Andrea Manzo
11/29/01
Bi 122 Set 7
1-1)
Three
gene transfer mechanisms are conjugation, transformation, and
transduction. Conjugation is the transfer of genetic
material from one bacterium to the other by direct cell-to-cell contact. Transformation is the transfer of exogenous
DNA segments from the environment into a bacterium’s chromosome. Transduction occurs when certain phages lyse
a bacterial cell and accidentally take up a piece of the bacterial DNA. Then when these phages inject their DNA into
the recipient cell, the donor bacterial DNA is transferred.
1-2)
An Hfr
strain is a strain of bacteria with a fertility factor that has been
incorporated
into the bacterial chromosome, instead of residing in the cytoplasm.
If the donor
strain was Hfr, then there would be high recombination frequency in the
progeny, and less genetic conversion of the recipient to a cell that could
donate DNA. High recombination
frequency is due to the fact that, in Hfr strains, it is a strand of the
bacterial chromosome that is transferred.
Therefore, there is likely to be lots of homology between the donor and
recipient bacterial chromosomes, and recombination is more likely to
occur. Lower likelihood of genetic
conversion is due to the fact that when the bacterial chromosmal strand is
transferred during conjugation, the F factor is always the last gene in
line. It is therefore very likely that
the chromosome will break before the full transfer is complete, and so genetic
conversion is unlikely.
If the donor
strain was F+, then there would be low recombination frequency, and high
instance of genetic conversion. Since
the genetic material that is being transferred is a plasmid that most likely
does not have any homology with the recipient chromosome, recombination is
unlikely to occur. The rate of genetic
conversion is high because it is only the one gene (the F factor) that needs to
be transferred instead of the whole chromosome in order to confer “maleness” to
the recipient.
1-3)
Wollman
and Jacob were able to prove that the E. coli chromosome was
circular by
analyzing linkage maps constructed from the results of interrupted-mating
experiments. At first glance, the
linkage maps of various Hfr strains seemed entirely different. However, in each strain, the genes had the
same neighbors on each flank (unless they were at the start or end of the
map). Also, the genes were not always
transferred in the same order. The only
way to account for these strange occurrences is if the Hfr chromosome was
circular, with the F factor inserted in a location that differed from strain to
strain. Since the Hfr chromosome is
merely the E. coli chromosome with the F factor, then the E. coli
chromosome is also circular.
1-4) Transformation is the method of genetic
transfer that is most likely to explain this phenomenon.
A vector with
these characteristics would have to have some sort of antibiotic resistance
gene, as well as an F factor. An
antibiotic resistance gene would confer selective advantage to the strains, and
an F factor would insure that the genetic material would be transferred from
cell to cell (and not disappear after multiple cell divisions).
2-1) Two
events occur that allow the cell to sense the different substrate and
synthesize all the necessary enzymes.
In the absence of lactose, a repressor is bound to the operator (a
section of the lac operon downstream of the promoter), blocking the RNA
polymerase from transcribing the genes necessary for lactose metabolism. Lactose binds to the repressor and removes
it from the operator. Also, low levels
of glucose cause the concentration of cAMP to rise. cAMP binds with CAP, which binds the promoter and activates the
lac operon. Both these events occur
fairly quickly.
2-2-1)
It is
possible to create a partial diploid bacterial cell using F’ factor. An
F’ factor is a
plasmid that carries a part of the bacterial chromosome. To make a partial diploid, first make a F’
factor that carries the part of the genome you are interested in (for example,
a lacI gene). Then you can insert this
F’ plasmid into the bacterium you are interested in, creating a partial
diploid.
2-2-2)
The gene
product of lac I acts as a repressor of the lac operon. Mutant
strains that
are lac I- show constitutive expression of all lac enzymes, while wildtype
strains show controlled expression of lac enzymes. Therefore, lac I must function as a regulator of transcription
such as a repressor. The fact that lac
I+ is dominant to lac I- shows that its product is a trans acting factor. (Even though one allele is nonfunctional,
the lac I+ allele still produces enough repressors to bind to the lac
operons). The lac O gene is the
operator for the lac operon; i.e., it is the section of DNA that the repressor
binds to in order to prevent transcription of the downstream genes. Since lac Oc is dominant over lac
O+, it must be a cis-acting factor. lac
Oc strains have some sort of mutation in the operator that prevents
the repressor from binding it; hence the unregulated expression of the lac
enzymes which are adjacent to lac O. If
lac Oc was a trans-acting factor, then it would not be dominant to
lac O+, since lac O+ would be able to produce enough of the product to make the
organism phenotypically wildtype.
2-2-3)
The lac Is repressor remains bound to the lac
operator even in the
presence of IPTG. IPTG will cause the dissociation of the lac
I+ product from the operator. However, there are still lac Is
repressors will still bind to the operator, and will do so if the lac I+
repressors do not. Therefore, in a lac
I+/lac I- cell, no lac enzymes will be produced. In a lac I-/ lac Is cell, the lac I- gene will not
produce any functional repressors.
However, the lac Is does produce repressors; furthermore,
they do not dissociate from the operator even in the presence of IPTG. No lac enzymes will be produced. Hence, lac Is is dominant to both
lac I- and lac I+.
The lac Is
protein repressor will not dissociate from the lac operator because it has a
mutation that alters its inducer-binding site, preventing it from binding to
lactose or any other normal inducer.
Normally, the association of the lac I repressor with lactose causes it
to release itself from the operator.
Since it cannot associate with lactose, the lac Is repressor
always stays bound to the operator.
2-3)
A cell
has a finite amount of energy it can afford to expend. Producing the enzymes necessary for both
glucose metabolism and lactose metabolism would be costly in terms of
energy. It is more efficient for the
cell to use up the surrounding glucose before synthesizing new enzymes to metabolize
lactose. Also, the bacteria must break
down lactose into glucose and galactose, which is an extra step in sugar
metabolism.
First,
cAMP binds to CAP. Then this cAMP-CAP
complex can bind to the CAP site of the operon within the lac promoter, upstream
of the RNA polymerase binding site. The
cAMP-CAP makes physical contact with the RNA polymerase and increases the
affinity of the RNA polymerase for the DNA, increasing the likelihood that the
RNA polymerase will transcribe the genes instead of just fall off the DNA. Also, when CAP binds to its site, it creates
a bend that is greater than 90 degrees in the DNA which increases the affinity
of RNA polymerase for the DNA.