PV92 (Chromosome 16 ALU insertion) genotypes of the Biology 1407 class were dete
ID: 144665 • Letter: P
Question
PV92 (Chromosome 16 ALU insertion) genotypes of the Biology 1407 class were determined by PCR and gel electrophoresis. From the data below, determine the frequency of the+ and- allele in the class population. 4) # of + alleles # of-alleles Total Frequency of + alleles (p) Frequency of alleles (a) From the data above, predict the expected frequency of each genotype: # of students expected frequency(%) Genotype l expected Total Now determine if the class is in Hardy Weinberg Equilibrium. State a reasonable null hypothesis:Explanation / Answer
Frequency of + allele: 22/40 = 0.55 = p
Frequency of - allele: 18/40 = 0.45 = q
(p + q = 0.55 + 0.45 = 1)
Null Hypothesis: There is no statistically significant difference between the observed phenotypic frequency of the test population and that expected under the Hardy-Weinberg law and any difference observed may have been due sampling error.
Alternative Hypothesis: There is statistically significant difference between the observed and expected phenotypic frequencies of the test population which cannot be attributed solely to sampling error.
Chi-square value: X 2 = 7.103
Degrees of freedom: 1
p value: 0.007696
This value is smaller than 0.05 and 0.01, the usual significance level taken for rejecting null hypothesis.
The class is NOT in Hardy-Weinberg equilibrium.
We reject the null hypothesis.
Note: Normally the degrees of freedom is taken as n-1 which in this case would have been 3-1 = 2. But degrees of freedom is the number of independent values in the final calculations. In this case, the value of p has been calculated from the observations themselves. There is no theoretical expected value for p (For example 9:3:3:1 is a theoretically expected ratio) So value of p is dependent on the observations. That is one degree of freedom lost. Once value of p is found, q is calculated by default. But of the 3 classes (p^2, 2pq, q^2) which we have taken for chi-square test, since they are dependent on the formula p^2 + 2pq + q^2 = 1, the value of the last class (any one of the three) is constrained by the values of the the other 2 classes. (Example: if the other two classes are 0.5 and 0.3, the third has to be 0.2). So it has no "freedom" to take just about any value. That is another dependent value. So one more degree of freedom is lost. Hence degree of freedom will be 3-2= 1.
But just in case you need the p value for degrees of freedom 2 it is 0.028682.
Genotype Students (o) + alleles - alleles +/+ 9 18 0 +/- 4 4 4 -/- 7 0 14 Total 20 22 18Related Questions
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