LIFESCI 4 Lecture Notes - Lecture 11: Meiosis, Escherichia Coli, Bacteriophage

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LIFESCI 4 - Lecture 11 - Recombination and Complementation Analysis
Review from online material:
- Difference between B and K:
- B: permissive host
- Phages can infect and lyse E.coli B
- K: non-permissive (restricve)
- There are 2 phenotypes:
- Morphology (shape) of plaque
- Wild type: small plaque
- Mutant: large plaque
Recombination Analysis
- see how far 2 genes/mutaons are
- Whether or not they are overlapping
- The distance between the 2 genes/mutaons
- In diploids, must look at meiosis
- In T4 phage (haploid, no meiosis) → need to co-infect them
- High M.O.I (mulplicity of infecon)
- High: the concentraon of the phage is high enough for both plates
- Cannot do low M.O.I because the (rII- 1) and (rII- 2) would not enter cell at the
same me
- Recombinaon must be done with high M.O.I and permissive strain
- In the second and third trial (rII- 1) and (rII- 2), you should not see any progeny grow on
K since it is non-permissive
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- The number that does end up growing on K is the reversion rate
- These 2 trials are meant to find the reversion rate
- Question: How many map units separate RII1 and RII2?
-0
# total
# wild type reversion =10 pfu mL
3 /
10 pfu mL
9/= 1 6
Complementation Analysis
- The different colors represent different genes (RIIA and RIIB)
- rII locus composes of A and B
- For T4 to lyse E.coli K (which is non-permissive), need (A) and (B) protein
- rII1 (red) has mutaon in A → can make protein B but not protein A
- No growth in E.coli K because it lacks A (cannot lyse)
- rII2 (blue) has mutaon in B → can make protein A
- No growth in E.coli K → unable to lyse due to lack of B
- Last trial which is high M.O.I: (rII1 and rII2 combined)
- rII1 can make B, rII2 can make A → combined, it is able to lyse
- Can share protein
- The DNA are mutants in this last trial
Clicker 1: RII Locus
Mutations rll#j and rll#k are separated by:
A) 0.1 mu
B) 0.2 mu
C) 0.5 mu
D) 1.0 mu
E) None of the above
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