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QUANTIFYING GENETIC CHANGE
Changes in phenotype frequencies do not always indicate evolution
P (parent gen)
0.5 B/0.5b
Bb
Bb
Bb
BB
Bb
bb
F1 (first gen)
0.5B/0.5b
Allele ‘shuffling’ in sexual reproduction changes phenotypic ratios
but not allele frequencies.
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
P
0.5 B/0.5b
Bb
Bb
Bb
F1
0.67B/0.33b
BB
BB
bb
Dominant/recessive allele relationships add to the challenge!
HARDY-WEINBERG EQUILIBRIUM
Populations 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
“Wild gray wolf still roaming California”
Random mating
HARDY-WEINBERG EQUILIBRIUM
No new
mutations
HARDY-WEINBERG EQUILIBRIUM
Using phenotype to determine genotype and allele
frequencies…
p+q=1
to find allele frequencies
where
p = dominant allele
q = recessive allele
p2 + 2pq + q2 = 1
to find genotype frequencies
*If the heterozygote
cannot be distinguished
from the homozygote
…will determine if a population is evolving
Why? http://www.uic.edu/classes/bms/bms655/lesson13.html Scroll to Fig 20
p+q=1
p2 + 2pq + q2 = 1
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 frequency
q = 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
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 =.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
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 = .0001
q = .01
p = .99
Calculate the frequency of
carriers.
2(.99)(.01) = .0198 ~ 2%
198 are carriers
PRACTICE
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.
Panaxia dominula
Hint: since it’s incomplete dominance, count alleles, then divide, to find p, q
PRACTICE
In a population of 1612 individuals 1469 are white- Panaxia dominula
spotted (AA), 138 are intermediate (Aa) and 5
have little spotting (aa)
Determine the frequency of both
the A and the a allele.
2(1469) + 138 = A alleles in population
3076/3224 = .954
2(5) + 138 = a alleles
148/3224 = .046