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Colorblindness problem #81 Destroyer

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Awuah29

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Guys, why are 50% of the daughters colorblind. I came up with 25%. Maybe I am missing something here. :confused:

Let see C= normal vision
c =colorblind

using the punnett square

Xc Y
XC XC Xc XC Y

Xc Xc Xc Xc Y

First girl carrier = normal 25%
Second girl = colorblind 25%
 

slashnroses19

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you did the punnett square correctly but you didn't interpret the results correctly. one daughter is color blind and the other one is a carrier. So, half the daughters are colorblind. The answer would be 25% if they asked the probability of their first CHILD being a colorblind daughter. check out the other punnett square problems, you'll see what i mean.
 

Awuah29

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you did the punnett square correctly but you didn't interpret the results correctly. one daughter is color blind and the other one is a carrier. So, half the daughters are colorblind. The answer would be 25% if they asked the probability of their first CHILD being a colorblind daughter. check out the other punnett square problems, you'll see what i mean.

wait a minute, why is half the daughters colorblind . One is carrier Cc (normal)
and the other one is cc colorblind . I understand with the probability scenario, but how do we get half of the daughters colorblind.


ok, gonna give jim and nancy a call:love:
 

Streetwolf

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Because the other 2 children in your square are SONS and the question asks about DAUGHTERS!

It wants to know the % of colorblind DAUGHTERS not children. So you only look at the two daughters. Of the two daughters, one is colorblind. 1 out of 2 = 50%.
 

dantheman2007

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one daughter is colorblind, one is a carrier
so the probability of a daughter being color blind is one half or 50%
that is all there is to it.
 

markimark

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50% of the DAUGHTERS will be colorblind; that is, 25% of ALL THE KIDS will be colorblind.

watch out for the wording and what you're being asked.
 
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