Microsatellites and their use in
conservation biology
Cheetah
(Acinonyx jubatus)
Black rhino
(Diceros bicornus)
70000
65000
35000
96% decline since 1970
2475
0
1970
1980
1990
Black rhino trend in Africa
Black rhino
(Diceros bicornus)
California Condor
(Gymnogyps californicus)
Black-footed ferret
(Mustela nigripes)
Black-footed ferret
(Mustela nigripes)
Prairie dog
(Cynomys leucurus)
Conservation genetics
Investigates
Consequences of loss of genetic variation
How genetic variation is lost
How we can minimize the loss of genetic
variation
Conservation genetics
Consequences of loss of genetic variation
Increases probability of
extinction
 Increased susceptibility to disease
 Increased inbreeding
 Inability to respond to environmental change
Conservation genetics
How is genetic variation lost?
Natural Selection
Genetic Drift
 Random change in allele frequencies through
time
 Rate of loss = 1/2Ne per generation
 Alleles added by mutation = alleles lost to drift
Conservation genetics
Allele frequencies change more rapidly
Rare alleles lost more quickly
9 alleles
1.0
1 allele
90 alleles
10 alleles
N = 10 alleles
0.0
N = 100 alleles
Conservation genetics
Allele frequencies change more rapidly
Rare alleles lost more quickly
10 alleles
1.0
10% change
1% change
0 alleles
N = 10 alleles
91 alleles
9 alleles
0.0
N = 100 alleles
Conservation Genetics
Effective Population Size - Ne
Size of an ideal population that loses genetic
variation at the same rate as the focal population.
Conservation Genetics
Ideal Population
Equal sex ratio
Poisson variance in reproductive success
Non-overlapping generation
Constant population size
Genetic drift
What increases the rate of loss of genetic
variation?
Small populations
Genetic drift
Why are populations small?
Natural condition
Bottleneck
 A drastic reduction in population size - either
temporary or permanent
Kirtland warbler
(Dendroica kirtlandii)
Genetic drift
What increases the rate of loss of genetic
variation?
Small populations
Isolated populations
Genetic drift
Why are populations isolated?
Natural condition
Habitat fragmentation
Human modification of the environment
prevents interpopulation dispersal
Cougar
(Felis concolour)
Genetic drift
What increases the rate of loss of genetic
variation?
Small populations
Isolated populations
Breeding system
Elephant seal
(Mirounga angustirostris)
Microsatellites
What are they?
DNA sequences consisting of repeated base
sequences.
Example
ACACACACAC
AGTAGTAGTAGT
Microsatellites
Characteristics
Mostly found in non-coding regions
Neutral - No selective pressures
High mutation rates
Easy to analyze with the Polymerase Chain Reaction PCR
Everyone has a unique microsatellite fingerprint
Mutation
Stepwise mutation
Every mutation changes the number of repeat
units in the microsatellite by 1
CACACA -> CACA
CACACA-> CACACACA
PCR (Polymerase Chain Reaction)
Makes large quantities of specific pieces
of DNA
Requires only small amounts of tissue,
blood, scat, feathers, hair, etc.
PCR (Polymerase Chain Reaction)
Typically run for 33 cycles = 8,589,934,592
PCR (Polymerase Chain Reaction)
Typically run for 33 cycles = 8,589,934,592
Analysis method
PCR output
Sharp-tailed grouse
Tympanuchus phasianellus
Question
What is the
influence of
habitat
fragmentation and
isolation on
genetic variation
in a lek breeding
bird?
Collection sites
Continuous populations
Isolated populations
Sample information
Location
N
Estimated Male
Population Size
KM to nearest
sample
KM to nearest
population
Sandhill
4
17
170
117
Seney
6
31
435
100
Solon
30
245
14
14
Moquah
6
34
39
14
Crex
30
132
53
13
Gordon
9
132
14
11
NBWA
8
74
34
5
Halsey
30
>1000
71
0
McKelvie
17
>1000
40
0
Valentine
30
>1000
40
0
Medina
4
>1000
101
0
Pierre
39
>1000
164
0
Sample information
Location
N
ADL2
3
ADL
44
ADL
230
LLST
1
LLSD
3
LLSD
4
Ho
Isolated
Populations
88
7
8
6
6
6
6
0.200
Continuous 121
Populations
11
12
11
10
14
13
0.629
11
12
12
10
15
13
Total
209
Observed heterozygosity as a function of
interpopulation distance
0.8
Observed Heterozygosity
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
0
20
40
60
80
Distance (km)
100
120
140
Average number of alleles in a population as
a function of interpopulation distance
12
Average Number of Alleles
10
8
6
4
2
0
0
20
40
60
80
Distance (km)
100
120
140
Misassignments
Location Halsey McKelvie
Halsey
14
6
Valentin
e
8
McKelvi
e
4
8
3
0
Valentin
e
Medina
7
6
16
1
42.4% of animals
Pierre
3 are missassigned
2
Medin Pierre
a
2
4
4
1
29
Misassignments
Location Seney Crex Gordon Moquah
Seney
NBWA
Sandhill
Solon
6
Crex
25
2
3
Gordon
1
6
1
Moquah
1
NBWA
1
5
1
6
Sandhill
Solon
3
3
2
17.5% of animals are missassigned
1
24
Population Subdivision
 Fst
Nebraska
Fst = 0.0021
Nm = 115.7
Nebraska, North Dakota,
South Dakota
Fst = 0.0027
Nm = 93
Wisconsin
Fst = 0.45
Nm = 0.3
All Populations
Fst = 0.71
Nm = 0.1
Conclusions
Habitat fragmentation
reduces allelic
diversity and
heterozygosity even
with adjacent
populations.
Conclusions
 Sharp-tailed grouse do
not appear to disperse
between populations
separated by poor
quality habitat.
Conclusions
 Maintenance of
genetic variation in
sharp-tailed grouse
seems to occur
through dispersal
between leks
Conclusions
 Translocations are likely to
be required to maintain
appropriate amounts of
genetic variation in
fragmented habitat.
Other uses for microsatellite data
Designating true populations
Thwarting poachers
Designating true population boundaries
Polar bear populations
Polar bear populations
Thalarctos maritimus
Thwarting poachers
Bobcat (Lynx rufus)
Management areas
Conservation problem in Michigan
Hunting bag limits differ in the upper
peninsula and the lower peninsula
Bobcats
Limit 3 in the UP; 1 in the lower
Bears
Chance drawing
Locus 1
A = 0.75
B = 0.17
C = 0.05
D = 0.03
Locus 2
W = 0.05
X = 0.05
Y = 0.40
Z = 0.50
Population 1
Locus 1
A = 0.03
B = 0.05
C = 0.17
D = 0.75
Locus 2
W = 0.50
X = 0.40
Y = 0.05
Z = 0.05
Population 2
How do you determine from which population an
animal is most likely to have been born in?
Locus 1
A = 0.75
B = 0.17
C = 0.05
D = 0.03
Locus 2
W = 0.05
X = 0.05
Y = 0.40
Z = 0.50
Locus 1
A = 0.03
B = 0.05
C = 0.17
D = 0.75
Locus 2
W = 0.50
X = 0.40
Y = 0.05
Z = 0.05
Population 1
Population 2
Say the animal’s genotype is AAZY
0.75*0.75*0.40*0.50
= 0.1125
0.03*0.03*0.05*0.05
= 0.000002
It is 50,000 times more likely that the animal was born in
population 1
Bear management areas
You are outta here!
Sharp-tailed grouse
(Tympanchus pasianellus)