What is the Hardy–Weinberg principle? Applying it to Practical Scenarios.
Today I’ll explain what the Hardy–Weinberg principle is.
The concept of genetic equilibrium, also known as Hardy-Weinberg equilibrium, describes the idealized conditions in a population where the allele frequencies remain constant over time. It assumes that certain conditions are met, including a large population size, random mating, no mutation, no migration, no selection, and no genetic drift. However, in real populations, these conditions would be often not fully met, and various factors can lead to changes in allele frequencies and genotypic proportions such as mutation, genetic drift. Therefore, while genetic equilibrium describes a theoretical scenario, real populations are subject to various evolutionary forces that can cause changes in allele and genotype frequencies over time.
https://en.wikipedia.org/wiki/Hardy%E2%80%93Weinberg_principle
I will explain taking this into consideration. Let’s illustrate it with a simple example. There is a population of hemp being cultivated, consisting of 2 individuals with the genotype: AA, 1 individual with the genotype: Aa, and 1 individual with the genotype: aa.
In that case, how will the genotype ratio be in the next generation? First, let’s examine the allele frequencies in the current population of cultivated beans
| A | a | Total | |
| No. Allele | 5 | 3 | 8 |
| Allele frequency | 0.625 | 0.375 | 1 |
Now, one question remains:
If we assume that A is dominant over a (and a is recessive to A), and if continuous generational progress occurs, wouldn’t a gradually decrease in number and eventually disappear completely?
However, according to the equations proposed by British mathematician Hardy and German physicist Weinberg, recessive genes like a do not disappear even as generations pass. For example, let’s consider one population with two contrasting alleles, A and a, and assume that the proportions of these alleles in the population are represented by p and q, respectively. In the initial generation, the allele proportions sum up to p+q=1. In the next generation, the genotypes AA will be present in a proportion of p, Aa in a proportion of 2pq, and aa in a proportion of q.
In other words, in the next generation, the proportion of contrasting alleles can be represented by the equation p2 + 2pq + q2 = 1.
That is, as we progress to the next generation, the frequencies of gene types remain constant. Now, let’s examine the genotype ratio in the next generation of the bean population mentioned in the example above.
In the population with 2 individuals of genotype: AA, and 1 individual each of genotypes: Aa and aa, the possible combinations that can occur in the next generation are shown in the above diagram.
The first row represents selfing, and the rest represents crosses with other individuals. Now, let’s calculate the 64 possible genotypes that can result from the 16 combinations such as AA * AA, AA * Aa, AA * aa, and so on.
Looking at each genotype, there are 25 of AA, 30 of Aa, and 9 of aa. Therefore, when calculating the genotype ratios for the total of 64 combinations, we have the following:
AA : 25/64 = 0.391 Aa : 30/64 = 0.469 aa : 9/64 = 0.141
| AA | Aa | aa | Total | |
| No. genotype | 25 | 30 | 9 | 64 |
| Genotype frequency | 0.391 | 0.469 | 0.141 | 1 |
When we initially calculated the frequency of the contrasting allele A, it was 0.625, and the frequency of allele a was 0.375. Now, in the next generation, the frequency of genotype AA is 0.391. This value is equal to the square of the initial frequency of allele A (0.625 * 0.625 = 0.391). The same applies to genotype aa (0.375 * 0.375 = 0.141).
So, what about Aa? The frequency of Aa is equal to twice the product of the frequencies of alleles A and a (2 * 0.625 * 0.375 = 0.469).
As explained earlier, in the initial generation, the proportion of contrasting alleles is represented by p + q = 1 [0.625 + 0.375 = 1]. In the next generation, the frequency of genotype AA is represented by p2 [0.625 * 0.625 = 0.391], Aa by 2pq [2 * 0.625 * 0.375 = 0.469], and aa by q2 [0.375 * 0.375 = 0.141].
Therefore, in the next generation, the frequencies of gene types are represented by p2 + 2pq + q2 = 1 [0.391 + 0.469 + 0.141 = 1]. In conclusion, even as we progress to the next generation, the frequencies of gene types remain constant.
Practical examples
Let’s further understand the Hardy-Weinberg equations through a few examples.
[Example 1] In a cabbage field, there are 80 cabbages planted, with the following genotypes: [24 AA, 4 Aa, and 52 aa]. Assuming random mating, what would be the frequencies of the AA, Aa, and aa genotypes in the next generation?
If we let the frequency of the contrasting allele A be represented by p and the frequency of allele a be represented by q, then we have p + q = 1 according to the Hardy-Weinberg equation. Furthermore, according to the equation, the frequencies of genotypes in the next generation would be p2 + 2pq + q2 = 1. In the cabbage field, out of the 80 cabbages, there are 52 cabbages with the allele A. Since AA is a diploid genotype, having 24 cabbages with genotype AA implies that there are 48 copies of allele A. Additionally, with 4 cabbages being Aa, we can conclude that there are 4 copies of allele A.
The number of contrasting alleles a is 108 (52 x 2 + 4 = 108). Among the 80 diploid cabbages, the total number of contrasting alleles is 160, with 52 being allele A and 108 being allele a. The frequencies of A and a can be calculated as A: 0.325 (52/160 = 0.325) and a: 0.675 (108/160 = 0.675), respectively. We can represent these frequencies as p = 0.325 and q = 0.675, and of course, p + q = 1.
Let’s examine the proportions of genotypes in the next generation. We are already familiar with the Hardy-Weinberg equation, p2 + 2pq + q2 = 1. Plugging in the values, we have (0.325)2 + 2*(0.325)*(0.675) + (0.675)2 = 1.
In other words, in the next generation, the frequency of genotype AA is 0.106 (= 0.325 * 0.325 = 0.106), the frequency of Aa is 0.439 (= 2 * 0.325 * 0.675 = 0.439), and the frequency of aa is 0.456 (= 0.675 * 0.675 = 0.456).
[Example 2] What would be the number of cabbages representing each genotype if there are 80 cabbages in the cabbage field above?
AA : 80 * 0.106 = 8.5 Aa : 80 * 0.439 = 35.1 aa : 80 * 0.456 = 36.5
| Genotype | 1st generation | 2nd generation |
| AA | 24 | 8.5 |
| Aa | 4 | 35.1 |
| aa | 52 | 36.5 |
[Example 3] When the next generation of 80 cabbages undergoes further generational advancement and becomes the generation after the next, what would be the frequencies of the genotypes AA, Aa, and aa in that generation?
According to the Hardy-Weinberg principle, which states that the frequencies of genotypes remain constant across generations in a population, the frequencies of genotypes in the generation after the next will also be the same. Thus, the frequencies in the next generation are as follows: AA 0.106, Aa 0.439, aa 0.456.
| Genotype | 1st generation | 2nd generation | 3rd generation |
| AA | 0.3 (=48/160) | 0.106 | 0.106 |
| Aa | 0.05 (=8/160) | 0.438 | 0.438 |
| aa | 0.65 (=8/160) | 0.456 | 0.456 |
[Example 4] When the second-generation advancement takes place in the cabbage field, assuming there are 200 individuals, what would be the frequencies of the genotypes AA, Aa, and aa at that time?
AA : 200 * 0.106 = 21.2 Aa : 200 * 0.439 = 87.8 aa : 200 * 0.456 = 91.2
[Example 5] According to the Hardy-Weinberg principle, which states that the frequencies of genotypes remain in equilibrium, in a cabbage cultivation area where the genotypic frequencies are in balance, if only the aa genotype is eliminated before flowering, what would be the frequencies of the possible genotypes in the next generation?
The genotype aa with a frequency of 0.456 has been eliminated from the original population with genotypic frequencies of AA 0.106, Aa 0.439, and aa 0.456. This necessitates calculating the frequencies of genotypes in the new equilibrium population first.
The frequencies of the new population’s alleles are as follows:
p = 0.194 + 1/2 * 0.806 = 0.597
q = 1/2 * 0.806 = 0.403
The changing allele frequency for the alternative allele A is obtained by adding half of the frequency of Aa, which contributes one A allele, to the existing frequency of AA. The frequency of the alternative allele a is obtained by taking half of the Aa frequency, which contributes one a allele.”
The frequencies of genotypes in the next generation are as follows:
AA: (0.5972)² = 0.357 Aa: 2 * (0.5972) * (0.4028) = 0.481 aa: (0.4028)² = 0.162
| Genotype | 1st generation | 2nd generation | after aa eliminated |
| AA | 0.3 (=48/160) | 0.106 | 0.357 |
| Aa | 0.05 (=8/160) | 0.438 | 0.481 |
| aa | 0.65 (=8/160) | 0.456 | 0.162 |
[Example 6] According to the Hardy-Weinberg principle, which states that the frequencies of genotypes remain in equilibrium, in a cabbage cultivation area where the genotypic frequencies are in balance, if only the aa genotype is eliminated after flowering, what would be the frequencies of the possible genotypes in the next generation?
This problem is easier to understand with some knowledge about seed reproduction.
Let’s consider yourself as a seed production specialist in a seed company. You are currently growing cabbage for cabbage seed production. To eliminate off-types, you visit the cabbage field. The flower color of the parent generation for F1 seed production is all white, and pink color is a recessive trait that should be eliminated before flowering. Due to work commitments, you visited the cabbage field late and discovered that flowering had already begun. In a hurry, you quickly removed the pink flowers (aa) and discarded them.
Now, let’s analyze the possible alleles in the next generation. Pollen has already been dispersed, indicating that pollen from the pink flowers (aa) may have fertilized somewhere, leading to potential contamination. Now, let’s calculate the frequencies of alleles in the paternal parent generation.
p = 0.106 + 1/2 * 0.439 = 0.325
q = 0.456 + 1/2 * 0.439 = 0.675
One thing to note when calculating the frequencies of alleles in the paternal parent generation is that the new subgroup [aa eliminated] has not yet been formed. This means that the probabilities should be calculated based on the original equilibrium population frequencies of AA 0.106, Aa 0.439, and aa 0.456. [i.e., The pollen from the paternal parent has already been dispersed. Even if we remove the paternal aa individuals, the pollen has already been carried away, making the act of removing aa meaningless].
However, let’s consider the frequencies of alleles in the maternal parent generation. The maternal parent’s aa individuals have already been removed completely. In other words, the frequencies in the maternal parent generation should be calculated based on the formation of the new subgroup [i.e., The maternal parent needs to have the aa individuals removed. The reason is that F1 seeds originate from the maternal parent. If there are remaining aa individuals in the maternal parent, the seeds produced from them will be contaminated. Therefore, the maternal aa individuals need to be eliminated, leading to the formation of a new subgroup].
| Genotype | AA | Aa | Total | |
| frequency | 0.106 | 0.439 | ……….. | 0.545 |
| new frequency | 0.194 | 0.806 | ……….. | 1 |
Frequencies of alleles in the new subgroup:
p = 0.194 + 1/2 * 0.806 = 0.597
q = 1/2 * 0.806 = 0.403
Next generation genotype frequencies:
AA : 0.325 * 0.403 = 0.194 Aa : (0.675)*(0.597) + (0.325)*(0.403) = 0.403 + 0.131 = 0.534 aa : 0.675 * 0.403 = 0.272
How can the Hardy-Weinberg principle be applied in practical scenarios within seed companies?
In the context of the Hardy-Weinberg theory, numerous logical principles can be applied in practical scenarios by seed companies. Let’s consider a few examples:
1) Watermelon cultivation is currently taking place in South Korea. During a visit to promote seed sales, a farmer expressed their intention to utilize the seeds from this year’s watermelon harvest for cultivation in the next season. The watermelons grown by the farmer this year displayed an elongated shape phenotype, with AA representing 65% of the population, Aa accounting for 34%, and the round shape phenotype making up only 1%. This indicates a 99% pure watermelon variety.
p= 0.65 + 1/2 * 0.34 = 0.82 q= 0.01 + 1/2 * 0.34 = 0.18
If the farmer plants 10,000 watermelon seeds again, what results can be expected?
p2 +2pq + q2 =1
AA : 0.82 * 0.82 = 0.6724 [67%]
Aa : 2 * 0.82 * 0.18 = 0.2952 [30%]
aa : 0.18 * 0.18 = 0.0324 [3%]
If the farmer chooses not to purchase seeds and instead cultivates 10,000 seeds from their own harvest, approximately 300 seeds [10,000 * 3% = 300] will yield the round shape phenotype, making them unsuitable for selling as elongated watermelons, which are in high demand from January to February. These watermelons require increased boiler operation and significant labor during the winter season, resulting in a cost of around $20 per watermelon. This would lead to a loss of approximately $6,000. Considering that purchasing 12,000 watermelon seeds would cost around $1000 – 1200, it is much more advantageous to buy the seeds.
