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
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?
1 answer
1 answer