Independent Alleles

13 Jun 2024 in Study / Rosalind on Rosalind problem

A problem from rosalind “Bioinformatics Stronghold” category, Probability Calculation

• A problem from rosalind “Bioinformatics Stronghold” category
  Rosalind Problem Link

• Description

  • 0^th generation: AaBb
  • 1^st generation: AaBb (0^th generation) x AaBb
  • n^th generation: ? ((n-1)^th generation) x AaBb
  • Each individual gets 2 offsprings
  • Goal: The probability that at least N Aa Bb organisms will belong to the k^th generation
  • The genes satisfy the law of independent assorment

My Solution

• Sketch
  • AaBb (0^th generation) x AaBb

    • Offspring genotype ratio

      
      

                  AA       Aa      aa
            BB     1        2       1
            Bb     2        4       2
            bb     1        2       1
      

  • Critical Point: In all of generations, the genotype ratio is maintained
    • That is, AaBb’s ratio in all of generations is 4/16 (0.75)
    • Reason
      • The ingredient (gamete’s genotype) for making offspring’s genotype in 0^th generation is AA, Ab, aB, ab
      • The ratio of each ingredient is maintained because the new ingredient is supplied as the same amount with AaBb (0^th generation)
      • For the same reason
        • When 0^th generation’s genotype is α and repeat to do cross with genotype α -> First offspring’s genotype ratio is maintained in all of generations
  • “The probability that at least N Aa Bb organisms will belong to the k^th generation”
    • Number of offsprings in k^th generation = 2^k
    • If the number of AaBb is n in k^th generation
      • Probability = _2^kC_n * 0.75^{(2k - n)} * 0.25ⁿ
    • Answer_prob = 1 - _2^kC₀ * 0.75^{(2k - 0)} * 0.25⁰ - _2^kC₁ * 0.75^{(2k - 1)} * 0.25¹ - _2^kC₂ * 0.75^{(2k - 2)} * 0.25² - ….. - _2^kC_{N - 1} * 0.75^{(2k - (N - 1))} * 0.25^{N - 1}

    
    

• Code

def factorial(n):
    if n == 1 or n == 0:
        return 1
    return n * factorial(n - 1)

def combination(n, r):
    return factorial(n)/(factorial(r) * factorial(n - r))

name = input()
f = open(name, 'r')
k, N = list(map(int, f.readline().split()))

prob = 1
num = 2 ** k

for i in range(N):
    sub_prob = combination(num, i) * (0.75 ** (num - i)) * (0.25 ** i)
    prob -= sub_prob

print(prob)