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Relationship Between Gene Frequency and Component Variances


population so the variance is just the average of the squares of these values. The specific calculation is as follows:

6.2. Calculation of component variances

Calculate cumulative variance (variance of seed values):

Genotype Frequency Cumulative value (breeding value) Frequency x (Breeding value) 2



2 2 2 2

A 1 A 1 p 2q 4p q

2 2

A 1 A 2 2pq (qp) 2pq(qp)

2 2 2 2

A 2 A 2 q -2p 4p q


2 2 2 2 2 2 2 2

Cong: 4p 2 q 2 2 + 2pq(qp) 2 2 + 4p 2 q 2 2 V A = 4p q + 2pq(qp) + 4p q

= 2pq 2 2 (2pq + q 2 - 2pq + p 2 +2pq)

= 2pq 2 2 (p 2 + 2pq + q 2 )

= 2pq 2 2 (p+q) 2

= 2pq 2 2 [2.12]

Replace the value of according to expression [2.7]:

2 2

VA = 2pq [a + d(qp)] [2.13]

Calculate the dominant variance:

Genotype Frequency Dominant Deviation Frequency x (Dominant Deviation) 2



2 2 2 4 2

A 1 A 1 p -2q d 4p qd

3 3 2

A 1 A 2 2pq 2pqd 8p qd

2 2 4 2 2

A 2 A 2 q -2p d 4p qd



Cong: 4p 2 q 4 d 2 + 8p 3 q 3 d 2 + 4p 4 q 2 d 2 V D = 4p 2 q 4 d 2 + 8p 3 q 3 d 3 + 4p 4 q 2 d 2

= 4p 2 q 2 d 2 (q 2 + 2pq + p 2 )

V D = (2pqd) 2 [2.14]

Calculate genetic variance:

V G = VA + VD + 2Cov AD

in which: Cov AD is the covariance between the breeding value and the dominant deviation. Also, because the values ​​have been calculated based on the difference from the population mean, the covariance is just the average of the squares of these values. The specific calculation is as follows:


Cov AD = Frequency x Breeding Value x Dominant Deviation

2

2q

-2q 2 d


p 2 (2q )(-2q 2 d)

(qp)

2pqd


2pq(qp) (2pqd)

-2p

-2p 2d​


q 2 (-2p )(-2p 2 d)

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Genotype Frequency Breeding value Dominant deviation Frequency x Breeding value xDominant deviation A 1 A 1 p

A 1 A 2 2pq

2

A 2 A 2 q



Ceng: -4p 2 q 3 d + 4p 2 q 3 d - 4p 3 q 2 d + 4p 3 q 2 d = 0 Since Cov AD = 0, V G = V A + V D

Substituting the values ​​of VA and VD in expressions [2.13] and [2.14], we have:

2 2 2

V G = 2pq [a + d(qp)] + (2pqd) [2.15]

For example: A mouse line has the dwarf gene pg, with frequency q=0.1; a=4 and d=2


2 2

V A = 2 x 0.9 x 0.1 [4 + 2 (0.1 - 0.9)] = 0.18 x (2.4) = 1.0368

2 2

V D = (2 x 0.9 x 0.1 x 2) = (0.36) = 0.1296 V G = V A + V D = 1.0368 + 0.1296 = 1.1664

If frequency q=0.4, then:

2 2

V A = 2 x 0.4 x 0.6 [4 + 2 (0.4 - 0.6)] = 0.48 x (3.6) = 6.2208

2 2

V D = (2 x 0.4 x 0.6 x 2) = (0.96) = 0.9216

V G = V A + V D = 6.2208 + 0.9216 = 7.1424


Summary: q = 0.1 q = 0.4


Variance of the same value

V A

1.0368

6,2208

Dominant heteroskedasticity

V D

0.1296

0.9216

Variance of genotype values

V G

1,1164

7,1424


Thus, the variance in breed value accounts for the largest proportion of the total genotype variance :

If q=0.1; VA ( %) accounts for 1.0368/1.1164 = 88.89% If q=0.4; VA ( %) accounts for 6.2208/7.1424 = 87.10%

The above observations will be applied in estimating genetic parameters.


The relationship between gene frequencies and component variances is depicted in Figure 2.1.


Gene frequency, q


Figure 2.1. Relationship between gene frequency and component variances

Note: The solid line is the genotypic variance, the thinner line is the additive variance, and the solid line is the dominance variance. The gene frequency q corresponds to the recessive allele. (a) Non-dominant, meaning d=0. (b) Completely dominant, meaning d=a. (c) Pure super-dominant, meaning a=0.


In practice, to estimate component variances, the following methods are used:

- Tracking and analyzing the similarities between related individuals will estimate

There are two types of component variances: VA : ( VD + VI + VE )

- Monitoring and analyzing the similarity between inbred lines will estimate 2 types of component variance: V G : V E

- Combining the two methods above will estimate 3 types of component variances: VA : (VD + VI ) : V E

Applications of component variance estimation will be presented in the estimation section.

genetic parameters

Chapter 3

Genetic relationships between individuals


Only by monitoring the livestock's gene pool can one determine the genetic relationship between individuals and assess the degree of inbreeding of an individual, on the basis of

This prevents the possibility of inbreeding. A full understanding of the genetic relationships between some related animals helps us to grasp the essence of the methods for determining genetic parameters discussed in the next chapter.


1. Vanity

Pedigree , also known as pedigree , is a diagram of an animal's bloodline.

To record a spectrum, one can use several different methods, thus forming several different types of spectrum:

- Vertical spectrum: Recorded according to the principle that each row is a generation, the previous generation is recorded in the lower row, the next generation is recorded in the upper row; in the same row, males are recorded on the right, females are recorded on the left.

For example: The family tree of individual X has the previous generation of X as parents (generation I) with father (B), mother (M). The previous generation of parents as grandparents (generation II) has father's father, paternal grandfather (BB), father's mother, paternal grandmother (MB), mother's father, maternal grandfather (BM), mother's mother, maternal grandmother (MM). The previous generation of grandparents (great-grandparents, generation III) also follows the same principle. The diagram is as follows:

M

B

MM

BM

MB

BB

MMM

BMM

MBM

BBM

MMB

BMB

MBB

BBB

X

I II III


- Horizontal spectrum: Recorded according to the principle that each column is a generation, the previous generation is recorded in the right column, the next generation is recorded in the left column; in the same column, males are recorded in the upper row, females are recorded in the lower row.

For example: The same spectrum of individual X and the symbols of related individuals as above, the diagram is as follows:

I II III


BBB

BB

MBB

B

BMB

MB

MMB

X

BBM

BM

MBM

M

BMM

MM

MMM

At the positions of related animals in the family tree, people record the animal's number or name. Each individual is numbered according to prescribed methods such as: cutting off the number of ears (for

Pigs), tattooing numbers on the ears or wearing plastic tags (with numbers written on them) on the ears (for pigs or cows), wearing aluminum tags (with numbers written on them) at the base of the wings or on the legs (for poultry) ...


- In practice, spectra are often recorded in a horizontal spectral format, but do not fully comply with the recording principles of this spectral format. For example:


1

1 2 3 4

2

S

SD

X

1 X

D

3


There can be three types of spectrum systems:

+ Full spectrum: Records all animals in different generations

+ Summary genealogy: Only records animals that are directly related to a certain ancestor.

+ Compact spectrum: Similar to the summary spectrum, but each animal appears only once in the spectrum.

Examples of 3 types of spectral systems:

S

1

SS 1 XSS 2

XXDSX 1

1

DD 2 1 D

2

2


2. Genetic factors

To describe the genetic relationship between two individuals but only consider the cumulative effect of genes, people use the concept of kinship coefficient and cumulative genetic relationship.


2.1. Familiarity coefficient

This concept was proposed by Malecot (1948). The relatedness coefficient between two individuals is the probability that a gene randomly drawn from any locus of this individual is of the same origin (a chemical copy of an original gene, not caused by mutation) as a gene randomly drawn from the corresponding locus of the other individual.

Suppose, 2 individuals X and Y at any locus have the following corresponding genes: XY

A i A j A Ý' A j'

From the locus of X, gene A i is randomly drawn , and from Y, gene A j is randomly drawn . The probability that 2 genes A i and A j have the same origin (both are copied from 1 original gene, not caused by mutation) is denoted as: P(A i =A j ).

The affinity coefficient between X and Y, denoted by f XY :


f XY = 1/4 [P(A i =A i' ) + P(A i =A j' ) + P(A j =A i' ) + P(A j =A j' ) ]


If X and Y are two siblings, and their parents are not inbred, then the probability of each case is 1/4. So:

f XY = 1/4 (1/4 + 1/4 + 1/4 + 1/4) = 1/4


The animal's own affinity coefficient will be calculated based on the affinity coefficient of X with X itself:


XX

A i A j A Ý A j


f XY = 1/4 [P(A i =A i ) + P(A i =A j ) + P(A j =A j ) + P(A j =A i ) ] f XY = 1/4 ( 1 + 0 + 1 + 0) = 1/2


2.2. Combined genetic factors

The cumulative genetic relationship between two individuals is twice the probability that a gene randomly drawn from a locus of this individual is of the same origin (a chemical copy of an original gene, not caused by mutation) with a gene randomly drawn from the corresponding locus of the other individual . Thus, the cumulative genetic relationship is twice the kinship coefficient, the cumulative genetic relationship between X and Y is denoted as a XY :

a XY = 2 f XY

It is possible to use probability calculus to calculate the cumulative genetic relationships of a number of relatives. Here are a few examples:


- Calculate the cumulative genetic relationship between father (or mother) and child:

Suppose at any locus, father (X) and mother (Y) have the corresponding genes i,j and k,l. Therefore, the child can have genes i, k or i, l or j, k or j, l:

Raft - MÑ X ij Y klThe Z ik

il jk jl


The probability of genes randomly drawn from this locus of the father and the child is as follows:

Gene drawn randomly from Z (probability of each event is 1/4)



i

j

k

l

Gene randomly drawn from X

i

1/8

1/8

1/8

1/8

(probability of each event is 1/2)

j

1/8

1/8

1/8

1/8



therefore:


Therefore:

In the table above, there are only 2 cases where the extracted genes have the same origin (i=i and j=j),


1/8 + 1/8

f XZ = = 1/4 1/8 + 1/8 + 1/8 + 1/8 + 1/8 + 1/8 + 1/8 + 1/8


a XZ = 2 f XZ = 2 x 1/4 = 1/2


- Cumulative genetic relationship between two siblings (same father and same mother):

Suppose at any locus, father (X) and mother (Y) have the corresponding genes i,j and k,l. Therefore, two siblings Z and W can both have the genes i,k or i,l or j,k or j,l:


Raft - MÑ

X ij

Yes​

siblings

Z i kil

WIKIL​


j,k

jl

jk

j,l


The probability of randomly drawing genes from this locus for Z and W is as follows:

Genotype Z



yes

il

jk

jl

yes

1/2

1/4

1/4

0

Genotype

il

1/4

1/2

0

1/4

cđà W

jk

1/4

0

1/2

1/4


jl

0

1/4

1/4

1/2

The average probability of the above cases is:

(4 x 1/2 + 4 x 0 + 8 x 1/4): 16 = 1/4

Therefore: f ZW = 1/4

a ZW = 2 f ZW = 2 x 1/4 = 1/2

Similarly, the cumulative genetic relationship between a number of relatives can be calculated as follows:

RelativesAdditive genetic relationship (a XY ) Father or mother – Child 1/2

Grandparents - Grandchildren 1/4

Previous generation - Next generation (n generations apart) (1/2) n Siblings 1/2

Half-siblings 1/4

(same father but different mother or same mother but different father)


2.3. Inbreeding coefficient

The inbreeding coefficient of an individual is the probability that two randomly drawn genes at any locus of that individual have the same origin (both are chemical copies of an original gene, not identical due to mutation).

Consider individual X, X's father and mother are S and D, the cumulative genetic relationship between S and D is a SD . The inbreeding coefficient of individual X is denoted as F X , therefore:

F X = 1/2 a SD


2.4. How to calculate additive genetic relationship and inbreeding coefficient

Either of the following two methods can be used to calculate additive genetic relationships and inbreeding coefficients.


2.4.1. Using the formula (Wright, 1922)

Consider the spectrum of individual V as follows:

W


Z

XY


V


X and Y have a common ancestor W.

Randomly draw a gene from X, call this gene i. Randomly draw a gene from Y, call this gene j.

According to the definition of additive genetic relationship between X and Y, we have:

a XY = 2 P(i=j)

Common ancestor W randomly transmits 1 gene to Z, randomly transmits 1 gene to Y. If W is not inbred, the probability that these 2 genes have the same origin (assuming both are i) will be 1/2. If W is inbred with an inbreeding coefficient of F W , this probability will be 1/2 + F W /2 = (1 + F W )/2.

Given that these two genes have the same origin and are both i, the probability that Z transmits gene i to X is

1/2. Given that X receives gene i, the probability of randomly drawing gene i from X is 1/2, the probability of randomly drawing gene i from Y is also 1/2.

All the above events are independent of each other, so the probability that genes i and j have the same origin is

after:


If:


P(i=j) = (1 + F W )/2 . 1/2 . 1/2 . 1/2

n is the number of generations (number of connections) from common ancestor W to X (V's father), the probability that W transmits gene i to X will be (1/2) n

p is the number of generations (number of connections) from common ancestor W to Y (mother of V), the probability that W transmits gene i to Y will be (1/2) p

Therefore:


and:

P(i=j) = (1 + F W )/2 . (1/2) . (1/2)

W

np

P(i=j) = (1/2) n+p (1 + F )/2


aXY = 2 P(i=j) = (1/2) n+p (1 + FW) FV = 1/2 aXY = (1/2) n+p (1 + FW)/2

= (1/2) n+p+1 (1 + FW) or:

= 1/2 [(1/2) n+p (1 + FW)]

If Z has multiple common ancestors, or a common ancestor has multiple different links

to Z's parents:

The cumulative genetic relationship between X and Y is calculated by the formula:


n + p

a XY = (1/2) kk (1 + F k ) [3.1]

k

where: n k , p k is the number of generations (number of links) from the common ancestor to X and YF k is the inbreeding coefficient of the common ancestor


The inbreeding coefficient of individual V is calculated according to the formula:

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