5 - Chi Square Test

Define key terms such as "null hypothesis," "Chi-Square test," and "p-value."

Definitions of Key Terms 1. Null Hypothesis (H₀)

• Definition: A statistical hypothesis that assumes no real difference between observed and expected results.

• Example in Genetics: If we expect a 3:1 phenotypic ratio in a monohybrid cross, the null hypothesis states that any deviations from this ratio are due to chance, not a real effect.

• Interpretation:

  • If the null hypothesis is rejected, the deviation is statistically significant and suggests a factor (e.g., linked genes) affecting inheritance.

  • If we fail to reject the null hypothesis, the observed results fit expected Mendelian ratios.

2. Chi-Square (χ²) Test

• Definition: A statistical test used to compare observed and expected genetic data to determine whether deviations are due to chance or a real effect.

• Formula: χ² = Σ [(O - E)² / E]

  • O = observed frequency

  • E = expected frequency

• Use in Genetics: Determines if experimental phenotypic ratios match predicted Mendelian ratios (e.g., 3:1 or 9:3:3:1).

• Interpretation:

  • A low χ² value → Observed data closely matches expected data → Deviations are due to chance.

  • A high χ² value → Observed data is significantly different from expected → Suggests a violation of Mendelian inheritance assumptions.

3. P-Value

• Definition: The probability that the observed deviations from expected genetic ratios occurred by chance.

• Interpretation in Genetics:

  • p > 0.05: Fail to reject the null hypothesis → Observed data fits expected ratios (Mendelian inheritance is supported).

  • p < 0.05: Reject the null hypothesis → Observed data is significantly different from expected ratios (suggesting non-Mendelian effects like linkage).

Explain the significance of deviations from expected genotypic or phenotypic ratios

Significance of Deviations from Expected Genotypic or Phenotypic Ratios 1. Expected Ratios in Mendelian Inheritance

• Monohybrid cross (Aa × Aa): 3:1 phenotypic ratio

• Dihybrid cross (AaBb × AaBb): 9:3:3:1 phenotypic ratio

• These ratios are based on Mendel’s four assumptions:

  1. Each allele is dominant or recessive.

  2. Segregation is unimpeded.

  3. Independent assortment occurs.

  4. Fertilization is random.

2. Why Do Deviations Occur?

When observed ratios deviate significantly from expected ratios, it suggests that at least one of Mendel’s assumptions is violated. Possible explanations include:

1. Small Sample Size (Chance Deviations)

   • In small populations, random chance can lead to fluctuations in observed ratios.

   • Solution: A larger sample size helps minimize random deviations.

2. Linkage Between Genes

   • If two genes are on the same chromosome, they may not assort independently, leading to deviations from expected 9:3:3:1 dihybrid ratios.

   • Solution: A recombination test (linkage map) can confirm gene linkage.

3. Lethal Alleles or Epistasis

   • Some genotypes may be lethal, preventing their expected frequency in offspring.

   • Example: A 2:1 ratio instead of 3:1 suggests a lethal recessive allele.

4. Non-Random Mating or Selection

   • If certain genotypes have a survival or reproductive advantage, the observed ratios may shift from Mendelian predictions.

3. Using the Chi-Square (χ²) Test to Assess Significance

• The χ² test determines whether deviations from expected ratios are due to chance or a real effect.

• p-value Interpretation:

  • p > 0.05: Deviations are not significant → Inheritance follows Mendelian ratios.

  • p < 0.05: Deviations are significant → Suggests a violation of Mendelian inheritance assumptions.

Key Takeaway

Small deviations = Likely due to chance.

• Large deviations = Indicate genetic linkage, epistasis, or other biological factors.

• The Chi-Square test helps distinguish between these possibilities.

Calculate the Chi-Square value for given observed and expected genetic data

How to Calculate the Chi-Square (χ²) Value for Genetic Data1. Chi-Square (χ²) Formula:

χ² = Σ [(O - E)² / E]
Where:

• O = Observed values (actual counts from an experiment).

• E = Expected values (theoretical counts based on Mendelian ratios).

2. Example Calculation

Scenario: A monohybrid cross (Aa × Aa) should result in a 3:1 phenotypic ratio in the F₂ generation.
You observed the following offspring in a sample of 100 individuals:

• Dominant phenotype (A_)

  • Observed (O) = 76

  • Expected ratio = 3/4 (75%)

  • Expected (E) = 75

• Recessive phenotype (aa)

  • Observed (O) = 24

  • Expected ratio = 1/4 (25%)

  • Expected (E) = 25

3. Calculate χ² for Each Category:

For dominant (A_) phenotype:
χ²₁ = (76 - 75)² / 75 = 0.013

For recessive (aa) phenotype:
χ²₂ = (24 - 25)² / 25 = 0.04

Total χ² value:
χ² = 0.013 + 0.04 = 0.053

4. Interpretation of the χ² Value

• Degrees of Freedom (df) = Number of categories - 1 = 2 - 1 = 1

• Using a Chi-Square table, compare χ² = 0.053 with the critical value at df = 1.

• If p > 0.05 → Fail to reject the null hypothesis (observed ratio fits Mendelian predictions).

• If p < 0.05 → Reject the null hypothesis (significant deviation from expected ratios).

Assess the validity of a null hypothesis based on Chi-Square test results and p-values

Assessing the Validity of a Null Hypothesis Using Chi-Square Test Results and P-Values

1. Null Hypothesis (H₀) in Genetics

• The null hypothesis (H₀) states that there is no significant difference between the observed and expected genetic ratios.

• Example: In a monohybrid cross (Aa × Aa), we expect a 3:1 phenotypic ratio in the F₂ generation. The null hypothesis assumes any deviations are due to chance.

2. Chi-Square Test and P-Value Interpretation

1. Calculate the Chi-Square (χ²) value:

   • Formula: χ² = Σ [(O - E)² / E]

   • This measures how much the observed data deviates from expected ratios.

2. Determine the Degrees of Freedom (df):

   • df = Number of categories - 1

   • Example: A monohybrid cross (3:1 ratio) has df = 2 - 1 = 1

3. Find the p-value using a Chi-Square table:

   • p-value = Probability that deviations occurred by chance

   • Compare χ² value to a critical Chi-Square value for the given df

3. Decision Rule for the Null Hypothesis

p-Value	Interpretation	Decision on H₀
p > 0.05	Deviation is not significant (expected ratio holds).	Fail to reject H₀ (Inheritance follows Mendelian predictions).
p < 0.05	Deviation is significant (observed ratio does not match expectations).	Reject H₀ (Non-Mendelian factors may be influencing inheritance)

4. Example Interpretation

• Observed ratio: 3.15:1 instead of 3:1

• Calculated χ² = 0.39 with df = 1

• p-value between 0.5 and 0.8 → Deviation is due to chance

• Conclusion: Fail to reject H₀ → Data supports Mendelian inheritance.

Key Takeaways

• If p > 0.05, the null hypothesis is valid (observed ratios match expected ratios).

• If p < 0.05, the null hypothesis is rejected, meaning factors like linkage or selection may be influencing inheritance.

Distinguish between cases where deviations from Mendelian ratios are due to chance or a violation of assumptions like independent assortment

Distinguishing Between Deviations Due to Chance vs. Violations of Mendelian Assumptions

1. Expected Mendelian Ratios

• Monohybrid Cross (Aa × Aa) → 3:1 phenotypic ratio

• Dihybrid Cross (AaBb × AaBb) → 9:3:3:1 phenotypic ratio

• These ratios rely on Mendel’s four key assumptions:

  1. Each allele is either dominant or recessive.

  2. Segregation occurs normally.

  3. Independent assortment applies (genes are not linked).

  4. Fertilization is random.

2. Deviations Due to Chance (Small Sample Size)

• Cause: Random fluctuations in small populations.

• Effect: Slight differences between observed and expected ratios.

• Example: In a 3:1 ratio, expected results for 100 offspring are 75 dominant : 25 recessive, but you might observe 76:24 instead.

• Test: A Chi-Square (χ²) test will typically show a high p-value (p > 0.05), meaning no significant difference from expected ratios.

3. Deviations Due to Violations of Mendelian Assumptions A. Linked Genes (Independent Assortment Violated)

• Cause: Genes located close together on the same chromosome do not assort independently.

• Effect: Ratios deviate from 9:3:3:1 (in a dihybrid cross, fewer recombinants are observed).

• Example: Instead of 9:3:3:1, you might see 7:1:1:7, suggesting linkage.

• Test: A Chi-Square test will give p < 0.05, indicating a significant deviation from expected ratios.

B. Epistasis (Gene Interactions)

• Cause: One gene masks the effect of another.

• Effect: Ratios shift from 9:3:3:1 to modified ratios, such as 9:7 (complementary genes) or 12:3:1 (dominant epistasis).

• Example: In labrador retrievers, coat color depends on two genes, leading to a 9:3:4 ratio.

C. Lethal Alleles

• Cause: Certain genotypes are lethal, preventing expected offspring ratios.

• Effect: Instead of 3:1, a 2:1 ratio is observed because homozygous recessive individuals die early.

• Example: The AY allele in mice causes a lethal phenotype in homozygous (AYAY) individuals.

4. How to Distinguish the Cause of Deviations

Cause	Observed Effect	Chi-Square p-Value	Conclusion
Chance (Small Sample Size)	Minor deviations from expected ratios	p > 0.05	Fail to reject H₀ (Mendelian inheritance holds).
Gene Linkage	Fewer recombinant phenotypes in a dihybrid cross	p < 0.05	Reject H₀ (Independent assortment violated).
Epistasis	Modified Mendelian ratios (e.g., 9:7 or 12:3:1)	p < 0.05	Reject H₀ (Gene interactions affect phenotype).
Lethal Alleles	Missing genotypic category (e.g., 2:1 ratio instead of 3:1)	p < 0.05	Reject H₀ (Lethal allele affects survival).

Key Takeaways

• Small deviations = Chance (p > 0.05, Mendelian ratios hold).

• Large deviations (p < 0.05) suggest genetic interactions, linkage, or lethal alleles.

• A Chi-Square test helps determine if the deviation is statistically significant.

Design a hypothetical breeding experiment to test whether alleles segregate randomly and specify the expected and observed outcomes.

Hypothetical Breeding Experiment to Test Whether Alleles Segregate Randomly

1. Research Question

• Do alleles for a given gene segregate randomly according to Mendel’s Law of Segregation?

2. Experimental Design

Organism: Pea plants (Pisum sativum)
Trait Studied: Flower color

• Dominant allele (P) = Purple flowers

• Recessive allele (p) = White flowers

Cross Setup: Perform a monohybrid cross between two heterozygous (Pp × Pp) pea plants.

• If Mendel’s Law of Segregation holds, the expected phenotypic ratio should be 3:1 (purple:white).

3. Expected Outcomes

If alleles segregate randomly during gamete formation, then the offspring’s phenotypes should follow Mendelian ratios:

Genotype	Expected Proportion	Expected Phenotype	Expected Count (100 plants)
PP	1/4 (25%)	Purple	25
Pp	1/2 (50%)	Purple	50
pp	1/4 (25%)	White	25

4. Observed Outcomes (Hypothetical Data)

After growing 100 offspring, we collect the following data:

Phenotype	Observed Count (O)	Expected Count (E)
Purple	78	75
White	22	25

5. Statistical Test (Chi-Square Test)

To determine if the deviation from expected ratios is due to chance or a violation of Mendel’s Law, we calculate χ² using the formula:

χ² = Σ [(O - E)² / E]

Step 1: Calculate χ² for Each Category

For purple (P_) phenotype:

χ²₁ = (78 - 75)² / 75 = 9 / 75 = 0.12

For white (pp) phenotype:

χ²₂ = (22 - 25)² / 25 = 9 / 25 = 0.36

Step 2: Sum the Chi-Square Values

χ² = 0.12 + 0.36 = 0.48

6. Interpretation of Results

• Degrees of Freedom (df) = 2 - 1 = 1

• Using a Chi-Square table, for χ² = 0.48 and df = 1, the p-value is > 0.05.

• Conclusion:

  • We fail to reject the null hypothesis.

  • The observed data fits Mendelian ratios, meaning alleles segregate randomly as expected.

Key Takeaways

1. If p > 0.05, Mendelian segregation is valid (random segregation confirmed).

2. If p < 0.05, deviations are statistically significant, indicating a possible violation of Mendel’s assumptions (e.g., gene linkage or selection effects).

3. A Chi-Square test helps determine whether deviations are due to chance or non-Mendelian inheritance.

List the assumptions underlying Mendel's predicted ratios

Assumptions Underlying Mendel's Predicted Ratios

Mendel’s 3:1 monohybrid and 9:3:3:1 dihybrid ratios are based on the following four key assumptions:

1. Each allele is either dominant or recessive.

   • One allele completely masks the other in heterozygous individuals (e.g., Pp shows the dominant phenotype).

2. Segregation is unimpeded.

   • During gamete formation, paired alleles separate randomly, ensuring each gamete receives one allele from each gene.

3. Independent assortment occurs.

   • Genes located on different chromosomes assort independently during meiosis (no genetic linkage).

   • This leads to the expected 9:3:3:1 ratio in dihybrid crosses.

4. Fertilization is random.

   • Any sperm has an equal chance of fertilizing any egg, ensuring Mendelian ratios hold.

Significance:

• If any of these assumptions are violated, the observed ratios will deviate from the expected ratios.

• Example: If two genes are linked, independent assortment is violated, altering dihybrid ratios.

Use the product law and sum law to determine probabilities in genetic crosses before performing a Chi-Square test

Using the Product Law and Sum Law to Determine Probabilities in Genetic Crosses

Before performing a Chi-Square test, we use probability laws to predict expected genetic outcomes.

1. Product Law (Multiplication Rule)

• Definition: The probability of two or more independent events occurring together is the product of their individual probabilities.

• Formula:
  P(A and B) = P(A) × P(B)

Example: Probability of Producing an AaBb Offspring

• Cross: Aa × Aa (for one gene)

  • Probability of Aa genotype = (½ A from one parent) × (½ a from the other) = ¼

• Cross: Bb × Bb (for another gene)

  • Probability of Bb genotype = (½ B from one parent) × (½ b from the other) = ¼

• Final probability of AaBb offspring:
  P(AaBb) = (¼) × (¼) = 1/16

2. Sum Law (Addition Rule)

• Definition: The probability of obtaining one of two or more mutually exclusive events is the sum of their individual probabilities.

• Formula:
  P(A or B) = P(A) + P(B)

Example: Probability of Heterozygous (Aa) Offspring in Aa × Aa Cross

• Probability of getting Aa = (½ A from one parent, ½ a from the other) = ¼

• Probability of getting aA = (½ a from one parent, ½ A from the other) = ¼

• Final probability of heterozygous (Aa) offspring:
  P(Aa) = ¼ + ¼ = ½

3. Using These Laws Before a Chi-Square Test

• Why?

  • Before performing a Chi-Square test, we need to calculate expected values based on probabilities.

  • Using Product and Sum Laws, we determine the theoretical Mendelian ratios for a given cross.

Example Before Chi-Square Calculation (AaBb × AaBb Cross)

• Expected 9:3:3:1 phenotypic ratio in a dihybrid cross is calculated using the Product Law.

• If total offspring = 160, then:

  • 9/16 × 160 = 90 (dominant for both traits)

  • 3/16 × 160 = 30 (dominant for one, recessive for the other)

  • 3/16 × 160 = 30 (opposite dominant-recessive pattern)

  • 1/16 × 160 = 10 (recessive for both traits)

These expected values are then compared with observed values in the Chi-Square test.

Key Takeaways

• Use the Product Law to calculate probabilities of independent events (e.g., two genes segregating independently).

• Use the Sum Law to determine probabilities of mutually exclusive events (e.g., Aa forming from either A × a or a × A).

• These calculations provide the "expected values" for the Chi-Square test to determine if observed data fits Mendelian predictions.

Justify rejecting or failing to reject a null hypothesis in light of a statistical analysis of genetic data

Justifying Rejecting or Failing to Reject a Null Hypothesis in Genetic Data Analysis

1. Null Hypothesis (H₀) in Genetics

• The null hypothesis (H₀) states that there is no significant difference between the observed and expected genetic ratios.

• In Mendelian genetics, this means that alleles segregate and assort independently as predicted (e.g., 3:1 or 9:3:3:1 ratios).

2. Decision Rule Based on Chi-Square (χ²) and p-Value

To assess whether to reject or fail to reject H₀, we use the Chi-Square test:

1. Calculate χ² using the formula:
   χ² = Σ [(O - E)² / E]

   • O = Observed frequency

   • E = Expected frequency

2. Determine Degrees of Freedom (df):
   df = Number of categories - 1

3. Compare χ² to the Critical Value (Using a p-Value):

   • If p > 0.05 → Fail to reject H₀ (differences are due to chance).

   • If p < 0.05 → Reject H₀ (differences are statistically significant).

3. Interpretation of Statistical Results

Scenario	Chi-Square Test Result	Decision	Conclusion
Observed ratios closely match expected ratios	χ² is low, p > 0.05	Fail to reject H₀	The deviation is likely due to random chance, and Mendelian inheritance holds.
Observed ratios significantly differ from expected ratios	χ² is high, p < 0.05	Reject H₀	The deviation is not due to chance, suggesting a violation of Mendelian assumptions (e.g., gene linkage, selection effects).

4. Example Case Study: Monohybrid Cross (Pp × Pp)

Expected 3:1 ratio for 100 offspring:

• Expected: 75 purple (P_), 25 white (pp)

• Observed: 78 purple, 22 white

• χ² Calculation:

  • χ² = (78 - 75)² / 75 + (22 - 25)² / 25 = 0.12 + 0.36 = 0.48

  • df = 1, p-value > 0.05

Conclusion:

Since p > 0.05, we fail to reject H₀ → The observed deviation is due to chance, and Mendelian segregation holds.

Key Takeaways

• Fail to reject H₀ (p > 0.05): The observed ratios match expected ratios, meaning deviations are due to random chance.

• Reject H₀ (p < 0.05): The observed ratios differ significantly from expected ratios, suggesting a non-Mendelian factor (e.g., linkage, selection, or epistasis).