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Chapter 15: Q3ITD (page 294)

The\({\chi ^2}\)value means nothing on its own- it is used to find the probability that, assuming the hypothesis is true, the observed data set could have resulted from random fluctuations. A low probability suggests that the observed data are consistent with the hypothesis, and thus the hypothesis should be rejected, A standard cutoff point used by biologists is a probability of 0.05(5%). If the probability corresponding to the\({\chi ^2}\)value is 0.05or considered statistically significant, the hypothesis (that the genes are unlinked) should be rejected. If the probability is above 0.05, the results are not statistically significant: the observed data are consistent with the hypothesis.

To find the probability, locate your\({\chi ^2}\)value in the\({\chi ^2}\)Distribution table in Appendix F. The 鈥渄egree of freedom鈥� (pdf) of your data set is the number of categories (here,4 phenotypes), minus 1, so df=3.

(a). Determines which values on the df =3 line of the table your calculated\({\chi ^2}\)value lies between.

(b). The column headings for these values show the probability range for your\({\chi ^2}\)number. Based on whether there is non-significant (p\( \le \)0.05) or significant (p>0.05) difference between the observed and expected values, are the data consistent with the hypothesis that the two genes are unlinked and assorting independently, or is there enough evidence to reject this hypothesis?

Short Answer

Expert verified

(a) The \({\chi ^2}\)values lie between 1.42 and 2.37 following the reference with appendix F.

(b) The hypothesis given in the question is supported since the probability values are statistically significant

Step by step solution

01

Chi-square

The chi-square value is calculated to compare with the degree of freedom distribution to find if the hypothesis is supported or not. The standard value of probability is 0.05.If the probability value is greater than 0.05, then the hypothesis is significant. If the probability value is less than or equal to 0.05, then the hypothesis is not significant.

02

Explanation of part “a”

The sum value is calculated as 2.14 from the previous subdivision.

The degree of freedom given in the question is 3. By comparing the degrees of freedom, the\({\chi ^2}\)value lies in between 1.42 and 2.37. The value 2.14 lies between the value of 1.42 and 2.37 by appendix F.

03

Step 3: Explanation of part “b”

The genes that possess unlinked alleles get assorted independently. The unlinked alleles are present in different chromosomes. The phenotypic ratio of the F1dihybrid cross is 1:1:1:1.

The hypothesis in the given question is supported since the p-value lies between 0.5 and 0.70. The p value is greater than 0.05, so the hypothesis is supported, and the p value is significant by the given condition.

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Most popular questions from this chapter

The gene that is activated on the Philadelphia chromosome codes for an intracellular tyrosine kinase. Review the discussion of cell cycle control in Concept 2.3 and explain how the activation of this gene could contribute to the development of cancer.

Assume you are mapping genes A, B, C, and D in Drosophila. You know that these genes are linked on the same chromosome, and you determine the recombination frequencies between each pair of genes to be as follows: A-B, 8%; A-C, 28%; A-D, 25%; B-C, 20%; B-D, 33%.

  1. Describe how you determined the recombination frequency for each pair of genes.

  2. Draw a chromosome map based on your data.

Crossing over is thought to be evolutionarily advantageous because it continually shuffles genetic alleles into novel combinations. Until recently, it was thought that the genes on the Y chromosome might degenerate because they lack homologous genes on the X chromosome with which to pair up prior to crossing over. However, when the Y chromosome was sequenced, eight large regions were found to be internally homologous to each other, and quite a few of the 78 genes represent duplicates. (Y chromosome researcher David Page has called it a "hall of mirrors.鈥�). Explain what might be a benefit of these regions.

A wild-type fruit fly (heterozygous for the gray body color and red eyes) is mated with a black fruit fly with purple eyes. The offspring are wild-type, 721; black purple, 751; gray purple, 49; black red, 45. What is the recombination frequency between these genes for the body color and eye color? Using information for problem 3, what fruit flies (genotypes and phenotypes) would you mate to determine the order of the body color, wing size, and eye color genes on the chromosome?

The goodness of fit is measured by\({\chi ^{^2}}\). This statistic measures the amounts by which the observed values differ from their respective predictions to indicate how closely the two sets of values match. The formula for calculating this value is

\({\chi ^{}} = \sum \frac{{{{\left( {o - e} \right)}^2}}}{e}\)

Where o=observed and e= expected. Calculate the\({\chi ^{^2}}\)value for the data using the table below. Fill out the table, carrying out the operations indicated in the top row. Then add up the entries in the last column to find the\({\chi ^{^2}}\)value.

Testcross Offspring

Expected

(e)

Observed

(o)

Deviation

(o-e)

(o-e)2

(o-e)2/e

(A-B-)

220

(aaB-)

210

(A-bb)

231

(aabb)

239

\({\chi ^2}\) =sum
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