Hardy Weinberg Equilibrium Flashcards
1
Q
p
A
freq of dom allele, expressed decimal fraction
p + q = 1
2
Q
q
A
freq of rec allele, expressed decimal fraction
p + q = 1
3
Q
Allele freq. (p,q) do not evolve
given these assumptions (5)
A
*Mating random (i.e. no sexual selection)
*No net mutation
*No migration
*Infinite pop size (math world)
*No differential success of phenotypes
4
Q
Hardy Weinberg Equilibrium
A
p2 + 2pq + q2 =1 (genotype)
5
Q
p2 =
A
freq. of homozygous dom.
6
Q
2pq =
A
freq. of heterozygous
7
Q
q2 =
A
freq. of homozygous rec.
8
Q
(p2 + 2pq)
A
Dominant Phenotypic Frequency
9
Q
Selection
A
Differential survival and
reproduction of phenotypes in a
population
10
Q
Natural Selection’s key factors acting
against HW Eq. (4)
A
*All pheno. not equal
*non-random mating
*Sexual selection
*Fecundity not random
11
Q
chi squared test
A
(O-E)^2 / E
