Login

Gene FrequencyModerator: BioTeam
9 posts • Page 1 of 1
Gene FrequencyTwo separate populations of equal size are in equilibrium for the same pair of alleles because of random mating within each. In population I, pA=0.6, while in population II, pA=0.2, with q=1p in each population. If a random sample of females from one population is crossed to a random sample of males from the other population, what would be the progeny genotype frequncies? If these progeny are then allowed to mate at random, what would be the expected gene and genotypic frequencies in the next generation? What happens to the heterozygote frequencies between F1 and F2?
I need help in this problem! All I can figure out is the p2, q2 and 2pq. For example,in population I : 36% AA, 48% Aa and 16% aa. In population II: 4% AA, 32% Aa and 64% aa. But I don't know what to do from there!
Then you have to make all the Punnet squares including the frequencies, so e.g. AA × AA would be in 0.36x0.04 cases (hopefully:). But I don't know, whether there is some easier way.
http://www.biolib.cz/en/main/
Cis or trans? That's what matters.
Genetic equilibrium means that the ratio of alleles is not changing between generations, but it has nothing to do with what that ratio is for the population.
All you have to do is modify Mendel's punnet square, which is essentially a probability grid. In Mednel's punnet square, you give a 50%50 chance to each allele (two rows/columns; one for each allele). However, you're dealing with populations. Imagine that population A has p=0.2 and q=0.8. You decide to put this population's allele frequencies on the horizontal axis of the probability grid (rows). Population B has p=0.25 and q=0.75, and you're putting it on the vertical axis (columns). This is how your probability grid should look. This grid shows the 20% chance of any offspring inheriting allele A from the population A parent and the 25% chance of inheriting it from the population B parent. Of course, if we knew the genotypes of the parents, we would give different probabilities. But that's not the information we have. We only know which population each parent was from, so we must use the allele frequencies of those populations to determine the odds. NOTE: probability grids only work when the two variables are independent of each other.
Notice that calculating the probabilities from a probability grid uses the same mathematics as calculating the area of a square.
20 ft × 25 ft = 500 ft² (homozygous for A) 20 ft × 75 ft = 1500 ft² (AB) 80 ft × 25 ft = 2000 ft² (BA) AB + BA = 3500 ft² (heterozygous) 80 ft × 75 ft = 6000 ft² (homozygous for B) 500:3500:6000 ∝ 5:35:60
First cross to get F1:
Females: A (0.6) a (0.4) crossed with Males: A (0.2) a (0.8 ) Frequency of AA = 0.6 x 0.2 = 0.12 Frequency of Aa = (0.6 x 0.8 ) + (0.4 x 0.2) = 0.48 + 0.08 = 0.56 Frequency of aa = 0.4 x 0.8 = 0.32 Than your new allele frequencies (call them p1 and q1 for F1 generation) derived from this: p1 = Frequency of AA + ½ Frequency of Aa and q1= Frequency of aa + ½ Frequency of Aa So, p1=0.4 and q1=0.6 Now mate Female A (0.4) a (0.6) with Male A (0.4) a (0.6) to get F2 generation: Frequency of AA = 0.4^2 = 0.16 Frequency of Aa = 2 x 0.4 x 0.6 = 0.48 Frequency of aa = 0.6^2 = 0.36 Your heterozygote (Aa) frequency decreased from F1 to F2…
Re:This is turning out to be an interesting thread (I love mathematics).
The following method for calculating p and q usually gives the same results as finding the square roots of p^2 and q^2. However, it does not give the same values in this case. sqaure root of 0.12 is approximately 0.35 and the square root of 0.32 is approximately 0.57 Not only are these values different, they do not add up to 1. With another equation, it can be shown that the values 0.4 and 0.6 are indeed correct. Remember, the P generation dominant allele frequencies are 0.6 and 0.2. (0.6 + 0.2) ÷ 2 = 0.4 (0.4 + 0.8) ÷ 2 = 0.6 So, it seems as if F1 somehow defies the Hardy Weinberg equation by having genotype frequencies that one would not expect from the allele frequencies. This effect is evened back out when the F1 interbreed to produce the F2 generation. The F2 has genotype frequencies that one would expect from the allele frequencies. I understand that this happened, but can anyone explain why this happened?
Re: Re:
Simple, Hardy Weinberg equation applies to large populations. In this case, two distinct populations (both of which achieved different equilibrium points) are mixed. Thus, Frequency of the resulting F1 AA genotype is not p^2, but p1 (of the 1st population) times p2 (that of the 2nd population), etc. The resulting frequency, however, is not equal to p^2 as the population in this cross is NOT at equilibrium. Thus, the "math discrepancy" so to speak. Now, next cross will result in establishment of new equilibrium and Hardy Weinberg equation will apply again...
That's right, and I guess the factor that puts it out of equilibrium is that the mating is not random. Of course, the mating isn't nonrandom due to sexual selection, it's nonrandom due to human interference ("breed these with those").
I've been workiong on a more mathematical explanation, but my understanding still isn't very intuitive. When interbreeding two distinct populations, one multiplies the allele frequencies of the interbreeding populations to determine the genotype frequencies for the resulting generation (see my probability grid above). However, the resulting allele frequencies are simply the averages of the interbreeding populations' two allele frequencies. e.g.
Now, imagine that p1 belongs to population one, p2 belongs to population two, and p3 belongs to the population resulting from the interbreeding. Then, in order for the resulting population to fit the HardyWeinberg model, p1×p2 (or q1×q2) must be equal to p3^2 when p3 is equal to (p1+p2)÷2 (the average or 'middle' of p1 and p2). It doesn't take long to realize that this will never happen unless p1=p2 is true. Furthermore, the further apart p1 and p2 are, the lower the square root of p1×p2 will be in comparison to (p1+p2)÷2. When taking any two numbers (n1 and n2) and comparing their average to the square root of their product, it's easy to see why this 'happens'. If one solves n1×z=n2, they see that the square root of the product of n1 and n2 is equal to n1 multiplied by the square root of z. For example, the average of 2 and 50 is 26. The square root of their product (100) is 10. 50 divided by 2 is 25. The square root of 25 is 5. 2 multiplied by 5 is 10.
9 posts • Page 1 of 1
Who is onlineUsers browsing this forum: No registered users and 1 guest 
© BiologyOnline.org. All Rights Reserved. Register  Login  About Us  Contact Us  Link to Us  Disclaimer & Privacy  Powered by CASPION