30% of 2500 calories = 0.30 * 2500 = 750 calories.

To maintain weight, the weekly calorie intake should be 2500 calories/day * 7 days = 17500 calories.

300 grams * 4 calories/gram = 1200 calories.

Calories from carbohydrates = 300g * 4 cal/g = 1200 calories.

Calories from carbohydrates = 300g * 4 = 1200 calories. Percentage = (1200/2400) * 100 = 50%.

25% of 3000 calories = 0.25 * 3000 = 750 calories from proteins. 750 calories / 4 calories per gram = 187.5 grams, rounded to 188 grams.

Probability of contracting and not recovering = Probability of contracting × (1 - Probability of recovery) = 0.20 × (1 - 0.30) = 0.20 × 0.70 = 0.14.

Percentage = (300 mg / 1500 mg) * 100 = 20%.

20% of 2500 calories = 0.20 * 2500 = 500 calories from fats. 500 calories / 9 calories per gram = 55.56 grams, rounded to 56 grams.

300 mg needed / 100 mg per glass = 3 glasses.

70 grams * 9 calories/gram = 630 calories.

Weekly protein requirement = 70 grams/day * 7 days = 490 grams.

20% of 2500 = 0.20 * 2500 = 500 calories.

To lose 1 pound per week, they need to reduce their intake by 500 calories/day. So, 2500 - 500 = 2000 calories.

The expected genotypic ratio from Aa x aa is 1 Aa to 1 aa, which is 1:1.

The cross Aa x aa will produce offspring in a 1:1 phenotypic ratio (A:a).

The cross Aa x Aa results in a 1/4 chance of AA (homozygous dominant), which is 25%.

The cross AaBb x Aabb results in 1/4 Aabb, so the probability of obtaining Aabb is 25%.

The cross Bb x bb results in a phenotypic ratio of 1 Bb (dominant) to 1 bb (recessive), which is 1:1.

The cross Tt x tt results in 50% Tt (tall) and 50% tt (short), so the fraction of tall offspring is 1/2.

Expected affected individuals = Prevalence × Population = 0.03 × 1000 = 30.

The F1 generation will all be Rr, which expresses the red phenotype since red is dominant over white.

The F1 generation will all be red (Rr) since red is dominant over white.

True positive rate = Sensitivity × Prevalence = 0.9 × 0.01 = 0.009.

Using Bayes' theorem, P(Disease | Positive) = (Sensitivity × Prevalence) / ((Sensitivity × Prevalence) + (1 - Specificity) × (1 - Prevalence)) = (0.9 × 0.01) / ((0.9 × 0.01) + (0.05 × 0.99)) = 0.018.

The cross Aa x aa will produce offspring in a 1:1 phenotypic ratio (A:a).

The son will inherit the X chromosome from his mother. If she is a carrier (X^HX^h), there is a 50% chance he will inherit the X^h and express the trait.

Using the binomial probability formula: P(X=k) = C(n,k) * p^k * (1-p)^(n-k). Here, n=5, k=3, p=0.8. Thus, P(3 out of 5) = C(5,3) * (0.8)^3 * (0.2)^2 = 10 * 0.512 * 0.04 = 0.2048.

Probability that treatment is not effective for one patient = 1 - 0.8 = 0.2. For three patients, the probability that it is not effective for all = 0.2^3 = 0.008. Therefore, the probability that it is effective for at least one = 1 - 0.008 = 0.992.

The probability of infection after vaccination = Initial probability × (1 - reduction) = 0.2 × (1 - 0.75) = 0.2 × 0.25 = 0.05.