Last week we derived a formula for calculating change in gene frequencies, given the initial gene frequency, the degree of dominance (if any), and the difference in fitness values (if any) of the various genotypes.

This week we’ll use that formula as it could apply in real-life.

Let’s revisit the following graph from Effectiveness of Selection: Initial Gene Frequency and Fitness Differences:

Change in gene frequency of undesired allele over generations, with different degrees of dominance.
Solid line: no dominance. Broken line: complete dominance. Dotted line: complete recessiveness.
© Optimate Group Pty Ltd

Assume that the ‘A₂‘ allele is undesired at the ‘A’ locus, and has a gene frequency q of 0.3. Assume too that the ‘A₂‘ allele is completely dominant, and with respect to fitness, that the A₁A₂ and A₂A₂ genotypes produce half as many offspring relative to the A₁A₁ genotype. Thus:

s₁ = 0
s₂ = 0.5
s₃ = 0.5

and

Let’s now assume that there is no dominance at the ‘A’ locus. With respect to fitness, the A₁A₂ genotype produces 25% fewer offspring relative to the A₁A₁ genotype, while the A₂A₂ genotype produces half as many offspring relative to the A₁A₁ genotype.

s₁ = 0
s₂ = 0.25
s₃ = 0.5

and

But what if ‘A₂‘ was completely recessive? Assume that the A₁A₂ genotype produces as many offspring relative to the A₁A₁ genotype, while the A₂A₂ genotype produces half as many offspring relative to the A₁A₁ genotype.

s₁ = 0
s₂ = 0
s₃ = 0.5

and

We have shown how, in one generation of selection, that the frequency q of the ‘A₂‘ allele drops noticeably from 0.3 to 0.2 if completely dominant. This is less marked if there is no dominance at the ‘A’ locus, where the frequency drops slightly from 0.3 to 0.24. The change in frequency is even less if the allele is completely recessive, from 0.3 to 0.27. These values can be seen graphically — below is the same graph as above, but with the inital gene frequency q = 0.3 marked with a red line. You can see how the steepness (change in frequency) differs for each line between where each begins at q = 0.3 (read this as generation 0) and one generation later (further to the right along the x-axis).

Change in gene frequency of undesired allele over generations, with different degrees of dominance.
Red line: gene frequency q = 0.3 for A2 allele.
Solid line: no dominance. Broken line: complete dominance. Dotted line: complete recessiveness.
© Optimate Group Pty Ltd

This concludes the Selection Strategies: Simply Inherited Traits section.

Assume a locus A with alleles A₁ and A₂. Either could be, but is not necessarily, dominant or recessive to the other.

p is the gene frequency of the A₁ allele and q is the gene frequency of the A₂ allele. (Similarly, p and q as used here do not necessarily ascribe dominance or recessiveness to either allele.)

P is the genotypic frequency of the A₁A₁ genotype.
H is the genotypic frequency of the A₁A₂ genotype.
Q is the genotypic frequency of the A₂A₂ genotype.

Remember that P + H + Q = 1. If there are three genotypes in a population, the proportion of each as a percentage (its frequency) must add up to 100%, or 1.

We can express the proportions of P, H and Q as:

Assume these frequencies occur in a population from which parents have not yet been selected to produce the next generation.
If their parents were randomly mated, then this population should be in a Hardy-Weinberg equilibrium, thus:

P = p²
H = 2pq
Q = q²

Substituting these values into the formulae above, we get:

But we need to take into account the degree of dominance with respect to fitness for each of the genotypes A₁A₁, A₁A₂ and A₂A₂.
Let s₁ be the relative fitness difference for A₁A₁.
Let s₂ be the relative fitness difference for A₁A₂.
Let s₃ be the relative fitness difference for A₂A₂.

If A₁A₁ is the fittest genotype (produces the most offspring), then its fitness difference, relative to itself, is s₁ = 0, and its relative fitness is (1 - s₁) = (1 - 0) = 1.

If, hypothetically, the A₁A₂ genotype produces 25% fewer offspring than the A₁A₁ genotype, then the fitness difference, relative to the fittest genotype A₁A₁, is s₂ = 0.25, and the relative fitness is (1 - s₂) = (1 - 0.25) = 0.75. We can say that the A₁A₂ genotype is 75%, or 0.75 as fit as the A₁A₁ genotype.

More generally, we can say that:
the relative fitness value for A₁A₁ is (1 - s₁),
the relative fitness value for A₁A₂ is (1 - s₂), and
the relative fitness value for A₂A₂ is (1 - s₃).

In other words:
the relative fitness of A₁A₁ is (1 - s₁) of its genotypic frequency P, or
(1 - s₁) × P = (1 - s₁) × p² = (1 - s₁) p²

Similarly, we can state the relative fitness of A₁A₂ as (1 - s₂)2pq, and the relative fitness of A₂A₂ as (1 - s₃)q².

We wish to select animals from this population to be the parents of the next generation. From this, their progeny become the next parents, with genotypic frequencies of P₁, H₁ and Q₁. Can you see how the following formulae are the same as above, but this time we have taken into consideration the relative fitness values for each of P₁, H₁ and Q₁:

In The Effect of Mating Systems on Gene and Genotypic Frequencies: Outbreeding, we saw how the gene frequency q = Q + ½H.

From this, the frequency of the A₂ allele after selection is:

Substituting for Q₁ and H₁, we get:

[Here we have ’simply’ added two (elaborate) fractions with a common denominator. This is just ^a/_c + ^b/_c = ^a + b/_c on steroids!]

As there are just two alleles A₁ and A₂, with the respective gene frequencies p  and q , these must add up to one, as p + q = 1. We can rewrite this as p = 1 - q.
We can substitute (1 - q)  for p, cancel some terms, and rearrange the rest to get:

We now have a formula with which we can calculate the new gene frequency in the next population, given the initial gene frequency, the degree of dominance (if any), and the difference in fitness values (if any) of the various genotypes.

Next week we’ll run through some scenarios to see this in practice!

This week we go over a third influence: the degree of dominance with respect to fitness.

Think of ‘fitness’ as a trait like any other. Complete dominance of ‘fitness’ at a locus has both homozygous and heterozygous dominant animals equally fit, and each would be expected to produce the same numbers of offspring on average.

Should there be no dominance of ‘fitness’ at a locus, homozygous dominants would produce the most number of offpsring, homozygous recessives the least, and heterozygous dominants a number exactly halfway between these two.

Consider the following graph:

Change in gene frequency of undesired allele over generations, with different degrees of dominance.
Solid line: no dominance. Broken line: complete dominance. Dotted line: complete recessiveness.
© Optimate Group Pty Ltd

Here are represented the changes in frequency of an undesired A₂ allele with respect to three degrees of dominance: no dominance (solid line), complete dominance (broken line), and complete recessiveness.

The solid line is the same as in the graphs here. With no dominance, gene frequency changes are fastest through the intermediate frequencies, and A₂ can be completely removed from the population with time.

Selection against completely dominant alleles at very high frequencies, however, shows a very slow rate of change at the beginning, but does pick up and those alleles too can be removed from the population.

Selection against completely recessive alleles is markedly different to the first two scenarios. While the change in gene frequencies begins rapidly, this slows until the allele continues to persist at some lower frequency and is never removed completely.

This persistence is because completely recessive alleles have completely dominant counterparts — the dominant allele in a heterozygous animal is the one expressed, thus completely masking any recessive allele that may be present. This is compounded by the fact that recessive alleles are ’spread out’ thinly amongst a population when at very low frequencies. These two factors make completely recessive alleles extremely difficult to detect and select against.

Let’s look again at this graph from The Effect of Selection on Gene and Genotypic Frequencies:

Relationships between gene and genotypic frequencies in a population
© Optimate Group Pty Ltd

q, the frequency of the recessive allele ‘a’ is very low on the left hand side of the graph, and there are very few homozygous recessive genotypes compared to heterozygous ones. At these low gene frequencies, the completely recessive ‘a’ allele is more likely to be ‘hidden’ in these heterozygous animals than be visible in any homozygous recessive ones, simply because there are more heterozygotes in the population.

For example, assume the recessive allele ‘a’ has a frequency of 0.1 (10%), and that there are 100 animals in a herd, none of which is homozygous recessive. This means of the 200 ‘A’ loci in that herd, 20 carry the ‘a’ allele and 180 carry the ‘A’ allele. As no animal is homozygous recessive, this means twenty animals must be the heterozygous ‘Aa’ genotype. There is no way to detect which twenty are carriers unless they are mated to another carrier, or there is a genetic test for the ‘a’ allele, and the ‘a’ allele will likely be passed on undetected for many generations.

This elusiveness of completely recessive alleles is why to this day coloured spots still appear from time to time in elite merino sheep despite many decades of culling and selective breeding. Many other examples apply to other breeds and species. The allele could simply be ‘undesirable’, such as attempting to eliminate red coat from black Angus cattle, or very much unwanted, such as genetic defects that are lethal when homozygous recessive.

Next week, some calculations of changes in gene and genotypic frequencies caused by selection!

One influence on the effectiveness of selection on gene frequency changes is the initial gene frequency in a population. Consider two alleles at locus A: A₁ is wanted and A₂ is not. This graph plots the frequency of A₂ in each successive generation, showing the effect of selection against that unwanted allele A₂ over many generations:

Change in gene frequency of undesired allele over generations
© Optimate Group Pty Ltd

The steeper the slope of the line, the faster the change in the frequency of A₂ from one generation to the next, and the quicker it is removed from the population.

In the above graph, A₂ exists at a very high frequency at the beginning of the selection programme — it occupies over 90% of the A locus in the entire population. It is a slow process to remove this allele at first, as so many animals carry it. This is shown by the slow-to-get going slope of the line from generations 1 through 5.

Once A₂ does drop in frequency, A₁ by default increases in frequency, and as more and more animals inherit A₁, selection against A₂ accelerates as more and more A₁ alleles in the population become available to select for.

Another influence on the effectiveness of selection on gene frequencies is the fitness of individuals within a population. ‘Fitness’ is the ability of an animal to have offspring. The fittest animals are those that are not only selected for their genotypes and phenotypes, but which also produce the most offspring with those genotypes and phenotypes.

Consider the following graph:

Change in gene frequency of undesired allele over generations, with different fitness differences.
Solid line: large fitness differences. Broken line: small fitness differences
© Optimate Group Pty Ltd

This is the same graph as above, but with an extra line. Here, the solid line represents large differences in fitness amongst the genotypes of the breeding population. The broken line represents small differences in fitness amongst the genotypes of the breeding population.

For the solid line, if A₁A₁ genotypes are the fittest relatively, A₁A₂ genotypes may produce only three-quarters as many offspring in comparison, and A₂A₂ genotypes may produce only half as many offspring as A₁A₁ again. The differences in fitness are large.

For the broken line, if A₁A₁ genotypes are again the fittest relatively, A₁A₂ genotypes may produce almost as many — seven-eights perhaps — while A₂A₂ genotypes may produce three-quarters as many offspring as A₁A₁ animals. The differences in fitness are small.

Changes in gene frequencies are fastest when the differences in fitness are largest. This is because the most fit animals — the A₁A₁ genotypes in this example — produce the most offspring that in turn are selected to produce the next generation of offspring. The more A₁A₁ genotypes in each generation, the higher the frequency of the A₁ allele at the A locus amongst the next generation.

Consider the extreme case of a lethal allele. Here the fitness difference is very large — homozygous and heterozygous animals with either no or just one copy of the allele survive and reproduce, while homozygous recessive animals with two copies die. Natural selection against that allele is very effective, in that the allele will exist at a very low frequency in the population, and homozygous recessive animals will be rare occurrences.

Compare that example with a gene responsible for smaller fitness differences — one that has a very small effect on the ability to conceive for example. Even the negatively-affected animals will still grow and reproduce at almost the level of fitness as the unaffected ones. There is very little difference in fitness between the genotypes and the allele will continue in the population with little influence from natural selection on its frequency. It is harder to effect change when there are only small differences in fitness amongst the genotypes.

Next week will cover the third influence on selection and gene frequency change: the degree of dominance with respect to fitness.

[Tabulating confidence levels here for a mix of mates from different groups (eg where some are known carriers, some are daughters, and still others are randomly picked) isn’t practical. There are two many combinations of some number of  known carriers, for some other number of daughters, for some other number again of randomly picked individuals. Formula 4a here can be used manually for any specific scenario however. (There is no formula for the reverse, determining the number of matings with mixed groups for a required level of confidence, as these cannot be easily calculated for more than one n.)]

1. Number of Matings Required to Detect a Completely Recessive Allele When Litter Size m = 1

2. Number of Matings Required to Detect a Completely Recessive Allele When Litter Size m = 5

3. Number of Matings Required to Detect a Completely Recessive Allele When Litter Size m = 10

These tables show very well how much the type of mate determines the number of matings that will be required for a particular level of confidence. Always choose as best you can with what you have to keep the number of matings as low as possible!

Also evident is how larger litter sizes require so fewer matings to achieve higher levels of confidence.

Always factor in more matings than your calculations would indicate, to compensate for a (likely) less than 100% successful birth rate.

1. The formulae for one offspring per mating and with mates from the same group
(eg all are known carriers, or all are daughters of the tested sire, or all are randomly selected from a population) are:

a. Confidence level for known number of matings

b. Number of matings needed for known confidence level

Let’s now explore other mating and birthing scenarios.

2. The formula for one offspring per mating and with mates from the same group
(eg multiple groups of mates, such as a group of known carriers as well as a group of daughters)

We need to modify the formula in 1a above so as to calculate the level of confidence:

Breaking this down, we have:
i is a counter referring to the individual groups of mates (can you see this subscript to the P_AA, P_Aa, P_aa and n?)
k is the number of groups of mates
Π, the capital Greek letter pi, stands for ‘product’, a word in maths that means ‘the result obtained when multiplying numbers together’

Thus means ‘calculate the stuff inside the parentheses for i = 1 (the first group of mates), i = 2 (the second group of mates), and so on for all i = k groups of mates, and then multiply the results together’.

Let’s step through a calculation so this is clearer. Assume a group of four known carrier mates and a group of ten daughter mates, half of which are expected to be ‘Aa’ carriers and the other half expected to be ‘AA’ homozygous dominant.

If not one recessive animal is born, we can be 92% confident that this particular sire won’t be a carrier. It would be worth adding a few more matings to increase this level of confidence a bit more, especially if some of these don’t go to full gestation.

Can we do the reverse, and calculate the number of matings needed for a known confidence level? Unfortunately as there is more than one n (one for each mating group), we cannot easily solve for n though we may have a level of confidence we are aiming for.

3. The formulae for multiple offspring per mating and with mates from the same group
(eg in species that twin or have litters)

a. Confidence level for known number of matings
For species with multiple births that average m per mating. Let’s test a ram by mating him to six daughters that average twin births:

b. Number of matings needed for known confidence level
How many matings would need to be done to get the level of confidence up to 95%?

4. The formula for multiple offspring per mating and with mates from multiple groups
(this is a generalised formula that can be used for any test mating scenario)

a. Confidence level for known number of matings

For species with multiple births that average m per mating. Going back to the test ram above, there aren’t another six daughters available, but there is access to a known carrier that averages a single birth:

Should she too average twins, the level of confidence would increase again:

Next week we’ll wrap up all this maths with some tables summarising levels of confidence and numbers of matings required for various scenarios.

Firstly, what are logs?

‘log’ is short for ‘logarithm’. And ‘logarithm’ can be defined as ‘how many times must one number be multipled by itself to get another number?’

For example, writing ‘log₁₀(100)’ means ‘how many times must we multiply 10 by itself to get to 100?’
The answer is 2, as 10 × 10 = 100.

Similarly, log₁₀(1000) is 3 and log₁₀(10000) is 4.

The ‘10′ part refers to ‘base 10′. You could have logs of other bases, like log₂(8) = 3 (2 × 2 × 2 = 8), or log₄(256) = 4 (4 × 4 × 4 × 4 = 256), and so on, but we really don’t need to worry about these. This is because base 10 is our regular everyday number system, and log₁₀ is the one most often used in maths and science. To save writing it all the time, people drop the ‘10′ part altogether and write it as ‘log’. If you see ‘log’ by itself, know it to automatically mean log₁₀.

Thus we could equally write log(100) = 2 or log(1000) = 3.

Before the days of calculators and electronic computers, people routinely used logarithms to perform quite involved mathematical calculations. These people were called ‘computers’, as in ‘one who computes’. Mathematicians were hiring ‘computers’ as early as the 1600s! It is only since the 20th century that the word ‘computer’ became synonymous with machines who took over that (often tedious and laborious) work.

Logarithms made working with difficult numbers by hand much easier, and knowing the rules made things easier again. One such logarithmic rule is:
log(x)ⁿ = n × log(x)

In algebra, it is common to use a ‘.’ (full stop or period) instead of the multiplication sign ‘×’.
Thus log(x)ⁿ = n × log(x) can also be written as log(x)ⁿ = n.log(x)

Going back to the formula from last week:

Line 2 is a simple swapping to have  by itself on the left.
Line 3 takes the log of both sides to remove the exponent n. (Can you recognise the application of the rule log(x)ⁿ = n.log(x) here?)
(If the log is taken on one side it must also be taken on the other side so both sides balance.)

Line 4 is a simple rearrangement, and from this we were able to solve for n as in the examples last week.

Let’s now calculate confidence levels and the required number of test matings to be statistically confident that a tested animal is not a carrier of a recessive allele.

Everything below assumes that the tested animal is a sire, that there is one offspring from one mating, and that all mates (dams) are of the same type for the allele of interest. That is, they are either all known carriers, or are all daughters of the tested sire, or are all randomly selected from the same population.

Now let:

n = number of matings that produce an offspring
P[D_n] = probability (P) of detection (D) that at least one homozygous recessive offspring is born for n number of matings. This is our level of confidence in the test.
P_AA = probability that a mate is homozygous dominant (’AA’) at the locus being tested.
P_Aa = probability that a mate is heterozygous (’Aa’) at the locus being tested.
P_aa = probability that a mate is homozygous recessive (’aa’) at the locus being tested.

From all this:

OK, so where did this come from, you may well be asking?! Let’s break this down bit by bit.

P[D_n], as stated above, is our level of confidence.
From last week, our level of confidence can also be written as 1 - α.

Thus P[D_n] = 1 - α.

Substitute α (alpha) with , and you have 1 - 

But what is that sum inside the parentheses?

Firstly, does it make sense to you that the (probability of homozygous recessive born) + (probability of no homozygous recessive born) must equal 1, or 100%?
In other words, that it is 100% certain that all offspring born must have some combination of alleles, regardless of what that combination is?

And from that, does it also make sense that to calculate the probability of at least one homozygous recessive offspring being born, that you could calculate the probabilities of no homozygous recessive offspring being born, and subtract that from 1 (100%)? We could write that as:
(probability of homozygous recessive born) = 1 - (probability of no homozygous recessive born)

From this, can you see how is the probability of no homozygous recessive born?

Going further:

P_AA is the probability that a dam is homozygous dominant (’AA’) at the locus being tested. This animal by definition can only contribute ‘A’ alleles.
The probability that she will birth a non-recessive offspring is 1 (100%), whether or not the sire is a carrier.
The P_AA is thus written as if it were 1 × P_AA.

P_Aa is the probability that a dam is heterozygous (’Aa’) at the locus being tested.
If both the sire and female are carriers, there are three chances in four that a dominant phenotype will be born, hence we write ³/₄P_Aa.

P_aa is the probability that a dam is homozygous recessive (’aa’) at the locus being tested.
If the sire is a carrier, and the female is homozygous recessive, there are two chances in four that a dominant phenotype will be born, hence we write ¹/₂P_aa.

To calculate the number of matings to give us enough confidence that a sire is not a carrier, we need to know (mating outcome 1) and (mating outcome 2), and so on for some number n such that the confidence level is high.

(mating outcome 1) and (mating outcome 2) and (mating outcome 3) and so on can be rewritten
(mating outcome 1) × (mating outcome 2) × (mating outcome 3) × (…), as in statistics, when we say ‘and’ we write ‘×’.

And (mating outcome 1) × (mating outcome 2) × (mating outcome 3) × (…) is the same as writing
(mating outcome)ⁿ, where n is the number of matings.

Hopefully isn’t so scarey-looking now?

Time to use it and work with some numbers!

Assume we have five known carriers to test a sire with. This means the probabilities of P_AA and P_aa must both be 0, and that of P_Aa is 1.

That isn’t a very high level of confidence, but let’s take an extreme and assume 30 known carriers:

Now that is a very high level of confidence that the sire is not a carrier! But it would appear we could use a smaller number of females for almost as good a result. That would certainly be a more preferable use of dams. Rather than just punching in numbers at random to find the most favourable values for n and [D_n], the most practical approach is to solve for n outright.

How are you with logs?! (I don’t want to dwell on logs too much here but I have written a little supplementary post here to explain how the formula was rearranged below.)

Let’s rearrange to solve for n:

The number of matings required to give us a 95% level of confidence that no homozygous recessives will be born, and that the sire is not a carrier, using known carrier dams is:

This is an interesting example to make a point with. The solution for n, to two decimal places, was actually 10.42. Hence the rounding down to 10. But it’s so close to 10.5 that maybe it should be rounded up to 11 for an extra margin of safety.

Indeed, you should always factor in additional matings anyway, to account for matings that don’t take or aborted foetuses. The calculated number of matings assumes a live birth for every mating, ie it assumes n number of successful matings.

The above assumed one offspring per mating and that all mates were of the same type. We’ll stop there as that was quite a lot of information to absorb! Next week we’ll go over the maths for other scenarios involving multiple groups of mates and multiple births per mate.