Analysis : Chi Square Analysis

A) Their F1 offspring were 97 wild type quahaug flies. What is the genotype of these F1 flies??

For one of the monohybrid crosses you performed in this Investigation, describe how to use the phenotype ratios to determine

---If given traits and parents, be able to use a Punnett square or patterns to predict the probability of offspring for a given cross and express it as a fraction, percent, or ratio.---

6. Gather data: On the DESCRIPTION tab, click Reset. Set DD and dd to any values you like. Fill in the initial values in the table below, and then run the Gizmo for five generations. Record the allele and genotype percentages for each generation in the table below.

Testing allows the p-value that represents the probability showing that results are unlikely to occur by chance. A p-value of 5% or lower is statistically significant. The p value helps in minimizing Type I or Type II errors in the dataset that can often occur when the p value is more than the significance level. The p value can help in stopping the positive and negative correlation between the dataset to reject the null hypothesis and to determine if there is statistical significance in the hypothesis. Understanding the p value is very important in helping researchers to determine the significance of the effect of their experiment and variables for other researchers

Apply your understanding of how alleles assort and combine during reproduction to evaluate a scenario involving a monohybrid cross.

It was decided that there would be 80 vestigial flies and 20 wild type flies to total to an initial population of 100 drosophila. Next, the flies were anesthetized flies using Fly Nap. The flies were counted out to reach desired ratio, sexing the flies making sure there are equal amounts of males and females to be sure there is ample individuals to allow successful mating. The fly’s food was prepared by taking a frozen rotten banana, cutting it in half, mashing up the banana meat, and mixing yeast into it. The

with a frequency of 25%, and crossover gametes between genes Y and Z are observed with a

It would be expected that the mutant F1 flies would be heterozygous for the allele responsible for the grounded trait. If two F1 flies were mated, the percentage of flies that would be expected to be wildtype in the F2 generation would be 25% mutants given that the mutant allele (ap) is predicted to be recessive and, leaving 75% to be wildtype (ap+).

This lab had 2 exercises. Exercise 9.1 involved observing pictures of 60 F2 offspring and recording the phenotypes for 6 different traits. Exercise 9.2 required us to perform the “chi-square test” to determine whether the data we collected matches the standard Mendelian ratio.

For our first generation (F1) of flies we chose to cross apterous (+) females and white-eye (w) males. We predicted that the mutation would be sex linked recessive. So if the female was the sex with the mutation then all females would be wild type heterozygous. Heterozygous is a term used when the two genes for a trait are opposite. The males would all be white eye since they only have one X chromosome. If the males were the sex that had the mutation then all the flies would be wild type but the females would be heterozygous.

The second step is defining the significance level, determining the degrees of freedom and finding the critical value. The a-level shows that for a result to be statistically significant, it cannot occur more than the a-level percentage of time by chance. The critical value can be obtained by using the t-test table. The degrees of freedom is

(1) Chi-square tests. This nonparametric test procedure operates by tabulating a variable into different categories and then computing a corresponding chi-square statistic (Chi-square tests, 2007). This so-called "goodness-of-fit test" is used to compare the observed and expected frequencies in each of the different categories in order to determine whether all of the categories contain the same proportion of values or that each category contains a proportion of values that are specified by the user (SPSS, 2007). According to York (2002), "The chi square is the appropriate statistical test for significant differences in selection ratios" (p. 253). The chi-square test does not require assumptions concerning the shape of the underlying distribution that is involved and all data are assumed to be random samples (SPSS, 2007).

Then determine the ratios in which each of the character traits is found and also what possible genotypes the parents might be.(stallsmith)

The pairs of alternative traits examined segregated among the progeny of a particular cross, some individuals exhibiting one traits, some the other