Snp Polimorfismo

Páginas: 2 (438 palabras) Publicado: 12 de agosto de 2012
SNP | Tests for deviation from Hardy-Weinberg equilibrium | Tests for association (C.I.: 95% convidence interval) |
| Controls | Cases | allele freq. difference | heterozygous | homozygous |allele positivity | Armitage's trend test |
SNP1 | n11=110 (111.39)
n12=157 (154.23)
n22=52 (53.39)
f_a1=0.59 +/-0.019
F=-0.01798
p=0.748134 (Pearson)
p=0.748026 (Llr)
p=0.817038 (Exact) | n11=52(50.31)
n12=68 (71.37)
n22=27 (25.31)
f_a1=0.59 +/-0.029
F=0.04727
p=0.566531 (Pearson)
p=0.566806 (Llr)
p=0.610485 (Exact) | Risk allele 2 |
| | | [1]<->[2] | [11]<->[12] |[11+]<->[22] | [11]<->[12+22] | common odds ratio |
| | | Odds_ratio=1.025
C.I.=[0.774-1.356]
chi2=0.03
p=0.86548 (P) | Odds_ratio=0.916
C.I.=[0.593-1.416]
chi2=0.16p=0.69377 | Odds_ratio=1.098
C.I.=[0.621-1.942]
chi2=0.10
p=0.74695 | Odds_ratio=0.962
C.I.=[0.639-1.448]
chi2=0.04
p=0.85106 | Odds_ratio=1.035

chi2=0.03
p=0.86566 |
| | | Risk allele1 |
| | | [2]<->[1] | [22]<->[12] | [22]<->[11] | [11+12]<->[22] | common odds ratio |
| | | Odds_ratio=0.976
C.I.=[0.737-1.292]
chi2=0.03
p=0.86548 (P) |Odds_ratio=0.834
C.I.=[0.484-1.439]
chi2=0.43
p=0.51411 | Odds_ratio=0.910
C.I.=[0.515-1.610]
chi2=0.10
p=0.74695 | Odds_ratio=0.866
C.I.=[0.519-1.445]
chi2=0.31
p=0.58064 | Odds_ratio=0.966chi2=0.03
p=0.86566 |

Legend: |
The tests for association are adapted from Sasieni PD (1997). |
n11(e): | Genotype 11 (expected) |
n12(e): | Genotype 12 (expected) |
n22(e): |Genotype 22 (expected) |
f a1: | Frequency of allele 1 +/- standard deviation |
F: | Inbreeding coefficient |
p (Pearson): | Pearson's goodness-of-fit chi-square (degree of freedom = 1) |
p (Llr): |Log likelihood ratio chi-square (degree of freedom = 1) |
p (Exact): | Exact test |
The following equations correspond to risk allele 2. |
Odds ratio (allele freq. difference): | (Case_a2 *...
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