Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Genetics and archaeogenetics of South Asia wikipedia , lookup
SNP genotyping wikipedia , lookup
Pharmacogenomics wikipedia , lookup
Human genetic variation wikipedia , lookup
Polymorphism (biology) wikipedia , lookup
Genome-wide association study wikipedia , lookup
Population genetics wikipedia , lookup
Microevolution wikipedia , lookup
Genetic drift wikipedia , lookup
Hardy-Weinberg Equilibrium Life is a fine balance between shared commonalities and amazing diversity. Hardy and Weinberg Godfrey Hardy mathematician Wilhelm Weinberg physician •Proposed in 1908 •Use mathematics to describe allele frequencies in gene pools •Based on following assumptions • (places to start from) HW Model assumptions pt 1 1. Organisms are diploid (2 genes/trait) 2. Reproduction is sexual 3. Generations do not overlap 4. Parents die before children reproduce No mutations HW model assumptions pt 2 5. Gametes unite at random Mating is random One allele is not favored over another for reproduction 6. NO migration in or out of population 7. Populations size is very large Remember: probability works best with large sample sizes HW First Prediction p p= allele frequency of dominant allele q= allele frequency of recessive allele p2 +q=1 X 2pq X q2 = 1 p2 (p X p) =frequency of homozygous dominant 2pq= frequency of heterozygotes q2 (q X q) = frequency of homozygous recessive HW Model Second Prediction Allele frequencies will remain stable over time - The Population is said to be in; “Hardy Weinberg Equilibrium” Genetic variation does not disappear Polymorphic gene pools stay polymorphic This is an ideal model.. Extremely rare for any population to meet all requirements perfectly – to be in HW equilibrium Genetic frequencies still tend to stay close to stable between generations Using HW to calculate frequences Frequencies p+q =1 Frequencies of alleles: of genotypes: p squared= homozygous dominant 2pq = heterozygous q squared = homozygous recessive Example: Brown vs. Blue eyes Often, B allele(dominant) b allele(recessive) Our only can look at phenotypes class has 18 students 17 brown eyes 1 non-brown eyes Step 1 Step 1: Calculate frequency of homozygous recessive genotype (aa) Why? q2 is homozygous recessive (bb) • The only ones you can tell their genotype just by looking at them q2 = 1/18 q2 = 0.0555 (we’re saying that around 5 percent of class has nonbrown eyes and are bb) Step 2 Step 2: calculate allele frequency of q q2 = (1/18)=(0.0555) q = the square root of (0.0555) q = 0.2356 (we’re saying that around 23 % of all the eye color alleles in our class population are b) Step 3 Step 3: calculate allele frequency of p We know: p + q = 1 So p=1–q p = 1- 0.2356 p = 0.7644 Step 4: Step 4: You know the frequencies of the alleles. You can now calculate the frequency of p2 and 2 pq p2 (homozygous dominant) = • p2 =0.584 • (around 58 % is homozygous BB) 2pq (heterozygous)= • 2 pq= 0.3554 • (around 35 % is Bb) Step 5: Double check your genotype frequencies (p2 ) + (2pq) + (q2 ) =1 (0.584) + (0.3554) + (0.0555) = 1 1=1 Yah, we’re correct!!!