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The blood type distribution in the US is as follows (according to this link):

O 45%
A 40%
B 11%
AB 4%

However the blood type is a result of the 2 alleles (in lower case) that a person gets from her parents.

aa -> A
oa -> A
ao -> A
oo -> O
ob -> B
bo -> B
bb -> B
ba -> AB
ab -> AB

How can I extract the distribution of the alleles {a,b,o} from the the known distribution of the blood types {O,A,B,AB} ?

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2 Answers 2

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The probability of the blood types can be defined in terms of the alleles: $$O = o^2$$ $$A=a^2+2oa$$ $$B=b^2+2ob$$ $$AB=2ab$$

These are 4 equations with 3 variables, and thus a solution is not guaranteed.

Solving for the first 3 equations we get: $$o=\sqrt{O}$$ $$a=\sqrt{A+O}-\sqrt{O}$$ $$b=\sqrt{B+O}-\sqrt{O}$$

We get that:

o=0.6708203932499369
a=0.25113405247935183
b=0.07751108410485141

Thus

AB=2ab=0.03893134532663842=3.9%

Which is pretty close to the number you described

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  • $\begingroup$ It would be nice to comment on where did the formulas come from. $\endgroup$
    – Tim
    Commented Feb 2, 2017 at 9:23
  • $\begingroup$ It comes from the Hardy-Weinberg principle $\endgroup$ Commented Jul 5, 2022 at 20:35
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EDIT: Wrong logic here, disregard this answer and focus on the accepted one.

Keeping the original one for reference of what no to do.


How about simply reversing the probability ? It sounds simple but it might be what you are looking for ?

if aa, oa and ao makes A and there are 40% of A, then the simplest assumption is that there are 40/3 % of each alleles pair, and so on

This gives, in %

aa -> 13.3 
oa -> 13.3
ao -> 13.3
oo -> 45
ob -> 3.7
bo -> 3.7
bb -> 3.7
ba -> 2
ab -> 2

Then, you sum the probability of the pairs that have an "a" pondered by their frequency in the pair (*0.5 in ao, *1.0 in aa for example)

Which gives, in %

a = 13.33 + 13.33 + 2  =  28.66
b = 3.66 + 3.66 + 2   =  9.33
o = 13.33 + 45 + 3.66 =  62.00
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  • $\begingroup$ I cannot agree with this idea as with do not have any information about the distribution of allele. Also allele distribution must satisfy Hardy-Weinberg equation. In this case $P(aa) = p^2 = P(a0) = p q$ which implies $P(bloodType = A|not B) = 2/3$ $\endgroup$ Commented Sep 7, 2020 at 8:38

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