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UH - BIOL 3301 - Class Notes - BIOL 3301 Ch. 6 &14

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Schools > University of Houston > Biology > BIOL 3301 > UH - BIOL 3301 - Class Notes - BIOL 3301 Ch. 6 &14

UH - BIOL 3301 - Class Notes - BIOL 3301 Ch. 6 &14

School: University of Houston
Department: Biology
Course: Genetics
Professor: Chin-Yo Cooper
Term: Fall 2016
Tags: Genetics
Name: BIOL 3301 Ch. 6 &14
Description: Chapters 6: Gene Mapping and Analysis in Bacteria and Bacteriophage. Chapter 14: Prokaryotic Gene Expression (IN FULL). Week 9 and first part of lecture for week 10. Lecture notes plus breakdown of components
Uploaded: 10/26/2016
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background image Week 9 Notes: ​ Chapters 6 (pg 1-6) & 14 (pg 7)  10/18/16  Gene Analysis and Mapping ​ in Bacteria and Bacteriophage    Why use bacteria and bacteriophage model systems?  ● Grow quickly→ Faster Results  
● Large population size (accuracy) 
● Defined environment, precise control of experiment 
  Viruses  Genetically streamlined→ carry genes only needed to reproduce  -Obligate parasites 
-Small Genomes→ circular or linear 
-Extremely Abundant,   10x number of bacteria    Prokaryotes:  Bacteria and Archaea   -No nucleus 
-Haploid→ Only homozygous genomes 
-Single Circular genome  
+Plasmids: nonessential, extra genes→ can be incorporated into chromosome    Prototroph ​: Able to synthesize everything needed from growth (AA, vitamins, nucleotides etc.)  Auxotroph ​: Unable to grow on its own, requires elements from media for growth   
Phenotypes Tested to distinguish Bacteria: 
-Antibiotic resistance 
-Metabolic capabilities 
-Auxotrophy, prototrophy 
-Colony growth, formation 
-Growth on Different Media Types 
 
  Metabolic Outputs-  Cell assembly of Vitamins, Amino Acids, Nucleotides   Metabolic Inputs-  Potential energy sources, Sugars    Mutation in Metabolic Output Genes  Mutations in metabolic output genes are detrimental  -Bacteria can only grow if the metabolic output is in media   
 
 
 
 
 
background image   Unless mutation is listed, assume gene is functional!   
Mutation in Metabolic Input Genes 
Mutations in metabolic input genes prevent bacteria from metabolizing that energy source  -Must be alternative energy source in media for survival      How Bacteria Transfer Genes  Vertically:​ ​duplication 
Parent→ Daughter cell 
 
Horizontally:​ ​recombination 
 Conjugation
​- Transfer of plasmid or partial genome  via sex pilus. Donor→ Recipient 
 Transduction
​- Transfer as part of viral genome, viral  insert 
 Transformation
​- DNA uptake from environment,  competence factors needed (NOT ON EXAM) 
 
Conjugation   
Lederberg's Conjugation Experiment 
Two complementary auxotrophic bacteria tested:  Bacteria grown separately on minimal media→  no growth  Complimentary bacteria  mixed  and  grown together  on minimal media→  growth   
Proved recombination existed among bacterial species 
→ But How?     Was the DNA transferred into broth then transformation brought DNA into recipient?  ● U-Tube experiment- fine filter in between allowed DNA to pass, not cells  → No growth  → Cell to Cell contact required 
 
 
 
 
 
 
 
 
 
 
 
background image  
 
 
 
 
 
 
 
 
 
 
 
 
 
 
Fertility ‘F’ Factor  Element of transfer of genes, F Plasmid  -Encodes all genes necessary to transfer itself 
between donor and recipient cells 
-Products of these genes (the 
tra  genes): facilitate  transfer by binding to the  oriT  DNA sequence and  unwinding one strand of DNA that is transferred into 
a recipient cell 
F+ : F plasmid present → donors  F- : F plasmid absent → recipients   
-Recipients then become ‘transconjugant’ or 
‘exconjugant’ 
After donation-
​ F+​ remains ​F+ and F- ​→​ F+   
 
 
 
 
 
 
 
 
 
 
 
 
 
 
background image   Hfr Strains  H ​igh ​F​requency of ​R​ecombination Strains    F+ transfer chromosomal genes to recipient bacteria at very high frequency  → ~1/10,000 to. ~1/10,000,000   F Plasmid is integrated into chromosomal DNA  → recombination mediated by similar genetic regions  → Insertion Sequence (IS) elements   
Only cells where the F plasmid is integrated into chromosome can transfer 
chromosomal genes 
 
Double Crossover Required✩  F Factor has ~4 IS elements 
E. Coli has ~10 IS elements 
● Transferred through normal plasmid transfer process → genes need to  recombine into recipients chromosome in order to maintained (Otherwise degraded by 
recipient cell) 
● Only DNA between the two crossovers is incorporated into recipient chromosome, the DNA  piece corresponding in recipient is then excised and degraded→ New DNA incorporated, 
ligase 
    F¹ Plasmid  Integrated F plasmids (Hfr strains) can excise from chromosome to reform a free plasmid  Imperfect excision→ F takes copy of the partial chromosome→  F Prime Plasmid   
 
← Single Crossover is sufficient since incoming DNA is circular 
 
← Integration creates 
duplicate sequences  on either side   
 
← Excision occurs by a second single crossover- if the original site was 
present in multiple copies 
→ this process can excise region of bacterial chromosome 
  
✧If F¹ -lac transfers to  lac- cell, it can compliment to give lac+ phenotype 
 
 
 
 
 
 
 

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School: University of Houston
Department: Biology
Course: Genetics
Professor: Chin-Yo Cooper
Term: Fall 2016
Tags: Genetics
Name: BIOL 3301 Ch. 6 &14
Description: Chapters 6: Gene Mapping and Analysis in Bacteria and Bacteriophage. Chapter 14: Prokaryotic Gene Expression (IN FULL). Week 9 and first part of lecture for week 10. Lecture notes plus breakdown of components
Uploaded: 10/26/2016
18 Pages 67 Views 53 Unlocks
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  • Notes, Study Guides, Flashcards + More!
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