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2/3 3/4 + 1/4 equals Probability of child carrying RR equals to 1/4 Probability of child carrying Rr equals to 1/2 1/4 Probab
R RR Rr


please answer question 1 and 2 that go together


What is the chance that the child of parents, who are both heterozygous for the Rand Talleles, carries at least one dominant
Required information Probability is the likelihood that an event will occur. If there is no chance of an event occurring the
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Answer #1

1) Aug = R : r | RRRr r Rr rr » Probability of RR - » Probability of Rre = + - 4 2 a probability of R. Ett (means R can be ho2) Aus- 1. R r 7 Probability of t = trung IRR/ Rr) rrrrr . T t τ τττε ttttt > Probability of at least one R homozygous trec

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