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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.

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Presentation on theme: "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."— Presentation transcript:

1 MCDB 4650 Developmental Genetics in C. elegans

2 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) WT cell will become the AC, mutant cell VU. b) Mutant cell will become the AC, WT a VU. c) Both cells will become ACs. d) Both cells will become VUs.

3 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.

4 You know the following information about mutants in the pathway of genes that help to specify the vulva: Mutant genePhenotype (lf mutations) let-60Vul; all Pnp cells make epidermis (3  fate) lin-1Muv; all Pnp cells 1  or 2  fate of vulva let-60; lin-1Muv; 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

5 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).

6 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

7 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 P1- loss embryos. Question: What phenotypes would you would predict for the two sets of embryos? AB-lossP1-loss a) DeadNormal b)NormalDead c) DeadDead d) NormalNormal Po AB P 1 ABa ABp EMS P 2

8 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

9 Pick the single best answer. See Wolpert Figure 5.13 and the explanation on p 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.

10 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

11 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

12 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

13 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.

14 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.

15 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.

16 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.

17 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

18 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

19 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

20 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

21 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.

22 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 egl- 30(+) 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.

23 Assume you have made a GFP reporter construct including both regulatory (promoter) and coding regions for each of four genes: a) lin- 3, 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

24 Assume you have made a GFP reporter construct including both regulatory (promoter) and coding regions for each of four genes: a) lin- 3, 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

25 Assume you have made a GFP reporter construct including both regulatory (promoter) and coding regions for each of four genes: a) lin- 3, 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

26 Assume you have made a GFP reporter construct including both regulatory (promoter) and coding regions for each of four genes: a) lin- 3, 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

27 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 skn- 1/+ 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-1- defective 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:

28 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 skn- 1/+ 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-1- defective 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:

29 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 Embryonic viability dem-1/+ hermaphrodite selfing: 100% dem-1/dem-1 hermaphrodite selfing:0% dem-1/dem-1 hermaphrodites x +/+ males100% dem-1/dem-1 hermaphrodites x dem-1/+ males100% dem-1/dem-1 hermaphrodites x dem-1/dem-1 males0% dem-1/+ hermaphrodites x dem-1/+ males100% dem-1/+ hermaphrodites x dem-1/dem-1 males0% +/+ hermaphrodites x dem-1/dem-1 males0% 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.

30 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 Embryonic viability dem-1/+ hermaphrodite selfing: 100% dem-1/dem-1 hermaphrodite selfing:0% dem-1/dem-1 hermaphrodites x +/+ males100% dem-1/dem-1 hermaphrodites x dem-1/+ males100% dem-1/dem-1 hermaphrodites x dem-1/dem-1 males0% dem-1/+ hermaphrodites x dem-1/+ males100% dem-1/+ hermaphrodites x dem-1/dem-1 males0% +/+ hermaphrodites x dem-1/dem-1 males0% When is the DEM-1 gene product most likely to function? 1) very early in embryogenesis 2) after gastrulation 3) after morphogenesis

31 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 ---| yum-1 ---| food perception 2) nos-1 ---| yum-1 ---> food perception 3) yum-1 ---| nos-1---> food perception 4) yum-1 ---| nos-1 ---| food perception

32 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 ---| yum-1 ---| food perception 2) nos-1 ---| yum-1 ---> food perception 3) yum-1 ---| nos-1---> food perception 4) yum-1 ---| 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?


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