What is the probability that mary and justin will have an affected child?

The correct answer would be 1/32

Explanation:

Mary and Justin can have affected child only when both of them are carriers that is, both of them should be Gg.

The probability of Justin's father to be a carrier is 2/4 or 1/2 [1 GG: 2Gg: 1gg].

Similarly, the probability of Jean to be a carrier would be 1/2.

Thus, the probability of Justin and his father to be a carrier would be equal to 1/2*1/2 ⇒ 1/4.

The probability of Mary to be a carrier = 1/2.

So, the probability of Mary and Justin to be the carrier of the disease would be 1/2*1/4 = 1/8.

Now, the probability of having an affected child would be 1/4 as it is a recessive disorder.

So, the probability of Mary and Justin to have an affected child = 1/4 * 1/8 ⇒ 1/32

The probability of Justin and Mary having an affected child will be 1/16.

Further Explanation:

The genotype of the Justin grandparents will be Gg ×Gg

The expected probability of their children will be 1/4GG, 1/2Gg, and 1/4gg.

So, the chance of Justin’s father getting a heterozygote is fifty percent or1/2 Gg. The possibility of having an affected child increases if Justin’s father has a Gg genotype.

So, Justin’s father is a heterozygote.  

Justin’s parents’ genotype will be: GG × Gg

The probability of their children will be: 1/2GG, 1/2Gg

The genotype of Mary’s parent will be: Gg × Gg

The probability of their parent will be: 1/4GG, 1/2Gg, and 1/4gg

There is a maximum probability of having an affected child if both Justin and Mary have heterozygote genotype.

Justin × Mary

                                                         Gg   × Gg

The probability of their children will be: ¼ GG, ½ Gg, ¼ gg

The probability that Mary is affected is fifty percent (Gg). Similarly, the probability of Justin is affected by fifty percent (Gg).

Therefore, the total probability of Justin and Mary have the affected children will be 1/16.

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Answer Details:

Grade: High School

Subjects: Biology

Chapter: Human genetic

Keywords:

Probability, Justin, Mary, affect, child, genotype, heterozygote, maximum, grandfather, parents.

answer: 21

step-by-step explanation:

divide 630 by 30. 30 goes into 630 21 times

they fuse to form they have half as many chromosomes as a somatic cell of the same individual. b