Help with math pls | calculating genetic prob of child from cousins with disorder (OP)
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
I played around with Excel to come up with a function for change over generations.
Assuming only 1st cousins are allowed to mate, the frequency of affected individuals would increase by the function:
(# of generations x inbreeding coefficient x "s" allele frequency in population) + {[1-(# of generations x inbreeding coefficient)] x "s" allele frequency in population x "s" allele frequency in population}
After 4 generations of only 1st cousins mating
(4 x 0.0625 x 0.05) + {[1 - (4 x 0.0625)] x 0.05 x 0.05} = 0.0144
affected individuals as compared to the 0.0025 initially. It would take 16 generations of only 1st cousins mating to max out at 0.05 frequency of "ss" individuals.
Assuming that only 1 in 100 matings are of 1st cousins, then it would take 400 generations to reach this same frequency of 0.0144.
Please, correct any errors that I made as it has been a while since I took population genetics and this is one of the rare times that I have actually used it.
Re: Help with math pls | calculating genetic prob of child from cousins with disorder
Selam aleykum
format_quote Originally Posted by MustafaMc
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 hardy Weinberg equilibrium doesn't apply here, I'm working under the assumption that people with the recessive genotype don't mate (early death or significant disability). I still remain convinced that you cannot calculate the percentage of carriers based on the percentage of infected people.
anyway, if you look at something for to long, you're staring blind on it. So I started over from scratch and found some more mistakes, it's still not error free though :s
(for example 0% still gives probability 2*10^-9)
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.
Calculaing: First generation
* Chance of person having (AB):
x/100
* Chance of person not having (AB):
1-(x/100) =
(100-x)/100
* p1: Chance of both partners having (AB):
(x/100)^2 =
x^2/10^4
* p2: Chance of none of the partners having (AB):
[(100-x)/100]^2 =
(100-x)^2/10^4
* p3: (Double) chance of one of the partners having (AB)
2(x/100)(100-x)/100 =
2(x)(100-x)/10^4
Second generation
* Chance of a child having (AB):
[(1/2)(p1) +(0)(p2) + (1/4)(p3)]/4 =
[(1/2)(x^2/10^4) + (1/4)(2x)(100-x)/10^4]/4 =
(x^2 +2x -x +100)/(8*10^4) =
(100 +x +x^2)/(8*10^4)
* Chance of a child not having (AB):
1-[(100 +x +x^2)/(8*10^4)]
(8*10^4 -100 -x -x^2)/(8*10^4)
(799*10^2 -x -x^2)/(8*10^4)
* q1: Chance of both children and both partners having (AB):
[(100 +x +x^2)/(8*10^4)]^2*[x^2/10^4]
(100 +x +x^2)^2(x^2)/(64*10^12)
* q2: Chance of both children and no partners having (AB):
[(100 +x +x^2)/(8*10^4)]^2*[(100-x)^2/10^4]
(100 +x +x^2)^2(100-x)^2/(64*10^12)
* q3: (Double) chance of both children and one partner having (AB):
[(100 +x +x^2)/(8*10^4)]^2*[2(x)(100-x)/10^4]
(100 +x +x^2)^2(2x)(100-x)/(64*10^12)
* q4: Chance of none of the children, but both partners having (AB):
[(799*10^2 -x -x^2)/(8*10^4)]^2*[x^2/10^4]
(799*10^2 -x -x^2)^2(x^2)/(64*10^12)
* q5: Chance of none of the children nor partners having (AB):
(moot issue)
* q6: (Double) chance of none of the children but one of the partners having (AB):
(moot issue)
* q7: (Double) Chance of one of the children but no partners having (AB):
(moot issue)
* q8: (Double) Chance of one of the children and both partners having (AB):
2[(799*10^2 -x -x^2)/(8*10^4)]*[(100 +x +x^2)/(8*10^4)]*[x^2/10^4]
2(799*10^2 -x -x^2)(100 +x +x^2)(x^2)/(64*10^12)
* q9: (Double) Chance of one of the children and only his/her partners having (AB):
(moot issue)
* q10: (Double) Chance of one of the children and only the other child's partners having (AB):
2[(799*10^2 -x -x^2)/(8*10^4)]*[(100 +x +x^2)/(8*10^4)]*[(x)(100-x)/10^4]
2(799*10^2 -x -x^2)(100 +x +x^2)(x)(100-x)/(64*10^12)
Third generation
* r: Chance that two cousins from the third generation have (AB):
[(1/4)(q1) +(1/16)(q2) +(1/8)(q3) +(1/16)(q4) +(0)(q5) +(0)(q6) +(0)(q7) +(1/8)(q8) +(0)(q9) +(1/16)(q10)]/16 =
[(4/16)(q1) +(1/16)(q2) +(2/16)(q3) +(1/16)(q4) +(2/16)(q8) +(1/16)(q10)]/16 =
[4(q1) +(q2) +2(q3) +(q4) +2(q8) +(q10)]/256
Fourth generation
* Chance that a child from the two cousins has (BB):
f(x) = r/4
f(x) = [4(q1) +(q2) +2(q3) +(q4) +2(q8) +(q10)]/1024
Re: Help with math pls | calculating genetic prob of child from cousins with disorder
format_quote Originally Posted by Abdul Fattah
Selam aleykum
The hardy Weinberg equilibrium doesn't apply here, I'm working under the assumption that people with the recessive genotype don't mate (early death or significant disability).
If the double recessive is lethal, then the gene will be eliminated from the population - unless there is a selective advantage for the carriers.
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