Compare and contrast the following.
Dominance and epistasis
Programmed cell death in C. elegans occurs when cells are engulfed and die during normal development. Mutations in several genes are known to affect this process. Two of these genes are ced-3 and ced-9. The following results were obtained with loss-of-function (lf) and gain-of-function (gf) mutations of these genes.
mutation phenotype
wildtype some cells die
ced-3(lf) all cells live
ced-9(gf) all cells live
ced-9(lf) more cells die than in wild-type animals
ced-9(lf); ced-3(lf) all cells live
What two possible genetic pathways are suggested by the single gene mutations? Explain your answer.
Which pathway is more likely given the double mutant result? Explain.
Explain the action of the ced-9(gf).
If newly hatched larvae of C. elegans are crowded and given little food, they do not grow into adults. Instead they develop into a dormancy stage called the dauer larva. Mutations in several genes affect dauer larva formation. Some mutations cause animals to always form dauer larvae [the dauer-constitutive (Daf-c) phenotype], and other prevent them from ever becoming dauer larvae [the dauer-defective (Daf-d) phenotype].
From the data given below (all mutations are recessive loss-of-function mutations) derive a genetic pathway for dauer larvae formation. Explain your reasoning.
Genotype Phenotype
daf-10 Daf-d
daf-11 Daf-c
daf-22 Daf-d
daf-10;daf-11 Daf-d
daf-11;daf-22 Daf-c
Since a pheromone (a chemical produced by the animals) is thought to induce dauer larva formation (i.e., be the signal for this pathway), predict which gene in the pathway might encode for the production of the pheromone. Explain your answer.