Homework Answers
This is the case of autosomal recessive inheritance. Because the parents (2-3,4) are unaffected but their child have disease.
Now the 4-4 individual is homozygous recessive aa. Because there is found disease.
Now 4-5 individual should be Aa. Because the parent 10 is affected. This suggests that 'a' allele pass from this parent. Where as parent 11 is normal.
Now when mating occurs between 4-4 and 4-5 occurs, then child is unaffected in nature.
Hence the genotype of parents : aa (4-4) × Aa (4-5)
Punnet square:
| Gametes | A | a |
| a | Aa ( carrier) | aa ( disease) |
| a | Aa ( carrier) | aa ( disease) |
Now all of the unaffecte child are carrier in nature. That's why the percentage of carrier individual that is unaffected is 100
Right answer is A ( 100%)
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