# Predicting Inheritance: Chi-squared Test ## What is the Chi-squared Test? - A **statistical test** that determines whether there is a **significant difference** between observed and expected results in an experiment - If the difference is **statistically significant**, this suggests a factor that isn't being accounted for - E.g. genes showing linkage rather than independent assortment - If the difference is **not significant**, any differences observed can be attributed to **chance alone** - The chi-squared test is used when data is **categorical** - Data falls into distinct groups or categories - E.g. blood group, flower colour, presence/absence of a characteristic

The chi-squared test compares observed (O) results with expected (E) results to determine if differences are due to chance or a significant biological factor

## Calculating Chi-squared Values - The chi-squared value is calculated using the formula: $$\chi^2 = \sum \frac{(O - E)^2}{E}$$ Where: - $\chi^2$ = chi-squared value - $\sum$ = sum of - $O$ = observed value - $E$ = expected value ### Steps for Calculation 1. **Obtain** the expected (E) and observed (O) results for the experiment 2. **Calculate the difference** between each set of results - Work out $(O - E)$ for each category 3. **Square each difference** - Calculate $(O - E)^2$ for each category - It doesn't matter whether the difference is positive or negative 4. **Divide** each squared difference by the expected value - Calculate $\frac{(O - E)^2}{E}$ for each category 5. **Add** all the resulting values together to obtain the chi-squared value - Sum all the $\frac{(O - E)^2}{E}$ values

Creating a table with columns for O, E, $(O - E)$, $(O - E)^2$, and $\frac{(O - E)^2}{E}$ helps organize calculations and reduces errors

## Analysing Chi-squared Values ### Critical Values - To interpret the chi-squared value, compare it to a **critical value** - The critical value is read from a **table of critical values** - The critical value depends on: - The **probability level** used - The **degrees of freedom** ### Probability Level - Biologists typically use a probability level of **0.05 (5%)** - This means there is only a 5% probability that any difference between O and E occurred by chance - Or a 95% probability that the difference is due to a real biological factor ### Degrees of Freedom - Calculated using the formula: $$\text{degrees of freedom} = \text{number of classes} - 1$$ - "Classes" refers to the number of categories or phenotypes being compared - E.g. if comparing 2 phenotypes: $2 - 1 = 1$ degree of freedom - E.g. if comparing 4 phenotypes: $4 - 1 = 3$ degrees of freedom

Degrees of freedom = number of categories - 1

### Critical Values Table | Degrees of freedom | Probability that difference is due to chance | | | | | |--------------------|----------------------------------------------|-----|------|------|-------| | | 0.1 | 0.05| 0.01 | 0.001| | | 1 | 2.71 | 3.84| 6.64 | 10.83| | | 2 | 4.60 | 5.99| 9.21 | 13.82| | | 3 | 6.25 | 7.82| 11.34| 16.27| | | 4 | 7.78 | 9.49| 13.28| 18.46| | ### Drawing Conclusions **If the chi-squared value is greater than or equal to the critical value:** - There **is** a significant difference between observed and expected results - A factor other than chance is causing the difference - The **null hypothesis can be rejected** **If the chi-squared value is smaller than the critical value:** - There is **no** significant difference between observed and expected results - Any differences are due to **chance** - The **null hypothesis is accepted**

The null hypothesis assumes there is no significant difference between observed and expected results

### Assessing Probability Levels - The critical values table can also assess the exact probability level at which differences become significant **Example assessments:** - If $\chi^2$ is smaller than the critical value at p = 0.05, but larger than the critical value at p = 0.1: - There is between 5-10% probability that the difference is due to chance - If $\chi^2$ is larger than the critical value at p = 0.001: - There is less than 0.1% probability that the difference is due to chance - The difference is highly significant ## Worked Example: Pea Plant Flower Colour An experiment investigated the inheritance of two genes in pea plants: one for flower colour (purple/white) and one for seed shape (round/wrinkled). In this dihybrid cross, the expected ratio of phenotypes was 9:3:3:1. Two heterozygous plants were crossed, and the phenotypes of all offspring were recorded. **Complete a chi-squared test to determine whether the difference between observed and expected offspring ratios is significant.** ### Step 1: Complete a table | Phenotypes of offspring | purple flower, round seed | purple flower, wrinkled seed | white flower, round seed | white flower, wrinkled seed | |-------------------------|---------------------------|------------------------------|--------------------------|----------------------------| | **Observed number (O)** | 152 | 39 | 53 | 16 | | **Expected ratio** | | 9:3:3:1 | | | | **Expected number (E)** | 146.25 | 48.75 | 48.75 | 16.25 | | **$O - E$** | 5.75 | -9.75 | 4.25 | -0.25 | | **$(O - E)^2$** | 33.06 | 95.06 | 18.06 | 0.06 | | **$\frac{(O-E)^2}{E}$** | 0.226 | 1.950 | 0.370 | 0.004 | **Calculating expected values:** - Total offspring: $152 + 39 + 53 + 16 = 260$ - Ratio total: $9 + 3 + 3 + 1 = 16$ - $260 \div 16 = 16.25$ - Purple round: $9 \times 16.25 = 146.25$ - Purple wrinkled: $3 \times 16.25 = 48.75$ - White round: $3 \times 16.25 = 48.75$ - White wrinkled: $1 \times 16.25 = 16.25$ ### Step 2: Calculate the chi-squared value $$\chi^2 = \sum \frac{(O - E)^2}{E}$$ $$\chi^2 = 0.226 + 1.950 + 0.370 + 0.004$$ $$\chi^2 = 2.55$$ ### Step 3: Compare to the critical value **Calculate degrees of freedom:** - Number of phenotypes = 4 - Degrees of freedom = $4 - 1 = 3$ **Probability level:** 0.05 (standard for biology) **Critical value:** 7.82 (from table: 3 degrees of freedom, p = 0.05) ### Step 4: Draw conclusions - The chi-squared value of **2.55** is **smaller than** the critical value of **7.82** - Therefore, there is **no significant difference** between the expected and observed results - Any differences that do occur are due to **chance** - Looking at the table horizontally, 2.55 is located to the left-hand side - This indicates there is a **greater than 10% probability** that the difference between O and E is due to chance - The **null hypothesis can be accepted**

Always show all working when calculating chi-squared values - completing the table and showing the formula used is essential for full marks

## Key Points to Remember - Always start by creating a clear table with all your data organized - Remember to state both the **phenotype** (what you observe) and **genotype** (the genetic makeup) where relevant - Show the calculation for degrees of freedom explicitly - State which probability level you are using (usually 0.05) - Clearly state whether you accept or reject the null hypothesis - Link your conclusion back to the biological context - E.g. "This suggests the genes are not linked" or "This supports the expected ratio for independent assortment"

If asked to suggest reasons why results might be significantly different, consider: gene linkage, epistasis, sex linkage, or environmental factors affecting survival of certain phenotypes

## Common Applications The chi-squared test is commonly used to test for: 1. **Independent assortment** in dihybrid crosses - Testing whether genes assort independently or show linkage - Expected ratio: 9:3:3:1 2. **Monohybrid inheritance** patterns - Testing whether a gene shows simple dominance - Expected ratio: 3:1 in F₂ generation 3. **Sex-linkage** - Testing whether males and females show equal frequencies of phenotypes - Expected ratios depend on the specific cross 4. **Epistasis** - Testing whether modified ratios (e.g. 9:3:4, 12:3:1) are present - Indicates interaction between genes