Hardy-Weinburg (4.5)

Question 1

Give the 5 conditions the Hardy-Weinburg formula is dependent upon

Answer 1

No mutation, ransom mating, no gene flow, infinite population size and no selection

Question 2

Explain why the H-W formula cannot be used for sex-linked conditions

Answer 2

Neither X or Y chromosome can be considered recessive or dominant therefore it cannot be applied to Weinburg’s equation which involves dominant and recessive alleles.

Question 3

Give the two formulas needed for Hardy-Weinburg calculations and provide a key

Answer 3

q= recessive p=dominant

q^2+2pq+p^2

Question 4

In a population of 600 individuals, 14 have Cystic fibrosis (i.e. they are homozygous recessive). Calculate the number of heterozygous individuals in this population

Answer 4

14/600= 0.02333…..
q^2= 0.02
q= 0.14
p= 1- 0.14 = 0.86
2pq = (2 x 0.14 x 0.86)
= 0.24

Question 5

Explain how reproductive isolation can result in speciation (the formation of new species)

Answer 5

Mechanisms of reproductive isolation act as barriers between closely related species, enabling them to diverge and exist as genetically independent species.