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fodog414 fodog414
wrote...
Posts: 279
10 years ago
Hey everyone, i'm confused on row III on how 2 and 3 are not heterozygous carriers, does it mean that since you don't know what the father is who is more then likely to be RR homozygous due to it being a rare inheritance, do we have to conclude that 2, and 3 are R_ for 4 i'm not sure i guess row II 3 and 4 could be incorrect.
thanks in advance
fodog
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wrote...
10 years ago
Hi fodog, can you update us on this? What was the answer?
Biology - The only science where multiplication and division mean the same thing.
fodog414 Author
wrote...
10 years ago
Alright yeah I figured it out...

So you can figure the genotype of an individual in a pedigree based on two types of information: the individual’s phenotype (affected or unaffected) and the phenotypes of his or her parents and/or children.

the pedigree shown is an autosomal recessive condition

Affected individuals are always homozygous recessive.

Unaffected children of an affected parent are always carriers (heterozygous).

Both parents of affected individuals must have at least one recessive allele.

If both parents are carriers, their unaffected children may be carriers or homozygous for the dominant, wild-type allele.

For rare conditions, you can assume that individuals marrying into a family do not carry the recessive allele if there is no evidence that they are carriers.


so for row II 2 we don't know what dad is which we have to conclude row III 2, and 3 are R_, R_

while in row II 3 and 4 we do know the genotype but she has half a chance of being a carrier so for row III 4 her genotype is as well R_

cheers
fodog
wrote...
10 years ago
Thanks man, highly appreciated.
Biology - The only science where multiplication and division mean the same thing.
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