site N These elements are structurally similar to retroviruses, such as HIV Z’“ These elements can produce mutations in genes A __ These elements require a reverse transcriptase to move N $.- £— Elements of this type are responsible for the movement of antibiotic resistance genes in bacteria Z2 This type of element is responsible for “hybrid dysgenesis” in fruit flies D _ Synthesis of the CDNA copy of this type of element occurs in the cytoplasm of eukaryotic cells There are repeated sequences at the ends of these elements 2 Q 11 NAME 4&5/4/6’16 (EX 8. (10 points) a.i. (4 points) The yeast pro-I mutation makes the cells auxotrophic for proline. How would you obtain revertants of yeast pro-1 mutation? ‘W 075 W'i/I'W 7 W Way/M iv: WWW/V? /acé/WJ fl /%Wd§/" af/a/wc7 7% W W m¢7w§x a.ii. (2 points) Is this an example of a genetic selection or a screen? gin/hm b. (4 points) In an organism, such as the nematode, C. elegans, in which it is not possible to generate “knockouts”, what technique could you use to determine the phenotype conferred by loss of a particular protein? [Mg « [W4 Maw/m» Mafia/601a a? KAM’ W/fl» flu 5W Wag M %’ Wm“ W/aggfib l2 NAME fl Mme E £2 9. Short answer (18 points). a. The genotypes A1A1, A ,A2, and A2A2 survive with probabilities 0.8, 0.6, and 02, respectively, What are the relative fitnesses of the three genotypes? Dir/Mle 57 J5“ 77W; /) &.?’{'MX42SP b. Fill in the integers X and y in this statement: Mutation rates (from normal to disease alleles) for human genetic disorders range from as high as 10" to as low or lower than 10". 2&4 y‘7 0. List four assumptions of the Hardy—Weinberg equilibrium. Answer in 25 words or less. [AM/0m MM? [/40 rmmnrafi‘oml 140 ran/z W Mmfl [/mg/aflfi/lfl/ 7%), flo flat/74141 £544,627» V,‘a./97’/,%7 7 904 0W0, 40 Mae/Mam. / [WM “24/ [mfmw‘yfle )fi/«Wcm- w: flu gcxvs. d. Define threshold trait, and give an example. Answer in 25 words or less. You may also include a diagram. 7,3,4} MM is WW3? /= [flaw/9r, £704 D/QWWSJW hm M an Wfifiqw: flaws/- /%// @4722? Mame a Mffl Mai, flu {742/7 ,9 MM, flat/r4 {(9. W/za/flren/fi/ away}? Me. an ex fiWfl/flo W 6 /b/ 75"” 73' e. Colorblindness results from an X-linked recessive allele, If, in a certain population, 10% of males are colorblind, what proportion of females will be colorblind, assuming Hardy—Weinberg equilibrium applies? Wan/5:7 fl/Ofl/was 40/ f. East (1916) crossed long and short—flowered varieties of a tobacco relative to produce F1 and F2 generations. What feature of his results provided evidence against blending inheritance? Answer in 10 words or less. % Mark/nae MW 447/ Z77”; f: Wage» l3 NAME éNSVl/éfl 156? 10. (12 points) In a population of mice, there are two alleles of the A locus, A1 and A2, with 385 mice of genotype AI/Al, 290 of Al/AZ, and 325 of A2/A2. a. Calculate the frequencies of the three genotypes. 729W : /oaa / : a 335 azqa’ WC A325" b. Calculate the frequencies of the two alleles in this population. ftflx/ifl/J: 0,3675% 5.270/é 7 0"(3 Z : fl. 9‘? ct Using your answer from (b), predict the genotype frequencies that you would expect if this population were in HWE. Aéazrazs’of 2): 0.4? xa¢77 0'9‘73’2— ayfzs 0,2207 (1. Does the population appear to be in HWE? If not, briefly suggest a possible explanation (25 words or M, Wm fao £4; Maw/QM fl Mam? Wow/9744' flm'rmm Mth [W flflce 5;; A74 IMMVIg/aj Mazda/4&4» W] m SW5 Wm fl/S/fl/W;/' [mfg/)4} iy/wp Jaw a 2 MM/ .42 Maw fosgyw éWMAZD-J p7112/ 2/? mw—rmm flOLfiMj. flmy/Mfléfyf X; WWW me WWW/Jam. NAME AZL/fil/VZEK ge Z 11. (15 points) In Dar es Salaam, Tanzania in the early 1950's, 24% of adults were found to be sickle-cell heterozygotes; the rest were normal homozygotes. No homozygotes for the sickle—cell allele were found, presumably because all such individuals died in childhood. 21. What is the frequency of the sickle—cell allele in this population? (Hint: the answer does not require taking a square root, or even using a calculator). [9, W/z : 0/5» b. What is the selection coefficient against homozygotes for the sickle-cell allele? “/ c. Assuming that the population is at equilibrium for a balanced polymorphism, what is the selection coefficient against normal homozygotes (51)? /’,d_95‘/; /f§/’2 5' Z88—/:0l%& f /(9,/5é/ d_ Suppose that heterozygosity for the sickle-cell allele did not confer malaria resistance, so that heterozygotes and normal homozygotes had equal fitness. What mutation rate would be needed to account for the observed frequency of the sickle-cell allele, assuming mutation—selection balance? 7 ’ “2 “form/st Sada) a c Wei = weed e, At around the same time, frequencies of sickle—cell heterozygotes among adults in other African populations ranged from 11% to 38%, Give an hypothesis for why the frequencies varied so much. Answer in 25 words or less. ._ amt/(W MJ/ (a? flmaa WM) 15 NAME J El 12. (15 points) a. Two inbred lines of beans are intercrossed. In the F1, the variance in bean weight is measured at 1.5. The F1 is selfed; in the F2, the variance in bean weight is 6.0. Estimate the broad—sense heritability (HZ) 0f bean weight in the F2 of this experiment. V9: a4§=44§ V/: 5 {712-5 3 b. Suppose that two triple heterozygotes (AaBch) are crossed. Assume the three loci are unlinked. What proportion of the offspring will be aabbcc? What proportion will have exactly one upper-case allele? 44 Mac -' (m3= V94 W W 4M6 024% .' 5'(Vi)(I/4fij(%/): 3/3; c. The table below gives the correlations of monozygotic (M2) and same-sex dizygotic (DZ) twins for three traits, from a study of children by Huntley (1966). The traits are the number of skin ridges on the fingertips, height (adjusted for age), and a measure of social maturity. Estimate the heritability (H2) of each trait. Finger-ridge count Height Social—maturity score MZ twins 0.96 0.90 0.97 DZ twins 0.47 0.57 0.89 r 0. é fil/é 2 (I711?— 02 ) 9' 6S! 6’ d. Referring to the data in (c) and your heritability calculations, which trait seems to be most influenced by the shared environment of twins? Explain in 25 words or less. ‘ w Vii/W Soda/“mahn” W6 m3 fl/zs [/me Ac/Wem firm?! flh/ZA (KM/wk MM fl W/ 34>?“ MK flak/71% exp/«7 As (9747 ad- E) In another study, the following concordance values and population frequencies were found for three threshold traits, quackophobia (crippling fear of waterfowl), quackophilia (unnatural attraction to waterfowl), and death metal addiction (DMA). Quackophobia Quackophilia DMA MZ twin concordance 80% 2% 20% DZ twin concordance 25% 2% 20% Population frequency l% 2% 5% Based on these data, which trait is heritable? Which trait is familial but not heritable? Which trait is neither heritable nor familial? - . . ' ‘ ' ‘ 0' hid/w (pafickQ/hoguza/wrrfdéfal [M410 Whalléa/fi f éL/ Wflfloflwau M M/MLM 55.2, NAME fiZQSWL/CZ ge Z 11. (15 points) In Dar es Salaam, Tanzania in the early 1950‘s, 24% of adults were found to be sickle—cell heterozygotes; the rest were normal homozygotes. No homozygotes for the sickle-cell allele were found, presumably because all such individuals died in childhood. a. What is the frequency of the sickle—cell allele in this population? (Hint: the answer does not require taking a square root, or even using a calculator). er/ng'/J/ b. What is the selection coefficient against homozygotes for the sick|e~cell allele? SF/ +3 f c. Assuming that the population is at equilibrium for a balanced polymorphism, What is the selection coefficient against normal homozygotes (sl)? /;a.s>9: ///rs/) /¢§/= //a,8? 5/ ’ %2g—/=0,/%g$ 5752/ +3 d. Suppose that heterozygosity for the sickle—cell allele did not confer malaria resistance, so that heterozygotes and normal homozygotes had equal fitness. What mutation rate would be needed to account for the observed frequency of the sickle~cell allele, assuming mutation—selection balance? g” : a/Z FSfy/fa/gz): szy/[M a s 4sz g afl/‘r’fié e. At around the same time, frequencies of sickle—cell heterozygotes among adults in other African populations ranged from 11% to 38%. Give an hypothesis for why the frequencies varied so much. Answer in 25 words or less. _, Whj S/MflWWMVmflj majme M; W W 15 ...
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This test prep was uploaded on 04/17/2008 for the course BIO 198 taught by Professor Sia during the Fall '08 term at Rochester.

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BIO 198 - Exam 4 - 2006 Key - NAME W5 (5 You may pick up...

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