The formula of the test is as follows: X2 = E(dle) X2 = chi-square %3D E= sum of d = difference between expected and observed results (most often termed the deviation) e = expected results Example In the cross of wild-type by vestigial wing, a 3:1 ratio is expected. Thus, if a student counts 160 flies, 40 are expected to have vestigial wings, while 120 are expected to have long wings, but if a student counts 44 vestigial-winged flies and 116 long-winged flies, then the value for the chi-square test would be as calculated in the Table 1. Note: You will have four different phenotype combinations tolanalyze. Table 1: Calculation of Chi-Square - X2 Expected Number (e) Partial Chi-Square dle Phenotype Observed Number Difference (d) Vestigial Wing 44 40 4 16 16/40 = 0.400 Long Wing 116 120 4 16 16/120 = 0.133 E(dle) = 0.533 X2 = 0.533

The formula of the test is as follows: X2 = E(dle) X2 = chi-square %3D E= sum of d = difference between expected and observed results (most often termed the deviation) e = expected results Example In the cross of wild-type by vestigial wing, a 3:1 ratio is expected. Thus, if a student counts 160 flies, 40 are expected to have vestigial wings, while 120 are expected to have long wings, but if a student counts 44 vestigial-winged flies and 116 long-winged flies, then the value for the chi-square test would be as calculated in the Table 1. Note: You will have four different phenotype combinations tolanalyze. Table 1: Calculation of Chi-Square - X2 Expected Number (e) Partial Chi-Square dle Phenotype Observed Number Difference (d) Vestigial Wing 44 40 4 16 16/40 = 0.400 Long Wing 116 120 4 16 16/120 = 0.133 E(dle) = 0.533 X2 = 0.533

Human Heredity: Principles and Issues (MindTap Course List)

11th Edition

ISBN:9781305251052

Author:Michael Cummings

Publisher:Michael Cummings

Chapter3: Transmission Of Genes From Generation To Generation

Section3.7: Mendelian Inheritance In Humans

Problem 2EG: For most cases, a p value of 0.05 is used to determine whether results fit expected values. Is this...

See similar textbooks

Similar questions

What slide preparation technique should you use for the following research goals? Briefly justify your answers. CHOOSE FROM THESE SLIDE PREPARATION TECHNIQUES: MACERATION, WHOLEMOUNT, SMEAR, SQUASH, and SECTIONa. To determine the chromosome number of a newly discovered species of flowering plantb. To depict the anatomical structure of the vegetative organs of an endemic fern speciesc. To determine whether 10 human subjects have abnormalities in red blood cellsd. To characterize the external morphology of a very small insect (ca. 1 mm long)e. To characterize the effects of a certain drug on the histology of mouse live
Purple and Smooth ________124__________ Purple and Wrinkled _______125________ Yellow and Smooth __________107__________ Yellow and Wrinkled _____121__________ Using chi-square test determine if the data fits the expected 1:1:1:1 ratio. Show all calculations and state the p value you obtain for each cross and what the p values indicate about your results.
PLEASE explain how to do this question step by tspe i am very confused! thank you! You have three independent mutant alleles in the Drosophila gene no legs (nlg): nlg1, nlg2, and nlg3. You assay the phenotype of Drosophila, which is an insect, that are heterozygous for the three allele (heterozygous for a wild-type and a mutant allele), and that are homozygous for the three alleles. Allele nlg1 nlg2 nlg3 heterozygous Wild-type Four pairs of legs Wild-type homozygous Stumpy legs Five pairs of legs No legs nlg1 is a ___X___ allele, nlg2 is a ___Y___ allele, and nlg3 is a ___Z___ allele. X Y Z A Gain of function Null Hypomorphic B Hypomorphic Gain of function Null C Null Gain of function Hypomorphic D Null Hypomorphic Gain of function E Hypomorphic Null Gain of function Referencing the table above, select the correct answer: 1 : A 2 : B 3 : C 4 : D 5 : E Correct answer is B
Interpret the two chromatograms below. Which of the two chromatograms will provide good sequencing data? Evaluate your answer using the following guide questions: How clear are the peaks? Do you observe baseline noise? Are there N calls or heterozygous peaks? Do peaks broaden?
List three (3) advantages and two (2) disadvantages for each of graphical presentation using 1.) Histograms, 2.) Line graphs and 3) Scatter plots
For most cases, a p value of 0.05 is used to determine whether results fit expected values. Is this a magic number, or could p values be set more stringently, say 0.01, or more lax, say 0.10? What effect would these values have on the reliability of the chi-square test?
Comment on the alignment? What is the root mean square (RMS)? Can you think of any potential issues?
Practice question: I understand how the F1 progeny is found (by punnett square), but I do not get how they were able to get the F2 progeny (can i use a punnet square? How do i find these results?)
Bradford Assay Fill in the average A595 and A595 of sample minus A595 of blank. Show some sample computation. Concentration (micrograms/ml) A595 - trial 1 A595 - trial 2 A595 - trial 3 average A595 A595 of sample minus A595 of blank 0 1.501 1.446 1.447 2 1.624 1.558 1.559 5 1.731 1.749 1.712 10 1.901 1.838 1.892 15 2.161 2.108 2.228 18 2.231 2.277 2.319
Download BLOSUM30 and BLOSUMB0 substitu- tion matrices and place them side by side on your computer screen. What are the differences between the two matrices? Why do you see these differences?
please help me! THANK YOU! In not more than 30 words explain the relevance of a chi-square test in genetics.
Complementation tests of distinct recessive mutants, 1 through 8, produce the data in the matrix below. A plus (+) indicates complementation, meaning the phenotype of the combined alleles is wild type, and a minus (-) indicates a failure to complement meaning that a mutant phenotype results. Assume that the missing mutant combinations would yield data consistent with the entries that are shown. How many complementation groups are formed by these eight mutants? (Picture attached) A) 2 B) 3 C) 4 D) 5 E) 6