Question

When solid red cattle (homozygous recessive for gene R) are crossed with Black-and-Red cattle (homozygous recessive for gene B) to produce Brindle (irregular narrow strips of black on top of a red background), the F1 progeny. The F1x F1 cross resulted in the following F2 progeny; 3615 Brindle (Black stripes on Red background) 1225 Black-and-Reds Patches 1565 Solid Reds 6405 = Total Based upon the above data, the pure breeding Black parent was of which genotype: O RRbb O rrBB O RRBB o none of these O rrbb

Answer 1

@6 rrBB (Pure breeding black and red patches should have genotyle gay BB.) Parent - bbRR X BBrr E BbRr. selfing. fra B_PR (3615) all *} abbroximate ratio B-ry: (1225) --- 31 - bbr. 20566) AJ 91314 Corecessive epistasis) bbro I Chi-square analysis Genotope observed (o) Expected cos Co-E % B-R- a 3615 -3602 8125 0.041 B-87 — 1225 1200.9375 0.482 bbR-} - 1565 1601, 25 0.821 bborf <=6405 1x2-1.8441 Since calculated chi-square value (x²= 1.344) is lower than tabular value (5.941) for 2 degree of freedom at b2005 Hence, we must accept our hypothests as population are under recessive epistasis accept.