5 You are studying the MN blood group in a small village pop

5. You are studying the MN blood group in a small village population in Borneo and want to know if the population is in Hardy–Weinberg (HW) equilibrium. MN blood type is due to a single-gene locus segregating the two co-dominant M and N blood type alleles (LM, LN). The following are sample data from this population: Phenotypes M MN N Total # genotypes 287 665 123 1075 A. Calculate the genotypic frequencies. (3 points) B. Calculate frequencies for the LM and LN alleles. (2 points) C. Use a 2 analysis and determine whether this population is in H-W equilibrium.

Solution

. You are studying the MN blood group in a small village population in Borneo and want to know if the population is in Hardy–Weinberg (HW) equilibrium. MN blood type is due to a single-gene locus segregating the two co-dominant M and N blood type alleles (LM, LN). The following are sample data from this population: Phenotypes M MN N Total # genotypes 287 665 123 1075 A. Calculate the genotypic frequencies. (3 points) B. Calculate frequencies for the LM and LN alleles. (2 points) C. Use a c2 analysis and determine whether this population is in H-W equilibrium.

Genotype frequencies:

L^ML^M= 287/1075 = 0.267

L^ML^N= 665/1075 =0.619

L^NL^N= 123/1075 = 0.114

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By using gene count method

b. Allele frequencies:

For the L^M allele: 0.267 +0.619/2=0.5765

Total number of all alleles = 1239 + 665 +2(123) = 2150.

For the L^N allele: =0.114 +0.619/2= 0.424

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c.

p= frequency of L^{M -}--------------^-(0.576),

and

q= frequency of   L^N-----------------^--(0.424)

.At equilibrium,p2+2pq+q2= 1.

The observed number of L^ML^Mindividuals = 287;

the expected number isp 2(1075) = (0.576)2(1075) = 357.

The observed number of L^NL^Nindividuals = 123;

the expected number =q2= (0.424)2(1075) = 193.

The observed number of L^ML^N individuals= 665;

the expected number = 2pq(1075) = 2(0.576)(0.424)(1075) = 525

Observed

expected

(O-E)^2/E

P^2

L^ML^M

287

357

(287 – 357)^2/(357)

13.7

2pq

L^ML^N

665

665

(665 – 525)^2/(525)

25.4

q^2

L^NL^N

123

193

(123 – 193)^2/(193)

37.3

1075

76.4

Degrees of freedom = 3 – 1 – 1 = 1.

Chi-square critical value = 3.814, which is much less than 76.4.

Therefore, this is a highly significant chi-square statistic, and the hypothesis must therefore be rejected.

We conclude that the M-N allele frequencies are not in equilibrium in this population

		Observed	expected	(O-E)^2/E
P^2	LMLM	287	357	(287 – 357)^2/(357)	13.7
2pq	LMLN	665	665	(665 – 525)^2/(525)	25.4
q^2	LNLN	123	193	(123 – 193)^2/(193)	37.3
		1075			76.4