 Back in the early 1900s, a couple of scientists named G. H. Hardy and Wilhelm Weinberg started thinking that Darwin's theory of natural selection, survival of the fittest, and Mendel's ideas about inheritance of genes could be combined to help us understand how populations evolve. They were interested in understanding, one, what is the frequency of specific alleles in a population, so what fraction of the total number of alleles are dominant, capital B, or recessive, lowercase b? And two, as the organisms mate and have offsprings over many generations, whether the frequency of those alleles changes over time. So if you start out with 60% of the alleles being capital B and 40% being lowercase b? After the people mate, are there still 60% capital B and 40% lowercase b? Notice also that the allele frequencies 0.6 and 0.4 add up to 1. This is a key aspect of learning how to solve Hardy-Wienberg equilibrium problems. So keep it in mind for later. Hardy and Weinberg postulated that if the allele frequencies did not change over time, then the population was in equilibrium. You would say that such a population is in Hardy-Wienberg equilibrium. In contrast, if the allele frequencies do change over time, the population is not in Hardy-Wienberg equilibrium. For a population to be in Hardy-Wienberg equilibrium, so for the allele frequencies to not change over time, there are five criteria that must be met. These are, number one, the population must be large. Number two, no mutations occur. Number three, no migration into and out of the population, also called gene flow. Number four, mating must be random. And number five, no natural selection can occur. If these five criteria are met, the population is in Hardy-Wienberg equilibrium. Evolution is not occurring at that locus, and the allele frequencies will remain the same over time. If you calculate allele frequencies and find that they have changed over time, you can say that the population is not in Hardy-Wienberg equilibrium, meaning that it has evolved. Let's go through two related examples of Hardy-Wienberg equilibrium problems and learn both how to calculate allele and genotype frequencies as well as to start thinking about why a population may or may not be evolving. You are studying the coat color locus for a population of 100 squirrels living in a forest along the coast of California. The dominant allele, B, gives black fur, while lowercase B, the recessive allele, gives white fur and is recessive. In population genetics, we call the frequency of the dominant allele P and the frequency of the recessive allele Q. The sum of these allele frequencies, all the alleles in the population, must equal 1. Now let's relate this to the genotypes, which is the probability of having an individual that is homozygous dominant, that is having one allele to be B, and the other also of B is P times P equals P squared. The probability of being heterozygous, having one capital B and one lowercase B, is 2 times P times Q, and the probability of being homozygous recessive is Q times Q equals Q squared. The sum of the genotype frequencies must also equal 1, which is this equation, P squared plus 2PQ plus Q squared equals 1. Now back to your squirrels. You find that 20 of the 100 squirrels have white fur, which corresponds to 0.2. White fur is homozygous recessive. So we know that these squirrels have the same genotype, little b, little b. Since we know that Q squared is the genotype frequency of homozygous recessive, little b, little b, we know that 0.2 equals Q squared. We can solve for Q by taking the square root of 0.2, which gives us Q equals 0.45. Since P plus Q equals 1, we can easily solve for P, which is 0.55. These squirrels breed and in a second generation, you go back to the forest and want to determine if the population is in Hardy-Weinberg equilibrium. You count and see that 40 of the 200 squirrels in this generation have white fur. This means that the frequency of the recessive genotype, Q squared, is 0.2. We can now easily solve for Q, the recessive allele frequency, so Q equals 0.45. We can easily solve for P, which is 0.55. Based on these equations, let's see if our population is in Hardy-Weinberg equilibrium. We started with P equals 0.55 and Q equals 0.45 in the first generation, and we calculated that P equals 0.55 and Q equals 0.45 in the second generation. So the allele frequencies did not change over time. This means we can accept the Hardy-Weinberg equilibrium null hypothesis and say that yes, this population is in Hardy-Weinberg equilibrium and has not undergone evolution. Example number two. You are also studying a population of squirrels that live on a nearby beach. In the first generation, they had P equals 0.55 and Q equals 0.45. In the second generation, you find that the frequency of the dominant allele P is 0.7 and the frequency of the recessive allele Q is 0.3. This population of squirrels on the beach is not in Hardy-Weinberg equilibrium, since the allele frequencies did change over generations. So we reject the null hypothesis and can say that this population evolved. Now what factors caused this population to undergo evolution? To answer this, think about the five criteria for Hardy-Weinberg equilibrium, and let's come up with some ideas as to which of these criteria were not met. Number three. No migration into and out of the population. Maybe the water level went down, so some white squirrels living on the coast could come from shore and start living on the island. Number five. No natural selection can occur. Maybe the white squirrels stood out more against the sand than the black squirrels, so they were more noticeable to predators and got eaten. Now let's do a third example to see how we can use the Hardy-Weinberg equilibrium idea to solve population genetic problems. You can use the Hardy-Weinberg equilibrium to solve population genetic problems if you assume that the population is in Hardy-Weinberg equilibrium. Let's say that you are now studying eye color in a population of squirrels. You have either brown eyes or blue eyes, and you know that blue is recessive. You look at all of your squirrels and find that 4% of the population has blue eyes. How can you figure out A, the frequency of the brown eye allele, and B, what percentage of the population is heterozygous for this trait? Assuming the population follows Hardy-Weinberg equilibrium, let's solve A and B. To solve A, let's start with the information that 4% of the population has blue eyes. This is telling you the frequency of the recessive genotype, blue eyes. So we know that 0.04 equals Q squared. We can easily solve for Q, the frequency of the recessive allele, which is 0.2. Therefore, given that P plus Q equals 1, we know that P equals 0.8. This is the allele frequency of brown eyes. To solve B, let's use the equation P squared plus 2PQ plus Q squared equals 1. The frequency of the heterozygous genotype is 2PQ. So 2 times 0.8 times 0.2 equals 0.32. To get the percentage, we can multiply by 100. So 32% of the population of squirrels are heterozygous at the eye-color locus. Overall, Hardy-Weinberg equilibrium is a way to apply Mendelian genetics, the idea that traits like coat and eye-color in squirrels get inherited to large populations and to figure out whether that trait is undergoing evolution. Through this lesson, you have learned how to calculate genotype and allele frequencies over several generations and use these calculations to determine if a population is or is not in Hardy-Weinberg equilibrium. If the population is in Hardy-Weinberg equilibrium, you can use Hardy-Weinberg equilibrium to solve population genetic problems. On the other hand, if the population is not in Hardy-Weinberg equilibrium, we can start thinking about what selective pressures might be in place that are causing evolution and anticipate how these selective pressures may shape future generations. This video has been provided to you by Eureka Science and Eye Biology, bringing the world's best biology to you.