Selam aleykum
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
If we now fill in the appropriate values:
q1 = (100 +x +x^2)^2(x^2)/(64*10^12)
q2 = (100 +x +x^2)^2(100-x)^2/(64*10^12)
q3 = (100 +x +x^2)^2(2x)(100-x)/(64*10^12)
q4 = (799*10^2 -x -x^2)^2(x^2)/(64*10^12)
q8 = 2(799*10^2 -x -x^2)(100 +x +x^2)(x^2)/(64*10^12)
q10 = 2(799*10^2 -x -x^2)(100 +x +x^2)(x)(100-x)/(64*10^12)
f(x) = [4(100 +x +x^2)^2(x^2)/(64*10^12) +(100 +x +x^2)^2(100-x)^2/(64*10^12) +2(100 +x +x^2)^2(2x)(100-x)/(64*10^12) +(799*10^2 -x -x^2)^2(x^2)/(64*10^12) +4(799*10^2 -x -x^2)(100 +x +x^2)(x^2)/(64*10^12) +2(799*10^2 -x -x^2)(100 +x +x^2)(x)(100-x)/(64*10^12)]/1024
Add fractions:
f(x) = [4(100 +x +x^2)^2(x^2) +(100 +x +x^2)^2(100-x)^2 +2(100 +x +x^2)^2(2x)(100-x) +(799*10^2 -x -x^2)^2(x^2) +4(799*10^2 -x -x^2)(100 +x +x^2)(x^2) +2(799*10^2 -x -x^2)(100 +x +x^2)(x)(100-x)]/(64*10^12)(1024)
Working out squares:
f(x) = [4(10^4 +2*10^2x +201x^2 +2x^3 +x^4)(x^2) +(10^4 +2*10^2x +201x^2 +2x^3 +x^4)(10^4 -200x +x^2) +2(10^4 +2*10^2x +201x^2 +2x^3 +x^4)(2x)(100-x) +(638401*10^4 -1598*10^2x -159798x^2 +2x^3 +x^4)(x^2) +4(799*10^2 -x -x^2)(100 +x +x^2)(x^2) +2(799*10^2 -x -x^2)(100 +x +x^2)(x)(100-x)]/(6.5536*10^16)
Working out:
f(x) = [(4*10^4x^2 +8*10^2x^3 +804x^4 +8x^5 +4x^6) +(10^8 +198*10^4x^2 -2*10^4x^3 +9801x^4 -198x^5 +x^6) +(4*10^6x +4*10^4x^2 +796*10^2x^3 -4x^4 +392x^5 -4x^6) +(638401*10^4x^2 -1598*10^2x^3 -159798x^4 +2x^3 +x^6) +(3196*10^4x^2 +3192*10^2x^3 +319196x^4 -8x^5 -4x^6) +(1598*10^6x -2*10^4x^2 +158002*10^2x^3 -159998x^4 -196x^5 +2x^6)]/(6.5536*10^16)
Adding up:
f(x) = (10^8 +1602*10^6x +641801*10^4x^2 +16020002x^3 +1001x^4)/(6.5536*10^16)
* Chance that a child from two unrelated people has (BB):
g(x) = (x/100)^2/4
g(x) = x^2/40000
