A Lebanese population of 1,000 members is tested for the autosomal recessive disease phenylketonuria (PKU). The testing results are shown here: AA 660
Aa 280
aa 60
a. Calculate the frequencies of the wild type allele and the PKU allele in this population.
b. What are the expected numbers of individuals of each genotypic class (i.e., AA, Aa, and aa) in this population? Assume Hardy-Weinberg conditions. c. Use the chi-squared test to determine if these data fit the Hardy-Weinberg equilibrium model. The degrees of freedom for this test should be 1. Why is this appropriate?
A Lebanese population of 1,000 members is tested for the autosomal recessive disease phenylketonuria (PKU). The testing results are shown here: AA 660
Aa 280
aa 60
a. Calculate the frequencies of the wild type allele and the PKU allele in this population.
b. What are the expected numbers of individuals of each genotypic class (i.e., AA, Aa, and aa) in this population? Assume Hardy-Weinberg conditions. c. Use the chi-squared test to determine if these data fit the Hardy-Weinberg equilibrium model. The degrees of freedom for this test should be 1. Why is this appropriate?
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1. Data from the caffeine metabolism experiment was pooled together using information from internal and external students across a number of years. The data collected is summarized in the table below. The genotypes are shown in the first column and the other two show the number of genotypes in the two categories of caffeine tolerance (tolerant/intolerant â the classes are derived from the questionnaire data). Use the data in this table to answer this question and others throughout the quiz.
| Genotype | Tolerant | Intolerant |
| AA | 148 | 37 |
| AC | 106 | 99 |
| CC | 83 | 137 |
What are the genotype frequencies using data from all the individuals measured in this experiment?
Select one:
a. AA=0.40 AC=0.44 CC=0.16
b. AA=0.165 AC=0.439 CC=0.396
c. AA=0.25 AC=0.50 CC=0.25
d. AA=0.303 AC=0.336 CC=0.361
2. What are the allele frequencies using all individuals in the experiment?
Select one:
a. A=0.5 C=0.5
b. A=0.471 C=0.529
c. A=0.371 C=0.629
d. A=0.25 C=0.75
3. What are the expected genotype frequencies under Hardy-Weinberg equilibrium?
Select one:
a. AA=0.25 AC=0.5 CC=0.25
b. AA=0.333 AC=0.333 CC=0.333
c. AA=0.222 AC=0.498 CC=0.280
d. AA=0.166 AC=0.668 CC=0.166
4. Test if the population is in equilibrium using a chi-square test. What is the chi-square value for this population?
5. What is the correct number of degrees of freedom for the chi-square test from the previous question?
6. Is this population in Hardy-Weinberg equilibrium?
Select one:
True
False
7. What is the genomic location of CYP1A2?
Select one:
a. 21p24.1
b. 7q24.1
c. 15p24.1
d. 15q24.1
8. What is the reported world-wide frequency for the C allele in human populations?
9. Test if the allele frequencies in our population are significantly different from the world-wide frequencies using a chi-square test.
Select one:
a. chi-square = 131.44; 1 degree of freedom, our population's allele frequencies are significantly different from the world-wide estimates
b. chi-square = 1.314; 1 degree of freedom, our population's allele frequencies are not significantly different from the world-wide estimates
c. chi-square = 1.314; 2 degrees of freedom, our population's allele frequencies are not significantly different from the world-wide estimates
d. chi-square = 131.44; 2 degrees of freedom, our population's allele frequencies are significantly different from the world-wide estimates
10. Use a chi-square independence test to determine if the genotypic frequencies differ between the tolerant and intolerant individuals. The chi-square value is:
