Question 3 A haemophilic man had a phenotypically normal daughter. She marries a man who does not have haemophilia. If they have four sons, what would be the probability that all will be haemophilic? 1)Zero 2)1/2 3)1/16 (answer) 4)1/8See answer
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Question 3 A haemophilic man had a phenotypically normal daughter She marries a man who does not have haemophilia If they have four sons what would be the probability that all will be haemophilic 1…
Question
Question 3
A haemophilic man had a phenotypically normal daughter. She marries a man who does not have haemophilia. If they have four sons, what would be the probability that all will be haemophilic?
1)Zero
2)1/2
3)1/16 (answer)
4)1/8
A haemophilic man had a phenotypically normal daughter. She marries a man who does not have haemophilia. If they have four sons, what would be the probability that all will be haemophilic?
1)Zero
2)1/2
3)1/16 (answer)
4)1/8
Basic Answer
Step 1: Determine the mother’s genotype.
The father is haemophilic, meaning he has the genotype XhY (where Xh represents the X chromosome carrying the haemophilia allele). Since the daughter is phenotypically normal, she must have received a normal X chromosome (X) from her mother. Therefore, the mother’s genotype must be XHXh (carrier).
Step 2: Determine the possible genotypes of the sons.
The mother (XHXh) and the father (XY) can produce the following offspring genotypes:
- XHX: Normal daughter
- XhX: Carrier daughter
- XHY: Normal son
- XhY: Haemophilic son
Step 3: Calculate the probability of a haemophilic son.
The probability of having a haemophilic son (XhY) is 1/2.
Step 4: Calculate the probability of all four sons being haemophilic.
Since each son’s genotype is independent of the others, the probability of all four sons being haemophilic is (1/2) * (1/2) * (1/2) * (1/2) = 1/16.
Final Answer
1/16