Genetic transfer via transformation can also be used to map genes along the bacterial chromosome. In this approach, fragments of chromosomal DNA are isolated from one bacterial strain and used to transform another strain. The experimenter examines the transformed bacteria to see if they have incorporated two or more different genes. For example, the DNA may be isolated from a donor E. coli bacterium that has functional copies of the araB and leuD genes. Let’s call these genes araB+ and leuD+ to indicate the genes are functional. These two genes are required for arabinose metabolismand leucine synthesis, respectively. To map the distance betweenthese two genes via transformation, a recipient bacterium is used that is araB− and leuD−. Following transformation, the recipient bacterium may become araB+ and leuD+. This phenomenon is calledcotransformation because two genes from the donor bacterium have been transferred to the recipient via transformation. In this type of experiment, the recipient cell is exposed to a fairly low concentrationof donor DNA, making it unlikely that the recipient bacterium will take up more than one fragment of DNA. Therefore, under these conditions, cotransformation is likely only when two genes are fairlyclose together and are found on one fragment of DNA. In a cotransformation experiment, a researcher has isolated DNA from an araB+ and leuD+ donor strain. This DNA was transformed intoa recipient strain that was araB− and leuD−. Following transformation, the cells were plated on a medium containing arabinose and leucine.On this medium, only bacteria that are araB+ can grow. The bacteriacan be either leuD+ or leuD− because leucine is provided in the medium. Colonies that grew on this medium were then restreaked on a medium that contained arabinose but lacked leucine. OnlyaraB + and leuD+ cells could grow on these secondary plates. Following this protocol, the researcher obtained the followingresults:Number of colonies growing on a medium containingarabinose plus leucine: 57Number of colonies that grew when restreaked on a mediumcontaining arabinose medium without leucine: 42What is the map distance between these two genes? Note: One way to calculate the map distance is to use the same equation that we used for cotransduction data, except that we substitute cotransformation frequency for cotransduction frequency.Cotransformation frequency = (1 − d/L)3The researcher needs to experimentally determine the valueof L by running the DNA on a gel and estimating the averagesize of the DNA fragments. Let’s assume they are about 2%of the bacterial chromosome, which, for E. coli, would beabout 80,000 bp in length. So L equals 2 minutes, which isthe same as 2%.
Bacterial Genomics
The study of the morphological, physiological, and evolutionary aspects of the bacterial genome is referred to as bacterial genomics. This subdisciplinary field aids in understanding how genes are assembled into genomes. Further, bacterial or microbial genomics has helped researchers in understanding the pathogenicity of bacteria and other microbes.
Transformation Experiment in Bacteria
In the discovery of genetic material, the experiment conducted by Frederick Griffith on Streptococcus pneumonia proved to be a stepping stone.
Plasmids and Vectors
The DNA molecule that exists in a circular shape and is smaller in size which is capable of its replication is called Plasmids. In other words, it is called extra-chromosomal plasmid DNA. Vectors are the molecule which is capable of carrying genetic material which can be transferred into another cell and further carry out replication and expression. Plasmids can act as vectors.
Genetic transfer via transformation can also be used to map genes along the bacterial chromosome. In this approach, fragments of chromosomal DNA are isolated from one bacterial strain and used to transform another strain. The experimenter examines the transformed bacteria to see if they have incorporated two or more different genes. For example, the DNA may be isolated from a donor E. coli bacterium that has functional copies of the araB and leuD genes. Let’s call these genes araB+ and leuD+ to indicate the genes are functional. These two genes are required for arabinose metabolismand leucine synthesis, respectively. To map the distance betweenthese two genes via transformation, a recipient bacterium is used that is araB− and leuD−. Following transformation, the recipient bacterium may become araB+ and leuD+. This phenomenon is called
cotransformation because two genes from the donor bacterium have been transferred to the recipient via transformation. In this type of experiment, the recipient cell is exposed to a fairly low concentration
of donor DNA, making it unlikely that the recipient bacterium will take up more than one fragment of DNA. Therefore, under these conditions, cotransformation is likely only when two genes are fairly
close together and are found on one fragment of DNA. In a cotransformation experiment, a researcher has isolated DNA from an araB+ and leuD+ donor strain. This DNA was transformed into
a recipient strain that was araB− and leuD−. Following transformation, the cells were plated on a medium containing arabinose and leucine.On this medium, only bacteria that are araB+ can grow. The bacteria
can be either leuD+ or leuD− because leucine is provided in the medium. Colonies that grew on this medium were then restreaked on a medium that contained arabinose but lacked leucine. Only
araB + and leuD+ cells could grow on these secondary plates. Following this protocol, the researcher obtained the following
results:
Number of colonies growing on a medium containing
arabinose plus leucine: 57
Number of colonies that grew when restreaked on a medium
containing arabinose medium without leucine: 42
What is the map distance between these two genes? Note: One way to calculate the map distance is to use the same equation that we used for cotransduction data, except that we substitute cotransformation frequency for cotransduction frequency.
Cotransformation frequency = (1 − d/L)3
The researcher needs to experimentally determine the value
of L by running the DNA on a gel and estimating the average
size of the DNA fragments. Let’s assume they are about 2%
of the bacterial chromosome, which, for E. coli, would be
about 80,000 bp in length. So L equals 2 minutes, which is
the same as 2%.
Trending now
This is a popular solution!
Step by step
Solved in 4 steps
