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 20PDQ
Summary Introduction
To determine: The stringency of p-value equal to 0.10 in rejecting null hypotheses.
Introduction: The chi-square test was developed by Karl Pearson to determine if the observed distribution of the values or individuals in
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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
Through 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…
What is the mis-application of probability or the simple but incorrect solution?
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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- What is probability, and how is it applied in genetic analysis?arrow_forwardYou perform a chi-square test to compare observed and expected values and obtain a chi-square value of 9.4 with 3 degrees of freedom. What do you conclude? it is not likely that the difference between observed and expected values is due to random chance, since p>0.05 it is impossible to conclude anything from this information it is likely that the difference between observed and expected values is due to random chance, since p>0.05 it is likely that the difference between observed and expected values is due to random chance, since p<0.05 the experiment was done incorrectly and must be repeated it is not likely that the difference between observed and expected values is due to random chance, since p<0.05arrow_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
- Pretend that you are comparing the actual genotype distribution for a population with the distribution of genotypes predicted by the Hardy-Weinberg theorem. So your hypothesis is that the population is in Hardy-Weinberg equilibrium (i.e. that actual population data fit the Hardy-Weinberg expectations). If you carry out a chisquare goodness of fit test and calculate a total chisquare value of 0.03 with 1 degree of freedom (see table), what does this mean? (select all true statements)a) The data do NOT fit the hypothesized distribution.b) The data do fit the hypothesized distribution well enough, so we accept the hypothesis at this time (i.e. we cannot reject the hypothesis). c) The probability that the data came from a population in Hardy-Weinberg equilibrium is too small, so we reject the hypothesis.d) The probability that the data came from a population in Hardy-Weinberg equilibrium is too big, so we reject the hypothesis.e) The data support Hardy-Weinberg expectations – there is no…arrow_forwardPretend that you are comparing the actual genotype distribution for a population with the distribution of genotypes predicted by the Hardy-Weinberg theorem. So you hypothesize that the population is in Hardy-Weinberg equilibrium (i.e. that actual population data fit the Hardy-Weinberg expectations). If you carry out chi-square goodness of fit test and calculate a total chi-square value of 0.03 with 1 degree of freedom (see table), what does this mean?arrow_forwardWhile 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_forward
- In assessing data that fell into two phenotypic classes, a geneticist observed values of 20:150. She decided to perform a Chi-Square (X) analysis by using the following two different null hypotheses: a) the data fit the 3:1 ratio, and b) the data fit the 1:1 ratio. Calculate the Chi-Square (x) values for each hypothesis. What can be concluded about each hypothesis?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 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_forwardSuppose we have the following table in our goodness of fit test results: Variable/Condition Value p-value 0.077 Significance level 0.05 Conclusion then what do we write in the cell corresponding to "Conclusion"? Suppose we have the following table in our goodness of fit test results: Variable/Condition Value p-value 0.077 Significance level 0.05 Conclusion then what do we write in the cell corresponding to "Conclusion"? We change the significance level to 0.95. We accept the null hypothesis that the disease possibly follows the specified mode of inheritance at the given gene. We reject the null hypothesis that the disease possibly follows the specified mode of inheritance at the given gene. We accept the null hypothesis that the disease almost certainly follows the specified mode of inheritance at the given gene. We collect more data.arrow_forward
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