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Complementation cloning: what, why and how?


I get that you're cloning a whole cDNA library, but why does E. coli need a complementary gene to allow the gene expression? Why is that complementary gene considered to be a mutant? AND how does one detect the gene of interest after cloning?

What is this kind of cloning used for?


I figured it out. You're complementation cloning when you know the function of a mutant sequence but not it's sequence. For example, your organism (human) gets skin cancer from too much UV light exposure.

With complementation, two mutants can restore the wild type phenotype (well functioning UV repair).

We know E. coli has a well studied UV repair gene so you transform the cDNA library from your human into an E. coli which we know has the mutation, then grow them all under intense UV light. The E. coli with the human mutant allele (the right cDNA fragment) will survive, so you can amplify that colony, sequence it, and then use it as a diagnostic tool in humans.

I'm wondering why that was vague. Is "complementation cloning" not a well used name for that technique? Was the question too broad or elementary? It's the first time I've used Stack and it seems I may be a little too undergrad for this crowd.


A general definition of complementation is the ability of two mutants in combination to restore a normal phenotype. Dominance observed in heterozygotes reflects the ability of wild-type alleles to complement loss-of-function alleles. You know that a dominant allele will determine the phenotype of a heterozygote composed of a dominant and a recessive allele. Often, recessive alleles are loss-of-function mutations, whereas the dominant allele is the wild type, encoding a functional enzyme. Using the example that led to Mendel's First Law, a cross between YY (yellow) peas and yy (green) peas yielded yellow peas in the F1 heterozygote (Yy). In this case the chromosome carrying the Y allele encodes the enzymatic function missing in the product of the recessive y allele, and the pathway for pigment biosynthesis continues on to make a yellow product. Thus you could say that the dominant Y allele complements the recessive y allele - it provides the missing function.

We can continue the analogy to the classic cross for Mendel's Second Law. Let's look at the same genes, but a different arrangement of alleles. Consider a cross between round green (RRyy) and wrinkled yellow (rrYY) peas in this case each parent is providing a dominant allele of one gene and a recessive allele of the other. The F1 heterozygote is round yellow (RrYy), i.e., the phenotypes of the dominant alleles are seen. But you could also describe this situation as the chromosomes from rrYYpeas complementing the deficiency in the RRyy chromosomes, and vice versa. In particular, the Yallele from the rrYYparent provides the function missing in the y allele from the RRyy parent, and the R allele from the RRyy parent provides the function missing in the rallele from the rrYY parent. If the phenotype you are looking for is a round yellow pea, you could conclude that mutants in the R-gene complement mutants in the Y-gene. Since in a heterozygote, the functional allele will provide the activity missing in the mutant allele (if the mutation is a loss-of-function), one could say that dominant alleles complement recessive alleles. Thus dominant alleles determine the phenotype in a heterozygote with both dominant and recessive alleles.

Complementation distinguishes between mutations in the same gene or in different genes

The ability of complementation analysis to determine whether mutations are in the same or different genes is the basis for genetic dissection. In this process, one finds the genes whose products are required in a pathway. In the examples from peas, the metabolic pathway to yellow pigments is distinctly different from the pathway to round peas, which is the starch biosynthesis pathway. Complementation analysis is useful in dissecting the steps in a pathway, starting with many mutants that generate the same phenotype. This is a more conventional example of complementation.

Many fungi can propagate as haploids but can also mate to form diploids prior to sporulation. Thus one can screen for mutants in haploids and obtain recessive mutants, and then test their behavior in combination with other mutants in the diploid state. Let's say that a haploid strain of a fungus was mutagenized and screened for arginine auxotrophs, i.e. mutants that require arginine to grow. Six of the mutants were mated to form all the possible diploid combinations, and tested for the ability of the diploids to grow in the absence of arginine (prototrophy). The results are tabulated below, with a + designating growth in the absence of arginine, and a - designating no growth.

Table 1.1. Growth of the diploids in the absence of arginine

As you would expect, when mutant 1 is mated with itself, the resulting diploid is still an auxotroph this is the same as being homozygous for the defective allele of a gene. But when mutant 1 is mated with mutant 2 (so their chromosomes are combined), the resulting diploid has prototrophy restored, i.e. it can make its own arginine. This is true for allthe progeny. We conclude that mutant 1 will complementmutant 2. If we say that mutant 1 has a mutation in gene 1 of the pathway for arginine biosynthesis, and mutant 2 has a mutation in gene 2 of this pathway, then the diagram in Figure 1.6 describes the situation in the haploids and the diploid. (Note that if the organism has more than one chromosome, then genes 1 and 2 need not be on the same chromosome.) Since the enzymes encoded by genes 1 and 2 are needed for arginine biosynthesis, neither mutant in the haploid state can make arginine. But when these chromosomes are combined in the diploid state, the chromosome from mutant 1 will provide a normal product of gene 2, and the chromosome from mutant 2 will provide a normal product of gene 1. Since each provides what is missing in the other, they complement. Mutant 1 will also complement mutant 3, and one concludes that these strains are carrying mutations in different genes required for arginine biosynthesis.

Figure 1.6. Complementation between two haploid mutants when combined in a diploid.

In contrast, the diploid resulting from mating mutant 1 with mutant 4 is still an auxotroph it will not grow in the absence of arginine. Assuming that both these mutants are recessive (i.e. contain loss-of-function alleles), then we conclude that the mutations are in the same gene (gene 1 in the above diagram). We place these mutants in the same complementation group. Likewise, mutants 2 and 3 fail to complement, and they are in the same complementation group. Thus mutant 2 and mutant 3 are carrying different mutant alleles of the same gene (gene 2).

Mutant 5 will complement all the other mutants, so it is in a different gene, and the same is true for mutant 6. Thus this mutation and complementation analysis shows that this fungus has at least 4 genes involved in arginine biosynthesis: gene 1 (defined by mutants alleles in strains 1 and 4), gene 2 (defined by mutants alleles in strains 2 and 3), and two other genes, one mutated in strain 5 and the other mutated in strain 6.

Genetic dissection by complementation is very powerful. An investigator can start with a large number of mutants, all of which have the same phenotype, and then group them into sets of mutant alleles of different genes. Groups of mutations that do not complement each other constitute a complementation group, which is equivalent to a gene. Each mutation in a given complementation group is a mutant allele of the gene. The product of each gene, whether a polypeptide or RNA, is needed for the cellular function that, when altered, generates the phenotype that was the basis for the initial screen. The number of different complementation groups, or genes, gives an approximation of the number of polypeptides or RNA molecules utilized in generating the cellular function.

Question 1.2. Consider the following complementation analysis. Five mutations in a biosynthetic pathway (producing auxotrophs in a haploid state) were placed pairwise in a cell in trans(diploid analysis). The diploid cells were then assayed for reconstitution of the biosynthetic pathway complementing mutations were able to grow in the absence of the end product of the pathway (i.e. they now had a functional biosynthetic pathway). A + indicates a complementing pair of mutations a - means that the two mutations did not complement.

a) Which mutations are in the same complementation group (representing mutant alleles of the same gene)?

b) What is the minimal number of enzymatic steps in the biosynthetic pathway?


Complementation cloning: what, why and how? - Biology

Example: Two white-flowered plants cross to produce purple flowers , although purple is dominant.
Each contains a mutation in a different gene, encoding a different enzyme needed to make the purple pigment.

Complementation analysis is easiest to do in bacteria, fungi, or C. elegans, where many mutants of a given phenotype can be obtained.

If we isolate a large number of strains with the same defective phenotype,we can cross them in all combinations, and figure out the number of complementation groups. Any two defective strains that FAIL to complement are in the same complementation group. Usually each complementation group represents one of the essential enzymes in the pathway.

Problem 1: Figure out how many complementation groups there are in these examples.

The concept of complementation is extremely important in molecular biology. For example, the sickle-cell mouse line could only be created because two strains with different defects (lack of mouse or human globin genes) could be mated to complement each other's defects. The fact that genes from different species can complement each other was one of the most significant conceptual advances in molecular biology. Complementation is now used routinely to answer more subtle questions of how genes are regulated. You absolutely need to understand complementation to understand molecular biology.

Bacterial complementation.
Bacterial genetic systems can show complementation in two important ways--each manipulating a natural process of bacterial genetics. These two processes have since been modified in biotechnology to provide most of the essential tools of gene cloning.

1. Specialized Transduction. A lysogenic bacteriophage can excise itself so as to carry a piece of host DNA by mistake. The phage will now carry a second copy of an allele (or linked alleles) into a host cell. The new bacterium is a partial diploid for the allele(s).

In biotechnology, a phage chromosome can have a piece of foreign DNA ligated into it in the test tube. Then the phage DNA is packaged into phage, and it can infect a new host where it either (1) produces many copies of the host gene or (2) lysogenizes the host, to express the cloned DNA.

2. F' plasmid. The F plasmid can recombine itself into the host chromosome, then recombine itself out again with some host DNA by mistake. When it enters the next host cell, it carries a second copy of several genes again, a partial diploid is created.

In biotechnology, a plasmid can have a piece of foreign DNA ligated into it in the test tube then the plasmid is transformed into E. coli. Then the plasmid makes many copies, including the cloned gene.

Suppressor mutation analysis.
A variation on complementation is suppressor mutations. A suppressor mutation corrects a defect in a different gene locus. A mutant version of gene A makes an altered gene product, which corrects the phenotype of a defective mutation in gene B.

Organization of Genes
In bacteria, a number of gene ORFs can be organized into an operon . All the gene sequences in a given operon are transcribed on a single mRNA, starting at one promoter . An example of an operon is shown, tuf-s10 , from Borrelia burgdorferi, which causes Lyme Disease:

  • elongation factor (tuf)
  • ribosomal proteins S10 (rpsJ)
  • L3 (rplC)
  • L4 (rplD)
  • L23 (rplW)
  • L2 (rplB)
  • S19 (rpsS)
  • L22 (rplV)
  • S3 (rpsC)

Human Growth Hormone Receptor

For other interesting operons, try searching GenBank, the international repository for all known DNA sequences. (Funded by the U.S. government--your tax dollars at work.)

  • DNA sequence--inversion or deletion
  • Transcription, sigma factor
  • Transcription, promoter sequence, repressor proteins
  • Translational repressor
  • Posttranslational modification

The Lac Operon
The Lac operon is the classic model for activation and repression of transcription. Concepts of analysis based on the Lac operon can be applied to other systems including animals and plants.

The following explanation of the Lac operon is modified from MIT Lac Operon.
Jacob and Monod were the first scientists to elucidate a transcriptionally regulated system. They worked on the lactose metabolism system in E. coli. When the bacterium is in an environment that contains lactose:

It should turn on the enzymes that are required for lactose degradation. These enzymes are: beta-galactosidase: This enzyme hydrolyzes the bond between the two sugars, glucose and galactose. It is coded for by the gene LacZ. Lactose Permease: This enzyme spans the cell membrane and brings lactose into the cell from the outside environment. The membrane is otherwise essentially impermeable to lactose. It is coded for by the gene LacY. Thiogalactoside transacetylase: The function of this enzyme is not known. It is coded for by the gene LacA. The sequences encoding these enzymes are located sequentially on the E. coli genome. They are preceded by the LacI region which regulates expression of the lactose metabolic genes. You might expect that the cell would want to turn these genes on when there is lactose around and off when lactose is absent. But the story is more complicated than that. For instance, the permease gene always needs to be expressed at a low level, in order for any lactose to get into the cell. So a certain low level of expression is constitutive --that is, occurs all the time, even if "repressed." Most bacterial operons are partly or totally constitutive. LacI expression, for example, is totally constitutive its promoter is always "turned on," for a very low level of expression, just enough to make a few repressor molecules.

A bacterium's prime source of food is glucose, since it does not have to be modified to enter the repiratory pathway. So if both glucose and lactose are around, the bacterium wants to turn off lactose metabolism in favour of glucose metabolism. There are regulatory sites upstream of the Lac genes that respond to glucose concentration.

  • Lactose induces transcription by pulling the LacI repressor off.
  • Glucose prevents transcription by pulling the CAP activator off.

When lactose is present, it acts as an inducer of the operon. It enters the cell, rearranges slightly to form allolactose, then binds to the Lac repressor. A conformational change causes the repressor to fall off the DNA. Now the RNA polymerase is free to move along the DNA, and RNA can be made from the three structural genes . The mRNA will be translated to the proteins which transport and metabolize lactose.

When the inducer (lactose) is removed, the repressor returns to its original conformation and binds to the DNA, so that RNA polymerase can no longer get past the promoter. No RNA and no protein is made.

Note that RNA polymerase can still bind to the promoter though it is unable to move past it. That means that when the cell is ready to use the operon, RNA polymerase is already there and waiting to begin transcription the promoter doesn't have to wait for the holoenzyme to bind. Catabolite Repression, with an Activator Protein
When levels of glucose (a catabolite) in the cell are high, a molecule called cyclic AMP is inhibited from forming. But when glucose levels drop, ATP phosphates are released until at last forming cAMP:

ATP --> ADP + Pi --> AMP + Pi --> cAMP

cAMP binds to a protein called CAP (catabolite activator protein), which is then activated to bind to the CAP binding site. This activates transcription, perhaps by increasing the affinity of the site for RNA polymerase. This phenomenon is called catabolite repression , a misnomer since it involves an activator protein , but understandable since it seemed that the presence of glucose repressed all the other sugar metabolism operons.

This image shows a "close-up" view of CAP regulation:

Corepressor control
Other operons are controlled by their products, rather than their substrates for example, expression of biosynthetic enzymes to build amino acids. This is called feedback inhibition. In the Trp operon, for tryptophan biosynthesis, transcription of mRNA for five enzymes is prevented by binding of the Trp corepressor in the presence of tryptophan. When tryptophan levels fall, Trp comes off of the corepressor, and the corepressor comes off of the promoter/operator site. Transcription now occurs, so that the cell has enzymes to make more tryptophan.
Analysis of operon control
What experiments do we perform to figure out how operons are regulated?
We use partial diploid strains created by F' or specialized transduction. In either case, we test what happens when a strain is diploid for regulatory elements.

Regulatory mutants can have various kinds of mutant phenotypes. For example:

p- Promoter fails to bind RNA polymerase. No transcription occurs.
lacI- Repressor fails to bind promoter/operator. Transcription occurs constitutively
(in the presence or absence of lactose)
o-c Operator fails to bind repressor. Transcription is constitutive.
lacZ- Structural gene is defective. No enzyme is made.

What will happen? What kinds of complementation can occur?Does is matter if the two mutant alleles are adjacent on the same chromosome ( cis ) or separated ( trans )?


"Wild type" Mutant
LacZ+ Makes B-gal enzyme LacZ- Make NO enzyme
LacI+ Makes Repressor LacI- Makes NO repressor
Transcription can be constitutive
IF PROMOTER IS FUNCTIONAL
p + RNA Pol binds promoter p - RNA Pol does NOT bind promoter
No Transcription
o + Operator binds repressor o - c Operator does NOT bind repressor
Transcription can be constitutive
IF PROMOTER IS FUNCTIONAL

Problem 2. Predict whether the following diploids produce B-galactosidase, in the presence of lactose in the absence of lactose. Explain why. Explain in each case whether it matters if the two mutant alleles are located in cis or in trans.

p + lacZ - lacI +
----------------- ----------
p + lacZ + lacI -

p + lacZ - lacI -
----------------- ----------
p - lacZ + lacI -

p + lacZ - lacI +
----------------- ----------
p + lacZ + lacI -

p + o-c lacZ + lacI + (A constitutive operator NEVER binds repressor,
-------------------------- -------- with or without lactose.)
p + o + lacZ - lacI +

Problem 3. Explain two different genetic processes in bacteria that can create a "partial diploid" for a small part of the genome. Explain why these processes are useful for bacterial genetic analysis.

Problem 4. State whether B-galactosidase is expressed by each lac operon diploid, (1) and (2), and briefly state why (one sentence). Complete possible genotypes for (3) and (4).


Lactose
Absent
Lactose
Present
LacI- P+ O-c LacZ+
LacI+ P+ O+ LacZ-
LacI+ P- O-c LacZ+
LacI+ P+ O+ LacZ-
LacI- P+ O+ ___
____ ___ __ LacZ-
- +
__ P- O-c LacZ+
__ P+ __ ___
+ +

Quiz on Lac Operon -- Highly Recommended

Molecular Structure of Promoters
Promoters are defined by sequences of base pairs upstream of the transcription start site. The RNA polymerase tends to recognize promoter sequences in which most of the base pairs match the promoter consensus sequence. The consensus sequence is a composite defined by the most common base to occur at each position. Base-substitution mutations can decrease or increase the efficiency of the promoter.

Bacterial consensus promoters include two regions of six base pairs each, at -10 and -35 bases upstream. However, no two promoters are exactly alike, and no promoter exactly matches the consensus sequence. Additional sites for environmental regulators can be found as far as -50 to -300 bases upstream.

  • When bacteria use up their carbon sources, they express RpoS, the starvation sigma factor . (Review, what is a sigma factor?)
  • RpoS joins RNA polymerase to initiate transcription of different environmental stress genes--genes protecting against all the different stresses that the bacteria might encounter before they enter a new human intestine. This phenomenon is known as cross-protection.
  • The stress genes can be used for conditions as unrelated as acid or base resistance.
  • The stress genes activated may or may not be part of multi-gene operons. They may face in opposite directions, from many different promoters, at all different loci around the genomic map.
  • Arsenic Resistance Operons. How do bacteria resist arsenic? An environmental response regulon is turned on by arsenic. The molecular basis is related to how cancer cells develop resistance to anti-cancer drugs.
  • Environmental regulation in Yersinia pestis (bubonic plague bacteria). Several complex regulons of genes respond to specific environmental factors, particularly iron and temperature. At low temperature, the presence of iron tells the bacterium, "I am in blood that has been swallowed by a flea." The bacterium expresses proteins that upset the flea's digestion, forcing it to regurgitate the bacteria into the blood of its next victim. (Susan Straley & Robert Perry, Trends in Microbiol., 1995)
  • E. coli virulence regulator. How do virulent E. coli strains kill children? A regulator protein binds to an operon encoding "pilins," for E. coli to make pili which attach to the intestinal epithelium.
  • Tuberculosis model gene expression. How are genes regulated in a tuberculosis-related pathogen?

    M. Donnenberg, U. Maryland
  • Virulence regulator in "Flesh-eating bacteria." A gene activator protein in Staphylococcus aureus turns on the virulence regulon that makes the flesh-eating toxins. This activator can be used as a vaccine against S. aureus.
  • DNA Microarrays. We can now put most of the protein-encoding genes onto a microarray chip, using technology based on the DNA silicon chip industry. The chip can be used to hybridize to cellular RNA, and measure the expression rates of a large number of genes in a cell.

Researchers routinely use cloning techniques to make copies of genes that they wish to study. The procedure consists of inserting a gene from one organism, often referred to as "foreign DNA," into the genetic material of a carrier called a vector. Examples of vectors include bacteria, yeast cells, viruses or plasmids, which are small DNA circles carried by bacteria. After the gene is inserted, the vector is placed in laboratory conditions that prompt it to multiply, resulting in the gene being copied many times over.

Researchers routinely use cloning techniques to make copies of genes that they wish to study. The procedure consists of inserting a gene from one organism, often referred to as "foreign DNA," into the genetic material of a carrier called a vector. Examples of vectors include bacteria, yeast cells, viruses or plasmids, which are small DNA circles carried by bacteria. After the gene is inserted, the vector is placed in laboratory conditions that prompt it to multiply, resulting in the gene being copied many times over.


This chapter was completed after valuable feedback from members of the Hodgkin Lab, D. Williams, M. Best, K. L. Chow and an anonymous reviewer, all to whom I am very appreciative. I am indebted to O. Hobert, K. L. Chow and J. Kimble for sharing unpublished data. KJY was supported by a Ruth L. Kirchstein NRSA (F32A1050333) from NIH.

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* Edited by Jonathan Hodgkin and Philip Anderson. Last revised October 05, 2005. Published October 06, 2005. This chapter should be cited as: Yook, K. Complementation (October 06, 2005), WormBook , ed. The C. elegans Research Community, WormBook, doi/10.1895/wormbook.1.24.1, http://www.wormbook.org.

Copyright: © 2005 Karen Yook. This is an open-access article distributed under the terms of the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original author and source are credited.

§ To whom correspondence should be addressed. E-mail: [email protected]

All WormBook content, except where otherwise noted, is licensed under a Creative Commons Attribution License.


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Cloning Yeast Genes by Complementation

This unit presents a generalized protocol and describes the principles involved in cloning yeast genes by complementation in yeast. The protocol is presented using a hypothetical mutation of yeast, the cdc101-1 mutation. This mutation was isolated as a cell cycle mutant and is both recessive and temperature-sensitive for growth: it can grow relatively normally at 30 o C but is unable to make a colony at 37 o C. A genomic DNA clone that complements this mutation will be isolated by transforming the cdc101-1 strain with a yeast genomic library and subsequently screening for temperature-resistant colonies. Once isolated, two steps are necessary to prove that the insert present on the plasmid contains the wild-type CDC101 gene. First, segregation of the complementing plasmid must result in co-loss of both the plasmid-borne selectable marker and the complementing phenotype, demonstrating that the observed complementation is plasmid-specific and is not due to reversion of the cdc101-1 mutation. Second, it must be ruled out whether the cloned gene encodes a phenotypic suppressor of the mutation, rather than the wild-type gene. This is done via a complementation test, which demonstrates whether or not a disruption of the cloned gene that is integrated into the genome can complement the original mutation.


Alpha Complementation method for Bacterial Screening

15 comments:

whats the role of alpha and omega fragment in operon system

Alpha and omega fragments complement each other and are necessarilu required for function

They play a key role in the production of beta-galactosidase protein as alpha and omega fragments become functional after interaction and then beta-galactosidase can be produced.

the cells which recived the recombinant plasmid appears white or blue?? and why? because i am confused
the cells mutant in alphe frarment will appeare white .but if these cells got the r plasmid they will appeare blue.
this is right or no?
but my be we have to insert the gene in lac z portion of the plasmid so both the cells(mutant and recombinant )will appeare white as the cant produse galactosidase.
thank u

according to what i understood the cells with recombinant plasmid will appear white


Tips for blue-white screening

  • Use a good control: Transform the backbone plasmid without insert. All colonies on this plate should be blue, indicating that your IPTG and x-gal are working as they should be.
  • Don't rush the process: It is important to give your plates enough time for any intact β-galactosidase to be expressed and process x-gal into blue pigment (16-20 hours). A plate with only white colonies is very suspicious!
  • Refrigerate your plates: Placing plates at 4C for few hours after the initial overnight incubation increases pigment precipitation, enhances the blue color of negative colonies, and allows for better differentiation between blue and white colonies.
  • Take care in making your plates: X-gal is light and temperature sensitive and needs to be added to media after autoclaving. If spread on top of pre-made plates, make sure it is evenly distributed and allow sufficient drying time before use.
  • Beware of false positives: Blue-white screening only indicates the presence of AN insert, not necessarily YOUR insert. Any cloning artifact that disrupts the α-peptide DNA will also lead to a white colony.
  • And also false negatives: These are rare, but if a small fragment is inserted in-frame, read-through can lead to a functional β-galactosidase enzyme and a blue colony. Blue-white screening is a good way to narrow down candidates for more specific analysis, like PCR or restriction digest.
  • Make sure you use a proper E. coli strain (i.e. contains the lacZΔM15 mutation): XL1-Blue, DH5α, DH10B, JM109, STBL4, JM110, and Top10 are a few examples.
  • Make sure you use a proper plasmid (i.e. contains the α-complementation cloning MCS): pGEM-T, pUC18 and pUC19, and pBluescript are a few common vectors.

Blue-white screening is just that – a screening process. It does not select only those cells that have taken up a plasmid and thus should be used in conjunction with selection methods. Combining selection and screening ensures that the white colonies you see are white due to successful cloning and not because the cell failed to take up the α-complementation plasmid. This way you can quickly and easily identify colonies that not only have your plasmid (antibiotic resistance), but also confirm those plasmids have your insert (blue-white screening).


Abstract

Chlamydomonas reinhardtii is an excellent model system for plant biologists because of its ease of manipulation, facile genetics, and the ability to transform the nuclear, chloroplast, and mitochondrial genomes. Numerous forward genetics studies have been performed in Chlamydomonas, in many cases to elucidate the regulation of photosynthesis. One of the resultant challenges is moving from mutant phenotype to the gene mutation causing that phenotype. To date, complementation has been the primary method for gene cloning, but this is impractical in several situations, for example, when the complemented strain cannot be readily selected or in the case of recessive suppressors that restore photosynthesis. New tools, including a molecular map consisting of 506 markers and an 8X-draft nuclear genome sequence, are now available, making map-based cloning increasingly feasible. Here we discuss advances in map-based cloning developed using the strains mcd4 and mcd5, which carry recessive nuclear suppressors restoring photosynthesis to chloroplast mutants. Tools that have not been previously applied to Chlamydomonas, such as bulked segregant analysis and marker duplexing, are being implemented to increase the speed at which one can go from mutant phenotype to gene. In addition to assessing and applying current resources, we outline anticipated future developments in map-based cloning in the context of the newly extended Chlamydomonas genome initiative.

Sometimes called green yeast ( Goodenough, 1992), the unicellular, eukaryotic green alga Chlamydomonas reinhardtii (hereafter called Chlamydomonas) is a venerable model system for plant biology as well as for cell motility. The tag green yeast refers to its haploid vegetative state, the existence of two mating types, and the general similarity in applicable genetic techniques. These aspects of Chlamydomonas biology have been previously reviewed ( Rochaix, 1995).

Like many microorganisms, screening of Chlamydomonas strains for rare mutations is straightforward, since large numbers of cells can be plated on an appropriate selective medium, or nonswimmers, for example, can be selected from large numbers of swimming cells. At the same time, the ease of nuclear transformation in Chlamydomonas, coupled with the plant-like nonhomologous integration of transforming DNA, facilitates the creation of insertional mutant collections. Taken together, the assortment of techniques useable for Chlamydomonas indulges both the amateur and experienced geneticist, yielding sometimes overwhelming collections of mutant strains. In this report, we focus on mutants affecting photosynthesis, in keeping with the thrust of this journal, and the emphasis of the newly renewed and National Science Foundation-supported Chlamydomonas genome project (http://www.chlamy.org/). However, the map-based cloning tools described here are generally applicable to Chlamydomonas biology.

The use of Chlamydomonas to study the elaboration and regulation of the photosynthetic apparatus is long established and was recently reviewed ( Grossman, 2000 Dent et al., 2001). Key to this is the ability to maintain nonphotosynthetic (PS−) mutants on acetate-containing media, as well as the ability to use replica plating ±acetate and/or chlorophyll fluorescence to identify such mutants ( Bennoun and Béal, 1997 Niyogi et al., 1997). Furthermore, numerous photosynthetic (PS+) suppressors have been recovered from screening of PS− strains (e.g. Girard-Bascou et al., 1992 Levy et al., 1997 Bernd and Kohorn, 1998 Nickelsen, 2000 Esposito et al., 2001 Li et al., 2002).

Recovery of wild-type alleles of genes mutated in PS− strains has been successful, since both genomic complementation with selection on minimal medium or identification of DNA flanking an insertional mutant site are relatively straightforward although perhaps tedious methods (e.g. Gumpel et al., 1995 Boudreau et al., 2000 Vaistij et al., 2000 Auchincloss et al., 2002 Dauvillee et al., 2003). However, in cases where PS+ suppressors have been recovered from PS− strains or where a trait otherwise cannot be selected on minimal medium, recovery of the mutation requires other methods. In the case of PS+ suppressors, for example, a recessive suppressor would yield a PS− phenotype upon complementation, and even dominant suppressors such as mcd2, which suppresses the nuclear mcd1-2 mutation responsible for instability of the chloroplast petD mRNA ( Esposito et al., 2001), require construction of a genomic library from the suppressed strain if they are to be cloned by complementation. Another example is the xanthophyll cycle mutant npq1, which is defective in nonphotochemical quenching ( Niyogi et al., 1997). Although npq1 was generated in an insertional mutant population, its defect is not linked to the ARG7 insertional mutagen. In each of these cases, isolation of the gene of interest could be achieved through map-based cloning in a suitably developed system.

Map-based cloning relies on two basic principles, namely the existence of a genetic/physical map and the ability to generate progeny of sexual crosses that segregate for the trait of interest as well as phenotypic and/or molecular markers. In higher plants, such resources are most fully advanced in Arabidopsis (Arabidopsis thaliana) and rice (Oryza sativa), which, not coincidentally, have complete nuclear genome sequences. Furthermore, interfertile polymorphic ecotypes (Columbia and Landsberg erecta and indica and japonica, respectively) have been utilized as sources of genetic variation to introduce into selected mutant backgrounds. Resources for Arabidopsis and rice have been extensively publicized and have been utilized in numerous examples of successful gene isolation ( Chen et al., 2002 Garcia-Hernandez et al., 2002 Torjek et al., 2003) maize (Zea mays) is another subject of intensive efforts ( Coe et al., 2002 Cone et al., 2002).

Here we present a case-based study to describe existing and projected resources for map-based cloning in C. reinhardtii. Mutations generated in this species can be mapped by crossing to the interfertile strain known as Chlamydomonas grossii, S1-D2 or its culture collection designation of CC-2290, which has a suitable profusion of sequence tagged sites (STS), cleavable amplified polymorphic sequence (CAPS), single nucleotide polymorphism (SNP), and RFLP markers ( Gross et al., 1988 Vysotskaia et al., 2001 Grossman et al., 2003). Beginning with laborious RFLP-based mapping ( Gross et al., 1988), Chlamydomonas mapping has moved toward a PCR-based method ( Kathir et al., 2003) and now is poised to incorporate more high-throughput methods. This, in concert with an increasingly complete nuclear genome sequence ( Grossman et al., 2003), provides the necessary tools for studies of all classes of mutations.

The nuclear mutants mcd4 and mcd5 were derived from strains petDLS2 and petDLS6, respectively, in which mutations engineered into the 5′-untranslated region of the chloroplast petD gene caused RNA instability and thus a PS− phenotype ( Higgs et al., 1999). Both mcd4 and mcd5 are spontaneous PS+ mutants, do not carry a molecular tag, and genetic analysis (data not shown) showed them to be recessive. Thus, complementation of mcd4 or mcd5 with the wild-type genes would revert the PS+ phenotype to PS−, making genomic complementation an inappropriate approach. We therefore decided to map mcd4 and mcd5 by using available genomic resources and by developing new ones as opportunities or needs arose.


Alpha Complementation of LacZ in Mammalian Cells

The bacterial lactose converting enzyme beta-galactosidase (β-gal) and its gene (LacZ) have been studied for many years ( 1 and references therein) and are among the most utilized tools in molecular biology. The LacZ product, a polypeptide of 1029 amino acids, gives rise to the functional enzyme after tetramerization ( 2 and references therein) and is easily detected by chromogenic substrates either in cell lysates or directly on fixed cells in situ ( 3 and references therein). The tetramerization is dependent on the presence of the N-terminal region spanning the first 50 residues ( 2 and references therein). Deletions in the N-terminal sequence generate a so-called omega peptide that is unable to tetramerize and does not display enzymatic activity. The activity of the omega peptide can be fully restored either in bacteria or in vitro ( 4) if a small fragment (called alpha peptide) corresponding to the intact N-terminal portion of the β-gal is added in trans. The phenomenon is called alpha complementation and the small N-terminal peptide is called alpha peptide. This effect has been widely exploited for studies in procaryotes, where special strains that constitutively express omega peptide exist and allow the detection of expression of the small alpha peptide.

Aiming to study the stability of microsatellites in ageing eucaryotic cells, we have engineered (CA)n repeats at the N-terminal position of LacZ ( Fig. 1A). However, we were disappointed by the rapid loss of activity caused by prolonged repeats (see below). We reasoned that, analogously to procaryotic systems, the alpha peptide might be more stable towards such insertions. Surprisingly we could not find any report describing alpha complementation of LacZ in mammalian cells. Therefore, we constructed eucaryotic expression vectors that should produce either a lacZ omega peptide [ Fig. 1A, construct g, called Z(d)NC] or different alpha peptides (constructs b–d, called Z-N58, Z-N85 and Z-N150, respectively). We have also compared the tolerance towards N-terminal inserts for the whole enzyme [construct f, Z( 1)NC] or for an alpha peptide [construct e, Z(i)N58]. Transfection of these constructs in various combinations has produced the results shown in Figure 1B, where the length of the bars represents the relative enzymatic activity obtained from extracts of transiently expressing HeLa cells. The data clearly show that co-expression of alpha and omega peptides (lines 4–6 and 8–12) resuscitates enzymatic activity that is absent in the single components (see control lines 2 and 3). We find that the optimum length of the alpha complementing peptide is ∼85 amino acids (line 5), while a shorter peptide (truncated at position 58) that is otherwise active in most prokaryotic systems (see Fig. 1 legend) demonstrates only a very weak alpha complementing activity (line 4). We could also demonstrate that alpha peptides are more tolerant towards insertions of foreign sequences (compare lines 8–12 with lines 13–15). The tolerance was tested with the relatively weak N58 alpha complementing peptide and we expect the N85 construct to be even more tolerant. We are looking forward to using similar peptides to monitor stability of dinucleotide repeats in mammalian cells. Recently we learned that expression of Z-N85 and Z(d)NC in yeast cells does also result in efficient trans complementation (D. Picard, personal communication). We observed the same complementation behaviour in other human cell lines (S. B. Verca, unpublished). Therefore, we are confident that these constructs can be used in a variety of eucaryotic cells.

The availability of this approach opens new perspectives in gene expression studies in cell cultures and animals, such as (i) easy monitoring of various fusion proteins in eucaryotic systems (ii) facilitated packaging of small reporter peptides in size-constrained viral vectors (iii) establishment of dual expression monitoring systems in which omega and alpha peptides are brought under distinct genetic control (iv) study of protein trafficking, where only co-compartmentalization shall produce significant complementation. Thus, we believe that this system will become the basis of many experiments and applications in eucaryotic systems.


Watch the video: The complementation test (January 2022).