Lab 3: Forward Genetics Project: Linkage Analysis
Now that you have found evidence of a homozygous recessive gene defect that is associated with a phenotype alteration from wild type (true breeding mutant), you will determine on which chromosome this defective gene is located. First, you need to determine whether or not the mutation is x-linked and, if not, on which of the five autosomes (possible linkage groups) it is found. This task is a prerequisite to mapping the mutation (locating where on a particular chromosome the mutation is likely to be found). Linkage testing is accomplished by determining the segregation behavior of your unmapped mutation relative to standard reference markers (e.g., mutations whose locations are already known). Recall that unlinked genes will segregate independently (your basic dihybrid inheritance as first observed by Gregor Mendel) whereas mutations associated with linked genes will not.
In practice, linkage tests are performed using the following steps (where "d" (dpy) represents your recessive mutant tested with reference marker "u" (unc)). The markers d and u must be visually distinguishable. Since homozygous mutant males usually will not mate, the desired double heterozygote is constructed by mating males that are heterozygous for your dpy mutation [wild type for all other genes including the reference mutation (d/+;+/+)] with hermaphrodites that are homozygous for the reference mutation unc (+/+; u/u) and have no dpy mutation. The genotypes of the F1 hybrids will be (+/d;u/+) and (+/+;u/+). We are only interested in the double heterozygote (+/d;u/+). The F1 hybrids containing only u are not useful. To select the (+/d;u/+) heterozygotes, we let 2 F1's self fertilize on two separate plates (two animal on each plate for a total of 4 worms). We score the progeny of the F1 individuals (the F2) for linkage. Only F1 worms which produce d/d homozygotes in the F2 generation are scored, since those are the F1 (+/d;u/+). You should find d/d homozygotes on 50% of the plates. Why?
F2 progeny of each class are counted in the (+/d;u/+) plates: wild-type (+/+;+/+); d (d/d;+/+); u (+/+;u/u) and du double (d/d;u/u). If assortment is independent, progeny will be: 9/16 wild type; 3/16 d, 3/16 u; 1/16 du (that is the 9:3:3:1 ratio)!
On the other hand, if the markers are closely linked double homozygotes (d u/d u) will occur only through a very rare recombination event; therefore, you are not likely to observe the double mutant class.
To Do Today:
1. For linkage testing set up five different crosses. Each cross will contain 3 heterozygous males (d/+; +/+) from the cross you initiated using your mutant Dpy worms. Make sure that these are the only animals that you transfer from that plate by transferring the males to a transfer plate and letting them crawl around for a minute - away from any contaminating worms - then pick a second time to the mating plate.
2. Each heterozygous (d/+; +/+) male will be mated to three L4 hermaphrodites that are homozygous for one of 5 known unc(+/+; u/u) mutations on a mating plate. The strains and their reference mutations are:
3. Label your five plates with your PURPLE Sharpie. With the genotype of the strain - for example: +/+; unc-13/unc-13 (H) X d/+; +/+ (M) with your initials and date.
4. Incubate all of the worms at 23°C for 3 days in your team's worm box.
3-4 days after lab:
- For linkage testing, transfer 1 wild type worm to each of 20 plates. You need 4 cross-progeny L4 stage hermaphrodites (potentially heterozygous for both traits) from each of your 5 crosses.
- Label your 20 (5 sets of quadruplicate) plates with your PURPLE Sharpie. Label each plate with your initials, the genotype of your worms and the date. In each case, why is it important that you transfer L4’s and not adults? What is the genotype and phenotype of your expected F2 progeny?
- Incubate all worms at 23°C until the next lab period.
A full results section on your Series 1 Project is due at the beginning of Lab 4. Instructions found at BISC219/F13: Assignment Help- Data Analysis 1 Autosomal vs. Sex linked inheritance