Homework Assignment # 3-Genes in Populations A sample of 37 individuals were cap
ID: 150844 • Letter: H
Question
Homework Assignment # 3-Genes in Populations A sample of 37 individuals were captured from a wild population of the terrestrial slug Deroceras leave (Mollusca, Gastropoda, Limacidae). Each individual in this sample was examined at a polymorphic at gene locus that codes for the protein, Esterase. Three distinct genotypes were documented in this sample. The OBSERVED numbers of genotypes (AA, AB, and BB) are given in the table below Genotypes at an Esterase Locus AB # Observed (from wild) 21 13 37 From these data calculate 10.8% i(B)-37a % (.eog 1.343 1) The gene frequencies of Alleles A and B in the sampled population: f(A) A) 2) The numbers of Genotypes, AA, AB and BB, EXPECTED under the Hardy Weinberg Equilibrium Model (H-W): AB # Expected** (From H-W) 37 Interpret the results of your analyses, being certain to answer the following questions: Do the observed genotype frequencies conform to those expected under the conditions of 1) 2) Provide a minimum of 3 plausible explanations as to why this sampled population has the 3) From the observed and expected genotype frequencies, calculate the inbreeding the H-W model? genotype frequencies observed coefficient F for this population of terrestrial slugs Be sure to show your work for possible partial creditExplanation / Answer
According to hardy weinberg equilibrium
p + q =1
Where p is the frequency of allele A, and q is the allelic frequency of B
(p + q )2 =1
p2 + q2 + 2pq =1
The genotypic frequency of AA is px p = p2
The genotypic frequency of BB is qx q = q2
The genotypic frequency of AB is pq + pq = 2pq
The calculated p is 0.608 and q is 0.392 . p2 = 0.369 q2 = 0.153
1) To check if the observed and expected genotypic frequency differ significantly chi-square analysis will be performed.
Null hypothesis: There is no significant difference between the observed and expected genotypic frequencies
If p value is greater than 0.05 (5%) we will accept the null hypothesis, if p value is less than 0.05 we will reject the null hypothesis
Degree of freedom is number of genotypic classes minus 1. So degree of freedom is 3-1 =2 .
In the chi-square table look for p value in df=2 row, the p value is much less than 0.01 (1%). This means our p value is less than 0.05. Therefore we will reject the null hypothesis which means there is significant difference between the observed and expected genotypic frequencies.
2) Population of slug is mating non-randomly.. There is inbreeding as individuals with AA are mating more with AA individual as the observed frequency of AA individuals is high. Similar pattern is seen for BB individual. This is the reason why the frequency of heterozygotes is low (3) as compared to expected frequency of 17.63
It may be due to sampling error as sample size is small.
The observed frequency may be due to natural selection. Homozygous individual (AA or BB) may be reproducing more than heterozygotes.
3) Inbreeding coefficient = number of expected heterozygotes(AB) - number of observed heterozygotes (AB) / number of expected heterozygotes(AB)
= 17.63 - 3 / 17.63
= 0.83
Genotype observed expected AA 21 p2 x 37 = 0.369 x 37 = 13.65 BB 13 q2 x 37 = 0.153 x 37 = 5.6 AB 3 2pq x 37 = 2 x 0.608 x 0.392 x 37 = 17.63 total = 37Related Questions
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