Population Genetics

Abstract

Population genetics studies how a trait varies over time and space in a group of individuals. To achieve this aim, the genetic structure of a population is monitored in different generations. The genetic structure is determined by the allele frequencies and genotypic frequencies of the population for the trait under study. The way to determine these parameters depends on the allelic interaction occurring in the gene under consideration, basically if it is dominance-recessiveness or not. The theoretical basis of population genetics is given by the principle of Hardy–Weinberg equilibrium, which states that in any population in panmixia, in the absence of mutation, migration, selection, and genetic drift, allele and genotypic frequencies remain the same generation after generation. It follows from this principle that the genetic structure of a particular gene in a population can change only if it is acted upon by one of the mentioned processes: mutation, migration, selection, or genetic drift. In the event that this occurs, the allelic frequencies and therefore the genotypic frequencies are altered. In relation to the panmixia, if panmixia does not occur (as in self-pollinated plants), genotypic frequencies will be altered but allele frequencies remain constant, something similar occurs when closely related animals mate, causing what is called inbreeding. Inbreeding is measured by an inbreeding coefficient and by a coancestry coefficient; this last one can be used to plan matings in any herd. Genetic variability in a population will be a consequence of mutation or migration, and it will be the process of selection (whether natural or artificial) or genetic drift that will define the prevalence or elimination of such changes in the next generation of the population. The activities of genetic improvement (for example, plant or animal breeding), seek is to alter the allele frequencies of a population with the purpose of obtaining the prevalence of individuals with the desired genotypes.

Solved Problems

• 1. Suppose you are assessing the genetic diversity of 20 genotypes of sesame. For this purpose, biochemical markers are used, specifically isoenzymes, and color of the flower (M encodes for purple flower, m encodes for white flower, M is dominant over m). The phenograms of three systems are presented below, one of the isozyme systems have 4 alleles, and two isozyme systems have 2 alleles., and one has 4 alleles. Also, the observed phenotypic proportions for flower color are presented.

Isoenzyme system 1

Isoenzyme system 2

Isoenzyme system 3

Number of individuals with purple flowers	Number of individuals with white flowers
4	16

• (a) Determine for each isoenzyme system, and for flower color, the allele and genotypic frequencies

Isozyme system 1: Of 20 individuals, 9 are A1A1, 8 are A1A2 and 3 are A1A2.

A2A2, therefore the genotypic frequencies are:

$$ {\displaystyle \begin{array}{l}\mathrm{A}1\mathrm{A}1=9/20=0.45\\ {}\mathrm{A}1\mathrm{A}2=8/20=0.40\\ {}\mathrm{A}2\mathrm{A}2=3/20=0.15\end{array}} $$

Allele frequencies are determined as follows:

$$ {\displaystyle \begin{array}{l}\mathrm{Frequency}\ \mathrm{A}1=\mathrm{p}=0.45+\left(0.50\kern0.5em \cdot \kern0.5em 0.40\right)=0,65\\ {}\mathrm{Frequency}\ \mathrm{A}2=\mathrm{q}=0.15+\left(0.50\kern0.5em \cdot \kern0.5em 0.40\right)=0,35\end{array}} $$

Isoenzyme system 2

The genotypic frequencies are:

$$ {\displaystyle \begin{array}{l}\mathrm{B}1\mathrm{B}1=11/20=0.55\\ {}\mathrm{B}1\mathrm{B}2=5/20=0.25\\ {}\mathrm{B}2\mathrm{B}2=4/20=0.20\end{array}} $$

The allele frequencies are:

$$ {\displaystyle \begin{array}{l}\mathrm{Frequency}\ \mathrm{B}1=p=0.55+\left(0.50\kern0.5em \cdot \kern0.5em 0.25\right)=0,675\\ {}\mathrm{Frequency}\ \mathrm{B}2=q=0.20+\left(0.50\kern0.5em \cdot \kern0.5em 0.25\right)=0,325\end{array}} $$

Isoenzyme system 3

The genotypic frequencies are:

$$ {\displaystyle \begin{array}{l}\mathrm{C}1\mathrm{C}1=4/20=0.20\\ {}\mathrm{C}1\mathrm{C}2=3/20=0.15\\ {}\mathrm{C}1\mathrm{C}3=0/20=0.00\\ {}\mathrm{C}1\mathrm{C}4=1/20=0.05\\ {}\mathrm{C}2\mathrm{C}2=5/20=0.25\\ {}\mathrm{C}2\mathrm{C}3=2/20=0.10\\ {}\mathrm{C}2\mathrm{C}4=1/20=0.05\\ {}\mathrm{C}3\mathrm{C}3=2/20=0.10\\ {}\mathrm{C}3\mathrm{C}4=1/20=0.05\\ {}\mathrm{C}4\mathrm{C}4=1/20=0.05\end{array}} $$

The allele frequencies are:

$$ {\displaystyle \begin{array}{l}\mathrm{Frequency}\ \mathrm{C}1=p=0.20+\left[\left(0.50\kern0.5em \cdot \kern0.5em 0.15\right)+\left(0.50\kern0.5em \cdot \kern0.5em 0.00\right)+\left(0.50\kern0.5em \cdot \kern0.5em 0.05\right)\right]=0,300\\ {}\mathrm{Frequency}\ \mathrm{C}2=q=0.25+\left[\left(0.50\kern0.5em \cdot \kern0.5em 0.15\right)+\left(0.50\kern0.5em \cdot \kern0.5em 0.10\right)+\left(0.50\kern0.5em \cdot \kern0.5em 0.05\right)\right]=0,400\\ {}\mathrm{Frequency}\ \mathrm{C}3=r=0.10+\left[\left(0.50\kern0.5em \cdot \kern0.5em 0.00\right)+\left(0.50\kern0.5em \cdot \kern0.5em 0.10\right)+\left(0.50\kern0.5em \cdot \kern0.5em 0.05\right)\right]=0,175\\ {}\mathrm{Frequency}\ \mathrm{C}4=s=0.05+\left[\left(0.50\kern0.5em \cdot \kern0.5em 0.05\right)+\left(0.50\kern0.5em \cdot \kern0.5em 0.05\right)+\left(0.50\kern0.5em \cdot \kern0.5em 0.05\right)\right]=0,125\end{array}} $$

Flower color

This trait is dominant, so it is not possible to determine by simple observation of all genotypic frequencies (homozygous dominant and heterozygous have the same phenotype). The genotypic frequency of the recessive mm can be calculated as 16/20 (80%). For the calculation of the other genotypic frequencies, allele frequencies must first be calculated.

To do this, it is assumed that the trait is in Hardy–Weinberg equilibrium.

The frequency of the rr genotype = 0.80; therefore, the frequency of the r allele is

$$ {\displaystyle \begin{array}{c}q={\left(0,80\right)}^{1/2}\\ {}=0,8944\end{array}} $$

As p + q = 1, p = 1–0.8944, p = 0.1056

Assuming Hardy–Weinberg equilibrium the genotypic frequencies are:

(p + q)² = p² + 2pq + q², where p² is the frequency of MM, 2pq is the frequency of Mm, and q² is the frequency of mm. Developing the binomial we get:

$$ {\left(p+q\right)}^2={p}^2+2 pq+{q}^2=0,0111+0,1889+0,8000. $$

The genotypic frequencies are:

$$ {\displaystyle \begin{array}{l} MM=0.0111\\ {}\mathrm{Mm}=0.1889\\ {}\mathrm{mm}=0.8000\end{array}} $$

• (b) Determine the equilibrium condition for each isoenzymatic system and for flower color The equilibrium condition is determined by comparing frequencies observed with the expected frequencies. This comparison can be made only when there is co-dominance since, as already mentioned, when there is dominance, the Hardy–Weinberg equilibrium is assumed and therefore it is not possible to know whether or not these traits are in equilibrium.

The expected frequencies for the three isoenzyme systems are:

Isoenzyme system 1

$$ {\displaystyle \begin{array}{c}{\left(p+q\right)}^2={\left(0.65+0.35\right)}^2\\ {}=0,4225+0,4550+0,1225\end{array}} $$

$$ {\displaystyle \begin{array}{l}\mathrm{A}1\mathrm{A}1=0.4225\\ {}\mathrm{A}1\mathrm{A}2=0.4550\\ {}\mathrm{A}2\mathrm{A}2=0.1225\end{array}} $$

When comparing these frequencies with the observed frequencies (point “a” of the present problem), they are observed to be different. Therefore, this isoenzyme system has not reached Hardy–Weinberg equilibrium.

Isoenzyme system 2

$$ {\displaystyle \begin{array}{c}{\left(p+q\right)}^2={\left(0.675+0.325\right)}^2\\ {}=0,4556+0,4388+0,1056\end{array}} $$

$$ {\displaystyle \begin{array}{l}\mathrm{A}1\mathrm{A}1=0.4556\\ {}\mathrm{A}1\mathrm{A}2=0.4388\\ {}\mathrm{A}2\mathrm{A}2=0.1056\end{array}} $$

When comparing these frequencies with the observed frequencies (point “a”), they are observed to be different. Therefore, this isoenzyme system has not reached Hardy–Weinberg equilibrium.

Isoenzyme system 3

$$ {\displaystyle \begin{array}{c}{\left(p+q+r+s\right)}^2=\Big({0.300}^2+{0.400}^2+{0.175}^2+{0.125}^2+2\kern0.5em \cdot \kern0.5em 0.300\kern0.5em \cdot \kern0.5em 0.400+2\kern0.5em \cdot \kern0.5em 0.300\ .0,175\\ {}+2\kern0.5em \cdot \kern0.5em 0,300\ .0,125+2\kern0.5em \cdot \kern0.5em 0,400\kern0.5em \cdot \kern0.5em 0,175+2\kern0.5em \cdot \kern0.5em 0,400\kern0.5em \cdot \kern0.5em 0,125+2\kern0.5em \cdot \kern0.5em 0,175+0,125\Big)\\ {}=0,0900+0,1600+0,0310+0,0156+0,2400+0,1050+0,0750+0,1400\\ {}+0,1000+0,0438\end{array}} $$

$$ {\displaystyle \begin{array}{l}\mathrm{A}1\mathrm{A}1=0.0090\\ {}\mathrm{A}1\mathrm{A}2=0.2400\\ {}\mathrm{A}1\mathrm{A}3=0.1050\\ {}\mathrm{A}1\mathrm{A}4=0.0750\\ {}\mathrm{A}2\mathrm{A}2=0.1600\\ {}\mathrm{A}2\mathrm{A}3=0.1400\\ {}\mathrm{A}2\mathrm{A}4=0.1000\\ {}\mathrm{A}3\mathrm{A}3=0.0310\\ {}\mathrm{A}3\mathrm{A}4=0.0438\\ {}\mathrm{A}4\mathrm{A}4=0.0156\end{array}} $$

When comparing these frequencies with the observed frequencies (point “a” of the present present), they are observed to be different. Therefore, this isoenzyme system has not reached Hardy–Weinberg equilibrium.

• (c) Determine the expected heterozygosity level for each isozyme system and for flower color. Indicate which trait is more variable.

$$ {\displaystyle \begin{array}{c}\mathrm{Isoenzyme}\ \mathrm{system}\ 1: EH=1-{0.65}^2-{0.35}^2\\ {}=0,45500\end{array}} $$

$$ {\displaystyle \begin{array}{c}\mathrm{Isoenzyme}\ \mathrm{system}\ 2: EH=1-{0.675}^2-{0.325}^2\\ {}=0,43875\end{array}} $$

$$ {\displaystyle \begin{array}{c}\mathrm{Isoenzyme}\ \mathrm{system}\ 3: EH=1-{0.300}^2-{0.400}^2-{0.175}^2-{0.125}^2\\ {}=0,70375\end{array}} $$

$$ {\displaystyle \begin{array}{c}\mathrm{Flower}\ \mathrm{color}: EH=1-{0.1056}^2-{0.8944}^2\\ {}=0,18900\end{array}} $$

The isoenzyme system 3 is the one with the highest heterozygosity and thus is considered the most variable locus among the four studied. This result was expected in view of the fact that the isoenzyme system 3 has 4 alleles, while other isoenzyme systems have 2 alleles. The greater the number of alleles within a locus, the greater the likelihood of obtaining a greater heterozygosis.

• (d) Indicate the possible reasons for the lack of equilibrium, if this is the case. In section “b” of this problem, it was concluded that the three isoenzymes systems are not in Hardy–Weinberg equilibrium. Possible causes are the presence of mutation, migration, or selection on the sample studied. Since the sample size is relatively small, there may be a case of problems with the sampling, i.e., that these 20 individuals do not represent the population under study (this is called genetic drift). However, it may also be the case that these 20 individuals do not belong to a particular population, but are simply samples that are not related to one another, and are of any interest to the researcher. The other possible cause of lack of equilibrium is autogamy. Sesame is a self-pollinating plant, which means that in the absence of mutation, migration and selection, it maintains the allele frequencies but changes its genotypic frequencies.

• 2. One of the problems in the production of papaya (Carica papaya) is the presence of plants with “male” flowers, which is determined by just one gene in the recessive state. Suppose you are advising a farmer, and the farmer tells you that season by season (taking seed from its own sowing), problems present with a large number of “male” plants. When performing an evaluation of the planting, 21 male plants are quantified out of a total of 420 plants. What technical recommendation would you give to the producer? Quantify how much would be the improvement in relation to the number of male plants after 6 generations.

The proportion of plants with male flowers is 21/420 = 0.05. Since the characteristic is recessive, the 21 plants with male flowers can be referred to have genotype “aa,” i.e., the genotypic frequency of aa will be 0.05. With this value, the frequency of the recessive allele can be calculated:

$$ {\displaystyle \begin{array}{c}q=0,{05}^{1/2}\\ {}=0,2236\end{array}} $$

Having the value of q, we calculate the value of p

$$ {\displaystyle \begin{array}{l}p+q=1\\ {}p=1-q\\ {}p=1-0,2236\\ {}\kern1.1em =0,7764\end{array}} $$

The genotypic frequencies will be:

$$ {\displaystyle \begin{array}{l}{\left(p+q\right)}^2={p}^2+2 pq+{q}^2\\ {}{\left(0,7764+0,2236\right)}^2=0,6028+0,3472+0,0500\end{array}} $$

$$ {\displaystyle \begin{array}{l}\mathrm{Frequency}\ \mathrm{AA}=0.6025\\ {}\mathrm{Frequency}\ \mathrm{Aa}=0.3472\\ {}\mathrm{Frequency}\ \mathrm{aa}=0.0500\end{array}} $$

The calculated allele and genotypic frequencies will remain constant in the absence of mutation, migration, and selection. In this case, what is of interest is to reduce the frequency of the recessive allele which can be achieved by selection. If the plants (aa) are removed from the field before they release their pollen, the field will be free of these genotype in this planting. However, it is important to consider that it is not the recessive allele “a” that is being eliminated since it is contained in the heterozygous plants; they are phenotypically normal. Therefore, when seed is obtained from this field and sown again, the aa genotype will appear, but with the advantage that it will be less frequent. After several selection processes (eliminating plants with the recessive genotype), a significant reduction can be achieved in the presence of male plants. This reduction can be quantified as follows:

The sum of the proportions of AA and Aa is equal to 0.9497, which will be 100% of the plants in the field, resulting in the following genotypic proportions:

$$ {\displaystyle \begin{array}{c}\mathrm{Frequency}\ \mathrm{AA}=0.6025/0.9497\\ {}=0,6344\end{array}} $$

$$ {\displaystyle \begin{array}{c}\mathrm{Frequency}\ \mathrm{Aa}=0.3472/0.9497\\ {}=0,3656\end{array}} $$

With these genotypic ratios, the new allele frequencies can be calculated:

$$ {\displaystyle \begin{array}{c}p=0,6344+\left(0,50\kern0.5em \cdot \kern0.5em 0,3656\right)\\ {}=0,8172\end{array}} $$

$$ {\displaystyle \begin{array}{c}q=0+\left(0,50\kern0.5em \cdot \kern0.5em 0,3656\right)\\ {}=0,1828\end{array}} $$

This population of plants, when crossed with each other with the new allele frequencies will originate a new population with the following genotypic frequencies:

$$ {\displaystyle \begin{array}{l}\mathrm{AA}={p}^2=0.6678\\ {}\mathrm{Aa}=2 pq=0.2987\\ {}\mathrm{aa}={q}^2=0.0334\end{array}} $$

Which will be displayed when seeds are sown in the next generation (first generation after selection). Repeating the process, eliminating the male plants, the genotypic frequencies change as follows:

$$ {\displaystyle \begin{array}{c}\mathrm{Frequency}\ \mathrm{AA}=0.6678/0.9665\\ {}=0,6909\end{array}} $$

$$ {\displaystyle \begin{array}{c}\mathrm{Frequency}\ \mathrm{Aa}=0.2987/0.9665\\ {}=0,3091\end{array}} $$

Therefore, the allele frequencies will be:

$$ {\displaystyle \begin{array}{c}p=0,6909+\left(0,50\kern0.5em \cdot \kern0.5em 0,3091\right)\\ {}=0,8455\end{array}} $$

$$ {\displaystyle \begin{array}{c}q=0+\left(0,50\kern0.5em \cdot \kern0.5em 0,3091\right)\\ {}=0,1545\end{array}} $$

This population of plants, when crossed with each other with the new allele frequencies will originate a new population with the following genotypic frequencies:

$$ {\displaystyle \begin{array}{l}\mathrm{AA}=p2=0.7149\\ {}\mathrm{Aa}=2 pq=0.2613\\ {}\mathrm{aa}={q}^2=0.0239\end{array}} $$

Which will be displayed when seeds are sown in the next generation (second generation after first selection). Repeating the process, eliminating the male plants, the genotypic frequencies change as follows:

$$ {\displaystyle \begin{array}{c}\mathrm{Frequency}\ \mathrm{AA}=0.7149/0.9762\\ {}=0,7323\end{array}} $$

$$ {\displaystyle \begin{array}{c}\mathrm{Frequency}\ \mathrm{Aa}=0.2613/0.9762\\ {}=0,2677\end{array}} $$

Therefore, the allele frequencies will be:

$$ {\displaystyle \begin{array}{c}p=0,7323+\left(0,50\kern0.5em \cdot \kern0.5em 0,2677\right)\\ {}=0,8662\end{array}} $$

$$ {\displaystyle \begin{array}{c}\mathrm{q}=0+\left(0,50\kern0.5em \cdot \kern0.5em 0,2677\right)\\ {}=0,1339\end{array}} $$

This population of plants, when plants are crossed with each other with the new allele frequencies will originate a new population with the following genotypic frequencies:

$$ {\displaystyle \begin{array}{l}\mathrm{AA}={p}^2=0.7503\\ {}\mathrm{Aa}=2 pq=0.2320\\ {}\mathrm{aa}={q}^2=0.0179\end{array}} $$

Which will be displayed when seeds are sown in the next generation (third generation after first selection). By repeating the process, eliminating the male plants, the genotypic frequencies change as follows:

$$ {\displaystyle \begin{array}{c}\mathrm{Frequency}\ \mathrm{AA}=0.7503/0.9823\\ {}=0,7638\end{array}} $$

$$ {\displaystyle \begin{array}{c}\mathrm{Frequency}\ \mathrm{Aa}=0.2320/0.9823\\ {}=0,2362\end{array}} $$

Therefore, the allele frequencies will be:

$$ {\displaystyle \begin{array}{c}p=0,7638+\left(0,50\kern0.5em \cdot \kern0.5em 0,2362\right)\\ {}=0,8819\end{array}} $$

$$ {\displaystyle \begin{array}{c}q=0+\left(0,50\kern0.5em \cdot \kern0.5em 0,2362\right)\\ {}=0,1181\end{array}} $$

This population of plants, when plants are crossed with each other with the new allele frequencies will originate a new population with the following genotypic frequencies:

$$ {\displaystyle \begin{array}{l}\mathrm{AA}={p}^2=0.7777\\ {}\mathrm{Aa}=2 pq=0.2083\\ {}\mathrm{aa}={q}^2=0.0139\end{array}} $$

Which will be displayed when seeds are sown in the next generation (fourth generation after first selection). Repeating the process, eliminating the male plants, the genotypic frequencies change as follows:

$$ {\displaystyle \begin{array}{c}\mathrm{Frequency}\ \mathrm{AA}=0.7777/0.9860\\ {}=0,7887\end{array}} $$

$$ {\displaystyle \begin{array}{c}\mathrm{Frequency}\ \mathrm{Aa}=0.2083/0.9860\\ {}=0,2113\end{array}} $$

Therefore, the allele frequencies will be:

$$ {\displaystyle \begin{array}{c}p=0,7887+\left(0,50\kern0.5em \cdot \kern0.5em 0,2113\right)\\ {}=0,8944\end{array}} $$

$$ {\displaystyle \begin{array}{c}q=0+\left(0,50\kern0.5em \cdot \kern0.5em 0,2113\right)\\ {}=0,1057\end{array}} $$

This population of plants, when plants are crossed with each other with the new allele frequencies will originate a new population with the following genotypic frequencies:

$$ {\displaystyle \begin{array}{l}\mathrm{AA}={p}^2=0.8000\\ {}\mathrm{Aa}=2 pq=0.1891\\ {}\mathrm{aa}={q}^2=0.0112\end{array}} $$

Which will be displayed when seeds are sown in the next generation (fifth generation after first selection). Repeating the process, eliminating the male plants, genotypic frequencies change as follows:

$$ {\displaystyle \begin{array}{c}\mathrm{Frequency}\ \mathrm{AA}=0.8000/0.9891\\ {}=0,8088\end{array}} $$

$$ {\displaystyle \begin{array}{c}\mathrm{Frequency}\ \mathrm{Aa}=0.1891/0.9891\\ {}=0,1912\end{array}} $$

Therefore, the allele frequencies will be:

$$ {\displaystyle \begin{array}{c}p=0,8088+\left(0,50\kern0.5em \cdot \kern0.5em 0,1912\right)\\ {}=0,9044\end{array}} $$

$$ {\displaystyle \begin{array}{c}q=0+\left(0,50\kern0.5em \cdot \kern0.5em 0,1912\right)\\ {}=0,0956\end{array}} $$

This population of plants, when plants are crossed with each other with the new allele frequencies, will originate a new population with the following genotypic frequencies:

$$ {\displaystyle \begin{array}{l}\mathrm{AA}={p}^2=0.8179\\ {}\mathrm{Aa}=2 pq=0.1729\\ {}\mathrm{aa}={q}^2=0.0956\end{array}} $$

which will be visualized when these seeds are sown in the next generation (sixth generation after first selection).

After six generations making selection against “males” plants, reduction of frequency of unfavorable allele is achieved, from 0.2236 to 0.0091, reducing male plants in the field from 5% to less than 1%.

• 3. The genealogy in a herd of cattle is as follows:

From the following crosses, organize in ascending order the most appropriate crossing to maintain the good genetic condition in the herd: H × I, H × B, F × I.

• 1. This problem can be solved by calculating the coancestry coefficient between H and I, between H and B, between F and I, and between D and I. The coancestry coefficient between two individuals is the same as the inbreeding coefficient of the offspring of these two individuals; therefore, the inbreeding coefficient can be calculated for the offspring coming from H and I, offspring coming from H and B, offspring coming from F and I, and offspring coming from D and I. The higher the coancestry coefficient between a pair of individuals, the less adequate the mating between those two individuals because the probability of obtaining the same allele in the descendant of the offspring is higher, which promotes homozygous individuals, which itself is deleterious for the herd.

• 2. The following steps address how to calculate the coancestry coefficient:

  • 2.1 Identify the common ancestors of the parents in each mating. To do that, you can go through the genealogy, in the diagram, from one parent to the other. You will ascend in the genealogy, and when you start to descend, and you can find a path to the other parent. You identify this animal by going down the genealogy after having gone up, like for a common ancestor. Some “rules” must be considered: First, you may ascend only from the parent to the common ancestor, and you may descend only from the common ancestor to the other parent. Second, you may use a path only one time; for example, if you use the path from E to A while ascending, you cannot use this path while descending. If you doubt this “method,” you can systematically write out the pedigree of the offspring of each mating in what is calling a speaking code. If for any offspring you write down the same animal twice, this is a common ancestor. For example, for mating HI, you can write out the pedigree of a hypothetical offspring of these two parents—whom we’ll name X—by writing out first the female parent and then the male parent, then the female parent of the female parent of X, then the male parent of the female parent of X, then the female parent of the male parent of X, then the male parent of the male parent of X, and so on. For example, for X, its pedigree would be IHCDEFBA--GA--. I is the female parent of X, H is the male parent of X, C is the female parent of I, D is the male parent of I, E is the female parent of H, F is the male parent of H, B is the female parent of C, A is the male parent of C, − represents that there is no information about the female parent of D, − represents that there is no information about the male parent of D, G is the female parent of E, A is the male parent of E, − represents that there is no information about the female parent of F, and – represents that there is no information about the male parent of F. If you detect that one letter in the pedigree has been repeated, it is a common ancestor. For example, in the pedigree HIEFDCGA----BA, A is repeated, which means that A is a potential common ancestor. Definitive proof requires descending the pedigree from A: If you’re using different paths, you’ll reach both parents, which marks a common ancestor. For A, this is the situation: It is repeated in the code, and you can reach H from A and can also reach I from A. To determine the coancestry coefficient between H and I, for any of the “methods” explained, A will serve as the common ancestor.

  • 2.2. All the possible pathways from one parent to the other parent represent the different ways that an allele can become a gamete of an animal. For example, to go from H to I, there is only one way: HEACI. This means that the only way for H and I to have the same alleles for one gene of A (a common ancestor) is for A to pass its allele on to E and C; later for E to pass its allele on to H; and then for C to pass its allele (the same allele that A passed on to E) on to I.

  • 2.3. For each path, the inbreeding coefficient of the potential offspring of two parents can be determined. Finding the ancestry coefficient between H and I is the same as finding the inbreeding coefficient of the offspring between H and I; let’s call this offspring X. The inbreeding coefficient of X will be (1 ÷ 2). The following tables can be used to determine this coefficient (Tables 11.1, 11.2, and 11.3):

Table 11.1 Mating H × I

Table 11.2 Mating H × B

Table 11.3 Mating F × I

Table 11.4 displays the coancestry coefficients, as follows:

Table 11.4 Coancestry coefficients of mating pairings

The following list ranks the mating pairings from worst to best on the basis of maintaining the good genetic condition in the herd:

• 1. H × I

• 2. H × B and F × I