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.)

  1. Describe the nature of the Pro- mutation in each strain (i.e., is it a null mutant, a temperature sensitive mutant, a cold sensitive mutant, etc).

    ANSWER: pro-1 causes a temperature sensitive phenotype; pro-2 causes a non-conditional, null phenotype; pro-3 causes a cold sensitive phenotype.

  2. Indicate the expected growth of the double mutants with a + or - in the table below.

  3. Which enzymatic step in the following pathway is affected in each mutant?

    ANSWER: a = pro-1; b = pro-2; c = pro-3

  4. Why could pro-3 crossfeed pro-2 at 22C but not at 30C or 42C?

    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:

  1. Based upon these results, what is the predicted order of the bioA, bioB, bioC, and bioD gene products in the metabolic pathway?

  2. These bio mutants were then tested to see if they could grow on several compounds believed to be intermediates in biotin biosynthesis. The results are shown below (+ indicates growth and - indicates no growth):


  3. Based upon biochemical studies, the inferred pathway of biotin biosynthesis was believed to be that 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.]

ANSWER: after SP (not fed by any other mutants in pathway), but cannot tell which of the two enzymatic steps

ANSWER: before DHQ (fed by all mutants later in pathway), but cannot tell which of the two enzymatic steps

ANSWER: between DHQ and DHS

ANSWER: between DHS and S

ANSWER: between S and SP

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.

  1. Based upon these crossfeeding results, indicate the order of the G, E, and F genes in the arginine biosynthesis pathway shown below.

  2. Briefly explain your rationale for the proposed order.

    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.

  1. Tails alone can bind to the surface of the bacteria. If tails are first mixed with cells, and the excess tails are removed by centrifugation, then heads (without tails) added subsequently, the heads will stick to the tails pre-bound to the surface of the cells, allowing the injection of DNA from the heads into the recipient bacterium. However, this scheme will only work if tails are added first, and then heads are added. If heads are added first, they will not bind to the cells and so removed during centrifugation. Thus, if P22-1 mutants are added followed by P22-2 mutants and the reciprocal order experiment performed where P22-2 is followed by P22-1, only one of the two orders will produce plaques. There are a couple of important points that must be considered: (i) the phages used for the ordering experiment must have been grown on a non-permissive strain so that they express the mutant phenotypes (i.e. one has no heads and one no tails); (ii) the ordering experiment must use a permissive host strain to score for plaques, because the reconstitued particles that inject DNA are still genetically tail minus, so progeny viruses in a non-permissive host are still tail-less and unable to form plaques, which require multiple rounds of infection to be seen in a lawn of sensitive bacteria.

  2. Since the DNA from phage P22 is packaged only in the head and not in the tail, only the tail-less defective particles will contain DNA. Thus if a portion of each defective phage is mixed together, a tailed head is produced which contains DNA which is genetically still tail minus. These reconstituted phages are plated out on a lawn of permissive host cells so that plaques form. The phages in these plaques are exclusively tail minus genetically but phenotypically are tailed because of growth in a permissive strain. One of the plaques produced in this manner is picked and grown in a non-permissive host because we need phage which are both genotypically and phenotypically tail minus for the next step. These phages are then mixed with either P22-1 or P22-2 separately and then plated onto a lawn of permissive bacteria so that plaque formation can be scored. Only one of the two mutant phage will produce tails and thus add tails to the tail minus phages to allow infection of the recipient cells. This event is seen as plaque formation.

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Last modified October 18, 2004