罗莎琳德：孟德尔冷杉

Question

我想在http://rosalind.info/problems/iprb/上解决这个问题

Given: Three positive integers k, m, and n, representing a population containing k+m+n organisms: k individuals are homozygous dominant for a factor, m are heterozygous, and n are homozygous recessive.

Return: The probability that two randomly selected mating organisms will produce an individual possessing a dominant allele (and thus displaying the dominant phenotype). Assume that any two organisms can mate.

我的解决方案适用于示例，但不适用于生成的任何问题。在进一步的研究之后，我似乎应该找到随机选择任何一个有机体的概率，找到选择第二个有机体的概率，然后找到配对产生具有显性等位基因后代的概率。在

我的问题是：我下面的代码找到的概率是多少？它能找到所有可能的交配中带有显性等位基因的后代所占的百分比吗？因此，我的代码不是计算一次随机交配的概率，而是求解所有配对都被测试时具有显性等位基因的后代的百分比？在

f = open('rosalind_iprb.txt', 'r')
r = f.read()
s = r.split()
############# k = # homozygotes dominant, m = #heterozygotes, n = # homozygotes recessive
k = float(s[0])
m = float(s[1])
n = float(s[2])
############# Counts for pairing between each group and within groups
k_k = 0
k_m = 0
k_n = 0

m_m = 0
m_n = 0

n_n = 0


##############
if k > 1:
    k_k = 1.0 + (k-2) * 2.0

k_m = k * m

k_n = k * n

if m > 1:
    m_m = 1.0 + (m-2) * 2.0

m_n = m * n

if n> 1:
    n_n = 1.0 + (n-2) * 2.0
#################
dom = k_k + k_m + k_n + 0.75*m_m + 0.5*m_n
total = k_k + k_m + k_n + m_m + m_n + n_n

chance = dom/total
print chance

Answer 1

我在这个问题上花了一些时间，所以，用python澄清一下：

lst = ['2', '2', '2']
k, m, n = map(float, lst)
t = sum(map(float, lst))
# organize a list with allele one * allele two (possibles) * dominant probability
# multiplications by one were ignored
# remember to substract the haplotype from the total when they're the same for the second haplotype choosed
couples = [
            k*(k-1),  # AA x AA
            k*m,  # AA x Aa
            k*n,  # AA x aa
            m*k,  # Aa x AA
            m*(m-1)*0.75,  # Aa x Aa
            m*n*0.5,  # Aa x aa
            n*k,  # aa x AA
            n*m*0.5,  # aa x Aa
            n*(n-1)*0  # aa x aa
]
# (t-1) indicate that the first haplotype was select
print(round(sum(couples)/t/(t-1), 5))