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

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 CHANGEP
0.5 B/0.5b
Bb Bb Bb F1
0.33B/0.67b
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 immigration/emmigration
“Wild gray wolf still roaming California”
Random mating
No selection

6. HARDY-WEINBERG EQUILIBRIUM
No new
mutations

7. Hardy-Weinberg Equilibrium
Using phenotype to determine genotype and allele frequencies…
p + q = 1
to find allele frequencies
p2 + 2pq + q2 = 1
to find genotype frequencies
where
p = dominant allele
q = recessive allele
*If the heterozygote cannot be distinguished from the homozygote
…to determine if a population is evolving
Why? Scroll to Fig 20

8. p + q = p2 + 2pq + q2 = 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

9. p + q = p2 + 2pq + q2 = 1
Use p, q values to determine the number of each
genotype in the population
p2 = homozygous dominant frequency
2pq - heterozygote frequency
q2 = homozygous recessive frequency

10. 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

11. PRACTICE 2(1469) + 138 = A alleles in population 3076/3224 = .954
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
2(1469) = A alleles in population
3076/3224 = .954
2(5) = a alleles
148/3224 = .046

12. 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?
Most people — about 85% — are Rh positive. But if a woman who is Rh negative and a man who is Rh positive conceive a baby, there is the potential for a baby to have a health problem. The baby growing inside the Rh-negative mother may have Rh-positive blood, inherited from the father. Approximately half of the children born to an Rh-negative mother and Rh-positive father will be Rh positive.
Rh incompatibility usually isn't a problem if it's the mother's first pregnancy because, unless there's some sort of abnormality, the fetus's blood does not normally enter the mother's circulatory system during the course of the pregnancy.
However, during delivery, the mother's and baby's blood can intermingle. If this happens, the mother's body recognizes the Rh protein as a foreign substance and can begin producing antibodies (protein molecules in the immune system that recognize, and later work to destroy, foreign substances) against the Rh proteins introduced into her blood.

13. 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) = frequency
.14 (400) = in the population
Most people — about 85% — are Rh positive. But if a woman who is Rh negative and a man who is Rh positive conceive a baby, there is the potential for a baby to have a health problem. The baby growing inside the Rh-negative mother may have Rh-positive blood, inherited from the father. Approximately half of the children born to an Rh-negative mother and Rh-positive father will be Rh positive.
Rh incompatibility usually isn't a problem if it's the mother's first pregnancy because, unless there's some sort of abnormality, the fetus's blood does not normally enter the mother's circulatory system during the course of the pregnancy.
However, during delivery, the mother's and baby's blood can intermingle. If this happens, the mother's body recognizes the Rh protein as a foreign substance and can begin producing antibodies (protein molecules in the immune system that recognize, and later work to destroy, foreign substances) against the Rh proteins introduced into her blood.

14. 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.

15. 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) = .02 2%