BISC 219/2009: Mod 1 Lab 3

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(Complementation)
(Complementation)
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It is not unusual to have series of mutations that confer similar phenotypes and also map to a identical or similar location on a chromosome.  In such cases, the practicing geneticist performs a complementation test to determine if the mutations are allelic (that is, in the same gene) or non-allelic. If the mutations are allelic there should be no complementation whereas you could recover the wild type phenotype (though complementation) if the two mutations are on different genes. The specifics of strain construction vary depending on the experimental organism.  However, the basic strategy in all cases is to construct a double heterozygote and then examine the phenotype of this organism.  As mentioned above, a wild-type phenotype indicates that the two mutations complement one another and are therefore in different genes.  Conversely, a mutant phenotype suggests the mutations are allelic to one another (that is, they fail to complement).  We will construct double hets as follows:<br>
It is not unusual to have series of mutations that confer similar phenotypes and also map to a identical or similar location on a chromosome.  In such cases, the practicing geneticist performs a complementation test to determine if the mutations are allelic (that is, in the same gene) or non-allelic. If the mutations are allelic there should be no complementation whereas you could recover the wild type phenotype (though complementation) if the two mutations are on different genes. The specifics of strain construction vary depending on the experimental organism.  However, the basic strategy in all cases is to construct a double heterozygote and then examine the phenotype of this organism.  As mentioned above, a wild-type phenotype indicates that the two mutations complement one another and are therefore in different genes.  Conversely, a mutant phenotype suggests the mutations are allelic to one another (that is, they fail to complement).  We will construct double hets as follows:<br>
<br>
<br>
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Cross #1:  unknown Dpy (dpy-u/dpy-u) x N2 males (+/+) yields dpy-u/+ progeny<br>
+
Cross #1:  unknown Dpy hermaphodites (dpy-u/dpy-u) x N2 males (+/+) yields dpy-u/+ progeny<br>
<br>
<br>
Cross #2:  dpy-u/+ males  x known dpy hermaphrodites (dpy-k/dpy-k)  yields  dpy-u +/+ dpy-k progeny<br>
Cross #2:  dpy-u/+ males  x known dpy hermaphrodites (dpy-k/dpy-k)  yields  dpy-u +/+ dpy-k progeny<br>

Revision as of 15:31, 4 September 2009

Wellesley College BISC 219 Genetics

Lab 3: Linkage Testing, Backcrossing and Mapping

Backcrossing
Confirm phenotype of the backcrossed strain; specifically, your backcrossed mutant should breed true meaning that it should produce again a homozygote for the mutation: Genotype d/d; phenotype mutant; the crosses are the following (P: d/d x +/+, F1: d/+ self fertilizes to produce F2: +/+, d/+ and d/d). Confirm that you get the expected proportion of homozygote recessive mutants

Linkage Testing
To determine in what chromosome or linkage group your dpy mutation is located record the number of dpy, unc and dpy unc mutants by examining, scoring the phenotype and removing that animal from the plate for each of your five crosses with reference unc mutations. If you see no double mutants (d u/d u), it could be taken as an indication that your dpy mutation is on the same chromosome or linkage group as that unc mutation. Remember that your mutation will only reside on a single chromosome; therefore, the ideal result is that your mutation segregates independently with respect to 4 of the 5 reference mutations. In which linkage group is your dpy mutation located?

Mapping
Assuming that you have determined the linkage group on which your mutation resides, you will continue working with that strain only where you suspect that the two mutations are one the same chromosome. To calculate distance between the two markers separate 5 unc mutants to 5 individual plates. As described above, most of these are wild type for the Dpy mutation (+ u/+ u), but hopefully some are heterozygous for the Dpy mutation (d u/+ u).


3-4 days later
Mapping: screen your 5 plates for double mutants. Pick 3 such putative double mutants to separate plates.

Complementation

It is not unusual to have series of mutations that confer similar phenotypes and also map to a identical or similar location on a chromosome. In such cases, the practicing geneticist performs a complementation test to determine if the mutations are allelic (that is, in the same gene) or non-allelic. If the mutations are allelic there should be no complementation whereas you could recover the wild type phenotype (though complementation) if the two mutations are on different genes. The specifics of strain construction vary depending on the experimental organism. However, the basic strategy in all cases is to construct a double heterozygote and then examine the phenotype of this organism. As mentioned above, a wild-type phenotype indicates that the two mutations complement one another and are therefore in different genes. Conversely, a mutant phenotype suggests the mutations are allelic to one another (that is, they fail to complement). We will construct double hets as follows:

Cross #1: unknown Dpy hermaphodites (dpy-u/dpy-u) x N2 males (+/+) yields dpy-u/+ progeny

Cross #2: dpy-u/+ males x known dpy hermaphrodites (dpy-k/dpy-k) yields dpy-u +/+ dpy-k progeny

The phenotype of the trans double heterozygote is then scored.

In this experiment, you will determine the identity of your “unknown” Dpy mutation and place these mutations in trans to five mutations in different Dpy genes that have been previously mapped (i.e., known mutations). We are attempting to determine if, our “unknown” Dpy mutations are allelic to any of the five mutations.

3-4 days after lab: Cross your unknown Dpy times N2 males by placing three to five L4 Dpy's on a plate in the presence of 5 to 8 N2 males. It is essential that the only wild-type animals present on this plate are males (because you desire +/dpy males from this plate, and not the +/+ males that would result of wild-type hermaphrodites were included on the plate).

Incubate the worms at 23°C until next lab period.

C. elegans General Information
Tools and Techniques
Lab 1: Welcome to C. elegans and Mutant Hunt
Lab 2: Linkage Test and Backcross
Lab 4: Mapping Part 2

Lab 5: Score!
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