Lecture 3

Question 1

Genotype to Phenotype (variations in CYP2C19)

Answer 1

-ultrarapid metabolizer (UM): alleles w inc activity
-extensive metabolizer (EM): alleles w normal activity
-Intermediate metabolizer (IM): allele w reduced and normal activity
-Poor metabolizer (PM): alleles w reduced activity

Question 2

how genotype determine phenotype

Answer 2

-variant can change PK or PD

Question 3

genetic variant effect on PK

Answer 3

-alter enzymatic activity during drug ADME*
-inter-patient difference in concentration, dose, etc
-inter-pt difference in toxicity profiling and efficacy

Question 4

Genetic variant effect on PD

Answer 4

-alter drug target activity
-create new drug target*
-alter protein structure
-change drug-receptor binding
-inter-pt difference in toxicity and efficacy

Question 6

CYP2C19?

Answer 6

no star

Question 7

Minor Allele Frequency (MAF) *

Answer 7

-

Question 8

Genotype

Answer 8

-combination of alleles read in PGx testing

Question 9

Genotype Frequency

Answer 9

-2 homologous chromosomes are read at same time
-can only calculate allele freq from population based on number of genotypes observed

Question 10

Allele freq

Answer 10

-calculated from population based on number of genotypes observed
-for a given population of N individuals, number of homo chromos or alleles is 2N

Question 11

Allele Frew based on observed genotype data **

Answer 11

-for a population of N individuals w:
-Q number of ppl w T/T
-R # w T/C
-S # w C/C

N = Q+R+S

-number of T allele: 2Q + R
-number of C allele: 2S + R

-divide by 2N for %

Question 12

Calculate allele Freq practice

Answer 12

-slide 11

Question 13

How to figure out MAF from known genotype freq *

Answer 13

-allele% = homozygote % + 1/2 heterozygote

-T% = (2Q+R)/2N = (Q/N + R/2N)

-Q/N = freq for T/T
-R/2N = T/C
-S/N = C/C

Question 14

slide 13 practive ***

Answer 14

slide 13 practice

Question 15

Common and rare allele *

Answer 15

-allele freq usually stay stable in population
-minor allele freq (MAF)
-Rare allele freq (RAF)

-T>C
-T is common allele, C is rare

Question 16

Rare allele *

Answer 16

-can be a common allele in another population

Question 17

Crossover and Recombination

Answer 17

-phenomenon between homologous chromos during meiosis to produce gametes
-important way to exchange genetic info

Question 18

Recombination

Answer 18

-result of crossover during meiosis that leads to recombinant chromosomes

Question 19

Haplotype *

Answer 19

-group of genes within organism that were inherited together from a single parent
-refer to inheritance of a cluster of SNPs

Question 20

The application of haplotype in PGx

Answer 20

-function of gene is combined result of all functional alleles
-C-A-T will have higher enzyme activity (example)

Question 21

Linkage Disequilibrium (LD) *

Answer 21

-non-random association of alleles at different loci on same chromosome
-no LD when infinite recombination
-complete LD when no recombination
-incomplete LD when recombination occurs in a portion of chromosomes

Question 22

If the distance between two loci is SHORT enough

Answer 22

-there will be NO recombination in a population
-leads to co-segregation of both loci into next generation = complete linkage
-DNA inherited by segments called haplotype block

Question 23

If distance between 2 loci is LONG enough

Answer 23

-there will be a large number of recombination in a population
-leads to independent segregation of the two loci into the next generation, a situation same as that 2 loci are located on two different chromosomes
-there is NO linkage between the two loci
-not all loci on the same chromosome are in linkage

-no star

Question 24

Measures of LD *

Answer 24

-R^2 = how strong the correlation between 2 variables
-R = 0: No LD
-R = 1: complete LD
-R >0.8: strong LD

Question 25

R^2. *

Answer 25

-measures extent of correlation between a pair of variables or the extent of concordance in genotypic association between loci
-varies between 0 and 1
-measures strength of LD

Question 26

Application of LD in PGx

Answer 26

-LD useful in PGx research and testing
-tagSNP

Question 27

LD example

Answer 27

-if R between locus 1,2, and 3 and also 4,5, and 6 is >/= 0.8, then we can choose any SNP from 1,2,3 and any SNP from 4-6 as a tagSNP
-this simplifies research and reduces cost
-can be used in clinical practice to select a tagSNP for PGx testing

Question 28

tagSNP

Answer 28

-SNP that can be representative for other SNPs due to strong LD
-whole genome sequencing is more popular so tagSNP less common but still useful as an anchor for reference