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### Course: AP®︎/College Biology>Unit 7

Lesson 4: Hardy-Weinberg equilibrium

# Hardy-Weinberg equation

The Hardy-Weinberg equilibrium is a principle that helps to predict allele frequencies in a population. It assumes no selection, no mutation, no geneflow, random mating, and large populations for stable allele frequencies. The equation p² + 2pq + q² = 1 calculates probabilities of homozygous dominant, heterozygous, and homozygous recessive genotypes. Created by Sal Khan.

## Want to join the conversation?

• why do you need to square the equation? (p + q = 1)
• It is p2+ 2pq+q2 because you are talking about the frequency of alleles and we are diploids i.e. we have 2 alleles for each trait, one allele we receive from our father and another from our mother. Thus,
p2= dominant allele i.e when we have both 'p' from parents
q2= recessive allele i.e when we have bothe 'q' from parents
and 2pq= heterozygote i.e. when we have say'p' from one parent and 'q' from another parent.
Thus in order to understand the equation p+q=1 in terms of diploid organisms we need to square the contents.
• Plus, shouldn't it be equilibrium. The powers balance the alleles...I do understand it IS an equation. Its just nomenclature right?
• The Hardy Weinberg equation describes a hypothetical "ideal" population in perfect equilibrium. It can't truly exist in nature, simply because there's always some force acting on a population. It's used as a reference point.
• Why can't you use the Hardy-Weinberg Principle for a small population?
• Its due to random sampling's effect w/ small population sizes. Imagine the percentages of resulting "heads" on coin flips between 2 people flipping coins vs 2000 people.
• If you choose your population wisely, and the observed genotype frequencies do not match the expected, what must be the case (according to H-W)?
• If a population's frequencies do not match those expected from hardy-weinberg, then the population is not in Hardy-Weinberg. Perhaps there is natural selection or non-random mating.

The Hardy-Weinberg population is a scientific ideal, though, it doesn't actually exist. Almost every single population on earth will differe from H-W expectied values.
• In the example given in the video, I am not sure why p^2 represents the probability of an individual being homozygous dominant because it looks like from the example, the two genotypes in question are Bb or bb. so BB would't even be possible would it? if the probability of getting just the B allele is 1/4, then p^2 should equal 1/16 or the probability of a homozygous dominant individual... but again is a BB individual even possible in the example of the two individuals given.

Hope my question was clear! thanks in advance!
• The example numerically is not related to the equation as the Hardy Weinberg Equation requires that a population be large and in the previous example, the population is very small, so the requirements for the Hardy Weinberg Principle are not met. Sal is simply using the alleles from the previous example to demonstrate the Hardy Weinberg Equation, but not the numbers. If it helps, just think of BB, Bb, and bb, and ignore everything else. But if the population did met the Hardy Weinberg Equation theoretically, the 3 Genotypes possible would be Bb, BB , and bb.
• Where did he get the 'q' from?
• P and Q are just letters that are used to show a specific allele but in a generic form. They are used to denote the 2 different alleles used in the problem.
• what does mean for the equation
given by hardy-wein berg
• It is a conceptual idea of population equilibrium that was developed by 2 scientists G.H. Hardy and William Weinberg, who suggested some assumptions for stable, non evolving population in which "allele frequencies do not change and therefore evolution does not occur". theses assumptions are :
1. No mutation
2.No small population
3.No sexual selection
4. No gene flow
5. No natural selection
In order to express Hardy Weinberg principle mathematically , suppose "p" represents the frequency of the dominant allele in gene pool and "q" represents the frequency of recessive allele. p+q=1 since the sum of both frequencies is 100% . In gene pool that include allele p and q the possible genotypes are
.................. Females
.................. (p) (q)
Males (p) 25% (pp) 25%(pq)
..... (q) 25%(qp) 25%(qq)
The total of all genotypes should be equal to 1 so
p^2+2pq+q^2=1
where
p^2=freq. of the HOMOZYGOUS DOMINANT GENOTYPE
2pq=freq. of HETEROZYGOUS GENOTYPE
q^2=freq. of the HOMOZYGOUS RECESSIVE GENOTYPE
p^2+2pq= freq. of the DOMINANT PHENOTYPE
q^2=freq. of the RECESSIVE PHENOTYPE
• In a Hardy Weinberg question, if they give you the # of Homozygous dominant, # of heterozygous and the # of homozygous recessive. You can calculate the p and q by using the total number of alleles of p or q divided by the total number of alleles in the population or finding q^2 to find q. however in certain questions you do not get the same p and q when you you do it both ways. Which is the best way to solve for p and q if they give u all of the individuals and their genotype? Also why does q from q^2 not equal to q when you solve for q using total number of alleles? For ex) 210 homo dom, 245 hetero and 45 homo rec. q from q^2 is 0.3. however (245 + 90)/1000= 0.67. could you explain the difference? please.
• I suggest reading over the assumptions behind the Hardy-Weinberg equation — do those have to be true?

If the two ways of calculating don't give the same answer, what can you conclude?

Do you think it could be evidence for something? If so, what?

Do these hints help?
• At 0.37, what is a heterozygote?