## Answers

**2. a)** As 556 individuals are studied, total number of alleles will be (556 x 2) or 1112 (Because each individual will have 2 alleles).

Now, **frequency of LM allele** = Number of LM allele / Total number of alleles = (167 x 2 + 280) / 1112 = 614/1112 = **0.5522** (Up to 4 decimals)

So, **frequency of LN allele** = Number of LN allele / Total number of alleles = (109 x 2 + 280) / 1112 = 498/1112 = **0.4478** (Up to 4 decimals)

**b)** Under Hardy-Weinberg equilibrium-

**Expected number of MM** = (Frequency of LM allele)^{2} x Total population = (0.5522)^{2} x 556 = **169.54** **170**

**Expected number of NN** = (Frequency of LN allele)^{2} x Total population = (0.4478)^{2} x 556 = **111.49** **111**

**Expected number of MN** = 2 x Frequency of LM allele x Frequency of LN allele x Total population = 2 x 0.5522 x 0.4478 x 556 = **274.97** **275**

Now, = + + = 0.1799 (Up to 4 decimals)

Now, degrees of freedom = 1

From the distribution table we find that probability of = 0.1799 with degrees of freedom 1 falls between 0.1 & 0.9. As the probability is greater than 0.05, we can say that difference between observed & expected values are due to chance alone & no significant difference exist between them. So, the genotype frequencies fit to the Hardy-Weinberg distribution.