Genetics and probability (the fun stuff?)

meh I would have picked B...

any help would be appreciated.

you sure C is the right answer? and the question is stated like that?

matth87 said:

meh I would have picked B...

any help would be appreciated.

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Punnet square gave me 25%. lol. And c makes no sense...it's a recessive trait.

pookiez88 said:

Punnet square gave me 25%. lol. Help!

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I didn't think it was feasible but you might actually be on this site more then me haha

If she has 2 girls

Alleles for girl 1) XX or XX'
Alleles for girl 2) XX or XX'

We only need to look at one of them right? I don't see why it wouldn't be 50%...

matth87 said:

I didn't think it was feasible but you might actually be on this site more then me haha

If she has 2 girls

Alleles for girl 1) XX or XX'
Alleles for girl 2) XX or XX'

We only need to look at one of them right? I don't see why it wouldn't be 50%...

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haha, that's quite possible! I got XcXc (colorblind), XXc (heterozygous carrier) XcY (colorblind), and XY (normal) so 1 affected female and 1 affected male. I found this genetics website which contains a similar problem:

Mate a female carrier with a color blind man

(1)For female offspring, ½ will be heterozygous normal, and ½ will be homozygous color blind

(2)For male off spring, ½ will be hemizygous normal, and ½ will be hemizygous color blind

(3)½ the offspring have normal vision, ½ are color blind

http://www.humboldt.edu/~zool110/Lectures/Lecture8.htm

The answer should be 25% out of all the children (to get a colorblind female). So, 50% chance either female is colorblind, 25% out of all the children. Did they add this to get 75% lol. I think the answer should be 25%, since the question is "what is the probability that at least one of the girls will be colourblind?" 1/2 chance of total kids being colorblind x 1/2 chance of them being girls=1/4=25%.

I don't see where you get 25%. You have two girls and two boys.
You just want to see how likely it is that one of the two girls is colorblind, so the boys do not matter at all in this calculation.

I thought the answer was an error and should be 50%.

The likelihood of the first one is 50%, while the likelihood of the second one is also 50%. Since they are not dependent on each other, you should not have to do anything to the numbers.

EDIT - pookie, I don't think you have to factor in the chance of them being girls, since the problem tells us there are already two girls.

Dr Gerrard said:

I don't see where you get 25%. You have two girls and two boys.
You just want to see how likely it is that one of the two girls is colorblind, so the boys do not matter at all in this calculation.

I thought the answer was an error and should be 50%.

The likelihood of the first one is 50%, while the likelihood of the second one is also 50%. Since they are not dependent on each other, you should not have to do anything to the numbers.

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Don't you have to account for the probability of them actually having a girl? Or is that a given?

So the parents are x'x and x'y.
If they have one girl the probability is 50% x'x' and 50% x'x
So conceptually you know that if there is one kid and the odds of her being colorblind is 50%, if you have two kids the odds of at least of one them being colorblind is greater than 50%. (Just like you are more likely to get heads at least once if you flip a coin twice than if you flip it once.)

Another way to look at it is that the odds of them both NOT being colorblind is 1/2 * 1/2 is 1/4. So the odds of at least one being colorblind is 1-1/4 = 0.75.

Ahh, I got it.

If the question stated, what are the odds of ONLY one of them being colorblind, then the answer would be 50%. But there is also a 25% chance that BOTH of them will be colorblind. So you add these two.

Dr Gerrard said:

Ahh, I got it.

If the question stated, what are the odds of ONLY one of them being colorblind, then the answer would be 50%. But there is also a 25% chance that BOTH of them will be colorblind. So you add these two.

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But the question says "what is the probability that at least one of the girls will be colourblind?"

Why are you accounting for both?

pookiez88 said:

But the question says "what is the probability that at least one of the girls will be colourblind?"

Why are you accounting for both?

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I hope this helps.
If those same parents have 1000 girls, what are the odds that at least one is colorblind?

wanderer100 said:

I hope this helps.
If those same parents have 1000 girls, what are the odds that at least one is colorblind?

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Ah, so do you consider half of that population/500 girls when solving the problem?

pookiez88 said:

Ah, so do you consider half of that population/500 girls when solving the problem?

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No, I just used that example to show that having a larger sample does affect the likelihood of specific outcomes.

wanderer100 said:

No, I just used that example to show that having a larger sample does affect the likelihood of specific outcomes.

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So because that's a large sample, the probability of at least 1 being colorblind is greatly increased? Is that what you were trying to get across?

ahhhhh so this would be an example of the rule of addition right?

Change for Girl 1 To be Color Blind - 50%
Chance for Girl 2 to be color Blind - 50%

Chance for Girl 1 or Girl 2 to be color blind

(1/2 + 1/2) - (1/2 x 1/2)

1 - 1/4 = 3/4?

matth87 said:

ahhhhh so this would be an example of the rule of addition right?

Change for Girl 1 To be Color Blind - 50%
Chance for Girl 2 to be color Blind - 50%

Chance for Girl 1 or Girl 2 to be color blind

(1/2 + 1/2) - (1/2 x 1/2)

1 - 1/4 = 3/4?

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YES! Omg thank you. Rule of addition-sum of either event occuring-product of them occuring/happening at the same time. Awesome.

pookiez88 said:

So because that's a large sample, the probability of at least 1 being colorblind is greatly increased? Is that what you were trying to get across?

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Yes.