Question

2. 10-points: The MN blood group in humans is under the control of a pair of co-dominant alleles LM and LN. I a group of 556 individuals, the following genotypic frequencies are found: 167 MM 280 MN 109 NN a) Calculate the frequency of the LM and LN alleles b) Test Using the Chi-squared Test whether the genotype frequencies fit to the Hardy-Weinberg distribution. Note: the degrees of freedom is interestingly "1" since while there are genotypes, these are determined by only 2 allelic frequencies. PCX SX 0.010 0.025 0.050 0.100 0.900 0.950 0.975 0.990 X ) Xans(r) xaus(r) An (). Xájom) Xas(n) Roces() () 0.001 0.004 0.016 1841 5.024 6.615 0.020 0.051 0.103 0.211 4.005 5.991 7.378 9.210 0.115 0.216 0.352 O.SN4 7.815 OLEN 0.297 0.484 0.711 11.14 13.28 0.554 O.R31 1.145 1.610 92.16 11.07 12.3 15.09 0.872 1.237 1.635 2.2014 1064 12.59 14.45 16.81 1219 1.00 2.167 12.02 14.07 16.01 18.48 2INO 2.733 3.4 1116 15.51 17.54 20.09 2.ORN 1125 16.92 19.02 21.67 2.55 3.9-40 15.09 18:31 20.48 212 2.700

Answer 1

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,
(Observed - Expected)? Expected
=
(167-170) 170
+
(109-11152 111
+
(280-275) 25
= 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.