4, and 5. Make a five-stage tree diagram of the outcomes (al- lergic or not alle

Question

4, and 5. Make a five-stage tree diagram of the outcomes (al- lergic or not allergic) for the 5 individuals, and use it to find the probability distribution of X. (We will see in Chapter 12 that X follows a binomial distribution.) 10.46 Mendelian inheritance. Some traits of plants and an- imals depend on inheritance of a single gene. This is called Mendelian inheritance, after Gregor Mendel (1822-1884). Each of us has an ABO blood type, which describes whether two characteristics called A and B are present. Every human being has two blood-type alleles (gene forms), one inherited from our mother and one from our father. Each of these alleles can be A, B, or O. Which two we inherit determines our blood

Explanation / Answer

(a)

We can construct the table below to calculate all combinations and probabilities.

So, blood types of their children are A with probability 0.25, B with probability 0.25 and AB with probability 0.25+0.25 = 0.5

(b)

We can construct the table below to calculate all combinations and probabilities.

So, blood types of their children are B with probability 0.25+0.25+0.25 = 0.75 and O with probability 0.25

(c)

We can construct the table below to calculate all combinations and probabilities.

So, blood types of their children are AB with probability 0.25, A with probability 0.25, B with probability 0.25 and O with probability 0.25

So, the probability that a child has a blood type of O is 0.25

As all children inherit independently from mother and father, the probability that all 3 children have a blood type of O = the probability that 1st child has a blood type of O * the probability that 2nd child has a blood type of O * the probability that 3rd child has a blood type of O = 0.25 * 0.25 * 0.25 = 0.015625

The proability that a child do not has a blood type of O = 1 - 0.25 = 0.75

As all children inherit independently from mother and father, the probability that first child has a blood type of O and other two do not = the probability that 1st child has a blood type of O * the probability that 2nd child do not has a blood type of O * the probability that 3rd child do not has a blood type of O

= 0.25 * 0.75 * 0.75 = 0.140625

Rachel Jonathan A (0.5) B (0.5) A (0.5) A (0.25) AB (0.25) B (0.5) AB (0.25) B (0.25)

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