Invert dev - exam questions

Question 1

1. What is a stem cell niche? Explain on the germline of Drosophila with cell names and pathways.
   
   1. probably need to explain this one better - X

Answer 1

1. The stem cell niche is a permissive environment that allows stem cell self-renewal
2. Cell Types:
   
   1. Terminal filament cells (TF)
   2. Cap cells (CC)
   3. Escort Cells (EC)
   4. Escort Stem cells (ESC)
   5. Follicle stem cells (FSC)
   6. Cytoblast (CB)

Question 2

1. You have a lin-3 egf(0) lin-12(gof) mutant: what phenotype does it have and how can you explain it with the function of the two genes in the wild type worm?

Answer 2

1. Likely phenotype would be vulvaless (Vul).
2. This is because lin-3 is required for inductive signal from anchor cell (AC) for VPCs to differentiate into 1° or 2° cells
3. Lin-12 (gof) does not make a difference here, because lin-12 (lof) means cells become 1° and lin-12 (gof) means they become 2° assuming that lin-3 egf is wt. But in this case all cells are 3° due to lin-3 mutation.
4. Shows the way in which lin-3 not working properly will suppresses a Muv phenotype.

Question 3

1. What relationship is there between mutation in a gene and a phenotype and how does this relate to the wild type worm? How is a mutant helpful?

Answer 3

1. A mutation in a gene can lead to changes in how the gene is expresses. For example it can be always turned on (gof) or always off (lof).
2. This mutation can then change the phenotype of the worm by changing which proteins are expressed.
3. For example, if the lin-3 gene is knocked out, then the worm does not develop any vulval cells due to the loss of the inductive signal. This may be observed by looking a the worm and seeing lots of eggs inside of it, and eventually lots of baby worms inside it because they can’t leave the worm due to lack of a vulva.
4. Mutants are helpful because they allow us to observe which genes affects which cells, and allow us to determine pathways by which these genes function.
5. For example, if one knows that gain of function in a gene such as lin-12 leads to all VPCs becoming 2°. One can then do an additional experiment with the mutant worms, where you also knock out lin-3. This would suppress the lin-12 phenotype, showing that lin-3 is upstream of lin-12.

Question 4

1. Mex-5 and mex-6: when and where are they expressed? What biological function do they have? What encodes the gene (gene product)? Where is it localized and what is its function?

Answer 4

1. Are expressed in the 1-cell stage embryo
2. Are localized to the anterior pole
3. Function to establish Soma/germline asymmetry in early C. elegans embryos.
4. Do this via inhibiting anterior expression of germline proteins through translational inhibition.

Question 5

1. Oskar: what is the largest protein of Oskar? What is the function of Oskar? Where is it expressed?

Answer 5

1. OSKAR is largest protein of oskar
2. Oskar mRNA is localized to the anterior pole of the drosophila oocyte
3. Induces germ plasm and ectopic PGC formation
4. OSKAR protein has central duel role in germ plasm formation:
   
   1. RNA binding protein that crosslinks nanos, pgc, and gcl mRNAs
   2. Localizes nanos mRNA at extreme posterior pole of unfertilized egg

Question 6

1. You have the segmentation mutant drosophila: it is missing two segments (one in the front and one in the back). What kind of mutation is this and explain.

Answer 6

Likely a GAP gene mutation

Probably tailless

Gap gene mutations are missing entire segments, while pair rule would be missing every other segment, and segment polarity are not missing any segments

Question 7

1. You have two mutants: mutant A and mutant B. they both show the same phenotype: how can you find out whether they have a mutation in the same gene?

Answer 7

1. Complementation analysis
2. Cross the two mutants together
3. If offspring have same phenotype as parents, they have failed to complement, meaning that mutation is on same gene
4. If heterozygous offspring have restored phenotype, genes have complemented, meaning that the mutations are on different genes

Question 8

1. The gene of mutant C is sequences and mapped and cloned. How can you find out where this gene is expressed?

Answer 8

1. Can use fluorescence in situ hybridization (FISH)
2. Fluorescent probes bind only to mRNA with high degree of complementarity
3. Can use it to see where the mRNA for this gene is localized

Question 9

1. The gene C is expressed in stripes over the body: how is it regulated, suggest and underlying molecular pathway that explains is morphology.

Answer 9

1. Can be explained by the reaction-diffusion model (Turing model)
   
   1. 2 substances in organism: activator and inhibitor
   2. Activator stimulates production of both, inhibitor slows production of activator
   3. Relative abundance of activator vs. inhibitor, as well as starting amounts for each can change the phenotype of stripes seen.
2. Another method is via Cis regulatory elements (CRE) modularity
   
   1. A morphogen activates expression of target gene
   2. Patterning output will then depend of morphogen concentration and binding site affinity for target genes
   3. This allows a gradient concentration of morphogen to control the activation of genes that might control pattern of stripes on a body

Question 10

1. Ced-1 is necessary to engulf dead cells, explain how this helped to find ced-3 and ced-4 who are necessary for killing and how this helped.

Answer 10

1. Ced-1 (lf) helpful because can see that if cells die they stay around (due to lack of engulfment)
2. Thus, in a ced-3 mutant we know that nothing died due to lack of corpses, rather than them being engulfed

Question 11

1. Two functions of PIE-1

Answer 11

a. Regulates maternal and zygotic gene expression in embryonic germ line
b. Keeps P2 from turning into EMS
c. Represses transcription by binding to RNA Pol II

Question 12

1. Main apoptotic pathway and function of each gene. How engulfment mutants allowed to detect mutations in the killing step (10 pts)

Answer 12

a. In ced-1 (lf) mutants, cells which died in the embryo phase would not be engulfed, and thus would still be visible in the larval stage, compared to wild-type where no cell corpses would be visible in the larval stage.
b. Thus, if killing step did not occur (such as in ced-3 (lf) mutants), no cell corpses would be visible in the larval stage.
c. A double mutant (ced-1 (lf); ced-3 (lf)) has a larval phenotype without cell corpses. A ced-1 (lf) mutant has a larval phenotype with cell corpses. A ced-3 (lf) mutant has a larval phenotype without cell corpses. This indicates that cells did not die in the double mutant and the ced-3 (lf) mutant. Because if they had died, cell corpses would be present in the double mutant larva due to ced-1 (lf) mutation.

Question 13

1. What happens when anchor cell is Mutated and let60 has a gain of function mutation?

Answer 13

a. Meaning that let60 is always on. Would be similar to Lin-3 (gf) mutation in which there is excess vulval induction. So would cause a Muv phenotype in the worm.

Question 14

1. What happens when the anchor cell is depleted in vulva development?

Answer 14

a. Assuming this is the same as experimentally ablating the anchor cell, then the result would be that if done early, no vulval cells would develop because the anchor cell sends out the inductive signal. So all VPCs would adopt the 3° cell fate.

Question 15

1. Two main functions of the niche of the germ line stem cells

Answer 15

a. Occupancy: is promoted by stimulating adherent junctions between niche cells and cells within, but not outside, the niche
b. Fate regulation: by repressing differentiation genes in cells inside, but not outside the niche.

Question 16

Name of technique when screening mutations for induced mutants

Answer 16

a. F2 forward genetic screen: screen for mutants in F2 phase.
b. Do in F2 and not F1 because most induced mutations reduce gene function by 50%.
c. This means that they will only show up as loss of function in F2.

Question 17

1. Which is the lateral signal in vulval development and why it was difficult to detect the gene responsible?

Answer 17

• Explanation:
  
  • Also called lateral inhibition
  • VPCs have natural cell fate of 3°
  • AC sends out inductive signal to P5.p, P6.p, and P7.p (is strongest to P6.p) telling them to adopt 1° or 2° fates
  • P6.p sends out lateral signal, inhibiting P5.p and P7.p from adopting 1° cell fate
• Difficult to detect gene responsible because:
  
  • LIN-12 Notch-like receptor, which tells cells not to become 1° cells (lin-12 (lf) –> all cells adopt 1° cell fate; lin-12 (gf) –> all cells adopt 2° cell fate)
  • There are multiple signaling molecule ligands which can bind the LIN-12 receptor. Thus, they are partially redundant (LAG-2, APX-1, DSL-1). This is a reason why it was difficult to determine the gene responsible, because knocking out one of these genes did not prevent the lateral signal.

Question 18

• Relation between phenotype, mutation, and development
• difference between wild-type phenotype and mutant phenotype.
• How development can be studied with phenotypes

Answer 18

• Relation between phenotype, mutation, and development
  
  • A mutation in a gene will cause an organism to develop in a certain way, leading to a specific phenotype (the physical characteristics of the organism which may be observed)
• difference between wild-type phenotype and mutant phenotype.
  
  • An WT phenotype is the phenotype that would develop if no changes were made to the genes of an organismt
  • the mutant phenotype is the phenotype that would develop if a gene is changed in some way (such as being mutated on purpose or by accident)
• How development can be studied with phenotypes
  
  • Can study development using model organisms such as C. elegans where the developmental path of the embryo to adult is completely known.
  • Because of this, the effect of mutations in the C. elegans genome may be observed by looking at the phenotype of the larva or adult worms. The mutant phenotypes may be compared to the Wild type worms to see what the effect of the mutation was.

Question 19

1. Which segmentation genes are mutated if the posterior and anterior segments are lost and explain why. WT has the 8 stripes and mutant 6

Answer 19

a. Likely a tailless gap gene mutation
b. Can’t be pair rule, because are not losing even or odd segments
c. Can’t be segment polarity, because that would not be losing segments

Question 20

1. How to study when and where the gene is expressed. Explain the procedure you would use.

Answer 20

1. Use fluorescence in situ hybridization (FISH)
2. In this procedure, fluorescently labelled probes with high degree of complementarity to mRNA are injected into the embryo
3. Then visualize the location of mRNA (and thus gene expression) at different (time) points in the development of the embryo

Question 21

1. The mutant is identified as a mutant of the product of gene A, which normally distributes in the posterior and anterior region. Mutant B only shows the posterior region. Explain the relation between A and B

Answer 21

1. The wording of this is confusing
2. However, an answer could potentially be that the distribution is ubiquitous, but expression is normally inhibited in the middle region by a different morphogen.
3. In mutant B the morphogen is expressed also in anterior region, accounting for the mutant phenotype?
4. I feel I need more information or different wording to give a good answer here

Question 22

1. Implications of a mutation in the 3UTR of lin-14. Explain how development timing is affected.

Answer 22

1. Lin-14 inhibits lin-29, which when activated causes transition to adult form of C. elegans.
   
   1. So as long as lin-14 is active, worm will stay in larval form, thus regulating timing of development
2. Lin-4 encodes small RNAs with antisense complementarity to the 3UTR of lin-14. Lin-4 thus regulates lin-14, inhibiting it.
3. If there is a mutation in the 3UTR of lin-14, small RNAs coded by lin-4 potentially cannot bind it, and lin-14 will remain active.
4. This means that lin-29 will remain inhibited, and the worm will not be able to transition out of larval form.