Female iii-4 is pregnant via male iii-5. the owner of this breeding pair wants to know the probabilities of several possible outcomes for their offspring (iv-3). answer each question by dragging the correct label to the appropriate location. labels can be used once, more than once, or not at all.

Question

16-04-2018
Biology

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Female iii-4 is pregnant via male iii-5. the owner of this breeding pair wants to know the probabilities of several possible outcomes for their offspring (iv-3). answer each question by dragging the correct label to the appropriate location. labels can be used once, more than once, or not at all.

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JABARBS JABARBS · Answer 1 · 2018-04-26T03:20:08+02:00

We can say that female 111-4 is being pregnant to a male 111-5. The breeding owner of this pairs wants to know various outcomes for their iv-3(offsprings). Let's say we have condition A and condition B. The probability that iv-3 will have condition A is one out of thirty-six (1/36).
The probability that iv-3 will have condition B is one out of eight (1/8).Then the probability that iv-3 will have both conditions A and B is 1/288.We have 245/288 that iv-3 will have neither probability at all. We can also say the probability that iv-3 will have either one of the two conditions is 43/288.

avantikar avantikar · Answer 2 · 2019-09-16T14:49:57+02:00

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 https://brainly.com/question/334927

2. Learn more about base pairing https://brainly.com/question/2491455

3. Learn more about RNA base pairing https://brainly.com/question/2416343

Answer Details:

Grade: High School

Subject: Biology

Chapter: Genetics

Keywords:

Genotype, probability, product rule, condition.