Download Forward-Screen-and-Backcrossing-mini-lecture

Forward Screen and Backcrossing: C. elegans
Designing a genetic screen allows you to answer questions about certain biological
processes. In an ideal situation, you can identify mutations with strong, observable
phenotypes that affect the biological process you are interested in. In this lesson I
am going to walk you through a scenario in which you perform a forward screen,
starting with a strong, observable phenotype affecting a certain biological process.
I will then go on to show you what you should do once you have identified a mutant
strain to be able to assure that you are looking at the phenotype of one mutant locus.
How do you perform a forward screen with C. elegans?
Let’s walk through the scenario of a forward screen to identify mutants that, instead
of gliding in a sinusoidal movement, continuously roll in place.

Mutagenize hermaphrodites.
In case they ask why hermaphrodites: mutagenizing hermaphrodite P0’s,
you can go on to single them in the next F1 generation and allow them to self,
therefore being able to observe homozygous mutations in the F2 progeny.

Allow the hermaphrodites to self-fertilize.
o In the F1 progeny of this cross, if a worm has a rolling phenotype,
it is likely a dominant mutation.
Why? Because you observe the mutant phenotype even though there
is a wild type copy of the gene.

Separate out the F1 progeny and let them grow up and self-fertilize on
separate plates.
o In the F2 progeny of this cross, if a worm has a rolling phenotype,
it is likely a recessive mutation.
Why? You observe the mutant phenotype only when there is no wild
type copy of the gene.
In the F2 progeny of this third cross, one worm has a weak rolling phenotype.
You separate out this worm and let it self-fertilize.
o In this F3 progeny, all worms have a strong rolling phenotype.
What explains both the weak rolling phenotype of the F2 parent
and the strong rolling phenotype of its F3 progeny? Maternal
rescue; the genotype of the weak rolling F2 worm was homozygous
mutant but since the F1 parent was heterozygous, there was wild type
mRNA deposited in the oocyte and was enough to weakly rescue the
rolling phenotype. But when the F2 worm self-fertilized its progeny it
did not have any wild type mRNA hanging around and so they were
not rescued, and they therefore had strong rolling phenotypes.
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
Separate out the rest of the F2 progeny and let them grow up and selffertilize on separate plates.
o In the F3 progeny of this third cross, one worm has a rolling
phenotype. What type of mutation is this? Maternal effect
mutation; same as maternal rescue but the rolling phenotype of the F2
parent had been completely rescued by the wild type copy and so had
appeared wild type.

Now that you’ve identified mutant strains, you are still not convinced that the
phenotypes you see are due to one locus or multiple loci. You therefore want
to clean up your genotype.
o What do you do next? Backcross.

Let’s draw out a backcross for the recessive mutant that arose in the F2
generation to remove 88% of the unlinked genes. (For purposes of this
problem, let’s labeled the mutant allele as r and wild type copies as +)
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Isolate your homozygous mutant that appeared in the F2 generation.
Let’s now call this PO.
Cross to wt male (1st backcross to remove 50% of unlinked genes)
F1: What genotype will the progeny will be? (r/+)
a. Will you see the phenotype? (No, because it’s a recessive mutant).
Single some heterozygotes and let them self.
F2: What will the phenotypic ratio be? 1:2:1 r/r : r/+ : +/+
a. What fraction of the progeny will have the rolled phenotype? ¼
r/r
Cross worms with the phenotype (r/r) to wt male (2nd backcross to
remove 50% of unlinked genes left, or 25% of originally present unlinked
genes)
F3: What will the phenotypic ratio be? 1:2:1 r/r : r/+ : +/+
a. What fraction of the progeny will have the rolled phenotype? ¼
r/r
Cross worms with the phenotype (r/r) to wt male (3rd backcross to
remove 50% of unlinked genes left, or 12.5% of originally present
unlinked genes)
F4: What will the phenotypic ratio be? 1:2:1 r/r : r/+ : +/+
a. What fraction of the progeny will have the rolled phenotype? ¼
r/r
Take the rolled phenotype (r/r) worms – this can now be your cleaned-up
stock.

Now you want to carry out a backcross for the dominant mutant that
arose in the F1 generation. First, however, you need to verify that it’s a
dominant mutation. How do you do this? Crossing to wild type. If you
cross to wild type and half of your progeny are rolling, it is a dominant
mutation. If none are rolling, it is a recessive mutation.
o But first, what would be useful in the background to be able to tell the
difference between self and cross progeny (remember that you can’t
tell the difference with a dominant mutation…)? Background recessive
marker to be able to tell the difference between self and cross progeny when
crossing to wild type.
o Let’s cross our F1 dominant mutant with wild type, this time with a
recessive homozygous marker in the background. (Let’s use a (madeup) mutant background which, when recessive, causes worms to visibly
tremble, and let’s write it as t for purposes of this problem.)
F1
t,r/t,+
X
+/+
F2
1:1
t,r/+ : t/+ CROSS PROGENY – HALF SHOULD BE ROLLING
NOTE: Here you can use the trembling background to make sure you aren’t
observing self-progeny. Self progeny would be
t,r/t,+ X
t,r/t,+  1:2:1 t,r/t,r : t,m/t,+ : t,+/t,+
So they would all be trembling. Just ignore the trembling progeny.
o Now let’s draw out the backcross for the dominant mutant that arose in
the F1 generation and that you have already tested for dominance (such
that you will start your backcross with the F2 generation). We can
utilize the trembling background in our backcross, again to be able to
tell the difference between self and cross progeny. Let’s assume that
the trembling locus is linked to the rolling locus so we don’t have to
worry about recombination.
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Allow your dominant rolling F2 worms to self.
F2
t,r/+
X
t,r/+
F3: 1:2:1 t,r/t,r : t,r/+ : +/+.
o Phenotypic ratio: ¼ trembling + rolling, ½ rolling, ¼ wt
o In case they DO ask: What if the trembling allele recombined such
that you ended crossing in your F1 self-cross t,r/t ,+ X t,r/t,+?
o F3 generation would then be 1:2:1 t,r/t,r : t,r/t : t,t
 Phenotype would be ¾ trembling + rolling, ¼ trembling, no
wild type. So, ignore plates that have no wild type worms.

Isolate all your rolling + trembling worms from this F3 generation; they
should be t,r/t,r.
Now, backcross 3 times to remove 88% of unlinked genes (50% each time):
 We’ll show the first backcross.
 Let’s now call the F3 t,r/t,r worms you ended up with above your P0
generation.
 P0:
t,r/t,r
X
+/+ male (1st backcross)
 F1: all t,r/+, all rolling, none trembling. Ignore the trembling if you do get
them (these are self-progeny).
 Single some t,r/+ worms and allow to self.
 F2: 1:2:1 t,r/t,r : t,r/+ : +/+
a. Phenotypic ratio: ¼ trembling + rolling, ½ rolling, ¼ wt
 Repeat 2 more backcross steps with the trembling + rolling (t,r/t,r) worms