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Download BIO 208 Homework: Bacterial Genetics 2011 17.1 constitutive gene
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BIO 208 Homework: Bacterial Genetics 2011 17.1 constitutive gene expression – certain genes are always “on” because the gene products (proteins) are required by the cell all of the time. Regulated gene expression for gene active in response to cellular needs. 17.3 polycistronic mRNA – coding information from more than one gene on one mRNA molecule. A number of genes may be under regulatory control of a single promoter. The mRNA is translated into the appropriate proteins. 17.4 Lactose triggers the coordinate induction of the synthesis of beta galactosidase, permase, and a transacetylase. Therefore, lactose is an inducer as is results in the turning on of gene expression. If glucose is present, the metabolism of lactose is not necessary and the lac operon remains off. 17.9 a Cis dominant mutations affect genes that are on the same stretch of DNA. For example, promoter, operator mutations. Trans dominant mutations affect the expression of genes on non local stretches of DNA. 17.10 note, the permease gene, Y is not part of the assignment a. I+ P+ O+ Z+ In the absence of lactose: The I+ gene encodes the repressor protein which binds to the operator and blocks transcription of Z (beta galactosidase structural gene) Inducer present (in the presence of lactose): Lactose binds the repressor protein causing a conformational change. The repressor can no longer bind the operator. Therefore, RNA polymerase can attach to the promoter and transcribe the Z gene. Betagalactosidase is synthesized and lactose is metabolized. b. I+ P+ O+ ZIn the absence of lactose: The I+ gene encodes the repressor protein which binds to the operator and blocks transcription of the Z gene In the presence of lactose: Lactose binds the repressor protein causing a conformational change. The repressor can no longer bind the operatore and RNA polymerase can attach to the promoter and transcribe the Z gene. BUT, the Z gene is mutant, so no betagalactosidase will be made. This cell cannot metabolize lactose . d. I- P+ O+ Z+ In the absence of lactose: The I gene cannot encode a repressor because the gene is mutant. Therefore, no repressor will bind the operator. RNA polymerase will transcribe the Z gene and betagalactosidase is made whether or not lactose is present. This is called a constitutive mutation because the operon is always on. In the presence of lactose: No change. Lactose cant bind a repressor, because the repressor protein is not there, but it does not matter because the operon is on anyway f. I+ P+ Oc Z+ In the absence of lactose: The I gene encodes the repressor protein. But, the repressor protein cannot bind to the operator because the operator is mutant. Because the operator is free, RNA polymerase can bind the promoter and transcribe the Z gene. This operon is always on, so the Oc mutation is considered a constitutive mutation In the presence of lactose: Lactose will bind the repressor protein BUT, it does not matter because the repressor could not bind the operator anyway j. I- P+ O+ Z+ ____________ I+ P+ O+ Z- This represents a partial diploid cell – usually bacteria only have one copy of each gene because there is a single chromosome. But, a plasmid with the same genes as the E. coli chromosome results in a partial diploid with respect to those genes. In the absence of lactose: The I+ gene on the one stretch of DNA encodes a normal repressor protein. Because this is a protein, it is diffusible and repressor molecules can interact with both operators. So, in the absence of lactose, both operons are off In the presence of lactose: Lactose binds the repressor, preventing its binding to the operators. In the top stretch of DNA, RNA polymerase will be made and a functional betagalactosidase mRNA is transcribed from the Z gene. The cell will metabolize lactose. But, in the bottom stretch of DNA, the Z gene is mutant, so even if a transcript is made, it would not be translated into functional betagalactosidase protein.