G.H. Hardy (the English mathematician) and W. Weinberg (the German physician) independently worked out the mathematical basis of population genetics in 1908. Their formula predicts the expected genotype frequencies using the allele frequencies in a diploid Mendelian population. They were concerned with questions like &quot;what happens to the frequencies of alleles in a population over time?&quot; and &quot;would you expect to see alleles disappear or become more frequent over time?&quot;
Sampled data 1: Hybrids are in some way weaker. Immigration in from an external population that is predomiantly homozygous B Non-random mating... white cats tend to mate with white cats and black cats tend to mate with black cats. Sampled data 2: Heterozygote advantage. What’s preventing this population from being in equilibrium.
Sickle Cell: In tropical Africa, where malaria is common, the sickle-cell allele is both an advantage & disadvantage. Reduces infection by malaria parasite. Cystic fibrosis: Cystic fibrosis carriers are thought to be more resistant to cholera: 1:25, or 4% of Caucasians are carriers Cc
Measuring Evolution of Populations SLIDE SHOW MODIFIED FROM KIM FOGLIA@explorebiology.com
Using Hardy-Weinberg equation What are the genotype frequencies? q 2 (bb): 16/100 = .16 q (b): √.16 = 0.4 p (B): 1 - 0.4 = 0.6 population: 100 cats 84 black, 16 white How many of each genotype? bb Bb BB p 2 =.36 2pq =.48 q 2 =.16 Must assume population is in H-W equilibrium!
Using Hardy-Weinberg equation bb Bb BB p 2 =.36 2pq =.48 q 2 =.16 Assuming H-W equilibrium Sampled data p 2 =.74 2pq =.10 q 2 =.16 How do you explain the data? p 2 =.20 2pq =.64 q 2 =.16 How do you explain the data? Null hypothesis bb Bb BB
Bb and BB pigs “look alike” so can’t tell their alleles by observing their phenotype. ALWAYS START WITH RECESSIVE alleles. p= dominant allele q = recessive allele 4/16 are black. So bb or q 2 = 4/16 or 0.25 q = 0.25 = 0.5
p + q = 1 p + 0.5 = 1 p = 0.5 Now you know the allele frequencies. The frequency of the recessive (b) allele q = 0.5 The frequency of the dominant (B) allele p = 0.5 http://www.phschool.com/science/biology_place/labbench/lab8/samprob1.html
1/1700 have cystic fibrosis q 2 = 1/1700 q = 0.00059 q = 0.024 p = 1 – 0.024 = 0.976 Frequency of C = 97.6% Frequency of c = 2.4% NOW FIND THE GENOTYPIC FREQUENCIES
CC or p 2 = (0.976) 2 = .953 Cc or 2pq = 2 (0.976) (0.024) = 0.0468 cc = 1/1700 = 0.00059 CC = 95.3% of population Cc = 4.68% of population cc = .06% of population
Now you can answer questions about the population: How many people in this population are heterozygous? It has been found that a carrier is better able to survive diseases with severe diarrhea. What would happen to the frequency of the "c" if there was a epidemic of cholera or other type of diarrhea producing disease? 0.0468 (1700) = 79.5 ~ 80 people are Cc Cc more likely to survive than CC. c will increase in population
The gene for albinism is known to be a recessive allele. In Michigan, 9 people in a sample of 10,000 were found to have albino phenotypes. The other 9,991 had skin pigmentation normal for their ethnic group. Assuming hardy-Weinberg equilibrium, what is the allele frequency for the dominant pigmentation allele in this population? q 2 = 9/10000 q = 0.0009 q = 0.03 p = 1 – 0.03= 0.97 Frequency of C = 97% Frequency of c = 3%
CC or q 2 = (0.976) 2 = .953 Cc or 2pq = 2 (0.976) (0.024) = 0.0468 cc = 1/1700 = 0.00059 CC = 95.3% of population Cc = 4.68% of population cc = .06% of population