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.
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