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In a dihybrid cross TtRr × TtRr in pea, the **probability** of producing a homozygous recessive (ttrr) offspring is:
A1/4
B1/8
C1/16
D9/16
Answer & Solution
Correct answer: C. 1/16
Tt × Tt yields tt with probability 1/4. Independently, Rr × Rr yields rr with probability 1/4. By independent assortment, P(ttrr) = (1/4)(1/4) = 1/16 — which matches the single '1' in the 9:3:3:1 ratio.
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