Question

1. 525 jackrabbits in the desert. Big ears (B) are dominant among jackrabbits, while stubby ears (b) are recessive. Of the study of this population, 189 rabbits have stubby ears.

Calculate:
1.The frequency of the "bb" genotype
2.The frequency of the "b" allele
3.The frequency of the "B" allele
4.The frequencies of the genotypes "BB" and "Bb"
5.The frequencies of the two possible phenotypes in your study population.

Answer 1

B+b=1,

(B+b)^2=BB+2Bb+bb=1

genotype BB=Big ears

genotype Bb=Big ears

genotype bb=Stubby ears

Since each rabbit has two alleles for ear type, 525*2=1,050 total alleles

Since 189 rabbits have the genotype bb, 189/525= 0.36, the phenotype frequency for stubby ears. This means that 1-0.36=0.64, the phenotype frequency for big ears.

plugging in the value for bb that we obtained into BB+2Bb+bb=1, we get

BB+2Bb+0.36=1 rearrangement gives,

BB+2Bb=0.64

Since the genotype frequency is 2;1 for Bb:BB, I will let BB=x, and Bb=2x, and solve for x.

x+2x=0.64
3x=0.64
x=0.64/3

x=0.213, meaning 21.3% of the remaining population of jackrabbits have the genotype BB (112 jackrabbits), and the other 42.6% have the genotype Bb (224 jackrabbits). Converting to decimals gives you the genotype frequencies:

Genotype BB frequencies= 0.213
Genotype Bb frequencies= 0.426


To determine the frequency of B alleles, multiply [(112*2)+(224*(1/2))]/total number of alleles(1,050)=0.32

Since B+b=1, solving for b

b=1-0.32=0.68

Answer 2

Remember, the 0.36 that wal calculated for the phenotype frequency of bb is the same as the genotype frequency.