From Wikionchus

Pristionchus pacificus has a genetic linkage map with more than 500 SNP markers. With the help of this map you can assign mutants to a chromosome, then to a specific region on that chromosome. Finally you can look for interesting genes in that region and test if they are mutated (single base pair exchanges or deletions). The following section is intended for beginners in genetics; if you are familiar with the procedure, you can read a more technical (also more detailed) protocol here.

Contents 1 Understanding the principle 1.1 Try to answer the following questions: 1.2 Determination of genetic distance: Recombination frequencies 1.3 Genetic linkage map & molecular markers 2 To do 3 To read

Understanding the principle

First recapitulate what you know about Meiosis and good-old Mendel's rules. How are they connected? Then consider a cross between a double mutant (e.g. a dumpy /egl hermaphrodite) with a genotypically wild type male. The F1 from this outcross will be phenotypically wildtype and genotypically heterozygous for both mutants (assuming that both mutations are recessive and different genes are mutated in the dumpy and the egl). There are two possible situations:

1) the two mutants are on different chromosomes

2) they are on the same chromosome

If they are on the same chromosome, there are two extreme possibilities, either they are

a) next to each other or

b) on opposite arms of the chromosome

Of course, they can also be in c) "intermediate" distance.

Try to answer the following questions:

• Assuming a random distribution of the mutations and equal chromosome sizes, how likely is possibility 1) versus 2) ?

( Pristionchus pacificus is a regular diploid organism with 2x6 chromosomes.)

• When the F1 makes its gametes, how likely is it that a particular oocyte (or sperm) inherits

- no mutation,

- only the dpy,

- only the egl,

- both mutations?

• How does this translate to the phenotype frequencies that you see in an F2 resulting from self-fertilization?

For the moment, only consider cases 1) and 2a). The answers are here.

Now assume that you have six different dpy mutants, one on each chromosome (They have the same phenotype but are mutated in different genes). In a mind-experiment, cross your egl mutant to each of the dpys. You get 6 different F1s. Mind-self-fertilize these F1s and count the phenotype frequencies in the six F2 plates.

• One of your F2 plates should be different. Why?

Determination of genetic distance: Recombination frequencies

Let's assume you found a double mutant (egl/egl dpy/dpy) in the case where the dpy is on the same chromosome as your egl. Mind-cross this double mutant with a wild type male(+/+ +/+), isolate oucrossed F1 males(egl/+ dpy/+) and back-cross them to the double mutant.

Let's say you get the following result when counting the offspring of the last cross (egl/egl dpy/dpy X egl/+ dpy/+):

(For simplicity we assume that there are only cross-fertilized and no selfed progeny)

phenotype	total nr.	frequency
wild type	77	48%
dpy & egl	73	46%
dpy, not egl	6	3,8%
egl, not dpy	4	2,5%

• What is the genotype of "wild type" looking animals?

• What happened, and where?

• r, the recombination fraction, can be determined by dividing "recombinant" gametes by the total number of gametes. What is the value of r in this case?

• The 1 million € question: Why can r never be higher than 0.5?

• Let's extend this logic to three mutants: egl, dpy, unc (all recessive). In a similar experiment the following recombination fractions are found:

egl-dpy r=0.15

dpy-unc r=0.2

egl-unc r=0.28

Obviously, the dpy-locus is in between egl and unc. Why, then, is the recombination fraction egl-unc not the sum of the two others, r=0.15 + 0.2 =0.35?

• Answers are here

Genetic linkage map & molecular markers

A genetic linkage map is constructed by sorting loci according to their genetic distance to each other. Genetic distance always refers to recombination frequencies (measured in centiMorgan), whereas physical distance is measured in (kilo/mega) basepairs (bp). A linkage group is a group of markers that are all significantly linked (r <0.5) to at least one other marker of the group. With a dense set of markers the number of linkage groups equals the number of chromosomes.

You can inspect the Pristionchus pacificus linkage map here: [1]

In principle one could build a linkage map entirely with morphological mutants (in fact, the first maps of e.g.Drosophila melanogaster were done that way). However, a much higher resolution can be achieved with molecular markers. In Pristionchus pacificus these are SNPs between the standard laboratory strain PS312California and the "mapping strain" PS1843Washington. For constructing the map the strain PS312 was crossed to PS1843. Genotypes ("maternal(California)", "paternal(Washington)" or "heterozygote") for random loci were determined for 42 F2 animals. This information is stored in the meiotic mapping panel, which then is used to determine genetic distances (this is done by the program mapmaker).

To do

To map a recessive, homozygously viable mutant you should set up a cross with Washington males. Isolate around 10-20 (phenotypically wild type) F1s. In the F2, pick around 50 mutant animals to single plates. When the plate is full with F3/F4 animals, do a preparation of genomic DNA. Do a PCR with the so-called Cocktail markers with the DNA of 21 animals. Determine genotypes by SSCP. For each marker, calculate the ratio of "paternal/Washington" to "maternal/California" markers. If this ratio is significatly smaller than 1/2 (ideally: 0) you found a marker that is probably linked to your mutant (Re-read the first section if this does not make sense to you [or change this text if I wrote non-sense]).

The next step is to refine the position of your mutant. Strategies can differ here, depending on the individual situation. Usually it makes sense to test more F2 animals with the marker that shows the lowest paternal/maternal-ratio. Keep DNA of all animals that are heterozygous at this position (these are your informative animals).

• Why are these animals called informative? Well, you know already that your mutated locus is on that particular chromosome. What you want to know now is, where exactly it is. In your informative animals a recombination happened between the loci of mutation and marker which allows you to infer where the locus is not.

Now you exclude all regions that are heterozygous and test more markers in the vicinity until you find some that stay "maternal" in all animals. Then increase the number of animals, until you find heterozygotes again. The closer you are to your locus, the less likely a recombination event occurs between marker and locus, and the more animals you have to test. Do this until only 1-2 markers remain.

To read

J. Srinivasan et al., A Bacterial Artificial Chromosome-based genetic linkage map of the nematode Pristionchus pacificus Genetics 162: 129-134 (2002)

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