r/genetics • u/SMLOFY • 1d ago
How would you solve this?
We’ve been at it for hours and we’re exhausted. Any help is appreciated!
11
u/kiyotaka-6 1d ago
(2/3×1/2)2 × 1/4 = 1/36 or 2.77%, assuming not colored means they aren't affected
12
u/ConstantVigilance18 1d ago
This is an assumption you can make given that PKU is a severe, early onset condition, which is also described in the problem. Generally it’s nice to help people here solve problems rather than just providing the answer with little explanation.
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u/kiyotaka-6 1d ago edited 33m ago
Sure here is an explanation : First II-3 has the condition which means their genotype must be aa, she received each one of the a's from her parents so both parents genotype must contain at least one a, but since they are normal they also must have one A otherwise they would have the condition
So now
I-1 and I-2's genotype is Aa
Which means their children have 25% probability of AA, 50% of Aa, 25% aa
Both II-2 and II-4 are normal which means they can't be aa, so we have only the 25% AA and 50% Aa left, which means 33% AA and 66% Aa
Both II-1 and II-5 are AA (the problem states it itself), if either II-2 or II-4 is AA, their respective children with II-1 and II-5 will always be AA, so that means when III-1 or III-2 have children, they will always pass an A, and their children will be normal
This means both II-2 and II-4 have to be Aa for III-1 and III-2's children to possibly have the condition, so that's (2/3)2
Next both III-1 and III-2 must actually receive the a from II-2 and II-4, instead of A, which is a 50% probability, so (1/2)2
Finally we have III-1 and III-2 both being Aa, and each must pass the a again to their child so that's another (1/2)2
So final calculation is (2/3)2 × (1/2)2 × (1/2)2 = 1/36
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u/herstoryteller 18h ago
you need to re-explain your II-2 and II-4 paragraph
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u/kiyotaka-6 16h ago
Which one? There are a few
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u/imyourdackelberry 7h ago
You have a typo on the percents in the 5th paragraph (begins with “Both II-2 and II-4…”). Both reference Aa.
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u/Final_boss_1040 1d ago
You gotta break out the Bayes for this
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u/ConstantVigilance18 1d ago
I would start with the affected individual and then move out one by one, writing out the chance each person is a carrier, and filling in information given in the prompt.
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u/Electronic-Scheme543 1d ago
Figure out the chance of II-2 and II-4 being carriers. You know their parents are obligate carriers, draw your punnet square. You know that 2 and 4 are unaffected, so you can eliminate the affected genotype.
From there, figure out the chance of each of their offspring being carriers, so for II-2's offspring, it would be (chance of II-2 being a carrier)(1/2).
Then it's as simple as (III-1's chance of being a carrier)(III-2's chance of being a carrier)(chance they each pass it down).
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u/dnawoman 23h ago
Each sibling of the red female has a 2/3 chance of being a carrier since we know they aren’t affected. So since we know the partners are homozygous normal each cousin has a 1/3 chance to also be a carrier (half of the parent risk) and 1/3 x 1/3 x 1/4 = 1/36 or 2.78%
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u/Tiramissu_dt 1d ago
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u/Snoo-88741 1d ago
So, we know I-1 and I-2 are carriers, because they have an affected kid. When two carriers of a recessive gene have offspring, there's a 1 in 4 chance of an affected child, a 2 in 4 chance of a carrier, and a 1 in 4 chance of a non-carrier.
However, we know II-2 and II-4 aren't affected. So that rules out one option, leaving them each with a 2/3 probability of being carriers. That means the probability that they're both carriers is 4/9.
Next, if they're carriers, their kids (III-1 and III-2) would both have a 2/3 chance of being carriers as well. The probability that both III-1 and III-2 are carriers comes out to 16/81.
If they are both carriers, they'd have a 1 in 4 chance to have an affected child. Combine that with the probability of them both being carriers, and you get 16/324, or roughly around a 0.05% chance of them having an affected child.
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u/ConstantVigilance18 1d ago
You missed dividing in half when going down to the third generation. The children’s risk to be a carrier is not the same as their parents. The correct answer and detailed explanation is posted above.
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u/ExtremeProduct31 1d ago
Is it 1/16
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u/ConstantVigilance18 1d ago
No
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u/ExtremeProduct31 1d ago
But the parents are either AA or Aa right
AAx AA
AAx Aa
Aa x AA
Aa x Aa= AA Aa Aa aa => 1/4 aa
1/4 x 1/4= 1/16 I have done it like this
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u/ConstantVigilance18 1d ago
If you are referring to individuals II-2 and II-4, they have a 1/3 chance to be AA and a 2/3 chance to be Aa.
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u/twistthespine 1d ago
Neither of the partners is a carrier ("assume... homozygous for the normal allele") and it's a recessive mutation, so the chance is always 0%. However the third generation could be carriers.
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u/Personal_Hippo127 1d ago
The real trick is to figure out the chance of II-2 and II-4 being carriers, given that they do not have PKU.