Question

Hardy Weinberg assignment

• P + Q = 1
  • In which P represents frequency of dominant allele and Q represents frequency of the recessive allele
• P² + 2PQ + Q² =1
  • P² represents frequency of homozygous dominant
  • 2PQ represents frequency of heterozygous
  • Q² represents frequency of homozygous recessive

1. Consider a population of beetles on an island. There are 1000 beetles and they have different colored wings. Black wings are dominant over silver wings. Calculate the allele and the genotypic frequencies in this population if you are able to observe 64 silver winged beetles

2. Consider a population of mice in your home. 50 mice are present and they have different abilities. The dominant trait is the ability to squeeze into small holes and the recessive ability is to jump up onto cabinets. You have managed to determine that the dominant allele is present 25% of the time. Calculate the remaining recessive allele and the genotypic frequencies

3. Snub nosed monkeys live in the southern region of China. Dominant traits are blue faces and recessive traits are pink faces. You notice in the population of 52 monkeys, 39 of them are blue faced. Calculate the allele frequencies and the genotypic frequencies

4. Bobby pins, strangely enough like other living things, reproduce at a rapid rate when unobserved. Homozygous dominant bobby pin traits are metal while homozygous recessive bobby pins are plastic. Heterozygous bobby pins are a mix of plastic and metal. You find in your drawer 80 bobby pins, of which you’ve noted that 8 are plastic. Give the remining allele, genotypic, and phenotypic frequencies

Answer 1

Hardy Weinberg equations :

1 = p + q

1 = p^2 + 2pq + q^2

1)

Total number of beetles = 1000

Number of silver winged beetles = 64

Frequency of a genotype = Number of individuals with the genotype / Total population

Frequency of silver winged beetles (q^2) = 64/1000 = 0.064

q = 0.253

p = 1 - 0.253 = 0.747

Frequency of homozygous black wings (p^2 ) = 0.747 * 0.747 = 0.558

Frequency of heterozygous black wings (2pq) = 2 * 0.747 * 0.253 = 0.378

Frequency of homozygous recessive (silver wings) = q^2 = 0.253 * 0.253 = 0.064

0.558 + 0.378 + 0.064 = 1

2)

Total mice = 50

Frequency of dominant allele (p) = 25% = 0.25

Frequency of recessive allele (q) = 1 - 0.25 = 0.75

Frequency of homozygous dominant (p^2) = 0.25 * 0.25 = 0.0625

Frequency of heterozygous (2pq) = 2 * 0.25 *0.75 = 0.375

Frequency of homozygous recessive (q^2) = 0.75 *0.75 = 0.5625

0.0625+ 0.375 + 0.5625 = 1

3)

Total population = 52

Number of blue faced monkeys = 39

Number of pink faces = 52 - 39 = 13

Frequency of pink faces (q^2) = 13/52 = 0.25

Frequency of recessive allele (q) = 0.5

Frequency of dominant allele (p) = 1 - 0.5 = 0.5

Frequency of homozygous dominant (blue faces) = p^2 = 0.5 *0.5 = 0.25

Frequency of heterozygous (blue faces) = 2pq = 2 * 0.5 * 0.5 = 0.5

Frequency of homozygous recessive (pink faces) = q^2 = 0.5 *0.5 = 0.25

0.25 + 0.5 + 0.25 =1

4)

Total bobby pins = 8

Number of plastic bobby pins = 8

Frequency of plastic bobby pins (q^2) = 8/80 = 0.1

Frequency of recessive allele (q) = 0.32

Frequency of dominant allele (p) = 1 - 0.32 = 0.68

Frequency of homozygous dominant (metal) = p^2= 0.68 *0.68 = 0.46

Frequency of heterozygous (mix of metal and plastic) = 2pq = 2 *0.68 * 0.32 =0.44

Frequency of homozygous recessive (plastic) = q^2 = 0.32 *0.32 = 0.1

1 = 0.46 + 0.44 + 0.1