text: # Thread: Hardy Weinberg law in genetics
  
  1. ## Hardy Weinberg law in genetics
  
  This law concerns the genetic make-up of a population from one generation to the next. It states that in sexually reproducing organisms, in the absence of genetic mutation, factors(called alleles) determining inherited traits are passed down unchanged from generation to generation. We want to show that the law is true. Consider the simple case of only two alleles, A and B, in a gene. The probability of occurrence of the A gene in a population in generation n is Pn, and that of the B gene is Qn. Pn + Qn = 1. These two alleles combine to form in the next generation AA or BB, with probability (P^2)n, (Q^2)n, and 2PnQn, respectively.
  
  a. The probability of the occurence of the A alleles in the n+1 generation is denoted by Pn+1 and that of the B alleles by Qn+1. We write
  Pn+1 = f(Pn, Qn), Qn+1 = g(Pn, Qn).
  
  Find the functions f and g.
  Hint: The probability of occurence of AA in generation n+1 from generation n is (P^2)n. The probability is 100% that the individual with the AA gene has the A alleles. The probability is only 50% that an individual with the AB gene will contribute an A allele to the next generation.
  
  b. Show that f = Pn and g = Qn, and therefore
  Pn+1 = Pn = P and Qn+1 = Qn = Q,
  
  where P and Q are independent of n.
  
  I really need help !!!
  
  2. OK, my knowledge of genetics is infinitesimal, but here is my take.
  
  An organism contains two copies of the gene in question, and each copy can be in one of two variants, or forms, or alleles: A and B. Suppose there is a population of organisms. Take a random organism and then take one of the two copies of the gene; the probability that this copy is A is p, and the probability that the copy is B is q = 1 - p. In other words, the probability that any given gene has allele (form) A is p and the probability that is has form B is q.
  
  During reproduction, the child randomly receives one copy of the gene from each parent, so the child again has two copies. It is stated as a hint that if in the parent population the probabilities of alleles A and B are p_n and q_n, respectively, then the probability that a child has two gene copies of allele A is $p_n^2$, the probability that a child has one A and one B is $2p_nq_n$, and the probability of BB is $q_n^2$.
  
  Suppose the (n+1)st generation has N organisms. Then there are $Np_n^2$ organisms with two A genes, $2Np_nq_n$ with one A and one B gene, and $Nq_n^2$ organisms with two B genes. The total number of genes is $2N$. The number of A genes is $(Np_n^2\cdot2+2Np_nq_n\cdot1)$; therefore, the probability of the A gene in the (n+1)st generation is $\displaystyle\frac{(2Np_n^2+2Np_nq_n)}{2N}=p_n^2+p _nq_n=p_n(p_n+q_n)=p_n$. Similarly, the probability of the B gene is $q_n$. Therefore, $f(p_n,q_n)=p_n$ and $g(p_n,q_n)=q_n$.
  
  3. So this question is not really two problems, a) 's answer is b) and all I need to do is just b) huh?
  
  Thanks !!!
  
  4. This is so great. I am also stuck here. UW Amath 383 sucks.
  
  5. So this question is not really two problems, a) 's answer is b) and all I need to do is just b) huh?
  Well, yes!