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This question is all about understanding gene mixing without evolutionary pressure, and keeping track of long, tedious calculations.
Hardy-Weinberg says that gene frequencies do not change without some outside pressure causing them to change. This ignores random drift, but allows some calculations to be easier, and is a good approximation in most situations.
So, if gene frequencies don't change, then shouldn't the genotypes of the offspring be the same as the genotypes of the parents? Yes, if and only if the population is in equilibrium. If the population is not, then genotypes will change over time until they reach equilibrium, but the gene frequencies themselves will remain constant.
Why would a population not be in equilibrium? Who knows. Maybe you have lots of orange cats and your neighbor has lots of black cats, and you decide one day to let your populations interbreed. The important thing is that you are given the gene distributions in your initial population.
Assume random breeding, then your three genotypes can breed with anyone with any of the genotypes mentioned. If you assume your population size is small, then calculation gets hard to impossible, so assume your population size is large. This allows you say that for anyone in the population, the chance of breeding with someone from any genotype is the same as the frequency of that genotype in the population. Therefore, you have a matrix of mating probabilities.
AA Aa aa
AA .6*.6 .6*.2 .6*.2
Aa .2*.6 .2*.2 .2*.2
aa .2*.6 .2*.2 .2*.2
or:
AA Aa aa
AA .36 .12 .12
Aa .12 .04 .04
aa .12 .04 .04
You could work the rest of the problem with matrices like that, or you could lump like groups together. If you lump the like groups together, then you have:
1. AA, AA = .36
2. AA, Aa = .24
3. AA, aa = .24
4. Aa, Aa = .04
5. Aa, aa = .08
6. aa, aa = .04
So, for groups 1, 3 and 6, the answers are easy. Group 1 will deliver AA offspring, group 3 will deliver Aa offspring, and group 6 will produce aa offspring. But we still need to figure out the offspring mix of the other groups, and add it all up.
We can make another matrix to indicate how each of our 6 groups above would be distributed in terms of the offspring they produce. Hopefully you can understand how these percentages come about easily enough:
Offspring type:
Group: AA Aa aa
1 100% 0% 0%
2 50% 50% 0%
3 0% 100% 0%
4 25% 50% 25%
5 0% 50% 50%
6 0% 0% 100%
We can then multiply those percentages above by the percent of our initial population which falls into each of our 6 groups, and we have the following matrix:
Offspring type:
Group: AA Aa aa
1 0.36 0 0
2 0.12 0.12 0
3 0 0.24 0
4 0.01 0.02 0.01
5 0 0.04 0.04
6 0 0 0.04
We can add those columns up to find the distribution of offspring, and it is:
AA -- 0.49
Aa -- 0.42
aa -- 0.09
So there is the answer. You can also check it to be sure that the percentages of individual genes have not changed. In the parents, you had:
AA -- 0.6
Aa -- 0.2
aa -- 0.2
Or, for each individual you had an average of 1.4 copies of gene A, and 0.6 copies of gene a.
In the offspring, you have:
AA -- .49
Aa -- .42
aa -- .09
Which gives you 2 * 0.49 + 0.42 = 1.4 copies of gene A, and 2 * 0.09 + 0.42 = 0.6 copies of gene a.
Therefore the answer checks out.
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