Genetics Population

Genetics Population

Data :

In a certain area there is bird population, amount: 3629 species,  which is found the following genotype distribution among the bird sampled : 1888 were BB, 1212 were Bb, and 529 werebb:

1.    Calculate the allele frequencies of B and b,

2.    Expected numbers of the three genotypic classes (assuming random mating).

3.    Using X², determine whether or not this population is in Hardy-Weinberg equilibrium.

·        p to the frequency of the dominant allele : BB

·        q to the frequency of the recessive allele : bb

Ø  OBSERVED GENOTYPE FREQUENCIES

–       BB(p²)         : 1888/3629 = 0.520

–       Bb(2pq)         : 1212/3629 = 0.334

–       bb(q2  )           :529/3629 = 0.146

Ø  ALLELE FREQUENCIES

-       Freq of B = p = p² + 1/2 (2pq) = 0.520 + 1/2 (0.334) = 0.520  + 0.167  = 0.687

-       Freq of b = q = 1-p = 1 -  = 0.313.

Ø  EXPECTED GENOTYPE FREQUENCIES (assuming Hardy-Weinberg): IF population is in HWE, then we’d expected the following frequencies:

-       BB (p²) = (0.687)² = 0.472

-       Bb (2pq) = 2 (0.687)(0.313) = 0.430

-       bb (q²) = (0.313)² = 0.098

Ø  EXPECTED NUMBER OF INDIVIDUALS of EACH GENOTYPE:

-       # BB = 0.472 X 3629 = 1713

-       # Bb = 0.430 X 3629 = 1560

-       # bb = 0.098 X 3629 = 356

Genotype	Expected Numbers	Observed Numbers
BB	0.472 X 3629 = 1713	0.520 X 3629 = 1887.8
Bb	0.430 X 3629 = 1560	0.334 X 3629 = 1212
bb	0.098 X 3629 = 356	0.146 X 3629 = 529.8

X² = [(1887.8 - 1713)²/ 1713] + [(1212 - 1560)² / 1560] + [(529.8 - 356)²/ 356]

= 17.837 +85.661+84.849

= 188.347

X²_(Calculated)>X²_(table) [3.841, 1 df, 0.05 ls]

The critical chi-square value for 1 degree of freedom is 3.841.  Since 188.347is greater than this, we absolutely reject the null hypothesis that the population is in Hardy-Weinberg equilibrium.

Nb: kalo hasil X²kalian lebih kecil dari 3.841 berarti:

X²_(calculated)<X²_(table) [3.841, 1 df at 0.05 ls).

Therefore, fail to reject null hypothesis and conclude that the population is in HWE