Chapter 3 Mendelian Genetics

Date: 21/22 September 2021

Topic: Probability theory, hypothesis testing

Duration: 30–90 min.

In this chapter you will learn how to compare observed to expected frequencies using a chi-squared test.

3.1 Corn Example

In the basic practicals you will count the number of differently colored corn kernels.

Figure 3.1: A cob of corn with differently colored kernels.

According to Mendelian genetics, if both parents were heterozygous for one color-causing gene, then we would expect a $3 : 1$ ratio of kernels with dominant to recessive trait. For example, consider heterozygous parents for the red color-causing gene R:

Table 3.1: Crossing of two heterozygous parents.
	R	r
R	RR (red)	Rr (red)
r	Rr (red)	rr (yellow)

If both parents were indeed Rr, then we would expected $\text{red} : \text{yellow} = 3 : 1$ in the offspring.

A statistical test works by taking on a statement of no difference and then collecting evidence against it. For the chi-squared test, that means that we compare the observed frequencies to the frequencies we would get under a $3:1$ ratio. You can calculate these expected frequencies simply by multiplying the total kernels counted with the expected proportion. (Take care to convert the ratio to a proportion first.)

Exercise 1

Suppose you count $361$ red kernels and $139$ yellow ones. What are the expected number of red and yellow kernels under the null-hypothesis of a $3 : 1$ ratio?

Exercise 2

The chi-squared test is defined as follows:

$\begin{equation} \displaystyle\chi^2 = \sum_{i = 1}^k \frac{(\text{observed}_i - \text{expected}_i)^2}{\text{expected}_i} \tag{3.1} \end{equation}$

Where $k$ is the number of groups. Since there are only two groups (red, yellow), this becomes:

$\displaystyle\chi^2 = \frac{(\text{observed}_\text{red} - \text{expected}_\text{red})^2}{\text{expected}_\text{red}} + \frac{(\text{observed}_\text{yellow} - \text{expected}_\text{yellow})^2}{\text{expected}_\text{yellow}}$

Calculate $\chi^2$ using the observed and expected numbers from exercise 1. Show your calculation.

3.2 Chi-Squared Test

In R you can compare observed to expected frequencies using chisq.test:

example <- c(361, 139)
chisq.test(example, p = c(3/4, 1/4)) # Test against a 3:1 ratio

## 
##  Chi-squared test for given probabilities
## 
## data:  example
## X-squared = 2.0907, df = 1, p-value = 0.1482

This test calculates $\chi^2$ and computes a corresponding $p$ -value that has the following meaning:

If the population has a $3 : 1$ ratio of red to yellow kernels, what is the chance of observing at least this large a deviation?¹

If this chance is very small, then perhaps the null-hypothesis of a $3 : 1$ ratio is not realistic and we reject it. If it is large, then this corn cob might as well have arisen from parents that yield a $3 : 1$ ratio, and we don’t reject the null-hypothesis.

So what is small and large? That is a matter of opinion, but in biology a value of $0.05$ is often used as a boundary. That means there is a $\frac{1}{20}$ chance of incorrectly concluding a significant deviation from $3 : 1$ .

Exercise 3

Look at the output from the chi-squared test. With a threshold of $0.05$ , would you reject the null-hypothesis? What do you conclude?

Exercise 4

Perform the same $\chi^2$ -test for your own counts of a corn cob and report the conclusion. Use the code from the example and adapt it for your own counts.

Exercise 5 (optional)

In the basic practicals, it was explained that all of four different dominant mutations are required for there to be red kernels: C, R, A1 and A2.

Since these are all dominant mutations, let’s assume that each mutant protein has a $\frac{3}{4}$ chance of being present in the offspring (due to a $3 : 1$ ratio). What is then the chance of a red kernel?

(For simplicity, ignore the possibility of C-inhibitor.)

Population here does not refer to a biological population, but a statistical population. It is the population of all possible red and yellow corn kernels that could have formed from a $3 : 1$ ratio.↩︎