Natural Selection Online Lab

This means that there will be no changes in gene/allele frequency in a population and thus no evolution in that population. However, as many populations do show changes in allele frequency over time, scientists can attempt to determine which of the H-W conditions has been altered. Natural selection is often an important factor in the change of allele frequency in a population. Individuals with different phenotypes (and thus genotypes) may experience differential survival and reproductive rates that is dependent on their environment. If one phenotype has better reproductive and survival rates than another, one would observe a change in allele frequencies over several generations for the population. This phenotype exhibits better survival and reproductive rates, thus has better fitness. In humans, cystic fibrosis (CF) is a genetic disorder where the affected individuals produce a thick mucus in their respiratory and digestive systems. These individuals are subject to respiratory infections and lowered nutrition because of the disease. Individuals with CF generally die young and do not reproduce. CF is a recessive allele thus homozygous recessive individuals exhibit the disease, but heterozygous and homozygous dominant individuals do not exhibit the disease. New cases of the disease arise when heterozygous parents have children or when there is a spontaneous mutation in a “normal” parent. Because of the dire ramifications of the disease, some heterozygous (carriers) individuals voluntarily refrain from having children to avoid passing the allele to their children. The deliberate attempt to reduce the continuation of a harmful allele in future generations is an example of negative eugenics. Unfortunately, negative eugenics has been utilized on people unknowingly or against their will to prevent the “unfit” individuals from reproducing. This practice was somewhat common in the 1900s in the United States and elsewhere and taken to the extreme by Nazi Germany. This experiment will compare the gene frequencies in two populations. In the first population, homozygous recessive individuals will not reproduce (selection pressure). In the second population, both homozygous recessive individuals and half of the heterozygous individuals will not reproduce (selection pressure and negative eugenics).

F = normal allele f = recessive allele (Cystic Fibrosis, CF) There are 100 alleles in the gene pool at the start of each generation. Population 1: Selection Pressure against Homozygous Recessive Genotypes Generation 1 Total Initial Gene Pool 75 alleles (F) 25 alleles (f) 100 alleles Percentage (%) 75% 25% 100% Frequency 0.75 0.25 1.0

1.75 dominant alleles (F) and 25 alleles (f) were placed in a container. These 100 alleles will represent the initial gene pool. The dominant allele represents a normal condition and the recessive allele is the Cystic Fibrosis allele. This gene pool has been generated by partnerships between heterozygous and homozygous dominant individuals. 2.The container with the alleles was shaken to simulate random mixing of the gametes during the first generation of reproduction. 3.Then without looking, two alleles were randomly removed from the container. These 2 alleles represent the first individual. If 2 dominant alleles were selected then the individual would be________ (circle one) a. Homozygous dominant b. Heterozygous c. Homozygous recessive (Cystic Fibrosis) If a dominant allele and a recessive allele were selected then the individual would be________ (circle one) a. Homozygous dominant b. Heterozygous c. Homozygous recessive (Cystic Fibrosis) If 2 recessive alleles were selected then the individual would be________ (circle one) a. Homozygous dominant b. Heterozygous c. Homozygous recessive (Cystic Fibrosis) 4.The rest of the alleles are selected in random pairs until all alleles have been removed from the container (total of 50 individuals). The number of homozygous dominant, heterozygous, and homozygous recessive individuals have been tallied up and recorded in the table below under Line 1. 5.We will select against all the homozygous recessive individuals (ff). This means that the alleles that represent the (ff) genotype will be removed from the gene pool. They have cystic fibrosis and will not survive to reproductive maturity and therefore will not reproduce and pass their genes to the next generation. 6.The number of individuals that were removed is shown on Line 2. In the total column, the number of ALLELES that were removed has been recorded as well. Recall that a homozygous recessive individual has 2 recessive allele. Record the number of ALLELES that were removed. 7.The number of surviving individuals has been recorded on Line 3.

Total Line 1 Generation 1 Number of individuals that are (FF) = 28 Number of individuals that are (Ff) = 19 Number of individuals that are (ff) = 3 100 alleles Line 2 Remove 0 (FF) 0 (Ff) 3 (ff) -6 alleles Line 3 Survivors 28(FF) 19(Ff) 0 (ff) alleles 8.Using the information in the table above calculate the percentages and frequencies for Generation 1. Find the number of (F) alleles (from Line 3): 2 x (FF) individuals (Ff) individuals = __________(F) Find the number of (f) alleles (from Line 3) : (Ff) individuals = ___________ (f) Total number of alleles = (F) (f) = __________ Find the frequency of (F) allele = (F) ÷ total alleles = ____________ or __________% Find the frequency of (f) allele = (f) ÷ total alleles = _____________ or ___________% Record this information for Generation 1 on Table 1.

Generation 2 Using the same procedure described in Generation 1, the next generation of individuals were selected for Generation 2. Total Line 1 Generation 2 Number of individuals that are (FF) = 34 Number of individuals that are (Ff) =12 Number of individuals that are (ff) = 4 100 alleles Line 2 Remove 0 (FF) 0 (Ff) 4 (ff) -8 alleles Line 3 Survivors 34 (FF) 12 (Ff) 0 (ff) alleles 9.Calculate the percentages and frequencies for Generation 2. Find the number of (F) alleles (from Line 3) 2 x (FF) individuals (Ff) individuals = __________(F) Find the number of (f) alleles (from Line 3): (Ff) individuals = ___________ (f) Total number of alleles = (F) (f) = __________ Find the frequency of (F) allele = (F) ÷ total alleles = ________________ or ___________% Find the frequency of (f) allele = (f) ÷ total alleles = _____________ or ___________% Record this information for Generation 2 on Table 1. _____________________________________________________________________________________Generation 3 Total Line 1 Generation 3 Number of individuals that are (FF) = 38 Number of individuals that are (Ff) = 11 Number of individuals that are (ff) = 11 100 alleles Line 2 Remove 0 (FF) 0 (Ff) 1(ff) -2 alleles Line 3 Survivors 38 (FF) 11 (Ff) 0 (ff) alleles 1.Calculate the percentages and frequencies for Generation 3. Find the number of (F) alleles (from Line 3): 2 x (FF) individuals (Ff) individuals = __________(F) Find the number of (f) alleles (from Line 3): (Ff) individuals = ___________ (f) Total number of alleles = (F) (f) = __________ Find the frequency of (F) allele = (F) ÷ total alleles = ________________ or _____________% Find the frequency of (f) allele = (f) ÷ total alleles = _____________ or ___________% Record this information for Generation 3 on Table 1. Generation 4 Total Line 1 Generation 4 Number of individuals that are (FF) = 40 Number of individuals that are (Ff) = 9 Number of individuals that are (ff) = 1 100 alleles Line 2 Remove 0 (FF) 0 (Ff) 1 (ff) -2 alleles Line 3 Survivors 40 (FF) 9 (Ff) 0 (ff) alleles 2.Calculate the percentages and frequencies for Generation 4. Find the number of (F) alleles (from Line 3): 2 x (FF) individuals (Ff) individuals = __________(F) Find the number of (f) alleles (from Line 3): (Ff) individuals = ___________ (f) Total number of alleles = (F) (f) = __________ Find the frequency of (F) allele = (F) ÷ total alleles = _______________ or _____________% Find the frequency of (f) allele = (f) ÷ total alleles = _____________ or ___________% Record this information for Generation 4 on Table 1. _____________________________________________________________________________________ Generation 5 Total Line 1 Generation 5 Number of individuals that are (FF) = 42 Number of individuals that are (Ff) = 7 Number of individuals that are (ff) = 1 100 alleles Line 2 Remove 0 (FF) 0 (Ff) 1 (ff) -2 alleles Line 3 Survivors 42 (FF) 7 (Ff) 0 (ff) alleles 3.Calculate the percentages and frequencies for Generation 5. Find the number of (F) alleles (from Line 3): 2 x (FF) individuals (Ff) individuals = __________(F) Find the number of (f) alleles (from Line 3): (Ff) individuals = ___________ (f) Total number of alleles = (F) (f) = __________ Find the frequency of (F) allele = (F) ÷ total alleles = _______________ or _____________% Find the frequency of (f) allele = (f) ÷ total alleles = _____________ or ___________% Record this information for Generation 5 on Table 1. You now have enough data to produce a graph of the decline in frequency of the (f) allele over five generations. Use the provided graph (Figure 1) to plot the frequency of (f) against the number of generations. Table 1. Change in Allele Frequency due to Selection Pressure (Population 1) F f Rounds # % Frequency # % Frequency Initial Generation 0 Generation 1 Generation 2 Generation 3 Generation 4 Generation 5 Figure 1: Change in Allele Frequency due to Selection Pressure against the f allele Population 2: Selection Pressure and Negative Eugenics Generation 1 Total Initial Gene Pool 75 alleles (F) 25 alleles (f) 100 alleles Percentage (%) 75% 25% 100% Frequency 0.75 0.25 1.0

1.Using the same procedure as the first experiment we will track the allele frequencies under a different type of selection pressure: Negative Eugenics. 2.Using the textbook or external resources provide a definition and a historical example of Negative Eugenics. 3.75 dominant alleles (F) and 25 alleles (f) were placed in a container. These 100 alleles will represent the initial gene pool. The dominant allele represents a normal condition and the recessive allele is the Cystic Fibrosis allele. This gene pool has been generated by partnerships between heterozygous and homozygous dominant individuals. 4.The container with the alleles was shaken to simulate random mixing of the gametes during the first generation of reproduction. 5.Then without looking, two alleles were randomly removed from the container. These 2 alleles represent the first individual. 6.The rest of the alleles are selected in random pairs until all alleles have been removed from the container (total of 50 individuals). The number of homozygous dominant, heterozygous, and homozygous recessive individuals have been tallied up and recorded in the table below under Line 1. 7.We will select against all the homozygous recessive individuals (ff). This means that the alleles that represent the (ff) genotype will be removed from the gene pool. They have cystic fibrosis and will not survive to reproductive maturity and therefore will not reproduce and pass their genes to the next generation. 8.Then select against the heterozygous individuals (Ff) by removing one half of them from the gene pool. If there is an odd number of Ff individuals, we will round up. For example, if there are 3 such individuals , then 2 will be removed. 9.The number of individuals that were removed is shown on Line 2. In the total column, the number of ALLELES that were removed has been recorded as well. Recall that one individual has 2 alleles. 10.The number of surviving individuals has been recorded on Line 3. Total Line 1 Generation 1 Number of individuals that are (FF) = 27 Number of individuals that are (Ff) = 21 Number of individuals that are (ff) = 2 100 alleles Line 2 Remove 0 (FF) ½ (Ff) Remove 11 individuals 2 (ff) -26 alleles Line 3 Survivors 27 (FF) 10 (Ff) 0 (ff) alleles 11.Calculate the percentages and frequencies for Generation 1. Find the number of (F) alleles (from Line 3): 2 x (FF) individuals (Ff) individuals = __________(F) Find the number of (f) alleles (from Line 3): (Ff) individuals = ___________ (f) Total number of alleles = (F) (f) = __________ Find the frequency of (F) allele = (F) ÷ total alleles = ___________ or __________% Find the frequency of (f) allele = (f) ÷ total alleles = ____________ or ___________% Record this information for Generation 1 on Table 2. ___________________________________________________________________________________ Generation 2 Total # Line 1 Generation 2 Number of individuals that are (FF) = 37 Number of individuals that are (Ff) = 11 Number of individuals that are (ff) =2 100 alleles Line 2 Remove 0 (FF) ½ (Ff) Remove 6 individuals 2 (ff) -16 alleles Line 3 Survivors 37 (FF) 5 (Ff) 0 (ff) alleles 12.Calculate the percentages and frequencies for Generation 2. Find the number of (F) alleles (from Line 3): 2 x (FF) individuals (Ff) individuals = __________(F) Find the number of (f) alleles (from Line 3): (Ff) individuals = ___________ (f) Total number of alleles = (F) (f) = __________ Find the frequency of (F) allele = (F) ÷ total alleles = __________________ or ___________% Find the frequency of (f) allele = (f) ÷ total alleles = _____________ or ___________% Record this information for Generation 2 on Table 2. Generation 3 Total Line 1 Generation 3 Number of individuals that are (FF) = 44 Number of individuals that are (Ff) = 6 Number of individuals that are (ff) = 0 100 alleles Line 2 Remove 0 (FF) ½ (Ff) Remove 3 individuals 0 (ff) -6 alleles Line 3 Survivors 44 (FF) 3 (Ff) 0 (ff) alleles 13.Calculate the percentages and frequencies for Generation 3. Find the number of (F) alleles (from Line 3): 2 x (FF) individuals (Ff) individuals = __________(F) Find the number of (f) alleles (from Line 3): (Ff) individuals = ___________ (f) Total number of alleles = (F) (f) = __________ Find the frequency of (F) allele = (F) ÷ total alleles = ________________ or ___________% Find the frequency of (f) allele = (f) ÷ total alleles = _____________ or ___________% Record this information for Generation 3 on Table 2. ___________________________________________________________________________________ Generation 4 Total Line 1 Generation 4 Number of individuals that are (FF) = 47 Number of individuals that are (Ff) = 3 Number of individuals that are (ff) = 0 100 alleles Line 2 Remove 0 (FF) ½ (Ff) Remove 2 individuals 0 (ff) -4 alleles Line 3 Survivors 47 (FF) 1 (Ff) 0 (ff) alleles 14.Calculate the percentages and frequencies for Generation 4. Find the number of (F) alleles (from Line 3): 2 x (FF) individuals (Ff) individuals = __________(F) Find the number of (f) alleles (from Line 3): (Ff) individuals = ___________ (f) Total number of alleles = (F) (f) = __________ Find the frequency of (F) allele = (F) ÷ total alleles = _____________ or _____________% Find the frequency of (f) allele = (f) ÷ total alleles = _____________ or ___________% Record this information for Generation 4 on Table 2. Generation 5 Total Line 1 Generation 5 Number of individuals that are (FF) = 49 Number of individuals that are (Ff) = 1 Number of individuals that are (ff) = 0 100 alleles Line 2 Remove 0 (FF) ½ (Ff) Remove 1 individuals 0 (ff) -2 alleles Line 3 Survivors 49 (FF) 0 (Ff) 0 (ff) alleles 15.Calculate the percentages and frequencies for Generation 5. Find the number of (F) alleles (from Line 3): 2 x (FF) individuals (Ff) individuals = __________(F) Find the number of (f) alleles (from Line 3): (Ff) individuals = ___________ (f) Total number of alleles = (F) (f) = __________ Find the frequency of (F) allele = (F) ÷ total alleles = _____________ or _____________% Find the frequency of (f) allele = (f) ÷ total alleles = _____________ or ___________% Record this information for Generation 5 on Table 2. You now have enough data to produce a graph of the decline in frequency of the (f) allele over four generations. Use the provided graph (Figure 2) to plot the frequency of (f) against the number of generations. Table 2. Change in Allele Frequency due to Selection Pressure and Negative Eugenics F f Rounds # % Frequency # % Frequency Initial Generation 0 Generation 1 Generation 2 Generation 3 Generation 4 Generation 5 Figure 2: Change in Allele Frequency due to Selection Pressure and Negative Eugenics against the (f) allele End of Exercise Questions 1.Did you observe a rapid or gradual change in the (f) allele frequency during the Population 1 experiment? Why doesn’t cystic fibrosis disappear in this population? 2.In Population 2, how did negative eugenics change the ( f) allele frequency as compared to Population 1? Discuss why this occurred. 3.Hardy-Weinberg equilibrium predicts that the genotype frequencies of the offspring will be the same as those of the parent generation as long as all H-W conditions are met. A) Did you observe H-W equilibrium in either of the populations of this experiment? B) If the genotype frequencies are different, hypothesize which H-W condition(s) were not met. Explain why you think they were not met. You are a genetic counselor working with a couple, both carriers for CF. What would you tell the couple about the probability of producing an affected child and of voluntary negative eugenics? Use Punnett squ

The post Natural Selection Online Lab appeared first on .
CLICK HERE TO GET THE BEST PAPERS. HIGH QUALITY.
No Plagiarism in Our Company.

Calculate the price of your order

550 words
We'll send you the first draft for approval by September 11, 2018 at 10:52 AM
Total price:
$26
The price is based on these factors:
Academic level
Number of pages
Urgency
Basic features
  • Free title page and bibliography
  • Unlimited revisions
  • Plagiarism-free guarantee
  • Money-back guarantee
  • 24/7 support
On-demand options
  • Writer’s samples
  • Part-by-part delivery
  • Overnight delivery
  • Copies of used sources
  • Expert Proofreading
Paper format
  • 275 words per page
  • 12 pt Arial/Times New Roman
  • Double line spacing
  • Any citation style (APA, MLA, Chicago/Turabian, Harvard)

Our guarantees

Delivering a high-quality product at a reasonable price is not enough anymore.
That’s why we have developed 5 beneficial guarantees that will make your experience with our service enjoyable, easy, and safe.

Money-back guarantee

You have to be 100% sure of the quality of your product to give a money-back guarantee. This describes us perfectly. Make sure that this guarantee is totally transparent.

Read more

Zero-plagiarism guarantee

Each paper is composed from scratch, according to your instructions. It is then checked by our plagiarism-detection software. There is no gap where plagiarism could squeeze in.

Read more

Free-revision policy

Thanks to our free revisions, there is no way for you to be unsatisfied. We will work on your paper until you are completely happy with the result.

Read more

Privacy policy

Your email is safe, as we store it according to international data protection rules. Your bank details are secure, as we use only reliable payment systems.

Read more

Fair-cooperation guarantee

By sending us your money, you buy the service we provide. Check out our terms and conditions if you prefer business talks to be laid out in official language.

Read more