Basics on animal genetic resources

Farm animal genetic resources, or simply AGR, refers to the genetic material we currently have in live animals and frozen in sperm banks, and which is of economical, cultural and scientific importance. Genetic diversity, both within and between breeds, is vital. Diversity

• helps the animals to adapt to their environment
• is the basis for animal breeding
• allows adaptation to changes and new breeding targets
• prevents inbreeding depression
• keeps the frequency of harmful allelels low.

Domestication has already reduced genetic diversity in farm animal species (and artifical insemination has reduced it even further). Domestication is the process by which captive animals adapt to man and the environment provided, and it is achieved through genetic changes. Genetic factors affecting domestication are inbreeding, genetic drift and selection.  

Impact of domestication on milk yield of dairy cows.

Inbreeding is necessary when selecting a desired trait, but it reduces heterozygosity and thus diversity, although it does not affect allele frequencies. Genetic drift causes alleles to become fixed or deleted randomly, and its direction cannot be estimated. Selection, both natural and artificial, has altered the fitness of certain traits. For example, in domestication species the distance to flee and fearfulness have been decreased, even though they are vital for a wild animal. These are both behavioral changes. Physiological changes involve changes in hormone levels, reproduction cycle and production traits (e.g. the increased milk yield of cows and the all-year farrowing of sows). Morphological changes have also occurred, as animals have grown larger and developed colors unseen in the wild (especially white). For an example of scientific study on domestication, see Giuffra et. al (2000) or Kantanen et al. (1999).

Studying domestication

Domestication can be studied in several methods. Paternal transmission of Y-chromosome shows traits have developed from sire to offspring. It's counterpart is the study of mtDNA (mitochondrial DNA), which is always inherited from the dam to all her offspring. mtDNA haplotypes are sequenced and aligned, and dendrograms or cladograms are drawn based on the multiple-sequence-alignment (MSA) results. The haplotypes can be further divided into groups, which helps to draw a network. For more information visit the blog The Genealogical World of Phylogenetic Networks and their post on interpreting rooted networks.

The most common method by far is still studying microsatellite markers. Usually 20-30 microsatellites are studied, but FAO has published recommendations for each animal species (FAO). Polymorphic loci with 4 or more alleles are recommended to eliminate false positives by identical-by-state -alleles. Unlike mtDNA and Y-chromosome, microsatellites are inherited from both parents to all offspring. Genotyping and aligning SNP-markers is similar to microsatellies, but requires the use of thousands or hundreds of thousands of SNP-markers. Microchip arrays are readily available for several species for SNP-analysis.

Practical application of genetic domestication studies.
(c) ILRI 2006

Studying ancestral DNA is tedious, but can yield valuable information on extinct species. Ancestral DNA can be thousands or tens of thousands of years old. It is collected either from animal remains such as fossils, teeth, wools, hides or bone pieces, or from the ground ("dirty DNA"). Ancestral DNA also helps to chart the spread of different animal and plant species and to determine temporal changes in their genetics. The problem with ancestral DNA is that the concentration of desired DNA is often low, while the concentration of microbial DNA is high. Contamination risk is very high indeed. Sterile environment must be maintained whenever possible when working with ancestral DNA. Another problem is that the ancestral DNA has been fragmented, and many chemical bonds have been broken. C > T and G > A mutations in PCR are common due to deamination. The results must be confirmed in several independent laboratories. In addition, all results derived from ancestral DNA must fit in to earlier context.

Retroviruses have been used as a study method on sheep. Retroviruses are viruses, which insert their RNA to the sheep's system, and with reverse-transcriptase produce DNA from the RNA. The viral DNA is then integrated as a part of the sheep's own genome. The viral DNAs which have infected germline cells are hereditary, and thus provide a tool for studying the evolution of sheep. The original virus infections happened 5-7 million years ago, and have continued to branch even during the last 10 000 years. Studies of retrovirus-DNA has shown that originally all sheep in Europe were used for meat production. A meat-and-wool producing breed was introduced later,  and replaced the old breed nearly completely. Still existing breeds originating from the first migration are Soay sheep, Gutesheep and Finnsheep.  

Determining the level of endangerment and the value of a breed

There are thousands of animal breeds in the world. FAO's DAD-IS -information system classifies breeds into four categories:

• local breeds, which exists in one country or area only
• transboundary breeds, which exists in several countries
• regional transboundary breeds, which exists in several countries but only in one continent
• international transboundary breeds, which exists in many continents.

Each breed is also classified based on the level of endangerment. There are five levels, which are determined by the population size and number of breeding males and females. Other classification systems also consider the direction of population size (growing or decreasing), the purity of the species and the number of populations (e.g. herds). FAO's five levels are

1. Extinct
2. Critical (with or without a conservation program)
3. Endangered
4. Not at risk
5. No information on the population size.

Currently DAD-IS lists (among others) 3093 cattle breeds, 2558 chicken breeds/lines and 1283 pig breeds. Altogether there are 14544 breeds listed for 38 animal species. Of the listed local cattle breeds, 181 are extinct (209 in 2006) and only 399 are not at risk. For pigs there are 110 extinct breeds (140 in 2006) and 206 not at risk. The numbers from the year 2006 are larger, probably due to renewal of the concept of breed or improved methods of separating breeds and collecting information. Below are a few examples of the tables created from DAD-IS system, showing the status of cattle, pig and sheep breeds in different regions.

The level of endangerment is the likelihood of the breed going extinct in the current circumstances within a certain time period. It can be used to estimate how much there is time to save a breed. The level depends on demographic factors (population size and its changes) and genetic variation. Genetic variation is calculated from effective population size N_e, which again is deduced from the change of inbreeding (Ne = 1/2 ΔF). Growth factor can be calculated from r = anti-log (( logN2–logN1) / t ), where N1 and N2 are the population size at two different measurements (generation 1 and 1+n), and t is time in years. The growth factor depends on animal births and deaths, cullings, changes in market prices and agricultural politics and on epidemics.

Method of estimation impacts the value of  ΔF. Pedigree-based studies give consistently lower estimated of ΔF than SNP-based evaluation. In one study, pedigree analysis found that 15 % of animals had F > 6.25 %, while in an SNP-study the percentage was 25 %. For pairwise kinship coefficients both methods are equally reliable for close relatives. For more distant relatives the pedigree analysis gives higher estimates than SNP-analysis. (Li et al. 2011)

However, N_e, ΔF and the growth factor are only single meters. To estimate the value of a breed for genetic conservation requires a more holistic approach. In addition to the meters mentioned before, the breed value depends on several factors. The factors, and examples related to them, are listed below.

•  its ability to adapt to a certain environment (Yakutian cattle, goat breeds in arid African countries)
• economically important traits (the excellent cheese-making qualities of the milk of Finncattle)
• unique traits (breed-specific mutations, alleles and gene combinations)
• cultural heritage, historical value (Yakutian cattle)
• unique genetics.

One must remember that the breed must be able to cope even in the future, and to continue being useful for the herders. It is not a viable option to maintain breeds which cannot survive for example after global warming or if their surroundings change due to industrialization. 

Yakutian cattle (c) EPFL / Anu Osva

Genetic analysis

XKCD's take on genetic analysis.

Genetic analysis may sound complicated, but it relies on very simple principles of heredity, statistics and probabilities. Sometimes the genotypes are not known at all, and we need to look at pedigrees to determine models of heritance. Here some basic principles of genetic analysis are discussed. Later on a post on bioinformatics will make a deeper analysis on how to analyze known genes an genomes.

Calculating probabilities

Inheritance, or which genes each parent passes on to the offspring, is always partially random. This is why probabilities are important. For example if we know the absolute frequency, i.e. the number, of a certain allele in a sample, we can deduce the probability of one randomly picked individual having that particular allele. Set the absolute frequency of the allele a to 118. Now we know that there are 118 a alleles in our sample of 500 chromosomes. The probability of a random individual to have a is simply the amount of a divided by the number of all possible alleles: 118 / 500 = 0,236. This is also the relative frequency of a.

So, the probability of A, whatever A is, is P(A) = the number of favorable results / the number of all possibilities. The probability of A's complement, of A not happening, is 1- P(A). Combinations of two or more independent variables are calculated as follows: P(A and B) = P(A) * P(B) and P(A or B) = P(A) + P(B). An example of a complement: if a chromosome has the allele a, it cannot have the allele A. Only either one or the other (barring some rare genetic mutations, which are not considered here).

The probability of two separate outcomes depends on if the outcomes are related or not. A union means that either (A or B) or (A and B) happen. For independent variables the union would be P(A) + P(B) - P(A u B). For dependent variables the union is zero: if A has happened, B cannot happen, or vice versa.

To ease your stress, here is a cute animal.

The conditional probability, the probability of A happening when we know that B has already happened and A and B are dependent, is (P(A) * P(B)) / P(B). This is often marked as P(A|B).

Permutations and combinations deal with the order of several possibilities. Permutations are used when the order of the events is important, and is counted as n!. Combinations take all orders into consideration, and are counted as n! / (r! (n-r!)).

Binomial probability is a bit more complex. It is used when we want to determine the probability of getting exactly r favorable results, each with the probability of p, out of n repeats. The magic word here is exactly - when that is used in an exercise, think of binomials.

 

More on probabilities:

Khan Academy

Wolfram

For the mathematically gifted: Wikipedia 

Cut the Knot (also covers Bayesian methods)

Statistics

Statistics have been covered before in a post named Variance Analysis (one of the most popular posts in this blog!), so I'll just remind you of the formulae we will need when doing genetic analysis.

There are thousands of helpful websites you can look up for more information on statistics. Here are just a few:

Khan Academy

x2 (khi squared) test and genotypes

x2 test and plant breeding

Statistics.com
Mendelian genetics by Phillip McLean

Examples

Right, let's get down to the real deal and do some analysis! The examples are from University lecture materials for course in genetic material, but I unfortunately cannot share the entire material due to copyright restrictions and a language barrier - the materials are not in English :)

Example 1. Parents are heterozygotic concerning their eye color. They both have the allele s for blue eyes and the allele S for brown eyes. Calculate the probability that they will have 

a) a child with blue eyes  b) five children with blue eyes.

From the way the alleles are written we see that S is the dominant allele. The genotypes of the parents are Ss and Ss, so the possible genotypes of their children are 

Now we can see that 3/4 of the offspring have the dominant allele, so only 1/4 has blue eyes (homozygote ss). Therefore the P(a child has blue eyes) is 1/4 = 25 %. The probability for each subsequent child is similar, so P (five of five children have blue eyes) is 0,25 * 0,25 * 0,25 * 0,25 * 0,25 = 0,000977.

Example 2. 32 % of people infected with a rare illness have mutation A, and 16 % have mutation B. 10 % of those infected have both A and B. Calculate the probability of randomly selected person to have at least one of the mutations? 

The selected person must now have either A or B or both. What is needed is the union of A and B: P(A) + P(B) - P(A u B) = 0,32 + 0,16 - 0,1 = 0,38. 38 % have either A, B or both mutations. Note that the union P(A u B) is NOT A*B in this case, but it is given as 10 %.

Example 3. There are 2 boys and 5 girls in a family. In how many different sequences could the children have been born? 

Because the order is not important, we'll use combinations: n over r, i.e. n! / r!(n-r)!. The n now is 7, the number of all kids. The r can be either 2 or 5: both give the same result. If r = 2, then 

7! / 2!(7-2)! = 7! / 2! 5! = 5040 / 240 = 21.

Example 4. Parents are heterozygotes concerning a rare recessive illness. Calculate the probability that out of three children

a) all are healthy

b) two are ill

c) at least two are ill.

a) Recessive heterozygotes produce 25 % of recessive homozygotic alleles (see example 1). The probability for each child to be healthy is thus 1 - 0,25. P(all are healthy) = 0,753 = 0,422.

b) Two out of three must be ill, so we need the binomial distribution. Now r = 2, n = 3 and p = 0,25. The first factorial, n over r, gives 3!/2!(3-2)! = 3. Continuing from there we have 3 * 0,25² * (1-0,25)^3-2 =3 * 0,0625 * 0,75 = 0,141.

c) If at least two must be ill, then the probability is P(two are ill) + P(three are ill). The first part is calculated like in part b: 3 * 0,25² * (1-0,25)^3-2 =3 * 0,0625 * 0,75 = 0,141. P(three are ill) is simply 0,253 = 0,015625. So P(at least two are ill) = 0,141 + 0,015625 = 0,156. 

Example 5. The penetrance of a certain illness varies between genotypes. The penetrances are 0,01 for AA, 0,05 for Aa and 0,5 for aa. In a population the allele frequencies are  f(a) = 0,05 and f(A) = 0,95. The population is in Hardy-Weinberg equilibrium. Count the prevalence of the illness in the whole population.

Whoa, lots of terms here! Penetrance is the probability of expressing a certain trait. Here the trait is the illness. H-W balance means that in an ideal population, where q and p are the relative frequences of a alleles, there are p² dominant homozygotes, 2pq heterozygotes and q² recessive homozygotes. What are they actually askingfor is the prevalence, i.e. the probability of a random member of the population to have the illness.

First we need to calculate the frequencies of the genotypes in the population. With the H-W equilibrium and the given frequencies we know that p = 0,05 and q = 0,95. Now the genotype frequencies are  
AA  = dominant homozygotes = p² = 0,95² = 0,9025
Aa = heterozygotes = 2pq = 2*0,95*0,05 = 0,095
aa = recessive homozygotes = q² = 0,05² = 0,0025

Now each genotype has its own probability of actually expressing the illness. To make it easier to understand we can build a table of  a tree of probabilities:

So now the random person we select has a 0,9025 % chance of having the genotype AA, and then 0,01 % chance of being sick.  Note that these variables are independent (a person can only have one genotype and be either sick or healthy). The possibility of expressing the illness is thus P(has a certain genotype) * P(is sick).

P(is sick) = P (AA and sick) + (P Aa and sick) + P(aa and sick) = P(0,9025 * 0,01) + P(0,095*0,05) + P(0,0025*0,5) = 0,009025 + 0,00475 + 0,00125 = 0,015.

Example 6. The observed genotype frequencies in a population are f(AA) = 31, f(Aa) = 89 and f(aa) = 122). Is the population in Hardy-Weinberg equilibrium?

If we were mean about it, we'd say no because no actual population is ever in H-W equilibrium. However in an exam we'd get 0 points for that, so let's calculate this. Now we need to use x2 or khi squared test. We already have the observed frequencies. Now we need the expected frequencies, i.e. the frequencies if the population was in H-W equilibrium.

To get there we calculate the allele frequencies in the population. There are 31 creatures with two A alleles and 89 with one. In total we then have 2 * 31 + 89 = 151 A-alleles. The recessive allele a is calculated similarly: 2*122 + 89 = 333. In total there are 151 + 333 alleles, so the relative frequencies are 151/484 = 0,312 for A and 1-0,312 = 0,688 for a. So now p = 0,312 and q = 0,688.The expected absolute genotype frequencies are now

AA  = p² = 0,3122 * 24² (the size of population) = 23.56
Aa = 2pq = 2 * 0.312 * 0.688 * 242 = 103.89
aa = q² = 0.688² * 242 = 114.55

We need to calculate the chi squared test value using the formula

To make it easier, let's put our values to a table. Then we can use the formula and calculate the x2 test variable.

The value 4.97 is not the answer. Remember the question: is the population in H-W equilibrium? The answer hides in a x2 distribution table. To use that we need degrees of freedom (df), which in an chi squared good of fit test is the number of classes minus 1. Here we have three classes (three genotypes), so our df = 3-2 = 1. Now we look at a x2 distribution table such as this.

With df = 2 we find that our  x2-value, 4.97, goes between 4.605 and 5.991. The corresponding P-values are 0.10 and 0.05. So our p is 0.1 - 0.05. Without going too far into interpreting p-values we can just note that it is higher than 0.05 which means that we abadon the hypothesis 0 (the population is in H-W equilibrium). P > 0,05 shows us that the population is NOT in H-W equilibrium.