Question

1B. In a hypothetical small population of only 100 frogs there exists a gene with two alleles, H and h. A scientists determines that 13 of the frogs are homozygous dominant (HH), and 37 homozygous recessive (hh).

calculate the allele frequencies for the next generation of frogs, and assume NO homozygous individuals were able to reproduce.

Answer 1

Total population of frog=100

Population consists of 2 alleles i.e. H and h. We have been given population of frog having different genotypes. Homozygous dominant (HH)=13; Homozygous recessive (hh) = 37;

We can find the number of heterozygotes (Hh)=

HH + hh = 13+37 = 50

Number of heterozygotes= total population - ( HH + hh)

= 100 - 50

= 50 (Hh)

We have been given that Homozygous individuals are not able to reproduce. This means heterozygous individuals reproduce and their population is 50.

Hh cross Hh

	H	h
H	HH	Hh
h	Hh	hh

HH = 1/4 ; Hh= 2/4 or 1/2 ; hh = 1/4

Population with

A) HH = 1/4 *50 = 12.5

B) Hh = 2/4 * 50 = 25

C) hh = 1/4*50 = 12.5

To find frequency of allele = number of individuals produced / total population

A) HH = 12.5/50 = 0.25

H^2= 0.25 =p^2

H =√0.25 = 0.5 = p

B) Hh = 25/50 = 0.5

2pq = 0.5 ( According to Hardy-Weinberg principle)

C) hh = 12.5/50 = 0.25

h ^2= 0.25 = q^2

h = √0.25 =0.5 =q

We can check our answer by putting values in Hardy-Weinberg equation = p^2 + 2pq + q^2 =1

Put values and it will be equal to 1.

0.25 + 2*(0.25+0.25) + 0.25

= 0.25. + 0.5 + 0.25

= 1

Thank you...