Jordan Holmes- Hardy Weinberg Lab

Hardy-Weinberg Lab
Created by Jordan Holmes

2014-2015


 *  Example Video 
 * Hardy Weinberg
 *  Example Data 
 * ​ The data from this lab are displayed in charts such as the one below.  The bold-italicized genotypes indicate the genotypic data collected while running the lab.  The class data section is a compilation of all the students' data collected while running the lab.


 * Table 1


 *  Example Calculations 
 * The letter "p" represents the frequency of the dominant allele.


 * The letter "q" represents the frequency of the recessive allele.


 * "p^2" represents the frequency of the dominant phenotype.


 * "2pq" represents the frequency of heterozygous individuals.


 * "q^2" represents the frequency of the recessive phenotype.


 *  Table 2 

​ The initial data for generation 0 are given in the lab packet. The data for generation 5 are found using the class data from Table 1.

The Hardy-Weinberg Equation states that p+q=1 and p^2+2pq+q^2= 1.

In this lab, the total number of alleles in the population equals the number of students in the class times 2. Therefore, the calculation used in this lab for the number of A alleles present in the fifth generation is:


 * 1) of offspring with genotype AA: 5 x 2 = 10 A alleles


 * 1) of offspring with genotype Aa:  12 x 1 = 12 A alleles

Total = 22 A alleles

p= (Total # of A alleles: 22 ) / (Total # of alleles in the population: 48 ) = 0.46

This process is repeated in order to find the frequency of the a allele which you will find to be 0.54.

With these frequencies, the entire data table can now be filled in using what you know about populations in Hardy-Weinberg Equilibrium.

Sample problem: 

If the frequency of the a allele in a population that is in Hardy-weinberg Equilibrium is 0.21, what are the frequncies of the A allele, dominant phenotype, recessive phenotype, and heterozygous individuals?

 

https://quizlet.com/79880400/hardy-weinberg-lab-by-jordan-holmes-flash-cards/
 *  Quizlet 


 * C onnections 
 * 1) Natural Selection- This lab relates to natural selection as seen in Case II.  In Case II, the individuals with genotype aa did not reproduce.  This lowers the frequency of the recessive allele.  The recessive allele was selected against in this case.  This type of selection can be seen in nature with sickle-cell anemia.  This disease is caused by a mutation that does not allow homozygous recessive individuals to reach reproductive age.
 * 2) Genetic Drift: This lab relates to genetic drift as seen in Case IV.  In Case IV, the large class population was divided into 2 smaller populations that were isolated from each other.  When populations become split and isolated from each other, it is possible that one allele may disappear from one population and be more prevalent in another due to chance.  For example, in a population of squirrels, the dominant phenotype is brown and the recessive allele is white.  If a road is built between the population of squirrels so that they become isolated, it is possible that all of the white squirrels will end up on one side of the road.  This will eliminate the brown allele from that population.
 * 3) Sexual Reproduction: This lab relates to sexual reproduction as seen in the recombination of alleles. Sexual reproduction results in offspring that are not genetically identical to the parents. Mechanisms for sexual reproduction include independent assortment, crossing over, and fertilization. Independent assortment can be seen in this lab when the cards are set up on the desk. The allele cards are assorted randomly; they are not lined up based on any factors other than randomness.


 * Answer to sample problem:



