quiz pool 10

Quiz 10 Question Pool 2026

Question 1

a. What are p & q? (2 points)
  - Definitions:
    - p: Frequency of the dominant allele in a population.
    - q: Frequency of the recessive allele in a population.
b. What is their sum?
  - Calculation:
    - $p + q = 1$
    - This equation states that the total frequency of alleles in a given population equals 1.

Question 2

a. Complete the following:
  - Formula:
    - (p + q)^2 = p^2 + 2pq + q^2 = ________
    - This formula represents the expansion of the binomial (p + q) squared, which relates to the genotype frequencies in a population
      - p^2: Frequency of homozygous dominant genotype.
      - 2pq: Frequency of heterozygous genotype.
      - q^2: Frequency of homozygous recessive genotype.
b. Which term represents the frequency of heterozygotes?
- Answer:
- The term 2pq represents the frequency of heterozygotes in a population.

Question 3

a. Does the dominant phenotype always have to be the prevalent one in a population?
- Answer:
- No, the dominant phenotype does not always have to be the most common in a population.
b. Give an example which illustrates your answer.
- Example:
- Consider a population where the dominant allele is for brown fur (B), while the recessive allele is for white fur (b). If the majority of the population carries the recessive allele (e.g., BB = 30%, Bb = 10%, bb = 60%), the dominant phenotype (brown fur) can be less prevalent despite having a dominant allele.

Question 4

What conditions have to be met for a population to be in Hardy-Weinberg equilibrium?
  - The following conditions must be satisfied:
    1. No mutations: Genetic changes should not occur.
    2. Random mating: All individuals must have an equal opportunity to reproduce.
    3. No natural selection: All phenotypes must have equal survival and reproductive rates.
    4. Large population size: To minimize genetic drift.
    5. No migration: No individuals can enter or leave the population, ensuring allele frequencies remain stable.

Question 5

In Hardy-Weinberg equilibrium:
  - a. What is the frequency (probability) of homozygous dominant individuals?
    - Answer:
    - The frequency of homozygous dominant individuals is given by $p^2$ .

  - b. What is the frequency (probability) of heterozygotes?
    - Answer:
    - The frequency of heterozygotes is given by $2pq$ .

  - c. What is the frequency (probability) of homozygous recessive individuals?
    - Answer:
    - The frequency of homozygous recessive individuals is given by $q^2$ .

  - d. What is the frequency (probability) of the dominant phenotype?
    - Answer:
    - The frequency of the dominant phenotype can be expressed as:
      - $ext{Frequency of dominant phenotype} = p^2 + 2pq$ .

Question 6

Genetic testing with PTC tasting:
  - The ability to taste PTC is controlled by a dominant allele T, while individuals homozygous for the recessive allele t cannot taste it.
  - In a class of 125 students:
    - Tasters (T): 88 students can taste PTC.
    - Non-tasters (t): 37 students cannot taste PTC.
  - Calculate the frequency of T and t alleles:
    1. Let p = frequency of T allele and q = frequency of t allele.
    2. Since $p + q = 1$ , we find q using the recessive phenotype as follows:
       - $q^2 = rac{37}{125} = 0.296$ (the frequency of non-tasters = homozygous recessive).
       - $q = ext{sqrt}(0.296) ext{ and } p = 1 - q$ .

  - Frequency of genotypes:
    - Use the allele frequencies to calculate:
      - Homozygous dominant (TT): $p^2$ ,
      - Heterozygous (Tt): $2pq$ ,
      - Homozygous recessive (tt): $q^2$ .

Problems from the 7th edition of the textbook:

Problem 24-4: Frog Population Analysis

Population Characteristics:
    - Green Frogs: 120
    - Brownish Green Frogs: 60
    - Brown Frogs: 20
    - Alleles: Brown allele ( $GB$ ), Green allele ( $GG$ )
a. What are the genotypic frequencies in the population?
b. What are the allelic frequencies of GB and GG in this population? (2 points)
c. What are the expected frequencies of the genotypes if the population is at Hardy-Weinberg equilibrium? (2 points)

Problem 24-5: Population Equilibrium Check

Evaluate the following populations for Hardy-Weinberg equilibrium:

Population	AA	Aa	aa
A	0.25	0.50	0.25
B	0.10	0.74	0.16
C	0.64	0.27	0.09
D	0.46	0.50	0.04
E	0.81	0.18	0.01

Problem 24-7: Blood Type Frequencies

Starting Proportions for Blood Types:
   - M: 0.5
   - MN: 0.2
   - N: 0.3
   - Note: Blood type determined by codominant alleles M and N.
a. Is this population at Hardy-Weinberg equilibrium? (3 points)
b. What will be the allele and genotype frequencies after one generation? (2 points)
c. What will be the allele and genotype frequencies after two generations? (2 points)

Problems from the 7th edition of the textbook:

Problem 24-4: Frog Population Analysis

Population Characteristics:
   - Green Frogs: 120
   - Brownish Green Frogs: 60
   - Brown Frogs: 20
   - Alleles: Brown allele ( $GB$ ), Green allele ( $GG$ )
a. What are the genotypic frequencies in the population?
   - Total frogs: 200
   - Genotype frequencies:
     - Frequency of $GB$ (Brown) = $\frac{20}{200} = 0.1$
     - Frequency of $GG$ (Green) = $\frac{120 + 60}{200} = 0.9$
b. What are the allelic frequencies of GB and GG in this population? (2 points)
   - Allelic frequencies:
     - Frequency of $GB$ = $\frac{GB}{Total alleles} = \frac{20}{400} = 0.05$
     - Frequency of $GG$ = $\frac{200}{400} = 0.5$
c. What are the expected frequencies of the genotypes if the population is at Hardy-Weinberg equilibrium? (2 points)
   - Expected frequencies:
     - Frequency of homozygous $GB GB$ = $p^2 = (0.05)^2 = 0.0025$
     - Frequency of heterozygotes $2pq = 2(0.05)(0.5) = 0.05$
     - Frequency of homozygous $GG$ = $q^2 = (0.5)^2 = 0.25$

Problem 24-5: Population Equilibrium Check

Evaluate the following populations for Hardy-Weinberg equilibrium:

Population	AA	Aa	aa	Equilibrium?
A	0.25	0.50	0.25	Yes
B	0.10	0.74	0.16	No
C	0.64	0.27	0.09	Yes
D	0.46	0.50	0.04	No
E	0.81	0.18	0.01	Yes

Problem 24-7: Blood Type Frequencies

Starting Proportions for Blood Types:
   - M: 0.5
   - MN: 0.2
   - N: 0.3
   - Note: Blood type determined by codominant alleles M and N.
a. Is this population at Hardy-Weinberg equilibrium? (3 points)
   - Calculate the expected genotype frequencies:
   - Frequency of M allele = $p = 0.5$
   - Frequency of N allele = $q = 0.5$
   - Expected proportions:
     - MM = $p^2 = 0.25$
     - MN = $2pq = 0.5$
     - NN = $q^2 = 0.25$
   - Since the observed proportions (M: 0.5, MN: 0.2, N: 0.3) do not match the expected proportions, it is not in equilibrium.
b. What will be the allele and genotype frequencies after one generation? (2 points)
- Assuming random mating and effective wash of allele frequencies:
- Allele frequencies remain the same in the next generation, unless acted on by evolutionary processes.
c. What will be the allele and genotype frequencies after two generations? (2 points)
- After two generations, assuming equilibrium, allele frequencies remain constant, as genetic drift, gene flow, or selection isn't acting upon the population.