The next four problems are part of a series. For these problems, see your book for a detailedexplanation 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 15black, 4 brown, and 6 yellow labs. You conduct a x² test of the null hypothesis that the offspringmatch the expected ratios for recessive epistasis, and any differences are due to chance. The xvalue you calculate is 0.190.Part I: How many yellow labs do you expect? Answer to one decimal place (e.g., 0.8).6.3Part II: What are your degrees of freedom? Your answer should be a round number (e.g., 9).2Part II: Assuming a p-value cut-off of 0.05, what is the critical value that you will use to evaluateyour hypothesis?O 3.841O 7.8155.991O 9.488 Part IV: Do you reject or fail to reject the null hypothesis? Why?Reject, because the p-value is greater than 0.05Fail to reject, because the p-value is greater than 0.05O Fail to reject, because the p-value is less than 0.05Reject, because the p-value is less than 0.05In 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 thefollowing best explains this result?each black fly had a mutation in a different genethe black body mutation is dominant to the wild typethe two black flies had mutations in the same genethe black body mutations are codominantMiniature 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 theF1 offspring have normal wings. When you intercross the F, offspring you get 200 F2 How many ofthese do you expect to have miniature wings? Enter your answer to one decimal point (e.g., 1.8).

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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

Biology (MindTap Course List)

11th Edition

ISBN:9781337392938

Author:Eldra Solomon, Charles Martin, Diana W. Martin, Linda R. Berg

Publisher:Eldra Solomon, Charles Martin, Diana W. Martin, Linda R. Berg

Chapter11: The Basic Principles Of Heredity

Section: Chapter Questions

Problem 2TYU: The F1 flies described in question 1 were mated with brown-eyed flies from a true-breeding line....

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