anjushree
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a women with cystic fibrosis married her first cousin ,what is the risk of their children to have cystic fibrosis
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| Geroo
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1/16?
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| msyamp
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i go with geroo
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| sturge_weber
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well is 1/16 not the genes that they share
dont we then do the 2/2 table
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| anjushree
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please explain
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| msyamp
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let us say the childer are generation 1. thier motherX is affected homologous diseased.ok
let us say that MOTHER X'S motherP and FATHER Y'S fatherQ are son and daughter of sisterQ and brothterP respectivly BORN TO Mr.ansestor AND MS.ANCECTOR
NOW FATHER y RISK TO BE KNOWN. MOTHER X TO GET DISEASE
hey this is gettng complicatd. now i did it in a diagram see that.
http://img57.imageshack.us/my.php?image=presentat...
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| msyamp
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see carefully and you will understand. its a mind teaser
the percentages showthat chances of having that alleles
Edited by msyamp on 01/11/06 - 07:41 PM
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| anjushree
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THANKS UR FIG IS GREAT
WHAT I UNDERSTAND IS - AA Aa Aa AA
=1/4 = 1/2
aa Aa
aa
1/2
1/4*1/2*1/2=1/16
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| anjushree
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it is just come close,making everything puzzled
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| msyamp
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if you understand it thats ok. you got it?
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| Geroo
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this is how i think about it,1st cousin shares 1/8 of the genes.so there is 1/8 possibility the the cousin is a carrier.so when they mates there is 1/2 possiblity that the child is affected.so the total risk is 1/2 X 1/8 = 1/16.is that logic??
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| Geroo
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anjushree is 1/16 the right answer??
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| anjushree
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yes,1/16 is right answer
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| linaorvos
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Wow, good Q, I got the concept from Geroo, thank you!
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| bluntknife
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cystic fibrosis is autosomal recessive...so the women will have genotype say aa....if the guy is a carrier(A,a) there will be 1/2 chance of the child havin cystic fibrosis and if hes homozygous normal(A,A)...then none of the childen will have cystic fibrosis...i cant get it how the answer cud be 1/16th...am i dumb or wat
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| bluntknife
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ok i got the concept from geroo's explanation...but now im confused wat would be the probability if it werent a autosomal recessive disease...wat if we had a autosomal dominant condition and the mother had it and she married her first cousin...if she was heterozygous then going by geroo's explanation the chance would be 1/12 right????
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| Geroo
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1/12? how ?
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| bluntknife
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heteozygous female for autosomal dominant condition A,a
first chance of being heterozygous 1/8....if both are heterozygous then 2/3rd of children will be affected from the condition...one homozygous and other two heterozygous...multiply 2/3 by 1/8 and u get 1/12....right???
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| bluntknife
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ooopsssss sprry not 2/3, multiply by 3/4.........so that makes 3/32....
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| Geroo
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if they are both heterozygous then the chance of the child being affected is 3/4.(do the 2/2 table).
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| Geroo
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yes 
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| bluntknife
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yeah i made the table.....but do u think that will be the probability...ie 3/32...or does this 1/8 chance of sharing her genes with cousin doesnt work for autosomal dominant???
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| Geroo
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u know I think the 1/8 doesn't work because she might have inherited the gen from the other parent(i.e it's not garanteed that the grandfather has the gen).right?
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| bluntknife
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does that mean we cant calculate the probabilt for an autosomal dominant condition...unless specific genotype is given?
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| Geroo
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i think so
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