Question

As you may remember from basic biology, the human A/B/O blood type system is controlled by...

As you may remember from basic biology, the human A/B/O blood type system is controlled by one gene for which 3 variants (“alleles”) are common in the human population – unsurprisingly called A, B, and O. As with most genes, everyone has 2 copies of this gene, one inherited from the mother and the other from the father, and everyone passes a randomly selected copy to each of their children (probability 1/2 for each copy, independently for each child). Focusing only on A and O, people with AA or AO gene pairs have type A blood; those with OO have type O blood. (A is “dominant”, O is “recessive”.)

Suppose Apple and both of her parents have type A blood, but her sister Olive has type O. Give exact answers as simplified fractions and provide a 1-2 sentence to explain your reasoning for each of them.

(a) What is the probability that Apple carries an O gene?

(b) Apple marries Oscar, who has type O blood. What is the probability that their first child will have type O blood?

(c) If Apple and Oscar’s first child had type A blood, what is the probability that Apple carries an O gene?

(d) If Apple and Oscar’s first child had type A blood, what is the probability that their second child will as well?

(e) Are the blood types of Apple and Oscar’s first two children independent? Justify your answer

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Answer #1

Part a)

Apple has A blood type , this means either it has gene pair AA or AO

Thus there is one pair that contains O gene out of two possible pairs

P(O gene) = ½

.

Part b)
Apple marries Oscar:

Possible gene pairs for Apple : AA or AO

Possible pairs for Oscar : OO

So in total in the three pairs there are three A genes and three O genes

Probability that baby has blood type O = 3/6 = ½

.

Part c)

If the child has blood type A : P(child A) = ½

Then the mother either has AA or AO pair

3 A genes and 1 O gene

                P(A contains O when child has A) = ¼

Part d)

The probability of second child is independent of first child

P(second child A) = P(first child A) = ½

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