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

2(5) + 138 = a alleles148/3224 = .046