Chapter 9 - An Introduction to Bacterial Genetics

9 - 1 Definitions in bacterial genetics

One of the major goals of genetic studies is to understand the organization and function of genes. To achieve this, alterations in DNA are created, either spontaneously or by the experimenter, fished out in some fashion, and the effects of these changes are observed. It is analogous to systematically breaking the pieces of a radio, seeing what happens, and from this information deducing the function of each component.

Bacterial genetics has helped to unravel the mysteries of DNA replication, the production of proteins from DNA, the genetic code, gene regulation and more! Genetic analysis will impact your life in many ways. The human genome project (which will sequence all 23 of the human chromosomes) involves the use of sophisticated bacterial genetic techniques. A growing list of mammalian and other proteins that are naturally found in extremely low amounts, are being expressed in microorganisms and purified including 1) human insulin-produced in bacteria for diabetics, 2) human growth hormone for treating dwarfism, and 3) proteins for the production of vaccines against diseases. On a more familiar level, bacterial genetics is being employed to improve the strains used to make cheese, beer, yogurt, and other fermented food products.

Here are a few key terms to help you understand bacterial genetics. A mutant is a strain that has an altered growth property (termed the phenotype) when compared to a designated bench mark strain, which is referred to as the wild-type strain (wt). A mutation is an alteration in the DNA sequence of an organism. Some mutations will cause a detectable change in its phenotype, others will not and are called silent mutations. The genotype of a microorganism is its DNA sequence. For example, if the wild-type strain is mutated so that it is unable to synthesize its own tryptophan, it is referred to as a tryptophan minus mutant (its phenotype is termed Trp^-). Its genotype is the specific base pair change that has taken place and this can be determined by DNA sequencing. If this mutant is plated onto minimal medium lacking tryptophan, most of the organisms will be unable to grow. However, a small fraction of the plated bacteria will revert back to Trp⁺ and form colonies. These are referred to as revertants. They can arise from a conversion of the mutant genotype back to wild-type, or from another secondary mutation that counteracts the initial mutation. Do not confuse a revertant with the wild-type strain, since reversion can also be caused by secondary mutations. An auxotroph refers to a mutant strain that is unable to synthesize a needed nutrient that the wt stain is able to make. For example a Trp^- mutant is termed a tryptophan auxotroph. A prototroph is an organism not requiring the nutrient in question, a wild-type strain is prototrophic. For more information, a good microbiology textbook will have a chapter on Bacterial Genetics.

Two short exercises here will introduce you to the techniques and strategies used when performing genetics. In the first experiment we look at selection of mutants. Taking a wild-type strain and demonstrating the development of drug resistance. In the second experiment, the transfer of genetic traits, the transfer of DNA, from one strain to another by conjugation is demonstrated.

---

9 - 2 Selection of mutants

DNA replication in all organisms is very accurate, it has to be. (What would happen if many mistakes were made during replication?). However, mistakes do occur at a very low rate and this mutation frequency is formally defined as the probability that a particular gene will mutate each time there is cell division. On the average, there is about one chance in a million (10^-6) that a particular gene will mutate when there is DNA replication. Since bacteria can be grown to a very large concentration in artificial culture (10⁹ cells/ml or more), mutants for most genes should be present in a population of microbes larger than a few milliliters. Finding the mutants is the problem.

This experiment introduces the simple technique of direct selection for the detection and isolation of certain mutants. This technique involves plating bacteria on a selective medium on which only the desired mutants can grow. Specifically, the experiment involves taking a culture of microbes at a high density and plating them on a rich medium containing the antibiotic streptomycin. Streptomycin inhibits protein synthesis by binding to the small subunit of the ribosome and blocking entrance of initiator N-formyl methionine tRNA into the ribosome, thus preventing the start of protein synthesis. A single mutation in the S12 protein of the small subunit of the ribosome prevents streptomycin from binding, thus causing the microbe to become resistant to the antibiotic. By plating a strain onto a medium containing streptomycin, it is possible to fish out the microbes present that have a mutation in the S12 gene. Please appreciate the power of this selective technique. By performing a very simple experiment, it is possible to fish out the 100 or so cells, in a mixture containing billions, that have a change in a specific gene. Also, note how easy it is for a microbe to become resistant to an antibiotic. While the drug is killing 99.9999% of the microbes present, 0.0001% are able to survive, and these microbes are now resistant to the antibiotic. This explains why drug resistance in microbes can occur so rapidly and is a constant problem in medicine.

Period 1

Materials

Tube of a concentrated suspension of Staphylococcus epidermidis (approximately 1 X 10¹⁰ CFUs/ml) in saline

1 plate of Nutrient Agar (NA)

1 plate of NA+10 µg streptomycin (SM) per ml of medium

1 plate of NA+100 µg SM per ml of medium

Micropipettes and sterile tips

1. From the S. epidermidis suspension, pipette 0.1 ml onto each plate. Spread the inocula with a sterile hockey stick, proceeding from NA to NA + 10 µg/ml SM to NA + 100 µg/ml
2. Incubate at 37°C for 2-3 days or at 30°C for 4-5 days.

Period 2

Materials

1 plate of NA+100µg SM per ml of medium

Beaker of sterile toothpicks

1. Note the confluent growth on the plate without streptomycin and the colonies on the plates with streptomycin. On the plate containing 10µg SM/ml, there are usually seen two apparently different sizes of colonies. Do you see the smaller type of colony on the plate containing 100µg SM/ml? (If a size difference is not seen, do you at least see a reduction in the number of colonies on the 100µg SM/ml plate?)

What may these differences in colony size and growth characteristics at these streptomycin concentrations mean in terms of the mechanisms of resistance? That is, for the cells in each of the two types of colonies, what structures or functions may have been altered to become resistant (or inhibitory) to the action of streptomycin?

The instructor should pre-run this experiment to determine if lower or higher concentrations of streptomycin should be used. Results of this experiment can vary according to the batch of Nutrient Agar and streptomycin used as well as the particular strain of S. epidermidis. (Twenty µg SM/ml may give better results than 10.) The above method is based on what has worked at UW-Madison (with the use of Difco Nutrient Agar).

2. Taking into account the approximate number of CFUs inoculated onto each plate, determine the mutation frequency of streptomycin-resistant CFUs in the original suspension from the colony counts on each plate containing streptomycin.
3. Mark the bottom of the new plate of NA+100µg SM/ml plate to delineate four sectors:
• one for the colonies on NA+100µg SM/ml.
• one for the largest colonies seen on NA+10µg SM/ml.
• one for the smallest colonies seen on NA+10µg SM/ml.
• one for the growth on the plate of NA without streptomycin.
4. Using the sterile toothpicks and aseptic technique, pick 5 isolated colonies from the plate of NA+100µg SM/ml, and spot-inoculate them onto the appropriate sector of the new plate. Be sure to use a new toothpick for each transfer, and discard them into the disinfectant. Repeat for isolated colonies from the other plates. For the plate without streptomycin, simply pick from 5 sites in the confluent growth.
5. Incubate at 37°C for 2-3 days or at 30°C for 4-5 days.

Period 3

1. Observe and record the number of spot-inoculations showing growth in each sector. Discuss the significance of your results, noting the questions in Period 2, step 1.

---

9 - 3 DNA transfer by conjugation

Although bacteria do not undergo true sexual reproduction as seen for eukaryotic organisms, there are mechanisms for gene transfer and genetic recombination in bacteria. The means by which genes are transferred between cells are divided into three general categories. You can read more about these phenomena in the chapter on Bacterial Genetics.

1. Transformation: Genes are absorbed by a cell from the environment where DNA from lysed cells is present.
2. Transduction: Bacterial genes are transferred by a bacteriophage.
3. Conjugation: Genes are transferred directly from one living cell to another with the aid of a protein bridge (pilus) allowing cell-to-cell contact.

In all three mechanisms, the DNA is transferred from a donor cell to a recipient cell. In most instances, only a small fraction of a chromosome is transferred. Once inside the recipient cell, the donor DNA has two pathways. If the DNA is capable of autonomous replication, it can exist indefinitely in the host cell. Any DNA entering the cell to position itself alongside the homologous portion of the recipient chromosome. Recombination may occur where, due to enzymatic action, a segment of the recipient DNA is excised and replaced by the integration of the homologous segment of donor DNA. In either case, the recombinant cell contains DNA derived from both the donor and recipient cells. Acquisition of donor genes may confer to the recipient an advantage such as being able to grow in a particular habitat or medium where growth was before impossible.

In Escherichia coli and presumably many other species, distinct mating types may be found. For E. coli, the F⁺ and F^- mating types are distinguished by the presence of a special plasmid, the F factor, in the former. This plasmid includes genes involved in the transfer by conjugation of DNA into compatible recipient cells; included are genes responsible for the synthesis of the F pilus. When the F plasmid is transferred to an F^- cell, the donor F⁺ cell retains a complete copy of the plasmid while the recipient F^- cell becomes an F⁺ donor cell. If the F plasmid becomes integrated into the chromosome of a F⁺ cell, an Hfr cell results. A cell of this mating type will be a donor, able to transfer part of its regular chromosome during conjugation (with part of the F factor on the leading end). However, an Hfr donor rarely transfers its mating ability to a recipient, since it would require complete transfer of the entire chromosome. Recombination involving any part of the transferred Hfr chromosome may subsequently occur in the recipient cell as discussed above.

In order for us to demonstrate conjugation and recombination effectively in this experiment, we must choose parent cells which are not only different mating types but are also different in certain phenotypic properties such that we may detect the result of recombination easily. We could not demonstrate the end result of these processes with phenotypically-identical mating types! For this experiment, we have chosen an Hfr strain which is a methionine auxotroph and an F^- strain which is a threonine auxotroph. Neither should grow on a minimal medium. However, after mixing the two strains and allowing for conjugation and subsequent recombination to occur, the appearance of any colony on the minimal medium can be attributed to an F^- cell which acquired the ability, from an Hfr cell (via conjugation and recombination), to synthesize threonine and thus all of its amino acids.

The real starting point in this experiment is the mixture of Hfr and F^- cells where we hope to have conjugation proceed. Knowing the concentration of each mating type in the mixture, we can calculate the percentage of F^- cells which acquire the ability to synthesize threonine and therefore produce colonies on the minimal medium. This percentage we call the recombination frequency.

The growth characteristics of relevant E. coli strains can be summarized as follows:

strain of E. coli	genotype designation	growth on Minimal Medium	growth on an all-purpose Medium
typical prototroph	Thr+ Met+	+	+
our Hfr strain (auxotrophic for methionine)	Thr+ Met-	-	+
our F- strain (auxotrophic for threonine)	Thr- Met+	-	+
possible recombinant 1	Thr+ Met+	+	+
possible recombinant 2	Thr- Met-	-	+

Period 1

Materials

Broth culture of E. coli: Hfr (donor) mating type; Thr⁺ Met^-; approximately 10⁸ cells/ml

Broth culture of E. coli: F^- (recipient) mating type; Thr^- Met⁺; approximately 10⁸ cells/ml

2 plates of Tryptone Yeast Extract Glucose Agar (TYEG)

(Alternatively, any all-purpose medium can be substituted; Minimal Medium plus threonine and methionine may also be used.)

6 plates of Minimal Medium (MM)

1 empty, sterile test tube

3 dilution blanks (9 ml)

Micropipettes and sterile tips

2 sterile swabs

Procedure I (quantitative)

1. For an experiment such as this, it helps to diagram the procedure and have the plates and tubes labeled appropriately for the upcoming steps in the procedure: Label one of each plating medium (i.e., TYEG and MM) for the Hfr culture, one of each plating medium for the F^- culture, the three dilution blanks for the dilutions to be made of the mixture of the two mating types (i.e., 10^-1, 10^-2 and 10^-3), and three plates of MM for the plated dilutions (i.e., 10^-2, 10^-3 and 10^-4).
2. Using separate pipette tips, aseptically transfer 1 ml of each strain to the empty sterile tube and mix gently. Note the time of the addition of the strains to each other. Set the tube aside where it will not be disturbed, and allow 30 minutes for conjugation to take place.
1. Considering the number of cells transferred in each 1 ml amount (see under Materials), what would be the number of cells per ml of the mixture you just made in the sterile tube?
2. Then, what would be the number of cells of either mating type per ml of the mixture?
3. If the cells were to pair up, what would be the number of pairs per ml of the mixture?
4. The last two values should be the same! Therefore, among these pairs of cells, we will be able to determine how many pairings resulted in F^- cells that received the Thr⁺ gene (after conjugation and recombination), as each of these cells will be able to grow (produce a countable colony) on MM.
3. Prepare control plates of the donor and recipient strains by streaking each culture onto a plate each of TYEG and MM. Any streaking method is acceptable as long as it is done the same for each plate. What do we expect to happen on each plate?
4. After the 30 minute period is up, prepare three serial, decimal (1/10) dilutions of the mixture and plate 0.1 ml from each dilution onto a plate of MM. Remember proper aseptic technique. Then, starting from the most dilute inoculum, spread the plates with a flame-sterilized hockey stick.
5. Incubate all plates (from steps 3 and 4) at 37°C for 2-3 days or at 30°C for 4-5 days.

Procedure II (qualitative)

6. Draw two perpendicular lines across the bottom of a plate of MM. Label one line for the donor and the other line for the recipient.
7. Dip a sterile cotton swab into the F^- broth culture and make a single streak across the medium in one direction, following the previously-marked line on the bottom of the plate. Discard the swab into the disinfectant.
8. Repeat the procedure for the Hfr culture, streaking along the other line. Notice how you are mixing Hfr and F^- cells when you cross the center of the plate. Therefore, where on the plate will you expect to see colonies of recombinant cells?
9. Incubate the plate at 37°C for 2-3 days or at 30°C for 4-5 days.

Period 2

Procedure I

1. Examine the streak plates on TYEG and MM of the donor and recipient cultures. On which medium was each strain able to grow?
2. Of the three MM dilution plates, count the one containing between 30 and 300 colonies. Multiplying this number by the appropriate dilution factor, determine the number of recombinant colony-forming units per ml of the mixture.
3. Compare the value obtained in step 2 to the total number of cells per ml of the mixture which had the potential to pick up DNA and undergo recombination - i.e., the recipient (F^-) cells. To determine the recombination frequency, use the following formula. (For convenience in this experiment we will treat colony-forming units and cells as equivalent.)



Figure 8.7. Calculating %recombinants. The percent of recombinants that successfully mating with the Hfr strain and recombined the thr region into the chromosome. The number of recipient cells per ml of the mixture was 5 x 10⁷.

Procedure II

4. Note any colonial or confluent growth on the plate. Do your results indicate that genetic recombination has taken place between the donor and recipient strains? Remember what genotype an organism needs to have in order to grow on this medium. Might it be possible to observe colonies resulting from revertants (back-mutations)? If so, where on the plate would they be evident?

---

9 - 4 Summary of bacterial genetics

With these two experiments we have barely scratched the surface of bacterial genetics. There are many more interesting and exciting experiments that can be done in microbial DNA manipulations. We understand more about the genomes and behavior of microbes than any other creatures, and this is in part due to the sophisticated techniques and powerful experiments that can be done with bacterial genetics and molecular biology. Hopefully these experiments have given you a small appreciation for this field.

---