Concepts of Genetics (12th Edition)
12th Edition
ISBN: 9780134604718
Author: William S. Klug, Michael R. Cummings, Charlotte A. Spencer, Michael A. Palladino, Darrell Killian
Publisher: PEARSON
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Chapter 3, Problem 19PDQ
In assessing data that fell into two
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In assessing data that fell into two phenotypic classes, a geneticist observed values of 250:150. She decided to perform a x2 analysis by using the following two different null hypotheses: (a) the data fit a 3:1 ratio, and (b) the data fit a 1:1 ratio. Calculate the x2 values for each hypothesis. What can be concluded about each hypothesis?
A geneticist, in assessing data that fell into two phenotypic classes, observed values of 250:150. He decided to perform chisquare analysis using two different null hypotheses: (a) the data fit a 3:1 ratio; and (b) the data fit a 1:1 ratio. Calculate the x2 values for each hypothesis. What can you conclude about each hypothesis?
In a series of three experiments, M. G. Addo sought to use chi-square test to
determine the goodness of fit at 5% for each of his results below. Carry out the
calculations and determine whether the results obtained in each case are
consistent with the 3:1 or 1:1:1:1 ratio he predicted in his hypothesis. Comment
on your results.
CROSS
Tall x short
Purple x white
Round yellow (F1) x Wrinkled green (F1)
RESULTS
HΥΡΟΤHESIS
712:352
3:1
705 : 224
3:1
31:26:27:26
1:1:1:1
Chapter 3 Solutions
Concepts of Genetics (12th Edition)
Ch. 3 - Pigeons may exhibit a checkered or plain color...Ch. 3 - Considering the Mendelian traits round versus...Ch. 3 - Using the forked-line, or branch diagram, method,...Ch. 3 - In one of Mendels dihybrid crosses, he observed...Ch. 3 - The following pedigree is for myopia...Ch. 3 - If they seek genetic counseling, what issues would...Ch. 3 - If you were in Thomass position, would you want to...Ch. 3 - If Thomas tests positive for the HD allele, should...Ch. 3 - HOW DO WE KNOW? In this chapter, we focused on the...Ch. 3 - CONCEPT QUESTION Review the Chapter Concepts list...
Ch. 3 - Albinism in humans is inherited as a simple...Ch. 3 - Which of Mendels postulates are illustrated by the...Ch. 3 - Discuss how Mendels monohybrid results served as...Ch. 3 - What advantages were provided by Mendels choice of...Ch. 3 - Mendel crossed peas having round seeds and yellow...Ch. 3 - Based on the preceding cross, what is the...Ch. 3 - Which of Mendels postulates can only be...Ch. 3 - In a cross between a black and a white guinea pig,...Ch. 3 - What is the basis for homology among chromosomes?Ch. 3 - In Drosophila, gray body color is dominant to...Ch. 3 - How many different types of gametes can be formed...Ch. 3 - Mendel crossed peas having green seeds with peas...Ch. 3 - In a study of black guinea pigs and white guinea...Ch. 3 - Mendel crossed peas having round green seeds with...Ch. 3 - Prob. 17PDQCh. 3 - The following are F2 results of two of Mendels...Ch. 3 - In assessing data that fell into two phenotypic...Ch. 3 - Prob. 20PDQCh. 3 - Consider the following pedigree. Predict the mode...Ch. 3 - Draw all possible conclusions concerning the mode...Ch. 3 - For decades scientists have been perplexed by...Ch. 3 - A wrongful birth case was recently brought before...Ch. 3 - TaySachs disease (TSD) is an inborn error of...Ch. 3 - Datura stramonium (the Jimsonweed) expresses...Ch. 3 - The wild-type (normal) fruit fly, Drosophila...Ch. 3 - To assess Mendels law of segregation using...Ch. 3 - Albinism, caused by a mutational disruption in...Ch. 3 - (a) Assuming that Migaloos albinism is caused by a...
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- The basis for rejecting any null hypothesis is arbitrary. The researcher can set more or less stringent standards by deciding to raise or lower the p value used to reject or not reject the hypothesis. In the case of the chi-square analysis of genetic crosses, would the use of a standard of p = 0.10 be more or less stringent about not rejecting the null hypothesis? Explain.arrow_forwardThrough PCR, we have determined the PER3 genotypes for a class of students as follows: H4/H4 = 125 individuals; H4/H5 = 90 individuals and H5/H5=85 individuals. Considering the Hardy Weinberg equilibrium and calculating the Chi-square statistic, do you reject or fail to reject the null-hypothesis? 1. Comparing the value of Chi-square that I calculated with the critical value of Chi-square for the appropriate degrees of freedom I conclude that P<0.05. Hence, I reject the null hypothesis. 2. Comparing the value of Chi-square that I calculated with the critical value of Chi-square for the appropriate degrees of freedom I conclude that P>0.05. Hence, I reject the null hypothesis. 3. Comparing the value of Chi-square that I calculated with the critical value of Chi-square for the appropriate degrees of freedom I conclude that P<0.05. Hence, I fail to reject the null hypothesis. 4. Comparing the value of Chi-square that I calculated with the critical value of Chi-square for the…arrow_forwardWhat is the first variable that can be calculated given these data, and what is the final variable we are requesting you to calculate? "In Finland, 256 people out of 10,000 are homozygous for the CCR5 allele mutation resulting in HIV resistance. Assuming the locus is in Hardy-Weinberg equilibrium, what is the expected genotype frequency of heterozygous carriers for the CCR5 mutation?"arrow_forward
- A population of wookies is polymorphic for the fur color, which is controlled by two loci. 230 have black hair, 64 have brown hair, and 32 have yellow hair. You test suspect that the epistatic interactions between the two loci are reponsible for this ratio and wish to test that hypothesis using a X^2 test. a) What is the most likely epistasis? b) Complete the following X^2 table. For the first three columns, enter your answer as intergers with no demical points. For the last column, enter your answer to two deimal points e.g 4.36 Fur Observed Expected O-E (O-E)^2 (O-E)^2/E Black 240 Brown 64 yellow 32 total 336 X^2arrow_forwardThrough PCR, we have determined the PER3 genotypes for a class of students as follows: H4/H4 = 125 individuals; H4/H5 = 90 individuals and H5/H5=85 individuals. Considering the Hardy Weinberg equilibrium and comparing the observed and the expected number of individuals for the three genotypes, calculate the value of the Chi-square statistic. Select only one answer. 1. 14.59 2. 0.05 3. 3.84 4. 28.67 5. 22.31 6. 45.43 7. 0.50arrow_forwarda) To begin your analysis, state your null hypothesis. b) Perform a chi-square analysis. Make sure to analyze your traits to be consistent with the information provided. Clearly present your results in a table as shown in the attachement called Testing Genetics Ratios (Hint: should the traits be tested in males and females?) c) State the correct # of degrees of freedom, calculate the chi-square value and show the approximate p value. d) Make a statement of how the p value relates to whether you decided to reject or not reject, your null hypothesis. e) Finally, make a concluding statement about your original genetic hypothesis(es) based on chi-square analysis. Fully typed, organized correctly, correct spelling, and grammar. Genetic hypothesis for each allele: Trait: Leg length Hypothesis: The wild-type and mutant alleles for leg length are incomplete dominant relative to each other. Trait: Wing shape Hypothesis: There are two alleles for wing type, one dominant and one…arrow_forward
- While studying the frequency of sickle-cell disease ("sickle cell anemia") in a population living in sub-Saharan Africa, you obtain the following data from a sample of n= 100 people (note that I chose a simpler system for identifying the alleles rather than using "Hb S" for sickle cell allele), which is the actual name and what you used in lab). What is the frequency of the sickle cell allele (b) in the sample below from a human population? Sample Data BB-60 individuals (No sickle cell disease) Bb-30 individuals (No sickle cell disease) bb-10 individuals (Sickle cell disease) 1. 0.25 2. 0.10 3. 0.35 4. 0.60 5. 0.20arrow_forwardHotchkiss and Marmur noted that the percentage of cotransformation was higher than would be expected on a random basis. For example, the results show that 2.6% of the cells were transformed into M and 4% were transformed into S. If the M and S traits were inherited independently, the expected probability of cotransformation of M and S (M S) would be 0.026× 0.04 = 0.001, or 0.1%. However, they observed 0.41% M S cotransformants, four times more than they expected. What accounts for the relatively high frequency of cotransformation of the traits they observed?arrow_forwardI need explanation for the why the answer is correct? And why would the other options wrongarrow_forward
- A complementation analysis was performed using pea plants which normally have purple flowers. Twenty-five mutants for white flower colour were discovered across 10 populations. The mutants were bred in a complementation test which showed two complementation groups with 10 mutants in one group and 15 in the other. (a) How many genes contribute to flower colour in pea plants based on this data? (b) Is it likely that more genes contribute to pea plant flower colour? Explainarrow_forwardA researcher examines a locus, or marker, in which there is a particular C/T polymorphism in a population of interest. Let’s call this Locus 1. They obtain the following genotype counts in a sample of the population: CC:42, CT:16, TT:32. a) Calculate the genotype frequencies and the allele frequencies for Locus 1 in the sample.b) Calculate the observed heterozygosity (the frequency of heterozygotes) and the observed homozygosity (the total frequency of all homozygotes) in the sample. Ensure that these two frequencies add up to 1.arrow_forwardA genetic engineer was attempting to cross a tiger and a cheetah. She predicted a phenotypic outcome of the traits she was observing to be in the following ratio: 4 stripes only: 3 spots only: 9 both stripes and spots. When the cross was performed and she counted the individuals she found 50 with stripes only, 41 with spots only and 85 with both. What is the computed chi-square value? А. 3.126 В. 2.415 С. 4.001 D. 1.163arrow_forward
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