#note Population Genetics A Concise Guide (5)

Page 14

Problem 1.3 Hardy-Weinberg frequencies in dioecious species may be investigated in an entirely different way. Let the genotype frequencies in females be x₁₁, x₁₂, and x₂₂, and in males, y₁₁, y₁₂, and y₂₂. Enumerate all nine possible matings (A₁A₁ female by A₁A₁ male, A₁A₁ female by A₁A₂ male, etc.) and calculate the frequencies of genotypes produced by each one as a function of the x\’s and y\’s. Sum these frequencies, weighted by the frequencies of the matings, to obtain the genotype frequencies in the second generation. Now let these genotypes mate at random to produce the third generation, If all goes well, a morass of symbols will collapse into the satisfying simplicity of the Hardy-Weinberg law.

Note:

Female	Male	Frequency	A1A1	A1A2	A2A2
A1A1	A1A1	x11y11	1	0	0
A1A1	A1A2	x11y12	1/2	1/2	0
A1A1	A2A2	x11y22	0	1	0
A1A2	A1A1	x12y11	1/2	1/2	0
A1A2	A1A2	x12y12	1/4	1/2	1/4
A1A2	A2A2	x12y22	0	1/2	1/2
A2A2	A1A1	x22y11	0	1	0
A2A2	A1A2	x22y12	0	1/2	1/2
A2A2	A2A2	x22y22	0	0	1

Write a Java program to simulate the random mating to check the first 5 generations:

package de.yanzhou;

import java.util.Arrays;
import java.util.List;

public class RandomMating {

  public static void main(String[] args){

    List<Double> frequenciesInFemale = Arrays.asList(0.65, 0.25, 0.1); //A1A1, A1A2, A2A2
    List<Double> frequenciesInMale = Arrays.asList(0.15, 0.35, 0.5);//A1A1, A1A2, A2A2
    List<Double> r = calculate(frequenciesInFemale, frequenciesInMale);
    System.out.println("2. Generation : " + r.get(0) + " " + r.get(1) + " " +r.get(2));
    int i = 3;
    while(i<=5){
      r = calculate(r,r);
      System.out.println(i + ". Generation" + " : " + r.get(0) + " " + r.get(1) + " " +r.get(2));
      i++;
    }
  }
  private static List<Double> calculate(List<Double> l1, List<Double> l2){
    List<Double> r =
            l1.stream()
                    .flatMap(i -> l2.stream()
                            .map(j -> i * j)
                    ).toList();
    double frequencyOfA1A1 = 1*r.get(0) + 0.5*r.get(1) + 0.5*r.get(3) + 0.25*r.get(4);
    double frequencyOfA1A2 = 0.5*r.get(1) + 1*r.get(2) + 0.5*r.get(3) + 0.5*r.get(4) + 0.5*r.get(5) + 1*r.get(6) + 0.5*r.get(7);
    double frequencyOfA2A2 = 0.25*r.get(4) + 0.5*r.get(5) + 0.5*r.get(7) + 1*r.get(8);
    return Arrays.asList(frequencyOfA1A1, frequencyOfA1A2, frequencyOfA2A2);
  }
}

Output:

2. Generation : 0.25187499999999996 0.59625 0.151875
3. Generation : 0.30249999999999994 0.49499999999999994 0.20249999999999996
4. Generation : 0.3024999999999999 0.49499999999999983 0.20249999999999996
5. Generation : 0.30249999999999977 0.4949999999999997 0.20249999999999985

So it will in the third generation reach equilibrium.

By input:

List<Double> frequenciesInFemale = Arrays.asList(0.25, 0.5, 0.25); //A1A1, A1A2, A2A2
List<Double> frequenciesInMale = Arrays.asList(0.25, 0.5, 0.25);//A1A1, A1A2, A2A2

It can reach equilibrium in the second generation (frequency of A₁A₁ is equal to the frequency of A₂A₂). Output:

2. Generation : 0.25 0.5 0.25
3. Generation : 0.25 0.5 0.25
4. Generation : 0.25 0.5 0.25
5. Generation : 0.25 0.5 0.25