MCB164S10+Practice+problem+set+1+key

MCB164S10+Practice+problem+set+1+key - MCB164 Spring 2010...

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MCB164 Spring 2010 practice problem set #1 Assume all mutations are unlinked and autosomal unless otherwise specified 1. Sex is determined by the X chromosome to autosome ratio (XX:AA) in C. elegans . When the ratio is 1.0 the worm is hermaphrodite. When the ratio is 0.5 the worm is male. tra-1 is a gene required for hermaphrodite development. A recessive, loss of function mutation tra- 1(e1099) transforms XX animals to males. A dominant, gain of function mutation tra- 1(e1575) transforms XX and XO animal to female. All mutants are fertile. Allele Phenotype Dom./ Rec. compared with wild type tra-1(wt) XX hermaphrodite NA XO male tra-1(e1575) XX female dominant XO female dominant tra-1(e1099) XX male recessive XO male recessive ______________________________________________________ Consider the following cross: Po XX tra-1 (e1575)/tra-1(wild type) X XO tra-1 (e1099) / tra-1(e1099) A. (4) What is the phenotype of the XO tra-1(e1575)/tra-1(e1099 ) F1 animal? Circle one Male Female Hermaphrodite B. (4) What fraction of the total F1 progeny will be female? Circle one All 2/3 1/2 1/3 1/4 none C. (7) You wish to create a loss of function allele of tra-1. Your strategy is to mutagenize the tra-1(1575) allele and isolate revertants that give a loss-of-function phenotype (you call these tra-1(e1575) lof ) . You treat XX tra-1(e1575)/tra-1(e1575) females with EMS and cross them to wild type XO males. What phenotype will you score among F1 progeny to find tra-1(e1575) lof mutations? males that would be XO tra-1 (e1575) lof /+
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2. the “green eggs in Him” screen (see Wiki) was used to identify genes involved in the proper segregation of X chromosomes during meiosis. In WT animals males arise ~0.1% of the time by the chance occurrence of a chromosome nondisjunction event leading to a nullo-X gamete. Mutants that increase this frequency of this event (High Incidence of Males) were sought in order to identify genes involved in the process of homolog segregation. The mutagenized strain contained a stable transgenic construct, Pxo1-1::gfp ( a male-embryo-specific promoter element fused to the ORF for green fluorescent protein), which will cause only male (XO) embryos to glow fluorescent green at a particular developmental stage. F2 hermaphrodites are starved before screening, which inhibits egg laying, thereby allowing embryos in utero to accumulate and age. Therefore, wild type gravid (starved) hermaphrodites will have few or no green eggs, whereas him mutants will have numerous green eggs. The basic outline of the mutagenesis scheme is as follows: EMS Self Self XX Pxo1-1::gfp F1 F2 Screen for him A. (4) TRUE of FALSE The him/him homozygous mutants are phenotypically XX males while him/+ heterozygous mutants are XX hermaphrodites. B.
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This note was uploaded on 06/06/2011 for the course MCB 164 taught by Professor Burgess during the Spring '08 term at UC Davis.

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MCB164S10+Practice+problem+set+1+key - MCB164 Spring 2010...

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