Colorblindness problem #81 Destroyer

[Awuah29]

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

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%

---

Members don't see this ad.

 

[slashnroses19]

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]

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

 

[Streetwolf]

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]

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]

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.