2nd Midterm Exam Key version 2

2nd Midterm Exam Key version 2 - LS4.1 2nd Midterm Exam May...

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Unformatted text preview: LS4.1 2nd Midterm Exam May 19, 2008 Student Name K UID Discussion section (circle) Wed 11AM Wed 1PM Wed 3PM Thurs 9AM Thurs 11AM Thurs 1PM 1 2 3 4 5 Total 6 7 8 9 10 Each question is 10 points. Problem 1, Part A: MM Master Plate + proline + histidine Replica Plates: MM MM MM +proline + histidine + proline +histidine + lactose (no glucose) a) Fill out the table with + or“ for the genotype of each marker in the colony. 1.pro‘- lac + his + 2. pro_+_lac "" his "" 3. pro + lac his j: 4. pro “v lac + his "'" 5. pro + lac - his i 6. pro ‘— iac "" his + 7. pro +lac '1' his "- Part B: For a given bacterial culture. you observe the following results from replica plating: Plate Minimal Medium Supplement Number of colonies Master + Adenine + Threonine 500 Replica 1 none 100 Replica 2 + Adenine 300 Replica 3 + Threonine 200 b) What genotypes and how many of each are represented on the master plate? (Specify mutant and wild type alleles) Genotype adeit‘hr—‘L number M adeithr; number fl ad __""‘_thr_'L numberfig’w ad 6 Bill! i number ’00 fl-v— @‘WXZ‘YWéo-vlffi' Problem 2, Part A: You screen 50.000 Neurospora mutants for auxotrophs. Among mutants that grow on rich media but fail to grow on minimal medium. nine mutants are rescued by supplementation of arginine or a known arginine precursor. ("+" = growth; and "—" = no growth) Supplement to Minimal Medium Known Precursors Mutant none & g Q Q E Arginine arg- 1 — — — — — — + arg'z — — + + + + + erg-3 — — — + — — + erg-4 _ — + + + — + erg-5 — _ _ + + _ + erg-6 — — — + — — + erg-7 — — + + + _ + erg-8 — — — — — — + erg-9 — + + + + + + a) What is the biochemical pathway of arginine biosynthesis? Which step is interrupted by which mutant? ——-)A' "‘95—1—93—‘9b—‘9Cfiflrj l l l l l ‘f 51 if}? 5- :}IL II? Part B: Among many Neurospora crassa mutants that require tryptophan for growth, just four mutants fail to be rescued by supplement of indolyI—glycerol-3-phosphate (IGP) to minimal medium. However, conversion of [GP to tryptophan is a single enzymatic step. The following table gives the vegetative growth of Neurospora heterokaryons (hyphal fusion of twp fungal strains) on minimal medium: ("+" = growth; and "—" = no growth) Mutant trp-t' I‘m-2 ftp-3 trp-4 trp-t' — + + —~ trp-2 — — + trp-3 — + trp-4 _ b) Use the above data to propose two explanations why the four mutants could be blocked at the same enzymatic step. /57W MWroa/dsméw4 (iady) l; “K554? ggg w MMS c) If trp-t and trp-2 were opposite mating types A d a and could form ditéloid zygotes then spcrulate , how many of 1000 random ascospores would you expect to grow on minimal medium? i) 1-50 ii) 51-100 iii 101-250 choice i. ii. or iii v) none of the other choices t’ my) M file“ #05417»: éC/QVI% Problem 3: An interrupted mating experiment was performed using either of three Hfr donor strains (HfrA. HfrB or HfrC). The common recipient strain could be distinguished from the donor strains at eight genetic loci. The results of the three experiments are given below: Conjugal transfer time (minutes) Gene HIE mg i ale 1 5 — — gal — 5 5 his — ~ 20 lac 3O 40 — pro 35 35 — (hr 5 — 45 ton — — 25 trp — 1 5 — a) Construct an E. coli chromosome map with minutes from the above interrupted mating data. Set the HfrA integration site at 0 minutes and indicate the positions and orientations of the integrated F episomes. Assume the E. cell map is 100 minutes. b) You isolate a new Hfr strain (HfrD) which transfers ale at high frequency. but very rarely transfers lac, what gene order would you expect to be transferred in the first 40 minutes of mating? Make a rough genetic map for HfrD (without minutes). A45 M4 can «[4 > c) If HfrB excised from the E. coli chromosome as a F' plasmid and transferred tip at high frequency, what other gene(s) would you anticipate being transferred with the F' plasmid? Problem 4: You have three E. coli' strains (A1. A2 and A3) with genotype A. and two strains (B1 and 32) with genotype B. Genotype A: thr‘ pro" aia“ Genotype B: st!“ thr' pro+ ala+ Frequency of prototrophic recombinants from conjugation between A and 8 strains: m i a: ff F4 A1 10'? o *' A2 0 10'F gfir- A3 10'3 o a) Fill-in the fertility state of each strain (F. F’ and Hfr) in the above table. b) In order to determine the gene order of auxotrophic markers in genotype A and B, what strain combination would you use? (Indicate the donor and recipient and their fertility state.) Also, how do you recovery exconjugants only and exclude the growth ofthe donor strain or the recipient strain that fails to recombine? From preliminary analysis. you know that thr is transferred after pro and ala. Thus. after 30 minutes, you put mating culture on a master plate -minima| medium supplemented with proline, alanine and antibiotic-w and then replica plate the master onto additional minimal medium plates with or without supplementation. A day later you count colonies on all plates: Master Re lica lates minimal medium Plate gone +Prolige +Alanine Number of colonies 549 245 249 289 c) In this experiment what are the possible genotypes for exconjugants? Each of the possible genotypes are represented by how many colonies on the master plate? pro+ aria+ é? j 915'- pro+ ala— ? y pro— ala+ .—‘/_ -—--""' pro_ aia_ 9‘20 2 $93 d) What is the gene order for the three auxotrophic markers? Provide a figure as yourjustification? Problem 5: A P1 tranducing phage lysate is made from arg‘ l‘ys+ trp’ E. coil (donor) and used to infect a arg— iys‘ trp‘ strain (recipient). The recipient culture is then plated on a master plate, from which three replica plates are made: Minimal Medium Plate Supplement Number of Colonies Master , +Tryptophan +Lysine 100 Replica 1 +Tryptophan 60 Replica 2 +Lysine 40 Replica 3 none 20 a) What marker(s) are selected on master plate? What markers are unselected? ZJZQW b) What are the co-transduction frequencies? ra7£r =7§o a? [09 J) M9“75 = [0% Anthat are the possible orders of the three genes? @ ’ x479 ff? —-——/7r ‘5? /7””’/7’ #fl] d) Propose one additional experiment with the same donor and recipient strains to distinguish between your answers to part (c). Problem 6: In a screen for tryptophan auxotrophic mutants in E. coli. you isolate 3 different alleles. In order to map their locations within the tryptophan synthase (trs) gene you carry out transduction experiments listed below using these 3 mutants and the flanking maltose utilization marker (mat maps to the left oftrs). + _ — _ Experiment1— Donor: mat trs—t' .recipientmai firs-2. Select irs+ recombinants on master plate with minimal media. Replica plate on minimal medium with maltose (and no glucose). Results: Master plate = 100 colonies Replica plate = 80 colonies a) What is the liker order of the mat, it's—1 and ire-2 markers? M #51 k/ + _ — _ ExperimentZ- Donor: mat ire-3 .recipientmai (rs-2. Select firs+ recombinants on master plate with minimal media. Replica plate on minimal medium with maltose (and no glucose). Results: Master plate = 120 colonies Replica plate = no colonies b) What is the liker order of the mat, trs-2 and (rs-3 markers? Mites 3' +51 0) What is the liker combined order of the mat, trs 1, trs2 and t‘rs3 markers? M 7Q») 3 94-52. 745/ d) Make a sketch showing the transduction recombinations necessary to produce mai+ trs+ colonies from mai+ trs-3_ donors and mai— trs- 1— recipients. W3 t Problem 7: A variegated plant has patches of white tissue on leaves and stems that are normally green. You have twa inbred cucumber plant lines. One plant is variegated and cold-sensitive, and the other plant is all green and cold-tolerant. You perform reciprocal crosses between the two parental cucumber lines by pollinating the pistil (female) of one parent with the stamen (male) of the second parent. When you cross a male, green, cold-tolerant cucumber plant and a female variegated, cold~sensitive, all F1 progeny are cold-tolerant plants. As for leaf color, however, many but not all F1 plants are also variegated: Some F1 plants are all white. and some more are all green. In subsequent generations, selfed variegated plants again produce cold-tolerant plants with a mixture of variegated plants, all white and all green plants. When you cross a male, variegated, cold-sensitive plant and a female, green, cold-tolerant plant, all F1 progeny are new green and cold sensitive. In subsequent generations, the green, cold—sensitive plants breed true. a) What is the inheritance pattern of the variegated leaf color? i) biparental cytoplasmic inheritance ii biparental nuclear inheritance aternal cytoplasmic inheritance iii) paternal cytoplasmic inheritance iv) sex chromosome linked inheritance b) What is the inheritance pattern of the co|d~tolerance? i) biparental cytoplasmic inheritance ii) biparental nuclear inheritance ii maternal cytoplasmic inheritance (in eternal cytoplasmic inheritance iv) sex chromosome linked inheritance c) How can a variegated plant produce a mixture of all white, variegated and all green plants? What is the relevant enoty e of each colorty e? i . 9 Wish 4 MM M M M” /f/. ’7. M M Ulrréj-ac ' , JAN/£179 pf/u/ 5‘7 ' ' i / f ‘ flemjg Kim-i 54.5%;- #4532243} IVA g 5WM49£§ 99/414“ d) If white color in variegation is due to a defect in the photosynt etic apparatus. what would explain the cold sensitive phenotype? Provide a diagram of organelleslgenotypes of each of the original parents to explain your ' answer. 04%;“ Probtem 8, Part A: A family comes to see you with a question about the chance their first born child will have cystic fibrosis. The husband's sister has cystic fibrosis. The wife‘s family has no history of this disorder. In the United States, the frequency of newborns with cystic fibrosis is about “2,500. a) What is the chance? (SHOW ALL YOUR WORK LEADING TO YOUR ANSWER.) 2/3xl/2 x Mix/L Zea "' 7/50 5% 3%xl/szfi/5'OKI/z. 9' ’50 Part B: A family comes to see you with a question about the paternity of their child. The mother is blood type A, the father is blood type B and the child is type 0. In this population, the blood type allele frequencies are 0.20 for 0.10 for I5 and 0.70 tori. b) What would the genotypes of the parents have to be for this child to be theirs? mother- f4 5 father~ I; Z. c) Assuming the mother is the mother, what is the chance of the father being the required genotype? / WM : W W " " ° n. J)! 4....) u 4 an“ 4')“ "" P"? ' d) Given the father’s phenotype, what other genotype is possible and what are the odds he is one genotype or the other? (SHOW YOUR WORK.) 3 :1”; at 0M5 {m 1* “57% 1. 5 .0/ {iii/1 ://7/5/ .0/ Problem 9: In the table below. nine ril mutants of phage T4 were used in pairwise mixed infections of E. coli strain B hosts and their progeny plated on E. coil strain K12(lambda) hosts. on n In the table "0" means the progeny do not produce any plaques on E. coil K12(lambda) cells; — means that only a very few progeny phages produce plaques, and "+" means that many progeny produce plaques (more than 10 times as many as in the — cases). Gwyn-aid C! The McGrau-Hul' Euripides. 1m Pemim required In! reproduction or display. l 2 3 4 S 6 7 8 CJ 1 — + + + + e — + + 2 —~ + + + + — + + a t] — + l) + + — 4 -— + —- + + + S w- + — + + 6 0 U — + T (l + + 8 a + t) .._ a) Use these data to map a map ofwhere the mutations are relative to each other. to the extent possible. Make clear the location andlor extent of each mutation. “I 4 Ci: 9‘ l 1:: 0‘2”) [E b) If mutant 9 complements mutants 1,2, 5, 7 and 8. and mutant 5 fails to complement all the other mutations except 3. 4 and 9, how many genes are represented in this sample? Which mutations are defective for each gene? c) If a new mutation is obtained which fails to recombine with mutants 3, 5 and 7. but recombines with all the other mutations, what conclusion might you reach about the structure of the HI locus? flit/‘2— '4 Problem 10, Part A: Pola_ty_ri LilLiLLiiiLi DNA 5(lflééfiiifiziifii 3I££££Liifiii11 :30 :5 U1». I: |<>\ [m |o lo I}; S: n E )9 J> Protein N T r' E I l e 3) Finish filling in the nucleotides, amino acids and polarity of DNA. mRNA and codons in the table ove. Part B: The N-terminal amino acid sequence ofa protein that you are studying is below. The wild type allele You have isolated an auxotrophic mutant allele of this same gene. You next select for prototrophy and isolate a revertant. When you sequence the N-terminus ofthe revertant, you find that the initial peptide sequence, before the first c steine residue, has chan ed: y 9 +5 AueAuanAduC/cheéél) iMld type: ...Me1' - Ile - A5n — Ser - Vol - Cys MG all I. "11"1 ' ilficfi' ‘ 'l x 1 | ‘l t ‘1 I. I 1 I u r ' l- .n ‘ i ‘1 I r ; l - . l : Revertant: ...Ile - Mp: Lyfi - Pine;— Commuhl '5 The nicGrow-Hiu Emma. Inc. POI'TI'II‘SW Wired for reproducum or uiaptay Second letter (—‘5 “E E z A 1.3%.. ‘z’ \_/ Fsrst letter Jana} pill-U. c) Indicate above the nucleotide coding sequence of the wild type mRNA (up to the first cysteine residue). ...
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2nd Midterm Exam Key version 2 - LS4.1 2nd Midterm Exam May...

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