The Hardy-Weinberg law states two things:
1. If no outside forces are acting on the population, then the allele frequency will remain constant from one generation to the next.
2. If no outside forces are acting on the population, then the genotype frequencies will be p^2, 2pq, and q^2.
The converses of these statements are not necessarily true; that is, the following statements are FALSE:
1. If the genotype frequencies are p^2, 2pq, and q^2, then no outside forces are acting on the population.
2. If the genotype frequencies remain constant from one generation to the next, then no outside forces are acting on the population.
The following two problems will illustrate the falsehood of these statements.
Genotype AA Aa aa
# of zygotes 1 .90 .81
a. Calculate the genotype frequencies in the population of zygotes (before selection).
b. What is the allele frequency in this population of zygotes? (No calculations allowed)
c. Calculate the genotype frequencies in the population of adults (after selection).
d. Calculate the allele frequency in this population of adults.
Now assume that you know nothing about the fitnesses or the allele frequencies, and that you have just sampled this population of adults. You find the following numbers:
AA Aa aa
14 25 11
e. Calculate the observed frequencies.
f. Calculate the expected genotype frequencies, assuming the Hardy-Weinberg conditions hold.
g. Compare e and f. Do the adult genotype frequencies fit in the Hardy-Weinberg equations?
h. Would you be able to detect natural selection if you sampled only the adults? What if you sampled both zygotes and adults? Explain.
2. Now consider another locus, with the following fitnesses. Let the initial q=.25, and assume random mating. Ignore mutation, migration, and genetic drift.
These problems are solved.
Genotype BB Bb bb
Fitness .90 1 .70
a. Calculate the allele frequency after one generation. After two generations.
b. Must natural selection cause the allele frequencies to change? Explain.
1. I think that you wrote # of zygotes, but actually this is the fitness value.
<br><br>Here I think its best to start with the adult population, and then work on the resulting zygotes.
<br><br>Since p=0.5, then q=0.5 (p+q=1). The fitness values are 1 for AA, 0.9 for Aa and 0.81 for aa. Since p and q = 0.5, you will have 1/4 AA, 1/2 Aa and 1/4 aa following the simple genetics rule. you multiply these frequencies by the fitness value, to see you much AA, Aa and aa you will have in the progeny (for example, for Aa, you have (1/2)*0.9). These would be your genotype frequencies (after selection). Fitness and simple genetic frequency act together to give you this answer. This will affect ...
This job determines fitness value.