Maltose transport in E. coli is mediated by an ABC transporter. The transport requires a periplasmic maltose binding protein (MBP), and three integral membrane proteins encoded by the genes malF, malG, and malK. Mutants lacking any of these proteins do not transport maltose. Shuman and colleagues selected mutants that lack MBP but can transport maltose. All of the original malF mutants had two mutations. To study the interactions between the mutations, they constructed mutants with only one of the mutations or with different combinations of two mutations. The results are shown below.

¹Key: -, +/-, +, ++, +++ indicate increasing amounts of maltose transport by the mutant strain.
²Data from Covitz, K.-Y., C. Panagiotidis, L.-I. Hor, M. Reyes, N. Treptow, and H. Shuman. 1994. Mutations that alter the transmembrane signalling pathway in an ATP binding cassette (ABC) transporter. EMBO J. 13: 1752-1759.

1. Why were none of the strains with single mutations able to grow on maltose?

   ANSWER: Mutations in any of the single alleles results in an inactive protein -- only mutants with both the original mutation and the suppressor mutation were active.

2. Why did different combinations of double mutants show different amounts of maltose transport?

   ANSWER: The different suppressors show partial allele specificity -- that is, note of the suppressors interact with the wild-type allele at the primary site but some suppressors interact with certain primary mutations better than others. The simplest interpretation of these results is that the amino acids at the second-site directly interact with amino acids at the primary site.

Oosawa and Simon isolated numerous intragenic pseudorevertants of an Ala-19 to Lys mutation in the tar gene in E. coli. (Ala-19 indicates an alanine at position 19 in the protein.) Several different pseudorevertants obtained are shown below [data from Oosawa, K., and M. Simon. 1986. Analysis of mutations in the transmembrane region of the aspartate chemoreceptor in Escherichia coli. Proc. Natl. Acad. Sci. USA 83: 6930-6934.]

1. Based upon pseudorevertants 1-3, what can you conclude about the role of amino acid 19 in the Tar protein?

   ANSWER: Insertion of a variety of different amino acids (including Ala, Gln, Ile, and Thr) at position 19 of the Tar protein does not disrupt its function.

2. When the second site suppressors were backcrossed into the wild-type tar gene (with Ala-19), most of the resulting proteins (with only the suppressor mutation) were functional. What can you conclude about the allele specificity of the suppressor mutations 4-8 and what does this tell you about the effect of these mutations on the Tar protein?

   ANSWER: Since the protein remained functional when the suppressor mutation was present but the wild-type amino acid was present at position 19, the results indicate that the suppressor can function with both Lys-19 and Ala-19 (amino acids with very different properties) and hence they are not allele specific suppressors. This suggests that the effect of the suppressor mutation on the Tar protein is independent of the amino acid at position 19, and they probably act by causing increased protein activity (possibly by increasing stability of the protein).

Klig, Oxender, and Yanofsky (1988) isolated second-site revertants of Trp repressor mutants in E. coli [Genetics 120: 651-655].

1. How did they avoid true revertants?

   ANSWER: The initial mutagenesis was performed using hydroxylamine, which is a unidirectional mutagen causing GC to AT transitions. Reversion mutagenesis using hydroxylamine was not be able to revert the primary mutations because those base pairs were changed to AT which is not mutated by hydroxylamine.

2. Were the mutants obtained allele-specific?

   ANSWER: No. The revertant alleles were able to suppress more than one of the original mutant alleles and are therefore not "allele-specific". The reversion mutation increased the activity of the protein in general rather than compensating for a specific change in the original mutant.

The aceA gene product is required to grow on acetate as a sole carbon source. Temperature sensitive (Ts) mutants that affect this gene were desired, but it is not possible to directly select for aceA mutants. Instead, revertants of an aceA mutant were isolated and the resulting mutants checked for a Ts phenotype.

1. How would you obtain the Ts revertants? [Note: the original mutation is not Ts --you are looking for revertants that have a Ts phenotype!]

   ANSWER:

   • Select for Ace⁺ revertants by plating on minimal plates with acetate as the sole carbon source at 30 C.
   • Replicate the colonies onto minimal acetate plates incubated at 30 C and minimal acetate plates incubated at 42 C.
   • Pick any colonies that grow on minimal acetate plates at 30 C but not at 42 C and recheck their phenotype. Note an essential control is to make sure these are not simply temperature sensitive lethal mutations by checking for growth on some other carbon source at 42 C.

2. Why is this approach easier than simply screening for aceA(Ts) mutants?

   ANSWER: Intragenic pseudorevertants are often temperature sensitive. Because there is a direct selection for revertants of the Ace^- mutant, it is possible to obtain the revertants without heavy mutagenesis, so the cells are not likely to have secondary mutations (which might have an unrelated temperature sensitive lethal phenotype). Thus, by looking for revertants first then scoring for temperature sensitivity, many fewer colonies will need to be screened to find the desired mutants.

3. The Ts revertants were grown at a permissive temperature, then isocitrate lyase activity (the aceA gene product) was assayed at a non-permissive temperature. The cells retained high levels of isocitrate lyase activity. Suggest a potential explanations for this result?

   ANSWER: The best answer is that the mutation is temperature sensitive for synthesis (Tss), but once made it remains active at the nonpermissive temperature.

What might cause a trans-dominant negative mutation in an enzyme? What might cause a trans-dominant negative mutation in a structural protein?

ANSWER: In both cases a trans-dominant negative mutation could be due to an alterted protein subunit that interfers with multimerization by wild-type subunits (for example, the mutant/wild-type heterodimer might be inactive). For a structural protein (for example, phage head proteins) this would interfer with formation of the proper stucture. For an enzyme, this could either interfer with formation of the proper structure per se, or it could mess up the active site of the enzyme.

Jarvick and Botstein (1975. Proc. Natl. Acad. Sci. USA 72: 2738-2742) observed that there is a high probability that a missense mutant will yield cold-sensitive or temperature-sensitive revertants. What is the most likely explanation for this result?

ANSWER: Suppressor mutations themselves often make missense changes to the gene product of the gene in which they are located, this is especially true where a protein is mutated to restore protein-protein interaction with another mutant protein. If the suppressor mutation is crossed into a new genetic background so that it is paired with the wild-type allele of its interacting subunit, the suppressor mutant protein often displays a missense phenotype because it fails to interact with the wild type subunit. Missense phenotypes are sometimes quite subtle and can result in temperature conditional alleles (i.e., CS and TS mutants) due to misfolding of the resulting protein.

What types of revertants (i.e., phenotypically wild-type) might arise from a leaky base-substitution mutation in the structural gene for an enzyme? How would you distinguish the different types of revertants genetically and biochemically?

An elegant genetic experiment to determine the number of bases required to code for each amino acid took advantage of a large collection of frameshift mutations in the rII gene of phage T4 [Crick, F., L. Barnett, S. Brenner, and R. Watts-Tobin. 1961. Nature 192: 1227-1232]. The wild-type DNA and amino acid sequence corresponding to the first portion of the rII gene are shown below.

ATG TAC AAT ATT AAA TGC CTG ACC AAA AAC GAA CAA GCT GAA ATT GTT AAA CTG TAT TCA
Met tyr asn ile lys cys leu thr lys asn glu gln ala glu ile val lys leu tyr ser

AGT GGT AAT TAC ACC CAA CAG GAA TTG GCT
ser gly asn tyr thr gln gln glu leu ala

ANSWERS:

1. FC0/FC1: 1
   ATG TAC AAT ATT AAA TGC CTG ACC AAA AAC GAC AAG CTG AAA TTG TTA AAA CTG TAT TCA
   Met tyr asn ile lys cys leu thr lys asn asp lys leu lys leu leu lys leu tyr ser
   61
   AGT GGT AAT TAC ACC CAA CAG GAA TTG GCT
   ser gly asn tyr thr gln gln glu leu ala

2. FC1/FC40: 1
   ATG TAC AAT ATT AAA TGC CTG ACC AAA AAC GAC AAG CTG AAA TTG TTA AAC TGT ATT CAA
   Met tyr asn ile lys cys leu thr lys asn asp lys leu lys leu leu asn cys ile gln

   61
   AGT GGT AAT TAC ACC CAA CAG GAA TTG GCT
   ser gly asn tyr thr gln gln glu leu ala

3. FCO/FC88: 1
   ATG TAC AAT ATT AAA TGC CTG ACC AAA AAC GAA CAA GCT GAA ATT GTT AAA ACT GTA TTC
   Met tyr asn ile lys cys leu thr lys asn glu gln ala glu ile val lys thr leu phe

   61
   AAG TGG TAA TTA CAC CAA CAG GAA TTG GCT
   lys trp stop

4. FC40/FC88:1
   ATG TAC AAT ATT AAA TGC CTG ACC AAA AAC GAA CAA GCT GAA ATT GTT AAA CTG TAT TCA
   Met tyr asn ile lys cys leu thr lys asn glu gln ala glu ile val lys leu tyr ser

   61
   AAG TGG TAA TTA CAC CAA CAG GAA TTG GCT
   lys trp stop

5. The double mutants FC0/FC1 and FC1/FC40 have compensating frameshift mutations that restore the reading frame allowing production of a full length protein product with a short region of incorrect amino acids. Because the protein is functional, this result indicates that the amino acids between the two frameshift mutations are not essential for the structure and function of the protein. The double mutants FC0/FC88 and FC40/FC88 both result in a stop codon before the compensating frameshift mutation, hence a truncated protein is produced.

The amino acid sequences of a peptide from wild-type and two independent revertants of the same frameshift mutant are shown below. Aside from the regions shown, the wild-type and revertant proteins are identical.

1. Using the information shown above, write the best possible sequence of the wild-type mRNA for this peptide and indicate where a base was added or deleted in the original frameshift mutant and each of the two revertants.

   ANSWER:

   

2. Would you expect to find revertants of this frameshift mutant in which the compensating frameshift mutation mapped to the left of codon #21? Explain your answer.

   ANSWER: Reversion of the +1 frameshift mutation requires a -1 frameshift mutation to restore the reading frame. Such a reversion to the left of codon #21 will result in a UGA codon (STOP) at codon #21 and thus terminate translation, clearly preventing the recovery of functional revertants.

3. Would you expect to find revertants in which the compensating mutation mapped to the right of codon #25? Explain your answer.

   ANSWER: Maybe. You can't tell if there are any potential stops generated by -1 frameshift reversions at codon # 26 or further downstream. Also, it is not possible to predict the influence that the alteration in the amino acid sequence will have on the function of the protein -- remember that different amino acids will make up the region between the -1 and the +1 frameshift mutations relative to the wild-type protein.

Please send comments, suggestions, or questions to smaloy@sciences.sdsu.edu
Last modified October 18, 2004