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How to review genetic association studies Lavinia Paternoster 3rd year PhD student.

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Presentation on theme: "How to review genetic association studies Lavinia Paternoster 3rd year PhD student."— Presentation transcript:

1 How to review genetic association studies Lavinia Paternoster 3rd year PhD student

2 Outline Traditional meta-analyses Why are genetic studies unique? Methods –choosing a genetic model –Multiple testing –Overall association –Per-allele mean differences Other things to consider

3 Research Question Does having gene “X” increase the risk of disease/trait “Y”? Same as: Does intervention “X” increase the risk of outcome “Y”? BUT……….

4 Traditional meta-analysis InterventionControl (or intervention 2) Observe Outcome 1Outcome 2 Input variable to be tested Outcome to test success of intervention

5 Traditional meta-analysis Intervention e.g. beta-blockers Control (or intervention 2) Observe Outcome 1 e.g. cardiovascular disease Outcome 2 e.g. no cardiovascular disease Input variable to be tested Outcome to test success of intervention

6 Traditional meta- analysis Intervention (beta-blockers) Control Outcome 1 (e.g. CVD) nn Outcome 2 (e.g. no CVD) nn Calculate relative risk (or odds ratio) for each study Pool relative risks by using weighting methods

7 Beta-blockers & cardiovascular disease

8 Traditional meta-analysis InterventionControl Observe Mean value of those with intervention Mean value of controls Variable to be tested Outcome to test success of intervention

9 Traditional meta-analysis Intervention e.g. exercise Control Observe Mean value of those with intervention e.g. mean fatigue scale value Mean value of controls e.g. mean fatigue scale value Variable to be tested Outcome to test success of intervention Edmonds et al Exercise for chronic fatigue syndrome. Cochrane

10 Traditional meta- analysis Observations (e.g. fatigue scale) Intervention 1 (exercise) nmeansd controlnmeansd Calculate mean difference (and 95%CI) for each study Pool mean differences by using weighting methods

11 Exercise & Fatigue

12 Genetic Associations The simplest mutation (a→b) gives 3 genotypes: aa, ab, bb Comparing 3 groups not 2 Conventional meta-analysis methods not suitable

13 Traditional meta-analysis Intervention e.g. beta-blockers Control (or intervention 2) Observe Outcome 1 e.g. cardiovascular disease Outcome 2 e.g. no cardiovascular disease Input variable to be tested Outcome to test success of intervention

14 Traditional meta-analysis Genotype AA Observe Outcome 1 e.g. cardiovascular disease Outcome 2 e.g. no cardiovascular disease Input variable to be tested Outcome to test success of intervention Genotype ABGenotype BB

15 Traditional meta-analysis InterventionControl Observe Mean value of those with intervention Mean value of controls Variable to be tested Outcome to test success of intervention

16 Traditional meta-analysis AABB Observe Mean value of those with genotype AA Variable to be tested Outcome to test success of intervention AB Mean value of those with genotype AB Mean value of those with genotype BB

17 My Research Meta-analysis of association between Carotid intima-media thickness and several genes Here I’ll show MTHFR example CC / CT / TT

18 Data CCCTTT

19 Methods in the literature Collapse into 2 groups –Assume genetic model Dominant (tt+ct v cc) Recessive (tt v ct+cc) –Multiple pairwise comparisons tt v cc, tt v ct, ct v cc dominant and recessive

20 Methods in the literature Analyse as 3 groups –Analyse as co-dominant (per-allele difference) –Meta-ANOVA

21 My Method 3 stage approach –Meta-ANOVA Looks for overall association between gene and trait but does not indicate which alleles increase/decrease –Determine genetic model use linear regression –Estimate mean differences using chosen genetic model

22 Meta- ANOVA Analyse by carrying out ANOVA using ‘genotype’ and ‘study’ as categorical variables and weighting each observation Test whether ‘genotype’ is a significant variable P=0.026

23 Which genetic model? Recessive –TT shows effect, CT = CC –MD1 = 0, so λ =0 Dominant –TT = CT and both show effect –MD1 = MD2, so λ =1 Co-dominant –CT will be half way between CC and TT –MD1/MD2 = 0.5 λ = MD1/MD2 MD1 = CT – CC MD2 = TT - CC Can use a linear regression of MD1 against MD2, weighted by study to determine overall the most appropriate genetic model

24 MD1 MD2 0.2 (95%CI, 0 to 0.4) λ = 0, so recessive

25 Mean differences For dominant and recessive genetic models combine 2 genotypes and use methods previously described –Recessive combine CT and CC, compare with TT –Dominant Combine TT and CT, compare with CC For co-dominant models use per-allele difference –Assumes same difference between TT & CT, and CT & CC

26 Mean differences MTHFR was associated when analysed by meta-ANOVA (p = 0.026) MTHFR was recessive (λ = 0.2) Mean difference between TT and CT/CC is: 20μm (95%CI 10 to 30)

27 Summary Genetic association studies have at least 3 groups –Chose a model based on previous evidence –Multiple comparisons –Overall association –Novel 3 stage approach

28 Other issues Other genetic models? Different polymorphisms within gene LD between genes? Whole genome meta-analysis


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