Performing a Chi-Squared Test

The trait for smooth peas (R) is dominant over wrinkled peas (r) and yellow pea colour (Y) is dominant to green (y)

• Two heterozygous pea plants are crossed (RrYy × RrYy) and yield the following results:
  

  • 701 smooth yellow peas
    

  • 204 smooth green peas
    

  • 243 wrinkled yellow peas
    

  • 68 wrinkled green peas

Step 1: Calculate expected frequencies for an unlinked trait 

• Expected frequencies can be determined by completing a dihybrid cross (i.e. punnett grid)
  

• Phenotypic ratios = 9 smooth yellow : 3 smooth green : 3 wrinkled yellow : 1 wrinkled green
  

Step 2: Construct a table of frequencies

• Observed values are the actual values collected from crossing the pea plants
  

• Expected values = phenotypic ratio × total number of peas 
  

• Total peas = 701 + 204 + 243 + 68 = 1216

Step 3: Calculate a chi-squared value

• χ2 = ∑(O – E)2 ÷ E
  

• 0.42 + 2.53 + 0.99 + 0.84 = 4.76
  

Step 4: Identify the p value

• The p value indicates the probability that the results are due to chance (lower p value is more significant)
  

  • A p value of less than 5% chance (p<0.05) is considered to be statistically significant
    

• The degree of freedom (df) designates what range of values fall within each significance level
  

  • For this dihybrid cross, the degree of freedom should be 3 (number of phenotypes – 1)
    

• If p<0.05 the alternative hypothesis is accepted, otherwise the null hypothesis is accepted

  • Alternative hypothesis: There is a significant difference between observed and expected frequencies (genes are linked)
    

  • Null hypothesis: There is no significant difference between observed and expected frequencies (genes are unlinked)
    

Step 5: Determine statistical significance

• The chi-squared value (4.76) is less than the critical value for significance (7.82)
  

• Hence the results are not statistically significant (null hypothesis is accepted – genes are unlinked)