Select a Community
Are you sure you want to trigger topic in your Anconeus AI algorithm?
You are done for today with this topic.
Would you like to start learning session with this topic items scheduled for future?
1/200
18%
51/282
199/200
5%
14/282
1/100
14%
40/282
1/400
50%
141/282
99/100
7%
20/282
Select Answer to see Preferred Response
The probability of having a child with red-green color blindness is 1/400. Since males only have one sex X chromosome, the frequency of the mutant allele in the general population is 1/200, and the frequency of the normal allele is 199/200 (p + q = 1). Since 199/200 is close to 1, we will use 1 to simplify the calculations. Therefore, the carrier frequency (2pq) is 2 x 1/200 x 1 = 1/100. The frequency of the normal individuals is (199/200)^2 which is ~99/100. The father is phenotypically normal; thus, he must not have the mutant allele, and in order for the couple to have a child with the disorder, the mother must be a carrier and the child born must be a male. The probability that the child will be a male is 1/2, the mother must be a carrier (1/100), and there is a 1/2 chance the male child will get the mutant allele. The probability of having a child with the disorder is 1/2 x 1/100 x 1/2 = 1/400. Incorrect Answers: Answer 1: 1/200 is the frequency of the mutant allele (q). Answer 2: 199/200 is the frequency of the normal allele (p). Answer 3: The carrier frequency (2pq) can be calculated as 2 x 1/200 x 1 = 1/100. Answer 5: 99/100 is the frequency of genotypically normal individuals (p^2).
3.5
(18)
Please Login to add comment