Download pdf

Survey
yes no Was this document useful for you?
   Thank you for your participation!

* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project

Document related concepts

Inbreeding wikipedia , lookup

Genetics and archaeogenetics of South Asia wikipedia , lookup

Polymorphism (biology) wikipedia , lookup

Dominance (genetics) wikipedia , lookup

Population genetics wikipedia , lookup

Genetic drift wikipedia , lookup

Microevolution wikipedia , lookup

Hardy–Weinberg principle wikipedia , lookup

Transcript
http://www.yucatantravel.com/hotels/celestun/images/flamingos.jpg
Initially… scientists believed that natural selection
“should always favor an optimal form.”
-why does variation continue?
-shouldn’t natural selection choose the “best” allele
for that trait and then cause other alleles to disappear?
-Don’t dominant genes push recessive genes out of the
gene pool?
http://news.bbc.co.uk/media/images/39092000/jpg/_39092214_pg_squirrel_pa.jpg
1908- G.H. Hardy and G. Weinberg
Gene frequencies remain constant from generation to
generation (equilibrium) if:
1. The population is large
2. Mating is random
3. No new mutations
4. No migration from other distinct populations
5. All genotypes are reproductively equal (natural selection
not occurring)
When H-W equation does NOT accurately predict gene
frequencies for generation after generation, it is safe to assume
that the population is evolving (not in equilibrium).
“Frequency”=
# within a given category
total # in population
Example: What is the phenotype frequency of individuals in
a given population with attached or unattached earlobes?
200 =.2
1000
200
800
800 =.8
1000
http://www.cartage.org.lb/en/themes/Sciences/LifeScience/PhysicalAnthropology/HeredityGenetics/lobe.gif
The allele for attached earlobes (e) is recessive and for
unattached (E) is dominant.
What is the frequency of ee in this population?
200 =.2
1000
200
800
800 =.8
1000
What is the frequency of EE in this population?
That’s a little trickier because EE and Ee look the same…
So, Hardy and Weinberg provided us with a familiar equation
to calculate this.
The Hardy-Weinberg Equation:
(p + q)2 = p2 + 2pq + q2
when (p + q) = 1
Where:
p = frequency of the more common allele (E)
q = frequency of the less common allele (e)
Therefore:
p2 = frequency of homozygous dominant (EE), in this case
q2 = frequency of homozygous recessive (ee)
2pq = frequency of heterozygous (Ee)
So the question was, what is the frequency
of EE in this population?
(p + q)2 = p2 + 2pq + q2
(EE)
(Ee)
(ee)
200
What do we already know?
•ee = q2 = .2
•so, q = !(.2) = .447
We also know that p + q = 1
•so, 1-.447 = .553 = p
EE is p2, so (.553)2 = .305
800
It’s a smart idea to check your work…
p2 = EE = (. 553)2 = .305
2pq = Ee = 2(.553)(.447) = .494
q2 = (.447)2 = .2
These three numbers should add up to approx. 1
.305 + .494 + .2 = .999
So what did we just figure out???
30.5% of the population is EE
49.4% of the population is Ee
20% of the population is ee