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Changes in phenotype frequencies do not always indicate evolution QUANTIFYING GENETIC CHANGE Bb Bb Bb 0.5 B/0.5b BB Bb bb0.5B/0.5b Allele ‘shuffling’ in.

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Presentation on theme: "Changes in phenotype frequencies do not always indicate evolution QUANTIFYING GENETIC CHANGE Bb Bb Bb 0.5 B/0.5b BB Bb bb0.5B/0.5b Allele ‘shuffling’ in."— Presentation transcript:

1 Changes in phenotype frequencies do not always indicate evolution QUANTIFYING GENETIC CHANGE Bb Bb Bb 0.5 B/0.5b BB Bb bb0.5B/0.5b Allele ‘shuffling’ in sexual reproduction changes phenotypic ratios but not allele frequencies. P (parent gen) F1 (first gen)

2 QUANTIFYING GENETIC CHANGE Have to look to the root of phenotype - the genotype If the proportion of alleles in a population changes, then we know its evolving

3 QUANTIFYING GENETIC CHANGE Bb Bb Bb 0.5 B/0.5b 0.67B/0.33b P F1 BB BB bb Dominant/recessive allele relationships add to the challenge!

4 HARDY-WEINBERG EQUILIBRIUM Populations will NOT evolve as long as the following conditions are met: Large population Phoenicopterus sp.

5 HARDY-WEINBERG EQUILIBRIUM No selection

6 HARDY-WEINBERG EQUILIBRIUM No immigration/emmigration Random mating “Wild gray wolf still roaming California”

7 HARDY-WEINBERG EQUILIBRIUM No new mutations

8 HARDY-WEINBERG EQUILIBRIUM where p = dominant allele q = recessive allele Using phenotype to determine genotype and allele frequencies… p + q = 1 to find allele frequencies p 2 + 2pq + q 2 = 1 to find genotype frequencies …will determine if a population is evolving Why? http://www.uic.edu/classes/bms/bms655/lesson13.html Scroll to Fig 20http://www.uic.edu/classes/bms/bms655/lesson13.html *If the heterozygote cannot be distinguished from the homozygote

9 1.Determine number of individuals with homozygous recessive phenotype (q 2 ) 2. Take square root to solve for q 3. Solve for p (1-q) Now you know: p = dominant allele frequency q = recessive allele frequency p + q = 1 p 2 + 2pq + q 2 = 1

10 4.Use p, q values to determine the frequency of each genotype in the population p 2 = homozygous dominant frequency 2pq - heterozygote frequency q 2 = homozygous recessive frequency p + q = 1 p 2 + 2pq + q 2 = 1 5. Use genotype frequency to determine how many individuals in the population per genotype

11 PRACTICE An individual either has, or does not have, the "Rhesus factor" - aka Rh - on the surface of their red blood cells. The presence of Rh reflects a dominant allele. In a study of human blood groups, it was found that among a population of 400 individuals, 230 had the Rh protein (Rh+) and 170 did not (Rh-). For this population, calculate both allele frequencies. How many of the Rh+ individuals would be expected to be homozygous dominant?

12 PRACTICE Among a population of 400 individuals, 230 had the Rh protein (Rh+) and 170 did not (Rh-). For this population, calculate both allele frequencies (use R and r). q 2 = 170/400 =.425 q =.652 p =.348 How many of the Rh+ individuals would be expected to be homozygous dominant? p2 = (.348)(.348) =.121 ++ frequency.121 (400) = 48 ++ in the population

13 PRACTICE Phenylketonuria is a genetic condition that causes severe mental retardation due to a rare autosomal recessive allele. About 1 in 10,000 newborn Caucasians are affected with the disease. Calculate the frequency of carriers.

14 PRACTICE About 1 in 10,000 newborn Caucasians are affected with PKU q 2 =.0001 q =.01 p =.99 Calculate the frequency of carriers. 2(.99)(.01) =.0198 ~ 2% 198 are carriers

15 Wing coloration in the Scarlet Tiger Moth, behaves as a single-locus, two-allele system with incomplete dominance. In a population of 1612 individuals 1469 are white-spotted (AA), 138 are intermediate (Aa) and 5 have little spotting (aa) Determine the frequency of both the A and the a allele. PRACTICE Panaxia dominula Hint: since it’s incomplete dominance, count alleles, then divide, to find p, q

16 In a population of 1612 individuals 1469 are white- spotted (AA), 138 are intermediate (Aa) and 5 have little spotting (aa) Determine the frequency of both the A and the a allele. PRACTICE Panaxia dominula 2(1469) + 138 = A alleles in population 3076/3224 =.954 2(5) + 138 = a alleles 148/3224 =.046


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