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#Question id: 15655
#Unit 13. Methods in Biology
You are studying a new strain of E. coli that can utilize the disaccharide melibiose very efficiently. You find that utilization depends on the enzyme melibiase, which is encoded by the gene Mel1. Mel1 is not expressed unless melibiose is present in the growth medium. Next you isolate a mutation, designated MelB–, which gives uninducible melibiase activity. Mapping experiments show that MelB– is linked to Mel1. Using an F' factor that carries the chromosomal region surrounding Mel1, you perform the following genetic tests:
Describe the proposal for the type of regulatory functions affected by the MelB– mutation.
TLS Online TPP Program
#Question id: 15656
#Unit 13. Methods in Biology
You are studying a new strain of E. coli that can utilize the disaccharide melibiose very efficiently. You find that utilization depends on the enzyme melibiase, which is encoded by the gene Mel1. Mel1 is not expressed unless melibiose is present in the growth medium. Explain the two regulatory pathways for Mel1 that can explain the behavior of the MelA– and MelB– mutations. For each regulatory pathways what is a role for the inducer melibiose?
TLS Online TPP Program
#Question id: 15657
#Unit 13. Methods in Biology
You are studying a new strain of E. coli that can utilize the disaccharide melibiose very efficiently. You find that utilization depends on the enzyme melibiase, which is encoded by the gene Mel1. Mel1 is not expressed unless melibiose is present in the growth medium.
You next construct a MelA– MelB– double mutant, which gives the following behavior:
Which of your two models is consistent with this new data?
a) the MelA- MelB- double mutant is identical to the MelB- mutant
b) MelA- is epistatic to MelB-
c) MelB- is epistatic to MelA-
d) the MelA- MelB- double mutant is identical to the MelA- mutant
Which of the following is the correct prediction about the double mutants?
TLS Online TPP Program
#Question id: 15658
#Unit 13. Methods in Biology
You are studying a new strain of E. coli that can utilize the disaccharide melibiose very efficiently. You find that utilization depends on the enzyme melibiase, which is encoded by the gene Mel1. Mel1 is not expressed unless melibiose is present in the growth medium.
Next, you isolate a third mutant, MelC–, which gives constitutive melibiase expression. The MelC– mutation is closely linked to Mel1 and MelB–. Genetic tests of the MelC– mutation yield the following:
As above, classify the MelC– mutation in terms of its basic genetic properties and explain how you arrived at your conclusions.
TLS Online TPP Program
#Question id: 16125
#Unit 13. Methods in Biology
You are studying regulation of the yeast enzyme glutamine synthetase (GS), which is encoded by the GLN1 gene. You have isolated two mutants, designated gln2– and gln3–, that give decreased GS activity. Mating of either gln2– or gln3– haploids to wild type produces heterozygous diploids that show normal amounts of GS expression. When you cross either a gln2– or gln3– haploid to a gln1– strain the resulting diploids show normal expression of GS.
From these experiments, Classify the gln2– and gln3– mutations in terms of their basic genetic properties explaining the rationale behind your conclusions. Based on these properties make a proposal for the types of regulatory functions affected by the gln2– and gln3– mutations.
TLS Online TPP Program
#Question id: 16126
#Unit 13. Methods in Biology
2 You are studying regulation of the yeast enzyme glutamine synthetase (GS), which is encoded by the GLN1 gene. You have isolated two mutants, designated gln2– and gln3–, that give decreased GS activity. Mating of either gln2– or gln3– haploids to wild type produces heterozygous diploids that show normal amounts of GS expression. When you cross either a gln2– or gln3– haploid to a gln1– strain the resulting diploids show normal expression of GS, that could explain the effect of the gln2– and gln3– mutations on the regulation of GLN1.
The GLN1 gene shows a rather complex regulation in response to different amino acids. When either glutamate (glu) or glutamine (gln) is added to the medium, the amount of GS expression diminished and when both glutamate and glutamine are added to the medium GS expression is shut off completely. The effects of different mutants on the response to glu and gln are shown below.
