Divide the contents of a large bag of different-colored candies r...

Question

Divide the contents of a large bag of different-colored candies randomly and approximately equally among the members of the group. Do not pick specific candy colors, but simply empty the contents of the bag onto a table and quickly divide the pile. If you are doing this exercise by yourself, divide the contents of the bag into five piles.

Tabulate the total number of candies of each color in the original bag by combining the numbers from each person. Use these numbers to determine the frequency of each color in the original bag.

Accepted Answer

Hey, everyone. Let's take a look at this question together in a population of individuals. A gene has two alleles dominant A and recessive A. The frequency of the dominant A allele is 0.6. And the frequency of the recessive A allele is 0.4. What are the expected frequencies of the homozygous dominant A heterozygous and homozygous recessive a genotypes under the Hardy Weinberg equilibrium. And we have four different answer choices containing different frequencies or each of the three genotypes which we have to determine which of the following answer choices is correct. So first, we note we are talking about Hardy Weinberg equilibrium which involves the equation P squared plus two P Q plus Q squared is equal to one. And we can recall from that equation that the frequency of the homozygous dominant genotype is represented as P squared. The frequency of the heterozygous genotype is represented as two P Q. And the frequency of the homozygous recessive genotype is represented as Q squared. And we know that the frequency of the dominant A allele is 0.6. And the frequency of the recessive A allele is 0.4 And we know that the frequency of the dominant A allele is represented as P and the frequency of the recessive A allele is represented as Q, which means that the frequency of the homozygous dominant A allele, which is P squared is 0.6 squared, which is P squared. And then Q squared which is that frequency of the homozygous recessive genotype. So we have 0.4 squared which 0. squared is equal to 0.36 0.4 squared is equal to 0.16, which is the frequency of that homozygous dominant A and Homozygous recessive A genotypes. And looking at our answer choices, we only have one answer choice where the expected frequency of the homozygous dominant A is 0.36. And the Homozygous recessive A is 0.16 which is answer choice. A. So answer choice A is the correct answer. And we can also double check that the frequency of that heterozygous is 0.48. Since two P Q is that frequency of the heterozygous genotype and two P Q is equal to two multiplied by 0.6, multiplied by 0.4, which is equal to 0.48. Thus confirming answer choice A is the correct answer. I hope you found this video to be helpful. Thank you and goodbye.

Key Concepts

Random Sampling

Frequency Distribution

Equitable Division

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