The probability that IV-3 will have condition A is 1/36, will have the condition B is 1/8, will have both the conditions is 1/288, neither will have the condition is 245/288, and will have at least one of the conditions is 43/288.
Further Explanation:
For the first part, condition A to occur, II-5 genotype will be Aa and probability will be 2/3. In case, in case II-5 has the genotype Aa and probability will be 2/3. In case II-5 has the genotype Aa, then he passes an allele to III-4 and the probability will be 1/2.
In case III-4 has the genotype Aa (accounted for by the above probability), then she passes an allele to IV-3 (probability will be 1/2).
II-7 has the genotype Aa, the probability will be 2/3. In case II-7 has the genotype Aa, then he passes an allele to III-5 and probability will be 1/2. In case III-5 has the genotype Aa, then he passes an allele to IV-3, the probability will be 1/2.
Applying the product rule,
Similarly, IV-3 will have the condition B is 1/8, will have both the conditions is 1/288, neither will have the condition is 245/288, and will have at least one of the conditions is 43/288.
Learn more:
1. Learn more about component of DNA brainly.com/question/334927
2. Learn more about base pairing brainly.com/question/2491455
3. Learn more about RNA base pairing brainly.com/question/2416343
Answer Details:
Grade: High School
Subject: Biology
Chapter: Genetics
Keywords:
Genotype, probability, product rule, condition.