18) In your research on gene regulation in E. coli, you produce 2 mutant strains with the same phenotype: they can grow on lactose but not on glucose. To investigate these 2 mutants further, you perform a complementation test using them and discover that the diploid combination strain can grow on glucose or lactose. From this, you can deduce that:
a.)Both mutants have a mutation in the same gene
b.)The 2 mutants have mutations in different genes
c.)Glucose and lactose metabolic pathways in E. coli utilize at least one shared gene in common
d.)At least 2 different genes are required for glucose metabolism in E. coli
e.) b) and d)
19. In your research on the lac operon, you produce a unique regulatory mutatant strain of E. Coli where the lac l gene's repressor protein binding site for the operator is normal (wild-type) but its lactose binding site no longer binds lactose. The expected phenotype of this mutant will be:
a. inability to grow on lactose only medium
b. ability to grow on lactose only medium
c. blue colony color when grown on glycerol plus x-gal medium
d. blue colony oclor when grown on lactose plus x-gal medium
e. b & d
20.
20) In your research on the lac operon, you create a diploid E.coli strain with the following genotype (+) = wild type gene, (-) = mutant gene.
I- /O+ /Z- /Y+
I+ /Oc /Z+ /Y+
Can this diploid mutant grow on medium containing lactose as the only carbon source?
a) Yes
b) No
c) Maybe, it depends on if the Oc mutation is dominant or recessive
18) In your research on gene regulation in E. coli, you produce 2 mutant strains with the same phenotype: they can grow on lactose but not on glucose. To investigate these 2 mutants further, you perform a complementation test using them and discover that the diploid combination strain can grow on glucose or lactose. From this, you can deduce that:
a.)Both mutants have a mutation in the same gene
b.)The 2 mutants have mutations in different genes
c.)Glucose and lactose metabolic pathways in E. coli utilize at least one shared gene in common
d.)At least 2 different genes are required for glucose metabolism in E. coli
e.) b) and d)
19. In your research on the lac operon, you produce a unique regulatory mutatant strain of E. Coli where the lac l gene's repressor protein binding site for the operator is normal (wild-type) but its lactose binding site no longer binds lactose. The expected phenotype of this mutant will be:
a. inability to grow on lactose only medium
b. ability to grow on lactose only medium
c. blue colony color when grown on glycerol plus x-gal medium
d. blue colony oclor when grown on lactose plus x-gal medium
e. b & d
20.
20) In your research on the lac operon, you create a diploid E.coli strain with the following genotype (+) = wild type gene, (-) = mutant gene.
I- /O+ /Z- /Y+
I+ /Oc /Z+ /Y+
Can this diploid mutant grow on medium containing lactose as the only carbon source?
a) Yes
b) No
c) Maybe, it depends on if the Oc mutation is dominant or recessive
Related textbook solutions
Related questions
For this question, focus on four parts:
I, the gene that produces the repressor protein;
P, the promoter region where RNA polymerase binds to initiate transcription;
O, the Operator site, where the repressor binds. When the repressor binds, it prevents transcription; and
ZYA, the three structural genes, all of which are required for the successful utilization of lactose as a food source.
In the left column of the table below, we indicate the genetic composition at the lac operon of particular strains of E. coli cells. A "+" (plus sign) following the letter indicates the wild-type allele (or normally functioning version of the gene). For I, P, and ZYA, a "-" (negative sign) following the letter indicates the mutant allele (or non-functional version of the gene). For the O allele, a "C" indicates the mutant or constitutive allele. It is not functional in that the repressor cannot bind, but it is functional in that transcription can still occur.
Given the conditions stated in each column, please indicate for each genotype whether lac mRNA is produced and at what level by typing in one of the following words in each cell in the table:
"Abundant"
"Minimal"
"None"
The completely wild-type genotype has already been filled in.
| Cellular conditions: | |||
| Genotype | Glucose present (low cAMP) and no lactose | Glucose present (low cAMP) and lactose present | No glucose (high cAMP) and lactose present |
| I+ P+ O+ ZYA+ | None | Minimal | Abundant |
| I- P+ O+ ZYA+ | |||
| I+ P- O+ ZYA+ | |||
| I+ P+ Oc ZYA+ | |||
| I+ P+ O+ ZYA- | |||
