| << Chapter < Page | Chapter >> Page > |
The probability rule we used in another module to predict genotype frequencies in the offspring generation of a specific population:
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
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
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
And, of course, if we know q, the frequency of the a allele, then
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.
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.
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.
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?
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.
Notification Switch
Would you like to follow the 'Understanding the hardy-weinberg equation' conversation and receive update notifications?
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|