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given: 6 fingers is an x-linked trait and is recessive (5 fingers being dominant
question: if you have a father with 6 fingers and a mother that has 5 fingers, but carries the 6 finger trait, what is the chance that they will have a child with 6 fingers?

You can do this on a Punnett Square. just remember the normal male (paternal) is xy and maternal is xx so if you have a male with 6 fingers and you designate that as fy and the mother a fx then you can make the square and calculate the odds. For directions go to:
http://en.wikipedia.org/wiki/Punnett_square

i came up with a 3:1 ratio?
one child will have 5 fingers but will also be a carrier of the 6 finger trait (Ff). 3 will have the trait of 6 fingers because it will be recessive (ff). 3:1 becuase dad is recessive (ff = 6 fingers) and mom is dominant but a carrier of the recessive trait (Ff). am i right?

Close but not right. Please go here:
http://www.biology.arizona.edu/mendelian_genetics/problem_sets/sex_linked_inheritance/07t.html

Now on the punnet square just make the male also have a green line in the chromosome and then check your ratios. I'm making you go throught this so you understand.

so it'd be a 1:2:1 ratio right?
1 (1/2 blind): 2 (normal): 1 (1/2 carriers)

I think you have it? one female has it
one female is a carrier, one male has it, one male is normal.
But not blind in our problem. six fingers.

thanks dear ^_^ you're awesome!

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