Question

ackground information:

 

Through PCR, we have determined the PER3 genotypes for a class of students as follows:

H4/H4 = 125 individuals; H4/H5 = 90 individuals and H5/H5=85 individuals.

 

Questions that I need help understanding:

 

1) How many H4 alleles are in the class?

1. 90

2. 215 

3. 340 

4. 125 

5. 260 

6. 85 

7. 175

 

 

2) What is the frequency of the H5 allele in the class? 

1. 0.43 

2. 0.19 

3. 0.49 

4. 0.32 

5. 0.57 

6. 0.24 

7. 0.14

 

 

3) Using the allele frequencies experimentally derived, calculate the frequency of the H4/H5 genotype that would be expected if the class were a population in Hardy-Weinberg equilibrium. 

1. 0.32

2. 0.19 

3. 0.49 

4. 0.57 

5. 0.14 

6. 0.24 

7. 0.43

 

 

4) Using the genotype frequencies derived assuming that the class were a population in Hardy-Weinberg equilibrium, calculate the number of H4/H4 individuals that would be expected in the class (rounded numbers).

1. 96 

2. 147 

3. 171

4. 57 

5. 72 

6. 129 

7. 42

 

 

5) Considering the Hardy Weinberg equilibrium and comparing the observed and the expected number of individuals for the three genotypes, calculate the value of the Chi-square statistic. 

1. 28.67 

2. 45.43 

3. 0.05 

4. 22.31 

5. 0.50 

6. 3.84 

7. 14.59

 

6) Considering the Hardy Weinberg equilibrium, which is the (correct) null hypothesis tested by Chi-square? 

 

1. The whole class represents a population that is not in Hardy-Weinberg equilibrium 

2. The whole class represents a population that may not be in Hardy-Weinberg equilibrium 3. The whole class represents a population that is in Hardy-Weinberg equilibrium 

4. The whole class represents a population that may be in Hardy-Weinberg equilibrium

 

 

7)Considering the Hardy Weinberg equilibrium and calculating the Chi-square statistic, do you reject or fail to reject the null-hypothesis?

 

1. Comparing the value of Chi-square that I calculated with the critical value of Chi-square for the appropriate degrees of freedom I conclude that P>0.05. Hence, I reject the null hypothesis. 

2. Comparing the value of Chi-square that I calculated with the critical value of Chi-square for the appropriate degrees of freedom I conclude that P<0.05. Hence, I fail to reject the null hypothesis. 

3. Comparing the value of Chi-square that I calculated with the critical value of Chi-square for the appropriate degrees of freedom, I conclude that P>0.05. Hence, I fail to reject the null hypothesis. 

4. Comparing the value of Chi-square that I calculated with the critical value of Chi-square for the appropriate degrees of freedom I conclude that P<0.05. Hence, I reject the null hypothesis.

 

 

8) Considering the Hardy Weinberg equilibrium and calculating the Chi-square statistic, what is the overall conclusion based on the results of the Chi-square test? 

 

1. The observed number of individuals per genotype is compatible with the assumption that the class represents a population in Hardy Weinberg equilibrium 

2. The observed number of individuals per genotype is not compatible with the assumption that the class represents a population in Hardy Weinberg equilibrium 

3. The observed number of individuals per genotype is compatible with the assumption that the class may not represent a population in Hardy Weinberg equilibrium 

4. The observed number of individuals per genotype is compatible with the assumption that the class may represent a population in Hardy Weinberg equilibrium