BISC 219/2009: Mod 1 Lab 2

Lab 2: Scoring Autosomal vs. X-linked

Record in your lab notebook the number of WT, Dpy, Unc and Dpy Unc mutants by scoring the phenotype and removing that animal from the plate (flame the pick to remove the worm). If unlinked, you should see WT's, Dpy’s, Unc’s and Dpy Unc’s in a ratio of 9:3:3:1. If linked, you should see greater than 1/7th Dpy Unc’s among the mutant progeny.

You should now be able to conclude which strain is autosomal and linked, autosomal and unlinked and finally which strain has an x-linked gene and which one it is Dpy or Unc.

Lab 2: Mutant Hunt: Gene Mapping

In this exercise, you will progress through the normal sequence of events in forward (classical) genetics. You will first perform a mutant hunt, scanning a plate for rare mutants that occur among the background of wild-type animals. You will then pick your mutants to a separate plate, confirm their mutant phenotype, and begin genetic analyses that culminate with mapping the mutation. In truth, mutants will have been secreted on the plate by your devious instructor. Thus, while rare, they will be more frequently encountered than had you sifted through the second-generation progeny from mutagenized worms (as is the case with a real mutant hunt).

Once you have recovered your mutant and confirmed its phenotype (by examining its progeny) you will next perform two tasks in parallel. One of these is to backcross the mutation into wild type. This consists of crossing your mutant times wild type and recovering homozygous mutants in the F2 generation. Backcrossing is necessary to eliminate any secondary mutations that might modify the phenotypes you are studying. This is important because most mutant hunts employ heavily mutagenized P0’s. Backcrossing will also reveal X-lined mutations, as you recall from the previous exercise.

The second parallel task is linkage testing; you will determine which of the five autosomes (linkage groups) on which your mutation is located. This task is a prerequisite to mapping. It is accomplished by determining the segregation behavior of your unmapped mutation relative to standard reference markers (e.g., mutations whose location is already known). Recall that unlinked mutations will segregate independently (your basic dihybrid inheritance as first observed by Gregor Mendel) whereas linked mutations will not.

In practice, linkage tests are performed using the following steps (where "m" represents your recessive mutant tested with reference marker "r"). The markers m and r must be distinguishable. Since homozygous mutant males usually won't mate, the desired trans double heterozygote is constructed by mating males heterozygous for one mutation (m/+) with hermaphrodites homozygous for the reference mutation (r/r). The genotypes of the F1 hybrids will be +m/r+ and ++/r+. We are only interested in the double heterozygote. The F1 hybrids containing only r are not useful. To select the m/r heterozygotes, we clone 4 or 5 individual F1's on small plates. We score the progeny of the F1 individuals (the F2) for linkage. Only F1 clones which segregate m/m homozygotes are scored. This should be 50% of the plates.

F2 progeny of each class are counted: m, r, mr double, and wild-type. If assortment is independent, progeny will be:

In the latter case, double homozygotes (r m) would occur only in the event of a recombination event. Such an event might occur in either the sperm or the occyte. If the probability of a recombination event is p, and if the event produces wild-type and double mutant recombinant chromosomes, then the probability of getting the double mutant chromosome in an individual gamete is p/2. The chance of an individual gamete which is an m r recombinant combining with another (m r) recombinant is (p/2) x (p/2) = P 2/4. If the map distance between 2 mutations is 10%, then the probability (P) of a recombination’s occurring is P = 0.1. p/2 is 0.5 and P2 is 0.0025. Consequently only about 2 - 3 worms in a thousand will be double mutants if the genes were 10MU apart. That is significantly lower than the 63/1000 one would expect if the genes were not linked. Therefore, the test for linkage is usually the virtual if not complete absence of the double mutant class mr.

To be sure that our results are a significant test for linkage, it is important to count enough worms. There are a number of approaches to this, but the simplest is to ignore the wild-type animals and focus on the mutants. If no double mutants are found, we only need to count about 40 worms before the results become very significant statistically (when tested against the hypothesis that the mutants are unlinked). If the mutations were unlinked, we would expect to see doubles at a frequency of one in seven (remember the 9:3:3:1 ratio). Seventy or more worms have to be scored to distinguish two weakly linked mutations from unlinked mutations.

There are a number of approaches (e.g., 2F crosses, 3F crosses, deletion mapping) to mapping a mutation once its linkage has been established. We will perform a 2F cross, which in C. elegans requires that we first construct a double mutant (r m). As discussed above, the appearance of a double mutant in the F2 requires the union of 2 recombinant gametes -- a very rare event. It is unlikely therefore that we'll find a double among the F2 progeny if the mutants are linked. The probability of a recombination event's having occurred on one of the two homologues is much better; that is, there will be many more progeny that are genotypically mr/+r than mr/mr. It would exhibit the reference mutation's phenotype and would segregate double mutants (r m) as one quarter of its progeny. To find a double mutant for mapping, we choose individuals of the r phenotype to clone and look for the segregation of the mr double mutant among their progeny. If two mutations recombine with a frequency of 10%, roughly one r homozygote in five will segregate the desired double mutant.

Two factor crosses involve scoring recombination between two genetic loci. With your mutation (m) trans to a reference mutation (r) [that is, m+/+r], the only phenotype that can be positively identified as recombinant among the segregants is that of the double mutant. As discussed, the appearance of this phenotype requires that both chromosomes be recombinant. Therefore, if mapping were done with two mutants in trans, one would have to count a ridiculously large number of worms before a statistically reliable map distance could be obtained. Consequently, mapping in C. elegans is done with the two mutations cis. This is the reason for your previous search for a double homozygote.

To map, one crosses a homozygous double mutant hermaphrodite with wild type males and clones the heterozygous F1 hermaphrodites. The F1 heterozygotes are allowed to segregate progeny, and the number of F2 individuals of each different phenotype are then counted. As discussed above, four phenotypes will be observed: wild, single mutant m, single mutant r, and double mr. The map distance (p) can be calculated any number of ways. However, we’ll now avoid a bunch of math and cut to the chase, which is that you will determine map distances using the formula:

In which p equals the map distance and R equals the number of single mutants (both m and r single mutants) divided by the total.

To Do Today Scan “mutagenized” plates for visible mutants and transfer 3 putative mutants to separate plates. Take care not to transfer any other animals except your putative mutant.

Incubate your worms at 23°C for 3 days.

C. elegans General Information
Tools and Techniques
Lab 1: Welcome to C. elegans
Lab 3: Linkage Test and Backcross
Lab 4: Linkage Test, Backcross and Mapping
Lab 5: Mapping Part 2
Lab 6: Score!