1. (4 pts). A female mouse with a white patterning called “splotch” was mated to
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Question
1. (4 pts). A female mouse with a white patterning called “splotch” was mated to a male “splotchy” mouse to produce the progeny below. Note “tottering” describes the atypical gait of mice with a particular single-gene neurological disorder. Subsequent crosses within each phenotypic class revealed that only the tottering progeny below are true-breeding.27 wild type
73 splotchy
22 splotchy, tottering
8 tottering
Indicate the genotypes of the parents and the genotypes of each phenotypic class of the progeny.
HINT: If phenotype isn’t specified, assume it’s wild type (i.e. 64 splotchy mice are not tottering). Also, note that the parents were not described as the “parental generation”.
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I believe the genotypes of the splotchy,tottering and tottering genotypes are sstt and s+s+tt respectively because they are true-breeding, but I am unsure of how to figure out the other genotypes.
Explanation / Answer
Given phenotypes are 27 wild type73 splotchy
22 splotchy, tottering
8 tottering
The genes corresponding to the traits are S - splotchy t- toter T-wild type s- wild type 27 wild type - ssT
73 splotchy - S T
22 splotchy, tottering - S tt
8 tottering - sstt To obtain these the genotypic cross is SsTt * SsTt ST St sT st ST ST ST ST ST ST ST Stt ST Stt sT ST ST ssT ssT st ST ssT ssT ssstt ST = 9 ssT=4 Stt=2 sstt=1 The given phenotypes are near to these ratios ssT=4 (73) ST = 9 (27) Stt=2 (22) sstt=1 (8) Stt=2 (22) sstt=1 (8)
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