PROBLEM 1

If 9% of an African population is born with a severe form of sickle-cell anemia (ss), what percentage of the population will be more resistant to malaria because they are heterozygous(Ss) for the sickle-cell gene?

f(ss)= 0.09

se a população está em HW, então

f(ss)=p2

p=SQR (0.09)=0.3; q=1-0.3=0.7

f(Ss)=2pq=2*0.3*0.7=0.42

PROBLEM 2

This is a classic data set on wing coloration in the scarlet tiger moth (Panaxia dominula). Coloration in this species had been previously shown to behave as a single-locus, two-allele system with incomplete dominance. Data for 1612 individuals are: White-spotted (AA) =1469 Intermediate (Aa) = 138 Little spotting (aa) =5. Calculate all genotipic and allelic frequencies.

f(AA) = 1469/1612 = 0.911

f(Aa) = 138/1612 = 0.086

f(aa) = 5/1612 = 0.003

f(A) = (1469 + 138/2)/1612 = 0.954

f(a) = (5 + 138/2)/1612 = 0.046

PROBLEM 3

After graduation, you and 19 friends build a raft, sail to a deserted island, and start a new population, totally isolated from the world. Two of your friends carry (that is, are heterozygous for) the recessive “a” allele, which in homozygotes causes cystic fibrosis. N=19+1=20; f(Aa) = 2/20=0.1

Assuming that the frequency of this allele does not change as the population grows, what will be the frequency of the cystic fibrosis allele on your island?

One needs to assume that if there is information on the carriers there would be information on any sick person as well. Therefore, at the arrival, f(aa)=0.

So f(AA)+f(aa) = 1-f(Aa) = 0.9, and because were are told that there are no changes as the population grows, then f(A)=√0.9=0.95. Therefore:

f(a) = 1- 0.95 = 0.05

Cystic fibrosis births on the island is how many times greater than the original mainland. The frequency of births on the mainland is 0.059%. f(aa)=0.052 = 0.0025; 0.059%=0.00059, 0.0025/0.00059=4.2

PROBLEM 4

For a human blood, there are two alleles (called S and s) and three distinct phenotypes that can be identified by means of the appropriate reagents. The following data was taken from people in Britain. Among the 1000 people sampled, the following genotype frequencies were observed SS = 99, Ss = 418 and ss = 483. Calculate the frequency of S and s in this population and carry out a X2 test. Is there any reason to reject the hypothesis of Hardy-Weinberg proportions in this population?

f(SS)=99/1000=0.099; f(Ss)=418/1000=0.418; f(ss)=483/1000=0.483

f(S)= [99+(418/2)]/1000= 0.308; f(s)= [483+(418/2)]/1000= 0.692

Observed Expected

SS 99 [0.3082*1000 = ] 94.864

Ss 418 [2*0.308*0.692*1000 = ] 426.272

ss 483 [0.6922*1000 = ] 478.864

Chi2= [(99 - 94.864)2/94.864] + [(418 - 426.272)2/426.272] + [(483 - 478.864)2/478.864] = 0.180 + 0.161 + 0.036

Chi2=0.377

Chi2t(0.05, 1)=3.84

Conclusion: The null hypothesis (the population is in HWE because there is no significant difference between the observed genotypic frequencies and the genotypic frequencies expected under HWE) cannot be rejected.

PROBLEM 5

A scientist has studied the amount of polymorphism in the alleles controlling the enzyme Lactate Dehydrogenase (LDH) in a species of minnow. From one population, 1000 individuals were sampled. The scientist found the following genotype frequencies: AA = 0.080, Aa = 0.280; aa = 0.640. From these data calculate the allele frequencies of the "A" and "a" alleles in this population. Use the appropriate statistical test to help you decide whether or not this population was in Hardy-Weinberg equilibrium.

f(AA) = 0.08; f(Aa) = 0.28; f(aa) = 0.64

f(A)=0.08+0.28/2= 0.22; f(a)=0.64+0.28/2=0.78

Observed Expected

SS 0.08*1000=80 [0.222*1000 = ] 48.4

Ss 0.28*1000=280 [2*0.22*0.78 *1000 = ] 343.2

ss 0.64*1000=640 [0782*1000 = ] 608.4

��2= �� [(80 - 48.4)2/48.4] + [(280 - 343.2)2/343.2] + [(640 - 608.4)2/608.4] = 20.631 + 11.638 + 1.641

��2=33.911

��2t(0.05, 1)=3.84