The Hardy-Weinberg-

Castle PrincipleHenry B. Sergio Jr.

The Hardy-Weinberg-Castle Principle

-states that under the following conditions both phenotypic and allelic frequencies remain constant from generation to generation in sexually reproducing populations. The law predicts how gene frequencies will be transmitted from generation to generation given a specific set of assumptions.

Conditions

1. large population size2. no mutation3. no immigration or emigration4. random mating5. random reproductive success

But….

Natural populations rarely meet all of these conditions. Large populations rarely occur in isolation, all populations experience some degree of random mutation, mating is seldom random, but rather is the result of careful selection of mates. Most importantly, selective pressures favor individuals whose alleles give them the greatest fitness, so survival and reproductive success are never random. Because of these factors inherent in natural selection, allelic frequencies do not remain constant and evolution occurs.

So why study the Hardy-Weinberg Law?

1. The law proves that natural selection is necessary for evolution to occur. Darwin's theory of evolution states that for evolution to occur, populations must be variable, there must be inheritance between generations, and natural selection must make survival and reproductive success non-random. The conditions set up by the Hardy-Weinberg Law allow for variability (the existence of different alleles) and inheritance, but they eliminate natural selection. The fact that no evolution occurs in a population meeting these conditions proves that evolution can only occur through natural selection.

So why study the Hardy-Weinberg-Castle Law?

2. The Hardy-Weinberg Law allows us to estimate the effect of selection pressures by measuring the difference between actual and expected allelic frequencies or phenotypes. In order to make such measurements, we must first write a Hardy-Weinberg equation for the frequencies of alleles in the population.

Equation

Phenotype Frequency: p+q = 1Genotype Frequency: p2 +2pq + q2 = 1

Equation

Sample Problem 1Albinism is a rare genetically inherited trait that is only expressed in the phenotype of homozygous recessive individuals (aa). The most characteristic symptom is a marked deficiency in the skin and hair pigment melanin. This condition can occur among any human group as well as among other animal species. The average human frequency of albinism in North America is only about 1 in 20,000.

Answer: Sample Problem 1

Referring back to the Hardy-Weinberg equation (p² + 2pq + q² = 1), the frequency of homozygous recessive individuals (aa) in a population is q². Therefore, in North America the following must be true for albinism:

q² = 1/20,000 = .00005

Answer: Sample Problem 1By taking the square root of both sides of this equation, we get: (Note: the numbers in this example are rounded off for simplification.)

q = .007

In other words, the frequency of the recessive albinism allele (a) is .00707 or about 1 in 140. Knowing one of the two variables (q) in the Hardy-Weinberg equation, it is easy to solve for the other (p).

p = 1 - qp = 1 - .007

p = .993

Answer: Sample Problem 1

The frequency of the dominant, normal allele (A) is, therefore, .99293 or about 99 in 100.

The next step is to plug the frequencies of p and q into the Hardy-Weinberg equation:

p² + 2pq + q² = 1(.993)² + 2 (.993)(.007) + (.007)² = 1

.986 + .014 + .00005 = 1

Answer: Sample Problem 1

p² = predicted frequencyof homozygousdominant individuals

= .986 = 98.6%

2pq = predicted frequencyof heterozygousindividuals

= .014 = 1.4%

q² = predicted frequencyof homozygousrecessive individuals (the albinos)

= .00005 = .005%

This gives us the frequencies for each of the three genotypes for this trait in the population:

Answer: Sample Problem 1

With a frequency of .005% (about 1 in 20,000), albinos are extremely rare. However, heterozygous carriers for this trait, with a predicted frequency of 1.4% (about 1 in 72), are far more common than most people imagine. There are roughly 278 times more carriers than albinos. Clearly, though, the vast majority of humans (98.6%) probably are homozygous dominant and do not have the albinism allele.

Seatwork 1

In a population of giraffes, the gene that determines neck length has two alleles, N for longer necks and n for shorter necks. Only giraffes with two N alleles will have long necks. In one generation, the frequency of N is p=0.4 and the frequency of n is q=0.6. This generation experienced a large forest fire that kills all low-growing plants, leaving tall trees with high branches as the only food source. In the next generation, 49% of giraffes have long necks. How much has selection contributed to the allele frequencies in the second generation?

Seatwork 2

In a population of giraffes, the gene that determines spot size has two alleles, S for larger spots and s for smaller spots. Giraffes with one of each allele have medium-sized spots. In one generation, the frequency of S is p=0.4 and the frequency of s is q=0.6. Since spots help the giraffes blend in with their surroundings, they are acted upon by natural selection. In the next generation, 64% of giraffes have small spots. What percentage of giraffes in the second generation will have medium spots? Large spots?

Ans: 1

The Hardy-Weinberg equation for a two-allele system is (p + q)^2 = 1 or p^2 +2pq +q^2 = 1. The frequency of giraffes with two N alleles is determined by p^2. In the first generation, this is (0.4)^2 = .16 or 16%. Since 49% of giraffes in the second generation have two N alleles, selection has increased the frequency of N in the population. Working backwards we find that 49% = 0.49 = p^2, so p = 0.7 in the second generation. Selection has caused the frequency of N to change from 0.4 to 0.7 in one generation.

Ans. 2

The Hardy-Weinberg equation for a two-allele system is (p + q)^2 = 1 or p^2 +2pq +q^2 = 1. The frequency of giraffes with twos alleles is determined by q^2. From the problem, we see that q^2 is 64% or 0.64. From this we find that the frequency of s in the second generation is q=0.8. Using the original equation, we plug in this value of q and solve for p, which we find is 0.2. The frequency of giraffes with large spots is p^2 = 0.04 or 4%. The frequency of giraffes with medium spots is 2pq = 0.32 or 32%.