If a population meets the Hardy-Weinburg conditions it is in genetic equilibrium, so there will be no change in gene frequency and therefore no evolution. The usefulness of the concept of genetic equilibrium is that we can measure changes in gene frequency that do occur and try to discover which of the stipulated conditions has been altered to give that change. Is the observed change in gene frequency over a period of time or in different populations due to mutation, small population size, selective mating, migrating individuals, or natural selection? For most genes, natural selection is almost always by far the most important factor in bringing about changes in gene frequency.
Natural selection is the differential reproduction of individuals with different phenotypes and genotypes. It can act only on phenotypes and will result in a change in allele frequencies if the selection against one of the phenotypes is stronger, so that another phenotype reproduces more successfully. The selection coefficient (S) measure the elimination of a phenotype (0 = minimal elimination; 1.0 = maximum elimination). The fitness (W) of the more successful phenotype is 1 – S. In the most extreme case, where one allele is lethal, selection against the corresponding phenotype will be 100% (1.0) since the affected individuals never reproduce. The frequency of a recessive lethal allele will drop rapidly, but it will persist in the population: (1) because there is no selection against the heterozygote carriers and (2) because of new mutations.
In humans, cystic fibrosis is an inherited disease due to an autosomal recessive ff gene located on chromosome #7. In the most common defective allele, three base pairs are deleted, and single phenylalanine is missing. Affected individuals carry two of the recessive alleles for the disease (genotype ff) and as a result form extremely thick mucus in their respiratory systems and elsewhere in body. Their lungs are susceptible to frequent infections and the disease is progressive and eventually fatal. Usually, the victims die in their teens or early twenties, and so do not reproduce. Among Caucasians, 1 in 20 is a carrier, that is, an Ff heterozygote for cystic fibrosis.
Severe natural selection has operated on the gene pool for the cystic fibrosis (f) and nonaffected (F) alleles over the centuries: The affected individuals do not reproduce and do not pass on their genes (S = 1.0). New cases arise only when both parents are heterozygous or through new mutations in a nonaffected parent.
What is the chance that two heterozygotes will produce an affected child? Make a Punnett square to show your results.
Why might the frequency of the f allele fail to drop overtime?
In the following exercise, you will graph the effect of natural selection on the frequency of the f allele in a model system. Among the children of heterozygous carriers of the f allele and genetically non-affected individuals, the frequency of the f allele should be 25%, the same as in the parents. We will use that frequency at the start of this experiment.
First Round:
There is a complete election against the recessive genotype that is expressed as the affected cystic fibrosis phenotype. The homozygous FF and the heterozygous Ff genotypes both have the normal phenotype and are not selected against. Both will therefore contribute to the next generation.
Obtain a small plastic bag with 75 blue beads and 25 orange beads. These 100 beads represent your initial gene pool which are 100 alleles give from the starting population of 50 individuals. This is the gene pool generated by heterozygous carriers and homozygous nonaffected individuals.
Shake up the beads to simulate random mixing of gametes during the first generation of reproduction.
Reach into the bag and without looking take out two beads: this is the first individual.
Set the beads aside and repeat this 49 times until you have used up all the beads, arranging the FF, Ff, and ff genotypes in groups as you draw.
Count the number of blue and orange beads left in the pool and calculate the percentage and frequency of each.
Enter the results in the table.
Please see example below:
Second Round:
Begin the second round by replenishing your beads with the same percentages of blue and orange beads that you had at the end of round 1. In our example we had 80% blue and 20% orange alleles. To keep this percentage, we will add 5 blue allele beads for a total of 75+5 = 80 blue allele beads: and 1 orange allele bead for a total of 19+1 = 20 orange allele beads.
*Note the 3 homozygous recessive individuals do not reproduce because reduced fitness (S = 1). This means that the alleles of those individuals were removed from the gene pool. The homozygous dominate and heterozygous individuals will reproduce as the are not affected by cystic fibrosis. The blue and orange beads added back to the population will represent offspring produced by non-affected and carriers of cystic fibrosis for a total population of 60 individuals. Each round you will replace the number of homozygous recessive individuals that were removed from the population due to reduced fitness.
Shake up the beads
Select 50 individuals
Remove all the ff individuals as before in round one.
Calculate the percentages and frequencies of blue and orange beads left after natural selection and enter the results in the table under generation 2.
Third Round and etc.:
Begin the third round by replenishing your beads so that you have the new percentages as you did in round 2.
Repeat the previous steps until you have finished 5 generations to complete the table.
When you have finished five rounds, you will be tired of drawing beads, but you will have enough data to make a meaningful graph of the decline in the frequency of the f allele over five generations. Below create a graph representing the frequency of the f allele on the y axis and the number of generations of the x axis. Make this line from the f allele one specific color. Next, graph the frequency of the F allele using a different color. This graph shows what would happen to the f and F alleles over 5 generations when recessive phenotypes do not reproduce. *Note: you will have two separate lines on the same graph.
Make sure to give the graph a title
Label both the x and y axis