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Functional genomics
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Functional genomics

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  1. Functional genomics Practice

  2. Real-Time PCR I. Pair the bands (2-6 on the gel) with the melting curves!

  3. Real-Time PCR I. 5 2 6 3 4 Pair the bands (2-6 on the gel) with the melting curves!

  4. Real-Time PCR II. 5 2 6 3 4 What can be said about the genotype of the 5 patient if the examined samples are STR sequences ?

  5. Real-Time PCR II. 5 2 6 3 4 2 types of STR in each pearson. 2nd, 6th: homozygote, many repeat 3rds, 4th: heterozygote 5th: homozygote, few repeat

  6. 3 patients; „A” gene expression analysis Real-Time PCR III.detection of point mutation What is the conclusion?

  7. 3 patients; „A” gene expression analysis Real-Time PCR gyakorlati alkalmazások I. pontmutáció detektálása What is the conclusion? 3 genotypes

  8. Homozygote wild type 3 patients; „A” gene expression analysis Real-Time PCR III. detection of point mutation What is the conclusion? 3 genotypes

  9. Homozygote wild type 3 patients; „A” gene expression analysis Real-Time PCR III. detection of point mutation Heterozygote What is the conclusion? 3 genotypes

  10. Homozygote wild type 3 patients; „A” gene expression analysis Homozygote mutant Real-Time PCR III. detection of point mutation Heterozygote What is the conclusion? 3 genotypes

  11. Healthy Diseased What is the correlation between the disease and the rate of gene expression?

  12. Healthy Diseased Decreased mRNA copy number in diseased patient.

  13. Healthy Diseased Setting of the threshold cycle

  14. Healthy Diseased

  15. Healthy Diseased Define Ct value!

  16. Healthy Diseased

  17. Healthy Disease Healthy: 21 Diseased: 31 ∆Ct=2 (Ctb-Cte)

  18. DNA chipmeasuringgeneexpression What do you see? A B C D v 0h 2h 4h 6h 8h 10h

  19. DNA chipmeasuring gene expression A gene continous, steady state expression (house keeping gene), B gene decreases, C increases, D no expression A B C D v 0h 2h 4h 6h 8h 10h

  20. Measuringsimilaritybetweenexpressionpatterns Q and T genes similar because of the same ratios at each time point. How similar it its response to that of genes S and Q?

  21. Measuring similarity between expression patterns Pearson-correlation coefficient It quantifies the extent to which the expression patterns of two genes go up together and down together over several time points.

  22. Measuringsimilaritybetweenexpressionpatterns Pearson-correlation coefficient It quantifies the extent to which the expression patterns of two genes go up together and down together over several time points. =1: expression patterns of the 2 genes track perfectly =-1: expression patterns of the 2 genes track perfectly, but in opposition to the another. 0: expression patterns of the two genes do not track each other at all.

  23. Measuringsimilaritybetweenexpressionpatterns Find the correlation between Q and S genes: 1: compute the sample mean and sample standard deviation of the expression values of each genes. Xs=2,83 SS=1,067 Xq= 2,5 Sq= 0,957

  24. Measuringsimilaritybetweenexpressionpatterns Find the correlation between Q and S genes: 2: substract Xs from each value in the S row and divide each result by Ss. Do the same in the Q row, to produce the following normalized row. Snorm: -1,7; 0,16; 1,1; 1,1; 0,16; -0,5 Qnorm: -1,5; -0,5; 0,5; 1,5; 0,5; -0,5

  25. Measuringsimilaritybetweenexpressionpatterns Find the correlation between Q and S genes: Multiply the first number in Snorm by the first number in Q norm, the second number in Snorm….. And so on, keeping a running sum of these products. Divide this sum by the number of elements in each row (6) to get the correlation coefficient.

  26. Measuringsimilaritybetweenexpressionpatterns Find the correlation between Q and S genes: ρ(Q,S)=0,897

  27. Measuringsimilaritybetweenexpressionpatterns Home work: find the correlation numbe r between T and V genes Ρ (T,V)= -1