Can you be 1/12th Cherokee?

Question

I was watching an old Daily Show clip and someone self-identified as "one twelfth Cherokee". It sounded peculiar, as people usually state they're "$1/16$th", or generally $1/2^n, n \in \mathbb{N}$.

Obviously you could also be some summation of same to achieve $3/32$nds, etc., but will an irreducible fraction with the numerator $1$ always need a power-of-two denominator? More generally does it always require a power-of-two denominator?

Assumptions (as per comments):

1. Nobody can trace their lineage infinitely
2. Incest is OK, including transgenerational, but someone can't be their own parent.

Answer 1

This depends on the model. Instead of arguing that we have only $46$ chromosomes and cross-overs or whatever the mechanism is called are not that common, let us assume a continuous model. That is, a priori, everybody can be $\alpha$ Cherokee for any $\alpha\in[0,1]$ and the rules are as follows

• Everybody has exactly two parents.
• If the parents have Cherokee coefficients $\alpha_m, \alpha_f$, then the child has $\alpha=\frac{\alpha_m+\alpha_f}2$
• In a sufficiently large but finite number of generations ago, people had $\alpha\in\{0,1\}$

It follows by induction, that $\alpha$ can always be expressed as $\alpha=\frac{k}{2^n}$ with $k,n\in\mathbb N_0$ and $0\le k\le 2^n$. Consequently $\alpha=\frac1{12}$ is not possible exactly (though for example $\frac{85}{1024}\approx\frac1{12}$ would be possible). It doesn't matter if there is any type of inbreeding taking place anywhere in the tree (or then not-tree) of ancestors. The only way to obtain $\alpha$ not of this form would involve time-travel and genealogical paradoxes: If you travel back in time and paradoxically become your own grandparent and one of the other three grandparents is $\frac14$-Cherokee and the others are $0$-Cherokee, you end up as a solution to $\alpha=\frac{\frac14+\alpha}4$, i.e. $\alpha=\frac1{12}$.

Answer 2

Answer:

Base 2 numbers! If someone is $a$ of race A ($a\in[0,1]$) and someone else is $b\in[0,1]$ race A, their offspring is $\frac{1}{2}(a+b)=\frac{1}{2}a+\frac{1}{2}b$.

This immediately screams "base 2" because you'll get this recursive half pattern.

So let's write something in base 2, take 101 this is 1/2+1/8 (there's a 0 in the 1/4 column) which is 5/8ths as binary, we halve it, which in base 2 is just shifting everything right once, to get 0101 and then add it with the other parent (after shifting theirs).

For example: 1x0 (where x means "have baby with") is 01+00=01=0.5.

The operation of x is closed with finite binary strings - which I will call "race strings".

Adding two terminating numbers is a terminating number.

That means this is closed. It's like the integers in the real numbers, using adding you cannot escape the integers, from inside them. Same sort of closure.

To be 1/3rd is not a finite race-string, so cannot have come from two finite race strings. QED.

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Here's how I got to the answer

It could be a fraction so close to 1/12 it's easier to say 1/12 than say "21/256ths"

Lets take "race A", if someone who is x/y race A and someone who is a/b of race A, their offspring is $0.5\frac{x}{y} + 0.5\frac{a}{b} = \frac{1}{2}(\frac{x}{y}+\frac{a}{b})$

But $\frac{x}{y}$ and $\frac{a}{b}$ must also come from this relation.

Now 1/3 (1/12 = 1/4 * 1/3) is a recurring number expressed in base 2 (in base 10 it goes 0.1s then 0.01s, then 0.001s and so forth, in base 2 it goes 1/2, 1/4, 1/8...)

So say we wanted someone who was 1/2 + 1/8 of race A, that'd be "101" in binary in this form, it terminates, the 3/4 used in the comments, that's "110" it terminates.

Remember the relation above, if someone is "0110" (3/8) and someone else is "1100" (3/4) say, we get the result by shifting one right and adding, in this case

 "00110"   
+"01100" which is "01001" or 9/32, 

So to be 1/12th would mean someone who was a quarter, and someone who is a third, but as you can see no one can be a third (in finite steps) starting from 1 or 0 of race A

To sum up! To be 1/3rd something (an infinite string of 0s and 1s) you can't have come from the "product" of two people who have finite strings representing their race. We've seen that "finite strings" are closed (2 people of finite-race string produce someone of finite race string) and thus can't have been produced by two people of finite strings.

Answer 3

No, nobody can be $1/12th$ Cherokee.

I'll prove a stronger statement:

For any natural $p$ and $q$, one can't be a $p/q$ Cherokee if $p$ is not divisible by $3$, and $q$ is divisible by $3$.

First, the statement that a person can be only "rational number of a Cherokee" can be easily proven - I won't waste space for the proof here.

Now, let's suppose a person was such $p/q$ Cherokee, then, supposing that his/hers parents were $p_1/q_1$ and $p_2/q_2$ Cherokee (fractions are reduced to their minimal form), the following would be valid

$$p/q = (p_1/q_1 + p_2/q_2)/2$$

or

$$2\times p\times q_1\times q_2=q\times (p_1\times q_2+p_2\times q_1)$$

Since $q$ is divisible by $3$, and $p$ is not divisible by $3$, this means that one of $q_1$ and $q_2$ must be divisible by $3$. In other words, one of the parents have the same property as the original person.

And than you can go ad infinitum - but since the oldest ancestor can be either 100% or 0% Cherokee, and that person does not satisfy the property of being "$p/q$ Cherokee if $p$ is not divisible by $3$, and q is divisible by $3$", this is a contradiction.

This means nobody can be $1/12th$ Cherokee.

Answer 4

There is more than one way to count genetics. Suppose that 120 years ago all four pairs of your mother's great grand parents (8 in all) were pregnant with their first child, and both pairs of your father's grandparents (4 in all) were pregnant with their first child.

Thus at that moment in time you have 12 ancestors, of mixed generations back, just entering child-bearing years.

If exactly one of those twelve ancestors is full-blooded Cherokee, and the other eleven have no Cherokee blood, then:

Yes, you are clearly 1/12th Cherokee.

Whether you are counting back in even generations or in even years is an arbitrary choice, and it is incorrect to state that only one makes sense. Perhaps there are other valid ways to count ancestry back that would enable other fractions and counting.

Answer 5

I believe that strictly from a statistical-genetic perspective, the most we can do is to compare genes and give a probabilistic value of somebody belonging to a certain genetic pool. Just consider mutations and any inter-race breeding that has happened during milleniums to highlight the fuzziness of the notion of defining somebody's race in any exact quantifiable terms. In addition, the notion of being exactly 1/2 Cherokee from statistical-genetic perspective does not make sense since the distribution of genes is not 50%:50% as very insightfully pointed out in the comments by @SteveJessop. There seems to be no mathematically sensical way to say somebody is 100% or 0% Cherokee, even less 1/n Cherokees for any choice of integer value n > 1, by just looking at their genetic markup.

Thus I would say the definition of being a proud member of Cherokee is a matter of social and personal identification. This definition is gray, but at least it gives us sample members of Human Race that we can meaningfully call fully Cherokee or totally not being a Cherokee (while ignoring the grey area cases), which validates the use of fractions in defining somebody's degree of "Cherokeeness".

After thus laying foundational definitions, we refer to the curious cases of children with three parents http://en.wikipedia.org/wiki/Three-parent_baby:

Three-parent babies are human offspring with three genetic parents, created through a specialized form of In vitro fertilisation in which the future baby's mitochondrial DNA comes from a third party.

See also http://www.bbc.com/news/magazine-28986843 for the young case of Mrs. Saarinen.

Such a child would have DNA from three people, and in a sense have three genetic parents. Let's assume one of the parents of such a child would be a Cherokee and two others "totally not Cherokees", and furthermore to make the case stronger, let's assume all three genetic parents would also participate in the upbringing and support of the child so that there really would be three parents in the strongest imaginable and possible sense. We could argue (though not with any mathematical rigor) that the resulting offspring would be 1/3 Cherokee. Now if she or he would produce an offspring, it would make some sense (even if not mathematically very rigorous) to say that offspring would be 1/6 Cherokee. Paring 1/6 Cherokee with non-Cherokee parent would give us an offspring that might want to call himself/herself 1/12 Cherokee.

Logically speaking, though, the mentioned case cannot be 1/12 Cherokee by this avenue because this controversial treatment option has not been in existence long enough for any such three parent child to have a grandchild.

Answer 6

You can get as close as you like to any fraction. For example, you can be between 1 / 12.5 and 1 / 11.5 Cherokee, and then claiming "I'm one 12th Cherokee" in every day language would be correct.

Also, the fractions that we usually use are approximations. If one parent is pure white and one parent is pure black, the child isn't half white and half black. It will be statistically close to 1/2 each, but in reality the genes are not taken exactly half from each parent, so that child will be a bit more in one of the two directions and not exactly in the middle.

Answer 7

It depends on how you define being Cherokee.

As others have shown, no possible normal breeding sequence can produce someone who is 1/12th.

However, we now have three-parent children (26 chromosomes from a male, 26 chromosomes from a female, an ovum with only the mitochondrial DNA from another female.) If one of those three parents is 1/4th Cherokee you could call the child 1/12th Cherokee.

Answer 8

Assuming that we define your heritage as 1/2 (father + mother), then it is not possible to come up with exactly 1/12. Assuming we start with people who are full Cherokee or not Cherokee at all, i.e. 0 or 1, then all their descendants are going to be some integer over a power of 2. A power of 2 can never reduce to 12, as 12 is a multiple of 3 and a power of 2 can never be a multiple of 3.

That said, someone could conceivably be a fraction that rounds to something close to 1/12. For example, suppose a Cherokee marries a non-Cherokee. The result is 1/2 Cherokee. This person marries a full Cherokee, result 3/4. Marries non, result 3/8. Marries full, result 11/16. Marries non, result 11/32. Marries non, result 11/64. Marries non, result 11/64. Marries non, result 11/128. That's pretty close to 1/12.

But THAT said, to come up with a number that is even approximately 1/12 you have to have detailed knowledge of your heritage with many inter-marriages going back at least 6 or 7 generations, and few people would have that much detail. So if you were (approximately) 1/12 Cherokee, I'd guess you wouldn't know it.

Years ago I helped develop a computer system for an Indian reservation, and they measured "bloodedness" in 8ths, which requires knowing your ancestry for 3 generations. They didn't cut it any finer than that.

Answer 9

Not without inbreeding. If we assume not shared ancestry paths, then each generation you go back you double the number of ancestors, thus 2^n where n is the generation. taking the previous 95 generations and dividing each by 12 reveals that there are no members of (this part) of the series that are divisible by 12. Interestingly, all moduli are alternately 4 or 8. While this is not a complete mathematical proof, beyond a few generations it is hard to determine race with much certainty anyway.

Answer 10

Genealogy and ancestry is fun. My sister did one of those DNA tests that places your geographic location genetically. Here are her results:

30% Great Britain, 29% Scandinavia, 20% Ireland, 8% Europe West, 6% Finland/NW Russia, 4% Europe East, 3% Iberian Peninsula.

Obviously, those don't fit nicely into the 1/(2^x) model. Part of this is because they are larger region (Scandinavia covers multiple countries for example). Another part of this is the fact that, genetically, these regions are similar enough to have some crossovers.

Answer 11

In a "continuous inheritance" model, no, because of the answers above.

In a naive chromosomal model, no, because 46 is not divisible by 12.

In a model which includes recombination, yes. Recombination (aka crossing over) between the parent's chromosomes results in a new chromosome type which has parts from one grandparent and parts from the other (usually one section from each). It is frequent and random enough that people by heritage 1/8th Cherokee would actually have a pretty big variation in their amount of genetic relatedness to their one 100% Cherokee g-grandparent. The diagrams here illustrate actual inheritance of one grandparent's genes: http://www.dnainheritance.kahikatea.net/autosomal.html - which shows that the variation in relatedness to a particular grandparent is huge.

The way real recombination works is actually pretty complex, though, and not purely random. Females recombine more than males (it's not just 50% of the time - females are ~45% something percent, males ~35%), and there are recombination hotspots which are locations which "prefer" to be the site of a cross somehow. Plus, there is lots of other selection going on even after fertilization. But these don't make it impossible to be 1/12, since they probably aren't monitoring the whole genotype.

On the actual "active gene" level, though, while you can do it, it may not mean much (since naively in terms of genes only, we are already extremely closely related to all other humans) It would make more sense to look for people who are 1/12 of the way genetically between two groups. i.e. take a Cherokee variant for 1/12 of all the genes which are not fixed in either Cherokee or the target group (being careful to spread them out). If you did the selection in a way which locally (within one chromosome) obeyed a plausible recombination history, this would be indistinguishable from somehow who just got lucky to be exactly 1/12 Cherokee.

Answer 12

Mathematically you can't. Let $a$ and $b$ mark the number of generations since the last time each of your parents had exactly one 100% Cherokee ancestor. Let's assume that

$\frac{\frac{1}{2^a} + \frac{1}{2^b}}{2} = \frac{1}{12}$

$\frac{1}{2^a} + \frac{1}{2^b} = \frac{1}{6}$

$\frac{6}{2^a} + \frac{6}{2^b} = 1$

From now on, just for the sake of simplicity, let's presume $a > b$.

$6 + 6*2^{a - b} = 2^a$

$3 + 3*2^{a - b} = 2^{a - 1}$

$2^{a - 1} - 3*2^{a - b} = 3$

This last equation is true if you manage to find two different powers of 2 (both even numbers), either one a multiple of 3 (it's still even), whose difference is an odd number. You won't find one, we have a contradiction.

But as stated, this is a mathematical statement. Genetics work quite different.