Genetics: 1. What are the observed genotype frequencies in this population? 2. H

Question

Genetics:

1. What are the observed genotype frequencies in this population?

2. How many M alleles are in this population and total number of alleles at the M/N locus?

3. What are the allele frequencies in this population?

4. What are the expected genotype frequencies in this population based on the Hardy-Weinberg model?

CASE 1 In a population of 200 people, testing for M/N blood type reveals the genetics and the model, provide the information requested below in the information provided below. Using your knowledge of Mendelian Ho riatebalbfiummdelproveeste appropriate blanks. 140 people are blood type M. 50 people are blood type MN. 50 people are blood type N

Explanation / Answer

First, the number don’t add up as given in the question. There are 240 people (not 200) showing either blood groups M, N, or MN. We will continue solving the problem assuming there is a population of 240 people.

Allelic frequency is the degree of occurrence of an allele in a given population.

[(Number of homozygous x 2) + Number of heterozygous x 1)] / (Total number of individuals x 2)

The genotypic frequency is the occurrence of a genotype in a given population.

Number of individuals exhibiting a genotype / Total number of individuals in the population=genotypic frequency.

According to Hardy-Weinberg law, in a population that is not evolving the sum of the allelic frequencies should be 1. So, if there are two alleles A and B:

f(A) + f(B) = 1.

Squaring the allelic frequency gives the genotypic frequencies as expected by the Hardy-Weinberg principle –

[f(A)+f(B)]2 = f(A)2 + f(B)2 + 2f(A)f(B) = 1.

Since we understand the terms, we can solve the give problem. We must assume that only two alleles M and N are present in order to get an accurate answer. It must be noted that the pattern of expression here is a co-dominant expression where there is no recessive allele.

1.       Observed genotypic frequencies:

a.       For MM:

Number of individuals exhibiting the Genotype MM / Total number of people =140/240

= 0.583

Genotype MM means a phenotype of blood group M. Similarly, genotype of NN means a phenotype of blood N. A genotype of MN signifies the blood group M/N.

b.      For MN:

Using the same formula above but substituting MM with MN = 50/240 = 0.208

c.       For NN:

Using the same formula and plugging in NN = 40/240 = 0.167

2.       There are (2 x homozygous M) + (1 x heterozygous M as in MN) alleles for M. This means (2 x 140) + (1 x 50) alleles of M = 280 + 50 + 330.

While total number of alleles in this locus is (2 x homozygous MM) + (2 x heterozygous MN) + (2 x homozygous NN) = 280 + 100 + 100 = 480

You can also solve this my taking the number of people (240) and since each individual has two alleles, multiple this number by 2 = 240 x 2 = 480.

3.       Allele frequencies are as follows:

a.       For M:

Number of M alleles in the population / Total number of alleles = 330 / 480 = 0.688 = f(M).

Note: we found these values in the previous answer section.

b.      For N:

There are (2 x homozygous N) + (1 x heterozygous N) alleles for N. This equals (2 x 50) + (1 x 50) = 100 + 50 = 150.

Total number of alleles remains 480

Number of M alleles in the population / Total number of alleles = 150 / 480 = 0.312 = f(N).

4.       According to Hardy-Weinberg principle, = 1 = 0.688 + 0.312

f(M) + f(N)

The expected genotypic frequency is [f(M)+f(N)]2 = f(M)2 + f(N)2 + 2f(M)f(N).

f(M)2 = 0.6882 = 0.473 (Homozygous M)

f(N)2 = 0.3122= 0.097 (Homozygous N)

2[f(M)xf(N)] = 2 (0.312x0.688) = 0.430 (Heterozygous MN)

Observed Genotypic Frequency:

                MM=0.583

                NN=0.167

                MN=0.208

Expected Genotypic frequency:

                MM=0.473

                NN=0.097

                MN=0.430

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