Given the expected phenotype probability that you determined above, what is the minimum number of offspring required to give a 95% chance of obtaining an individual with the A B phenotype?
Given the expected phenotype probability that you determined above, what is the minimum number of offspring required to give a 95% chance of obtaining an individual with the A B phenotype?
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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
Transcribed Image Text:To test for gene linkage a dihybrid testcross of A/a;B/b x a/a;b/b is carried out
resulting in 100 progeny with the following phenotypes:
AB - 30
Ab-22
a B - 20
ab - 28
Your null hypothesis is that genes A and B assort independently and are not linked.
How many offspring would you expect to see with the A B phenotype?
To test the null hypothesis that genes A and B are unlinked, you perform an
exhaustive set of calculations and arrive at a chi-square value of 2.72. Based on the
table below, do you accept or reject your null hypothesis with 95% confidence?
Table 1: Critical values of the chi-square distribution with df degrees of freedom.
0.5
0.02
0.01
0.1 0.05
2.706 3.841
0.455
5.412
6.635
4.605 5.991
7.824
9.21
1.386
2.366 6.251 7.815
9.837 11.345
3.357 7.779 9.488 11.668 13.277
df
1
2
3
4
Write "accept" or "reject" in the blank.
Given the expected phenotype probability that you determined above, what is the
minimum number of offspring required to give a 95% chance of obtaining an
individual with the A B phenotype?
4
Expert Solution

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Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.
Step 1: Calculate the expected number of offspring with the AB phenotype under the assumption of independent
VIEWStep 2: Perform a chi-square test to determine whether to accept or reject the null hypothesis:
VIEWStep 3: Compare the calculated chi-square value to the critical value for 1 degree of freedom (df) at a 95%
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