1.  Three genotypes were observed at the Pgm-1 locus in a human population.  In a sample of 1110 individuals, the three genotypes occurred as follows:
 

Genotypes	1/1	1/2	2/2
Numbers	634	391	85

Calculate the gene frequencies and genotype frequencies.
 

2. Human serum haptoglobins are determined by three alleles at a single locus.  In a sample of 285 Egyptians, the six genotypes occurred as follows:
 

Genotypes	1/1	1/2	2/2	1/3	2/3	3/3
Numbers	9	135	75	2	39	25

Calculate gene frequencies and genotype frequencies.
 

3. Calculate the expected frequencies for all genotypes in problems 1 and 2 assuming Hardy-Weinberg equilibrium.  Use the chi-square test to determine whether observed and expected numbers of individuals are significantly different in either case.
 

4. The following table gives the number of individuals for each of the M-N blood groups in samples from various human populations.  Test whether each population appears to be in Hardy-Weinberg equilibrium.
 

Population	M	MN	N
Eskimos	475	89	5
Russians	195	215	79
Belgians	896	1559	645
Papuans	14	48	138

5. Several chromosomal arrangements, differing by a series of overlapping inversions, are known in the third chromosome of Drosophila pseudoobscura.  Four arrangements (ST=standard, AR=arrowhead, CH=Chiricahua, TL=tree line) were found in three natural populations.  The observed numbers of each genotype were as follows:
 

Locality	St/Ar	St/Ch	St/Tl	Ar/Ch	Ar/Tl	Ch/Tl	St/St	Ar/Ar	Ch/Ch
Keen Camp	53	66	3	48	3	6	30	11	44
Pinyon Flat	40	53	5	37	3	7	31	11	21
Andres Canyon	87	47	12	20	4	2	89	18	4

For each population, calculate the frequency of each of the four chromosomal arrangements and
the expected frequency of heterozygotes (combined).
 

6.  An experimental population of Drosophila is started with 100 bw/bw females and 100 bw⁺/bw⁺ males.  What will be the genotype frequencies in the F1 and at Hardy-Weinberg equilibrium?
 

7.  The frequency of red-green color blindness in the men in a certain population is 0.08.  This form of color blindness is caused by a sex-linked recessive allele.  What are the expected frequencies of the three genotypes among women?
 

8. The most common form of hemophilia is due to a sex-linked recessive allele with a frequency of 0.0001.  What are the expected frequencies of the two male genotypes and the three female genotypes in the population?
 

9.  Tay-Sachs disease is caused by an autosomal recessive allele.  The disease is characterized by mental deficiency and blindness, with death occurring by four years of age.  The incidence of the disease among newborns is about 10 per million births.  Assuming Hardy-Weinberg equilibrium, estimate the frequency of the allele and of the heterozygotes.
 

10.  Acatalasia is a recessive condition first discovered in Japan.  Heterozygotes can be identified by the intermediate level of catalase in their blood.  The frequency of heterozygotes is 0.09% in Hiroshima, but 1.4% in the rest of Japan.  Assuming Hardy-Weinberg equilibrium, calculate the allelic frequencies in both Hiroshima and the rest of Japan. 

11.  A populations begins with the following genotypic composition at a sex-linked locus:

Answers:

1) X₁₁= 0.571,  Y₁₂= 0.352,  Z₂₂= 0.077,  p₁= 0.747,  q₂ = 0.253
2) U₁₁= 0.032,  V₁₂= 0.474,  W₂₂= 0.263,  X₁₃= 0.007,  Y₂₃= 0.137,  Z₃₃= 0.088,  p₁= 0.272,  q₂= 0.568,  r₃= 0.160
3) For problem (1), chi square = 4.99.   For problem (2), chi square = 102.3.  Yes, significantly different in both cases.
4) All populations are in Hardy-Weinberg equilibrium except the Papuans.
5)
 

Locality	St	Ar	Ch	Tl	He
Keen	0.345	0.239	0.394	0.023	0.670
Pinyon	0.385	0.245	0.334	0.036	0.679
Andres	0.572	0.260	0.136	0.031	0.584

6) F₁: X_bb= 0.0, Y_b+= 1.00, Z₊₊= 0.0.   Hardy-Weinberg: X_bb= 0.25, Y_b+= 0.50, Z₊₊= 0.25
7) X₊₊= 0.846,  Y_+b= 0.147,  Z_bb= 0.006
8) Men: p = 0.9999, q = 0.0001.   Women: X₊₊= 0.9998, Y_+h= 0.0002, Z_hh= 1 x 10^-8
9) q_dd= 0.003, Y_+d= 0.006
10) Hiroshima q_aa = 0.0004502,  Japan q_aa= 0.007  (both approximate) 
11) Males: X_A = 0.60, Y_a = 0.40.   Females: X_AA = 0.36, Y_Aa = 0.48,  Z_aa = 0.16