LIFESCI 4 Lecture Notes - Lecture 8: Arabinose, Arginine, Methionine
LIFESCI 4 - Lecture 8 - Bacterial Genetics and Review
Clicker 1:
Various strains of bacteria were incubated on plates containing minimal media plus several
amino acids as indicated. From the pattern of growth, predict genotypes of colonies 1-4.
Which colony (a, b, c, d) has the genotype arg-, met+, asp-, ile+?
Answer: (a)
- First class of mutation: metabolize mutants
- Mutation in ability to make something metabolize (amino acids)
- Arg-: (-) indicates the non-ability to synthesize arginine
- Met+: (+) indicates ability to synthesize methionine
- Genotype has 4 types → first look at colony (a):
- Plate 5:
- No arginine → (a) does not grow
- (a) indicates arg(-), which means that (a) cannot synthesize arginine by
itself
- Plate 4:
- No methionine → (a) grows
- (a) indicates met(+), which means that (a) can synthesize methionine by
itself
- Plate 3:
- No asp → (a) does not grow
- (a) indicates asp(-), it cannot synthesize asp by itself
- Plate 2:
- No ile → (a) grows
- (a) indicates ile(+), it can synthesize ile by itself
Clicker 2:
The plates shown below contain different sugars as carbon sources as indicated. Cells were first
plated on minimal media with glucose (master plate) and then replica plated onto plates with
lactose, mannose, or arabinose (no glucose).
1. Which colony has the genotype lac- man- ara+?
Answer: (e)
- Plate 1: (a) (b) (c) indicates lac(+), so (d) (e) must indicate lac(-)
- Plate 2: (a) (b) (e) indicates man(+), so (d) can be eliminated from the choices
- Plate 3 verifies (e) as the correct answer
2. What is the genotype of colony c?
A) Lac+ man+ ara+
B) Lac+ man- ara-
C) Lac- man- ara+
D) Lac- man+ ara+
E) None of the above
Explanation:
- Looking at colony (c) → go through each plate
- Lactose plate → (c) grows → must be lac+
- Mannose plate → (c) does not grow → man-
- Arabinose plate → (c) grows → ara+
- (c) genotype = lac+ man- ara+
Clicker 3:
Hfr strain is crossed with F- strain . You mate and plate ontr trp lys hiss S+ + + tr trp lys hiss r− − −
minimal media plus streptomycin and histidine and get colonies.
Which genes are selected for on the master plate?
A) Strep and histidine
B) Strep, tryptophan, lysine, and histidine
C) Strep, lysine, and histidine
D) Strep, tryptophan and lysine
E) None of the above
Explanation:
-Conjugation experiment
: mating between bacteria
- Need one strain of /Hfr and one strain of F-
- Hfr donates DNA while F- receives DNA → recombination occurs in F-
- After mating, there is selection for the exconjugant
Document Summary
Lifesci 4 - lecture 8 - bacterial genetics and review. Various strains of bacteria were incubated on plates containing minimal media plus several amino acids as indicated. From the pattern of growth, predict genotypes of colonies 1-4. Mutation in ability to make something metabolize (amino acids) Arg-: (-) indicates the non-ability to synthesize arginine. Met+: (+) indicates ability to synthesize methionine. Genotype has 4 types first look at colony (a): No arginine (a) does not grow (a) indicates arg(-), which means that (a) cannot synthesize arginine by itself. No methionine (a) grows (a) indicates met(+), which means that (a) can synthesize methionine by itself. No asp (a) does not grow (a) indicates asp(-), it cannot synthesize asp by itself. No ile (a) grows (a) indicates ile(+), it can synthesize ile by itself. The plates shown below contain different sugars as carbon sources as indicated. Plate 1: (a) (b) (c) indicates lac(+), so (d) (e) must indicate lac(-)
