Genotypes of leopard frogs from a population in central Kansas were determined f
ID: 38587 • Letter: G
Question
Genotypes of leopard frogs from a population in central Kansas were determined for a locus (M) that encodes the enzyme malate dehydrogenase. The following #'s of genotypes were observed: Total sample =107
b. What are the actual genotypic frequencies in the above populations? What would the expected numbers of genotypes be if the population were in Hardy- Weinberg equilibrium?
Actual genotypic frequencies:
M1M1 = 20/107 = 0.1869 --> 0.2 M1M2 = 45/107 = 0.42056 -->0.42 M2M2 = 42/107 = 0.3925 --> 0.39
For population in Hardy- Weinberg equilibrium:
M1M1 = p2 = (0.186)2 = 0.034 x 107 = 3.73 --> 4 frogs with M1M1
M1M2 = 2(pq) = (0.420) = 0.841 x 107 = 89.99 --> 90 frogs with M1M2
M2M2 = q2 = (0.392)2 = 0.154 x 107 = 16.4 --> 16 frogs with M1M1 This equals 110....due to rounding but is all of this correct?
Genotype Number M1M1 20 M1M2 45 M2M2 42Explanation / Answer
Actual genotypic frequencies:
M1M1 = 20/107 = 0.1869 --> 0.2 M1M2 = 45/107 = 0.42056 -->0.42 M2M2 = 42/107 = 0.3925 --> 0.39
For population in Hardy- Weinberg equilibrium:
M1M1 = p2 = (0.186)2 = 0.034 x 107 = 3.73 --> 4 frogs with M1M1
M1M2 = 2(pq) = (0.420) = 0.841 x 107 = 89.99 --> 90 frogs with M1M2
M2M2 = q2 = (0.392)2 = 0.154 x 107 = 16.4 --> 16 frogs with M1M1
According to HWE, the allele frequencies of observed population, (p+q)2 = 1. But, the actual genotype frequencies are equal to 1.
M1M1 + M1M2+ M2M2 = 0.2 + 0.42 + 0.39 = 1.01
According to HWE, the allele frequencies of expected population is = 0.034 + 0.841 + 0.154 = 1.029
Therefore, the observed frequency is skightly lower than the expected frequency and this can be referred to as segregation of gametes independently during parental crosses and allele frequencies are near to 1 and are in Hardy - Weinberg equilibrium.
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