Quick question for lac operon! Genotype expression for the different situation!
ID: 300364 • Letter: Q
Question
Quick question for lac operon! Genotype expression for the different situation! Really need help! (It would be wonderful if you provide step by step explanation)
1. For each genotype and situation listed below, describe which genes would be expressed and explain your reasoning. Assume there is no glucose in the media unless otherwise specified. A (3 points). Is P"?"Z+ Y+ / i+ P+ Oc Z+ Y. (Conditions : lactose present)T B. (3 points) i+ P+ ?'Z-Y+ , l+ P. Oc Zry. (Conditions : lactose present C. (3 points) l + P+ ?"Z-Y+ , Is P. Oc Zry. (Conditions: say what happens with and without lactose, both in the absence of glucose)fIExplanation / Answer
Is shows the trans effect which means that it can bind to the other operator present on the other operon.
Oc shows the cis effect which means it can only regulate the expression of the same operon
A- IsP+O+Z+Y+/ I+P+OcZ+Y-
In this case, one operon will make super-repressor which can bind to the operator even if lactose sugar is present. But in the second operon Constitutive operon is present which is mutated in such a way that even the super-repressor cannot bind to it. So, in this case, there will be transcription and only Beta-galactosidase will be expressed.
B- I+P+O+Z+Y-/ I+P-OcZ+Y-
In this case, the first operon will make normal repressor which can bind to the lactose and there will be the production Permease will be expressed. In the case of the second operon, there is a mutation in the promoter os that RNA polymerase will not be able to bind operon and there will not be any transcription. So, no production of either Beta-galactosidase or permease from the second operon. So only the production of Permease will take place.
C- I+P+O+Z-Y+/ IsP-OcZ+Y-
In this case, there will be the production of super-repressor. Lactose doesn't bind to super-repressor as its protein is modified in such a way that its binding affinity towards the lactose is totally lost. So super-repressor will bind to the operator. So even in presence of lactose, there will not be any transcription. So nothing will be produced. In the second operon promoter is mutated so it will not be able to transcribe the gene.
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