Answer:
1 in 16
Explanation:
When two heterozygotes (AaBb x AaBb) for two autosomal genes are crossed the expected probability for the offspring is 9 A-B-, 3 A-bb, 3 aaB- and 1 aabb. In other words, out of sixteen offspring, 9 are expected to be dominant on both genes (either homozygous AA or BB or heterozygous Aa or Bb), 3 are expected to be dominant on the A gene but recessive on the b gene, 3 are expected to be recessive on the a gene but dominant on the A gene, and only 1 is expected to be recessive on both genes aabb.