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 Fail to reject, because the p-value is less than 0.05 Reject, because the p-value is less than 0.05 Question 13 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 O the black body mutations are codominant Question 14 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 F, offspring have normal wings. When you intercross the F1 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).
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 Fail to reject, because the p-value is less than 0.05 Reject, because the p-value is less than 0.05 Question 13 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 O the black body mutations are codominant Question 14 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 F, offspring have normal wings. When you intercross the F1 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).
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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Question
can you do 12, 13, 14 please
![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 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
Question 10
Part II: What are your degrees of freedom? Your answer should be a round number (e.g., 9).
2
Question 11
Part III: 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
7.815
O 5.991
O 9.488](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe93a29d4-c02a-4172-814f-b43a1b577321%2F041305e8-d52c-4251-a1c2-01fb0f6c56ab%2Fj7t742_processed.jpeg&w=3840&q=75)
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 x? 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
Question 10
Part II: What are your degrees of freedom? Your answer should be a round number (e.g., 9).
2
Question 11
Part III: 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
7.815
O 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
Fail to reject, because the p-value is less than 0.05
Reject, because the p-value is less than 0.05
Question 13
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
O the black body mutations are codominant
Question 14
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
F, offspring have normal wings. When you intercross the F1 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).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe93a29d4-c02a-4172-814f-b43a1b577321%2F041305e8-d52c-4251-a1c2-01fb0f6c56ab%2Fuu3xy3s_processed.jpeg&w=3840&q=75)
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
Fail to reject, because the p-value is less than 0.05
Reject, because the p-value is less than 0.05
Question 13
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
O the black body mutations are codominant
Question 14
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
F, offspring have normal wings. When you intercross the F1 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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