Question

 An Amazon rainforest population of leaf beetles has 100,000 individuals. The beetles exist in two color morphs, purple and green. Purple beetles live on epiphytic orchid blooms whereas the green beetles inhabit the leaves of the orchid plants. The color is controlled by two alleles at a single gene locus. The purple allele (P) is dominant to the green allele (p) 36% of the beetle population is green and the other 64% of the population is purple. Assuming the population is in Hardy-Weinberg answer the following questions. (Show all your calculations below) 

(a) What is the frequency of the recessive genotype, pp? _______ The homozygous dominant genotype, PP? _______ The heterozygous genotype, Pp? _______ 

(b) How many beetles in the population are green? _______ purple?_______  (express as actual numbers of beetles, not percentages) 

(c) What would be the percentages of the body color phenotypes after 10 generations if the population maintains the conditions of Hardy-Weinberg equilibrium? 

Percentage of green phenotype? _______ 

Percentage of purple phenotype? _______  

(d) What would the allele frequencies be if a population bottleneck occurred and only 1000 individuals survived: 250 green, 500 heterozygous purple and 250 homozygous purple beetles? 

(Show all your calculations below) 

Frequency of P? _______ Frequency of p?

(f) Has the population evolved?___ Explain_______ 

Answer 1

The question should can be solved using the Hardy-Weinberg Principle.

The principle states that allele frequencies in a population are stable and is constant from generation to generation. The gene pool ( total genes and their alleles in a population) remains a constant in the absence of disturbing factors.

However the two information you need from this principle to solve this question are:

• The sum of total allelic frequencies is 1 (
  p+q=1
  ) ., where p is the allele frequency of P(purple) and q is the allele frequency of p(green).
• In a diploid, p and q represent the frequency of allele T and allele t respectively. The frequency of PP(homozygous purple) is represented by $p^2$ . The frequency of Pp(heterozygous purple) is represented by 2pq. And the frequency of pp(Short) is represented by $q^2$ .

Given,

Percentage of population that is green = 36%

Percentage of population that is purple= 64%

Therefore,

Frequency of green beetles= $q^2$ = 36/100= 0.36

Therefore, frequency of recessive(p) allele= q= = 0.6.

0.361/2

As from the first bullet point we know,

p+q=1

or, p= 1-q

or, p= 1-0.6

or, p= 0.4

a)

The frequency of recessive genotype pp=  $q^2$= 0.6x0.6 = 0.36

The frequency of homozygous dominant genotype PP= $p^2$ = 0.4 x0.4 = 0.16

The frequency of heterozygous genotype Pp= 2pq= 2x0.4x0.6= 0.48

b)

36% of 100,000 beetles are green

= 36/100 x 100,000 = 36,000 green beetles.

64% of 100,00 beetles are purple

=64/100 x 100,000 = 64,000 purple beetles.

c)

Hardy-Weinberg principle states that allele frequencies are constant from generation to generation. Therefore the phenotype percentages stay the same.

So in 10th generation also,

36% green phenotype

64% purple phenotype

d)

Total number of living individuals= 1000

Number of homozygous purple(PP)=250

Number of heterozygous purple(Pp)= 500

Number of homozygous green(pp)=250

Therefore,

Frequency of green(pp) individuals =  $q^2$= 250/1000 = 0.25

Therefore

Frequency of allele p = q= 0.5

we know, p+q=1

Therefore, frequency of allele P = p = 1-q

= 1- 0.5 = 0.5

Frequency of allele P = 0.5

Frequency of allele p= 0.5

f)

Yes, the population has evolved as the allele frequencies have changed from the initial value so we can conclude that evolutionary forces are at play.