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Provided by the Academic Center for Excellence 1 Hardy-Weinberg Equilibrium
Hardy-Weinberg Equilibrium
Hardy-Weinberg Equilibrium, also referred to as the Hardy-Weinberg principle, is used to compare allele frequencies in a given population over a period of time. A population of alleles must meet five rules in order to be considered โin equilibriumโ:
- No gene mutations may occur and therefore allele changes do not occur.
- There must be no migration of individuals either into or out of the population.
- Random mating must occur, meaning individuals mate by chance.
- No genetic drift, a chance change in allele frequency, may occur.
- No natural selection, a change in allele frequency due to environment, may occur.
Hardy-Weinberg Equilibrium never occurs in nature because there is always at least one rule being violated. Hardy-Weinberg Equilibrium is an ideal state that provides a baseline against which scientists measure gene evolution in a given population. The Hardy-Weinberg equations can be used for any population; the population does not need to be in equilibrium.
There are two equations necessary to solve a Hardy-Weinberg Equilibrium question:
๐๐ + ๐๐ = 1 ๐๐ยฒ + 2๐๐๐๐ + ๐๐ยฒ = 1
๐๐ is the frequency of the dominant allele. ๐๐ is the frequency of the recessive allele. ๐๐ยฒ is the frequency of individuals with the homozygous dominant genotype. 2 ๐๐๐๐ is the frequency of individuals with the heterozygous genotype. ๐๐ยฒ is the frequency of individuals with the homozygous recessive genotype.
Example 1a: A population of cats can be either black or white; the black allele (B) has complete dominance over the white allele (b). Given a population of 1,000 cats, 840 black and 160 white, determine the allele frequency, the frequency of individuals per genotype, and number of individuals per genotype.
To solve this problem, solve for all the preceding variables (๐๐, ๐๐, ๐๐ยฒ, 2๐๐๐๐, ๐๐๐๐๐๐ ๐๐ยฒ).
Step 1 : Find the frequency of white cats, the homozygous recessive genotype, as they have only one genotype, bb. Black cats can have either the genotype Bb or the genotype BB, and therefore, the frequency cannot be directly determined.
๐น๐น๐น๐น๐น๐น๐๐๐น๐น๐น๐น๐๐๐น๐น๐น๐น ๐๐๐๐ ๐๐๐๐๐๐๐๐๐๐๐๐๐๐๐น๐น๐๐๐๐๐๐ =
๐ผ๐ผ๐๐๐๐๐๐๐๐๐๐๐๐๐น๐น๐๐๐๐๐๐ ๐๐๐๐๐๐๐๐๐๐ ๐๐๐๐๐๐๐น๐น๐๐๐๐๐๐๐๐๐๐๐๐
Provided by the Academic Center for Excellence 2 Hardy-Weinberg Equilibrium
๐ผ๐ผ๐ผ๐ผ๐ผ๐ผ๐ผ๐ผ๐ผ๐ผ๐ผ๐ผ๐ผ๐ผ๐ผ๐ผ๐ผ๐ผ๐ผ๐ผ๐ผ๐ผ ๐๐๐๐๐๐๐ผ๐ผ๐ผ๐ผ ๐๐๐๐๐๐๐ผ๐ผ๐ผ๐ผ๐ผ๐ผ๐๐๐ผ๐ผ๐๐๐ผ๐ผ
160 1 , 000
Frequency of white cats = 0.16; therefore, ๐๐ยฒ = 0.
Step 2 : Find ๐๐ by taking the square root of ๐๐ยฒ. โ(๐๐ยฒ) = โ(0.16) ๐๐ = 0.
Step 3 : Use the first Hardy-Weinberg equation (๐๐ + ๐๐ = 1) to solve for ๐๐. ๐๐ + ๐๐ = 1 ๐๐ = 1 โ ๐๐ ๐๐ = 1 โ (0.4) ๐๐ = 0.
Now that the allele frequencies in the population are known, solve for the remaining frequency of individuals by using ๐๐ยฒ + 2๐๐๐๐ + ๐๐ยฒ = 1.
Step 4 : Square ๐๐ to find ๐๐ยฒ. ๐๐ = 0. ๐๐ยฒ = (0.6)ยฒ ๐๐ยฒ = 0.
Step 5 : Multiply 2 ร ๐๐ ร ๐๐ to get 2 ๐๐๐๐. 2 ๐๐๐๐ = 2(0.6)(0.4) 2 ๐๐๐๐ = 0. Therefore : The frequency of the dominant alleles: ๐๐ = 0. The frequency of the recessive alleles: ๐๐ = 0. The frequency of individuals with the dominant genotype: ๐๐ยฒ = 0. The frequency of individuals with the heterozygous genotype: 2 ๐๐๐๐ = 0. The frequency of individuals with the recessive genotype: ๐๐ยฒ = 0.
Remember: Frequencies can be checked by substituting the values above back into the Hardy-Weinberg equations. 0.6 + 0.4 = 1 0.36 + 0.48 + 0.16 = 1 Step 6 : Multiply the frequency of individuals (๐๐ยฒ, 2๐๐๐๐, ๐๐๐๐๐๐ ๐๐ยฒ) by the total population to get the number of individuals with that given genotype.
๐๐ยฒ ร ๐๐๐๐๐๐๐๐๐๐ ๐๐๐๐๐๐๐น๐น๐๐๐๐๐๐๐๐๐๐๐๐ = 0.36 ร 1,000 = 360 black cats, BB genotype. 2 ๐๐๐๐ ร ๐๐๐๐๐๐๐๐๐๐ ๐๐๐๐๐๐๐น๐น๐๐๐๐๐๐๐๐๐๐๐๐ = 0.48 ร 1,000 = 480 black cats, Bb genotype.
Provided by the Academic Center for Excellence 4 Hardy-Weinberg Equilibrium
Step 4 : Use the first Hardy-Weinberg equation (๐๐ + ๐๐ = 1) to solve for ๐๐. ๐๐ + ๐๐ = 1 ๐๐ = 1 โ ๐๐ ๐๐ = 1 โ (0.3) ๐๐ = 0. Now that the allele frequencies in the population are known, solve for the frequency of all individuals by using ๐๐ยฒ + 2๐๐๐๐ + ๐๐ยฒ = 1.
Step 5 : Square ๐๐ to find ๐๐ยฒ. ๐๐ = 0. ๐๐ยฒ = (0.7)ยฒ ๐๐ยฒ = 0.
Step 6 : Multiply 2 ร ๐๐ ร ๐๐ to get 2 ๐๐๐๐. 2 ๐๐๐๐ = 2 ร 0.7 ร 0. 2 ๐๐๐๐ = 0.
Therefore : The frequency of the dominant alleles: ๐๐ = 0. The frequency of the recessive alleles: ๐๐ = 0. The frequency of individuals with the dominant genotype: ๐๐ยฒ = 0. The frequency of individuals with the heterozygous genotype: 2 ๐๐๐๐ = 0. The frequency of individuals with the recessive genotype: ๐๐ยฒ = 0.
Example 2b: The next generation of finches has a population of 400. There are 336 with black beaks and 64 with yellow beaks. Is this population in Hardy-Weinberg Equilibrium?
Step 1 : Solve for ๐๐ยฒ.
๐ผ๐ผ๐ผ๐ผ๐ผ๐ผ๐ผ๐ผ๐ผ๐ผ๐ผ๐ผ๐ผ๐ผ๐ผ๐ผ๐ผ๐ผ๐ผ๐ผ๐ผ๐ผ ๐ค๐ค๐ผ๐ผ๐๐โ ๐๐โ๐๐ ๐
๐
๐๐๐
๐
๐๐๐ผ๐ผ๐ผ๐ผ๐ผ๐ผ๐ผ๐ผ๐๐ ๐บ๐บ๐๐๐ผ๐ผ๐๐๐๐๐บ๐บ๐๐๐๐
๐๐๐๐๐๐๐ผ๐ผ๐ผ๐ผ ๐๐๐๐๐๐๐ผ๐ผ๐ผ๐ผ๐ผ๐ผ๐๐๐ผ๐ผ๐๐๐ผ๐ผ =^
64 400 = 0.
Step 2 : Take the square root of ๐๐ยฒ to find ๐๐. ๐๐ยฒ = 0. โ(๐๐ยฒ) = โ(0.16) ๐๐ = 0.
Because the recessive allele frequency (๐๐) has changed, the population is NOT in a state of Hardy-Weinberg Equilibrium.
Provided by the Academic Center for Excellence 5 Hardy-Weinberg Equilibrium
Practice Problems
- Scale coloration of lizards has a complete dominance relationship where green scales are dominant over blue scales. There are 1,024 individuals with the genotype GG, 512 individuals with the genotype Gg, and 64 individuals with the genotype gg. Find: the frequency of the dominant and recessive alleles and the frequency of individuals with dominant, heterozygous, and recessive genotype.
- The next generation of lizards has 1092 individuals with green scales and 108 individuals with blue scales. Is the population in Hardy-Weinberg Equilibrium? Solve for p and q.
- Rabbitโs ears can be either short or floppy, where short ears are dominant over floppy ears. There are 653 individuals in a population. 104 rabbits have floppy ears and 549 have short ears. Find: the frequency of the dominant and recessive alleles and the frequency of individuals with dominant, heterozygous, and recessive genotypes.
- The next generation of rabbits has 560 individuals with short ears and 840 individuals with floppy ears. Is the population in Hardy-Weinberg Equilibrium? Solve for p and q.
- Petal coloration of pea plants has a complete dominance relationship where purple petals are dominant over white petals. There are 276 plants, 273 have purple petals. Find: the frequency of the dominant and recessive alleles and the frequency of individuals with the dominant, heterozygous, and recessive genotype.
- The next generation of pea plants has 552 plants, 546 have purple petals. Is the population in Hardy-Weinberg Equilibrium? Solve for p and q.
Provided by the Academic Center for Excellence 7 Hardy-Weinberg Equilibrium
McGraw-Hill. 306-307. Mader, Sylvia S. (2001). Biology (10th^ ed.). New York, NY: McGraw-Hill. 285-286. Sal. (Sep 30, 2009). Khan Academy. Hardy-Weinberg Principle. Retrieved from http://www.khanacademy.org/science/biology/v/hardy-weinberg-principle.
