Disease Model

Project Member

Hu Ye,first year graduate student from the ACCESS program.

Project Description

Common disease models include:
Dominant – One mutation on either chromosome causes the disease.
Recessive – Two mutations, one on each chromosome are required to cause the disease. A single mutation will not cause the disease (called a carrier).
Additive – Each mutation contributes to disease risk.

To take these factors into association model

Goal for the quarter

Complete the medium project

Weekly Progression

First Week 5/7
Background Study
Self Evaluation A-

Second Week 5/14
Figure out the statstics for recessive and dominant disease
1. Classify the three types of Genotype AA Aa aa into two categories for statistics
2. Take relative risk into account
Self Evaluation A+

Third Week 5/21
Try to figure out the statstics for additive disease
1. Three null hypothesis
2. Three statistics
3. Evaluation of three statistics
Self Evaluation A

Forth Week 5/28
Prepare for Presentation
Self Evaluation A

Fifth Week 6/4
1.Prepare the website
2.Add mono-chromosome model for additive effect
Self Evaluation A

Monochromosome Disease Model

Our statistics in class assume that individuals have only 1 chromosome. However, in actual, human beings is diploid:
22 pairs of auto chromosomes
1 pair of sex chromosomes
2 alleles at one locus
01300000337727123240805521113_s.gif

Common disease models include:
1. Dominant – One mutation on either chromosome causes the disease.
2. Recessive – Two mutations, one on each chromosome are required to cause the disease. A single mutation will not cause the disease (called a carrier).
3. Additive – Each mutation contributes to disease risk.

Genotype

For a single SNP a Assume:

Major Homozygous Genotype: AA with Frequency PAA
Heterozygous Genotype: Aa with Frequency PAa
Minor Homozygous Genotype aa with Frequency Paa

Modify the statistics to take into account recessive and dominant disease models

Recessive Disease:

We can classify genotype into :
Type I Without Minor Allele: A* with Frequency PA*, PA*=PAa+PAA
Type II Minor homozygous Genotype: aa with Frequency Paa

In cases, the frequency of Type I P+aa
In control, the frequency of Type I P-aa

Null Hypothesis: P+aa=P-aa

Statistics:

(1)
\begin{align} S_{A^*}=\frac{P_{aa}^+-P_{aa}^-}{\sqrt{2/N}\sqrt{P_{aa}(1-P_{aa})}} \end{align}
(2)
\begin{align} power=\phi(\phi^{-1}(\alpha/2)+S_{aa})+1-\phi(-\phi^{-1}(\alpha/2)+S_{aa}) \end{align}

Considering Relative Risk $\gamma$

Assume

(3)
\begin{align} P(+|aa)=\gamma P(+|A^*) (\gamma \geq 1) \end{align}

We'll have

(4)
\begin{align} P^+_{aa}=\gamma P_{aa}/(P_{aa}(\gamma-1)+1) \end{align}
(5)
\begin{align} \hat{P}=\frac{P^+_{aa}+P^-_{aa}}{2} \end{align}
(6)
\begin{align} S_{aa}=\frac{(\gamma -1)P_{aa}(1-P_{aa})}{{(P_{aa}(\gamma -1)+1)\sqrt{\hat{P}_{aa}(1-\hat{P}_{aa})}\sqrt{2/N}}} \end{align}

An Example Of Recessive Disease

Assume a population with:
$F$ or $P(+)=0.02$
$P_{aa}=0.1, P_{Aa}=0.2, P_{AA}=0.7$
$\gamma =2.0$
Case:$N^+ =500$
Control: $N^- =500$
Results: $S_{aa}=3.718183 \gg 1.96(\alpha=0.05)$
Power=96.06%

Dominant Disease:

We can classify genotype into
Type I With Minor Allele: a* with Frequency $P_{a^*}=P_{aa}+P_{Aa}$
Type II Major homozygous Genotype: AA with Frequency $P_{aa}$
In cases, the frequency of Type II $P^+_{AA}$
In control, the frequency of Type II $P^-_{AA}$
Null Hypothesis: $P^+_{AA}=P^-_{AA}$
Statistics:

(7)
\begin{align} S_{AA}=\frac{P_{AA}^--P_{AA}^+}{\sqrt{2/N}\sqrt{P_{AA}(1-P_{AA})}} \end{align}
(8)
\begin{align} power=\phi(\phi^{-1}(\alpha/2)+S_{AA})+1-\phi(-\phi^{-1}(\alpha/2)+S_{AA}) \end{align}

Considering Relative Risk $\gamma$

Assume

(9)
\begin{align} P(+|a^*)=\gamma P(+|AA) (\gamma \geq 1) \end{align}

We'll have

(10)
\begin{align} P^+_{AA}=(P_{AA}/(P_{AA}-\gamma P_{AA} + \gamma) \end{align}
(11)
\begin{align} \hat{P}_{aa}=\frac{P^+_{aa}+P^-_{aa}}{2} \end{align}
(12)
\begin{align} S_{AA}=\frac{(\gamma -1)P_{a^*}(1-P_{a^*})}{{(P_{a^*}(\gamma -1)+1)\sqrt{\hat{P}_{a^*}(1-\hat{P}_{a^*})}\sqrt{2/N}}}=\frac{(\gamma -1)P_{AA}(1-P_{AA})}{{(\gamma - \gamma P_{AA} +P_{AA} )\sqrt{\hat{P}_{AA}(1-\hat{P}_{AA})}\sqrt{2/N}}} \end{align}

An Example Of Dominant Disease

Assume a population with:
$F$ or $P(+)=0.02$
$P_{aa}=0.1, P_{Aa}=0.2, P_{AA}=0.7$
$\gamma =2.0$
Case:$N^+ =500$
Control: $N^- =500$
Results: $S_{AA}=5.260037 \gg 1.96(\alpha=0.05)$
Power=99.95%

Modify the statistics to take into account additive effect

Assume:

(13)
\begin{align} P(+|aa)=\gamma_1P(+|Aa) (\gamma_1 \geq 1, \gamma_1=1:Dominant) \end{align}
(14)
\begin{align} P(+|Aa)=\gamma_2P(+|AA) (\gamma_2 \geq 1, \gamma_2=1:Recessive) \end{align}

So:

(15)
\begin{align} P(+|aa)=\gamma_1\gamma_2P(+|AA) \end{align}
(16)
\begin{align} \frac{1}{\gamma_1\gamma_2}P(+|aa)=P(+|AA) \end{align}

Null Hypothesis:

(17)
\begin{equation} P^+_{AA}=P^-_{AA} \end{equation}
(18)
\begin{equation} P^+_{Aa}=P^-_{Aa} \end{equation}
(19)
\begin{equation} P^+_{aa}=P^-_{aa} \end{equation}
(20)
\begin{equation} P(+)=P_{aa}P(+|aa)+P_{Aa}P(+|Aa)+P_{AA}P(+|AA) \end{equation}
(21)
\begin{align} P(+)=P_{aa}P(+|aa)+P_{Aa}P(+|aa)/\gamma_1+P_{AA}P(+|aa)/(\gamma_1\gamma_2) \end{align}
(22)
\begin{align} P(+)=P(+|aa)(P_{aa}+P_{Aa}/\gamma_1+P_{AA}/(\gamma_1\gamma_2)) \end{align}
(23)
\begin{align} P(+|aa)=P(+)/(P_{aa}+P_{Aa}/\gamma_1+P_{AA}/(\gamma_1\gamma_2)) \end{align}
(24)
\begin{equation} P(aa|+)=P(+|aa)P_{aa}/P(+) \end{equation}
(25)
\begin{align} P^+_{aa}=P_{aa}/(P_{aa}+P_{Aa}/\gamma_1+P_{AA}/(\gamma_1\gamma_2))=\gamma_1\gamma_2P_{aa}/(\gamma_1\gamma_2P_{aa}+\gamma_2P_{Aa}+P_{AA}) \end{align}

For Null Hypothesis:

(26)
\begin{equation} P^+_{aa}=P^-_{aa} \end{equation}
(27)
\begin{align} S_{aa}=\frac{\gamma _1\gamma_2P_{aa}(1/\gamma _1\gamma_2-\gamma _1\gamma_2P_{aa}-\gamma_2P_{Aa}-P_{AA})}{(\gamma _1\gamma_2P_{aa}+\gamma_2P_{Aa}+P_{AA})\sqrt{\hat{P}_{aa}(1-\hat{P}_{aa})}\sqrt{2/N}} \end{align}
(28)
\begin{equation} P(+)=P_{aa}P(+|aa)+P_{Aa}P(+|Aa)+P_{AA}P(+|AA) \end{equation}
(29)
\begin{align} P(+)=P_{aa}P(+|Aa)\gamma_1+P_{Aa}P(+|Aa)+P_{AA}P(+|Aa)/\gamma_2 \end{align}
(30)
\begin{align} P(+)=P(+|Aa)(P_{Aa}+P_{aa}\gamma_1+P_{AA}/\gamma_2) \end{align}
(31)
\begin{align} P(+|Aa)=P(+)/(P_{Aa}+P_{aa}/\gamma_1+P_{AA}/\gamma_2) \end{align}
(32)
\begin{equation} P(Aa|+)=P(+|Aa)P_{Aa}/P(+) \end{equation}
(33)
\begin{align} P^+_{Aa}=P_{Aa}/(P_{Aa}+P_{Aa}\gamma_1+P_{Aa}/\gamma_2)=\gamma_2P_{Aa}/(\gamma_1\gamma_2P_{aa}+\gamma_2P_{Aa}+P_{AA}) \end{align}

For Null Hypothesis:

(34)
\begin{equation} P^+_{Aa}=P^-_{Aa} \end{equation}
(35)
\begin{align} S_{Aa}=\frac{P_{Aa}(\gamma _1\gamma_2P_{aa}+\gamma_2P_{Aa}+P_{AA}-\gamma_2)}{(\gamma _1\gamma_2P_{aa}+\gamma_2P_{Aa}+P_{AA})\sqrt{\hat{P}_{Aa}(1-\hat{P}_{Aa})}\sqrt{2/N}} \end{align}
(36)
\begin{equation} P(+)=P_{aa}P(+|aa)+P_{Aa}P(+|Aa)+P_{AA}P(+|AA) \end{equation}
(37)
\begin{align} P(+)=\gamma_1\gamma_2P_{aa}P(+|AA)+\gamma_2P_{Aa}P(+|AA)+P_{AA}P(+|AA) \end{align}
(38)
\begin{align} P(+)=P(+|AA)(P_{AA}+P_{Aa}\gamma_2+P_{aa}\gamma_1\gamma_2) \end{align}
(39)
\begin{align} P(+|AA)=P(+)/(P_{AA}+P_{Aa}\gamma_2+P_{aa}\gamma_1\gamma_2) \end{align}
(40)
\begin{equation} P(AA|+)=P(+|AA)P_{AA}/P(+) \end{equation}
(41)
\begin{align} P^+_{AA}=P_{AA}/(P_{AA}+P_{Aa}\gamma_2+P_{aa}\gamma1\gamma_2) \end{align}

For Null Hypothesis:

(42)
\begin{equation} P^+_{AA}=P^-_{AA} \end{equation}
(43)
\begin{align} S_{AA}=\frac{P_{AA}(\gamma _1\gamma_2P_{aa}+\gamma_2P_{Aa}+P_{AA}-1)}{(\gamma _1\gamma_2P_{aa}+\gamma_2P_{Aa}+P_{AA})\sqrt{\hat{P}_{AA}(1-\hat{P}_{AA})}\sqrt{2/N}} \end{align}

Power

(44)
\begin{align} power_{AA}=\phi(\phi^{-1}(\alpha/2)+S_{AA})+1-\phi(-\phi^{-1}(\alpha/2)+S_{AA}) \end{align}
(45)
\begin{align} power_{Aa}=\phi(\phi^{-1}(\alpha/2)+S_{Aa})+1-\phi(-\phi^{-1}(\alpha/2)+S_{Aa}) \end{align}
(46)
\begin{align} power_{aa}=\phi(\phi^{-1}(\alpha/2)+S_{aa})+1-\phi(-\phi^{-1}(\alpha/2)+S_{aa}) \end{align}

Considering Mono Chromosome Model for Addtive Disease(Modified after Presentation)

(47)
\begin{equation} P(+)=P(+|AA)P_{AA}+P(+|Aa)P_{Aa}+P(+|aa)P_{aa} \end{equation}
(48)
\begin{align} P(+)=P(+|AA)P_{AA}+\gamma_2P(+|AA)P_{Aa}+\gamma_1\gamma_2P(+|AA)P_{aa} \end{align}
(49)
\begin{align} P(+&a)=0.5P(+|Aa)P_{Aa}+P(+|aa)P_{aa} \end{align}
(50)
\begin{align} P(+&a)=0.5\gamma_2P(+|AA)P_{Aa}+\gamma_1\gamma_2P(+|AA)P_{aa} \end{align}
(51)
\begin{align} P(+|a)=P(+&a)/P(a)=\frac{0.5\gamma_2P(+|AA)P_{Aa}+\gamma_1\gamma_2P(+|AA)P_{aa}}{0.5P_{Aa}+P_{aa}} \end{align}
(52)
\begin{align} P(+&A)=0.5P(+|Aa)P_{Aa}+P(+|AA)P_{AA} \end{align}
(53)
\begin{align} P(+&a)=0.5\gamma_2P(+|AA)P_{Aa}+P(+|AA)P_{AA} \end{align}
(54)
\begin{align} P(+|A)=P(+&A)/P(A)=\frac{0.5\gamma_2P(+|AA)P_{Aa}+P(+|AA)P_{AA}}{P_{AA}+0.5P_{Aa}} \end{align}

Relative risk of Allele a

(55)
\begin{align} \gamma_a=P(+|a)/P(+|A)=\frac{0.5\gamma_2P(+|AA)P_{Aa}+\gamma_1\gamma_2P(+|AA)P_{aa}}{0.5P_{Aa}+P_{aa}}/\frac{0.5\gamma_2P(+|AA)P_{Aa}+P(+|AA)P_{AA}}{P_{AA}+0.5P_{Aa}} \end{align}
(56)
\begin{align} \gamma_a=\frac{(\gamma_2P_{Aa}+2\gamma_1\gamma_2P_{aa})(P_{Aa}+2P_{AA})}{(\gamma_2P_{Aa}+2P_{AA})(P_{Aa}+2P_{aa})} \end{align}

Statics:

(57)
\begin{equation} P_a=0.5P_{Aa}+P_{aa} \end{equation}
(58)
\begin{align} P^+_a=\gamma_a P_a/(P_a(\gamma_a-1)+1) \end{align}
(59)
\begin{align} \hat{P_a}=\frac{P^+_a+P^-_a}{2} \end{align}
(60)
\begin{align} S_a=\frac{(\gamma_a -1)P_a(1-P_a)}{{(P_a(\gamma -1)+1)\sqrt{\hat{P_a}(1-\hat{P_a})}\sqrt{2/N}}} \end{align}

Power:

(61)
\begin{align} power_a=\phi(\phi^{-1}(\alpha/2)+S_a)+1-\phi(-\phi^{-1}(\alpha/2)+S_a) \end{align}

Examples Of Additive Disease

Assume a population with:
$F$ or $P(+)=0.02$
$P_{aa}=0.1, P_{Aa}=0.2, P_{AA}=0.7$
$\gamma_1\gamma_2 =2.0$
Dominant: $1\leq\gamma_1\leq 2$ Recessive
Case:$N^+ =200$
Control: $N^- =200$

When $\gamma_1 =1.41$ (additive)
$\gamma_2 =1.41$
$S_aa=3.20$
$S_Aa=0.9323$ (not significant at alpha=0.05)
$S_AA=2.2541$
$Power_aa=52.7\%$
$Power_Aa=15.4\%$
$Power_AA=65.61\%$

N=200

200.png

N=500

500.png

Future Work

Try to figure out a combination of statistics for additive models

Consider multiple variants in the same gene causing effects and consider different interaction models between them.

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