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Genes & Gene Expression

by: Charity Daniel

Genes & Gene Expression BIS 101

Charity Daniel
GPA 3.76


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This 4 page Class Notes was uploaded by Charity Daniel on Tuesday September 8, 2015. The Class Notes belongs to BIS 101 at University of California - Davis taught by Staff in Fall. Since its upload, it has received 10 views. For similar materials see /class/187632/bis-101-university-of-california-davis in Biological Sciences at University of California - Davis.

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Date Created: 09/08/15
Another conjugation problem 1 In E coli the ability to utilize the sugar lactose synthesize the amino acid methionine and resist death by the antibiotic streptomycin is conferred by the alleles lac met and strR respectively Mutants for all three alleles ie lac39met39strs cannot tolerate streptomycin and require glucose and methionine in their growth medium E coli strains A B C and D are characterized by being donors or recipients and their ability to grow in certain media as indicated below DOES IT FORM A GROWTH IN MEDIA CONTAINING CONJUGATION STRAIN TUBE woose lactose glucosemet glucosemelslr lactosemelslr A yes yes yes yes no no B yes yes yes yes no no C yes yes yes yes no no D no no no yes yes no In separate experiments male strains A B and C are mated with strain D for 15 minutes mating is then interrupted and the resulting cells are checked for their ability to produce colonies onthe following nutrient plates Are GROWTH IN MEDIA CONTAINING These Cells MATING glucosesir glucosemelslr lactosesir lactosemelslr Donors A X D yes yes no no yes B X D no yes no no yes C X D no yes no yes no A Identify the genotype of each strain at the LAC MET and STR loci and identify each strain as either F F F or Hfr Hint Determine the genotypes from the first chart and confirm the F plasmid status from the second STRAIN A STRAIN B STRAIN C STRAIN D B What allele does the F strain donate C What allele does the Hfr strain donate D What was the purpose of adding the streptomycin to all the mating selection plates Answers A Inspection of table 1 allows us to determine the genotype of the strains but not the F plasmid status Strain A It is a donor forms conjugation tube met grows in glucose or minimal media without the need of any aminoacid lac has the ability to grow in lactose StrS unable to grow in presence of antibiotic streptomycin Strain B Same as strain A donor met lac StrS Strain C also same as strains A and B donor met lac StrS Strain D recipient no tube so must be F met needs glucose and methionine to grow lac does not grow in lactose StrR grows in presence of antibiotic We inspect now table 2 which provides data on the exconj ugants recipient strain D after conjugation A x D conjugation changes recipient D to met since it grows in minimal media glucose otherwise its genotype remains the same It is also changed from recipient to donor The met allele was donated by a piece of chromosome of strain A Therefore donor strain A and resulting exconjugant must be F to be a donor and to carry an allele from strain A in the newly acquired F plasmid B x D conjugation changes recipient D to donor However its genotype remains the same met lac strR Since it is a donor and does not carry any allele from B similar to donor strain E the exconjugant must be F C x D conjugation changes the recipient D to lac and conserves its recipient status Therefore the donor strain C is Hfr since the resulting exconjugant is not donor and carries a gene from strain C The purpose of adding streptomycin is to kill all the donor strains in the three mating experiments Three Point Mapping Problem Suppose you analyze three phenotypes in the commo fruit y Drosophila crossveinless absence of crossveins in the wings echinus an eye abnormality and cut wing a wing shape abnormality The recessive alleles causing the mutations are represented by av EC and ct respectively A test cross is performed between a trihybrid female with the wild type phenotype and a male with the three recessive phenotypes The following progeny are obtained Phenotype Number 1 echinus 73 2 echinus cut 140 3 wild type 3 4 crossveinless echinus 766 5 crossveinless cut 80 6 cut 756 7 crossveinless echinus cut 1 8 crossveinless 150 1977 a Why do these data indicate genetic linkage of the three loci The data deviates from expectations for independent assortment 18 for each class Furthermore the data can be classified in pairs on the basis approximate number of individuals per class class 48 6 are the highest likely to be parentals class 3 amp 7 are the lowest likely to be double cross overs class 2 amp 8 1amp5 intermediate likely single cross overs b What is the order of loci There are three possibilities F1 Double cross over cv ec cv 1 gt ct ec ct This order does not agree with data parentals since double cross over classes are wild type 3 and triple mutant 7 F1 double co 2 ec cv ec gt ct mt This order does not agree either 3 F1 double co ec cv 23 gt ct ec ct cv This order certainly agrees with the data so it must be it c What are the alleles on each of the homologous chromosomes of the trihybrid female As in the third possibility above loci 1 and 2 and 2 and 3 in trans loci 1 and 3 in cis d What are the recombination distances between each adjacent pair of loci ecct interval number of ies in classes 2 8 recombinants at this interval classes 3 7 double co total number of ies 140 150 3 1 2941977 015 recombination fraction RF between ec and ct 15 recombination or 15 map units ct cv interval classes 1 5 3 1 73 80 31 1571977 0079 RF between ct and cv 79 recombination or 79 map units e What is the interference value for these data I 1 c c coef of coincidence observedexpected number of double co expected dco product of RF for each interval multiplied by the total number of gametes expected dco 015 x 0079 0012 x 1977 237 c 3 1237 0167 1 1 0167 0833 Therefore 83 of the double cross overs failed to materialize in this cross due to mechanical interference


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