Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 3: Problem 19

In assessing data that fell into two phenotypic classes, a geneticist observed values of \(250: 150 .\) She decided to perform a \(\chi^{2}\) analysis by using the following two different null hypotheses: (a) the data fit a 3: 1 ratio, and (b) the data fit a 1: 1 ratio. Calculate the \(\chi^{2}\) values for each hypothesis. What can be concluded about each hypothesis?

Short Answer

Expert verified
Answer: Based on the calculated \(\chi^{2}\) values (\(33.333\) for hypothesis (a) and \(25\) for hypothesis (b)), hypothesis (b), the 1:1 ratio, better fits the given data. However, a formal conclusion should be drawn by using \(\chi^{2}\) critical values or p-values, along with the level of significance and degrees of freedom.

Step by step solution

01

Calculate the Expected values for each hypothesis

For hypothesis (a), the expected ratio is 3:1 and for hypothesis (b), the expected ratio is 1:1. We need to calculate the expected values for each phenotype class according to these ratios. First, we find the total number of individuals: \(250 + 150 = 400\). For hypothesis (a), the expected ratio is 3:1, so we have: - Expected value for the first class: \(\frac{3}{(3+1)} \times 400 = 300\) - Expected value for the second class: \(\frac{1}{(3+1)} \times 400 = 100\) For hypothesis (b), the expected ratio is 1:1, so we have: - Expected value for the first class: \(\frac{1}{(1+1)} \times 400 = 200\) - Expected value for the second class: \(\frac{1}{(1+1)} \times 400 = 200\)
02

Calculate the \(\chi^{2}\) values for each hypothesis

For both hypotheses, we will use the \(\chi^{2}\) formula: \(\chi^{2} = \sum \frac{(O - E)^2}{E}\) Where \(O\) is the observed value, and \(E\) is the expected value. For hypothesis (a): - \(\chi^{2} = \frac{(250-300)^{2}}{300} + \frac{(150-100)^{2}}{100}\) - \(\chi^{2} = \frac{(-50)^{2}}{300} + \frac{50^{2}}{100}\) - \(\chi^{2} = \frac{2500}{300} + \frac{2500}{100} = 8.333 + 25 = 33.333\) For hypothesis (b): - \(\chi^{2} = \frac{(250-200)^{2}}{200} + \frac{(150-200)^{2}}{200}\) - \(\chi^{2} = \frac{50^{2}}{200} + \frac{(-50)^{2}}{200}\) - \(\chi^{2} = \frac{2500}{200} + \frac{2500}{200} = 12.5+12.5 = 25\)
03

Interpret the results

The calculated \(\chi^{2}\) values are \(33.333\) for hypothesis (a) and \(25\) for hypothesis (b). The lower the \(\chi^2\) value, the better the fit between the observed data and the expected values under a given null hypothesis. Therefore, based on our calculations, hypothesis (b) (i.e., the data fits a 1:1 ratio) is more likely to be true. However, to formally conclude which hypothesis best fits or clearly reject one of them, the analysis should be completed using \(\chi^2\) critical values or p-values, which requires the level of significance (such as \(0.05\) or \(0.01\)), and a degree of freedom (in this case, \(1\)) in order to draw any definite conclusions.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Mendel crossed peas having round seeds and yellow cotyledons (seed leaves) with peas having wrinkled seeds and green cotyledons. All the \(F_{1}\) plants had round seeds with yellow cotyledons. Diagram this cross through the \(\mathrm{F}_{2}\) generation, using both the Punnett square and forked-line, or branch diagram, methods.

An alternative to using the expanded binomial equation and Pascal's triangle in determining probabilities of phenotypes in a subsequent generation when the parents' genotypes are known is to use the following equation: \(\frac{n !}{s ! t !} a^{s} b^{t}\) where \(n\) is the total number of offspring, \(s\) is the number of offspring in one phenotypic category, \(t\) is the number of offspring in the other phenotypic category, \(a\) is the probability of occurrence of the first phenotype, and \(b\) is the probability of the second phenotype. Using this equation, determine the probability of a family of 5 offspring having exactly 2 children afflicted with sickle-cell anemia (an autosomal recessive disease \()\) when both parents are heterozygous for the sickle-cell allele.

Draw all possible conclusions concerning the mode of inheritance of the trait portrayed in each of the following limited pedigrees. (Each of the four cases is based on a different trait.) a. b. c. d.

Mendel crossed peas having round green seeds with peas having wrinkled yellow seeds. All \(\mathrm{F}_{1}\) plants had seeds that were round and yellow. Predict the results of testcrossing these \(\mathrm{F}_{1}\) plants.

The basis for rejecting any null hypothesis is arbitrary. The researcher can set more or less stringent standards by deciding to raise or lower the \(p\) value used to reject or not reject the hypothesis. In the case of the chi- square analysis of genetic crosses, would the use of a standard of \(p=0.10\) be more or less stringent about not rejecting the null hypothesis? Explain.
See all solutions

Recommended explanations on Biology Textbooks

Genetic Information

Read Explanation

Communicable Diseases

Read Explanation

Chemistry of Life

Read Explanation

Cells

Read Explanation

Responding to Change

Read Explanation

Reproduction

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free