Answer:
Explanation:
Hardy and Weinberg described all the possible genotypes for a gene with two alleles. The binomial expansion representing this is, p2 + 2pq + q2 = 1.0
Where,
p2 = proportion of homozygous dominant individuals (BB) = 313
q2 = proportion of homozygous recessive individuals (RR) = 857
2pq = proportion of heterozygotes (BR) = 1820
The proportion of BB individuals in the population is = 313/2990 = 0.1046
The proportion of BR individuals in the population is = 1820/2990 = 0.6086
The proportion of RR individuals in the population is = 2/82 = 0.2866
I).
a). The frequency of p allele in the population is, (B) = p2 + 1/2(2pq) = 0.1046 + ½ (0.6086) = 0.4089
b). The frequency of q allele in the population is, (R) = q2 + 1/2(2pq) = 0.2866 + ½ (0.6086) = 0.5909.
II).
The expected number of individuals with the BB genotype = 0.4089* 0.4089 = 0.1671*2990 = 499.6 = 500
The expected number of individuals with the BR genotype = 2*0.4089* 0.5909 = 0.4832*2990 = 1444.768 = 1445
The expected number of individuals with the RR genotype = 0.5909*0.5909 = 0.3491*2990 = 1043.809 = 1044
CHI - SQUARE (X2):
X2 = Σ(O - E)2 / E
Where O = Observed frequency
E = Expected frequency
Phenotype O E (O-E) (O-E)^2 (O-E)^2/E
BB 313 500 -187 34969 69.938
BR 1820 1445 375 140625 97.31834
RR 857 1044 1.5 2.25 0.155172
2990 2989 189.5 167.4115
The calculated Chi-square value is = 167.4115
Degrees of freedom is = n-1 = 3-1 = 2
The P-value is < 0.00001, which is significant at p < 0.05. So, we reject the null hypothesis.
Conclusion: There is a significant difference between the observed and expected values.
