LIFESCI 4 Lecture Notes - Lecture 10: Bacteriophage, Microbiological Culture, Escherichia Coli
LIFESCI 4 - Lecture 10 - Gene Order in Bacteria
Clicker 1:
You infect a 10 mL bacterial culture with bacteriophage. At various mes you ter the01 3
culture. The results are shown in the graph. How long is the life cycle?
What is the average burst size for this phage?
A) 1
B) 10
C) 100
D) 1000
Explanaon :
- At t=0, there are 100 pfu/mL
- At t=40, there are 1000 pfu/mL
- → the concentraon of bacteriophage grew by 10 mes for each cycle0
100
1000 = 1
(40 minutes each)
Clicker 2:
What is a single pfu (plaque forming unit)?
A) A single bacteria
B) A single bacteria infected with a bacteriophage
C) A single bacteriophage
D) More than one of the above
Explanaon :
- A plaque forming unit means that a plaque will be formed → can be applied for a single
bacteria infected by a bacteriophage
- When a bacteria is infected by a bacteriophage, it will be lysed (killed) in the end
→ burst → make a hole (plaque)
- When a single bacteriophage infects a bacteria, a plaque is formed
- Both (B) and (C) are correct
Clicker 3:
A stock of phage T4 is diluted by a factor of and 0.1mL of it is mixed with 0.1mL of 01 4
−01 8
E.coli/mL. The mixture is plated onto an agar plate. The next day 1130 plaques are visible. What
is the concentration of T4 phages in the original T4 stock?
A) .13 0 pf u mL 1 × 1 5 /
B) .13 0 pf u mL 1 × 1 6 /
C) .13 0 pf u mL 1 × 1 7 /
D) .13 0 pf u mL 1 × 1 8 /
E) .13 0 pf u mL 1 × 1 9 /
Explanaon :
- Need to do diluon since otherwise there would be too many to count
- Serial diluon is used when dilung the phage T4
- = diluon factor01 4
−
- where represents concentraon0
M1
M2= 1 4
−M
-V VM1 1 =M2 2
- Queson is asking for value of M1
- Find = solute / solvent = M2130 pfu 0.1mL .13 0 pfu mL 1 /= 1 × 1 4/
-0 1.13 0 pfu mL)(10 ) .13 0 M1 =M2
10 4
−=M2× 1 4= ( × 1 4/4= 1 × 1 8
- For 0.1mL of E.coli, it does not affect the calculaons → simply allows for counng and
visualizaon
Clicker 4:
What is the map distance between and ?rarb
X ra−rb− → 80% Large, 20% Small
A) 20 m.u
B) 30 m.u
C) 40 m.u
D) 50 m.u
E) None of the above
- and is wild type (large)ra−rb−
- and is smallra+rb+
Parental
______ra−rb+
Large
0.3
Parental
______ra+rb−
Large
0.3
Recombinant
______ra−rb−
large
0.2
Recombinant
______ra+rb+
small
0.2
- The two recombinants adds up → 0.2 + 0.2 = 0.4 → 40 m.u
Document Summary
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