Problem Set 4

Problems for Classes 7 and 8 with answers 2/7/06 1. (3 points) In normal C. elegans hermaphrodite development, either of two adjacent AC/VU 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. In animals carrying a lf mutation in the lin-12 gene (encodes a Notch homolog), both cells develop as ACs. a) (1 pt) Knowing that Notch is a receptor, would you expect that the lin-12 gene would act (1) autonomously or (2) non-autonomously to determine AC/VU fates __1__ b) (1 pt) In a mosaic animals where one of the two precursors carries the lin-12 mutation and the other is lin-12(), what would be the fate of the mutant cell 1) Always VU. 2) Always AC. 3) Sometimes VU, sometimes AC. ___2___ c) (1 pt) Explain your reasoning briefly. The mutant animal shows that lin-12 (Notch) expression is required for one of the cells to become a VU cell if it acts autonomously, as a receptor would, then the fate of each cell will depend on its own genotype, by definition, so the mutant cell will always become the AC. Note that if the mutation were in the C. elegans delta homolog (the ligand), then in the mosaic animal the mutant cell would remain a VU and the nonmutant cell would become the AC i.e., the delta gene would act nonautonomously. 2. (3) You isolate new loss of function mutations in two different C. elegans genes that affect body 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. a) (1.5) 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 ______ and ______ Normal function of lon-8 must be to inhibit body growth, since a lon-8(lf) mutant is abnormally large. It could act directly, as in (1) or indirectly, as in (3). Therefore, these are the possible pathways consistent with the phenotypes. You could also use the same logic starting with normal function of sma-9, which must be to enhance body growth. b) (1.5) If you do an epistasis test and determine that a sma-9lon-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 Explain your reasoning. ___3___ Since sma-9 is epistatic to lon-8, it must be downstream in the pathway. 3. (3 points) You have isolated a strong loss-of-function mutation in the lin-15 gene that causes a Muv phenotype. To see if it can be located in the ras pathway (see text, slides, and Class 7 lecture notes) by epistasis tests, you make the following double mutants using other loss-of-function mutations in ras pathway genes, and find the phenotypes indicated Single mutant phenotypes Double mutant phenotypes let-60 Vul lin-15 let-60 Vul mek-2 Vul lin-15 mek-2 Vul let-23 Vul lin-15 let-23 Vul lin-1 Muv lin-15lin-1 Muv lin-3 Vul lin-15 lin-3 Vul a) (1.5) Is the lin-15 gene on the ras pathway What is the evidence Ans Yes, because there is clear epistasis. If it were not on the pathway, one might expect to see intermediate phenotypes. b) (1.5) If so, where does the lin-15 gene act in the ras pathway (Indicate in words, using upstream and downstream to indicate direction along the pathway). Ans Because all the mutations that cause Vul phenotype are epistatic to lin-15, it must be at the very start of the pathway, acting upstream of lin-3 to turn it off in the non-VPCs of the vulval equivalence group. 4. (1) Which concept(s) did you find most difficult in this weeks material If you didnt have much difficulty with it, then what did you find most interesting MCDB 4650 DEVELOPMENTAL BIOLOGY Spring 2006