AP Biologymediummcq1 pt
In a cross AaBbCc × AaBbCc (all unlinked), what fraction of offspring are AABBCC?
A.1/64
B.1/8
C.1/16
D.27/64
Explanation
Core Concept
Each gene's homozygous dominant probability is 1/4, and the three unlinked genes multiply: (1/4)³ = 1/64.
Correct Answer
A1/64
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