Help with math pls | calculating genetic prob of child from cousins with disorder

[Abdul Fattah]

Hi, in reference to a recent thread, I thought to calculate the chance of cousins having a child with a genetic defect based on the percentage of dormant disabilities within the populations. However I apparently made some mistakes, cause the results are nonsensical (ei: negative probability for a percentage of 100%)

Definitions:
*People without genetic disability: (AA)
*People with genetic disability: (BB)
*Percentage of people with dormant trait (AB) or (BA): x

Objective:
* Find function f(x) that gives the percentage of two cousins having a child with trait (BB)
* Find function g(x) that gives the percentage of two unrelated people having a child with trait (BB)
* Compare which of the two cases has the highest risk
Note1: In laws should be considered to have a chance of x/100 of having (AB)
Note2: The chance of people of the first generations having (BB) is not added, we assume that in such cases there will be no further marriages and children from that generation.

Calculating:
First generation:
* The chance that a person has (AB):
x/100
* The chance that a person doesn't has (AB):
1-(x/100) = (100/100)-(x/100) = (100-x)/100
* Scenario p1; the chance that both partners have (AB):
(x/100)^2 = x^2/10000
* Scenario p2; the chance that neither partner has (AB):
[(100-x)/100]^2 = (100-x)^2/10000
* Scenario p3; the chance that one of the two partners has (AB):
2(x/100)[(100-x)/100] = (100x-x^2)/5000

Second generation:
* Chance that a child of 2nd gen has (AB): (p1/2 + p2*0 + p3/4)/4
[(x^2/10000)/2 + 0 + ((100x-x^2)/5000)/4]/4 =
[(x^2)/20000 + (100x-x^2)/20000]/4 = (100x^3-x^4)/80000
* Chance that a child of sec gen doesn't have (AB): 1-[(p1/2 + p2*0 + p3/4)/3]
1 -[(100x^3-x^4)/80000] =
(80000-100x^3-x^4)/80000
* Chance that a partner of the 2nd gen has (AB)
x/100
* Chance that a partner of the 2nd gen doesn't have (AB)
(100-x)/100
* Scenario q1; the chance that both children of 2nd gen and their partners have (AB):
[[(100x^3-x^4)/80000]^2]*[(x/100)^2] =
[(100x^3-x^4)^2/(64*10^8)]*[x^2/10000] =
(10^4x^6-200x^7+x^4)(x^2)/(46*10^12) =
(10^4x^8-200x^9+x^6)/(46*10^12)
* Scenario q2; the chance that both children of 2nd gen and one of their partners have (AB):
[[(100x^3-x^4)/80000]^2]*[(100x-x^2)/5000] =
[(100x^3-x^4)^2/(64*10^8)]*[(100x-x^2)/5000] =
(10^4x^6-200x^7+x^4)(100x-x^2)/(64*10^12) =
(100x^5 -x^6 +10^6x^7 -3*10^4x^8 +200x^9)/(64*10^12)
* Scenario q3; the chance that both children of 2nd gen but none of their partners have (AB):
[[(100x^3-x^4)/60000]^2]*[(100-x)^2/10000] =
[(100x^3-x^4)^2/(64*10^8)]*[(100-x)^2/10000] =
(10^4x^6-200x^7+x^4)(10^4 -200x +x^2)/(64*10^12) =
(10^8x^6 -2*10^6x^7 +10^4x^4 -2*10^6x^7+ 4*10^4x^8 -200x^5 +10^4x^8 -200x^9 +x^6)/(64*10^12)
* Scenario q4; the chance that neither of the children of 2nd gen but both partners have (AB):
[[1-[(100x^3-x^4)/80000]]^2]*[(x/100)^2] =
[(80000-100x^3-x^4)^2/(64*10^8)]*[x^2/10000] =
(64*10^8 -16*10^6x^3 -16*10^4x^4 +10^4x^6 +200x^7 +x^8)(x^2)/(64*10^12) =
(64*10^8x^2 -16*10^6x^5 -16*10^4x^6 +10^4x^8 +200x^9 +x^10)/(64*10^12)
* Scenario q5; the chance that neither of the children of 2nd gen but one partners has (AB):
[[1-[(100x^3-x^4)/80000]]^2]*[(100x-x^2)/5000] =
[(80000-100x^3-x^4)^2/(64*10^8)]*[(100x-x^2)/5000] =
2(64*10^8 -16*10^6x^3 -16*10^4x^4 +10^4x^6 +200x^7 +x^8)(100x-x^2)/(32*10^12) =
4(64*10^10x -64*10^8x^2 -16*10^8x^4 +16*10^4x^6 +10^6x^7 +10^4x^8 -100x^9 -x^10)/(64*10^12)
* Scenario q6; the chance that nor the children of 2nd gen nor their partners have (AB):
(moot issue)
* Scenario q7; the chance that one of the children of second gen and their partners have (AB):
2[(100x^3-x^4)/80000][(60000-100x^3-x^4)/80000]*[(x/100)^2] =
2[(100x^3-x^4)(80000-100x^3-x^4)/(64*10^8)]*[x^2/10000] =
2[(8*10^6x^3 -8*10^4x^4 -10^4x^6 +x^8)(x^2)/(64*10^12) =
2[(8*10^6x^5 -8*10^4x^6 -10^4x^8 +x^10)/(64*10^12)
* Scenario q8; the chance that one of the children of second gen but neither partners have (AB):
2[(100x^3-x^4)/80000][(80000-100x^3-x^4)/80000]*[(100-x)^2/10000] =
2[(100x^3-x^4)(80000-100x^3-x^4)/(64*10^8)]*[(100-x)^2/10000] =
2(8*10^6x^3 -8*10^4x^4 -10^4x^6 +x^8)(10^4 -200x +x^2)/(64*10^12) =
2(8*10^10x^3 -16*10^8x^4 +16*10^6x^5 -10008*10^4x^6 +10^6x^7 -10^2x^9 +x^10)/(64*10^12)
* Scenario q9; the chance that one of the children of second gen and only his partners have (AB):
[(100x^3-x^4)/80000][(80000-100x^3-x^4)/80000]*[(100x-x^2)/5000] =
[(100x^3-x^4)(80000-100x^3-x^4)/(64*10^8)]*[(100x-x^2)/5000] =
(8*10^6x^3 -8*10^4x^4 -10^4x^6 +x^8)(100x-x^2)/(32*10^12) =
2(8*10^8x^4 -16*10^6x^5 +8*10^4x^6 -10^6x^7 +10^4x^8 +10^2x^9 -x^10)/(64*10^12)
* Scenario q10; the chance that one of the children of second gen and only the partner of the other child of the 2nd gen have (AB):
[(100x^3-x^4)/80000][(80000-100x^3-x^4)/80000]*[(100x-x^2)/5000] =
[(100x^3-x^4)(80000-100x^3-x^4)/(64*10^8)]*[(100x-x^2)/5000] =
(8*10^6x^3 -8*10^4x^4 -10^4x^6 +x^8)(100x-x^2)/(32*10^12) =
2(8*10^8x^4 -16*10^6x^5 +8*10^4x^6 -10^6x^7 +10^4x^8 +10^2x^9 -x^10)/(64*10^12)

Third generation:
* Chance that a child from the 3th gen has (AB):
[q1/2 + (q2/2 + q2/4)/2 + q3/4 + q4/4 + (q5*0 + q5/4)/2 + q6*0 + (q7/2 + q7/4)/2 + (q8*0 +q8/4)/2 + (q9*0 + q9/2)/2 + q10/4]/10 =
(q1/2 + 3q2/8 + q3/4 + q4/4 + q5/8 + 3q7/8 + q8/8 + q9/4 + q10/4)/10
* Scenario r; the chance that both cousins from the 3th gen have (AB):
[(q1/2 + 3q2/8 + q3/4 + q4/4 + q5/8 + 3q7/8 + q8/8 + q9/4 + q10/4)/10]^2
(q1/2 + 3q2/8 + q3/4 + q4/4 + q5/8 + 3q7/8 + q8/8 + q9/4 + q10/4)^2/100

Fourth generation:
* Chance that the child from our two cousins is born with a genetic defect (BB):
f(x) = r/4
f(x) = [(q1/2 + 3q2/8 + q3/4 + q4/4 + q5/8 + 3q7/8 + q8/8 + q9/4 + q10/4)^2/100]/4
f(x) = (4q1/8 + 3q2/8 + 2q3/8 + 2q4/8 + q5/8 + 3q7/8 + q8/8 + 2q9/8 + 2q10/8)^2/400
f(x) = [(4q1 +3q2 +2q3 +2q4 +q5 +3q7 +q8 +2q9 +2q10)/8]^2/400
f(x) = (4q1 +3q2 +2q3 +2q4 +q5 +3q7 +q8 +2q9 +2q10)^2/256*10^2

If we now fill in the values:
q1 = (10^4x^8-200x^9+x^6)/(64*10^12)
q2 = (100x^5 -x^6 +10^6x^7 -3*10^4x^8 +200x^9)/(64*10^12)
q3 = (10^8x^6 -2*10^6x^7 +10^4x^4 -2*10^6x^7+ 4*10^4x^8 -200x^5 +10^4x^8 -200x^9 +x^6)/(64*10^12)
q4 = (64*10^8x^2 -16*10^6x^5 -16*10^4x^6 +10^4x^8 +200x^9 +x^10)/(64*10^12)
q5 = 4(64*10^10x -64*10^8x^2 -16*10^8x^4 +16*10^4x^6 +10^6x^7 +10^4x^8 -100x^9 -x^10)/(64*10^12)
q7 = 2[(8*10^6x^5 -8*10^4x^6 -10^4x^8 +x^10)/(64*10^12)
q8 = 2(8*10^10x^3 -16*10^8x^4 +16*10^6x^5 -10008*10^4x^6 +10^6x^7 -10^2x^9 +x^10)/(64*10^12)
q9 = 2(8*10^8x^4 -16*10^6x^5 +8*10^4x^6 -10^6x^7 +10^4x^8 +10^2x^9 -x^10)/(64*10^12)
q10 = 2(8*10^8x^4 -16*10^6x^5 +8*10^4x^6 -10^6x^7 +10^4x^8 +10^2x^9 -x^10)/(64*10^12)
We get:
f(x) = [4(10^4x^8-200x^9+x^6)/(64*10^12) +3(100x^5 -x^6 +10^6x^7 -3*10^4x^8 +200x^9)/(64*10^12) +2(10^8x^6 -2*10^6x^7 +10^4x^4 -2*10^6x^7+ 4*10^4x^8 -200x^5 +10^4x^8 -200x^9 +x^6)/(64*10^12) +2(64*10^8x^2 -16*10^6x^5 -16*10^4x^6 +10^4x^8 +200x^9 +x^10)/(64*10^12) +4(64*10^10x -64*10^8x^2 -16*10^8x^4 +16*10^4x^6 +10^6x^7 +10^4x^8 -100x^9 -x^10)/(64*10^12) +6(8*10^6x^5 -8*10^4x^6 -10^4x^8 +x^10)/(64*10^12) +2(8*10^10x^3 -16*10^8x^4 +16*10^6x^5 -10008*10^4x^6 +10^6x^7 -10^2x^9 +x^10)/(64*10^12) +4(8*10^8x^4 -16*10^6x^5 +8*10^4x^6 -10^6x^7 +10^4x^8 +10^2x^9 -x^10)/(64*10^12) +4(8*10^8x^4 -16*10^6x^5 +8*10^4x^6 -10^6x^7 +10^4x^8 +10^2x^9 -x^10)/(64*10^12)]^2/256*10^2
Adding up the fractions:
f(x) = [[4(10^4x^8-200x^9+x^6) +3(100x^5 -x^6 +10^6x^7 -3*10^4x^8 +200x^9) +2(10^8x^6 -2*10^6x^7 +10^4x^4 -2*10^6x^7 +4*10^4x^8 -200x^5 +10^4x^8 -200x^9 +x^6) +2(64*10^8x^2 -16*10^6x^5 -16*10^4x^6 +10^4x^8 +200x^9 +x^10) +4(64*10^10x -64*10^8x^2 -16*10^8x^4 +16*10^4x^6 +10^6x^7 +10^4x^8 -100x^9 -x^10) +6(8*10^6x^5 -8*10^4x^6 -10^4x^8 +x^10) +2(8*10^10x^3 -16*10^8x^4 +16*10^6x^5 -10008*10^4x^6 +10^6x^7 -10^2x^9 +x^10) +4(8*10^8x^4 -16*10^6x^5 +8*10^4x^6 -10^6x^7 +10^4x^8 +10^2x^9 -x^10) +4(8*10^8x^4 -16*10^6x^5 +8*10^4x^6 -10^6x^7 +10^4x^8 +10^2x^9 -x^10)]^2/(64*10^12)^2]/256*10^2
Working out the numberator:
f(x) =[(4*10^4x^8-8*10^2x^9+4x^6 +3*10^2x^5 -3x^6 +3*10^6x^7 -9*10^4x^8 +6*10^2x^9 +2*10^8x^6 -4*10^6x^7 +2*10^4x^4 -4*10^6x^7 +8*10^4x^8 -400x^5 +2*10^4x^8 -400x^9 +2x^6 +128*10^8x^2 -32*10^6x^5 -32*10^4x^6 +2*10^4x^8 +400x^9 +2x^10 +256*10^10x -256*10^8x^2 -64*10^8x^4 +64*10^4x^6 +4*10^6x^7 +4*10^4x^8 -400x^9 -4x^10 +48*10^6x^5 -48*10^4x^6 -6*10^4x^8 +6x^10 +16*10^10x^3 -32*10^8x^4 +32*10^6x^5 -20016*10^4x^6 +2*10^6x^7 -2*10^2x^9 +2x^10 +32*10^8x^4 -64*10^6x^5 +32*10^4x^6 -4*10^6x^7 +4*10^4x^8 +4*10^2x^9 -4x^10 +32*10^8x^4 -64*10^6x^5 +32*10^4x^6 -4*10^6x^7 +4*10^4x^8 +4*10^2x^9 -4x^10)^2/(4.096*10^27)]/256*10^2
Adding up:
f(x) = (+256*10^10x -128*10^8x^2 +16*10^10x^3 -320002*10^4x^4 -800001*10^2x^5 +320003x^6 -7*10^6x^7 +13*10^4x^8 -2x^10)^2/(1.048576*10^32)

* The chance of two unrelated people having a child with trait (BB)
g(x) = [(x/100)^2]/4
g(x) = x^2/40000

 

[جوري]

If you'd tell me what the name of the disorder is, I will do the math for you, or try to help you do the math better.. whether dominant, recessive, x linked, mitochondrial all will influence the outcome differently..
Also many mutations occur de novo such as with achondroplasia..

 

[Abdul Fattah]

If you'd tell me what the name of the disorder is, I will do the math for you, or try to help you do the math better.. whether dominant, recessive, x linked, mitochondrial all will influence the outcome differently..
Also many mutations occur de novo such as with achondroplasia..

I wasn't thinking of a specific one. Just for the recessive type. If a disorder is dominant, wouldn't all the people who carry it have it? Also, whether dominant or regressive, isn't the chance of passing on the gene the same?

 

[جوري]

I wrote a long extensive post and then LI went dead, it really makes me so angry when that happens..

just a quick recap as I won't delve into as much detail as I did earlier -- and keep in mind I took genetics over 7 years ago
Dominant traits don't skip generations, if you do a punnet square you'll find that they are expressed in a 1:2:1 ratio, I.E
one expresses and carries the trait, 2 carry the traits but don't exhibit it, and one who neither carries not expresses the traits. so you'll have a 25% chance of having a completely healthy child irregardless of both you and your spouse exhibiting the trait.
Please use the concept of a coin being tossed when applying this concept... when you toss a coin you'll have a 50% chance of it being heads and 50 % chance of it being tales.. you may toss a coin 50 times and get heads every single time, it is true but it doesn't detract from the possibility that with every toss it could go either way. So when a couple has two kids expressing the traits they shouldn't feel, well I got the mutations out of the way, the next one is healthy.. or vice versa, I have four healthy kids meaning I am cured of this trait.

this also depends greatly on the state of heterozygosity vs homozygosity..
If a man had AA marries another AA, is obviously different outcome than Aa marrying AA or aa for that particular trait..
recessive skips around, dominant always shows in every generation.
X linked recessive affects mostly males given that only mothers pass the X chromosome to their sons. You ask why don't females express it, well because one X is always methylated and it is almost always the mutant x, that isn't to say that females never express an x linked trait, it is just very very rare and in a much lesser degree than males if they do
.Mitochondrial defects are inherited only from the mother.. girls pass it on the males express it, it can also be passed down in a form of autosomal recessive inheritance is possibly the most common model in mitochondrial disorders.
x-linked autosomal dominant are almost always lethal and die in utero
De Novo mutations are sporadic and affect everyone equally, there is no calculating it really, as no one knows what causes them, i.e whether environmental or chemical or or or
Br. MustafaMC is probably best to consult with these as he does this for a daily living so insha'Allah I too will be curious to read what he has to say..

http://books.google.com/books?id=3d...859-1&sig=nWeqC0RPIx3IWNkWNpLl-lESsMo#PPA8,M1

this is a concise look at medical genetics, if you are interested in reading? and here are a few quick schematic diagrams of mutations

recessive

x-linked dominant

mitochondrial

I think this diagram is a bit misleading, in medical genetics mitochoindrial inheritence is always maternal.. there is no mixing as the mitochondria of sperm is dropped off..

mitochondrial inheritence is really a world of genetics all its own, as the disorders can be so unusual and lethal, affecting the electron transport chain and krebs cycle etc have a very bad impact on muscle and nervous tissue.. I think you'll be doing calculations for a really long time because there is so much great variability.. I think you'll have an easier time doing it, if you knew what the disorder actually is..

anyhow best to review all of those from scratch I am very very rusty

 

[Abdul Fattah]

Yeah, thx for your post, but as far as I can tell that's irrelevant isn't it? In all of these cases when a person carries one bad and one good gene there is always a 50% chance of passing it on right?

Anyway, the calculations I was trying to make is to form a formula with an unknown parameter, so that you can formulate a percentage when you don't know what genes the parents have, based on the percentage of dormant carriers in the population

 

[جوري]

Yeah, thx for your post, but as far as I can tell that's irrelevant isn't it? In all of these cases when a person carries one bad and one good gene there is always a 50% chance of passing it on right?

It is irrelevant in a sense that unless you have a robertsonian translocation where you are guranteed to have an afflicted child, everything else falls on probability, with some probabilities more likely to occur than others.. Not everything is 50% no.. sometimes 75% sometimes 25% some are self-correcting.. some de novo..

Anyway, the calculations I was trying to make is to form a formula with an unknown parameter, so that you can formulate a percentage when you don't know what genes the parents have, based on the percentage of dormant carriers in the population

That will be quite expansive.. I can't claim to have your mathematical abilities which is quite impressive.. the numbers that you get will only be as good as putting a name on such probabilities...

Good luck with all that, I'd be very curious as to the results.. maybe one day you'll get to publish it insha'Allah

 

[Abdul Fattah]

Not everything is 50% no.. sometimes 75% sometimes 25% some are self-correcting.. some de novo..

Oh, that's news to me, could you explain on that some more?

That will be quite expansive.. I can't claim to have your mathematical abilities which is quite impressive.. the numbers that you get will only be as good as putting a name on such probabilities...

Well as you can see in the opening post, it's do-able. There did creep a small mistake in somewhere though. Driving me insane ^_^

 

[جوري]

Oh, that's news to me, could you explain on that some more?

are you asking about robertsonian translocation or the percentage difference?

Well as you can see in the opening post, it's do-able. There did creep a small mistake in somewhere though. Driving me insane ^_^

lol.. It is amazing what comes easily for some of us and rather daunting for others...

I am sure if you work on it some more it will come to you.. problem solving ability really comes when the brain is in alpha waves.. well for me anyhow, not sure about most other folks...

I will be very curious to discuss your work after you are done insha'Allah

 

[جوري]

anyhow let me tell you about robertsonian translocation quick before I go to dindin..
basically in very simple terms, is when you have a whole arm translocation of a centric fusion on a chromosome... say with downsyndrome it is a problem generally with non-disjunction and happens with advanced maternal age because the eggs are just tired.. and so in the process of meiosis you'll have three pairs of chromosomes because one lagged behind during one of those phases in meiosis II you know prophase, metphase teleophase, whichever, I really don't remember which, but after the daughter cells have split, you'll notice one chromosme missing in one cell and in another cell there is an extra one and in the baby that manfiests as an extra arm, I picked chromosome 21 as an example, it can really happen to any.. but most of these problems abort in utero, very few mutations actually make..

a robertsonian translocation, I am giving you this from wiki because it is fast and actually correct

In humans, when a Robertsonian translocation joins the long arm of chromosome 21 with the long arm of chromosome 14 (or 15), the heterozygous carrier is phenotypically normal because there are two copies of all major chromosome arms and hence two copies of all essential genes.[2] However, the progeny of this carrier may inherit an unbalanced trisomy 21, causing Down Syndrome.

so even though the mother is normal ALL her off spring will display downs..

 

[teen-omar]

LOOOL u lot make me laugh
you might as well make a scientific forum out of this now
masha'allah seems like u both know what you are talking about
makes us people look really dumb... *sneaks away from the smart people*

 

[Abdul Fattah]

are you asking about robertsonian translocation or the percentage difference?

What interests me isthe percentage difference. Translocation is neglectable here cause there's no indication that cousins have a higher probability for these mutations, right?

lol.. It is amazing what comes easily for some of us and rather daunting for others...

Alhamdoelillahi ^_^

I will be very curious to discuss your work after you are done insha'Allah

Found a small mistake, that could have been teh cause, calculating again....

 

[جوري]

What interests me isthe percentage difference. Translocation is neglectable here cause there's no indication that cousins have a higher probability for these mutations, right?

No this has nothing at all to do with cousins, once a person has a robertsonian translocation they could marry a martian and it wouldn't make a bit of a difference...but in AD, the odds are 75% not 50%, simply because one carries and exhibits and will pass it down in same fashion while two carry without exhibiting and they too will pass that down in some fashion of course contingent on whom they marry... there is also such a thing as incomplete penetrence where even though a person has the disease doesn't exhibit it in any severe form.. There are also what is called trinucleotide repeat expansion such as fragile X and that is inherited in X-linked Dominant and that gets worst with each generation as the CGG repeat gets longer and longer.. of course there are spinocerbellar ataxias that get passed down in same fashion, also huntington dz though that is autosomal dominant I believe? the de novo are very sporadic and in general happen at a certain percentage but doesn't discriminate..

Alhamdoelillahi ^_^

yes al7mdlilah

Found a small mistake, that could have been teh cause, calculating again....

Back to the drawing board

 

[Na7lah]

x) = [4(10^4x^8-200x^9+x^6)/(64*10^12) +3(100x^5 -x^6 +10^6x^7 -3*10^4x^8 +200x^9)/(64*10^12) +2(10^8x^6 -2*10^6x^7 +10^4x^4 -2*10^6x^7+ 4*10^4x^8 -200x^5 +10^4x^8 -200x^9 +x^6)/(64*10^12) +2(64*10^8x^2 -16*10^6x^5 -16*10^4x^6 +10^4x^8 +200x^9 +x^10)/(64*10^12) +4(64*10^10x -64*10^8x^2 -16*10^8x^4 +16*10^4x^6 +10^6x^7 +10^4x^8 -100x^9 -x^10)/(64*10^12) +6(8*10^6x^5 -8*10^4x^6 -10^4x^8 +x^10)/(64*10^12) +2(8*10^10x^3 -16*10^8x^4 +16*10^6x^5 -10008*10^4x^6 +10^6x^7 -10^2x^9 +x^10)/(64*10^12) +4(8*10^8x^4 -16*10^6x^5 +8*10^4x^6 -10^6x^7 +10^4x^8 +10^2x^9 -x^10)/(64*10^12) +4(8*10^8x^4 -16*10^6x^5 +8*10^4x^6 -10^6x^7 +10^4x^8 +10^2x^9 -x^10)/(64*10^12)]^2/256*10^2
Adding up the fractions:
f(x) = [[4(10^4x^8-200x^9+x^6) +3(100x^5 -x^6 +10^6x^7 -3*10^4x^8 +200x^9) +2(10^8x^6 -2*10^6x^7 +10^4x^4 -2*10^6x^7 +4*10^4x^8 -200x^5 +10^4x^8 -200x^9 +x^6) +2(64*10^8x^2 -16*10^6x^5 -16*10^4x^6 +10^4x^8 +200x^9 +x^10) +4(64*10^10x -64*10^8x^2 -16*10^8x^4 +16*10^4x^6 +10^6x^7 +10^4x^8 -100x^9 -x^10) +6(8*10^6x^5 -8*10^4x^6 -10^4x^8 +x^10) +2(8*10^10x^3 -16*10^8x^4 +16*10^6x^5 -10008*10^4x^6 +10^6x^7 -10^2x^9 +x^1

good luck brother can't help u though :X
*runz away*

 

[Güven]

^Lol I have never seen in my whole life so many numbers I cant even concentrate lookin at them :uuh:

 

[Abdul Fattah]

Second attempt, I had an error in scenario r and forgot some factors in the q-scenario's. There's still something wrong with it though...

Defenitions:
*People without genetic disabilty: (AA)
*People with genetic disability: (BB)
*Percantage of people with dormant trait (AB) or (BA): x

Objective:
* Find function f(x) that gives the percentage of two cousins having a child with trait (BB)
* Find function g(x) that gives the percentage of two unrelated people having a child with trait (BB)
* Compare which of the two cases has the highest risk
Note1: In laws should be considered to have a chance of x/100 of having (AB)
Note2: The chance of people of the first generations having (BB) is not added, we assume that in such cases there will be no further marriages and children from that generation.

Calculating:
First generation:
* The chance that a person has (AB):
x/100
* The chance that a person doesn't has (AB):
1-(x/100) = (100/100)-(x/100) = (100-x)/100
* Scenario p1; the chance that both partners have (AB):
(x/100)^2 = x^2/10000
* Scenario p2; the chance that neither partner has (AB):
[(100-x)/100]^2 = (100-x)^2/10000
* Scenario p3; the chance that one of the two partners has (AB):
2(x/100)[(100-x)/100] = (100x-x^2)/5000

Second generation:
* Chance that a child of 2nd gen has (AB): (p1/2 + p2*0 + p3/4)/4
[(x^2/10000)/2 + 0 + ((100x-x^2)/5000)/4]/4 =
[(x^2)/20000 + (100x-x^2)/20000]/4 = (100x^3-x^4)/80000
* Chance that a child of sec gen doesn't have (AB): 1-[(p1/2 + p2*0 + p3/4)/3]
1 -[(100x^3-x^4)/80000] =
(80000-100x^3-x^4)/80000
* Chance that a partner of the 2nd gen has (AB)
x/100
* Chance that a partner of the 2nd gen doesn't have (AB)
(100-x)/100
* Scenario q1; the chance that both children of 2nd gen and their partners have (AB):
[[(100x^3-x^4)/80000]^2]*[(x/100)^2] =
[(100x^3-x^4)^2/(64*10^8)]*[x^2/10000] =
(10^4x^6-200x^7+x^4)(x^2)/(46*10^12) =
(10^4x^8-200x^9+x^6)/(46*10^12)
* Scenario q2; the chance that both children of 2nd gen and one of their partners have (AB):
2[[(100x^3-x^4)/80000]^2]*[(100x-x^2)/5000] =
2[(100x^3-x^4)^2/(64*10^8)]*[(100x-x^2)/5000] =
2(10^4x^6-200x^7+x^4)(100x-x^2)/(64*10^12) =
2(100x^5 -x^6 +10^6x^7 -3*10^4x^8 +200x^9)/(64*10^12)
* Scenario q3; the chance that both children of 2nd gen but none of their partners have (AB):
[[(100x^3-x^4)/60000]^2]*[(100-x)^2/10000] =
[(100x^3-x^4)^2/(64*10^8)]*[(100-x)^2/10000] =
(10^4x^6-200x^7+x^4)(10^4 -200x +x^2)/(64*10^12) =
(10^8x^6 -2*10^6x^7 +10^4x^4 -2*10^6x^7+ 4*10^4x^8 -200x^5 +10^4x^8 -200x^9 +x^6)/(64*10^12)
* Scenario q4; the chance that neither of the children of 2nd gen but both partners have (AB):
[[1-[(100x^3-x^4)/80000]]^2]*[(x/100)^2] =
[(80000-100x^3-x^4)^2/(64*10^8)]*[x^2/10000] =
(64*10^8 -16*10^6x^3 -16*10^4x^4 +10^4x^6 +200x^7 +x^8)(x^2)/(64*10^12) =
(64*10^8x^2 -16*10^6x^5 -16*10^4x^6 +10^4x^8 +200x^9 +x^10)/(64*10^12)
* Scenario q5; the chance that neither of the children of 2nd gen but one partners has (AB):
2[[1-[(100x^3-x^4)/80000]]^2]*[(100x-x^2)/5000] =
2[(80000-100x^3-x^4)^2/(64*10^8)]*[(100x-x^2)/5000] =
2(64*10^8 -16*10^6x^3 -16*10^4x^4 +10^4x^6 +200x^7 +x^8)(100x-x^2)/(32*10^12) =
4(64*10^10x -64*10^8x^2 -16*10^8x^4 +16*10^4x^6 +10^6x^7 +10^4x^8 -100x^9 -x^10)/(64*10^12)
* Scenario q6; the chance that nor the children of 2nd gen nor their partners have (AB):
(moot issue)
* Scenario q7; the chance that one of the children of second gen but both their partners have (AB):
2[(100x^3-x^4)/80000][(60000-100x^3-x^4)/80000]*[(x/100)^2] =
2[(100x^3-x^4)(80000-100x^3-x^4)/(64*10^8)]*[x^2/10000] =
2[(8*10^6x^3 -8*10^4x^4 -10^4x^6 +x^8)(x^2)/(64*10^12) =
2[(8*10^6x^5 -8*10^4x^6 -10^4x^8 +x^10)/(64*10^12)
* Scenario q8; the chance that one of the children of second gen but neither partners have (AB):
2[(100x^3-x^4)/80000][(80000-100x^3-x^4)/80000]*[(100-x)^2/10000] =
2[(100x^3-x^4)(80000-100x^3-x^4)/(64*10^8)]*[(100-x)^2/10000] =
2(8*10^6x^3 -8*10^4x^4 -10^4x^6 +x^8)(10^4 -200x +x^2)/(64*10^12) =
2(8*10^10x^3 -16*10^8x^4 +16*10^6x^5 -10008*10^4x^6 +10^6x^7 -10^2x^9 +x^10)/(64*10^12)
* Scenario q9; the chance that one of the children of second gen and only his partners have (AB):
2[(100x^3-x^4)/80000][(80000-100x^3-x^4)/80000]*[(100x-x^2)/5000] =
2[(100x^3-x^4)(80000-100x^3-x^4)/(64*10^8)]*[(100x-x^2)/5000] =
2(8*10^6x^3 -8*10^4x^4 -10^4x^6 +x^8)(100x-x^2)/(32*10^12) =
4(8*10^8x^4 -16*10^6x^5 +8*10^4x^6 -10^6x^7 +10^4x^8 +10^2x^9 -x^10)/(64*10^12)
* Scenario q10; the chance that one of the children of second gen and only the partner of the other child of the 2nd gen have (AB):
2[(100x^3-x^4)/80000][(80000-100x^3-x^4)/80000]*[(100x-x^2)/5000] =
2[(100x^3-x^4)(80000-100x^3-x^4)/(64*10^8)]*[(100x-x^2)/5000] =
2(8*10^6x^3 -8*10^4x^4 -10^4x^6 +x^8)(100x-x^2)/(32*10^12) =
4(8*10^8x^4 -16*10^6x^5 +8*10^4x^6 -10^6x^7 +10^4x^8 +10^2x^9 -x^10)/(64*10^12)

Third generation:
* Chance that a child from the 3th gen has (AB):
[q1/2 + (q2/2 + q2/4)/2 + q3/4 + q4/4 + (q5*0 + q5/4)/2 + q6*0 + (q7/2 + q7/4)/2 + (q8*0 +q8/4)/2 + (q9*0 + q9/2)/2 + q10/4]/16 =
(q1/2 + 3q2/8 + q3/4 + q4/4 + q5/8 + 3q7/8 + q8/8 + q9/4 + q10/4)/16
* Scenario r; the chance that both cousins from the 3th gen have (AB):
[(q1/2 + 3q2/8 + q3/4 + q4/4 + q5/8 + 3q7/8 + q8/8 + q9/4 + q10/4)/16]^2
(q1/2 + 3q2/8 + q3/4 + q4/4 + q5/8 + 3q7/8 + q8/8 + q9/4 + q10/4)^2/256

Fourth generation:
* Chance that the child from our two cousins is born with a genetic defect (BB):
f(x) = r/4
f(x) = [(q1/2 + 3q2/8 + q3/4 + q4/4 + q5/8 + 3q7/8 + q8/8 + q9/4 + q10/4)^2/256]/4
f(x) = (4q1/8 + 3q2/8 + 2q3/8 + 2q4/8 + q5/8 + 3q7/8 + q8/8 + 2q9/8 + 2q10/8)^2/1024
f(x) = [(4q1 +3q2 +2q3 +2q4 +q5 +3q7 +q8 +2q9 +2q10)/8]^2/1024
f(x) = (4q1 +3q2 +2q3 +2q4 +q5 +3q7 +q8 +2q9 +2q10)^2/65536

If we now fill in the values:
q1 = (10^4x^8-200x^9+x^6)/(64*10^12)
q2 = 2(100x^5 -x^6 +10^6x^7 -3*10^4x^8 +200x^9)/(64*10^12)
q3 = (10^8x^6 -2*10^6x^7 +10^4x^4 -2*10^6x^7+ 4*10^4x^8 -200x^5 +10^4x^8 -200x^9 +x^6)/(64*10^12)
q4 = (64*10^8x^2 -16*10^6x^5 -16*10^4x^6 +10^4x^8 +200x^9 +x^10)/(64*10^12)
q5 = 4(64*10^10x -64*10^8x^2 -16*10^8x^4 +16*10^4x^6 +10^6x^7 +10^4x^8 -100x^9 -x^10)/(64*10^12)
q7 = 2[(8*10^6x^5 -8*10^4x^6 -10^4x^8 +x^10)/(64*10^12)
q8 = 2(8*10^10x^3 -16*10^8x^4 +16*10^6x^5 -10008*10^4x^6 +10^6x^7 -10^2x^9 +x^10)/(64*10^12)
q9 = 4(8*10^8x^4 -16*10^6x^5 +8*10^4x^6 -10^6x^7 +10^4x^8 +10^2x^9 -x^10)/(64*10^12)
q10 = 4(8*10^8x^4 -16*10^6x^5 +8*10^4x^6 -10^6x^7 +10^4x^8 +10^2x^9 -x^10)/(64*10^12)
We get:
f(x) = [4(10^4x^8-200x^9+x^6)/(64*10^12) +6(100x^5 -x^6 +10^6x^7 -3*10^4x^8 +200x^9)/(64*10^12) +2(10^8x^6 -2*10^6x^7 +10^4x^4 -2*10^6x^7+ 4*10^4x^8 -200x^5 +10^4x^8 -200x^9 +x^6)/(64*10^12) +2(64*10^8x^2 -16*10^6x^5 -16*10^4x^6 +10^4x^8 +200x^9 +x^10)/(64*10^12) +4(64*10^10x -64*10^8x^2 -16*10^8x^4 +16*10^4x^6 +10^6x^7 +10^4x^8 -100x^9 -x^10)/(64*10^12) +6(8*10^6x^5 -8*10^4x^6 -10^4x^8 +x^10)/(64*10^12) +2(8*10^10x^3 -16*10^8x^4 +16*10^6x^5 -10008*10^4x^6 +10^6x^7 -10^2x^9 +x^10)/(64*10^12) +8(8*10^8x^4 -16*10^6x^5 +8*10^4x^6 -10^6x^7 +10^4x^8 +10^2x^9 -x^10)/(64*10^12) +8(8*10^8x^4 -16*10^6x^5 +8*10^4x^6 -10^6x^7 +10^4x^8 +10^2x^9 -x^10)/(64*10^12)]^2/65536
Adding up the fractions:
f(x) = [[4(10^4x^8-200x^9+x^6) +6(100x^5 -x^6 +10^6x^7 -3*10^4x^8 +200x^9) +2(10^8x^6 -2*10^6x^7 +10^4x^4 -2*10^6x^7 +4*10^4x^8 -200x^5 +10^4x^8 -200x^9 +x^6) +2(64*10^8x^2 -16*10^6x^5 -16*10^4x^6 +10^4x^8 +200x^9 +x^10) +4(64*10^10x -64*10^8x^2 -16*10^8x^4 +16*10^4x^6 +10^6x^7 +10^4x^8 -100x^9 -x^10) +6(8*10^6x^5 -8*10^4x^6 -10^4x^8 +x^10) +2(8*10^10x^3 -16*10^8x^4 +16*10^6x^5 -10008*10^4x^6 +10^6x^7 -10^2x^9 +x^10) +8(8*10^8x^4 -16*10^6x^5 +8*10^4x^6 -10^6x^7 +10^4x^8 +10^2x^9 -x^10) +8(8*10^8x^4 -16*10^6x^5 +8*10^4x^6 -10^6x^7 +10^4x^8 +10^2x^9 -x^10)]^2/(64*10^12)^2]/65536
Working out the numberator:
f(x) =[(4*10^4x^8-8*10^2x^9+4x^6 +6*10^2x^5 -6x^6 +6*10^6x^7 -18*10^4x^8 +12*10^2x^9 +2*10^8x^6 -4*10^6x^7 +2*10^4x^4 -4*10^6x^7 +8*10^4x^8 -400x^5 +2*10^4x^8 -400x^9 +2x^6 +128*10^8x^2 -32*10^6x^5 -32*10^4x^6 +2*10^4x^8 +400x^9 +2x^10 +256*10^10x -256*10^8x^2 -64*10^8x^4 +64*10^4x^6 +4*10^6x^7 +4*10^4x^8 -400x^9 -4x^10 +48*10^6x^5 -48*10^4x^6 -6*10^4x^8 +6x^10 +16*10^10x^3 -32*10^8x^4 +32*10^6x^5 -20016*10^4x^6 +2*10^6x^7 -2*10^2x^9 +2x^10 +64*10^8x^4 -128*10^6x^5 +64*10^4x^6 -8*10^6x^7 +8*10^4x^8 +8*10^2x^9 -8x^10 +64*10^8x^4 -128*10^6x^5 +64*10^4x^6 -8*10^6x^7 +8*10^4x^8 +8*10^2x^9 -8x^10)^2/(4.096*10^27)]/65536
Adding up:
f(x) =[(256*10^10x -128*10^8x^2 +16*10^10x^3 +320002*10^4x^4 -2080002*10^2x^5 +96*10^4x^6 -12*10^6x^7 +12*10^4x^8 +14*10^2x^9 -10x^10)^2/(2.68435456*10^32)

* The chance of two unrelated people having a child with trait (BB)
g(x) = [(x/100)^2]/4
g(x) = x^2/40000

 

[جوري]

If you are sure of your results, you might want to forward them to appropriate Scientific Committee for review?...
Maybe you can mention me if you win a prize one day ey?

I can picture it now..

thanks mom and dad for all your support and love, my friends, family, my extended family on LI and especially sister Skye Ephémérine for all her support, review and encouragement..then raise your nobel and say this is one is for you..

 

[MustafaMc]

An approach with example from population genetics:

Frequency of Sickle-cell anemia in African Americans is 0.25% = 0.0025
Frequency of Sickle-cell gene (s) is square root of 0.0025 = 0.05
Frequency of normal gene (S) is 1 - freq of s = 1-0.05 = 0.95

Frequency in random mating population
Freq(SS) = Freq(S)*Freq(S) = 0.95x0.95 = 90.25%
Freq(Ss) = 2Freq(S)*Freq(s) = 2x0.95x0.05 = 9.5%
Freq(ss) = Freq(s)*Freq(s) = 0.05x0.05 = 0.25%

Frequency in 1st generation of cousins mating
Inbreeding coefficient for 1st cousins F=6.25%
Freq(SS) = F*Freq(S) + (1-F)*Freq(S)*Freq(S) = 0.0625x0.95 + (1-0.0625)x0.95x0.95 = 90.55%
Freq(Ss) = (1-F)*2Freq(S)*Freq(a) = (1-0.0625)x2x0.95x0.05 = 8.9%
Freq(ss) = F*Freq(s) + (1-F)*Freq(s)*Freq(s) = 0.0625x0.05 + (1-0.0625)x0.05x0.05 = 0.55%

Random mating = 0.25% anemia
1st generation of cousins mating = 0.55% anemia

 

[Abdul Fattah]

Selam aleykum mustafa

An approach with example from population genetics:
Frequency of Sickle-cell anemia in African Americans is 0.25% = 0.0025
Frequency of Sickle-cell gene (s) is square root of 0.0025 = 0.05
Frequency of normal gene (S) is 1 - freq of s = 1-0.05 = 0.95

The 0.25% represents people with (ss) You cannot just take a square of that and assume that you have the correct percentage of carriers. In reality, you'll find that the percentage of carriers is actually larger then the percentage of infected people! In general I would say there is no mathematical way to calculate the percentage of carriers based on infected people. The only way to know the percentage of carriers would be to check large random parts of the population for dormant genes and calculate the percentage based on the rules of statistics. That is why instead of calculating the frequency, I'm using an unknown parameter in my calculations, so that you could reach conclusions even without knowing the percentage of carriers.

 

[MustafaMc]

The 0.25% represents people with (ss) You cannot just take a square of that and assume that you have the correct percentage of carriers. In reality, you'll find that the percentage of carriers is actually larger then the percentage of infected people! In general I would say there is no mathematical way to calculate the percentage of carriers based on infected people. The only way to know the percentage of carriers would be to check large random parts of the population for dormant genes and calculate the percentage based on the rules of statistics.

wa alaikum salaam

Assuming a random mating population, the frequency of the recessive gene in that population is the square root of the frequency of individuals expressing the trait. This is a function of the Hardy-Weinberg Equilibrium: p^2 + 2pq + q^2

The frequency of carriers is:

Freq(Ss) = 2*Freq(S)*Freq(s) = 2 x 0.95 x 0.05 = 9.5%

Which is, as you noted, larger than the percentage of affected people (0.25%). Note that with additional generations of inbreeding the frequency of the carriers (Ss) decrease and the frequency of those expressing the trait (ss) increase up to the frequency of the gene itself (0.25% > 5%), of course assuming no negative selection pressure.

Since my mental dexterity is not what it used to be, I had some trouble following your calculations. Perhaps, a simplified version would be better for me.

 

[جوري]

^^ you are actually correct. 10% of African American are carriers of sickle cell trait which is probably rounded from your 9.5%

Sickle cell trait is relatively prevalent among African Americans in the United States, with carrier frequencies of 8–10%1 while rates as high as 25–30% in parts of West Africa have been recorded.2

http://www.greenjournal.org/cgi/content/full/109/4/870