For the first two exercises you will need the following: 50 red M&Ms and 50 gree
ID: 44433 • Letter: F
Question
For the first two exercises you will need the following:
50 red M&Ms and 50 green M&Ms or 50 each of two items that are distinguishable by color but are similar in size and texture (e.g., dimes and pennies, two different color beads).
Four containers large enough to hold the above items.
Experiment 7 Exercise 1: Evolutionary Change without Natural Selection
In this first exercise, we are going to look for evidence of evolutionary change in a population in the absence of natural selection by looking at the change in allele frequencies over time in a simulated population. We will start with a population of 50 individuals in which there are two alternate alleles (H and h) in equal proportions (each at a frequency of 0.5 or 50%). Individuals have the possible genotypes: HH, Hh or hh. These two alleles do not offer any selective advantage, so neither is selected for or against, meaning they are neutral. We will record the frequency of these alleles over 10 generations. Prior to advancing on to the next generation, six alleles (= three individuals) will be removed at random.
Before you begin, answer the following:
Question
What is your prediction as to what will happen to the frequencies (note that this is different than the number) of these two alleles over 10 generations? Word your prediction as an
Explanation / Answer
In the given experiment, generation 2 have 94 alleles, and the no. of alleles further decreases with each generation. Thus, the 10th generation has 46 alleles.
Their frequencies may increase or decrease depending on random 6 alleles, which are removed
After the generation 2 assume that
1) 3 Hh individuals are removed. Then we will have 47H and 47h alleles for next generation and frequency of alleles is 0.5
2) But if 3HH individuals are removed, then we will have 44H and 50 alleles and frequency of h allele increases over H allele. Frequency of H allele will be (44/94), whereas that of h allele will be (50/94)
Similarly, if 3hh individuals are removed, then frequency of H allele increases and that of h allele decreases.
3) Not only these, but there are total 9 possibilities to remove 3 individuals (there are 3 possibilities to remove each individual. so, total no. of possibilities 3*3*3=27, if we remove the repeating possibilities then we will have 9 possibilities)
The frequency of allele depends on the total no. of alleles present in generation 2, if there are more no. of H allele than h allele, then the frequency of H will be more than that of h allele. Frequency of an allele = (no. of that type allele)/total no. of alleles present in that generation.
If we removed 2HH, 1Hh, then frequency of H=(45/94), of h=(49/94)
If 1HH,1Hh,1hh are removed. then frequency of both alleles will be same as total no. of alleles present in that generation has same no. of H allele and h allele. frequency is (47/94) =0.5 and there are 4 other possibilities for gen 2.
similar things happens for gen 3. In 3rd gen, we will have 88 alleles. the no. of each type of allele should be noted and we should calculate the frequency. it depends on the individuals removed before3rd generation
For 4th generation frequency depends on total individuals removed before 4th generation.(i.e before 3rd generation and (before 4th generation and after 3rd generation))
At the end of 10 th generation total no. of alleles =46 and the frequency of the allele may be < or > or = to 0.5 depending on alleles present at the end. Thus, it depends on total individuals removed before current generations.