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Chapter 11: Problem 17

Mendel and the peas Gregor Mendel \((1822-1884)\) an Austrian monk, is considered the father of genetics. Mendel studied the inheritance of various traits in pea plants. One such trait is whether the pea is smooth or wrinkled. Mendel predicted a ratio of 3 smooth peas for every 1 wrinkled pea. In one experiment, he observed 423 smooth and 133 wrinkled peas. Assume that the conditions for inference were met. Carry out an appropriate test of the genetic model that Mendel predicted. What do you conclude?

Short Answer

Expert verified
Mendel's predicted 3:1 ratio is not significantly different from the observed data.

Step by step solution

01

State the Hypotheses

We need to set up our null and alternative hypotheses. The null hypothesis ( \( H_0 \) ) is that the proportion of smooth peas ( \( p_{smooth} \) ) is 3/4 and the proportion of wrinkled peas ( \( p_{wrinkled} \) ) is 1/4, as Mendel predicted. The alternative hypothesis ( \( H_1 \) ) is that these proportions are not 3/4 and 1/4, respectively.
02

Calculate Expected Counts

Based on Mendel's prediction, the expected number of smooth peas is \( \frac{3}{4} \times 556 = 417 \) and the expected number of wrinkled peas is \( \frac{1}{4} \times 556 = 139 \) . Here, 556 is the total number of peas observed.
03

Compute the Test Statistic

Use the chi-square test for goodness-of-fit: \[ \chi^2 = \sum \left( \frac{(O_i - E_i)^2}{E_i} \right) \] where \( O_i \) are the observed counts and \( E_i \) are the expected counts. For smooth peas, \( O_1 = 423 \) and \( E_1 = 417 \), and for wrinkled peas, \( O_2 = 133 \) and \( E_2 = 139 \). Hence, \[ \chi^2 = \left( \frac{(423 - 417)^2}{417} \right) + \left( \frac{(133 - 139)^2}{139} \right) = 0.086 + 0.259 = 0.345 \]
04

Determine the P-value

Using a chi-square distribution table with 1 degree of freedom (since we have 2 categories - 1 degree of freedom), find the p-value corresponding to our test statistic \( \chi^2 = 0.345 \) . The p-value is greater than 0.05.
05

Make a Conclusion

Since the p-value is greater than the significance level (typically 0.05), we fail to reject the null hypothesis. This means there is not enough evidence to conclude that the observed ratios significantly differ from Mendel's predicted 3:1 ratio.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Goodness-of-Fit
The chi-square test for goodness-of-fit is a statistical method used to determine if observed data fits a particular distribution. In our exercise, it is applied to test if the observed counts of smooth and wrinkled peas align with Mendel's genetic model.

This test compares the observed frequencies (actual counts) in your data to the expected frequencies that would occur if the model were correct. The idea is simple: if your data matches what you expected, any differences are likely due to random chance; if not, the model might not be a good fit for your data.

The test calculates a chi-square statistic (\( \chi^2 \)), which measures how much the observed data deviate from the expected data. A larger chi-square statistic implies a greater discrepancy between observed and expected data. However, interpreting this number requires further steps, like checking relevant p-values.
Genetic Model
A genetic model provides predictions on how traits are passed from generation to generation based on inheritance patterns. In Mendelian genetics, these models use ratios to describe how dominant and recessive traits manifest in offspring.

In this example, the genetic model is a 3:1 ratio of smooth to wrinkled peas, which is a classic representation of simple Mendelian inheritance. Mendel hypothesized that smooth peas dominate wrinkled ones in a specific way after cross-pollination.
  • Smooth peas (dominant trait) are predicted to appear three times more often than wrinkled peas (recessive trait).
This kind of ratio makes up the expected values against which one compares actual data, using hypothesis testing methods like the chi-square test.
Null and Alternative Hypotheses
In hypothesis testing, you'll frequently encounter two contrasting statements: the null hypothesis (\( H_0 \)) and the alternative hypothesis (\( H_1 \)). These are essential parts of setting up statistical tests, like our chi-square test.

The null hypothesis is the foundational assumption that there is no significant difference between observed and expected data. In our context, it represents Mendel's prediction: the ratios of smooth to wrinkled peas are exactly 3:1, or 75% smooth and 25% wrinkled.

The alternative hypothesis, on the other hand, states that there is a significant difference. It suggests that the actual proportion might not match Mendel's prediction. This hypothesis is only considered if there is enough evidence to reject the null.

Testing these hypotheses allows us to systematically evaluate whether the genetic model fits the observed data.
P-Value Analysis
The p-value is a measure that helps us decide if we can reject the null hypothesis. It's used to interpret the results of statistical tests like the chi-square test for goodness-of-fit.

A p-value is essentially the probability of observing data as extreme as, or more extreme than, the actual observed data, assuming the null hypothesis is true. In simpler terms, it helps check how well the observed data aligns with the expected data.
  • A small p-value (typically less than 0.05) suggests that the observed data significantly deviates from what's expected under the null hypothesis, prompting us to reject it.
  • A large p-value means we cannot reject the null hypothesis; thus, there's not enough evidence to state that the genetic model doesn't fit our data.
In our exercise, the p-value is greater than 0.05, indicating we retain the null hypothesis: the observed pea ratio adheres to Mendel's predicted model.

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Most popular questions from this chapter

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