Intracellular Complementation


A new proline auxotrophic mutant (pro-53) was isolated. By constructing partial diploid strains, a complementation table was constructed as shown below. One copy of the genes is shown on the top of the table and the other copy of the genes is shown down one side of the table. Only the mutant genes are shown in the table. "-" indicates failure to complement and "+" indicates complementation. (Note that a proA mutation cannot complement another proA- mutation, a proB mutation cannot complement another proB mutation, and a proC mutation cannot complement another proC mutation). What gene does the pro-53 mutation affect?

	proB	proA	proC
proB	  -	  +	  +
proA	  +	  -	  +
proC	  +	  +	  -
pro-53	  +	  -	  +
ANSWER: The pro-53 mutation is unable to complement a proA mutant but complements a proB mutant and proC mutant indicating that pro-53 affects the proA gene.


A collection of phage T4 mutants were isolated which were unable to lyse E. coli cells. The table below shows the results of coinfection experiments with pairs of these T4 mutants (+ indicates that the spot containing the two phage was lysed).

  1. How many different complementation groups are there?

    ANSWER: There are at least 2 complementation groups.

  2. What is the simplest interpretation of these results?

    ANSWER: There are at least two genes.

  3. Which mutations map in each complementation group?

    ANSWER:

  4. Explain any anomalies in the data.

    ANSWER: Although they appear to be in different complementation groups based upon complementation with other mutants, the mutants G and H fail to complement each other. The simplest explanation of these results is that there is a trans-dominant negative interaction between mutants G and H. Most likely this is the result of an interaction between the gene product of mutant H and the gene product of mutant G, since no dominance effects are seen between mutant H and mutants A, C, or E which are different alleles of the same gene as mutant H.


In addition, several temperature sensitive mutants of phage T4 were obtained. These mutants were also analyzed by complementation tests. The tests were performed by spotting a mixture of the two phage mutants to be tested (about 106 of each) onto a rich medium plate spread with E. coli and incubating the plate at 40 C. The results are shown below [+ = complete lysis; - = no lysis or only a few plaques in the spot].

  1. How many complementation groups of conditional mutants were obtained?

    ANSWER: There are three complementation groups and thus three genes.

  2. Which mutants fall into each complementation group?

    ANSWER:

  3. What results would you expect if the experiment was done at 30 C?

    ANSWER: All the phage would grow, so the diagonal would show + indicating the results are not valid.

  4. How would you determine if any of these mutations affected the same gene as the mutations described in the question above?

    ANSWER: Do complementation analysis with pairwise combinations of these mutations with the mutations shown in the question above. The complementation grid would thus be A-M across the top and A-M down the side.


Complementation tests were done on five mutants of phage T4 by co-infecting a supo strain of E. coli with two mutant phage. The results are shown below.

12345
1---++
2-+++
3-++
4-+
5-

How many complementation groups do these mutants represent? [Explain your logic.]

ANSWER: Four complementation groups:

Mutants 4 and 5 are clearly in separate complementation groups. In addition, mutant 2 can complement mutant 3 suggesting that they are in separate complementation groups. But, the surprising result is that mutant 1 cannot complement either mutant 2 or mutant 3. One explanation for this result is that mutant 2 or 3 interact to repair the phenotype. Alternatively, mutant 1 may be polar on mutant 2 and mutant 3.

For example, imagine that gene A and gene B are in an operon -- mutant 1 could be an amber mutation in gene A and mutant 3 a missense mutation in gene A -- if mutant 2 was in gene B then mutant 1 would be phenotypically A- B- while mutant 3 would be A- B+. (See figure below.)


Three conditional mutants of phage T4 were isolated. One is cold sensitive (Cs), one is temperature sensitive (Ts), and one is an amber mutant (Am). The ability of these mutants to infect and lyse two different E. coli strains is shown in the table below.

PhageStrain EC1Strain EC2
25C30C42C25C30C42C
Cs-++-++
Ts++-++-
Am---+++

Fill in the following table to show the expected phenotype if the bacteria were infected with the double mutant phage [i.e. each cell is infected with a single phage particle that has two mutations.]

PhageStrain EC1Strain EC2
25 C30 C42 C25 C30 C42 C
Cs Ts
Cs Am
Ts Am

ANSWER: see the table below.

PhageStrain EC1Strain EC2
25 C30 C42 C25 C30 C42 C
Cs Ts-+--+-
Cs Am----++
Ts Am---++-

Note that these results strongly suggest that strain EC2 has an amber suppressor!


Felix 0 (F0) is a virulent phage (that is, it cannot lysogenize its host). When wild-type phage F0 phage is spotted on a lawn of Salmonella typhi the cells are lysed (indicated by white spots on the bacterial lawn shown below). Three conditional mutants were isolated that prevent lysis of a nonpermissive host. All three of these mutants affect the synthesis and assembly of the phage head, a complex structure that requires proper interactions between several different proteins. To determine if these mutations affect different genes, cells were coinfected with two different mutant phage under nonpermissive conditions as shown in the figure below.

  1. Do the mutations in FO-1 and FO-3 map in different genes?

    ANSWER: The phenotype indicates that the two mutations complement each other -- thus, unless this is a rare example of intragenic complementation, then the two mutations affect different genes.

  2. Why is the amount of lysis much less in the spot containing FO-1 and FO-2 compared to the spot containing FO-1 and FO-3?

    ANSWER: As described above FO-1 and FO-3 complement each other, so every cell coinfected with both phage will be lysed. In contrast, the results indicate that FO-1 and FO-2 cannot complement each other, so the rare plaques observed are probably due to recombination between the two phage to produce wild-type phage. This requires both that the cell is coinfected with FO-1 and FO-2 and that a subsequent cross-over occurs in the region between the two mutations, and hence it is much rarer than simple complementation.


Resistance to the toxic proline analog azetidine-2-carboxylic acid can occur in two ways: (i) specific missense mutations in the proB gene (the first step in proline biosynthesis) which make it insensitive to feedback inhibition; and (ii) mutations that inactivate the putP gene (the permease which transports proline into the cell).

  1. Which class of mutants would you expect to be more common and why?

    ANSWER: The putP mutations would be most common because a null mutation anywhere within the gene could inactivate the protein. Only a limited number of sites in proB would be likely to produce a functional protein which is insensitive to feedback inhibition.

  2. Indicate whether each of the two types of mutations are dominant or recessive to the wild-type allele of that gene. [Explain your logic.]

    ANSWER: A putP null mutation is a simple loss of function mutation which would be recessive to the wild-type allele. That is, putP- / putP+ would be able to transport both proline and azetidine-2-carboxylic acid into the cell, resulting in sensitivity to the proline analog. A proB mutant insensitive to feedback inhibition ("proB*") would be dominant to wild-type. That is, proB* / proB+ would have one copy of the gene which is insensitive to feedback inhibition, allowing accumulation of a high intracellular proline concentration and thus resulting in resistance to the proline analog.


Following chemical mutagenesis of Salmonella typhimurium cells, Lawes and Maloy isolated host mutants which are conditionally resistant to phage P22. One class of these mutants affect the maturation pathway for phage P22 -- in these mutants DNA is injected into the host cell from phage heads but no progeny viruses are produced. Three cold-sensitive (CS) bacterial mutants with this maturation defective phenotype (Mat-) were mapped by recombination experiments. The three mutations mapped at three unlinked loci as follows:

matA (CS) 75.8 min
matB (CS) 76.4 min
matC (CS) unknown location but not linked to above loci
Complementing clones were then sought by screening through a S. typhimurium clone library, and one plasmid was isolated which restored the P22 sensitivity phenotype (Mat-) to the matC (CS) mutant. This plasmid clone (pMat+) was extracted and backcrossed into the matC (CS) mutant.

As a control, the pMat+ plasmid was also introduced into the matA (CS) and matB (CS) mutant strains and the P22 sensitivity phenotype scored at 30C. The observed results are shown below, along with the expected results:

StrainP22 Sensitivity at 30C (+= sensitive)
ObservedExpected
matA (CS)--
matA (CS)/pMat++-
matB (CS)--
matB (CS)/pMat++-
matC (CS)--
matC (CS)/pMat+++

  1. What is a backcross and why was a backcross done?

    ANSWER: A backcross is strictly speaking a cross between the F1 (recombinant) generation and the parental generation. In this case, the uncomplemented mutant is the "parental" generation and the complemented mutant is the "recombinant". A backcross involves removing the pMat+ plasmid back into the uncomplemented parent. This is done to demonstrate that the "complementing" activity is located on the plasmid and not due to a spontaneous reversion or second site suppressor in the recipient chromosome.

  2. Describe the complementation results for the the pMat+ plasmid and explain what these results mean.

    ANSWER: The expectation is that only matC would be complemented because the three genes tested represent dicrete loci. So if all three mutations are complemented, either all three genes are cloned on the plasmid (but that is NOT consistent with mapping data which places the genes in three separate locations) or some kind of overexpression suppression is occuring that rescues the phenotype of all three mutants. The overexpressed gene may be one of the three mat genes or a fourth gene that we do not know about. Note that it is overexpression on a plasmid producing this effect because the mutant strains contain a single copy of each of these genes in their chromosome and that is inadequate to produce a suppression of the Mat phenotype.

  3. How could you determine if the complementing activity on plasmid pMat+ is specific to S. typhimurium genes of the P22 maturation pathway, or simply provides an activity which restores function to cold-sensitive gene products in general? ANSWER: If the activity restores function to cold-sensitive proteins in general then it should be able to suppress cold-sensitive mutations in genes that play no role in the phage P22 life cycle, for example a his (CS) mutant strain could be tested for suppression to restore the ability to make histidine at the non-permissive temperature (30C) in the presence and the absence of pMat+.


Many antibiotics act by binding to bacterial ribosomes and inhibiting translation. Resistance to such antibiotics is often due to mutations that produce specific nucleotide changes in one of the ribosomal subunits which prevent the antibiotic from binding to the ribosome. In such cases, the antibiotic resistant alllele is often recessive to the antibiotic sensitive allele. Why?

ANSWER: If the antibiotic stops the ribosomes in their tracks, it will prevent subsequent ribosomes from transcribing that message. If cells have both resistant and sensitive ribosomes, both the resistant and sensitive ribosomes will bind to the mRNA -- however, the sensitive ribosomes will become stuck and block resistant ribosomes from translating the message. Thus, the sensitive phenotype is dominant to the resistant phenotype.


Briefly compare and contrast complementation vs recombination.

ANSWER: Complementation = mixing of gene products, changes phenotype not genotype, no breakage/covalent rejoining of DNA. Recombination = changes genotype, requires breakage/covalent rejoining of DNA.


The his operon encodes a group of genes required for the biosynthesis of the amino acid histidine. Thus, His- mutants are histidine auxotrophs (unable to grow on minimal medium but grow on minimal medium + histidine).

Five new His- auxotrophs were isolated. Complementation analysis of each of the mutants gave the following results (+ indicates growth on minimal medium without histidine).

12345
1-++++
2--++
3-++
4-+
5-

  1. How many complementation groups are there?

    ANSWER: There are 4 complementation groups: 1 = his-1; 2 = his-2 and his-3; 3 = his-4; 4 = his-5

  2. How many genes do the five His- mutants affect?

    ANSWER: Four genes.


Complementation analysis was done on six mutants that lack theonine synthetase activity. The order of the mutational sites is not known. The results are shown below:

thr-1thr-2thr-3thr-4thr-5thr-6
thr-1-++-0-
thr-2--++-
thr-3-++-
thr-4---
thr-5--
thr-6-

  1. How many complementation groups are represented?

    ANSWER: At least two complementation groups: Group 1 = thr-1, thr-4, thr-5; Group 2 = thr-2, thr-3; Ungrouped = thr-6

  2. Suggest an explanation for the results for thr-6.

    ANSWER: thr-6 fails to complement all of the other mutants. This could be either due to a trans-dominant negative phenotype of this mutant (e.g. due to a missense mutation that poisons threonine synthetase) or a cis-dominant negative phenotype caused by the mutation (e.g. due to an amber mutation that prevents expression of downstream genes). In either case, it is impossible to determine whether the thr-6 mutation is within one of the two complementation groups described by the other mutations or whether it is in a different complementation group.


Several E. coli mutations were isolated that produce a Lac- phenotype when present on the chromosome. The lac operon has three gene products that can be assayed independently. Complementation analysis was done with each of the new lac mutations on multicopy plasmids and known lac mutations on the chromosome. The results are shown in the following table. Based upon the results, indicate which gene the mutation affects and explain your conclusion.

  Plasmid Chromosome Mutant gene Explanation
    lacZ lacY lacA or genes  
1 lac-9001 + - + lacY Cannot complement a lacY mutant but can complement lacZ and lacA
2 lac-9002 + + - lacA Cannot complement a lacA mutant but can complement lacZ and lacY
3 lac-9003 - + + lacZ Cannot complement a lacZ mutant but can complement lacY and lacA
4 lac-9004 - - - lacZ,Y,A Cannot complement a lacY, lacZ, or lacA mutants
5 lac-9005 + + + unknown Complements lacZ, lacY, and lacA mutants -- the three known genes

  1. Assuming that all the results described are valid and that no mutations outside of the lac operon are required for lactose utilization, what could cause the results observed for lac-9005 in row 5?
    ANSWER: The lac-9005 mutation is known to have a Lac- phenotype when present on the chromosome. The simplest explanation for this complementation result is that the lac-9005 mutation results in overproduction suppression when present on a multicopy plasmid. [An alternative possibility is that the mutation affects some unknown gene that affects some new gene that results in a Lac- phenotype -- the statement "the lac operon has three genes ..." was the clue that there really are only three specific, structural genes that are required for lactose catabolism.

  2. How could you test your hypothesis?
    ANSWER: A simple test for overproduction suppression is to do the complementation analysis with a single copy of each lac mutant.


Complementation tests were done on five mutants of phage T4 by co-infecting a supo strain of E. coli with two mutant phage. The results are shown below.

  1 2 3 4 5
1 - - - + +
2   - + + +
3     - + +
4       - +
5         -

How many complementation groups do these mutants represent?

ANSWER: There are at least four complementation groups:

Mutants 4 and 5 are clearly in separate complementation groups. The logic for the first two complementations groups is described below. In addition, mutant 2 can complement mutant 3 suggesting that they are in separate complementation groups. But, the surprising result is that mutant 1 cannot complement either mutant 2 or mutant 3. The simplest explanation for this result is that mutant 1 and mutant 2 or 3 are in the same complementation group but mutant 1 has a negative dominant effect on the other mutant. For example, imagine that gene A and gene B are in an operon -- mutant 1 could be an amber mutation in gene A and mutant 3 a missense mutation in gene A -- if mutant 2 was in gene B then mutant 1 would be phenotypically A- B- while mutant 3 would be A- B+.



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Last modified October 18, 2004