LIFESCI 4 Lecture Notes - Lecture 9: Azide, Hfr Cell, Streptomycin
LIFESCI 4 - Lecture 9 - Bacterial Genetics: Transduction
Gene Transfer in Bacteria
- Bacteria can pick up DNA by infection
- Conjugation: DNA goes from F+ (donor) to F- (recipient)
- F+ has fertility factor
- After conjugation, F- becomes F+
- During conjugation, only the plasmid has the fertility factor
- Recombination:
- DNA is transferred through the pilus
- After entering F- cell, the fertility factor is an exogenote → must do double
crossover with F- endogenote to insert itself into donor DNA permanently
-Exconjugant: a bacteria that was formerly F- but has now acquired an extra piece
of DNA (the fertility factor)
How was conjugation discovered?
- When 2 bacteria with complementar genotypes are crossed:
[image provided by Professor Pham]
- Not a good experimental design because there is only 1 mutation (met- thr+)
- A higher probability that the results occurred by chance
- E.g. Out of 2x10^8 for (met- thr+), there are 25 mutants → there are
random mutations where F- revert back to F+
- Improved design → using 4 instead of 2 markers
[image provided by Professor Pham]
- Probability of two mutations simultaneously reverting back is very slim
- When doing conjugation, there are 20 that were mutated compared to 0
reversion in either parent
- Can conclude that when bacteria are mixed, they can exchange genes such that
recombinants will grown on minimal media
Interrupted mating
- Selection is done so that the only things remaining on the plate are the recombinants →
experiments can be done on these
- Hfr and F- are killed → add in elements that Hfr and F- are sensitive to
- To select against Hfr, add in streptomycin (str)
- To select against F-, add in either (azi) or (ton)
Clicker 1:
[provided by Professor Pham]
A) zi T on Lac GalA s s+−
B) zi T on Lac GalA r s−−
C) zi T on Lac GalA s r+−
D) zi T on Lac GalA r r−−
E) None of the above
Explanation:
- Master plate contains MM + strep + glucose → this kills the bacteria strains containing
and the remaining strains that survive must be trs Strs r
- Replica plate with MM + azide: colonies that grow must be resistant to azide since they
grow despite the presence of azide
- The colony indicated by red arrow must be zia r
- Replicate plate with MM + T1 phage: colonies that grow must be → colonies thatont r
don’t grow would be ont S
- The colony indicated by red arrow must be because it does not appear onont S
this replica plate
- Last 2 replica plates with MM + lactose and MM + glucose: the indicated colony does
not grow on either, which means it must be since it cannot utilize the sugarsac gall − −
given and grow
- Conclusion: add up the above analysis to find that the colony indicated by the red arrow
must have genotype zi T on Lac GalA rs− −
Clicker 2:
You have isolated 100 auxotrophic strains in E.coli. None of these 100 strains can grow on
minimal media, but al 100 can grow on minimal media plus methionine and arginine. Further
analysis shows that 30 can grow on minimal media plus arginine and 20 can grow on minimal
media plus methionine. Fill in the number of strains that have the following genotype and
identify how many have a genotype.et argm −−
- → 50et argm −−
- → 30et argm +−
- → 20et argm −+
- → 0et argm ++
- TOTAL → 100
Explanation:
- Example: If there is MM + arginine on the plate and 30 of them grow, then those
30 colonies must lack ability to synthesize arginine (arg-) and is (met+)
Document Summary
Lifesci 4 - lecture 9 - bacterial genetics: transduction. Conjugation : dna goes from f+ (donor) to f- (recipient) During conjugation, only the plasmid has the fertility factor. After entering f- cell, the fertility factor is an exogenote must do double crossover with f- endogenote to insert itself into donor dna permanently. Exconjugant : a bacteria that was formerly f- but has now acquired an extra piece of dna (the fertility factor) When 2 bacteria with complementar genotypes are crossed: Not a good experimental design because there is only 1 mutation (met- thr+) A higher probability that the results occurred by chance. Out of 2x10^8 for (met- thr+), there are 25 mutants there are random mutations where f- revert back to f+ Improved design using 4 instead of 2 markers. Probability of two mutations simultaneously reverting back is very slim. When doing conjugation, there are 20 that were mutated compared to 0 reversion in either parent.
