BISC 219/F10: Lab 5: Difference between revisions
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[[BISC 219/F10: Lab 6 | Lab 6: Score]]<br> | [[BISC 219/F10: Lab 6 | Lab 6: Score]]<br> | ||
== Lab 5: Mapping == | == Lab 5: Continue Mapping the Mutation == | ||
'''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.<br> | '''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.<br> | ||
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Revision as of 10:13, 14 August 2010
Lab 2: Gene Mapping
Lab 3: Linkage Test Part 1
Lab 4: Linkage Test Part 2 and Mapping
Lab 6: Score
Lab 5: Continue Mapping the Mutation
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 cross plate initiated last week onto three new mating plates (remember these are heterozygous for your dpy 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 3 known dpy reference strains to the mating plates with the males.
Available Dpy strains:
| Chromosome 1 | Chromosome 2 | Chromosome 3 | Chromsome 4 |
|---|---|---|---|
Label your plates, using your Orange Sharpie, with your initials and the date and the cross dpy/+ (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 3 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 and your known mutation under study are in different genes (not allelic): the heterozygote male that you incorporated into Cross #2 will produce two types of gametes: type 1) with a dpy-u mutation and a wild type copy of the dpy-k gene and type 2) with wild type dpy-u gene and wild type dpy-k gene. 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 dpy-u gene and a mutation in dpy-k.
These gametes can be combined so that you will have two possible genotypes. Combining gametes type 1 and 3 you get dpy-u from the male and wild type for the hermaphrodite so the progeny are heterozygous at the dpy-u locus (dpy-u/+) and for the dpy-k locus you get wild type from the male and dpy-k from the hermaphrodite (+/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 the dpy-u locus you get wild type alleles from the male and the hermaphrodite (+/+) and for dpy-k you get wild type from the male and dpy-k from the hermaphrodite (+/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 in different genes (not allelic) will you observe any mutants in the progeny of cross #2?
B) If your unknown mutation (dpy-u) and one of the previously identified mutations under study (dpy-k) are in the same gene (allelic): the heterozygote male that you incorporated into Cross #2 will produce two types of gametes: type 1) with a dpy-u mutation and type 2) with wild type gene. One the other hand, the hermaphrodite you added will produce only one type of gamete we'll call it gamete type 3) with the dpy-k mutation.
These gametes can be combined so that you will have two possible genotypes. Combining gametes type 1 and 3 you get dpy-u from the male and dpy-k from the hermaphrodite so the genotype is (dpy-u/dpy-k). These individuals are phenotypically mutant. Combining gametes 2 and 3 you get wild type from the male and dpy-k from the hermaphrodite so the genotype 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 in the same gene (allelic) will you observe mutants in the progeny of cross #2
