Lifesci 4 - lecture 7 - beadle and tatum. Beadle and tatum: one gene - one enzyme. The alternate alleles that give abnormal function of these genes are a, b, c, and d. a black aa bb cc dd is crossed with a colorless aa bb cc dd to give a black f1. The f1 is then backcrossed with a colorless aa bb cc dd. Assume that a, b, c, d act in a pathway as follows: Colorless colorless grey brown black: what percentage of the f2 are brown, 1/2, 1/4, 1/8, 1/16, none of the above. Aa bb cc dd x aa bb cc dd. From p( aa bb cc dd ) x aa bb cc dd. P(f2 = brown) = p(aa bb cc dd) = x x x = 1/16: what percentage of the f2 are colorless, 1/4, 3/8, 1/2, none of the above. Colorless if either enzyme a or b is mutated.