5 PTS! Which type of tissue suspends the intestines within the abdominal cavity?
2 answers:
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The genealogy tree which is missing from the question is in the attachment.
Answer: 1) P1 = ; 2) P2 =
; 3)P3 =
Explanation: 1) According to the genealogy tree, the individual III-3 has a P1 = 1/2, because each of the parents are heterozygous for the condition, since one of the ofspring is affected.
2) For individual III-4, one of the parents is heterozygous for the condition, so the probability of being a carrier is P2 = 1/2.
3) Now, for individual IV-1 be affected, the parents has to be carrier. So, this probability, depends on the probability of each parent being heterozygous and that the individual will be carrying the condition, which means:
P3 = 1/2 · 1/2 · 1/4 = 1/16.
The probability of having the condition is 1/16 or 6.25%.
