Download BIO 208 Homework: Bacterial Genetics 2011 17.1 constitutive gene

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.