Q46 - Epistatic Coat Color in Rodents
Refer to the question paper for figures and tables.
A. If a true-breeding brown rodent (SSBB) is crossed with a completely recessive rodent (ssbb), what is the F2 phenotypic ratio?
9 brown : 3 black : 4 pale: F1 from SSBB x ssbb = SsBb (all brown). F2 from SsBb selfing: 9 S_B_ (brown), 3 S_bb (black), 3 ssB_ (pale), 1 ssbb (pale). The ss genotype is epistatic to B/b (no pigment regardless of B allele), so 3 ssB_ + 1 ssbb = 4 pale. Final ratio: 9 brown : 3 black : 4 pale.
Bi. Cross (i): 7 brown : 7 pale offspring from brown female x ssbb male. Maternal genotype?
SsBB: The 1:1 ratio of brown:pale indicates segregation at the S locus only (Ss x ss gives 1/2 Ss : 1/2 ss). All pigmented offspring are brown (no black), meaning the mother must be BB (all pigmented offspring get at least one B). Maternal genotype: SsBB.
Bii. Cross (ii): 8 brown : 9 black offspring from brown female x ssbb male. Maternal genotype?
SSBb: All offspring are pigmented (no pale), meaning the mother is SS (all offspring get S). The 1:1 brown:black ratio indicates Bb x bb segregation. Maternal genotype: SSBb.
Biii. Cross (iii): 5 brown : 6 black : 12 pale offspring from brown female x ssbb male. Maternal genotype?
SsBb: The overall ratio approximates 1 brown : 1 black : 2 pale. Mother is Ss (half pigmented, half pale) and Bb (among pigmented, half brown, half black). Ss x ss gives 1/2 S_ : 1/2 ss. Among the S_ offspring, Bb x bb gives 1/2 brown : 1/2 black. So 1/4 brown : 1/4 black : 1/2 pale. Maternal genotype: SsBb.