Changes in phenotype frequencies do not always indicate evolution QUANTIFYING GENETIC CHANGE Bb Bb B 0.5 B/0.5b BB Bb b 0.5B/0.5b Allele ‘shuffling’ in sexual reproduction changes phenotypic ratios but not allele frequencies. P (parent gen) F1 (first gen)
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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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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)
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
QUANTIFYING GENETIC CHANGE
Bb Bb Bb0.5 B/0.5b
0.67B/0.33b
P
F1
BB BB bb
Dominant/recessive allele relationships add to the challenge!
HARDY-WEINBERG EQUILIBRIUMPopulations will NOT evolve as long as the following conditions are met:
Large population
Phoenicopterus sp.
HARDY-WEINBERG EQUILIBRIUM
No selection
HARDY-WEINBERG EQUILIBRIUM
No immigration/emmigration
Random mating
“Wild gray wolf still roaming California”
HARDY-WEINBERG EQUILIBRIUM
No new mutations
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
p2 + 2pq + q2 = 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 20
*If the heterozygote cannot be distinguished from the homozygote
1. Determine number of individuals with homozygous recessive phenotype (q2)
2. Take square root to solve for q
3. Solve for p (1-q)
Now you know:
p = dominant allele frequencyq = recessive allele frequency
p + q = 1 p2 + 2pq + q2 = 1
4. Use p, q values to determine the frequency of each genotype in the population
p2 = homozygous dominant frequency
2pq - heterozygote frequency
q2 = homozygous recessive frequency
p + q = 1 p2 + 2pq + q2 = 1
5. Use genotype frequency to determine how many individuals in the population per genotype
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?
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). q2 = 170/400 =.425q = .652p = .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
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.
PRACTICE
About 1 in 10,000 newborn Caucasians are affected with PKU
q2 = .0001q = .01p = .99
Calculate the frequency of carriers.
2(.99)(.01) = .0198 ~ 2%198 are carriers
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
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 population3076/3224 = .954