Download Chrom. I - ucsf biochemistry website

2010 C. elegans problem set
I. Basic concepts of diploid genetics and linkage
You cross a dpy-5(e61) male (“dumpy” body shape phenotype, recessive mutation)
with an unc-13(e51) hermaphrodite (“uncoordinated” movement phenotype,
recessive mutation).
a. What phenotype(s) do expect to see in F1 cross progeny?
100% nonUnc, nonDpy
b. What assumptions are implicit in your answer?
Mating has successfully occurred, neither gene is on X, there is no unanticipated
lethality/defect as a transheterozygote.
c. You single (i.e. move individual worms onto separate plates and let them self)
several of the F1 cross progeny at a stage that you are certain they have not mated
with their siblings. What genotypic and phenotypic ratios do you expect to see in the
F2 if the two mutations are unlinked? (Draw out the dihybrid cross). How do these
ratios change if the mutations are linked?
If unlinked: 3/16 Dpy, 3/16 Unc, 1/16 UncDpy, rest wild type looking. If on the same
chromosome: Unc Dpy will be very rare.
d. At this point, let’s assume that the specific Dpy and Unc genes are linked. From the
F2 progeny plate (see above) you single out several Dpy animals, 1 per plate (again,
you are certain that they have not mated). What do you expect to see phenotypically
in their progeny (F3 relative to the starting cross)?
Most F2 Dpy animals are likely dpy +/dpy +, so these will have all Dpy looking
progeny. Rare F2 Dpy are dpy, unc-13/dpy, +, so these will also be all Dpy but ¼ will
appear as DpyUnc.
e. Of the 32 Dpy F2 animals that you single, 2 of the worms give rise to progeny (F3)
that are ¼ DpyUnc. What is the genotype of these two F2 animals? Think about which
chromosomes are parental, and which are recombination products. How many map
units are Dpy and Unc apart?
2/32 animals are recombinant. The two chromosomes in each F2 are independent
and come from independent meiosis. Therefore, each F2 scans 2 recombination
events. So, 2/64 chromosomes recombined. RF=2/64=0.031 RFX100=map units.
3.1 map units away.
2. Mapping using polymorphisms (SNPs)
You have isolated a mutation that appears to be due to inactivation of a single locus.
Your screen was started in the N2-Bristol background strain. To identify the mutated
gene, you cross your mutant to the CB4856
1
Hawaiian mapping strain (polymorphic relative to starting strain). You pick F1 cross
progeny, allow them to self, and from their progeny (F2), you pick animals that once
again display the mutant phenotype. You pick 20 such animals. These animals are
then placed individually on separate plates and allowed to lay eggs. About a week
later, all 20 plates are filled with animals. From each plate you wash some of the
animals away for DNA extraction and retain the plates (each plate will still have lots of
animals).
You know of SNIP-SNPs that inform on each of the chromosomes
(polymorphisms that can be detected because they create/eliminate a restriction
cutting site, hence SNIP; In each case, you PCR amplify ~0.5-1 kb region of the DNA
in which the pre-determined SNIP-SNP is contained. Since the SNPs generate or
destroy restriction sites, by subjecting the fragments to specific restriction
endonucleases and running gels, you can distinguish the presence or absence of a
SNP). So, from the the collection of progeny of 10 of the mutant Fss that you have
picked, you extract DNA, subject each DNA to PCR amplification for the six regions
that contain the SNIP-SNPs that inform on the 6 chromosomes of the worm, perform
restriction digest, and run a gel. Based on SNIP-SNP cutting patterns you can score
each SNP as CB4856 (CB), N2, or both. A gel showing the SNIP-SNP patterns for
each chromosomal marker are shown. Lanes one and two show the band patterns
for the N2 and CB strains as controls, and lanes 1-10 show the band patterns for each
of 10 F2 clones that are tested.
Chrom. I
N2 CB
_
_
_
Chrom II
N2 CB
_
_
_
1
_
_
2
_
_
3
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4
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5
6
7
8
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1
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2
4
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5
6
7
8
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3
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1
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2
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3
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4
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5
6
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7
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9
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10
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9
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10
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9
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10
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Chrom III
N2
_
CB
_
_
_
2
Chrom IV
N2 CB
_
_
_
1
2
3
4
5
6
7
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8
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1
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2
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3
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4
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5
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1
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2
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3
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4
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5
9
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10
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6
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7
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9
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10
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6
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8
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9
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10
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Chrom V
N2
_
CB
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Chrom X
N2 CB
_
_
_
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_
_
_
a) Which chromosome does this mutation map to? Why? Chrom. V since all
others have 75% or more Hawaiian
b) What is a rough estimate of map position of your mutation in relation to the
SNP on that chromosome? ~5 map units away from the indicated SNP. (1/20
chromosomes are Hawaiian 5%).
Now, using SNPs along the chromosome you then get the following data (The SNPs
are shown in proper genetic order, ie. SNP1 is on the left arm and SNP5 is on the right
arm). Remember double recombination events on one chromosome occur but are
rare):
Starting
strain
CB4856
1
2
3
4
5
6
7
8
9
SNP1
N2
SNP2
N2
SNP3
N2
SNP4
N2
SNP5
N2
CB/CB
CB/CB
CB/CB
CB/CB
CB/CB
N2/N2
N2/CB
CB/CB
CB/CB
CB/CB
N2/N2
N2/CB
N2/CB
N2/CB
N2/N2
N2/CB
N2/CB
N2/CB
CB/CB
N2/N2
N2/CB
N2/CB
N2/CB
N2/N2
N2/CB
N2/N2
N2/N2
CB/CB
N2/N2
N2/N2
N2/CB
N2/CB
N2/N2
N2/N2
N2/N2
N2/N2
CB/CB
N2/N2
N2/N2
N2/N2
N2/N2
N2/CB
N2/CB
N2/N2
N2/N2
N2/CB
N2/CB
N2/N2
N2/CB
N2/N2
3
10
11
12
13
14
15
16
17
18
19
20
N2/CB
N2/N2
N2/CB
N2/N2
N2/N2
N2/N2
CB/CB
N2/CB
N2/CB
N2/CB
N2/N2
N2/CB
N2/N2
N2/CB
N2/N2
N2/N2
N2/N2
N2/CB
N2/CB
N2/CB
N2/CB
N2/N2
N2/CB
N2/N2
N2/CB
N2/N2
N2/N2
N2/N2
N2/CB
N2/N2
N2/CB
N2/CB
N2/N2
N2/N2
N2/N2
N2/N2
N2/N2
N2/CB
N2/N2
N2/N2
N2/N2
N2/N2
N2/N2
N2/N2
N2/N2
N2/N2
N2/N2
CB/CB
N2/CB
N2/CB
N2/N2
N2/N2
N2/N2
N2/N2
N2/CB
c) Which marker is most closely linked to the mutation? SNP-4
d) Which two markers does your mutation lie between? How do you know? To
the right of SNP-3 and left of SNP-4. Closer to 4 than to 3
e) Give one trivial explanation as to why worm #5 doesn't fit with the rest of the
data. Recombinant #5 does not make sense. It is either a double cross over,
which is quite rare and unlikely. More likely option is that it was a mis-pick
(not really F2 mutant phenotype), so go back to your F3 plate, chunck animals
out and re-phenotype, if you get a mixture of phenotypes, just throw out
recombinant #5 from further analysis. A bit of a tricky question but closer to
real life.
3. Epistasis analysis and pathway building:
During programmed cell death cells are killed off and the corpses are engulfed and
cleared by neighboring cells. Regulatory relationships between pathway components
may be inferred through epistasis. Consider the following results for ced (cell death
defective) mutants (cell counts refer to final tally in adult hermaphrodites excluding
the germ-line):
Genotype
Wild type
ced-3(lf)
ced-4(lf)
ced-9(lf)
ced-9(lf); ced-3(lf)
ced-9(lf); ced-4(lf)
live cells
959
1090
1090
0
1090
1090
dead cells
131
0
0
1090
0
0
A) What regulatory relationships can be inferred between ced-3, ced-4, ced-9
and their relationship to cell death?
All we can say from this data is: ced-3 and ced-4 are required for cell death and both
are epistatic to ced-9, which normally inhibits cell death. Relationship of ced-3 and
4
ced-4 to each other is ambiguous with what is presented (cannot tell if one or two
pathways; cannot tell whether ced-3 is upstream of ced-4 or the other way around).
One way of showing this would be:
ced-9
| ced-3
cell death
(you can replace ced-3 with ced-4, or put them on top of each other)
Corpse engulfment genes of programmed cell death are identified as mutations in
which cell death still occurs leaving a visibly identifiable cell corpse but one that is
not cleared away. Consider:
Genotype
ced-1(lf)
ced-3(lf)
ced-1(lf), ced-3(lf)
phenotype
cells die, but leave ugly corpses
cells do not die
cells do not die
B) What is the epistatic relationship of ced-3 and ced-1?
ced-3is epistatic to ced-1. This is the equivalent of a biochemical pathway. Remember
that in a precursor-product pathway the upstream step is epistatic:
precursor (undecided cell)  intermediate (dead cell corpse)  product (decent
burial)
ced-3
ced-1
(initial identification of engulfment mutants (e.g. ced-1) allowed for identification of
“killer” genes such as ced-3 since they emerged, as you can see, as suppressors of the
ced-1 phenotype).
It is quite difficult to count cell deaths through the lineage (tracing all 131). In
practice, one can focus on one anatomical area of the worm and focus on death
events there and, if needed, show that findings in that region are also mimicked by
phenotypes seen in other cell types/anatomical regions.
For instance, loss of
function mutations in ced-1, ced-2, ced-7, and ced-10 were identified as engulfment
genes as null mutations in each of these genes left, on average, one corpse per
pharynx (C. elegans feeding organ). However, more than one cell undergoes
programmed cell death in the pharynx, implying that some corpse engulfment is still
occurring in each of the single mutant backgrounds. The following results were
obtained from mutant combinations (all cases are null alleles):
ced-1; ced-2: average four corpses/pharynx
ced-1; ced-7: average one corpse/pharynx
ced-1. ced-10: average four corpses/pharynx
ced-2; ced-7: average four corpses/pharynx
ced-2; ced-10: average one corpse/pharynx
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ced-7; ced-10 average four corpses/pharynx
C) What can you conclude about relationships between these four ced genes?
Ced-1, ced-7 are in one pathway that is likely parallel to the ced-2, ced-10 pathway.
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