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The next four problems are part of a series. For these problems, see your book for a detailed explanation of epistasis in labs, and also for the x? table. You have crossed two black labs that are both BbEe genotypes, and the resulting litter consists of 15 black, 4 brown, and 6 yellow labs. You conduct a x test of the null hypothesis that the offspring match the expected ratios for recessive epistasis, and any differences are due to chance. The x² value you calculate is 0.190. Part I: How many yellow labs do you expect? Answer to one decimal place (e.g., 0.8). 6.3 Part II: What are your degrees of freedom? Your answer should be a round number (e.g., 9). 2 Part IlI: Assuming a p-value cut-off of 0.05, what is the critical value that you will use to evaluate your hypothesis? 3.841 O 7.815 O 5.991 O 9.488

The next four problems are part of a series. For these problems, see your book for a detailed explanation of epistasis in labs, and also for the x? table. You have crossed two black labs that are both BbEe genotypes, and the resulting litter consists of 15 black, 4 brown, and 6 yellow labs. You conduct a x test of the null hypothesis that the offspring match the expected ratios for recessive epistasis, and any differences are due to chance. The x² value you calculate is 0.190. Part I: How many yellow labs do you expect? Answer to one decimal place (e.g., 0.8). 6.3 Part II: What are your degrees of freedom? Your answer should be a round number (e.g., 9). 2 Part IlI: Assuming a p-value cut-off of 0.05, what is the critical value that you will use to evaluate your hypothesis? 3.841 O 7.815 O 5.991 O 9.488

Human Anatomy & Physiology (11th Edition)
Human Anatomy & Physiology (11th Edition)
11th Edition
ISBN:9780134580999
Author:Elaine N. Marieb, Katja N. Hoehn
Publisher:Elaine N. Marieb, Katja N. Hoehn
Chapter1: The Human Body: An Orientation
Section: Chapter Questions
Problem 1RQ: The correct sequence of levels forming the structural hierarchy is A. (a) organ, organ system,...
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The next four problems are part of a series. For these problems, see your book for a detailed explanation of epistasis in labs, and also for the x? table. You have crossed two black labs that are both BbEe genotypes, and the resulting litter consists of 15 black, 4 brown, and 6 yellow labs. You conduct a x² test of the null hypothesis that the offspring match the expected ratios for recessive epistasis, and any differences are due to chance. The x value you calculate is 0.190. Part I: How many yellow labs do you expect? Answer to one decimal place (e.g., 0.8). 6.3 Part II: What are your degrees of freedom? Your answer should be a round number (e.g., 9). 2 Part II: Assuming a p-value cut-off of 0.05, what is the critical value that you will use to evaluate your hypothesis? O 3.841 O 7.815 5.991 O 9.488
Transcribed Image Text:The next four problems are part of a series. For these problems, see your book for a detailed explanation of epistasis in labs, and also for the x? table. You have crossed two black labs that are both BbEe genotypes, and the resulting litter consists of 15 black, 4 brown, and 6 yellow labs. You conduct a x² test of the null hypothesis that the offspring match the expected ratios for recessive epistasis, and any differences are due to chance. The x value you calculate is 0.190. Part I: How many yellow labs do you expect? Answer to one decimal place (e.g., 0.8). 6.3 Part II: What are your degrees of freedom? Your answer should be a round number (e.g., 9). 2 Part II: Assuming a p-value cut-off of 0.05, what is the critical value that you will use to evaluate your hypothesis? O 3.841 O 7.815 5.991 O 9.488
Part IV: Do you reject or fail to reject the null hypothesis? Why? Reject, because the p-value is greater than 0.05 Fail to reject, because the p-value is greater than 0.05 O Fail to reject, because the p-value is less than 0.05 Reject, because the p-value is less than 0.05 In fruit flies, black body type is a recessive variant of the wild-type brown body. You cross two true- breeding black-bodied flies, and the resulting offspring all have brown bodies. Which of the following best explains this result? each black fly had a mutation in a different gene the black body mutation is dominant to the wild type the two black flies had mutations in the same gene the black body mutations are codominant Miniature wings are recessive to normal size (wild-type) wings in fruit flies. You have two true- breeding lines of miniature wing flies from two different locations, and when you cross them all the F1 offspring have normal wings. When you intercross the F, offspring you get 200 F2 How many of these do you expect to have miniature wings? Enter your answer to one decimal point (e.g., 1.8).
Transcribed Image Text:Part IV: Do you reject or fail to reject the null hypothesis? Why? Reject, because the p-value is greater than 0.05 Fail to reject, because the p-value is greater than 0.05 O Fail to reject, because the p-value is less than 0.05 Reject, because the p-value is less than 0.05 In fruit flies, black body type is a recessive variant of the wild-type brown body. You cross two true- breeding black-bodied flies, and the resulting offspring all have brown bodies. Which of the following best explains this result? each black fly had a mutation in a different gene the black body mutation is dominant to the wild type the two black flies had mutations in the same gene the black body mutations are codominant Miniature wings are recessive to normal size (wild-type) wings in fruit flies. You have two true- breeding lines of miniature wing flies from two different locations, and when you cross them all the F1 offspring have normal wings. When you intercross the F, offspring you get 200 F2 How many of these do you expect to have miniature wings? Enter your answer to one decimal point (e.g., 1.8).
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