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Will B.
Molecular Biology Research Assistant
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Biomedical Science
TutorMe
Question:

Name 5 parts that make up one generic gene and briefly explain their functions.

Will B.

Answers include but are not limited to: 1) Open reading frame (ORF): The different ways to view triplet nucleotides (codons) that could generate a protein product. 2) Exon: A nucleotide sequence that ends up in the messenger RNA (mRNA). 3) Intron: Nucleotide sequence(s) that exist in the genome but do not (typically) end up in the mature mRNA transcript. 4) Promoter: A non-coding nucleotide sequence that recruits proteins to initiate transcription of the DNA to mRNA. 5) Enhancer: A more distant regulatory DNA sequence that recruits proteins that can either promote or inhibit the promoters ability to recruit transcriptional machinery. 6) Start codon: An 'ATG' sequence in the mRNA transcript that initiated translation of the mRNA to protein. 7) Untranslated regions (UTRs): Regions that are in the mRNA transcript that are not translated into protein. Found at the 5' and 3' ends of the mRNA. 8) Transcription termination sequence: The sequence at the 3' end of the DNA which causes transcription of the gene to terminate. 9) Splice junctions: Region at the junctions of introns and exons which signals for splicing machinery to remove intronic regions. 10) Ribosomal binding site (RBS): The site on the mRNA which recruits the ribosome in order to initiate translation of the transcript into the protein product.

Biology
TutorMe
Question:

A researcher wants to know whether a species of frog is in violation of Hardy-Weinberg (HW) equilibrium at the "A" locus. He genotypes 100 frogs and finds: $$AA = 38$$ $$Aa = 44$$ $$aa = 18$$ Help him determine whether the frog species is in violation of the HW equilibrium. (Assume a difference of between plus or minus 5 between expected and observed qualifies as no difference)

Will B.

HW equilibrium is defined as: $$p^2 + 2pq + q^2 = 1$$ Where p and q are the frequency of the two alleles in the population. For our frog population, there are 200 total alleles (2 from each individual). Homozygotes (AA or aa) have 2 copies of the same allele, whereas heterozygotes (Aa) have one copy of each allele. So: Let p = frequency of allele A = $$\frac{(2*38)+44}{200} = 0.6$$ Let q = frequency of allele a = $$\frac{(2*18)+44}{200}$$ or $$1 - p = 0.4$$ To determine whether our frogs are in violation of HW equilibrium, we must determine whether our observed genotypes are different from what is expected based on our allelic frequencies: Expected genotypes: $$AA = p^2 = (0.6)^2 = 0.36$$ $$Aa = 2pq = 2(0.6)(0.4) = 0.48$$ $$aa = q^2 = (0.4)^2 = 0.16$$ Therefore, in a population of 100 frogs, our expected data should be: $$AA = 36$$ $$Aa = 48$$ $$aa = 16$$ All of which are within 5 of our observed data, meaning that we cannot conclude that the population of frogs is in violation of HW equilibrium.

Statistics
TutorMe
Question:

An undergraduate student in biology is trying to determine the feeding preferences of two different species of hummingbird. The two species of hummingbird are ruby-throated and black-chinned. The student watches and records the birds choose to feed on a sugar water dispenser or a honey and water dispenser. They found that ruby-throated hummingbirds chose the sugar water 28 times and the honey water 22 times. The black-chinned hummingbirds chose the sugar water 35 times and the honey water 15 times. Since both birds fed on both food sources, the student wanted to know whether one species had more of a preference for a food source than the other. How would they do this? (use an $$\alpha$$ of 0.05)

Will B.

One approach to answering this question would be a chi-squared contingency test. The students data look like this: By Species: Ruby-throated: 28 sugar + 22 honey = 50 total Black-chinned: 35 sugar + 15 honey = 50 total By Food Source: Sugar: 28 Ruby + 35 Black = 63 total Honey: 22 Ruby + 15 Black = 37 total To begin, we must determine our null and alternative hypotheses: $$H0 =$$ There are no differences in the food choice between the species. $$HA =$$ There are differences in the way the species choose their food source. If we assume the null hypothesis is true, we can determine the expected rates by which the hummingbirds visit each food source. As an example: How many ruby-throated hummingbirds should we expect to randomly visit the sugar water feeder? 50 out of 100 total visits were by ruby-throated hummingbirds 63 out of 100 total visits were to the sugar water feeder. $$\frac{50}{100} * \frac{63}{100} = 0.315$$ Which means for every 100 visits, we should expect 31.5 of them to be by ruby-throated hummingbirds to the sugar water feeder if there is no preference for a species to select a specific food source. The rest of our expected data is as follows (which can be calculated like above): By Species: Ruby-throated: 31.5 sugar + 18.5 honey = 50 total Black-chinned: 31.5 sugar + 18.5 honey = 50 total By Food Source: Sugar: 31.5 Ruby + 31.5 Black = 63 total Honey: 18.5 Ruby + 18.5 Black = 37 total Notice how the totals do not change from the observed data, but the rates in which each species visits a particular food source does change slightly. This is because we are assuming that differences in the observed data are due to general and not species-specific biases. Now we can do a chi-squared calculation to determine whether our observed data is significantly different from our expected data: $$\chi^{2} = \sum \frac{(observed - expected)^2}{expected}$$ $$\chi^{2} = \frac{(28 - 31.5)^2}{31.5} + \frac{(35 - 31.5)^2}{31.5} + \frac{(22 - 18.5)^2}{18.5} + \frac{(15 - 18.5)^2}{18.5} = 2.1021$$ The degrees of freedom for two categorical variables is 1. With $$df = 1$$ and $$\chi^2 = 2.1021$$, this returns a p-value of 0.14, which is larger than our $$\alpha$$ of 0.05, meaning we cannot reject our null hypothesis. Therefore there is no evidence for the two hummingbird species having different feeding preferences.

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