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Much of the old literature on quantitative genetics was based on knowledge at that time. A gene was known to control a particular chemical function or reaction, and a gene was a gene. The exact number of genes was not known, but thought to be very large. Genes were known to be composed of DNA, but probably most people thought genes were equal in length, and maybe evenly spaced on the chromosomes.
Today, much more is known about the molecular aspects of genes and chromosomes. The human genome has been completely mapped and there are known to be common regions of DNA between species. DNA has two complementary strands which are linear arrangements of nucleotides (A= adenine, T=thymine, G= guanine, and C=cytosine). A base pair is A-T or G-C, and there are 6 billion base pairs in a typical mammalian genome. Sets of 3 (triplets) base pairs code for an amino acid. There are start and stop triplets. Sequences of amino acids subsequently code for an enzyme or protein. A gene is now defined as all of the nucleotides that translate into a polypeptide.
A gene contains a promoter region ( to start the replication ), a number of exon regions (expressed regions), and a number of intron regions (intragenic regions) which appear to be extraneous DNA that does not code for anything. A gene can be 1,000 to 2 million base pairs in length, with the average gene being 100,000 base pairs. Different forms of the gene (differences in base pairs within the specific location) are known as alleles. If the base pair difference occurs in an intron region, then there may not be any difference in gene function, but differences in the exon and promoter regions could result in the production of a different polypeptide or lack of production of the usual polypeptide.
Much of quantitative genetics theory assumes genes have only two possible alleles. However, looking at the genes controlling blood antigens and blood proteins in dairy cattle, there are likely going to be many genes with several alleles. Some alleles are more prevalent than others. Some alleles are lethal is they occur together, such as in the case of white coat colour in horses ( Ww = white colour, ww = non white colour, and WW = lethal).
Blood Antigens | Blood Proteins | |||
Locus | Alleles | Protein | Locus | Alleles |
A | 10 | Hemoglobin | Hb | 5 |
B | >300 | Albumin | Al | 3 |
C | >35 | Post-albumin | Pa | 2 |
FV | 4 | Transferrin | Tf | 7 |
J | >4 | globulin | 2 | |
L | 2 | Alkaline phos. | F | 2 |
M | 3 | Amylase | Am | 3 |
S | >10 | Milk Proteins | ||
R'-S' | 2 | lactoglobulin | Lg | 6 |
Z | 3 | lactalbumin | La | 1 |
casein | Cn | 7 | ||
casein | Cn | 4 | ||
casein | Cn | 2 | ||
casein | Cn | 4 |
The site where a gene is located in the genome is known as a locus. In humans, there are at least 547 genes that control the functioning of the eye, and over 1,200 involved with the heart. Thus, many genes are involved in controlling each trait that we observe in most livestock species. Genes also interact with each other, and so if we try to change one gene to have a particular allele, then we will be changing the effects of other genes that depended on the gene we have changed.
The total number of genes in mammals is now estimated to be around 30,000 to 50,000. Work is underway to determine what polypeptides are produced by each gene, what organs or functions do these enzymes regulate, and where are these genes.
The following notes start with a single locus, a gene having two alleles. As you read notice that there is a population perspective about a locus, and also an individual perspective. Later, the notes progress to considering two loci simultaneously, and from there a big jump to an infinite number of loci. Keep this progression in mind.
Acknowledgement: These notes are largely based upon material prepared by Dr. Brian Kennedy, which he used in his Quantitative Genetics course. In turn, his notes were based on many sources of information too.
1. Introduction
An understanding of genetic improvement of animals begins with an understanding of inheritance at a single locus. Even the action of a single locus is being further defined by molecular geneticists into sub-sections with sites for turning the locus on or off, or controling the activity of that locus. However, the molecular level of genetics will not be covered. Only diploid species will be considered.
2. Hardy-Weinberg Equilibrium
The conditions under which Hardy-Weinberg Equilibrium applies are
Males | ||||
A_{1} | A_{2} | |||
p | q | |||
Females | A_{1} | A_{1}A_{1} | A_{1}A_{2} | |
p | p^{2} | pq | ||
A_{2} | A_{2}A_{1} | A_{2}A_{2} | ||
q | pq | q^{2} |
The results can be summarized as
Genotype | Frequency |
A_{1}A_{1} | p^{2} |
A_{1}A_{2} | 2pq |
A_{2}A_{2} | q^{2} |
In general, with n alleles, A_{i}, with frequency p_{i}, the genotypic frequency of A_{i}A_{j} in a population under Hardy-Weinberg equilibrium will be 2p_{i}p_{j} for , and is p^{2}_{i} for i=j.
2.1 Sex Differences in Frequencies
If the gene frequencies are different in the sexes, say p_{i}^{m} is the frequency of A_{i} allele in males, and p_{i}^{f} is the frequency of A_{i} allele in females, then the genotypic frequency of A_{i}A_{j} in the next generation of offspring will be p_{i}^{m}p_{j}^{f} + p_{j}^{m}p_{i}^{f}, for . To illustrate, assume n=2, p_{1}^{m}=0.2, p_{2}^{m}=0.8, p_{1}^{f}=0.4, and p_{2}^{f}=0.6, then the genotypic outcome is
Males | ||||
A_{1} | A_{2} | |||
0.2 | 0.8 | |||
Females | A_{1} | A_{1}A_{1} | A_{1}A_{2} | |
0.4 | 0.08 | 0.32 | ||
A_{2} | A_{2}A_{1} | A_{2}A_{2} | ||
0.6 | 0.12 | 0.48 |
The frequency of the A_{1} allele in the offspring is
2.2 Sex-Linked Locus
With a sex-linked locus, usually the male carries the Y-chromosome which is considered to be nonactive genetically. Assuming n=2 alleles, then genotypically, males will be either A_{1}- with frequency p_{1}^{m}(t), or A_{2}- with frequency p_{2}^{m}(t). Females, on the other hand, will be A_{i}A_{j} with frequency p_{i}^{f}(t)p_{j}^{f}(t) for i and j going from 1 to 2. The subscript t refers to generation number.
As an example, suppose the frequencies of alleles in males and females is unequal in generation 0, let p_{1}^{m}(0) = 0.2, and p_{1}^{f}(0) = 0.4, then the genotypic frequencies in the progeny will be as follows:
Males | |||||
- | A_{1} | A_{2} | |||
0.5 | 0.1 | 0.4 | |||
Females | A_{1} | A_{1}- | A_{1}A_{1} | A_{1}A_{2} | |
0.4 | 0.2 | 0.04 | 0.16 | ||
A_{2} | A_{2}- | A_{2}A_{1} | A_{2}A_{2} | ||
0.6 | 0.3 | 0.06 | 0.24 |
The frequency of A_{1} allele in the male progeny is equal to
The average frequency of A_{1} across males and females in
generation 1 is
As the number of generations of random mating increases,
the allele frequencies in
males and females approaches Hardy-Weinberg equilibrium.
Let d_{i} be the initial difference in allele frequencies between
males and females, then for t=1,
3. Means and Variances
Assume two alleles at a locus and the frequency of A_{1} is p, and of A_{2} is q. Genotypic values are assigned to each genotype as shown in the table below.
Genotype | Frequency | Value |
A_{1}A_{1} | p^{2} | a |
A_{1}A_{2} | 2pq | d |
A_{2}A_{2} | q^{2} | -a |
3.1 Genotypic Mean
The mean of the genotypes, ,
is calculated by weighting the
genotypic values by the genotypic frequencies.
3.2 Genotypic Variance
The variance of genotypes,
,
is calculated by
weighting the square of the genotypic values by the genotypic
frequencies and subtracting the square of the genotypic mean.
3.3 Partitions of Genotypic Variance
The genotypic variance can be partitioned into an additive
genetic variance and a dominance genetic variance.
To determine the additive genetic variance, the breeding values of
the genotypes must be determined. Consider the effects that are
transmitted from a parent to its offspring (i.e. either one allele
or the other are transmitted). If an A_{1} allele is
transmitted, then p offspring will have genotype
A_{1}A_{1}with genotypic value of a, and q offspring will have
genotype
A_{1}A_{2} with genotypic value of d. The total
effect of the A_{1} allele is then
Similarly, if an A_{2} allele is transmitted, then p offspring
will have genotype
A_{1}A_{2} with value d, and q offspring
will have genotype
A_{2}A_{2} with value -a, for a total
effect of A_{2} allele of pd-qa. The average effect of the
A_{2} allele is
The breeding values (BV) can be summarized as follows:
Genotype | Frequency | BV | |
A_{1}A_{1} | p^{2} | ||
A_{1}A_{2} | 2pq | ||
A_{2}A_{2} | q^{2} |
The obtain the dominance genetic variance, the dominance deviations
for each genotype need to be calculated. If
Genotype | Frequency | Dominance |
A_{1}A_{1} | p^{2} | -2q^{2}d |
A_{1}A_{2} | 2pq | 2pqd |
A_{2}A_{2} | q^{2} | -2p^{2}d |
Recall that
Generally, the additive genetic relationship between two individuals is the proportion of genes that they share in common. However, if only a single locus is considered, the possibility exists that none of the alleles are shared between two individuals, or that both of them are shared. The average over all genes in the genome should equal the additive genetic relationship.
Now that markers and eventually QTLs will be used, the need to compute relationships for a single locus is present, and the gametic relationships are the straightforward approach to this problem.
Consider two individuals X and Y, whose genotypes are A_{1}A_{1} and A_{1}A_{2}, respectively. They have an offpsring, Z, that has genotype A_{1}A_{2}. Clearly, the A_{2} allele in Zhas come from parent Y, and the A_{1} allele in Z could be either A_{1} allele of parent X with equal probability. This can be illustrated in the following table. The diagonals of this table are always equal to unity. Let the rows and columns of this table be numbered from 1 to 6.
X | Y | Z | |||||
X | Y | ||||||
A1 | A1 | A1 | A2 | A1 | A2 | ||
A1 | 1 | 0 | 0 | 0 | a | b | |
X | |||||||
A1 | 0 | 1 | 0 | 0 | c | d | |
A1 | 0 | 0 | 1 | 0 | e | f | |
Y | |||||||
A2 | 0 | 0 | 0 | 1 | g | h | |
A1 | a | c | e | g | 1 | i | |
Z | |||||||
A2 | b | d | f | h | i | 1 |
Situations 1 and 2
In the above example, the parent source of both alleles in animal Z are known. That is, we know which parent provided the A_{1}allele and we know which parent provided the A_{2} allele with absolute certainty ( P = 1.0 ). This is situation 1. Thus, the elements in column 6 should be identical to those of column 4.
In the case of the A_{1} allele, parent X has two such alleles and it could be either of these alleles with equal probability of .5. This is situation 2. Thus, the elements in column 5 are an average of columns 1 and 2 (the possible sources of A_{1}). Therefore, the missing values are computed as follows:
The end result is
X | Y | Z | |||||
X | Y | ||||||
A1 | A1 | A1 | A2 | A1 | A2 | ||
A1 | 1 | 0 | 0 | 0 | .5 | 0 | |
X | |||||||
A1 | 0 | 1 | 0 | 0 | .5 | 0 | |
A1 | 0 | 0 | 1 | 0 | 0 | 0 | |
Y | |||||||
A2 | 0 | 0 | 0 | 1 | 0 | 1 | |
A1 | .5 | .5 | 0 | 0 | 1 | 0 | |
Z | |||||||
A2 | 0 | 0 | 0 | 1 | 0 | 1 |
Situation 3
Suppose now that X and Y are both heterozygous A_{1}A_{2}, and their offspring is also heterozygous.
X | Y | Z | |||||
X,Y | X,Y | ||||||
A1 | A2 | A1 | A2 | A1 | A2 | ||
A1 | 1 | 0 | 0 | 0 | a | b | |
X | |||||||
A2 | 0 | 1 | 0 | 0 | c | d | |
A1 | 0 | 0 | 1 | 0 | e | f | |
Y | |||||||
A2 | 0 | 0 | 0 | 1 | g | h | |
A1 | a | c | e | g | 1 | i | |
Z | |||||||
A2 | b | d | f | h | i | 1 |
Now the parental source of the alleles can not be determined with absolute certainty. The A_{1} allele could be from either X or Y with equal probability (.5), and the A_{2} allele could be from either parent with equal probability (.5). Column 5 in the table above is therefore an average of columns 1 and 3 (the sources for the A_{1} allele) and column 6 is the average of columns 2 and 4 (the sources for the A_{2} allele).
The completed table is shown below.
X | Y | Z | |||||
X,Y | X,Y | ||||||
A1 | A2 | A1 | A2 | A1 | A2 | ||
A1 | 1 | 0 | 0 | 0 | .5 | 0 | |
X | |||||||
A2 | 0 | 1 | 0 | 0 | 0 | .5 | |
A1 | 0 | 0 | 1 | 0 | .5 | 0 | |
Y | |||||||
A2 | 0 | 0 | 0 | 1 | 0 | .5 | |
A1 | .5 | 0 | .5 | 0 | 1 | 0 | |
Z | |||||||
A2 | 0 | .5 | 0 | .5 | 0 | 1 |
There can be various combinations of these situations, but just keep clear for a particular allele the possible sources of that allele from the two parents. For example, suppose that X and Y were both homozygous A_{1}A_{1}, which would make Z also homozygous. Each A_{1} allele in Z would have a .25 probability of being one of the four A_{1}alleles in the parents.
Example Problem
Below are ten animals, their genotype at the A-locus, and their parents (with their genotypes).
Animal | Genotype | Sire | Genotype | Dam | Genotype |
X | 12 | - | - | ||
Y | 12 | - | - | ||
W | 11 | X | 12 | Y | 12 |
U | 12 | X | 12 | Y | 12 |
V | 22 | X | 12 | Y | 12 |
T | 22 | U | 12 | V | 22 |
S | 12 | U | 12 | V | 22 |
R | 12 | U | 12 | V | 22 |
P | 22 | U | 12 | V | 22 |
Q | 12 | S | 12 | T | 22 |
Construct the gametic relationship table for this locus and these individuals. Which animals are inbred? Which pairs of animals have a non-zero dominance relationship?
Suppose we have a cell (between two animals) as follows:
X | |||
A1 | A2 | ||
A1 | e | b | |
Y | |||
A2 | c | f |
Dominance relationships are computed by
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Larry Schaeffer