1. 9% of a human population are suffering from a genetic disease which is triggered by a recessive allele. 42% are healthy, but they are conductors. Might this be an ideal population? Why (not)?
The distributions of a two-allele gene in an ideal, diploid population fulfills the equations p + q = 1 and p^2 + 2 * p * q + q^2 = 1.
p^2 = 0.09
=> p = 0.3
p + q = 1
=> q = 1 - 0.3 = 0.7
2 * p * q = 0.42
=> q = 0.21 / 0.3 = 0.7 Both equations are fulfilled. Therefore it's possible that this is an ideal population.
2. Assume that a tiny human population of only 5 persons has the same distributions of genes as an ideal population. There are two allels of the gene in question, with the frequencies p and q. 48% of the population are heterozygotous. Calculate p and q, where p is the frequency of the more widely spread allele.
Since p and q can only be multiples of 0.2, there are only two possibilities: either p = 0.8 and q = 0.2 or p = 0.6 and q = 0.4. In the first case 2 * p * q = 2 * 4 / 25 = 32%, in the second case 2 * p * q = 2 * 6 / 25 = 48%. Therefore the second possibility is the correct solution.
3. 36% of an ideal, human population are suffering from a genetic disease which is triggered by a dominant allele. How many percent of the sick ones will definitely pass the illness on to the next generation?
A be the frequency of the dominant allele, a be the frequency of the recessive allele. Then the following calculations apply:
a^2 = 1 - 0.36
=> a = 0.8
A + a = 1
=> A = 1 - 0.8 = 0.2
=> A^2 = 0.04
0.04 / 0.36 = 1 / 9 = about 11.12%
4. There are three alleles of a gene, with the frequencies of p, q and r. In an ideal, human population a fifth of the people carry the allele p. The genotype qr has a frequency of 24%. Calculate q and r, where r is the frequency of the more widely spread allele.
p + q + r = 1
=> q = 1 - p - r = 1 - 0.2 - r = 0.8 - r
2 * q * r = 0.24
=> q = 0.12 / r
0.12 / r = 0.8 - r
=> r^2 - 0.8 * r + 0.12 = 0
=> r1 = (0,8 + sqrt(0.64 - 0.48)) / 2 = 0.6;
r2 = (0.8 - sqrt(0.64 - 0.48)) / 2 = 0.2
=> r = r1 = 0.6;
q = r2 = 0.2
5. Derive the Hardy-Weinberg law for a gene with two alleles and an ideal, triploid population.
1 = (p + q)^3 = p^3 + 3 * p^2 * q + 3 * p * q^2 + q^3
The correct solutions have been found by Boiled Brain and Tor G. J. Myklebust. Congratulations!