BISC 219/2009: Mod 1 Lab 4

Lab 4: Mapping and Complementation

Mapping: Cross N2 males X your double mutant (rm). Remember to follow the normal precautions to insure that all progeny are either self progeny (rm mutants) or cis double heterozygous cross progeny (rm/++). Set up duplicate crosses to be sure you get enough worms.

3-4 days after lab:
Pick 2 double heterozygotes to each of two plates for each of your crosses (4 plates total).

Complementation: You will now set up Cross #2: Pick 3-5 L4 and adult males from the plate initiated last week onto five new plates (remember these are heterozygotes for your dpy-u mutation so they are phenotypically wild type but they carry dpy (dpy-u/+). Again, it is essential that only males be transferred onto these plates. Add three L4’s from each of the known dpy strains to your plates.

3-4 days after lab:
Examine each of the five plates initiated last laboratory period. The main question is whether there are any progeny of Dpy phenotype present in the progeny of cross#2

For each plate consider that:

A) If your unknown mutation (dpy-u) and your known mutation under study (one of five) are on different genes (not allelic) we will call the two genes A and B respectively. The heterozygote male that you incorporated into Cross #2 will produce two types of gametes: type 1) with a mutant gene A (carrying the dpy-u mutation) and a wild type gene B and type 2) with wild type gene A and wild type gene B. One the other hand, the hermaphrodite you added will produce only one type of gamete we'll call it gamete type 3) with wild type gene A and mutant gene B (carrying the dpy-k mutation).

These gametes can be combined so that you will have two possible genotypes. Combining gametes type 1 and 3 for gene A you get dpy-u from the male and wild type for the female so gene A is (dyp-u/+) and for gene B you get wild type from the male and dpy-k from the female so gene B is (+/dpy-k). These individuals are phenotypically wild type since they have one wild type allele for each of the genes. Combining gametes 2 and 3 for gene A you get wild type alleles from the male and the female so gene A is (+/+) and for gene B you get wild type from the male and dpy-k from the female so gene B is (+/dpy-k). These individuals are phenotypically wild type since they have at least one wild type allele for each of the genes. In conclusion if your unknown mutation (dpy-u) and your known mutation are on different genes (not allelic) will you observe any mutants in the progeny of cross #2?

B) If your unknown mutation (dpy-u) and your known mutation under study (one of five) are on the same gene (allelic) we will call that gene A. The heterozygote male that you incorporated into Cross #2 will produce two types of gametes: type 1) with a mutant gene A (carrying the dpy-u mutation) and type 2) with wild type gene A. One the other hand, the hermaphrodite you added will produce only one type of gamete we'll call it gamete type 3) with mutant gene A (carrying the dpy-k mutation).

These gametes can be combined so that you will have two possible genotypes. Combining gametes type 1 and 3 for gene A you get dpy-u from the male and dpy-k from the female so gene A is (dpy-u/dpy-k). These individuals are phenotypically mutant. Combining gametes 2 and 3 for gene A you get wild type from the male and dpy-k from the female so gene A is (+/dpy-k). These individuals are phenotypically wild type since they have one wild type allele. In conclusion if your unknown mutation (dpy-u) and your known mutation are on the gene (allelic) will you observe mutants in the progeny of cross #2?

C. elegans General Information
Tools and Techniques
Lab 1: Welcome to C. elegans and Mutant Hunt
Lab 2: Linkage Test and Backcross
Lab 3: Linkage Test, Backcross and Mapping
Lab 5: Score!