LAB EXERCISE 5.3 continued Name Part 2 If we know all of the genotypes of a trai
ID: 207047 • Letter: L
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LAB EXERCISE 5.3 continued Name Part 2 If we know all of the genotypes of a trait in a population, we can take Hardy-Weinberg a step genotype frequencies in an actual population further and see how much the frequencies) differ from expected frequencies if the alleles were in a completely random manner (that is,if no evolutionary forces were acting, and if mating were random) The following example uses the MN blood group, whose alleles, M and N, are codominant. Thus, we know the genotypes automatically from the phenotypes (blood types). Frequencies Blood Type Genotype Frequency (number of individuals for each genotype divided by total number of individuals) Number of Individuals (phenotype)Genotype Type M 40 20 40 100 Type MN MN Type N Calculating Expected Frequencies Obtain the actual number of alleles present in the population to obtain allele frequency (for p and q). This population of 100 individuals has 200 total alleles MM individuals have 80 (2 M alleles each) MN individuals have 20 (1 M allele each) Number of M alleles: Total of M alleles 100 NN individuals have 80 (2 N alleles each) MN individuals have 20 (1 N allele each) Number of N alleles: Total of N alleles 100 Frequency of M allele (p)-100/200 (total alleles) - s Frequency of N allele (g) 100/200 (total alleles) 5 Now that we know the allele frequencies from the genotypes, we will plug them into the Hardy-Weinberg formula to see whether the actual genotype frequencies above conform to expectations if the population were in equilibriunm not evolving. and experiencing random mating) p2 + 2pq + q2 = 1 .52 + 2.5%5) + .52-.25 + .50 + .25-I Thus, the genotype frequencies expected if the alleles were to combine in a completely random manner would be: Expected Frequencies Number of GenotypeFrequency Individuals .25 .50 .25 25 (Remember this population has 100 individuals.) 50 25 100 MN Compare these expected numbers to those observed in our actual population. They appear very different from on another, but this can be verified with the use of a statistical test such as Chi-square. We would assume either that m s nonrandom in this population or that evolutionary forces are in action, or both.Explanation / Answer
1. genotype frequency-
GG-25/100=0.25
GA-40/100= 0.40
AA-35/100=0.35
2. total no of alleles
as the no of individual in this population is 100, then the total no of allele will be 200.
number of G alleles = GG + GA
= 50 ( 2 X 25) +40 =90
number of A allele = AA +GA
= 35 X 2 +40=110
total no of population= ( 90+110)= 200
3. frequency of G alleles (p)= 90/200=0.45
frequency of A alleles(q) = 110/200=0.55
4. p2+2pq+q2= (0.45)2 + 2 X 0.45 X 0.55 + (0.55)2
= 0.2025 +0.495+0.3025=1
5. Expected frequency
GG= 100 X 0.2025= 20
GA= 100X 0.495 = 50 ( approx)
AA=100 X0.3025 = 30
6.
GG( observed- expected)= (0.25-0.2025)=0.4075
GA ( observed-expected)= ( 0.40-0.495)=-0.095
AA ( observed-expected)= ( 0.35-0.3025)=0.0475
The original population is not in equilibrium.
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