Three different proline auxotrophs were streaked on minimal plates without proline and incubated at the following temperatures for two days. (The shaded areas indicate growth and white areas indicate where cells were streaked but did not grow.)
ANSWER: pro-1 causes a temperature sensitive phenotype; pro-2 causes a non-conditional, null phenotype; pro-3 causes a cold sensitive phenotype.
ANSWER: a = pro-1; b = pro-2; c = pro-3
ANSWER: At 22 C, the cold sensitive phenotype of pro-3 prevents the conversion of glutamyl-semialdehyde (GSA) into proline. Thus GSA accumulates and is able to crossfeed the pro-2 mutant which is blocked in an earlier step in the pathway. In contrast, at the permissive temperature (30 C), the pro-3 gene product is functional and so GSA is converted to proline in the pro-3 mutant strain. Since GSA is rapidly converted into proline, no precursors are accumulated. Proline does not crossfeed because it is rapidly incorporated into protein and is thus not accumulated.
Biotin is an essential coenzyme. Thus, mutants unable to synthesize biotin are auxotrophs which require exogenous biotin for growth. Cleary and Campbell (1971) characterized a large number of E. coli bio mutants by crossfeeding tests. The mutants fell into 4 classes as shown below:
None of the bio mutants were alanine auxotrophs but it was not possible to test whether or not the mutants could synthesize pimelic acid. Based upon the results shown in the table and the predicted pathway, indicate which steps of the biosynthetic pathway are blocked in each of the four bio mutants.
Five different aromatic amino acid auxotrophs were isolated. To determine if they affect different steps of the aromatic amino acid biosynthesis pathway, each of these mutants was tested for the ability to crossfeed the other mutants. The results are shown in the figure below. [The shaded arrows indicate growth.]
Based on the above results, indicate which step(s) of the following pathway are blocked in each of the five mutants and explain your logic below the figure. [Each arrow indicates an enzymatic step.]
If another Aro- mutant was isolated that behaved the same as the aroB mutant in crossfeeding experiments, describe a genetic test you could do to determine if both mutations affect the same enzymatic step. [No DNA sequencing or enzyme assays allowed.]
ANSWER: Note that if the mutations behave the same in crossfeeding experiments, doing additional crossfeeding experiments would not be a very good approach. Assuming that both mutations were recessive, then the best test would be an in vivo complementation test with two copies of the genes, one copy with a mutation in aroB and one copy with a mutation in the new Aro- mutant. This test must come with controls for each mutation against itself.
What results would you would obtain if the two mutations affected different enzymatic steps.
ANSWER: If the two mutations affected two different steps (for example, the one affected each arrow in the synthesis of DHQ), then the complementation test described above would allow growth without aromatic amino acids.
Three Arg- mutants were isolated and crossfeeding tests were done as shown below.
ANSWER: The F mutant accumulates the diffusible intermediate ornithine which is excreted and can be used by the E mutant. The G mutant accumulates the diffusible intermediate citrulline which excreted and can be used by the F and E mutants.
Using a permissive and nonpermissive host strain of Salmonella, you isolate two conditional lethal mutants of phage P22 (P22-1 and P22-2). Under permissive conditions one mutant only makes heads and the other mutant only makes tails, but you don't know which mutant has each phenotype. Devise a genetic experiment (using no electron microscopes, or other physical or chemical methods) to determine which mutant makes tails and which mutant makes heads. (Note that free heads and tails can be assembled in vitro to produce viable phage particles.)
ANSWER: There are two answers to this question, (a) the short version and (b) the connoisseur version.