BISC 219/2009: Mod 1 Lab 4

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Wellesley College BISC 219 Genetics

Lab 4: Mapping and Complementation

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

Label both plates with your initials, date and the cross set up + +/+ + (M) X d u/d u (H) with your PURPLE Sharpie.

Incubate your worms at 23°C for 3 days

3 days after lab:
To allow us to FINALLY determine the map distance between our two mutations, by determining the number of recombinant gametes which is related to the number of recombination events between the two genes, pick 3-4 wild type males (double heterozygous-d u/+ +) to each of two new mating plates and mate them to L4 hermaphrodite double mutants (homozygotes d u/d u) - this is called a test cross.

Label both plates with your initials, date and the cross d u/+ + males X d u/d u (H) with your PURPLE Sharpie.

Incubate your worms at 23°C for until next lab


Complementation: You will now set up Cross #2: Pick 3-5 males from the plate initiated last week onto four new mating plates (remember these are heterozygous for your dpy-MB? mutation so they are phenotypically wild type but they carry dpy (dpy-MB?/+). Again, it is essential that only males be transferred onto these plates. Add three L4’s from each of the 4 known dpy reference strains (dyp-k1-4)to the mating plates with the males.

Label your plates, using your Orange Sharpie, with your initials and the date and the cross dpy-MB?/+ (M) X dpy-k (replacing k with the gene number of each of the reference strains)/dpy-k. Incubate the worms at 23°C until next lab period.

3 days after lab:
Examine each of the four plates initiated last laboratory period. The main question is whether there are any MALE progeny of Dpy phenotype present in the progeny of cross #2
WHY male? Think about the hallmark of a sucessful cross! What else can hermaphrodites do other than mate with males - might be why you have dpys on more than one plate.

For each plate consider that:

A) If your unknown mutation (dpy-MB?) and your known mutation under study (one of five) are in different genes (not allelic) we will call the two genes responsible for the mutations A and B, respectively. The heterozygote male that you incorporated into Cross #2 will produce two types of gametes: type 1) with a defective gene A (carrying the dpy-MB? 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-MB? from the male and wild type for the female so gene A is (dyp-MB?/+) 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-MB?) 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-MB?) and your known mutation under study (one of four) 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-MB? from the male and dpy-k from the female so gene A is (dpy-MB?/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-MB?) and your known mutation are on the same gene (allelic) will you observe mutants in the progeny of cross #2?

In Lab 5 you will complete the Gene Mapping Series and move on to begin work in Series 2 - RNA interference. Please read all of the background information and protocols in the RNAi section of the wiki before coming to lab next time.

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!
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