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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.

Short Answer

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Answer: Using a p-value of 0.10 in a chi-square analysis of genetic crosses is less stringent about not rejecting the null hypothesis compared to lower p-values such as 0.05 or 0.01. It allows for a higher chance of accepting the null hypothesis even if there might be significant differences between the observed and expected ratios.

Step by step solution

01

Understanding p-value and null hypothesis

A p-value is a probability value that allows us to determine if there's enough evidence to reject the null hypothesis. The null hypothesis (\(H_0\)) states that there is no significant difference between the observed and expected values. The smaller the p-value, the more evidence we have that the observed and expected values are not equal, which means we can reject the null hypothesis. Typically, a p-value of \(0.05\) or lower is considered significant evidence to reject the null hypothesis.
02

Chi-square analysis in genetic crosses

In the context of genetic crosses, chi-square analysis is used to determine if the observed ratios in a genetic cross follow the expected ratios based on Mendelian inheritance. The null hypothesis, in this case, would be that the observed ratios follow the expected ratios. If the p-value is less than the chosen significance level (alpha), we reject the null hypothesis, indicating that the observed ratios do not follow the expected ratios and there might be some other factors influencing the genetic cross.
03

Stringency of p-value in chi-square analysis

Using a p-value of \(0.10\) is less stringent about not rejecting the null hypothesis than a lower p-value (such as 0.01 or 0.05). When the p-value is larger, it is harder to reject the null hypothesis, meaning there is a higher chance of accepting it even if there might be significant differences between the observed and expected values. More stringent standards imply using a smaller p-value, which increases the chances of rejecting the null hypothesis when there is a significant difference between the observed and expected values.
04

Conclusion

In the case of chi-square analysis of genetic crosses, using a p-value of \(0.10\) is less stringent about not rejecting the null hypothesis. This means that there will be a higher chance of accepting the null hypothesis (i.e., the ratios follow Mendelian inheritance) even if there might be significant differences between the observed and expected ratios.

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Most popular questions from this chapter

To assess Mendel's law of segregation using tomatoes, a truebreeding tall variety (SS) is crossed with a true-breeding short variety \((s s) .\) The heterozygous \(F_{1}\) tall plants \((S s)\) were crossed to produce two sets of \(\mathrm{F}_{2}\) data, as follows. \(\begin{array}{cc}\text { Set I } & \text { Set II } \\ 30 \text { tall } & 300 \text { tall } \\ 5 \text { short } & 50 \text { short }\end{array}\) (a) Using the \(\chi^{2}\) test, analyze the results for both datasets. Calculate \(\chi^{2}\) values and estimate the \(p\) values in both cases. (b) From the above analysis, what can you conclude about the importance of generating large datasets in experimental conditions?

Two true-breeding pea plants were crossed. One parent is round, terminal, violet, constricted, while the other expresses the respective contrasting phenotypes of wrinkled, axial, white, full. The four pairs of contrasting traits are controlled by four genes, each located on a separate chromosome. In the \(\mathrm{F}_{1}\) only round, axial, violet, and full were expressed. In the \(\mathrm{F}_{2},\) all possible combinations of these traits were expressed in ratios consistent with Mendelian inheritance. (a) What conclusion about the inheritance of the traits can be drawn based on the \(\mathrm{F}_{1}\) results? (b) In the \(\mathrm{F}_{2}\) results, which phenotype appeared most frequently? Write a mathematical expression that predicts the probability of occurrence of this phenotype. (c) Which \(\mathrm{F}_{2}\) phenotype is expected to occur least frequently? Write a mathematical expression that predicts this probability. (d) In the \(F_{2}\) generation, how often is either of the \(P_{1}\) phenotypes likely to occur? (e) If the \(F_{1}\) plants were testcrossed, how many different phenotypes would be produced? How does this number compare with the number of different phenotypes in the \(\mathrm{F}_{2}\) generation just discussed?

Assume that Migaloo's albinism is caused by a rare recessive gene. (a) In a mating of two heterozygous, normally pigmented whales, what is the probability that the first three offspring will all have normal pigmentation? (b) What is the probability that the first female offspring is normally pigmented? (c) What is the probability that the first offspring is a normally pigmented female?

Discuss how Mendel's monohybrid results served as the basis for all but one of his postulates. Which postulate was not based on these results? Why?

In a study of black guinea pigs and white guinea pigs, 100 black animals were crossed with 100 white animals, and each cross was carried to an \(\mathrm{F}_{2}\) generation. In 94 of the crosses, all the \(\mathrm{F}_{1}\) offspring were black and an \(\mathrm{F}_{2}\) ratio of 3 black: 1 white was obtained. In the other 6 cases, half of the \(\mathrm{F}_{1}\) animals were black and the other half were white. Why? Predict the results of crossing the black and white \(\mathrm{F}_{1}\) guinea pigs from the 6 exceptional cases.

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