Solutions For An Introduction to Genetic Analysis Chapter 5 Problem 41P

2. E. coli can be grown in liquid or on an agar medium. Discrete colonies can be identified when inoculated on agar medium. Each clone of descendants from a single cell is visible to the naked eye when it reaches more than 107 cells.

3. E coli are an enteric bacterium, in its natural habitat they live symbiotically within the intestinal gut region of host organisms like human.

4. Minimal medium consists of inorganic salts, a carbon source for energy, and water. Hence, these are the minimal requirements for E coli cells to divide.

5. Prototrophs are organisms that can grow on minimal media; they are referred as wild-type phenotype. Auxotrophs are mutants that can grow only on medium supplemented with one or more specific nutrients that are not required by the wild-type strain.

6. In this experiment, the Hfr (high-frequency recombination cell) and the exconjugants are protrotrophs, since they can grow on minimal medium, whereas the recipient F^¯ and the exconjugants that do not grow on minimal medium are auxotrophs.

7. On medium enriched with proline and thiamine, the unknown strains would be grown as individual colonies, and then cells from each colony could be picked using sterile inoculation loop and placed individually onto medium supplemented with either thiamine or proline or onto minimal medium. Proline and thiamine auxotrophs would be identified on the basis of growth patterns. For example, a pro¯ strain will grow only on medium supplemented with proline and thi¯ strain can grow only on medium supplemented with thiamine.

Picking up individual cells is a labor-intensive method; instead replica plating method can be used to transfer some cells of each colony from a master plate supplemented with proline and thiamine to plates that contain the various media. The physical arrangement and positional patterns of colonies is used to identify the various colonies as they are transferred from plate to plate.

8. The two chemicals proline and thiamine are essential nutrients for cell growth. Proline is an amino acid, and thiamine is vitamin B1. Their chemical nature does not matter to the experiment other than that are necessary chemicals for cell growth that prototrophs can synthesize from ingredients in minimal medium and specific auxotrophic mutants cannot.

9. Diagram showing the full set of manipulations performed in this experiment,

10. To roughly map genes onto the circular bacterial chromosomes, interrupted-mating experiments are used.

11. In a solution the Hfr and F¯ strains are mixed together, the samples are removed at various times and put into a kitchen blender, vortexed for a few seconds to disrupt conjugation, and then plated onto a medium containing the appropriate supplements. The amount of time that has passed from the mixing of the strains to mating disruption is used as a measurement for mapping. The time of first appearance of a specific gene form the Hfr in the F¯ cell gives the gene’s relative position in minutes. Typically, the F¯ cells are streptomycin-resistant and the Hfr cells are streptomycin-sensitive. In various media the antibiotic is used to kill the Hfr cells that are prototrophic and allow only those F¯ cells that have received the appropriate gene or genes from the Hfr to grow. In this case, it would be discovered that some of the F¯ cells would become thi⁺ in samples taken earlier in the experiment than samples taken when they first become pro⁺.

12. There is no attempt to disrupt conjugation in this experiment. The two strains are mixed and after sometimes it is plated onto medium containing thiamine. This selects for strains that are pro⁺, since proline is not present in this medium.

13. Exconjugants are recipient cells otherwise called F¯ cells that now contain alleles form the donor (Hfr) cells. Typically, the F¯ cells are streptomycin-resistant and the Hfr cells are streptomycin-sensitive. The streptomycin antibiotic used in various media allows only the appropriate F¯ exconjugants to grow and it kills Hfr cells.

14. The statement “pro enters after thi” is one of gene position and order relative to the transfer of the bacterial chromosome by a particular Hfr. For the Hfr in this experiment, transfer occurs such that the pro gene is transferred after the thi gene. Since, this Hfr is also pro⁺, it means that pro⁺ allele enters specifically.

15. In this experiment, “fully supplemented medium” indicates media with all nutrients, specifically rich in proline and thiamine.

16. All exconjugants are pro⁺, which does not grow on minimal medium because that is the way they were selected. Hence, they grow on medium supplemented with thiamine.

17. Genetic exchange in eukaryotes takes place between two whole genomes prokaryotes as it does not in eukaryotes. It takes place between one complete circular genome, the F¯, and an incomplete linear genomic fragment donated by the Hfr. In this way, exchange of genetic information is non-reciprocal (from Hfr to F¯). Only even numbers of crossovers are allowed between the two DNAs (deoxyribonucleic acid), because the circular chromosome would become linear otherwise. This results in unidirectional exchange, because part of the DNA of the recipient chromosome is replaced by the DNA of the donor, while the other product (the rest of the donor DNA now with some recombined recipient DNA) is nonviable and lost.

18. In this experiment, the map distance is calculated by selecting for the last marker pro⁺ to enter and then determining how often the earlier unselected marker thi⁺ is also present. Observe the following diagram:

For the F¯ cell to become pro⁺, two recombination events have to occur. One in the left region marked as A and one in either region to the right, marked as B or C. Thus, the percentage of pro⁺ (second recombination within either B or C) that are thi¯ (second events only within region B) can be used to determine map distance where 1% = 1 map unit. (The map units calculated in this manner cannot be combined with other map unit calculations from other experiments to build a larger genomic map. The map units obtained are just giving an estimate of the relative sizes of intervals B and C as targets for crossover.

a. The two genotypes being cultured are pro⁺ thi¯ that grows only on media supplemented with thiamine and pro⁺ thi⁺ that can grow on minimal media.

b. Two recombination events must occur, one on either side of pro, since exconjugants were plated on medium supplemented with thiamine, only pro⁺ cells would have grown. The pro⁺ thi⁺ strains would have had recombination in regions labeled A and B, and the pro⁺ thi⁺ strains would have had recombination in regions labeled A and C.

c. The distance between pro and thi is:

=

Hence, the distance between pro and thi is 11.1%