1. In a large laboratory population of Drosophila melanogaster, it was observed that the frequency of allele T was 0.6 and the frequency of allele t was 0.4. Calculate the expected Hardy-Weinberg equilibrium genotype frequencies.

The genotypic frequencies expected given Hardy-Weinberg equilibrium may be easily computed using the Hardy-Weinberg formula: p² + 2pq + q² = 1, where p² is the expected frequency of the TT genotype, 2pq is the expected frequency of the heterozygote Tt, and q² is the expected frequency of the tt genotype.

Expected: f(TT) = 0.6² = 0.36 f(Tt) = 2(0.6)(0.4) = 0.48 f(tt) = 0.4² = 0.16

Note: As a check of your arithmetic it is useful to sum the genotypic frequencies for all genotypes. The sum should always equal 1.

2. In a sample of 1000 frogs (Rana leopardus), these three genotypes occurred in the following numbers. Calculate the genotypic and allele frequencies.

Genotype	AA	Aa	aa
Number	490	420	90

The genotypic frequencies are simply the numbers of each genotype divided by the total number of individuals, or

f(AA)=490/1000=.49 f(Aa)=420/1000=.42 f(aa)=90/1000=.09

The correct way to calculate the allele frequencies is to divide the number of each allele present by the total number of all alleles, i.e.

f(AA)= [(2 x #Homozygotes, since each has two A alleles)+(#Heterozygotes)] divided by the (Total number of alleles (2 per individual)) ]

= [(2 x (490)) + 420]/2000 = p = 0.7

Alternatively, one may halve the number of heterozygotes rather than double the number of homozygotes and the denominator. Thus,

p = .49 + (.42/2) = p = 0.7. This is also perfectly valid.

q = 1 - p = 1 - .07 = 0.3. It is useful, given time, to calculate q as shown for p above to double-check your arithmetic. Remember, the sum of the allele frequencies must always equal 1!

3. In a sample of 100 Hawaiians, it was found that 35, 40, and 25 individuals had blood groups MM, MN, and NN, respectively. Calculate the allele frequencies and the expected Hardy-Weinberg genotypic frequencies. Is this population in Hardy-Weinberg equilibrium?

The calculation of allele frequencies is as with #2 above, i.e.

p = f(M) = .35 + (.40/2) = .55 q = f(N) = .25 + (.40/2) = .45

As a check, .55 + .45 = 1, as it should.

The genotypic frequencies expected if population is in H-W equilibrium:

Expected:

f(MM) = (.55)² = .3025

f(MN) = 2(.55)(.45) = .4950

f(NN) = (.45)² = .2025

For a population of 100 individuals, this corresponds to MM=30, MN = 50, and NN = 20.

Because the actual genotypic frequency in the population (.35, .40, and .25) is not equal to that expected (.30, .50, and .20) under Hardy-Weinberg for the given allele frequencies (.55, .45), and for no other reason, we determine that the population is not in Hardy-Weinberg equilibrium.

VERY IMPORTANT NOTE!!!!!

Many people miss these sorts of problems for the following reason: Given a frequency of MM homozygotes of 0.35, they computed p as the square root of 0.35. THIS IS NOT VALID!!!!!!! Why? Because the frequency of homozygotes equals p² and q² ONLY WHEN THE POPULATION IS IN HARDY-WEINBERG EQUILIBRIUM!!! Because you do not know that this is the case (indeed, you are asked to determine if this is true), you must determine the allele frequency the old fashioned way, you learn it, i.e. you compute it as above by adding the proportion of the homozygote to one-half the proportion of heterozygotes. Only use the short cut of the square root of the homozygote proportion when you know the population is in Hardy-Weinberg equilibrium.

4. In purebred Holstein cattle, about 1 calf in 100 is spotted red rather than black; red being due to a recessive gene. What is the proportion of red genes in the population? What are the frequencies of homozygous black and of heterozygous black individuals in this population? Assume that the population is in Hardy-Weinberg equilibrium!

In this case, because we can not distinguish black homozygotes from black heterozygotes, we must assume that the population is in Hardy-Weinberg to calculate p and q. Making this assumption, we can calculate the proportion of the recessive allele as the square root of the proportion of homozygous recessives (rr). Thus, q=square root of .01=.1. p= 1-q = .9. Again, because we are assuming H-W, we calculate expected genotypic frequencies as: f(RR) = p² = (.9)² = .81; f(Rr) = 2(.9)(.1) = .18.

Again, to check, .01 + .18 + .81 = 1.