Respuesta :
Answer: the answer to this question contains pictures and can be found in the attachment below.
Explanation:
a.
In the situation, the bacteria have the genotype I+ P+ OC Z+ Y+. The bacterial organism, has all functional genes and the operator is constitutive in nature, which will not allow the action of repressor and thus, the operon will be transcribed constitutively. However, in the absence of glucose, the CAP (catabolite activator protein)-cAMP (cyclic adenosine monophosphate) complex will be formed, which will further activate beta-galactosidase production. Hence, in absence of glucose, transcription rates will be higher.
The autoradiograph for this situation can be found in the attachment below:
b.
Here, The bacteria has the genotype I+ P+ OC Z- Y+. It is clear that this bacterium has a mutated beta-galactosidase and as such, no mRNA (messenger ribonucleic acid) for beta-galactosidase will be produced in presence of glucose or lactose.
The autoradiograph for this situation can be found in the attachment below:
c.
The genotypic composition of the bacteria given is I+ P- OC Z+ Y+. In this situation, the operator is constitutive but the promotor element is mutated, which will not allow binding of RNA (ribonucleic acid) polymerase and thus, no transcription will be observed in any condition. The autoradiograph for this situation can be found in the attachment below:
d.
The genotype composition here is I- P+ OC Z+ Y+. It is clear that the operator is constitutive and thus, operon will be expressed in presence of glucose as well as lactose. However, in presence of glucose, CAP-cAMP complex will not associate with the operon and thus, the only basal level of transcription will be observed.
The autoradiograph for this situation can be found in the attachment below:
e.
The genotype of the organism is I+ P+ O+ Z- Y-. The gene for beta-galactosidase is mutated and thus no mRNA will be produced for beta-galactosidase.
The autoradiograph for this situation can be found in the attachment below: