Alkaptonuria is a recessive trait. If tow parents are carriers(heterozygous, unaffected), what is the chance they would have:

A. 2 children, both with alkaptonuria?

B. 3 children, all 3 not having alkaptonuria?

C. 5 children, the first 3 not having alkaptonuria and the last 2 having alkaptonuria?

D. % children, 3 don't have Alkaptonuria, and 2 do have alkaptonuria?

Using the Punnett square, 1/4 children will have the disorder.

If the events are independent, the probability of both/all events occurring is determined by multiplying the probabilities of the individual events.

A. 1/4 * 1/4 = 1/16

B. (1-1/4)^3 = (1-1/4)(1-1/4)(1-1/4) = ?

Use similar method for the other problems.

https://www.google.com/search?client=safari&rls=en&q=punnett+square&ie=UTF-8&oe=UTF-8

To answer these questions, we need to understand how the inheritance of a recessive trait like alkaptonuria works. In this case, both parents are carriers (heterozygous) and unaffected. Let's analyze each scenario step by step:

A. To have two children with alkaptonuria, both parents must contribute the recessive allele (a) to each child. The chance of this happening is 25% for each child. Since we are looking for the probability of having two children with alkaptonuria, we multiply the probabilities together: 0.25 * 0.25 = 0.0625, or 6.25%.

B. To have three children without alkaptonuria, each child must receive at least one dominant allele (A) from either parent. The probability of this happening is 75% for each child. Since we want all three children to be unaffected, we multiply the probabilities together: 0.75 * 0.75 * 0.75 = 0.421875, or 42.19%.

C. To have the first three children without alkaptonuria and the last two with alkaptonuria, we need to calculate the probability for each case separately and then multiply them together.

For the first three children not having alkaptonuria, each child must receive at least one dominant allele (A) from either parent (probability 75%). Since we want all three children to be unaffected, we multiply the probabilities together: 0.75 * 0.75 * 0.75 = 0.421875, or 42.19%.

For the last two children to have alkaptonuria, both parents must contribute the recessive allele (a) to each child (probability 25%). Since we want both children to be affected, we multiply the probabilities together: 0.25 * 0.25 = 0.0625, or 6.25%.

Now we multiply the probabilities for the two cases together: 0.421875 * 0.0625 = 0.0263671875, or 2.64%.

D. The probability of having three children without alkaptonuria is the same as in scenario B (42.19%). The probability of having two children with alkaptonuria is the same as in scenario A (6.25%).

To calculate the overall probability of having 3 children unaffected and 2 affected, we multiply the probabilities together: 0.421875 * 0.421875 * 0.421875 * 0.0625 * 0.0625 = 0.003233909, or 0.3234%.