We need a mathematical tool to measure how much the population is evolving.
Numbers will enable us to evaluate, compare, and then
predict evolutionary processes.
Population GeneticsAnd
The Hardy-Weinberg Model
Suppose we could quantify, give numbers, how fast a population is evolving in terms on one single trait.
How would this be useful for medicine, studying the wildlife, other?
Let’s Define:–a Population is a localized group
of interbreeding individuals.–Gene pool is the entire collection
of alleles in the population.(What is the size of our class’ gene pool?)–Allele frequency is how common
(in %) an allele (H or h) is in the population. (What are the frequencies in our cards?)
Evolution = a change in the allele frequencies.
How can we tell if the allele frequencies are changing from a
single observation?
We predict how a gene pool should be like if there is no change. Then we compare to the actual gene pool.
W. Weinbergphysician
G.H. Hardy
Mathema-tician
Hardy and Weinberg: “At equilibrium (not evolving) the frequencies of the alleles do not change”.
Mix alleles, try again: What are the frequencies of H, h in our class?
5 Agents of evolutionary changeMutation Gene Flow
Genetic Drift
Selection
Non-random mating
Hypothetical conditions under which the allele frequencies do not change:
1. very large population size 2. no migration (no gene flow in or
out)3. no mutation (no genetic change)4. random mating (no sexual
selection)5. no natural selection (everyone is
equally fit)
Under the Hardy-Weinberg conditions:We assume 2 alleles = H, h–frequency of dominant allele (H) = p –frequency of recessive allele (h) = q 1) Frequencies must add up to 1 (100%), so: p + q = 1
hhHhHH
h (q)H (p)
Accordingly – what are the chances to select a certain genotype? Phenotype?
H (p)
h (q)pxq
qxpp2
q2
HH Hh
Hh hh q2
Hardy-Weinberg theoremCounting Individuals:–frequency of HH: pxp = p2 –frequency of hh: qxq = q2 –frequency of Hh:
(pxq) + (qxp) = 2pq2) Frequencies of all individuals must add to 1 (100%), so:
p2 + 2pq + q2 = 1
What are the genotype frequencies?
Using Hardy-Weinberg equation
q2 (bb): 16/100 = .16q (b): √.16 = 0.4p (B): 1 - 0.4 = 0.6
population: 100 cats84 black, 16 whiteHow many of each genotype?
bbBbBBp2=.36 2pq=.48 q2=.16
Must assume population is in H-W equilibrium!
Using Hardy-Weinberg equation
bbBbBBp2=.36 2pq=.48 q2=.16
Assuming H-W equilibrium
Actual data bbBbBB
p2=.74 2pq=.10 q2=.16
How do you explain the data?
p2=.20 2pq=.64 q2=.16
How do you explain the data?
Null hypothesis
In this population, we find that the frequency of heterozygotes is greater than expected under the H-W assumption.
Therefore: •The population is not at equilibrium (Why???)
•There might be an advantage to being a heterozygote. Why??
Let’s try it with alleles in ‘our hands’
h = albinoH = non-albino
Each student gets two alleles at random.
Will the allele frequencies agree with the H-W equation?
AP Biology 2008 writing question: Evolution involves change in the frequencies of alleles in a population. For a particular genetic locus in a population, the frequency of the recessive allele (a) is 0.4 and the frequency of the dominant allele (A) is 0.6.a. What is the frequency of each genotype (AA, Aa, aa) in this population? b. What is the frequency of the dominant phenotype?
To Conclude:* The H-W equation describes a rare population that is at equilibrium.* By Comparing an actual population to the H-W frequencies, we can detect an instability in allele frequency.We can then ask which of the five H-W assumptions is not met.
Hardy-Weinberg formulas
Alleles: p + q = 1
Individuals: p2 + 2pq + q2 = 1
hhHhHH
HH
H h
Hh hh
For a population at equilibrium:
In the scarlet tiger moth (Panaxia dominula), coloration had been previously shown to behave as a single-locus, two-allele system with incomplete dominance. Data for 1612 individuals are given below: White-spotted (AA) =1469 Intermediate (Aa) = 138 Little spotting (aa) =5 Calculate: (¦)A, (¦)a, (¦)AA, (¦)Aa, (¦)aa
Scarlet tiger moth (Panaxia dominula)
Application of H-W principle• Sickle cell anemia
– inherit a mutation in gene coding for hemoglobin• oxygen-carrying blood protein• recessive allele = HsHs
– normal allele = Hb
– low oxygen levels causes RBC to sickle• breakdown of RBC• clogging small blood vessels• damage to organs
– often lethal
Sickle cell frequency• High frequency of heterozygotes
– 1 in 5 in Central Africans = HbHs
– unusual for allele with severe detrimental effects in homozygotes• 1 in 100 = HsHs
• usually die before reproductive age
Why is the Hs allele maintained at such high levels in African populations?
Suggests some selective advantage of being heterozygous…
Malaria Single-celled eukaryote parasite (Plasmodium) spends part of its life cycle in red blood cells
1
2
3
Heterozygote Advantage• In tropical Africa, where malaria is common:
– homozygous dominant (normal)• die or reduced reproduction from malaria: HbHb
– homozygous recessive • die or reduced reproduction from sickle cell anemia: HsHs
– heterozygote carriers are relatively free of both: HbHs
• survive & reproduce more, more common in population
Hypothesis:In malaria-infected cells, the O2 level is lowered enough to cause sickling which kills the cell & destroys the parasite. Frequency of sickle cell allele &
distribution of malaria
Hardy-Weinberg equilibrium• Hypothetical, non-evolving population
– preserves allele frequencies• Serves as a model (null hypothesis)
– natural populations rarely in H-W equilibrium– useful model to measure if forces are acting on a population
• measuring evolutionary change
W. Weinbergphysician
G.H. Hardymathematician
We can count the phenotypes, and accordingly, calculate the frequency of the genotypes, under equilibrium conditions.
Suppose we know which allele is recessive. The chances for an offspring to inherit a b are qxq = q2