How do I calculate the P value? TSeed capsules of the Shepherd's pusse are either triangular or ovid. A cross blu a pure breeding plant withOvoid seed capsules and a pure breeding plant with triangular seed Capsules viewed Fl hybrids that all have triangularseed capsules. The Fl hybrid self-cross produced 50 F2 plants, 72 of which had triangular seed Capsules and 8of which had ovoid Seed Capswer You state your null hypothesis as follows: There is no significant differenblw the observed data and the expected ratio of 3:1 that would be predicted for a monohybrid cross.Any deviation abserved is simply due to random chance.a) Expected Ratio: 3:1 Fractions: 3/4 ib)Phena typic ClassObserved # Expected #72880TriangularOvaidTotalExpected Triangular:Expected Ovoid :80 × 3/4 = 6080 x 1/4= 20Triangular: (72-60)²( 60 )Ovoid: (8-20)2 = 7.2(20)d) P-valueP-value: 0.00194577= 2.4602080P<0.05= reject the null hypothesP>0.05 fail to reject1/4(0-E)/E2.47. 2x² = 9.6Total = 2.4+7.2c) Degrees of Freedom :Degrees of = (# of phenotypic class - 1) = (2-1) = 1Freedom5 Nitrogenous Bases > U, T, C, A, G= 9.6 Chi square vakeWhat is a P-valve = the probability that we would see the degree of deviation observed heregiven that the null hypothesis is true.

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How do I calculate the P value?

Human Heredity: Principles and Issues (MindTap Course List)

11th Edition

ISBN:9781305251052

Author:Michael Cummings

Publisher:Michael Cummings

Chapter3: Transmission Of Genes From Generation To Generation

Section: Chapter Questions

Problem 17QP

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Question

How do I calculate the P value?

T
Seed capsules of the Shepherd's pusse are either triangular or ovid. A cross blu a pure breeding plant with
Ovoid seed capsules and a pure breeding plant with triangular seed Capsules viewed Fl hybrids that all have triangular
seed capsules. The Fl hybrid self-cross produced 50 F2 plants, 72 of which had triangular seed Capsules and 8
of which had ovoid Seed Capswer You state your null hypothesis as follows: There is no significant differen
blw the observed data and the expected ratio of 3:1 that would be predicted for a monohybrid cross.
Any deviation abserved is simply due to random chance.
a) Expected Ratio: 3:1 Fractions: 3/4 i
b)
Phena typic Class
Observed # Expected #
72
8
80
Triangular
Ovaid
Total
Expected Triangular:
Expected Ovoid :
80 × 3/4 = 60
80 x 1/4= 20
Triangular: (72-60)²
( 60 )
Ovoid: (8-20)2 = 7.2
(20)
d) P-value
P-value: 0.00194577
= 2.4
60
20
80
P<0.05= reject the null hypothes
P>0.05 fail to reject
1/4
(0-E)/E
2.4
7. 2
x² = 9.6
Total = 2.4+7.2
c) Degrees of Freedom :
Degrees of = (# of phenotypic class - 1) = (2-1) = 1
Freedom
5 Nitrogenous Bases > U, T, C, A, G
= 9.6 Chi square vake
What is a P-valve = the probability that we would see the degree of deviation observed here
given that the null hypothesis is true.

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