Download Gene and Genotype frequencies

Genetic constitution of a
population
Gene and Genotype
Frequencies
Lecture 3
Aim
• To equip students with more
principles of population genetics as
basis for quantitative genetics
Objectives
• By the end of the topic, students
should be able to
• Define gene and genotype frequencies
• Relate gene and genotype frequencies
• Describe HW equilibrium
• Explain conditions required to fulfill HW
equilibrium
Genetic constitution of a
population
Can be described by
specifying their genotypes and stating the
number of genotypes in the group
Specifying the genes carried by the
population that influence a particular trait
and stating the number or proportions of
different alleles at a locus
For simplicity sake, we assume a single
locus A that has two alleles, designated
A1, A2; or A, a
Genotype frequency
From two alleles A, a , there are three
possible combinations (genotypes)
i.e. an individual can be
A A
Aa
aa
In a population
A
A
a
A
a
a
The genetic constitution of the group or
population would be described by the
proportion or percentage of individuals
that belong to each genotype
A population of sheep
Dorper and Red Masai sheep
Genotype frequency
These proportions are called Genotype
frequencies
Definition: Proportion or frequency of a
particular genotype among individuals in
a population is called Genotype
frequency
Genotype frequency
Assume there are 100 individuals in a
population and would like to describe the
genetic structure of the population
You count how many belong to AA Aa
aa In a population
Naturally the frequencies of all genotypes
must add up to unity or 100 %
Calculation of genotype
frequency
At the SAU, there are 57 chickens of three
different phenotypes
Yakuda (assume BB)
Chiphulusa (assume Bb)
Kawangi (assume bb)
30
17
10
If these represent three genotypes, their
frequencies are
Calculation of genotype
frequency
N = 57
BB (Yakuda)
Bb (Chiphulusa)
bb (Kawangi)
30/57 = 0.526
17/57 = 0.298
10/57 = 0.175
Two populations are different if they have
different genotype frequencies
Gene Frequency
This is a proportion of a particular gene in
a given pool of genes
In terms of a population, it is the
proportion in a given population of the
loci having a given allelic series occupied
by a particular gene
It is a fraction of all genes at a given locus
that are of specific type
Gene Frequency
The genetics of a population is
concerned with genetic constitution of
individuals and also with the transmission
of genes from one generation to the next
In this transmission, genotypes of parents
are broken down and it is genes that are
transmitted in the gametes
New set of genotypes are therefore
formed in the progeny
Gene Frequency
AA
A
Aa
A
AA
A
aa
a
Aa
a
Parents
a
aa
in gametes
Progeny
Gene Frequency
The genes carried in a population have
continuity from generation to generation
but not the genotypes in which they
appear
The genetic constitution of a population
referring to genes it carries is described by
Gene frequencies
Gene frequencies can be determined
from knowledge of genotype frequencies
Calculation of gene frequency
From chick genotypes
BB
Bb
bb
Total
Individuals
Genes
30
17
10
57
B
60
17
0
77
b
0
17
20
37
Calculation of gene frequency
 Each individual contains two genes at a locus
 In a population of N individuals, there are 2N loci
for any allelic series
 For a locus with only 2 alleles (B and b) the
frequency of
B
N
B
 B  b

B
2N
is the number of individuals in a
population
Calculation of gene frequency
In the chick example
There are 2N
f(B)
f(b)
= 114 genes
= 77 / 114 = 0.675
= 37 / 114 = 0.325
Gene frequency notation
The following designations are used
p
q
f(B)
f(b)
In a natural, random mating population,
p + q
=
1
Genotype frequency notation
 When the two parents mate and produce a
zygote, the following genotype is observed
p
q
p
p2
pq
q
pq
q2
 Genotype frequency will be p2
2pq
q2
Gene and Genotype frequencies
In a population with natural forces
p2 + 2pq + q2
p + q
=
=
1 or unity
1 or unity
If genotype frequencies are known, gene
frequencies can be estimated
Gene Frequency = (homozygous genotype
frequency + ½
1
heterozygous
f BB   f 2 AB
2
genotype frequency)
f B  
/ Total
Total
Gene and Genotype frequencies
Homozygous are BB (p2) and bb(q2)
Heterozygous are Bb (pq)
These appear in the ratio of 1:2:1
1
p  2 pq
2
2
p 2
 p  pq
2
p  2 pq  q
2
Similarly,
q  q  pq
2
Gene and Genotype frequencies
Genotype frequency in subsequent
generations depends on the gene
frequency in the gametes of the previous
generation
Expected or predicted genotype
frequency in the offspring is obtained by
multiplying the gene frequency in the
males by the gene frequency in the
females
An example
A group of cocks produced 1000 sperms
comprising 200 dwarf (dw) types, the rest
being normal type (N)
Then the frequency of

dw  200

 0.2  q
1000
And the frequency of

N  800

 0.8  p
1000
An example
A group of hens produced 1000 eggs
comprising 200 dwarf (dw) types, the rest
being normal type (N)
Then the frequency of

dw  200

 0.2  q
1000
And the frequency of

N  800

 0.8  p
1000
An example
We can use these frequencies to predict
genotype of possible offspring with
respect to this trait
0.8N
0.2dw
0.8N
0.64NN
0.16Ndw
0.2dw
0.16Ndw
0.04dwdw
So we would expect 0.64 NN, 0.32 Ndw and 0.04dwdw
Reading for More examples
• Please go through examples in
Falconer, first Chapter dealing with
genotype and gene frequencies
Assumptions behind examples
This assumes no mutations, migration and
selection was taking place in the
population
Under such conditions, the population is
said to be in Hardy-Weinberg equilibrium
An exercise
• For a herd of 100 Shorthorn cattle
there are:
Red
Rhone
White
Genotype
RR
Rr
rr
Number
30
50
20
Phenotype
frequencies
the f(R) = p =
and f(r) = q =
Total
100
The Hardy – Weinberg Law
In 1908, two scientists working
independently
G.H. Hardy (English Mathematician)
W. Weinberg (German Physician and
Geneticist)
Developed a relationship between gene
and genotype frequencies
These developed a law that states that
The Hardy – Weinberg Law
In a large random – mating population
with no forces disturbing the environment
ie
no selection
No mutation
No migration
Random genetic drift
Gene frequencies and genotype
frequencies remain unchanging from
generation to generation
 A population with constant gene and
genotype frequencies is therefore said to
be in Hardy – Weinberg equilibrium
The Hardy – Weinberg Law
In Hardy – Weinberg equilibrium, there is a
simple relationship between gene and
genotype frequencies
If gene frequencies of two alleles among
the parents are
p and q,
then the genotype frequencies in the
progeny will be
p2, 2pq, and q2
The Hardy – Weinberg Law
Genes in parents
A1 A2
p
Frequencies
q
Male
Female
A1
p
A2
q
A1
p
A1 A1
p2
A1 A2
pq
A2
q
A1 A2
pq
A2 A2
q2
Progeny
The Hardy – Weinberg Law
Genotypes in parents
A1 A1
A1 A2
A2 A2
Frequencies
p2
2pq
q2
Which is basically squaring the gamete array
(p + q)2
p2 is the frequency of A1A1 genotype
2pq is the frequency of A1A2 genotype
q2 is the frequency of A2A2 genotype
The Hardy – Weinberg Law
In a population in HW equilibrium, we
may predict
Gene frequency
Genotype frequency
Phenotype frequency
Frequency of various mating
Carriers of a certain recessive allele
We use HW Law to predict mating of sires
and dams
Conditions to fulfil HW equilibrium
The population is infinitely large
There is no or at least very infrequent
mutation
There is no selection
Mating is at random
There is normal gene segregation
Conditions to fulfil HW equilibrium
There is equal fertility of parents and
equal fertilising ability of gametes and
reproduction is sexual
Equal viability and equal gene
frequencies in male and female parents
Gene frequency is same in parents and
offspring
Generations are not overlapping
Gene and Genotype frequencies
See details in Falconer ~ check
excel
Gene frequency of b
Genotype frequency
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
1
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
Relationship between genotype and gene frequencies for
two alleles in a population in HW equilibrium
BB
Bb
bb
Testing a population in HW
Equilibrium
MM
MN
NN
Total
No observed
233
385
129
747
No expected
242.4
366.3
138.3
Frequency
p2
2pq
q2
•First calculate gene frequencies
•p = 0.57
•q = 0.43
•To calculate expected frequencies, find
frequency of each genotype, then multiply by
total
Testing a population in HW
Equilibrium
MM  p 2 * Total  0.57  * 747  242.4
2
MN  2 pq * Total  2 * 0.57 * 0.43 * 747  366.3
NN  q 2 * Total  0.43 * 747  138.3
2
• Chi-square goodness of fit method used
to test whether the population we work
with is in HW equilibrium or not
Chi-square goodness of fit
 
2
Observed  Expected
2
Expected
• Under 1 d.f., X2 test statistic = 1.96 (from tables)
• Larger test statistic than 1.96 means less
likelihood that deviations occur simply due to
chance deviations alone
• That is, observed numbers are not in satisfactory
agreement with expected numbers, hence
population is not in HW Equilibrium
Chi-square goodness of fit ~
check excel
MM
MN
NN
Total
No observed
233
385
129
No expected
242.4
366.3
138.3
Frequency
p
2
2pq
q
747
2
Sum
Chi-Square
0.3645
0.9546
0.6253
1.94