In this exercise, you will model natural selection actingon a pair of alleles in a population. You will sample amodel population and select only select individuals with thedominant allele to produce offspring for the next generation eachgeneration. You will calculate starting allelic frequenciesfor each generation and graph the results.
Procedure
Put on your safety goggles.
Verify that the plastic cup from Exercise 1 contains 50brown beans (B allele for brown fur) and 50 white beans (b allelefor white fur).
Write “BB,” “Bb,” and “bb” on a clean sheet of paper, which willbe used to record alleles represented by the beans.
Randomly, without looking, withdraw two beans(alleles) from the sample population in the cup.
Record a hash mark on the sheet of paper beside the appropriategenotype.
Return both beans (alleles) to the population and shake the cupto redistribute the alleles.
Note: returning the beans to the cup assures that eachindividual has an equal chance of receiving eitherallele.
Repeat steps 4-6 for 49 times so that a total of 50 individualshave been sampled.
Record the number of individuals for each of the three genotypesthat were sampled as Generation 1 in Data Table5.
Note: The bb genotype is recorded as deaths as theseindividuals are selected against in the population and do notproduce offspring. Check your work. The sumof the genotypic numbers (BB + Bb + bb) should equal the totalnumber of individuals in the population.
Calculate the final genetic frequency (% of livingindividuals) for each genotype using the equations:
p2=BB(BB +Bb)2pq=Bb(BB + bb)p2=BB(BB +Bb)2pq=Bb(BB + bb)
Note: The frequency of the genotype bb (q2)has been recorded as 0 because all homozygous recessive individualsdie without reproducing. Check your work. The sum of thegenotypic frequencies should equal one (p2+2pq + q2 =1).
Final B = 2(BB individuals) + Bb individualsFinal b = BbindividualsFinal B = 2(BB individuals) + Bb individualsFinal b = Bbindividuals
Record the final allelic numbers for Generation 1in Data Table 5.
Calculate the starting allelic frequencies for p and q forGeneration 2 using the final allelic numbers from Generation 1 andthe equations:
p=B(B + b)q=b(B + b)p=B(B + b)q=b(B + b)
Note: Check your work. The sum of the starting allelicfrequencies should equal one(p + q = 1).
B=p*100b=q*100B=p*100b=q*100
Note: Check your work. The sum of the starting allelicfrequencies should equal one(p + q = 1).
Record the p and q values for Generation 2 in Data Table 5.
Calculate the starting allelic numbers B and b forGeneration 2 using the equations:
Note: Check your work. The sum of the starting allelicnumbers should equal 100. Thus, there will be 100 alleles total inthe population.
Record the values for B and b for Generation 2in Data Table 5.
Adjust the numbers of each allele in the cup to match therecorded numbers for Generation 2.
Repeat steps 3-19 nine times, until data is complete for 10generations.
Create a line graph of the starting allelic frequencies for eachgeneration, plotting generation on the independent axis(x-axis) allelic frequency on the dependent axis (y-axis).
Note: The graph will have two lines; one for p and onefor q across 10 generations.
Upload an image of the graph into Graph2.
Cleanup:
- Exercise 2 Modeling Natural Selection In This Exercise You Will Model Natural Selection Acting On A Pair Of Alleles In 1 (45.02 KiB) Viewed 77 times
2. How did the allelic frequencies change over the 10generations modeled in response to natural selection? Do you thinkthe white allele could be eliminated from the population if themodel had continued for 10 more generations? Reference Graph 2 inyour explanation.
3. A mutation occurs in a population of rabbits affecting earlength. After multiple generations, 30% of the population exhibitthe new recessive phenotype for short ears. Use the Hardy-Weinbergequation to determine the genotypic frequencies of the rabbitpopulation. Show all work in your answer.
4. How would the allelic frequencies and resulting graph differfrom the results in Data Table 5 and Graph 2 for a population atHardy-Weinberg equilibrium for 10 generations?
Experiment 2 Data Table 5: Sampling Without Replacement Generation Starting Allelic Frequency 1 2 3 4 5 6 7 8 9 10 Average p 0.50 9 0.50 Starting Allelic Number B 50 50 Data Table 5 Genotypic Number BB Bb Graph 1 Deaths (bb) Final Genotypic Frequency 0² 2pq q² 0 0 0 0 0 0 0 0 0 0 0 Exercise 2 Final Allelic Number b B