Transcript Inbreeding and Kinship - West Virginia University

Lab 4: Inbreeding and Kinship
Inbreeding
• Causes departure from Hardy-Weinburg
Equilibrium
• Reduces heterozygosity
• Changes genotype frequencies
• Does not change allele frequencies
http://scienceblogs.com/notrocketscience/2
009/04/14/how-inbreeding-killed-off-a-lineof-kings/
Inbreeding: Breeding between closely related individuals.
The inbreeding coefficient (f) can be calculated by:
Hf  2 pq  2 pqf  2 pq(1  f )
Hf
f  1
.
2 pq
Hf= Heterozygosity observed in a population
experiencing inbreeding
Inbreeding coefficient (f)
1. Probability that two homologous alleles in an individual
are IBD.
2. Value of “f” ranges from 0 to 1.
A1A2
A2A2
A1A2
IBD
A1A2
A1A1
A1A2
A1A2
A1A2
A1A2
A1A1
Not IBD
The inbreeding coefficient (f) can be calculated using
the fixation index (F), assuming the departure from
HWE is entirely due to inbreeding.
H
H H
E
O
O
F 1 

H
H
E
E
HO > HE, negative F-value.
HO < HE, positive F-value.
Selfing: The most extreme form of inbreeding
•Many plants, and some animals, are capable of selffertilization
•Some only self, while others have a mixed mating system
oSelfing rate: S
oOutcrossing rate: T = (1 - S)
(http://www.life.illinois.edu/help/digitalflowers/BreedingSystems/5.htm)
At inbreeding equilibrium, there is no change in
heterozygosity i.e. Ht = Ht-1 = Heq
H t 1
H t  T 2 pq  S
.
2
H eq
f eq
T 4 pq (1  S )4 pq


.
2S
2S
H eq
S
 1

.
2 pq 2  S
Rate of self-fertilization (S) can be estimated
from the relationship:
H E  HO
S
F

HE
2S
Assumptions:
1.Population is in inbreeding equilibrium.
2.Deviation from HWE is entirely due to self-
fertilization.
Problem 1: Mountain dwarf pine (Pinus mugo) typically grows at high elevations in
Southern and Central Europe.
(Ukrainian
Carpathians;
Wikimedia
Commons)
Unlike most other pines, P. mugo forms multiple intertwined stems and one tree can
occupy areas as large as 50-100 m2. Relatively little is known about the population
genetics of this species, with most of the available information coming from several
studies based on allozyme markers.
Download the data (file pmugo_allozymes.xls), analyze them using GenAlEx, and
use the output of your analyses to answer the following questions:
a) Are most populations and loci in HWE? If not, are departures generally due to
heterozygote excess or deficiency?
b) How do you explain differences among loci in departures from HWE? Do some
loci tend to show more departures than others?
c) How do you explain differences among populations?
d) P. mugo has a mixed mating system. Assuming that the observed level of
inbreeding can be accounted for by self-fertilization alone, what is the estimated
rate of self-fertilization S?
e) The rate of self-fertilization can be estimated more reliably if the genotypes of
the progeny are compared to the genotypes of their mothers for multiple loci. An
estimate of the average rate of self-fertilization using this approach is S = 0.15.
How would you explain the difference between this estimate and the one you
calculated in d)? Please consider the biology of this organism in your response.
Example 1: Estimate the inbreeding coefficient of progeny
resulting from mating between half-first cousins.
Half first-cousins share one grandparent.
CA
CA
B
C
D
E
B
C
D
E
P
P
CA
A1A2
P(A1) = ½
CA
A1A2
P(A1) = ½
C
B
P(A1) = ½
P(A1) = ½
D
E
P(A2)= 1/2
P(A2)= 1/2
B
C
P(A2)= 1/2
P(A2)= 1/2
D
E
P(A2)= 1/2
P(A1) = ½
P
A2A2
P
A1A1
P( A1A1) 
P(A2)= 1/2
P(A1) = ½
1 1 1 1 1 1 1
      .
2 2 2 2 2 2 64
P( A2 A2) 
1 1 1 1 1 1 1
      .
2 2 2 2 2 2 64
Overall probability that the two alleles in the offspring will
be IBD is:
f = P(A1A1) + P(A2A2) = 1/64 + 1/64 = 1/32
Chain- Counting Technique:
Chain for half-first cousin: D-B-CA-C-E
3
2
4
1
5
1
f  
2
N
Where, N= # of individuals in the chain.
5
1
1
f   
32
2
Example 2: Estimate the inbreeding
coefficient of progeny P.
1
f   
i 1  2 
m
CA2
CA1
B
C
D
E
P
Ni
 m= # of common ancestors = 2
Chain 1: D-B-CA1-C-E
Chain 2: D-B-CA2-C-E
N1= 5
N2= 5
1
f   
i 1  2 
m
Ni
5
5
1
1 1
     
 2   2  16
When common ancestors are
inbred :
Ni
1
f      (1  f CA(i ) )
i 1  2 
m
Where, fCA(i) is the inbreeding coefficient
of the i- th common ancestor.
Estimation of Kinship coefficient
X
A1A2
A3A4
A1A3
A2A3
A3A3
A2A3
Inbreeding coefficient (f):
Probability that two homologous
alleles in an individual are IBD.
Y
Kinship coefficient (fxy): Probability that two alleles, one
randomly chosen from each individual are IBD.
Estimation of Kinship coefficient
A3A4
A1A2
Kinship coefficient between
two individuals X and Y (fXY)
A1A3
A2A3
Y
X
A3A3
A2A3
A3A3
H
=
inbreeding coefficient (f) of
a hypothetical offspring from
X and Y.
Problem 2: The kinship coefficient of two individuals is the probability that two
alleles, one randomly chosen from each individual, are IBD. If the two individuals
can reproduce, their kinship coefficient is equal to the inbreeding coefficient of the
resulting offspring. Assuming that all common ancestors have fCA = 0.01, determine
the kinship coefficients for the following relationships:
a) Half-sibs (i.e., siblings that share one parent).
b) Full first cousins (offspring of full siblings).
c) GRADUATE STUDENTS ONLY: Monozygotic twins (Hint: Do not use the chain
counting technique for this case. Think about the strict definition of the kinship
coefficient).
d) Parent and offspring.
e) Uncle and niece.
f) Grandmother and granddaughter.
g) GRADUATE STUDENTS ONLY: Double-first cousins (i.e., first cousins that
share all four grandparents).
http://www.babiestoday.com/graphics/ds027.jpg
http://z.about.com/d/multiples/1/0/i/E/blgal481.jpg
Problem 3. You have decided to do some
targeted sequencing to determine actual
genotype distributions for the locus controlling
flower color in the Mountain Laurel population
described in Lab 3. You obtain the following
results:
a) Quantitatively evaluate the null hypothesis
that this population does not deviate from
Hardy Weinberg expectations.
b) Assuming the departure from HWE results
entirely from inbreeding, what is the inbred
fraction of this population?
c) Develop a biological hypothesis to explain
your results.
HO
F  1
HE
Genotype
PP
Count
343
PR
87
PW
62
RR
248
RW
57
WW
223
IBD