Average Gene Effect and Breeding Value in Animal Genetics
ANIMAL GENETICS & BREEDING
UNIT – II
Principles of Animal & Population Genetics
Lecture – 6
Average Effect of Gene & Breeding Value
Dr K G Mandal
Department of Animal Genetics & Breeding
Bihar Veterinary College, Patna
Bihar Animal Sciences University, Patna
Introduction
•
There is relationship
between genetic properties of a
population and
transmission of value from parents to
offspring.
•
Genotype is not transmitted
from parents to offspring as
such, it is the genes which are being transmitted from
parents to offspring through haploid gametes.
•
These genes combine together to
form the new
genotype
of the progeny and they act
additively
or
non-
additively.
•
Therefore, it is imperative to measure the
average effect
of genes
which are transmitted from parents to offspring
and thereby the
breeding value
of the individual which
is carrying those genes.
Average effect of Gene
•
The average effect of a gene will enable us to determine the
breeding value of an individual.
•
B
reeding value
is the value associated with the genes carried by
an individual and transmitted to the offspring.
•
The average effect of a gene
is the mean deviation from the
population mean of individuals which have received that gene
from one parent and the gene received from other parent having
come at random from the population.
•
Let a number of gametes all carrying A1 gene united at random
with gametes carrying either A1 or A2 gene from a random
breeding population, then the mean deviation of the genotypes so
produced from the population mean will be referred as the
average effect of the gene A1.
Estimation of average effect of a gene
•
Let us see how the average effect of a gene is related
to the genotypic values
a
and
d,
in terms of which
the population mean was expressed.
•
Let us consider a locus with two alleles A1 & A2 with
their respective frequencies as
p
and
q.
•
Let us consider the average effect of A1 gene is
α
1
and to that of A2 is
α
2
.
Gene
Frequency
Average effect
A
1
p
α
1
A
2
q
α
2
_______________________________________________________
A
1
= A1A1
A
1
Gene
A
2
= A1A2
Genotype
Frequency
Genotypic Value
A
1
A
1
p
a
ap
A
1
A
2
q
d
qd
_________________________________________
•
Let the gametes carrying A1 gene united at random with the
gametes carrying either A1 or A2 gene taken from a random
mating population, then the frequencies of A1A1 and A1A2
genotypes will be
p
and
q
respectively, since the frequencies
of A1 and A2 genes in the whole gametic population are
p
and
q
respectively.
•
The assigned genotypic value of A1A1 is +
a
and to that of
A1A2 is
d.
Thus the genotypic value of A1A1 and A1A2 is
ap
and
qd
respectively. Thus, the mean genotypic value of
A1A1 and A1A2 genotypes is (ap + qd)/ (p + q).
=
ap + qd.
•
The difference between the mean of these two genotypes
so produced from the population mean is the average
effect of the gene A1.
•
Thus, the
average effect of the gene A1
is,
α
1
= (ap + qd) – [a(p – q) + 2pqd]
= ap + qd –ap + aq -2pqd
= aq + qd – 2pqd
= q[a + d - 2pd ]
= q[a + d(1 – 2p)]
= q[a + d (p + q – 2p)]
=
q[a + d(q – p)]
Gene
Frequency
Average effect
A
1
p
α
1
A
2
q
α
2
_______________________________________________________
A
1
= A1A2
A
2
Gene
A
2
= A2A2
Genotype
Frequency Assigned Value
GV
A
1
A
2
p
d
pd
A
2
A
2
q
- a
- aq
_________________________________________
•
Let the gametes carrying A2 gene united at random with
the gametes carrying either A1 or A2 gene taken from a
random mating population, then the frequencies of A1A2
and A2A2 genotypes will be
p
and
q
respectively, since
the frequencies of A1 and A2 genes in the whole gametic
population are
p
and
q
respectively.
•
The assigned genotypic value of A1A2 is
d
and to that of
A2A2 is
- a
.
Thus the genotypic value of A1A2 and A2A2
is
pd
and -
aq
respectively. Thus, the mean genotypic
value of A1A2 and A2A2 genotypes is (pd - aq)/ (p + q).
=
pd – aq.
Thus, the
average effect of gene A
2,
α
2
= pd - aq -[a(p-q)+2pqd]
= pd- aq – ap +aq - 2pqd
= p[-a+d-2qd]
= p[-a + d(1 – 2q)]
= -p[a-d(p+q-2q)]
= - p[a – d(p – q)]
=
-p[a+d(q-p)]
Thus,
α
1
=
q[a+d(q-p)]
α
2
=
- p[a+d(q-p)]
Average effect of gene in terms of gene substitution
•
The
average effect of a gene substitution
is the difference
between the average effect of two genes in question for a given
locus.
•
Suppose A2 gene is substituted or changed by A1, then A1A2
will be changed into A1A1, and A2A2 will be changed into
A1A2.
•
Changing from A1A2 to A1A1 there will be change of the
genotypic value from
d
to
a
, and the effect will be
(a – d)
. And
changing from A2A2 to A1A2 there will be change of
genotypic value from
– a
to
d.
Therefore, the effect will be {d –(-a)} =
(d + a)
.
•
Since A2 gene is taken at random from the
population, a proportion of A2 will be found in
A1A2, the frequency of which is equal to
p
, and a
proportion of the same gene i.e.,A2 will be found
in A2A2 which is equal to
q
.
•
Thus, due to change from A1A2 to A1A1, the
genotypic value
of
A1A1
will be
p(a – d)
and due
to change from A2A2 to A1A2, the
genotypic
value A1A2
will be
q(d + a).
•
The
average change
is, therefore,
[p(a – d) + q(d + a)]
= [pa – pd + qd + aq]
= [pa + aq + qd – pd]
= a(p + q) + d(q – p)
=
a + d( q – p) =
α
This is the average effect of gene substitution.
•
Relation between
α
,
α
1 and
α
2 can be seen
from the following equation:
α
1 –
α
2 =
q[a + d(q – p)] – [ - p{a + d(q – p)}]= q[a + d(q – p)] +p[a + d(q – p)]
= (q + p)[a + d(q – p)]
=
a + d(q – p)
=
α
α
1
=
q[a+d(q-p)]
= q
α
α
2
=
- p[a+d(q-p)]
= - p
α
Breeding Value
Parents pass on their genes and not their genotypes
to the progeny.
It is, therefore, the average effect of the parent’s
genes that determine the mean genotypic value of
the progeny.
The value of an individual is judged by the mean
value of its progeny, which is called the
breeding
value
of the individual.
Thus
Breeding value
of an individual is the mean
phenotypic value of its progeny.
If an individual is mated to a number of individuals
taken at random from the population, then its
breeding
value
is twice the mean deviation of the progeny from
the population mean.
The
deviation has to be doubled because the parent
in question provides only half of the genes of the
progeny,
the other half is coming at random from the
population.
Breeding values can be expressed in absolute units but
more conveniently expressed as deviation from the
population mean.
Breeding value
is also defined as the value associated
with the genes carried by an individual.
•
In terms of
average effect of gene
, the
breeding value
of an
individual is equal to sum of the average effect of genes it carries,
the summation being made over the pair of alleles at each locus
and over all loci.
Estimation of Breeding Value
For a single locus with two alleles A1 & A2, the breeding values
of three genotypes will be as follows:
Mean Breeding Value
•
If breeding values are expressed in absolute units
the mean breeding value must be equal to the
mean genotypic value and to the mean
phenotypic value.
•
In a Hardy – Weinberg Equilibrium population
the mean breeding value is equal to zero.
•
This can be verified by multiplying the breeding
value by the frequency of each genotype and
their summation gives the mean breeding value.
•
Thus, mean breeding value, expressed as
deviation from the population mean, as :
2p2q
α
+
2pq(q – p)
α
– 2q2p
α
= 2pq
α
[p + q – p – q]
= 0
The breeding value is also known as
additive
effect of gene
and is denoted by
A.
In absence of dominance effect, the breeding
value is equivalent to genotypic value.
Thus,
G = A
Dominance Deviation
•
Dominance deviation is the interaction between
alleles within a locus. It is also known as intra genic
gene interaction.
•
When a single locus is under consideration, the
difference between the genotypic value
G
and the
breeding value
A
of a particular genotype is known
as dominance deviation. So that,
G = A + D.
•
Since the average effect of genes and the breeding
values of genotypes are dependent on gene
frequency in the population, the dominance
deviation is also dependent on gene frequency.
•
The
dominance deviation
can be expressed in terms of
arbitrary assigned genotypic values
a
and
d
by subtraction of
breeding value from the genotypic value.
•
Considering a single locus with two alleles A1 and A2 with
respective frequency
p
and
q,
the dominance deviation of
three genotypes will be as follows:
Mean Dominance Deviation
•
Since dominance deviations are expressed as deviation
from the population mean, therefore, the
mean
dominance deviation will be equal zero.
•
This can be verified by multiplying the dominance
deviation with frequency of respective genotype and
summing over all the genotypes.
•
Thus, the mean dominance deviation is
- 2p2q2d + 4p2q2d – 2p2q2d = 0
•
All the dominance deviations are the function of
d,
if
d
is
zero
then
dominance deviation will also be zero.
•
Therefore,
in absence of dominance deviation the
breeding values and dominance deviations will be the
same.
Interaction Deviation
•
When a single locus is under consideration, the genotypic
value is made up of breeding value and dominance deviation
only.
•
When two or more than two loci are involved for the
expression of a character, then the genotypic value may
contain an additional deviation due to non-additive gene
combination.
•
Let GA be the genotypic value of an individual attributable to
a locus A and GB attributable to another locus B, then the
aggregate genotypic value of both the loci together
G = GA + GB + IAB,
•
Where, IAB is the interaction deviation , deviated from
additive combination of these genotypic values. This IAB is
also known as gene interaction or epistasis.
•
Therefore, for polygenic traits when several
loci interacted together for expression of a
trait then, genotypic value can be written as
G = A + D + I
Where ,
A = sum of the breeding values
attributable to several loci.
D = Sum of the dominance deviations
I = Sum of the Interaction deviations
•
The
mean interaction deviations
of all the loci
in a population is equal to
zero.
Thank You
In the field of animal genetics and breeding, it is crucial to comprehend the average effect of genes and breeding value of individuals. Genes are transmitted from parents to offspring through haploid gametes, influencing the genotype and breeding potential. Estimating the average effect of a gene involves considering genotypic values and frequencies of alleles. This process plays a significant role in determining the genetic properties of populations and the value passed on from generation to generation.
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ANIMAL GENETICS & BREEDING UNIT II Principles of Animal & Population Genetics Lecture 6 Average Effect of Gene & Breeding Value Dr K G Mandal Department of Animal Genetics & Breeding Bihar Veterinary College, Patna Bihar Animal Sciences University, Patna
Introduction There is relationship between genetic properties of a population and transmission of value from parents to offspring. Genotype is not transmitted from parents to offspring as such, it is the genes which are being transmitted from parents to offspring through haploid gametes. These genes combine together to form the new genotype of the progeny and they act additively or non- additively. Therefore, it is imperative to measure the average effect of genes which are transmitted from parents to offspring and thereby the breeding value of the individual which is carrying those genes.
Average effect of Gene The average effect of a gene will enable us to determine the breeding value of an individual. Breeding value is the value associated with the genes carried by an individual and transmitted to the offspring. The average effect of a gene is the mean deviation from the population mean of individuals which have received that gene from one parent and the gene received from other parent having come at random from the population. Let a number of gametes all carrying A1 gene united at random with gametes carrying either A1 or A2 gene from a random breeding population, then the mean deviation of the genotypes so produced from the population mean will be referred as the average effect of the gene A1.
Estimation of average effect of a gene Let us see how the average effect of a gene is related to the genotypic values a and d, in terms of which the population mean was expressed. Let us consider a locus with two alleles A1 & A2 with their respective frequencies as p and q. Let us consider the average effect of A1 gene is 1 and to that of A2 is 2.
Gene A1 A2 _______________________________________________________ A1= A1A1 A1 Gene A2= A1A2 Genotype Frequency Genotypic Value A1A1 p A1A2 q _________________________________________ Frequency p q Average effect 1 2 a d ap qd
Let the gametes carrying A1 gene united at random with the gametes carrying either A1 or A2 gene taken from a random mating population, then the frequencies of A1A1 and A1A2 genotypes will be p and q respectively, since the frequencies of A1 and A2 genes in the whole gametic population are p and q respectively. The assigned genotypic value of A1A1 is +a and to that of A1A2 is d. Thus the genotypic value of A1A1 and A1A2 is ap and qd respectively. Thus, the mean genotypic value of A1A1 and A1A2 genotypes is (ap + qd)/ (p + q).= ap + qd.
The difference between the mean of these two genotypes so produced from the population mean is the average effect of the gene A1. Thus, the average effect of the gene A1 is, 1 = (ap + qd) [a(p q) + 2pqd] = ap + qd ap + aq -2pqd = aq + qd 2pqd = q[a + d - 2pd ] = q[a + d(1 2p)] = q[a + d (p + q 2p)] = q[a + d(q p)]
Gene A1 A2 _______________________________________________________ A1= A1A2 A2 Gene A2= A2A2 GenotypeFrequency Assigned Value A1A2 p A2A2 q _________________________________________ Frequency p q Average effect 1 2 GV pd - aq d - a
Let the gametes carrying A2 gene united at random with the gametes carrying either A1 or A2 gene taken from a random mating population, then the frequencies of A1A2 and A2A2 genotypes will be p and q respectively, since the frequencies of A1 and A2 genes in the whole gametic population are p and q respectively. The assigned genotypic value of A1A2 is d and to that of A2A2 is - a. Thus the genotypic value of A1A2 and A2A2 is pd and - aq respectively. Thus, the mean genotypic value of A1A2 and A2A2 genotypes is (pd - aq)/ (p + q).= pd aq.
Thus, the average effect of gene A2, 2 = pd - aq -[a(p-q)+2pqd] = pd- aq ap +aq - 2pqd = p[-a+d-2qd] = p[-a + d(1 2q)] = -p[a-d(p+q-2q)] = - p[a d(p q)] = -p[a+d(q-p)] Thus, 1= q[a+d(q-p)] 2 = - p[a+d(q-p)]
Average effect of gene in terms of gene substitution The average effect of a gene substitution is the difference between the average effect of two genes in question for a given locus. Suppose A2 gene is substituted or changed by A1, then A1A2 will be changed into A1A1, and A2A2 will be changed into A1A2. Changing from A1A2 to A1A1 there will be change of the genotypic value from d to a, and the effect will be (a d). And changing from A2A2 to A1A2 there will be change of genotypic value from a to d. Therefore, the effect will be {d (-a)} = (d + a).
Since A2 gene is taken at random from the population, a proportion of A2 will be found in A1A2, the frequency of which is equal to p, and a proportion of the same gene i.e.,A2 will be found in A2A2 which is equal to q. Thus, due to change from A1A2 to A1A1, the genotypic value of A1A1 will be p(a d) and due to change from A2A2 to A1A2, the genotypic value A1A2 will be q(d + a).
The average change is, therefore, This is the average effect of gene substitution. [p(a d) + q(d + a)] = [pa pd + qd + aq] = [pa + aq + qd pd] = a(p + q) + d(q p) = a + d( q p) =
Relation between , 1 and 2 can be seen from the following equation: 1 2 = q[a + d(q p)] [ - p{a + d(q p)}] = q[a + d(q p)] +p[a + d(q p)] = (q + p)[a + d(q p)] = a + d(q p) = 2 = - p[a+d(q-p)] = - p 1= q[a+d(q-p)] = q
Breeding Value Parents pass on their genes and not their genotypes to the progeny. It is, therefore, the average effect of the parent s genes that determine the mean genotypic value of the progeny. The value of an individual is judged by the mean value of its progeny, which is called the breeding value of the individual. Thus Breeding value of an individual is the mean phenotypic value of its progeny.
If an individual is mated to a number of individuals taken at random from the population, then its breeding value is twice the mean deviation of the progeny from the population mean. The deviation has to be doubled because the parent in question provides only half of the genes of the progeny, the other half is coming at random from the population. Breeding values can be expressed in absolute units but more conveniently expressed as deviation from the population mean. Breeding value is also defined as the value associated with the genes carried by an individual.
In terms of average effect of gene, the breeding value of an individual is equal to sum of the average effect of genes it carries, the summation being made over the pair of alleles at each locus and over all loci. Estimation of Breeding Value For a single locus with two alleles A1 & A2, the breeding values of three genotypes will be as follows: Genotype Breeding value Breeding Value A1A1 A1A2 A2A2 2 1 1+ 2 2 2 2q (q-p) - 2p
Mean Breeding Value If breeding values are expressed in absolute units the mean breeding value must be equal to the mean genotypic value and to the mean phenotypic value. In a Hardy Weinberg Equilibrium population the mean breeding value is equal to zero. This can be verified by multiplying the breeding value by the frequency of each genotype and their summation gives the mean breeding value.
Thus, mean breeding value, expressed as deviation from the population mean, as : Genotype Frequency Breeding value Freq x Breeding value A1A1 P2 2q P2(2q ) A1A2 2pq (q p) 2pq(q p) A2A2 q2 - 2p q2(-2p ) 2p2q + 2pq(q p) 2q2p = 2pq [p + q p q] = 0
The breeding value is also known as additive effect of gene and is denoted by A. In absence of dominance effect, the breeding value is equivalent to genotypic value. Thus, G = A
Dominance Deviation Dominance deviation is the interaction between alleles within a locus. It is also known as intra genic gene interaction. When a single locus is under consideration, the difference between the genotypic value G and the breeding value A of a particular genotype is known as dominance deviation. So that, G = A + D. Since the average effect of genes and the breeding values of genotypes are dependent on gene frequency in the population, the dominance deviation is also dependent on gene frequency.
The dominance deviation can be expressed in terms of arbitrary assigned genotypic values a and d by subtraction of breeding value from the genotypic value. Considering a single locus with two alleles A1 and A2 with respective frequency p and q, the dominance deviation of three genotypes will be as follows: Genotype Frequency Assigned Value Breeding Value Dominance Deviation A1A1 P2 a 2q - 2q2d A1A2 2pq d (q p) 2pqd A2A2 q2 - a - 2p - 2p2d
Mean Dominance Deviation Since dominance deviations are expressed as deviation from the population mean, therefore, the mean dominance deviation will be equal zero. This can be verified by multiplying the dominance deviation with frequency of respective genotype and summing over all the genotypes. Thus, the mean dominance deviation is - 2p2q2d + 4p2q2d 2p2q2d = 0 All the dominance deviations are the function of d, if d is zero then dominance deviation will also be zero. Therefore, in absence of dominance deviation the breeding values and dominance deviations will be the same.
Interaction Deviation When a single locus is under consideration, the genotypic value is made up of breeding value and dominance deviation only. When two or more than two loci are involved for the expression of a character, then the genotypic value may contain an additional deviation due to non-additive gene combination. Let GA be the genotypic value of an individual attributable to a locus A and GB attributable to another locus B, then the aggregate genotypic value of both the loci together G = GA + GB + IAB, Where, IAB is the interaction deviation , deviated from additive combination of these genotypic values. This IAB is also known as gene interaction or epistasis.
Therefore, for polygenic traits when several loci interacted together for expression of a trait then, genotypic value can be written as G = A + D + I Where , A = sum of the breeding values attributable to several loci. D = Sum of the dominance deviations I = Sum of the Interaction deviations The mean interaction deviations of all the loci in a population is equal to zero.
