How to Read a Complementation Table in Genetics
Intracellular Complementation
A new proline auxotrophic mutant (pro-53) was isolated. By constructing partial diploid strains, a complementation table was constructed as shown beneath. One re-create of the genes is shown on the top of the table and the other re-create of the genes is shown down i side of the table. Merely the mutant genes are shown in the table. "-" indicates failure to complement and "+" indicates complementation. (Notation 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 touch?
ANSWER: The pro-53 mutation is unable to complement a proA mutant only complements a proB mutant and proC mutant indicating that pro-53 affects the proA gene.proB proA proC proB - + + proA + - + proC + + - pro-53 + - +
A collection of phage T4 mutants were isolated which were unable to lyse East. coli cells. The tabular array below shows the results of coinfection experiments with pairs of these T4 mutants (+ indicates that the spot containing the ii phage was lysed).
- How many dissimilar complementation groups are at that place?
Answer: In that location are at least 2 complementation groups.
- What is the simplest interpretation of these results?
ANSWER: There are at least 2 genes.
- Which mutations map in each complementation group?
Reply:
- Group 1: A, C, E, H
- Grouping 2: B, D, F
- Ungrouped: G could exist a separate group or could be an unusual allele of group 1
- Explain any anomalies in the data.
Answer: Although they appear to exist 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-ascendant negative interaction between mutants Thousand and H. Most likely this is the consequence of an interaction betwixt the gene product of mutant H and the gene product of mutant K, since no dominance effects are seen between mutant H and mutants A, C, or E which are dissimilar alleles of the same gene as mutant H.
In addition, several temperature sensitive mutants of phage T4 were obtained. These mutants were too analyzed by complementation tests. The tests were performed past spotting a mixture of the two phage mutants to exist tested (about 10six of each) onto a rich medium plate spread with E. coli and incubating the plate at forty C. The results are shown below [+ = complete lysis; - = no lysis or only a few plaques in the spot].
- How many complementation groups of conditional mutants were obtained?
Respond: There are iii complementation groups and thus three genes.
- Which mutants fall into each complementation group?
Answer:
- Gene 1: I, Fifty
- Gene 2: J
- Gene iii: K, M
- What results would yous expect if the experiment was done at thirty C?
ANSWER: All the phage would grow, and then the diagonal would show + indicating the results are non valid.
- How would y'all determine if any of these mutations affected the same factor as the mutations described in the question in a higher place?
Reply: Practice complementation assay 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 downwardly the side.
Complementation tests were washed on five mutants of phage T4 by co-infecting a supo strain of E. coli with 2 mutant phage. The results are shown beneath.
| 1 | 2 | iii | 4 | 5 | |
|---|---|---|---|---|---|
| 1 | - | - | - | + | + |
| two | - | + | + | + | |
| 3 | - | + | + | ||
| 4 | - | + | |||
| 5 | - | ||||
How many complementation groups do these mutants represent? [Explain your logic.]
Answer: Four complementation groups:
- Group 1: mutant 2
- Group 2: mutant three
- Group 3: mutant 4
- Grouping 4: mutant 5
- Ungrouped: Mutant 1 may be in a separate complementation group or may exist an unusual allele of group 1.
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. Ane is common cold sensitive (Cs), one is temperature sensitive (Ts), and one is an bister mutant (Am). The ability of these mutants to infect and lyse two different E. coli strains is shown in the table below.
| Phage | Strain EC1 | Strain EC2 | ||||
|---|---|---|---|---|---|---|
| 25�C | 30�C | 42�C | 25�C | xxx�C | 42�C | |
| Cs | - | + | + | - | + | + |
| Ts | + | + | - | + | + | - |
| Am | - | - | - | + | + | + |
Fill in the following table to prove the expected phenotype if the bacteria were infected with the double mutant phage [i.e. each jail cell is infected with a single phage particle that has 2 mutations.]
| Phage | Strain EC1 | Strain EC2 | ||||
|---|---|---|---|---|---|---|
| 25 C | thirty C | 42 C | 25 C | 30 C | 42 C | |
| Cs Ts | ||||||
| Cs Am | ||||||
| Ts Am | ||||||
ANSWER: encounter the table below.
| Phage | Strain EC1 | Strain EC2 | ||||
|---|---|---|---|---|---|---|
| 25 C | 30 C | 42 C | 25 C | 30 C | 42 C | |
| Cs Ts | - | + | - | - | + | - |
| Cs Am | - | - | - | - | + | + |
| Ts Am | - | - | - | + | + | - |
Notation 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-blazon phage F0 phage is spotted on a backyard 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 3 of these mutants affect the synthesis and assembly of the phage head, a complex structure that requires proper interactions betwixt several different proteins. To determine if these mutations affect different genes, cells were coinfected with 2 unlike mutant phage under nonpermissive conditions as shown in the effigy below.
- Practise 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, so the 2 mutations affect dissimilar genes.
- Why is the amount of lysis much less in the spot containing FO-1 and FO-2 compared to the spot containing FO-i and FO-three?
Reply: As described to a higher place FO-ane and FO-iii complement each other, so every cell coinfected with both phage will be lysed. In dissimilarity, the results indicate that FO-ane 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 betwixt the ii mutations, and hence information technology is much rarer than elementary complementation.
Resistance to the toxic proline analog azetidine-two-carboxylic acid can occur in two ways: (i) specific missense mutations in the proB factor (the first footstep in proline biosynthesis) which get in insensitive to feedback inhibition; and (ii) mutations that inactivate the putP gene (the permease which transports proline into the cell).
- Which class of mutants would you await to be more common and why?
Respond: The putP mutations would be nearly 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.
- Signal whether each of the ii types of mutations are ascendant or recessive to the wild-type allele of that cistron. [Explain your logic.]
Respond: A putP null mutation is a elementary loss of function mutation which would be recessive to the wild-type allele. That is, putP - / putP + would be able to ship 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 exist ascendant to wild-type. That is, proB* / proB + would have one copy of the gene which is insensitive to feedback inhibition, assuasive accumulation of a loftier 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 impact 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. 3 cold-sensitive (CS) bacterial mutants with this maturation defective phenotype (Mat-) were mapped by recombination experiments. The three mutations mapped at iii unlinked loci as follows:
matA (CS) 75.8 minComplementing clones were then sought by screening through a S. typhimurium clone library, and ane 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.
matB (CS) 76.4 min
matC (CS) unknown location but not linked to above loci
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 30�C. The observed results are shown below, along with the expected results:
| Strain | P22 Sensitivity at 30�C (+= sensitive) | |
|---|---|---|
| Observed | Expected | |
| matA (CS) | - | - |
| matA (CS)/pMat+ | + | - |
| matB (CS) | - | - |
| matB (CS)/pMat+ | + | - |
| matC (CS) | - | - |
| matC (CS)/pMat+ | + | + |
- What is a backcross and why was a backcross done?
ANSWER: A backcross is strictly speaking a cross betwixt 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" action is located on the plasmid and not due to a spontaneous reversion or second site suppressor in the recipient chromosome.
- Describe the complementation results for the the pMat+ plasmid and explain what these results mean.
Reply: The expectation is that simply matC would be complemented because the three genes tested correspond dicrete loci. So if all iii mutations are complemented, either all three genes are cloned on the plasmid (but that is NOT consistent with mapping information which places the genes in 3 separate locations) or some kind of overexpression suppression is occuring that rescues the phenotype of all 3 mutants. The overexpressed factor may be one of the 3 mat genes or a fourth cistron that nosotros do not know about. Note that it is overexpression on a plasmid producing this result because the mutant strains comprise a single copy of each of these genes in their chromosome and that is inadequate to produce a suppression of the Mat phenotype.
- How could you lot determine if the complementing activity on plasmid pMat+ is specific to S. typhimurium genes of the P22 maturation pathway, or simply provides an activeness which restores office to cold-sensitive factor products in general? Answer: If the activity restores function to cold-sensitive proteins in full general then it should exist able to suppress cold-sensitive mutations in genes that play no role in the phage P22 life cycle, for instance a his (CS) mutant strain could be tested for suppression to restore the ability to make histidine at the non-permissive temperature (30�C) in the presence and the absenteeism 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 forbid the antibiotic from bounden to the ribosome. In such cases, the antibody resistant alllele is often recessive to the antibody sensitive allele. Why?
ANSWER: If the antibiotic stops the ribosomes in their tracks, information technology will prevent subsequent ribosomes from transcribing that message. If cells accept both resistant and sensitive ribosomes, both the resistant and sensitive ribosomes will bind to the mRNA -- however, the sensitive ribosomes will go stuck and block resistant ribosomes from translating the message. Thus, the sensitive phenotype is ascendant to the resistant phenotype.
Briefly compare and dissimilarity complementation vs recombination.
Reply: Complementation = mixing of gene products, changes phenotype non genotype, no breakage/covalent rejoining of DNA. Recombination = changes genotype, requires breakage/covalent rejoining of DNA.
The his operon encodes a grouping of genes required for the biosynthesis of the amino acid histidine. Thus, His- mutants are histidine auxotrophs (unable to abound on minimal medium merely abound 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).
| one | 2 | 3 | 4 | 5 | |
|---|---|---|---|---|---|
| 1 | - | + | + | + | + |
| 2 | - | - | + | + | |
| 3 | - | + | + | ||
| four | - | + | |||
| 5 | - |
- How many complementation groups are there?
Reply: In that location are 4 complementation groups: one = his-1; 2 = his-two and his-3; iii = his-four; 4 = his-5
- How many genes do the v His- mutants impact?
Reply: Four genes.
Complementation analysis was washed on half dozen mutants that lack theonine synthetase activeness. The order of the mutational sites is not known. The results are shown beneath:
| thr-i | thr-2 | thr-3 | thr-4 | thr-five | thr-half dozen | |
|---|---|---|---|---|---|---|
| thr-1 | - | + | + | - | 0 | - |
| thr-ii | - | - | + | + | - | |
| thr-3 | - | + | + | - | ||
| thr-four | - | - | - | |||
| thr-5 | - | - | ||||
| thr-6 | - |
- How many complementation groups are represented?
Answer: At least two complementation groups: Grouping one = thr-1, thr-4, thr-5; Group 2 = thr-2, thr-iii; Ungrouped = thr-6
- Suggest an caption for the results for thr-half-dozen.
Reply: 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 acquired past the mutation (due east.g. due to an amber mutation that prevents expression of downstream genes). In either case, it is impossible to decide whether the thr-6 mutation is inside one of the two complementation groups described by the other mutations or whether information technology 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 tin 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 post-obit table. Based upon the results, indicate which gene the mutation affects and explicate your conclusion.
| Plasmid | Chromosome | Mutant factor | Explanation | |||
| lacZ | lacY | lacA | or genes | |||
| 1 | lac-9001 | + | - | + | lacY | Cannot complement a lacY mutant but tin can complement lacZ and lacA |
| two | lac-9002 | + | + | - | lacA | Cannot complement a lacA mutant but can complement lacZ and lacY |
| 3 | lac-9003 | - | + | + | lacZ | Cannot complement a lacZ mutant simply 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 |
- 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?
Respond: The lac-9005 mutation is known to accept a Lac- phenotype when present on the chromosome. The simplest explanation for this complementation upshot is that the lac-9005 mutation results in overproduction suppression when nowadays on a multicopy plasmid. [An alternative possibility is that the mutation affects some unknown gene that affects some new cistron that results in a Lac- phenotype -- the statement "the lac operon has three genes ..." was the inkling that there really are just three specific, structural genes that are required for lactose catabolism. - How could you examination your hypothesis?
ANSWER: A simple test for overproduction suppression is to do the complementation assay 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.
| ane | 2 | three | 4 | 5 | |
| ane | - | - | - | + | + |
| 2 | - | + | + | + | |
| iii | - | + | + | ||
| 4 | - | + | |||
| 5 | - |
How many complementation groups do these mutants correspond?
Respond: There are at least four complementation groups:
- Mutant 1 and either (2 or 3)
- Mutant 3 (if to a higher place is ane and two) or 2 (if above is ane and iii)
- Mutant four
- Mutant 5
Mutants iv and v are clearly in split up complementation groups. The logic for the first two complementations groups is described below. In improver, mutant 2 can complement mutant 3 suggesting that they are in separate complementation groups. Simply, the surprising result is that mutant 1 cannot complement either mutant ii or mutant 3. The simplest explanation for this result is that mutant ane and mutant 2 or iii are in the same complementation group but mutant 1 has a negative dominant outcome on the other mutant. For example, imagine that factor A and factor B are in an operon -- mutant 1 could be an amber mutation in gene A and mutant iii a missense mutation in gene A -- if mutant 2 was in gene B then mutant 1 would exist phenotypically A- B- while mutant 3 would be A- B+.
Please send comments, suggestions, or questions to smaloy@sciences.sdsu.edu
Last modified October 18, 2004
Source: http://www.sci.sdsu.edu/~smaloy/MicrobialGenetics/problems/complementation/intracellular/
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