The probability rule we used in another module to predict genotype frequencies in the offspring generation of a specific population:
Rule:
can be use to generate a general formula for doing the same thing. That is, we can create a formula that describes how frequently particular genotypes will appear in the offspring generation when a population is not subject to an agent of evolution. This formula is known as the Hardy-Weinberg equation.
To do this, we will first generate the individual elements of the Hardy-Weinberg equation.
As the boxed rule above says, offspring genotype frequencies are calculated using parental allele frequencies. So imagine a population that has only two alleles for a given locus, A and a, and that in this population
- the A allele occurs with a frequency equal to p
- the a allele occurs with a frequency equal to q
To make sure you understand these phrases, substitute a number for p or for q. For example, p might be 0.4 meaning that the A allele occurs in 40% of the population's loci for this gene.
Notice that, because only two alleles exist in the population for this locus, the frequencies of the A and a alleles, p and q respectively, must sum to 1, the equivalent of 100%. Or
- p + q = 1
Also notice that, if we know the frequency of only one of the two alleles in a population, we can use simple algebra to work out the frequency of the second. For example, if we know p, the frequency of the A allele, then
- q = 1 - p
And, of course, if we know q, the frequency of the a allele, then
- p = 1 - q
To confirm your understanding of this relationship between the frequencies with which two alleles occur in a population when only two alleles exist for a given locus, answer the following questions.
Exercise 1
Please explain in your own words why p + q must always equal 1 when only two alleles exist in a population for a given locus.
Solution
An example answer: I imagine a locus as slots for alleles. In diploid organisms, each individual will have two 'slots' or two loci for the particular gene of interest, one per chromosome. If only two alleles exist to fill every slot in this population, the slots that are not filled with one of those two alleles must be filled with the second. Consequently, if 50% of the slots are filled with A alleles, then the remaining 50% must be filled with a alleles to account for 100% of the population's loci. Because 50% is equivalent to a frequency of 0.5, then the frequency of the A allele or p equals 0.5 as does the frequency of the a allele or q so that p + q = 1.
Exercise 2
Let’s consider a real example of this. In 2005, Stefasson et al. reported the fascinating discovery of an allele in humans whose presence is associated with increased fertility in Icelandic and European populations. Females with at least one copy of the allele have approximately 3.5%, and males 2.9%, more children on average than non-carriers. The exact mechanism by which the allele, known as H2, affects fertility is unknown.
If we know that 21% of European loci for this gene house the H2 allele, then how frequently must the single alternative allele, H1, for that locus occur in this population? Why?
Solution
If 21% of the loci (equivalent to a frequency of 0.21) in a population contain the H2 allele and H1 is the only other possible allele for this locus, then 79% of the remaining loci (equivalent to a frequency of 0.79) must have this allele. No other alleles exist for this locus consequently, if H2 does not occur at a locus then H1 must be there instead.
Exercise 3
A colleague determines that the B1 and B2 alleles of the B locus both occur with a frequency equal 0.45. Surprised, she redoes her work and confirms her results.
a. What could be the cause of your colleague's surprise? Please explain.
b. Because your colleague confirms her results she now needs to explain them. She turns to you for assistance. What do you suggest? Please be sure to explain how your explanation accounts for her observations.
Solution
Your colleague was probably surprised because she thought that B1 and B2 were the only two alleles that occurred at this locus in this population. Consequently, the discovery that their frequencies, p and q, summed to 0.9 as opposed to 1 was startling. Your suggestion to look for at least one additional allele to account for the 10% of the alleles unaccounted for in her study is well taken. She realizes that the existence of one or more additional alleles would explain the missing 10% and enable her to bring the summed allele frequencies for the B locus to 1.
Now that we have designated p to represent the freqeuncy of A allele and q, the a allele, we are ready to move forward with our efforts to construct the elements of the Hardy-Weinberg equation. Imagine that every individual in a population, in which both copies of both the A and a allele occur, is equally likely to survive and to reproduce.
What possible genotypes could occur in the offspring of this population?
Exercise 4
To answer this question, determine all the possible genotypes that could be formed from a population of individuals whose loci collectively warehouse numerous copies of A and a alleles. Remember that, because these individuals are all equally likely to reproduce, all combinations of these two alleles have the potential to form. Visit this module if you have questions.
Solution
There are four possible genotypes:
- AA
- aa
- Aa
- aA
Now that we know what genotypes could form, we can use the rule highlighted at the very beginning of this module to predict how frequently each of these genotypes will appear in the offspring generation.
Exercise 5
What are these frequencies? Apply the highlighted (boxed) rule above to complete the phrases below using the symbols p and q.
If all individuals are equally likely to survive and to reproduce, then the
- frequency of genotype AA will equal ____________________
- frequency of genotype aa will equal _____________________
- frequency of genotype Aa will equal ____________________
- frequency of genotype aA will equal ____________________
Solution
Because the Aa and aA genotypes are genetically equivalent, we can summarize the relationships you articulated above as
- frequency of genotype AA will equal p x p = p2
- frequency of genotype aaaa will equal q x q = q2
- frequency of genotype Aa will equal 2 x (p x q) = 2pq
And there you have it, the three fundamental elements of the Hardy-Weinberg equation that describe how frequently the three possible genotypes will appear in the offspring generation of a population that is not subject to an agent of evolution! Remember that only three genotypes are possible because we are only working with a gene for which only two alleles exist in a population.
To test your understanding of these relationships, answer the following questions.
Exercise 6
Please explain in your own words what these three formulae tell us about the relationship between allele frequencies in the population and genotype frequencies in the offspring generation when all individuals are equally likely to survive and to reproduce.
Solution
In plain English, these three relationships tell us that
1. If we want to know the frequency of the homozygous genotype (AA or aa) in the offspring of a population in which all individuals are equally likely to survive and reproduce, then we simply square the frequency with which the appropriate allele (A or a) occurs population.
2. If we want to know the frequency of the heterozygous genotype (Aa) in the offspring of a population in which all individuals are equally likely to survive and reproduce, then we multiple the frequency with which each allele (A and a) occurs in the population and multiply this result by 2.
Exercise 7
Return to the scenario described in problem 2. How frequently do you expect to the H1H1, H1H2, and H2H2 genotypes to appear in Europeans if the population is not evolving with respect to this allele?
To solve this problem, review the section above and generate a list of the information you need to generate and describe how you plan to get it.
Solution
Check your outline by answering the questions below.
1. How frequently do the H1 and H2 alleles occur in this population? This can be found in solution to problem 2.
2. Calculate the expected frequency of the H1H1, H1H2 and H2H2 genotypes in offspring of this population. To do this, square the frequency with which the H1 allele occurs in the population (p2), multiply the frequency with which the H1 allele occurs with the frequency with which the H2 allele occurs and multiply this result by two (2pq), and finally square the frequency with which the H2 allele occurs in the population (q2).
