1). Flowering plants that establish on islands have higher incidence of self-pol
ID: 217920 • Letter: 1
Question
1). Flowering plants that establish on islands have higher incidence of self-pollination (selfing). The Mokeyflower, Mimulus gattatus, has recently become established on islands along the coast of California.
A researcher interested in how selfing might contribute to establishment of flowering plant populations has recently genotypes 500 different plants from Catalina Island at a A/G nucleotide polymorphism within the NAC1 gene. Here are the observed counts:
?AA homozygotes: 68 AG heterozygotes: 64 GG homozygotes: 368
a) Is the Catalina Island population at Hardy Weinberg equilibrium at the NAC1 locus? Perform a statistical test of this hypothesis. Make sure you state your conclusion in complete sentences and provide statstical support for this conclusion.
b) The inbreeding coefficient at the NAC1 locus of mainland Monkeyflower pupulations is zero. What is the inbreeding coefficient at the NAC1 locus in the Catalina Island population?
c) What does your answer to part (b) suggest about the biology of island establishment by flowering plants?
d) The same researcher is going to expand her research to investigate patterns of heterozygosity at 100 additional loci randomly sampled from across the genome. Why is this next step important for inferences about the role of selfing in establishment of populations on islands?
Explanation / Answer
AA Homozygotes = 68
AG Heterozygotes = 64
GG Homozygotes = 368
Total = 500
Let frequency of allele A = p, and frequency of allele G = q
Let f(AA) be genotype frequency of AA and so on
Therefore,
p = f(AA) + 0.5*f(AG) = 0.1360 + (0.5*0.1280) = 0.1360 + 0.0640 = 0.2
q = 1 - p = 0.8
Let the expected frequency of AA be f '(AA) and so on
Therefore,
f '(AA) = p2 = 0.2*0.2 = 0.04, hence expected AA genotypes = 0.04*500 = 20
f '(AG) = 2pq = 2*0.2*0.8 = 0.32. hence expected AG genotypes = 0.32*500 = 160
f '(GG) = q2 = 0.8*0.8 = 0.64, hence expected GG genotypes = 320
a)
The chi square table will look like follows with H0 = The population is in Hardy - Weinberg equilibrium.
Summation of chi square statistic is 180
Compare this value with a chi square distribution table for degree of freedom 1 (dF = 1).
for 99% confidence (in rejecting the null hypothesis) the total chi square value must be greater then or equal to 6.64 but our value 180 >>> 6.64
Hence we can reject the null hypothesis at very high level than 99%.
Therefore the given population is not in Hardy - Weinberg equilibrium.
b)
If the population is inbred then then
f(AA) = p2 + pqF
f(AG) = 2pq*(1 - F)
f(GG) = q2 + pqF
?Where F is inbreeding coefficient,
Using,
f(AA) = 0.1360 = p2 + pqF
?F = (0.1360 - p2) /pq
? = (0.1360 - 0.04)/0.16
? = 0.6
Hence inbreeding coefficient F = 0.6
c) Since F is non zero and 0.6 is a pretty high value, we can say that there is not much cross pollination happening on the island. One of the reason behind this can be that the flowers are growing in patches where the plants are dense but the patches are far away from each other hence there is very little cross pollination and most of the time self pollination is happening or another reason can be that cross pollination is happening but mostly inside the patch only leading to inbreeding.
d) Using multiple loci to check her hypothesis will always increase the (statistical) confidence in her results. In the study of only one locus it can be by chance that the F came out be large or F is large but not due to selfing but some other phenomenon such close kin cross pollination due to patches of flowers. In order figure these things out, multiple loci can be studied to confirm inbreeding and showing that the phenomenon responsible is selfing.
AA AG GG Genotype number 68 64 368 Genotype frequency 68/500 = 0.1360 64/500 = 0.1280 368/500 = 0.7360Related Questions
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