Question

3) Your parents' generation has 80MM, 240MN, and 180NN individuals in each genotype for the M & N blood group. Your generation (siblings, cousins, yourself) has 15 MM, 11MN, and 24 NN individuals. Is your generation in Hardy-Weinberg Equilibrium with your parents? 4) There is a place where blue and white squirrels live. The blue allele (B) is dominate over the white allele (w) and has a frequency of 0.8. That is, B -0.8. In a population of 100 individuals, how many white squirrels would you expect?

Answer 1

3)

Hardy-Weinberg Equations :

1 = P2 + 2pq + q2

1 = p + q

Generation of Parents :

Total population = 80+240+180= 500

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

Frequency of MM ( p2) = 80/500 = 0.16

p = 0.4

q = 1- 0.4 = 0.6

q2 = 0.36

2pq = 2*0.4*0.6 = 0.48

Applying Hardy-Weinberg equation,

1 = p2 + 2pq + q2

0.16 + 0.36 + 0.48 =1

Generation of childrens:

Total population = 15 + 11 +24 = 50

Frequency of MM (p2) = 15/50 = 0.3

p = 0.547

q = 1 - 0.547 = 0.453

q2 = 0.205

2pq =0.495

Applying Hardy - Weinberg equation,

1 = p2 + 2pq + q2

0.3 + 0.495 + 0.205 = 1

Both parents' generation and our generation are in Hardy - Weinberg equilibrium

4)

Given figures :

Total population = 100

frequency of dominant allele B (p) = 0.8

Hardy weinberg equations:

1 = p2 + 2pq + q2

1 = p + q

p = 0.8

q = 1 - 0.8 = 0.2

frequency of recessive allele w (q) = 0.2

genotype ww (q2) = 0.04

Frequency of a genotype = number of individuals with the genotype / total number of individuals in the population

therefore ,

Number of white squirrels (ww) = frequency of genotype ww (q2) * total individuals

= 0.04 * 100 = 4

therefore, Number of white squirells = 4