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Population Genetics Problem Solving Worksheet Key, Exercises of Genetics

The Hardy Weinberg Model worksheet questions with solutions

Typology: Exercises

2020/2021

Uploaded on 04/20/2021

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Population Genetics
Problem-Solving
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Population Genetics

Problem-Solving

Hardy-Weinberg Equilibrium

The Hardy-Weinberg Proof:

If the frequency of the dominant and recessive alleles change

from one generation to the next, then, by definition,

microevolution has occurred.

Genetic Term Hardy-Weinberg Term

Dominant allele (D)

Recessive allele (d)

Homozygous recessive (dd)

Homozygous dominant (DD)

Heterozygous (Dd)

p

q

q

2

p

2

2pq

D ( p ) d^ ( q )

D ( p )

d ( q )

Frequency of Dominant and recessive alleles: 0.

pโˆ™p = p

2

q โˆ™ q = q

2

pโˆ™q

pโˆ™q

p + q = 1 and p

2

+ 2 pq + q

2

0.5 + 0.5 = 1 and 0.25 + 0.5 + 0.25 = 1

Dark fur is dominant to light fur. 64 rock pocket mice have light & 36 have dark fur. **p + q = 1 and p 2

  • 2pq + q 2 = 1**
  1. What is the frequency of the homozygous recessive genotype ( q 2 =# of individuals with the recessive phenotype divided by the total )?
  2. What is the frequency of the homozygous recessive allele ( q )?
  3. What is the frequency of the dominant allele ( p = 1 โ€“ q )?
  4. What is the frequency of the homozygous dominant genotype ( p 2 )?
  5. What is the frequency of the heterozygous genotype (2 pq )?
  6. Thirty years ago p was 0.6 and q was 0.4. Has the population experienced microevolution? Why/why not?

Dark fur is dominant to light fur. 64 rock pocket mice have light & 36 have dark fur. **p + q = 1 and p 2

  • 2pq + q 2 = 1**
  1. What is the frequency of the homozygous recessive genotype ( q 2 =# of individuals with the recessive phenotype divided by the total )? q 2 = 64/100 = 0.
  2. What is the frequency of the homozygous recessive allele ( q )? q = ๐ŸŽ. ๐Ÿ”๐Ÿ’ = 0.
  3. What is the frequency of the dominant allele ( p = 1 โ€“ q )? p = 1 โ€“ 0.8 = 0.
  4. What is the frequency of the homozygous dominant genotype ( p 2 )? p 2 = (0.2) 2 = 0.
  5. What is the frequency of the heterozygous genotype (2 pq )? 2 pq = 2(0.2)(0.8) = 0.
  6. Thirty years ago p was 0.6 and q was 0.4. Has the population experienced microevolution? Why/why not? Yes, because the allele frequencies changed
  1. What frequency of the population is homozygous recessive? 81/100 = 81% or 0.81= q 2
  2. What is the frequency of the recessive allele ๐’’ ๐Ÿ = ๐ŸŽ. ๐Ÿ–๐Ÿ ; q = 0.
  3. What is the frequency of the dominant allele? p + 0.9 = 1; p = 0.
  4. What is the frequency of the dominant allele? p 2 = (0.1) 2 = 0.
  5. What frequency is heterozygous for fur-color? 2 pq = 2(0.1)(0.9) = 0.
  6. Thirty years ago q 2 was 0.9, p 2 was 0.01 and 2pq was 0.09. Has the population experienced microevolution? Why/why not? Yes, the population has experienced microevolution because allele frequencies have changed.

Dark fur is dominant to light fur. 1 rock pocket mouse has light- colored & 99 have dark fur.

  1. What is the frequency of the homozygous recessive genotype?
  2. What is the frequency of the recessive allele?
  3. What is the frequency of the dominant allele?
  4. What frequency is homozygous dominant?
  5. What frequency is heterozygous for fur-color?
  6. Thirty years ago q 2 was 0.16, p 2 was 0.34 and 2pq was 0.5. Has the population experienced microevolution? Why/why not?
  1. What is the frequency of light mice caught on light substrate? (6+80+34) / (6 + 85 + 77) = 0. What is the frequency of dark mice caught on light substrate? (5 + 43) / (6 + 85 + 77) = 0.
  2. What is the frequency of light mice caught on dark substrate? 3 / (7 + 5 + 45) = 0. What is the frequency of dark mice caught on a dark substrate? (7 + 5 + 42) / (7 + 5 + 45) = 0.
  1. Light granite substrate dd = (6 + 80 + 34) / (6 + 85 + 77) = 0. q 2 = 0.71, square root, q = 0.84, p + q = 1, so p = 0.16 and p 2 = (0.16) 2 = 0. 2pq = 2(0.16)(0.84) = 0.
  2. Dark substrate dd = 3 / (7 + 5 + 45) = 0. q 2 = 0.05, square root, q = 0.22, p + q = 1, so p = 0.78 and p 2 = (0.78) 2 = 0. 2pq = 2(.78)(.22) = 0.