Four independent lac⁻ mutants (mutants A to D) are isola...

Question

Four independent lac⁻ mutants (mutants A to D) are isolated in haploid strains of E. coli. The strains have the following phenotypic characteristics:

Mutant A is lac⁻, but transcription1 of operon genes is induced by lactose.
Mutant B is lac⁻ and has uninducible2 transcription of operon genes.
Mutant C is lac⁺ and has constitutive3 transcription of operon genes.
Mutant D is lac⁺ and has constitutive3 transcription of operon genes.

A microbiologist develops donor and recipient varieties of each mutant strain and crosses them with the results shown below. The table indicates whether inducible, constitutive, or noninducible transcription occurs, along with and growth habit for each partial diploid. Assume each strain has a single mutation.

Mating Transcription and Growth
A × B lac⁻
A × C lac⁺, inducible
A × D lac⁺, constitutive
B × C lac⁺, inducible
B × D lac⁺, constitutive
C × D lac⁺, constitutive

Use this information to identify which lac operon gene is mutated in each strain.

Accepted Answer

Welcome back, everyone. Let's look at our next question. A researcher wants to identify the LAC operon gene that is mutated in a lack mutant strain of E COLI that does not produce beta galactica to do this. They perform a complementation assay by transforming the lac mutant strain with a plasmid with the genotype lac I minus lac C plus to create a partial diploid strain. Which of the following statements accurately describes the expected results of this complementation. Essay choice. A the partial diploid strain will be LAC plus and produce functional beta galactica only if the mutation is in the LAC I gene. Choice B. The partial diploid strain will be lac plus and produce functional beta galactosidase. Only if the mutation is in the LAC Z gene. Choice C, the partial diploid strain will be lack minus and produce no functional beta galactica. If the mutation is in either the lack I gene or the LAC C gene and choice D, the partial diploid strain will be LAC plus and produce functional beta galactica regardless of which gene is mutated. Well, this is a little complicated on first glance, but let's just think through what these two genes do that. We're looking at black eye is the gene for the lac repressor. So a cell, we say therefore, a cell that is lack I minus will produce the lac genes including beta gal all the time. So, regardless of the presence of lactose, which is usually what activates the gene. If the lac repressor is mutated, it can't bind to the operator and can't block the expression of those lac genes when there's no lactose present. So whether the lack I gene is mutated or not mutated will not affect. So we'll put a little arrow here. No effect on whether the beta gal is functional. It will affect whether it's produced in the absence of lactose but not whether or not it's functional LA Z is a structural gene under control of the lac operon. And that is the gene for beta galactosidase or we write beta gal for short. So, a mutation in a lay gene, la Z minus means no functional beta gal. So with that in mind, let's go back to our problem. We have a mutant strain that does not produce beta galactic. A they do a complementation essay with a plasmid that is lack I minus but lack Z plus. So our plasmid has a functional beta gal gene and we've created a partial diploid strain. So that means if the original mutation was in that lax E gene and that's why it doesn't produce fatal lactic this complementation assay using this plasmid should produce functional beta galactosidase if lax Z gene is what was originally mutated. So that's choice B, the partial diploid strain will be lac plus and produce functional beta galactica only if the mutation is in a lac gene. Because then the lax C plus gene that comes in with the plasmid will be there to produce functional beta galactica. Just looking briefly at our other answer. Choices. Choice A says it will produce functional beta galactica only if the mutation is in a lack I gene. Again, we discussed that wouldn't affect the functionality of beta lactica since the lack I gene is the gene for the repressor protein. Choice C says that the diploid strain will be lack minus and produce no functional beta gal if the mutation is in either gene, well, that's not a correct answer because again, if the mutation was in the lax E gene, that plasmid would provide it with functional beta galactica. And finally, Joyce D, the partial diploid strain will be lac plus and produce functional beal lactic regardless of which gene is mutated. And that's incorrect. Uh because the lac Z gene is the gene that produces that beta galactic protein. So it's dip partial diploid state with a functional lax E gene will correct for that. So choice B again, is that correct answer? See you in the next video

Key Concepts

Lac Operon Structure and Function

Types of Mutations in the Lac Operon

Partial Diploids and Complementation Testing

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