MCDB 4650
Developmental Genetics in C.
elegans
Suppose you could make a genetic mosaic worm, in
which one of these two cells (i.e. the
prospective AC/VU pair) was homozygous for a
lf mutation in the gene for the Delta-like
protein, and the other cell was wild type (WT).
What do you predict will be the fates of the
two cells?
a)
b)
c)
d)
WT cell will become the AC, mutant cell VU.
Mutant cell will become the AC, WT a VU.
Both cells will become ACs.
Both cells will become VUs.
How would you go about trying to identify the
genes that control anchor-cell specification
and vulval development?
What would be the best of the following defective
phenotypes to choose for a saturation screen?
a) Inability to lay eggs.
b) Lack of a vulva.
c) Lack of an anchor cell.
d) Lack of Pnp cells on the ventral midline.
e) Larval arrest at the L4 stage.
You know the following information about mutants in the pathway
of genes that help to specify the vulva:
Mutant gene
(lf mutations)
let-60
lin-1
let-60; lin-1
Phenotype
Vul; all Pnp cells make epidermis (3 fate)
Muv; all Pnp cells 1 or 2 fate of vulva
Muv; all Pnp cells 1 or 2 fate of vulva
Of the pathways shown below, which best describes the above data?
(Pathways show a gene’s normal function.)
a) lin-1  let-60  vulval fate
b) lin-1  let-60  vulval fate
c) let-60  lin-1  vulval fate
d) let-60  lin-1  vulval fate
Which of the following statements is/are false? A genome-wide
RNAi screen for a particular phenotype in C. elegans, e.g. egg-laying
defective animals,
a) could not have been done until the entire genome had
been sequenced and genes identified.
b) would be much simpler and less time-consuming than the
original genetic screen.
c) would identify loss-of-function as well as gain-of-function
phenotypes of candidate genes.
d) would eliminate the need for positional cloning of
candidate genes.
e) would be able to identify genes that promote vulval
development (Vul mutant phenotype) but not genes that inhibit vulval
development (Muv mutant phenotype).
Suppose you have made a strain that is homozygous
mutant for a lin-3(lf) mutation and carries a small
unstable duplication that includes a lin-3(+) gene.
If the duplication fails to segregate correctly
during the first embryonic cleavage only, such that
the AB cell loses it and the P1 cell retains it, what
will be the phenotype of the resulting worm?
a) Vulva-less
b) Wild-type
c) Multi-vulva
A tricky mosaic problem:
The skn-1 gene product, SKN-1, acts in the 4-cell C. elegans
embryo in the EMS cell and its daughter MS, where its function is
required for MS to give rise to posterior pharynx and induce ABa to make
anterior pharynx. In the absence of SKN-1 protein, no pharynx is
produced and the embryo dies.
Suppose you have made a strain that is homozygous for a skn-1(lf)
mutation, but carries an unstable duplication that includes a skn-1(+) gene.
In the embryos produced by this strain, you can tell by means of a marker
on the duplication whether it is lost from either the AB-derived cells or
from the P1-derived cells.
You analyze first a set of AB-loss embryos and then a set of P1loss embryos.
Question: What phenotypes would you would predict for the two
sets of embryos?
a)
b)
c)
d)
AB-loss
Dead
Normal
Dead
Normal
P1-loss
Normal
Dead
Dead
Normal
Po
AB
P1
ABa ABp EMS P2
From among the possibilities listed below (1-5), choose the
correct predicted outcome(s) of the following laser
ablation and blastomere isolation experiments using
4-cell C. elegans embryos immediately after second
cleavage (may be more than one correct outcome).
Ablate the P2 cell immediately after division of P1.
1) No pharynx forms
2) Only anterior pharynx forms
3) Only posterior pharynx forms
4) ABp takes on the same fate as ABa
5) EMS descendants express no gut markers
Pick the single best answer. See Wolpert Figure 5.13 and
the explanation on p. 198. Loss-of-function (lf) mutations
in the C. elegans gene lin-4 have the same effect as
gain-of-function mutations in the gene lin-14 and the
opposite effect of lin-14(lf) mutations on timing of events
in larval development. This finding suggests a model in
which the products of these two genes interact such that
a) one stimulates the other to activate events in the first
larval stage.
b) one stimulates the other to activate events in later
larval stages.
c) one antagonizes the other to allow activation of
events in the first larval stage.
d) one antagonizes the other to allow activation of
events in later larval stages.
From among the possibilities listed below (1-5), choose the
correct predicted outcome(s) of the following laser ablation
and blastomere isolation experiments using 4-cell C.
elegans embryos immediately after second cleavage (may
be more than one correct outcome).
Remove the EMS cell from the embryo immediately after
P1 division and observe its progeny.
1) No pharynx forms
2) Only anterior pharynx forms
3) Only posterior pharynx forms
4) ABp takes on the same fate as ABa
5) EMS descendants express no gut markers
From among the possibilities listed below (1-5), choose the
correct predicted outcome(s) of the following laser ablation
and blastomere isolation experiments using 4-cell C.
elegans embryos immediately after second cleavage (may
be more than one correct outcome). Ablate the ABa cell.
1) No pharynx forms
2) Only anterior pharynx forms
3) Only posterior pharynx forms
4) ABp takes on the same fate as ABa
5) EMS descendants express no gut markers
From among the possibilities listed below (1-5), choose the
correct predicted outcome(s) of the following laser ablation
and blastomere isolation experiments using 4-cell C.
elegans embryos immediately after second cleavage (may
be more than one correct outcome). Ablate the EMS cell.
1) No pharynx forms
2) Only anterior pharynx forms
3) Only posterior pharynx forms
4) ABp takes on the same fate as ABa
5) EMS descendants express no gut markers
In normal C. elegans hermaphrodite development, either of
two adjacent precursor cells in the developing gonad can
become the anchor cell (AC; source of the signal for vulval
formation), with 50% probability. The other of the two
precursors remains a relatively undifferentiated ventral
uterine (VU) cell. This choice is mediated by the lin-12
gene, which encodes a Notch homolog.
In animals homozygous for a lin-12(lf ) mutation, how do
you predict these cells will develop?
1) Both cells will develop as ACs.
2) Both cells will develop as VUs
3) One cell will develop into an AC, the other into a VU.
4) Neither cell will differentiate.
In normal C. elegans hermaphrodite development, either of
two adjacent precursor cells in the developing gonad can
become the anchor cell (AC; source of the signal for vulval
formation), with 50% probability. The other of the two
precursors remains a relatively undifferentiated ventral
uterine (VU) cell. This choice is mediated by the lin-12
gene, which encodes a Notch homolog.
Given your knowledge of Notch-Delta signaling, does the
lin-12 gene act autonomously or non-autonomously to
determine AC/VU fates?
1) autonomously
2) non-autonomously.
In normal C. elegans hermaphrodite development, either of
two adjacent precursor cells in the developing gonad can
become the anchor cell (AC; source of the signal for vulval
formation), with 50% probability. The other of the two
precursors remains a relatively undifferentiated ventral
uterine (VU) cell. This choice is mediated by the lin-12
gene, which encodes a Notch homolog.
In animals carrying a lin-12(gf) mutation (in which the
mutant Notch homolog sends its normal signal to the
nucleus regardless of whether the Delta homolog is
complexed to it), how do you predict the two cells will
develop?
1) Both cells will develop as ACs.
2) Both cells will develop as VUs
3) One cell will develop into an AC, the other into a VU.
4) Neither cell will differentiate.
In normal C. elegans hermaphrodite development, either of
two adjacent precursor cells in the developing gonad can
become the anchor cell (AC; source of the signal for vulval
formation), with 50% probability. The other of the two
precursors remains a relatively undifferentiated ventral
uterine (VU) cell. This choice is mediated by the lin-12
gene, which encodes a Notch homolog.
In animals homozygous for a lf mutation in the gene for
the Delta homolog (gene not yet identified) that normally
interacts with LIN-12, how do you predict the two cells will
develop?
1) Both cells will develop as ACs.
2) Both cells will develop as VUs
3) One cell will develop into an AC, the other into a VU.
4) Neither cell will differentiate.
You isolate new loss-of-function mutations in
two different C. elegans genes that affect
body growth and final size. When the sma-9
gene is mutated, the adult animals are
abnormally short and small. When the lon-8
gene is mutated, the animals are abnormally
long and large.
The normal function of the lon-8 gene must
be to
1) stimulate body growth.
2) inhibit body growth
You isolate new loss-of-function mutations in
two different C. elegans genes that affect
body growth and final size. When the sma-9
gene is mutated, the adult animals are
abnormally short and small. When the lon-8
gene is mutated, the animals are abnormally
long and large.
The normal function of the sma-9 gene must
be to
1) stimulate body growth.
2) inhibit body growth
You isolate new loss-of-function mutations in
two different C. elegans genes that affect
body growth and final size. When the sma-9
gene is mutated, the adult animals are
abnormally short and small. When the lon-8
gene is mutated, the animals are abnormally
long and large.
Which two of the following possible regulatory pathways is
consistent with these phenotypes? Briefly explain your
reasoning.
1) sma-9  lon-8  body growth
2) sma-9  lon-8  body growth
3) lon-8  sma-9  body growth
4) lon-8  sma-9  body growth
You isolate new loss-of-function mutations in two
different C. elegans genes that affect body growth
and final size. When the sma-9 gene is mutated,
the adult animals are abnormally short and small.
When the lon-8 gene is mutated, the animals are
abnormally long and large.
If you do an epistasis test and determine that a
sma-9;lon-8 double mutant worms have the
same phenotype as sma-9 mutant worms (i.e.
they are abnormally small), which one of the
above regulatory pathways must be correct?
1) sma-9  lon-8  body growth
2) sma-9  lon-8  body growth
3) lon-8  sma-9  body growth
4) lon-8  sma-9  body growth
C. elegans homozygous for a recessive mutation in the ncl-1 gene have smaller
than normal nuclei that are easily recognizable in the adult animal. A
homozygous ncl-1 mutant animal carrying a small free duplication fragment that
includes a ncl-1(+) gene produces progeny in which all nuclei are normal. This
strain is commonly used for mosaic animals, since the ncl-1(-) cells are
distinguishable from the ncl-1(+) cells . You find one progeny animal in which
all muscle cells in the posterior pharynx and the body have small nuclei, while
the muscles of the anterior pharynx have normal nuclei. In the embryo that
gave rise to this animal, the duplication fragment must have been lost
a) in the first cleavage division, such that the AB daughter cell inherited
the fragment but the P1 daughter cell did not.
b) in the first cleavage division, such that the P1 daughter cell inherited
the fragment but the AB daughter cell did not.
c) in the second cleavage division of AB, such that the ABp daughter
cell inherited the fragment but the ABa daughter cell did not.
d) in the second cleavage division of AB, such that the ABa daughter
cell inherited the fragment but the ABp daughter cell did not.
You have isolated a loss-of-function mutation that causes a Vulvaless
(Vul) phenotype. Mapping and complementation tests show that it is
in a previously uncharacterized gene, which you call egl-30. To find
out where this gene might be acting, you make a homozygous
mutant strain that carries a small free duplication including the egl30(+) gene and a marker that allows you to tell which cells retain the
duplication in the adult animal. Among populations of this strain, you
find some hermaphrodites that lost the duplication during first
cleavage, a few in the entire AB lineage and a few in the entire P1
lineage. You find that the AB-loss animals have the Vul phenotype,
while the P1-loss animals have normal vulval development. This
result indicates to you that the egl-30 gene must normally function
a) in the anchor cell (AC), to produce the inducing signal for vulval
development.
b) in the ventral uterine (VU) sister of the AC, to respond to lateral
inhibition by the AC.
c) in all the vulval precursor cells (VPCs), to allow reception of or
response to the AC signal.
d) in the central cell of the VPC group, to allow lateral inhibition of
the nearby VPCs.
Assume you have made a GFP reporter construct including both
regulatory (promoter) and coding regions for each of four genes: a) lin3, b) lin-12, c) let-23, and d) a target gene inactivated by the inhibitory
lin-1 transcription factor. You then make a transgenic strain for each
construct by injection into wild-type C. elegans. In which of the cells
listed below (may be more than one) would you predict that each of
these genes would be expressed in larvae undergoing normal vulval
development, after the AC has fully differentiated?
a)
lin-3::GFP
(1) The AC but not its VU sister.
(2) The VU sister of the AC, but not the AC itself.
(3) The flanking cells of the vulval equivalence group P3p, P4p,
and P8p, but not the VPCs P5p, P6p, and P7p.
(4) The VPCs P5p, P6p, and P7p, but not the flanking cells P3p,
P4p, and P8p.
(5) All 6 cells of the vulval equivalence group, P3p through P8p
Assume you have made a GFP reporter construct including both
regulatory (promoter) and coding regions for each of four genes: a) lin3, b) lin-12, c) let-23, and d) a target gene inactivated by the inhibitory
lin-1 transcription factor. You then make a transgenic strain for each
construct by injection into wild-type C. elegans. In which of the cells
listed below (may be more than one) would you predict that each of
these genes would be expressed in larvae undergoing normal vulval
development, after the AC has fully differentiated?
b) lin-12::GFP
(1) The AC but not its VU sister.
(2) The VU sister of the AC, but not the AC itself.
(3) The flanking cells of the vulval equivalence group P3p, P4p, and
P8p, but not the VPCs P5p, P6p, and P7p.
(4) The VPCs P5p, P6p, and P7p, but not the flanking cells P3p,
P4p, and P8p.
(5) All 6 cells of the vulval equivalence group, P3p through P8p
Assume you have made a GFP reporter construct including both
regulatory (promoter) and coding regions for each of four genes: a) lin3, b) lin-12, c) let-23, and d) a target gene inactivated by the inhibitory
lin-1 transcription factor. You then make a transgenic strain for each
construct by injection into wild-type C. elegans. In which of the cells
listed below (may be more than one) would you predict that each of
these genes would be expressed in larvae undergoing normal vulval
development, after the AC has fully differentiated?
c) let-23::GFP
(1) The AC but not its VU sister.
(2) The VU sister of the AC, but not the AC itself.
(3) The flanking cells of the vulval equivalence group P3p, P4p, and
P8p, but not the VPCs P5p, P6p, and P7p.
(4) The VPCs P5p, P6p, and P7p, but not the flanking cells P3p,
P4p, and P8p.
(5) All 6 cells of the vulval equivalence group, P3p through P8p
Assume you have made a GFP reporter construct including both
regulatory (promoter) and coding regions for each of four genes: a) lin3, b) lin-12, c) let-23, and d) a target gene inactivated by the inhibitory
lin-1 transcription factor. You then make a transgenic strain for each
construct by injection into wild-type C. elegans. In which of the cells
listed below (may be more than one) would you predict that each of
these genes would be expressed in larvae undergoing normal vulval
development, after the AC has fully differentiated?
d) GFP reporter construct made with a target gene
regulated by the inhibitory Lin-1 transcription factor
(1) The AC but not its VU sister.
(2) The VU sister of the AC, but not the AC itself.
(3) The flanking cells of the vulval equivalence group P3p, P4p, and
P8p, but not the VPCs P5p, P6p, and P7p.
(4) The VPCs P5p, P6p, and P7p, but not the flanking cells P3p,
P4p, and P8p.
(5) All 6 cells of the vulval equivalence group, P3p through P8p
The gene skn-1, required for normal development in the C. elegans
early embryo, was discovered in screens for strict maternal-effect
embryonic lethal mutations. This gene is required for specification of
the EMS cell in the 4-cell embryo: mutant dead embryos have a
recognizable phenotype (no pharynx primordium forms).
You begin a C. elegans experiment by mating a skn1/+ male to a wild-type (+/+) hermaphrodite on a petri plate with plenty
of bacteria. Several days later, enough time has passed for two
generations of worms to be present, both males and hermaphrodites.
You find on the plate some dead F3 embryos (i.e. they were laid by F2
animals from the original cross) that exhibit the characteristic skn-1defective phenotype. Using the following list, indicate the correct
genotype(s) for questions (a) and (b) below.
(1) +/+
(4) skn-1/+ or skn-1/skn-1
(2) skn-1/+
(5) +/+ or skn-1/+ or skn-1/skn-1
(3) skn-1/skn-1
The genotype of the hermaphrodite parent of the dead embryo has to be:
The gene skn-1, required for normal development in the C. elegans
early embryo, was discovered in screens for strict maternal-effect
embryonic lethal mutations. This gene is required for specification of
the EMS cell in the 4-cell embryo: mutant dead embryos have a
recognizable phenotype (no pharynx primordium forms).
You begin a C. elegans experiment by mating a skn1/+ male to a wild-type (+/+) hermaphrodite on a petri plate with plenty
of bacteria. Several days later, enough time has passed for two
generations of worms to be present, both males and hermaphrodites.
You find on the plate some dead F3 embryos (i.e. they were laid by F2
animals from the original cross) that exhibit the characteristic skn-1defective phenotype. Using the following list, indicate the correct
genotype(s) for questions (a) and (b) below.
(1) +/+
(4) skn-1/+ or skn-1/skn-1
(2) skn-1/+
(5) +/+ or skn-1/+ or skn-1/skn-1
(3) skn-1/skn-1
The genotype of the embryo itself has to be:
You have isolated a new C. elegans mutation that causes
embryonic lethality, and you have named the gene it identifies as
dem-1, for dead embryos. To characterize it, you do the following
self-mating experiments and crosses, with the results shown.
Assume that there is no self-fertilization in mating crosses.
Experiment
dem-1/+ hermaphrodite selfing:
dem-1/dem-1 hermaphrodite selfing:
dem-1/dem-1 hermaphrodites x +/+ males
dem-1/dem-1 hermaphrodites x dem-1/+ males
dem-1/dem-1 hermaphrodites x dem-1/dem-1 males
dem-1/+ hermaphrodites x dem-1/+ males
dem-1/+ hermaphrodites x dem-1/dem-1 males
+/+ hermaphrodites x dem-1/dem-1 males
Embryonic viability
100%
0%
100%
100%
0%
100%
0%
0%
When during development must the dem-1 gene be transcribed, and in what
cells?
1) During oogenesis in maternal parent, in cells that give rise to oocytes.
2) During spermatogenesis in maternal or paternal parent, in cells that give rise
to sperm.
3) In the very early embryo, before gastrulation.
4) In the embryo, after activation of embryonic transcription at gastrulation.
You have isolated a new C. elegans mutation that causes
embryonic lethality, and you have named the gene it identifies as
dem-1, for dead embryos. To characterize it, you do the following
self-mating experiments and crosses, with the results shown.
Assume that there is no self-fertilization in mating crosses.
Experiment
dem-1/+ hermaphrodite selfing:
dem-1/dem-1 hermaphrodite selfing:
dem-1/dem-1 hermaphrodites x +/+ males
dem-1/dem-1 hermaphrodites x dem-1/+ males
dem-1/dem-1 hermaphrodites x dem-1/dem-1 males
dem-1/+ hermaphrodites x dem-1/+ males
dem-1/+ hermaphrodites x dem-1/dem-1 males
+/+ hermaphrodites x dem-1/dem-1 males
Embryonic viability
100%
0%
100%
100%
0%
100%
0%
0%
When is the DEM-1 gene product most likely to function?
1) very early in embryogenesis
2) after gastrulation
3) after morphogenesis
You isolate two new loss-of-function mutations in C. elegans that
affect the differentiation of sensory neurons. Worms homozygous
for one mutation, no-smell (nos-1), cannot sense food in their
environment. Worms homozygous for a mutation in a different
gene, yum-1, are hyper-attracted to food and get so distracted in
the presence of two different food sources that they can’t eat.
Which two of the following regulatory pathways is consistent with
the results so far?
1) nos-1 ---|
2) nos-1 ---|
3) yum-1 ---|
4) yum-1 ---|
yum-1 ---| food perception
yum-1 ---> food perception
nos-1---> food perception
nos-1 ---| food perception
You isolate two new loss-of-function mutations in C. elegans that affect
the differentiation of sensory neurons. Worms homozygous for one
mutation, no-smell (nos-1), cannot sense food in their environment.
Worms homozygous for a mutation in a different gene, yum-1, are
hyper-attracted to food and get so distracted in the presence of two
different food sources that they can’t eat.
1) nos-1 ---|
2) nos-1 ---|
3) yum-1 ---|
4) yum-1 ---|
yum-1 ---| food perception
yum-1 ---> food perception
nos-1---> food perception
nos-1 ---| food perception
If you do an epistasis test and determine that a nos-1; yum-1 double
mutant worm is hyper-attracted to food just like yum-1, which one of the
above pathways is correct?