Intergenic Suppression


What does allele-specific mean?

ANSWER: Allele-specific means that when one particular nucleotide or codon is at a specific site in a gene, another particular nucleotide or codon must be at a precise second site in that gene product or in a gene product it interacts with to produce a functional phenotype.


How can you determine if a suppressor is allele specific?

ANSWER: A simple initial approach is to test the function of the protein with only the suppressor mutation (i.e. the wild-type amino acid present at the position in the protein that was originally mutant when selecting for the suppressor). However, a variety of other amino acids at the two positions (i.e. original mutation site and suppressor site) must be tested before you can feel confident that the suppressor is truly allele specific.


What does it tell you if a suppressor is allele-specific?

ANSWER: If a suppressor is allele specific it indicates that the nucleotide or codon at the supressor site directly interacts with the nucleotide or codon at the site of the original mutation. Although a suppressor mutation that restores functional interactions with a mutant protein may suppress some alleles of the protein much better that others, very few interaction suppressors are absolutely allele-specific. Why? [Hint: think about the nature of protein-protein interactions.]


Although a suppressor mutation that restores functional interactions with a mutant protein may suppress some alleles of the protein much better that others, very few interaction suppressors are absolutely allele-specific. Why? [Hint: think about the nature of protein-protein interactions.]

ANSWER: If protein function requires specific interactions between amino acids, the charge and size of the interacting amino acids must be correct to allow the proper "fit". However, often substitution with a another amino acid that has similar charge and size properties will restore at least a partial interaction.


Two codons (CAG and CAA) are used for the amino acid glutamine.

  1. What is the minimal number of tRNA-Gln needed to read these codons? List the anticodon(s) expected for each predicted tRNA-Gln.

    ANSWER: The minimal number of tRNAGln needed is one because an anticodon of 5' UUG which can recognise 5' CAA codons in the mRNA (as expected) may also recognise 5' CAG codons because of "wobble" pairing on the 3rd base of the codon. The 5' U of the anticodon can pair with either A or G at the 3rd base of the codon.

  2. An amber suppressor (sup-2) is due to a mutation in a glutamine tRNA gene. What anticodon would you expect this suppressor tRNA to have?

    ANSWER: The codon 5' CAG for gln would be recognized by the anticodon 5' CUG. A single base substitution in the tRNA could change this anticodon to 5' CUA which would recognise the amber codon 5' UAG.

  3. An ochre suppressor (supE) is also due to a mutation in a glutamine tRNA gene. What anticodon would you expect this suppressor tRNA to have?

    ANSWER: 5' UUA to recognise the codon 5' UAA.

  4. A double-mutant carrying both suppressor mutations (sup-2 and supE) can be made. The double-mutant suppresses both the amber and ochre mutations and the two suppressor mutations are not linked. Given this information, revise your estimate of the minimal number of glutamine tRNA genes.

    ANSWER: From the above data, you should conclude that there are two different mutated tRNAGln genes, one is mutated to recognise amber codons and one is mutated to recognise ochre codons, however there is still the requirement to insert glutamine for CAG or CAA codons or else the cell will die from an inability to produce functional essential proteins. Therefore there must be at least one more tRNAGln working to recognise both Gln codons via wobble pairing, or else two tRNAGln genes code for different anticodons, one for each codon. So the final answer is either 3 ot 4 tRNAGln genes minimum.


An amber mutation in phage T4 can grow on strains carrying sup-1 but not on strains containing sup-2, even though both sup-1 and sup-2 are amber suppressors. Suggest an expanation for this result. [Hint -- think about the effect of suppression on the resulting protein.]

ANSWER: sup-1 and sup-2 are both amber suppressors -- due to a mutation in the gene encoding a tRNA which allows recognition of the UAG codon. However these two mutations affect two different tRNA genes such that, although both mutant tRNAs recognise amber codons, they insert different amino acids (because they are charged with the amino acid that charges each of the two different wild-type tRNAs). For example, sup-1 might be a mutated tRNA-leu gene whereas sup-2 might be a mutated tRNA-tyr gene. Thus, the supression in sup-1 would insert either leucine and the suppression in sup-2 would insert a tyrosine at the position in the protein corresponding to the amber codon. If the inserted amino acid is not similar in size and/or charge to the amino acid at that position in the wild type protein, the resulting amino acid substitution may interfer with the structure and function of the resulting protein.


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.

Amino acid     21    22    23    24    25    26
wild-type = met - asn - glu - met - thr - his
revertant #1 = met - asn - glu - asn - ala - his
revertant #2 = met - lys - lys - met - thr - his
  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.

  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 cannot tell if there are any potential stops generated by -1 frameshift reversions at codon # 26 or further right. Additionally, 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.


The Salmonella typhimurium strain TR248 is auxotrophic for both histidine and cysteine due to a mutation in the hisC gene and a mutation in the cysA gene. His+ revertants are found at a frequency of 1 per 107 cells. Cys+ revertants are also found with a frequency of 1 per 107 cells.

  1. How would you select for His+ revertants only or Cys+ revertants only? [What kind of medium would you plate the cells on.]

    ANSWER: Minimal media supplemented with cysteine will select for growth of His+ revertants (but will not select for Cys+ revertants). Minimal media supplemented with histidine will select for growth of Cys+ revertants (but will not select for His+ revertants).

  2. What frequency would you expect to find revertants that are both His+ and Cys+? How could you directly select for such double revertants?

    ANSWER: Double revertants will arise at a frequency of about 1 x 10-14 per cell [calculated as the product of each individual reversion frequency -- i.e. 1 x 10-7 X 1 x 10-7]. Double revertants can be selected on minimal media without supplements so that the survivors have to repair both the ability to make histidine and the ability to make cysteine.

  3. His+ Cys+ revertants are actually found at a frequency of 1 per 108 cells. Suggest an explanation for this observation.

    ANSWER: If both the his and the cys mutations are both nonsense mutations of the same kind (e.g. amber mutants) then a rare single hit mutation (1 x 10-8 frequency) to a particular tRNA gene could produce an amber suppressor tRNA that will suppress both the His- and the Cys- phenotypes.

  4. Describe a simple genetic experiment to test your hypothesis.

    ANSWER: If the "double revertant" strain is really an amber suppressor mutant (for example) then the ability to suppress amber mutants should allow at least some kinds of P22 amber mutant phage to form plaques, which they will not do on the parent strain TR248 or on revertants which are not nonsense suppressors.


A variety of suppressor mutations have been isolated in chromosomal tRNA genes in E. coli but tRNA suppressors which insert tryptophan are rare. Until recently it was believed that it is not possible to isolate such suppressors in haploid bacteria. Using the codon table, explain these results.

ANSWER: Consulting the codon table will show that there is only one tRNATrp gene, thus it is unlikely that mutations to the anticodon to change specificty will be viable because tryptophan will not be inserted into proteins that contain bona fide trp codons in their mRNA sequence. This would result in termination or mis-translation of a number of essential proteins and thus kill the cell. Hence, such suppressor mutations would have to alter the specificty of the tRNA so that it recognises both the wild type trp codon and the (nonsense) mutation being suppressed. A mutation such as a base substitution elsewhere in the tRNA molecule or a change in methylation of the molecule may affect the three dimensional structure of the tRNA such that it can accomplish both tasks with enough efficiency to allow the cell to live.


Various nonsense mutants of phage T4 were tested for their ability to grow on three different E. coli strains containing uncharacterized suppressor mutations. The results are summarized in the table below. [+ = ability to form plaques; 0 = no plaques are formed; supo = no nonsense suppressor]

Phage mutantHost strain
suposup-61sup-62sup-63
amber-10+0+
amber-20++ +
amber-30+0 0
amber-4000 0
amber-50++ +
ochre-100+ 0
ochre-200+ 0
ochre-3000 0

  1. Which of the host strains carry amber suppressors?

    ANSWER: sup-61; sup-63

  2. Which of the host strains carry ochre suppressors?

    ANSWER: sup-62 (can also suppress some amber mutations due to wobble, but not an amber suppressor per se)

  3. What anticodon would you expect for the suppressor tRNA of sup-62?

    ANSWER: 5' UUA

  4. Give two independent explanations for the slight differences in the suppression patterns of the sup-61 and sup-63 strains.

    ANSWER: sup-61 and sup-63 are different mutant tRNAs, so two possibilities seem most likely: (i) the two suppressors insert different amino acids at same nonsense codon, which will be tolerated differently depending upon the position in the protein into which they are inserted; or (ii) the efficiency of recognition is different (context etc.) between the two tRNAs so one suppresses well enough to allow functional product to be made at high enough levels, whereas the other does not.


A +1 frameshift mutation was obtained in the structural gene for an enzyme. The base sequence of the region of the wild-type mRNA where the mutation maps and the corresponding region of the mutant mRNA are shown below.

	
	wild-type =	5'  UUA AGC CUA AAU GAA AAU AGG    3'
	mutant =	5'  UUA AGC CCU AAA UGA AAA UAG G  3'

  1. Indicate the amino acid sequence of the wild-type protein encoded by this seqment of mRNA.

    ANSWER: The wild-type polypeptide sequence is determined by simply reading the amino acids from the codon table = Leu Ser Leu Asn Glu Asn Arg

  2. Indicate the amino acid sequence of the mutant protein produced by this segment of mRNA. What is the effect of the frameshift mutation on the protein?

    ANSWER: Note that the frameshift mutation causes translation termination at UGA codon, thus producing the mutant polypeptide sequence = Leu Ser Pro Lys Stop

  3. What protein products would be made by the mutant gene in the presence of each of the following tRNAs:

  4. Show the mRNA sequence of a second-site, single base deletion in the DNA that could potentially produce a functional pseudorevertant protein.

    ANSWER: There are multiple solutions possible. One potential solution is shown below.

    UUA AGC CCU AAU GAA AAU AGG
    Leu Ser Pro Asn Glu Asn Arg


Brenner, Stretton, and Kaplan (1965. Nature 206: 994-998) inferred the codon for nonsense mutations by studying suppression of mutants in the gene encoding the head protein of phage T4. The phage mutation H36(Am) is an amber mutation produced by mutagenesis with hydroxylamine in vitro, a condition which causes GC -> AT transitions. Explain their results shown in the table below. [Although they didn't have the benefit of knowing the "genetic code", you can use the genetic code table to answer this question.]

Host		Phage		Head protein amino acid sequence
Wild-type Wild-type Ala Gly Val Phe Asp Phe Gln Asp Pro Ile Asp Ile Arg ... Wild-type H36(Am) Ala Gly Val Phe Asp Phe Suppressor H36(Am) Ala Gly Val Phe Asp Phe Ser Asp Pro Ile Asp Ile Arg ...

ANSWER: The mutant T4 phage, H36(Am), has an amber mutation in the gene encoding the head protein. This results in a nonsense codon "UAG" in the mRNA of the protein. The phage produces a truncated protein in a wild-type host strain, but a complete protein in a host strain with a nonsense suppressor. The suppressor has a mutation in a tRNA-Ser gene such that the anticodon has been mutated from 5'-CGA-3' (Ser) to 5'-CUA-3' (Amber). The resulting tRNA is still charged with serine. The tRNASer recognizes 5'-UAG-3' in the mRNA and suppresses the premature truncation by the amber mutation.


Phage P1 efficiently infects and lyses both galE and galU mutants of Salmonella typhimurium, but not gal+ strains. When a culture of a galE62 mutant is infected with P1, most of the cells are killed but about 1 in 107 of the S. typhimurium cells is a Galsup>+ revertant that is resistant to P1. Similarly, when a culture of a galU14 mutant is infected with P1, about 1 in 107 of the S. typhimurium cells is a Gal+ revertant that is resistant to P1. When a culture of a galE62 galU14 double mutant is infected with P1, the reversion frequency is somewhat less -- about 1 in 108 of the S. typhimurium cells is a Galsup>+ revertant that is resistant to P1.

  1. If the reversion of the galE62 and galU14 mutations were independent events, what would be the predicted reversion frequency of the double mutant?

    ANSWER: If the two mutations reverted independently, the frequency of double reversion expected would equal the probability of the two separate events (10-7 x 10-7 = 10-14). Thus, it is very unlikely that reversion of the double mutant is due to two independent events.

  2. The frequency of reversion of both the galE62 and galU14 mutants is greatly increased by the mutagen ICR-191 but not alkylating agents. What does this suggest about the nature of these mutations? Explain your answer.

    ANSWER: Because the reversion is stimulated by a a frameshift mutagen but not by mutagens that cause base substitutions, the results suggest that both of the gal mutations are frameshift mutations.

  3. Given the above results, what is a likely reason that the frequency of double revertants is much higher than expected for two independent events?

    ANSWER: Single revertants of either gene can arise by either (i) a back mutation (true reversion), (ii) a mutation that produces an intragenic suppressor, or (iii) a mutation that produces an intergenic suppressor. However, only an intergenic suppressor is likely to simultaneously suppress both the galE and galU mutations. Because the mutagen results described in (b) suggest that both the galE and galU mutations are frameshift mutations, a reasonable explanation for these results is that reversion of the double mutant is due to a mutation that produces a frameshift suppressor (suf).

  4. How could you test this hypothesis?

    ANSWER: A simple test would be to assay for suppression of a characterized frameshift mutation in both the parent strain and the double revertant. For example, a lacZ frameshift mutation could be moved into both strains on a plasmid and B-galactosidase expression could be assayed using Lac indicator plates (e.g. Maconkey Lactose plates).


Wild-type phage P22 efficiently infects and lyses Salmonella Typhimurium strain LT2, releasing about 50 progeny phage per cell. Mutations in either the P22 gene 24 or P22 gene 23 prevent the phage from reproducing in LT2. (Note: phage genes are often indicated by numbers instead of three-letter mnemonics like in bacteria.)

When LT2 is infected with a P22 23 mutant, about 1 in 105 of the infecting phage is a revertant that can reproduce efficiently. When LT2 is infected with a P22 24 mutant, about 1 in 106 of the infecting phage is a revertant that can reproduce efficiently. [Note: you should NOT be surprised that the P22 23 and P22 24 mutations revert at somewhat different frequencies because the frequency of true reversion or intragenic pseudoreversion may depend upon the type of mutation in each mutant or on the DNA sequence near each mutation. Furthermore, reversion of point mutations often occurs at frequencies of between 10-5 and 10-7.] In contrast, when the S. typhimurium strain MS1363 is infected with a P22 24 mutant or with a P22 23 mutant, the infected cells are lysed by the phage. Furthermore, when MS1363 is infected with a P22 23 24 double mutant, the infected cells are lysed by the phage.

  1. Suggest a simple explanation for why the P22 23, the P22 24, and the P22 23 24 double mutant can infect strain MS1363 but not LT2.

    ANSWER: One potential explanation for the growth of the P22 mutants in strain MS1363 is that the phage mutations reverted. However, if the reversion of the P22 23 and P22 24 mutations were independent events, the frequency of double reversion expected would equal the probability of the two separate events: approximately10-6 x 10-5 = 10-11. Thus, it is very unlikely that such mutations would arise independently. However, if both mutations can be suppressed by a single informational suppressor (for example, an amber suppressor) then each of the three mutant phage would grow in a host strain that has an appropriate suppressor mutation. Thus, a reasonable explanation for these results is that strain MS1363 has a suppressor mutation. (Although the data shown above does not indicate what kind of suppressor is present in MS1363, we know that this strain has a supE mutation that produces an amber suppressor that inserts glutamine at UAG sites.)]

  2. How could you test this hypothesis?

    ANSWER: A simple test would be to infect both LT2 and MS1363 with a known amber mutant of P22. If the mutant phage reproduces in strain MS1363 but not LT3, the results would indicate that MS1363 has an amber suppressor. Alternatively, you could test for suppression of some other known amber mutant in the two strains. For example, you could move a known amber mutation into MS1363 and test the phenotype of the resulting strain -- a known lacZ(Am) mutation could be moved into MS1363 then tested for B-galactosidase expression on X-gal plates, or a trpA(Am) mutation could be moved into MS1363 and then tested to see if the resulting strain is a tryptophan auxotroph by testing for growth on minimal medium without tryptophan. (A potential biochemical test would be to infect LT2 and MS1363 with the mutant phage and then separate the proteins produced on a SDS-polyacrylamide gel to determine whether the P22 23 and 24 gene products were truncated proteins in LT2 and full length proteins in MS1363, however most truncated proteins (called "nonsense fragments") are rapidly degraded by cellular proteases so this trick rarely works.]


Phage P1 efficiently infects and lyses both galE and galU mutants of Salmonella typhimurium , but not gal+ strains. When a culture of a galE62 mutant is infected with P1, most of the cells are killed but about 1 in 107 of the S. typhimurium cells is a Gal+ revertant that is resistant to P1. Similarly, when a culture of a galU14 mutant is infected with P1, about 1 in 107 of the S. typhimurium cells is a Gal+ revertant that is resistant to P1. When a culture of a galE62 galU14 double mutant is infected with P1, the reversion frequency is somewhat less -- about 1 in 108 of the S. typhimurium cells is a Gal+ revertant that is resistant to P1.

  1. If the reversion of the galE62 and galU14 mutations were independent events, what would be the predicted reversion frequency of the double mutant?

    ANSWER: 10-7 x 10-7 = 10-14

  2. The frequency of reversion of both the galE62 and galU14 mutants is greatly increased by the mutagen ICR-191 but not alkylating agents. What does this suggest about the nature of these mutations? Explain your answer.

    ANSWER: Both mutants are probably due to frameshift mutations because ICR-191 is an intercalating agent that increases the frequency of +1 and -1 frameshift mutations.

  3. Given the above results, what is a likely reason that the frequency of double revertants is much higher than expected for two independent events?

    ANSWER: Reversion of the double mutants is probably due to formation of a frameshift suppressor tRNA which can suppress the independent frameshift mutations in the two different gal genes.


The putP gene encodes the major proline permease (transport protein) in S. typhimurium. A putP mutant is unable to transport sufficient proline for use as a sole nitrogen source. However, when a putP mutant is plated on medium with proline as a sole nitrogen source, rare colonies are obtained (Ekena, Liao, and Maloy. 1990. J. Bacteriol. 172 : 2940-2945).

  1. List potential ways that the mutant may have acquired the ability to use proline. Briefly explain your answers.

    ANSWER: Any of the classes of reversion events might give this phenotype: true revertants, intragenic suppressors, intergenic interaction suppressors, overexpression suppressors, bypass suppressors, etc.

  2. Similar results were observed if the original strain had a putP deletion mutation. What possible explanations are eliminated or favored by this observation.

    ANSWER: It is not possible to isolate a true revertant of a putP deletion. An intragenic suppressor is also unlikely because the deletion removes part of the structural gene. Thus, the most likely type of reversion is due to a bypass suppressor (e.g., a mutation in another permease that allows the transport of proline into the cell).


The Salmonella typhimurium strain MS1882 is a leucine auxotroph due to a base substitution mutation in the leuA gene. Prototrophic revertants of MS1882 arise at a high frequency, and many of the mutations are not in the leuA gene. (a) How would you select for Leu+ revertants? [What kind of medium would you plate the cells on.] (b) Describe three potential ways that Leu+ revertants might result from changes outside of the leuA gene. (c) Describe genetic experiments that would eliminate one of these three possibilities and explain the expected result for each of the three types of revertants. [Think genetics -- no DNA sequencing or in vitro experiments allowed.]

ANSWERS:


Both supD and supE mutants are efficient amber suppressors.

  1. When a supD mutation is moved into a strain with the lacZ9004 mutation, the phenotype becomes Lac+. What can you conclude about lacZ9004 based upon these results?

    ANSWER: lacZ9004 is due to an amber codon (UAG) in the the lacZ gene and it can be suppressed by insertion of the amino acid specified by the supD tRNA suppressor.

  2. When a supE mutation is moved into a strain with the lacZ9004 mutation, the phenotype remains Lac-. Suggest a potential reason for the difference between the suppression by supE and supD.

    ANSWER: supD inserts a different amino acid from supE. The amino acid inserted by supD is tolerated in the resulting protein, resulting in a functional gene product. However, the amino acid inserted by supE is probably not tolerated, so no functional gene product in formed. Because the gene product is required for the Lac+ phenotype, the lacZ9004 supE strain will remain Lac-.


Jarvick and Botstein (1975) developed an approach to using suppressor mutations determine the order of a developmental pathway [Proc. Natl. Acad. Sci. USA 72: 2738-2742]. When backcrossed to separate the suppressor mutation from the original mutation, many of the suppressor mutations had an independent phenotype. Briefly explain why.

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 such as CS and TS mutants, which is elegantly shown in this experiment.


Given a Salmonella phage with an amber mutation in a gene essential for lysis of host cells, how could you isolate a Salmonella amber suppressor mutant?

ANSWER:


The OmpA protein in the outer membrane of E. coli is the receptor for phage Ox2. It is possible to obtain mutants that are resistant to phage Ox2 by mixing 108 E. coli cells and 109 Ox2 and selecting for survivors. Typically 1-2 Ox2 resistant mutants will arise from this selection. Separation of the outer membrane proteins on SDS-polyacrylamide gels indicates that the OmpA protein is still produced in most of these mutants. Based upon these results, what is the most likely type of ompA mutation in the resistant mutants.

It is possible to isolate host-range mutants of Ox2 that infect the E. coli mutants which are resistant to wild-type Ox2. These phage host-range mutants have specific amino acid substitutions in product of Ox2 gene 38. Two classes of host-range mutants were obtained: class I only recognized the E. coli mutants which are resistant to wild-type Ox2; class II recognized both the resistant mutants and wild-type E. coli cells. Suggest a simple explanation for each class of Ox2 host-range mutants.



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Last modified October 18, 2004