Questions: Below are tables with data in them. The data were obtained from a previous BIO 340 class. Data obtained from reference papers regarding the USA population are also provided and will be used to determine the "Expected" frequency and counts. Some of the information in the table is missing. You will have to analyze the data and provide phenotype, genotype and allele frequencies as well as perform chi-square analyses. Remember, p+q=1. Complete every single empty square. For frequencies, (O-E)^2 / E and χ^2 round to two decimal places. Use whole numbers for counts... people are counted in whole numbers!

To solve the problem, we need to fill in the missing data in the table and perform the necessary calculations. Let's break down the steps:

Given:

• Tasters frequency = \( p^2 + 2pq = 0.5 \)
• Non-tasters frequency = \( q^2 = 0.25 \)

From \( q^2 = 0.25 \), we can find \( q \): \[ q = \sqrt{0.25} = 0.5 \]

Using \( p + q = 1 \): \[ p = 1 - q = 1 - 0.5 = 0.5 \]

Using the allele frequencies \( p = 0.5 \) and \( q = 0.5 \), we can calculate the expected genotype frequencies:

• \( p^2 = (0.5)^2 = 0.25 \)
• \( 2pq = 2 \times 0.5 \times 0.5 = 0.5 \)
• \( q^2 = (0.5)^2 = 0.25 \)

The total number of individuals in BIO 340 is the sum of tasters and non-tasters: \[ \text{Total} = 225 + 75 = 300 \]

Using the frequencies, calculate the expected counts:

• Expected tasters count = \( (p^2 + 2pq) \times \text{Total} = 0.5 \times 300 = 150 \)
• Expected non-tasters count = \( q^2 \times \text{Total} = 0.25 \times 300 = 75 \)

For the USA population, we have:

• Tasters frequency = 0.25
• Non-tasters frequency is missing.

Since \( p + q = 1 \), and assuming the same allele frequencies apply:

• \( q^2 = 0.25 \) (as given for BIO 340)
• Therefore, the non-tasters frequency for the USA is also 0.25.

To perform a chi-square analysis, we need to compare observed and expected counts for BIO 340:

Observed counts:

• Tasters: 225
• Non-tasters: 75

Expected counts:

• Tasters: 150
• Non-tasters: 75

Chi-square formula: \[ \chi^2 = \sum \frac{(O - E)^2}{E} \]

Calculations:

• For tasters: \( \frac{(225 - 150)^2}{150} = \frac{75^2}{150} = \frac{5625}{150} = 37.50 \)
• For non-tasters: \( \frac{(75 - 75)^2}{75} = 0 \)

Total chi-square value: \[ \chi^2 = 37.50 + 0 = 37.50 \]

• Allele frequencies for BIO 340: \( p = 0.5 \), \( q = 0.5 \)
• Expected counts for BIO 340: Tasters = 150, Non-tasters = 75
• USA non-tasters frequency: 0.25
• Chi-square value for BIO 340: 37.50

This completes the analysis and fills in the missing data in the table.