Population Genetics
 Mendelian genetics predicts the outcome of specific
matings between individuals
 What about the genetics of an entire population?
 Population = all individuals of one species living in a given area
 Population genetics works with the entire gene pool
 or all the alleles present in the whole population
How will alleles change in the population?
 Among a population of 2000 people:
 720 have blue eyes (recessive)
 1280 have brown eyes (dominant)
 DNA testing reveals:
 320 are homozygous for Brown Eyes(BB)
 960 are heterozygous for Brown Eyes (Bb)
Allele frequency
 Allele frequency is the fraction:
no. of a particular allele
no. of all alleles in population
 For these 2000 people, there are 4000 alleles in the
gene pool:
 720 bb
 960 Bb
 320 BB
how many B alleles?
how many b alleles?
f(B)= 960 + 320 + 320/ 4000 = 0.4
F(b) = 960 + 720 +720/4000 = 0.6
What happens in the next generation?
 In all the matings for this generation, what is the chance
that an egg with the B allele will be fertilized by a sperm
with the b allele and create a person with Bb genotype?
Recall:
 40% of all eggs will carry B
 60% of all sperm will carry b
Recall the Rule of Multiplication:
(prob. of event a) (prob of event b)=
probability of both events happening
0.4 x 0.6 = 0.24
24% of offspring Bb
*Assuming no difference between sexes
and no mating preferences!
What happens in the next generation?
 In all the matings for this generation:
40%
60%
B
b
0.4 x 0.4 = 16% BB
0.4 x 0.6 x 2 = 48% Bb
40%
B
BB
Bb
-Bb and bB
60%
b
Bb
bb
-(rule of addition)
0.6 x 0.6 = 36% bb
What happens in the next generation?
 In all the matings for this generation:
if 4000 offspring are born:
 0.4 x 0.4 = 16% BB
 0.4 x 0.6 x 2 = 48% Bb
 0.6 x 0.6 = 36% bb
640 BB
1920 Bb
1440 bb
2560 Brown
1440 Blue
What happens in the next generation?
 New allele frequencies:
If 4000 offspring
640 BB
3200 B allele/8000 = 0.4
1920 Bb
4800 b allele/8000 = 0.6
1440 bb
After 5 generations:
64,000 offspring
10,240 BB
30,720 Bb
23,040 bb
51,200 brown alleles / 128,000 = 0.4
76,800 blue alleles / 128,000 = 0.6
 After 5 generations (or any number):
 Allele frequencies do not change!
 Recessive alleles are maintained in the population
*If some specific assumptions are made
Hardy-Weinberg equilibrium
 Godfrey Hardy (mathematician) and Wilhelm
Weinberg (physician) (early 1900s):
 Given some assumptions, allele frequencies won’t
change:
 The population is large
 Mating is random
 No migration in or out
 No mutation
 No selection (no allele is advantageous)
 How often in nature are ALL of these assumptions met?
 Rarely, if ever. This is an “ideal” state.
Does Hardy-Weinberg work?
 In large populations, the Hardy-Weinberg equations
predict results quite well for many traits
 If a population is not in equilibrium:
 Allele frequencies are changing
 Evolution is occurring!
Hardy-Weinberg equations
 Allele frequency:
 Let p = frequency of the dominant allele
 Let q = frequency of the recessive allele
 Then, p + q = 1
 Genotype frequency:
 p2 = frequency of homozygous dominant genotype
 q2 = frequency of homozygous recessive genotype
 2pq = frequency of heterozygous genotype
 p2 + 2pq + q2 = 1
4 Steps to solving H-W Problems
1. set recessives = q2
2. Take square root of q2
3. 1-q = p
4. Plug into expanded equation
Example: 16% of the cat population is white:
1. q2= 0.16
2. square root = 0.4
so, 36% of population is TT
3. 1- 0.4 = p p = 0.6
48% of population is Tt
4. .16 + 2 (.6) (.4) + .36 = 1
Another Example:
 Fraggles are mythical, mouselike creatures that live
beneath flower gardens.
 Of the 100 fraggles in a population, 75 have green hair
(FF or Ff) and 25 have grey hair (ff).
 Assuming genetic equilibrium:
 What are the gene frequencies of F and f?
 What are the genotypic frequencies?
Answer to Fraggle Problems:
 Gene frequencies:
 q2= .25, so:
 q= .5
 p= .5
 Genotypic frequencies
 FF = .25
 Ff = .5
 f f = .25
Application of H-W principle
 Sickle cell anemia
 inherit a mutation in gene coding for hemoglobin


oxygen-carrying blood protein
recessive allele = s
 normal allele = S
 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 = Ss
 unusual for allele with severe
detrimental effects in homozygotes


1 in 100 =ss
usually die before reproductive age
Why is the s 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: SS
 homozygous recessive
 die or reduced reproduction from sickle cell anemia: ss
 heterozygote carriers are relatively free of both: Ss

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
Sickle Cell Example:
 If 9% of an African population is born with a severe
form of sickle-cell anemia (ss), what percentage of the
population will be more resistant to malaria?
 f(ss)= .09 = q2
 q= .3
 P= .7
 2pq= .42
 so 42% of the population is resistant to malaria.
Using Hardy-Weinberg
Cystic Fibrosis: 1 in 1700 US Caucasian newborns have
cystic fibrosis. Use an F for the normal allele and f for
recessive:
.00058 = q2
What percent of the above population have cystic fibrosis? q= .024
What percent are healthy, non carriers?
What percent are carriers of cyctic fibrosis?
In a population of 1700 people, how many would you
expect to be homozygous normal?
In a population of 1700 people, how many would you
expect to be heterozygous?
p= .976
P2= .9524
2pq= .0468
1700 x .9524= 1619
1700 x .0468= 80
Stem Cells
 Nova