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Problem 1:
C. elegans epistasis
You are interested in studying vulval development in C. elegans! Because you really truly
enjoy examining the bag-of-worms phenotype that results when eggs hatch inside their
mother, and the creepiness of multi-vulval worms delights you!
Yup, pretty cool, right?
You have the genotypes listed on the following table. gf = over-expression (you have a
line with a transgene containing multiple copies of each gene) lf= complete loss of
function)
Single mutants
Multivulval
lin-15(lf)
lin-3(gf)
let-60(gf)
lin-1(lf)
lin-3(lf)
let-23(lf)
let-60(lf)
lin-45(lf)
Vulvaless
lin-3(lf)
let-23(lf)
let-60(lf)
lin-45(lf)
lin-15(lf)
Multivulval
Vulvaless
Vulvaless
Vulvaless
lin-3(gf)
Multivulval
vulvaless
No data
No data
let-60(gf)
Multivulval
Multivulval
Multivulval
Vulvaless
lin-1(lf)
Multivulval
Multivulval
Multivulval
Multivulval
A) Which genes in this pathway are positive regulators of vulval formation? Which genes
are negative regulators of vulval formation?
Positive regulators: lin-3, let-23, let-60, and lin-45
Negative regulators: lin-15, lin-1
B) How would you order these genes in a pathway?
lin-3 --| lin-15 --| let-23  let-60  lin-45 --| lin-1 --| vulval formation
C) To complicate matters further – another lab has recently identified another mutant,
glp-1 which lacks a germline altogether. glp-1 mutants also don’t have a normal vulva.
Can you order glp-1 in your pathway?
Purely from an epistatis pathway, yes. glp-1 since it ablates the germline altogether is
technically epistatic to all the lin- genes. However, in reality because glp-1 specifies
cells that are precursors to normal germline development, glp-1; lin-15 double mutants
don’t really tell you much about the lin-15 pathway. This is similar to the original
example Jeff used when introducing epistatis (ie in a hairless mouse, talking about
which gene specifies coat color is not particularly informative)
Problem 2:
You decide to rotate with Cynthia Kenyon your second quarter, and of course, she has
you searching for the fountain of youth. She wants you to do a screen for new genes that
make worms live longer. Explain how you start your screen.
Mutagenize WT hermaphrodites.
Allow them to self.
What do you expect the phenotype of your F1s to be?
All hermaphrodites, all WT phenotype.
Wow, you get one worm in your F1 progeny that just won’t die, you call it liv-1. What do
you think is up with this worm? How do you verify this?
It has a dominant mutation. Cross it to a WT male to check. If dominant, half of the
progeny should live long and half are wild type.
If recessive, none should live long. liv/+ X +/+ à 1:1 liv/+ : +/+
Ok, now that you’ve got liv-1 all sorted out, you want to find more mutants. What do you
do?
Move your F1 progeny onto their own plates. Allow for self-mating. Examine the
phenotypes of the F2s.
Great, you’ve found a bunch of other mutants that live more than 2 times the average
worm life span! Cynthia’s tells you to make sure that all the phenotypes are each caused
by a single mutation. What do you do and why is this important?
Back-cross to WT males. You will get 50% WT males, and 50% WT hermaphrodites.
Self the hermaphrodites and pick mutants. Repeat. This is important to eliminate
random background mutations and make sure that you have isolated a single mutation.
Not outcrossing enough has actually caused big mistakes in the worm field! The sir2 gene
in worm was originally reported to have a 40% lifespan extension, but later it was
discovered that actually the reported lifespan extension was caused by a background
mutation because they didn’t outcross their strains enough. Some people still believe that
sir2 causes a 10% lifespan extension, but it’s a huge controversy in the field. Moral of the
story- backcross your strains – at least 5x.
Because this is 2012, we have lots of tools available to us that make things faster and
easier- like deep sequencing. To figure out what genes are causing lifespan extension we
might be able to deep sequence. Would you still want to outcross first? Why? If you were
going to identify the mutation via deep sequencing, would you still need to map at all?
You would still want to outcross to isolate a single mutation. Deep sequencing the
entire genome would still be really expensive. You would want to map to a region of the
chromosome, but you wouldn’t necessarily need to map to a small region
of the genome.
Problem 3:
You need to map a newly identified mutation (m) by the end of your rotation in the
Ashrafi lab. You decided to use a mapping strain homozygous for all three known genetic
markers a, b and c.
a is on Choromome I
b is on Chromosome II
c is on Chromosome III
You know that:
1) m is a recessive mutation
2) a, b and c have a BamHI restriction site
Before BamHI digestion
After BamHI digestion
A
300bp
300bp
a
300
230 and 70
B
100
100
b
100
50 and 50
C
200
200
c
200
150 and 50
What steps do you go through before before you get to the mapping portion? Will you
use F1s or F2s? Why?
1) After mating m/m strain with a test strain, you isolated F1 hermaphrodites.
2) You took F1 and put them on separate plates for self-fertilization.
3) You isolated m/m F2 progeny to figure out which of the three markers is linked to
your mutation.
4) You performed PCR to amplify all three marker loci.
5) You cut the PCR products with BamHI and ran the fragments on a gel.
You isolate F2s because F1s are hets and thus uninformative.
What fraction of the time would you expect your mutation to segregate with your marker
if it is unlinked to the chromosome that the marker is on?
75% of the time
What fraction of the time would you expect your mutation to segregate with your marker
if it is linked to the chromosome that the marker is on?
<75% of the time – dependent on how many map units apart you are from your
mutation of interest
You got the following result:
On which chromosome is the mutant allele found? (To make it easier to pick out, what is
the genotype of each worm?)
Chromosome I.
Since it rarely segregates with a.
What is the map position of the mutation? Remember that each lane on your gel
represents two chromosomes.
Map units (cM) = 100*(# of recombinant chromosomes/total # chromosomes)
2/24*100 = 8 cM