/sci/ - Science & Math » Thread #10908807

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Anonymous Thu 22 Aug 18:48:43 2019 No.10908807 View Reply Original Report

Quoted By: >>10908828 >>10908843 >>10908850 >>10909484 >>10909500 >>10911179

If I make a black clone of myself, and he tests his IQ to be 85, does that mean mine is 105?

Anonymous

Anonymous Thu 22 Aug 2019 18:53:05 No.10908819 Report

Quoted By:

>102
ftfy

Anonymous

Anonymous Thu 22 Aug 2019 18:54:54 No.10908828 Report

Quoted By:

>>10908807
No, it means that you're even dumber than your black self

Anonymous

Anonymous Thu 22 Aug 2019 19:01:39 No.10908843 Report

Quoted By:

>>10908807
no, 58

Anonymous

Anonymous Thu 22 Aug 2019 19:04:38 No.10908850 Report

Quoted By:

>>10908807
if you can make a black clone of yourself and it tests 85 IQ then your IQ is also 85 because you are also black

Anonymous

Anonymous Thu 22 Aug 2019 21:48:04 No.10909130 Report

Quoted By:

>another IQ thread

Anonymous

Anonymous Thu 22 Aug 2019 21:52:22 No.10909138 Report

Quoted By:

If you made a black clone and he was able to score 85, then I would call you a liar because your iq is clearly not high enough to bot just make a clone of yourself, but especially to make a black one.

Anonymous

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Anonymous Thu 22 Aug 2019 21:53:28 No.10909140 Report

Quoted By:

>another IQ thread

Anonymous

Anonymous Fri 23 Aug 2019 01:34:29 No.10909484 Report

Quoted By: >>10909497 >>10911630

>>10908807
a clone by definition has the same genes. If that wasn't the case, it would be wrong to call it a "clone". If IQ is purely hereditary, the clone would have your IQ. If IQ is at least partially environmental, it may differ to some degree.

Anonymous

Anonymous Fri 23 Aug 2019 01:37:30 No.10909497 Report

Quoted By: >>10909502 >>10909521

>>10909484
With our modern understanding of epigenetics it is not only possible but also extremely likely a clone version will be a lesser version. It's an issue they are finding with in vitro fertilization where the children are significantly more likely to suffer from things like anxiety, obesity, and so on due to the procedure inducing an extreme stress to the cell that turns on and off certain genes. Interesting thought.

Anonymous

Anonymous Fri 23 Aug 2019 01:38:34 No.10909500 Report

Quoted By:

>>10908807
It really depends on when you clamp his umbilical cord, etc.

Anonymous

Anonymous Fri 23 Aug 2019 01:39:16 No.10909502 Report

Quoted By: >>10909512

>>10909497
could I get a link to a paper for that? genuinely curious about dev bio with regard to IVF and had not heard of this

Anonymous

Anonymous Fri 23 Aug 2019 01:41:53 No.10909512 Report

Quoted By: >>10909516

>>10909502
I'm afraid I dont. I saw it in a documentary about the rise of our understanding the importance of epigenetics. Probably not the most reliable source but there you have it.

Anonymous

Anonymous Fri 23 Aug 2019 01:43:01 No.10909516 Report

Quoted By:

>>10909512
All good this is what google scholar is for I’ll do the leg work myself, thank you for the info

Anonymous

Anonymous Fri 23 Aug 2019 01:46:02 No.10909521 Report

Quoted By: >>10909534

>>10909497
You got me there but epigenetics probably won't turn a person black. Some variations is always there.

Anonymous

Anonymous Fri 23 Aug 2019 01:49:44 No.10909534 Report

Quoted By: >>10909550

>>10909521
Interesting question (I dont know much about the genes that code for skin color). IS it possible to make a clone that is an exact replica of yourself, EXCEPT for skin color? Maybe certain skeletal features are still off and all that, but is skin color so exclusive of a trait that it can be done? Or is it inherent to too many racial traits that you are forced to end up with a slightly different skull shape, and so on?

Anonymous

Anonymous Fri 23 Aug 2019 01:55:55 No.10909550 Report

Quoted By:

>>10909534
Considering that even though radiation causes skin to tan in most subjects, it does not cause permanent change of the original skin tone in humans. Without radiation, people's skin color will always revert back to said baseline which is apparently determined by melanocyte-inhibiting hormones. Also acknowleding that severe radiation or sunburns does not alter skin color either, it is safe to assume that epigenetical effects only havy very little influence on skin color.

Anonymous

Anonymous Fri 23 Aug 2019 20:09:21 No.10911179 Report

Quoted By:

>>10908807
no but you just have proven that your are without a doubt sub 85 IQ.

Anonymous

Anonymous Fri 23 Aug 2019 22:37:16 No.10911630 Report

Quoted By:

>>10909484
all right, black "version", how about that

Anonymous

Anonymous Sat 24 Aug 2019 15:22:04 No.10913312 Report

Quoted By:

$ \delta \, \epsilon \left ( 0,1 \right ) \\ \prod_{k=0}^{n} \left ( 1 + \delta ^{2^{k}} \right ) = (1+ \delta)(1+ \delta ^{2})(1+ \delta ^{4}) \cdots (1+ \delta ^{2^{n-1}})(1+ \delta ^{2^{n}}) \\ (1 - \delta)\prod_{k=0}^{n} \left ( 1+ \delta ^{2^{k}} \right ) = (1- \delta)(1+ \delta)(1+ \delta ^{2})(1+ \delta ^{4}) \cdots (1+ \delta ^{2^{n-1}})(1+ \delta ^{2^{n}}) \\ = (1- \delta ^{2})(1+ \delta ^{2})(1+ \delta ^{4}) \cdots (1+ \delta ^{2^{n-1}})(1+ \delta ^{2^{n}}) \\ = (1- \delta ^{4})(1+ \delta ^{4}) \cdots (1+ \delta ^{2^{n-1}})(1+ \delta ^{2^{n}}) = (1- \delta ^{2^{n}})(1+ \delta ^{2^{n}}) = 1- \delta ^{2^{n+1}} \\ \\ \boxed{(\delta ^{2^n})^2 = \delta ^{2 \cdot 2^n}=\delta^{2^{n+1}}} \\ \prod_{k=0}^{n} \left ( 1+ \delta ^{2^{k}} \right ) = \dfrac{1- \delta^{2^{n+1}}}{1- \delta} \\ \lim_{n \to \infty} \prod_{k=0}^{n} \left (1+ \delta ^{2^{k}} \right ) = \lim_{n \to \infty} \dfrac{1- \delta^{2^{n+1}}}{1- \delta} = \dfrac{1}{1- \delta} $

Anonymous

Anonymous Sat 24 Aug 2019 15:23:43 No.10913317 Report

Quoted By:

$\displaystyle \delta \, \epsilon \left ( 0,1 \right ) \\ \prod_{k=0}^{n} \left ( 1 + \delta ^{2^{k}} \right ) = (1+ \delta)(1+ \delta ^{2})(1+ \delta ^{4}) \cdots (1+ \delta ^{2^{n-1}})(1+ \delta ^{2^{n}}) \\ (1 - \delta)\prod_{k=0}^{n} \left ( 1+ \delta ^{2^{k}} \right ) = (1- \delta)(1+ \delta)(1+ \delta ^{2})(1+ \delta ^{4}) \cdots (1+ \delta ^{2^{n-1}})(1+ \delta ^{2^{n}}) \\ = (1- \delta ^{2})(1+ \delta ^{2})(1+ \delta ^{4}) \cdots (1+ \delta ^{2^{n-1}})(1+ \delta ^{2^{n}}) \\ = (1- \delta ^{4})(1+ \delta ^{4}) \cdots (1+ \delta ^{2^{n-1}})(1+ \delta ^{2^{n}}) = (1- \delta ^{2^{n}})(1+ \delta ^{2^{n}}) = 1- \delta ^{2^{n+1}} \\ \\ \boxed{(\delta ^{2^n})^2 = \delta ^{2 \cdot 2^n}=\delta^{2^{n+1}}} \\ \prod_{k=0}^{n} \left ( 1+ \delta ^{2^{k}} \right ) = \dfrac{1- \delta^{2^{n+1}}}{1- \delta} \\ \lim_{n \to \infty} \prod_{k=0}^{n} \left (1+ \delta ^{2^{k}} \right ) = \lim_{n \to \infty} \dfrac{1- \delta^{2^{n+1}}}{1- \delta} = \dfrac{1}{1- \delta} $

Anonymous

Anonymous Sat 24 Aug 2019 15:26:19 No.10913318 Report

Quoted By:

$ \delta \, \epsilon \left ( 0,1 \right ) \\ \displaystyle \prod_{k=0}^{n} \left ( 1 + \delta ^{2^{k}} \right ) = (1+ \delta)(1+ \delta ^{2})(1+ \delta ^{4}) \cdots (1+ \delta ^{2^{n-1}})(1+ \delta ^{2^{n}}) \\ (1 - \delta)\prod_{k=0}^{n} \left ( 1+ \delta ^{2^{k}} \right ) = (1- \delta)(1+ \delta)(1+ \delta ^{2})(1+ \delta ^{4}) \cdots (1+ \delta ^{2^{n-1}})(1+ \delta ^{2^{n}}) \\ = (1- \delta ^{2})(1+ \delta ^{2})(1+ \delta ^{4}) \cdots (1+ \delta ^{2^{n-1}})(1+ \delta ^{2^{n}}) \\ = (1- \delta ^{4})(1+ \delta ^{4}) \cdots (1+ \delta ^{2^{n-1}})(1+ \delta ^{2^{n}}) = (1- \delta ^{2^{n}})(1+ \delta ^{2^{n}}) = 1- \delta ^{2^{n+1}} \\ \\ \boxed{(\delta ^{2^n})^2 = \delta ^{2 \cdot 2^n}=\delta^{2^{n+1}}} \\ \displaystyle \prod_{k=0}^{n} \left ( 1+ \delta ^{2^{k}} \right ) = \dfrac{1- \delta^{2^{n+1}}}{1- \delta} \\ \displaystyle \lim_{n \to \infty} \prod_{k=0}^{n} \left (1+ \delta ^{2^{k}} \right ) = \lim_{n \to \infty} \dfrac{1- \delta^{2^{n+1}}}{1- \delta} = \dfrac{1}{1- \delta} $

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