You’re studying C.elegans and have isolated two homozygous mutant C. elegans lines, R1 and R2, that...
You’re studying C.elegans and have isolated two homozygous mutant C. elegans lines, R1 and R2, that each display the roller phenotype. When each mutant is crossed to wild type, the respective F1 progeny are normal. An F1 self results in F2 progeny that phenotypically segregate 3 wildtype:1 roller. Crossing R1 x R2 results in wildtype F1 worms. Selfing these F1 (the F1 from the R1 x R2 cross) will produce the following F2 phenotypic ratio:
Group of answer choices
all wild-type : none roller
9 wild-type : 7 roller
13 wild-type : 3 roller
7 wild-type : 9 roller
3 wild-type : 1 roller
Homework Answers
when R1 is crossed with R2 the progenies are wildtype so mutations in R1 and R2 are in different genes,
let the genes be A and B.
the genotype of R1 worms is aaBB and genotype of R2 worms is AAbb
R1 * R2
aaBB * AAbb
AaBb (F1)
F1 is wildtype it has normal alleles of both genes.
AaBb * AaBb
| AB | Ab | aB | ab | |
| AB | AABB ( wildtype) | AABb( wildtype) | AaBB( wildtype) | AaBb( wildtype) |
| Ab | AABb( wildtype) | AAbb (roller) | AaBb( wildtype) | Aabb(roller) |
| aB | AaBB( wildtype) | AaBb( wildtype) | aaBB(roller) | aaBb (roller) |
| ab | AaBb( wildtype) | Aabb(roller) | aaBb(roller) | aabb (roller) |
wildtype: roller=9:7
so the answer is b) 9 wild-type : 7 roller
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