Developmental Genetics - AndersenLab/Genetic-Analysis GitHub Wiki

Lecture 9

1. The cell lineage of C. elegans has been completely determined, which makes this model great for studying development!

2. The wild-type vulva of this nematode is composed of three vulval precursor cells (P5.p, P6.p, and P7.p) and three non-vulval precursor cells (P3.p, P4.p, and P8.p).

3. Bob Horvitz did a screen for Egl mutants (cannot lay eggs) and found several biological reasons why each mutant could not lay eggs.

4. Each mutant was characterized (Vul for vulvaless, Muv for multivulva)

5. Double mutants were phenotyped to create the pathway.

6. lin-39 was shown to be a Hox gene (a class of genes that are arranged along the chromosome in the same order that they should be acting in the body plan)

7. Ablation of the anchor cell led to a loss of vulval development. This result suggests that the AC is necessary for proper vulval development.

8. If you move the AC to a different location, a different precursor cell will receive the signal that was intended for P6.p and the vulva will be shifted.

9. The anchor cell ablation and loss of lin-3 seem to have the same phenotype and are both upstream of other genes in the vulval development pathway.

10. lin-3 expression in whole animal or in just the AC was sufficient to rescue the vulvaless phenotype of a lin-3 null mutant.

Pop Quiz: Why was the GFP plasmid injected along with the lin-3 plasmid? Why was this experiment done by injecting lin-3 WT into a lin-3 null animal and not the other way around?

11. Cell autonomy of the let-60 experiment:

    • The let-60 loss-of-function mutation makes a vulvaless animal. Question: Does let-60 act in the vulval cells (cell autonomous) or in other cells (cell non-autonomous) to make this mutant?
    • start with a let-60 knockout animal
    • insert let-60 rescue plasmids containing wild-type let-60 with GFP (to track which cells have the construct)
    • look for examples where a certain cell type is not GFP (meaning that let-60 is deficient in this particular cell) and see if that animal is able to develop a normal vulva
    • They found that lacking let-60 function in the anchor cell or in non-vulval cells did not matter. let-60 expression in the vulval cells was required to rescue the vulvaless phenotype and create a wild-type animal.

12. Cell autonomy of lin-3 experiment:

Pop Quiz: Given what you learned about lin-3 and what you have learned about cell autonomy experiments, can you tell what this experiment proved?

13. lin-2, lin-7, and lin-10 all work together to localize the LET-23 receptor to the proper side of the cell

Question: Why do we care about vulval development in C. elegans?

Answer: It taught us about Ras signaling and the basics about how cells decide to divide or not. This last point is exactly what goes wrong in human cancer. So tiny worms taught us A LOT about cancer!

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The Drosophila eye is made up of hundreds of ommatidia, hexagonal structures that protect the underlying neurons. Each ommatidium contains eight neuronal cells: R1-R8.

When wild-type flies are faced with the choice of white light or UV light, they will choose the UV light. Mutants that prefer white light were found to lack a certain neuron in their ommatidia -- R7. The gene responsible for this mutant phenotype was named sevenless.

To determine if sevenless was cell autonomous (necessary within the R7 cells) or cell non-autonomous (required in some other cell besides the R7), a mosaic analysis experiment was done.

In this case, we will be losing sevenless function in a small number of cells, whereas other cells will have wild-type sevenless function.

How can we track which cells are homozygous for the sevenless mutant allele?

• Start with a white-eyed (w^-/w^-); sev⁺/sev^- embryo.

• Introduce a P element that will insert a wild-type white gene w⁺ copy distal to the wild-type sev allele. And let it develop to adulthood.

• The cells in the eye of that adult fly will have the following genotype:

  • sev⁺ w⁺/sev^- ; w^-/w^- (they will look red and be heterozygous for the sevenless mutation)

• If you induce double-strand breaks in the DNA, mitotic recombination will occur between the centromere and the sev locus, and homology-driven repair will result in a new mitotic product. These cells will either be:

  • sev^-/sev^- ; w^-/w^- (they will look white and lack sevenless function)
  • sev⁺ w⁺/sev⁺ w⁺ ; w^-/w^- (they will look red and have wild-type sevenless function)

Pop Quiz: If we see a red cell, do we know if it is heterozygous or homozygous for sevenless?

Determine cell autonomy:

Look at all of the flies that have white spots in their eye. These animals have some cells that lack sevenless function.

Some ommatidia of those animals will have certain neurons that are lacking sevenless function.

Your question: In which neurons (R1-R8) does loss of sevenless function result in the lack of an R7 cell, and therefore our mutant phenotype?

Answer: If you lose sevenless function in R1, R2, R3, R4, R5, R6, or R8, R7 cells still have the ability to develop properly. Therefore, sevenless function is not required in those cells. However, when you lose sevenless function in the R7 cell itself, you have the mutant phenotype (R7 cell is not present and flies fail to go toward UV light). sevenless is cell-autonomous.

You have another gene, bride of sevenless (boss) that also results in lack of an R7 cell and flies that fail to go toward UV light. You perform a cell autonomy experiment with this mutant.

Pop Quiz: How would you generate a mosaic analysis experiment where you can track cells that lost boss function?

This gene was found to be non cell-autonomous (it is required in R8 to make a proper R7 cell). Think of all the possible outcomes of your mosaic analysis experiment and what phenotypes you would expect in each case. (e.g. what happens when you lose boss function in the R7?)

Using a virus gene mutant found in chickens, a temperature-sensitive reduction-of-function mutation of sev (sev^B4) was engineered.

At the permissive temperature, sev^B4 animals have 100% R7 development.

At the restrictive temperature, sev^B4 animals have 90% R7 development.

You decide to do an enhancer screen to find mutations on the third chromosome (sev is on the X chromosome) that decrease the R7 development at the permissive temperature. We will call these enhancers E(sev).

Pop Quiz: Given a sev loss-of-function strain and a strain with a TM3 balancer chromosome with a sev^B4 P element on the third chromosome, write out an enhancer screen for third chromosome enhancers of the sev^B4 mutation. (Hint: the answer is in lecture)

In the same manner as we did for the sev and boss mosaic analysis, cell autonomy for E(sev) can be established.

The opposite way of looking at the mosaic analysis:

• Use a P element with w⁺ distal to the mutant E(sev) allele

• The most common cells will have pigment and will have the mutant E(sev) allele

• Cells without pigment will have the wild-type allele of E(sev)

• Wormbook: Vulval Development - Excellent review for Lecture 9.
• Wikipedia: Cell Fate Determination
• Berkeley Review of Mosaic Analysis