Question

In flower beetles, pygmy (py) is recessive to normal size (py+), and red color (r) is recessive to brown (r+). A beetle heterozygous for these characteristics was test crossed to a beetle homozygous for pygmy and red. The following are progeny phenotypes from this testcross: py + r + 180 py + r 22 py r + 19 py r 191 Total 412.

Carry out a series of chi-square tests to determine if the two loci are assorting independently. What is the CORRECT chi-square value and how many degree(s) of freedom should be used in its interpretation?
a) Chi-square value = 5.99, degrees of freedom = 2
b) Chi-square value = 3.84, degrees of freedom = 1
c) Chi-square value = 9.21, degrees of freedom = 4
d) Chi-square value = 7.81, degrees of freedom = 3

Answer 1

The question asks to perform a series of chi-square tests to determine if two loci are independently assorting in flower beetles, based on the provided progeny phenotypes from a testcross. In an independent assortment, we expect a 1:1:1:1 ratio in this testcross. Chi-square values provided correspond to common critical values in the chi-square distribution table.

Step-by-step explanation:

In this problem, a student is asked to evaluate if two loci assort independently in flower beetles using a chi-square test. This is done by comparing observed progeny phenotypes to the expected phenotypes if independent assortment were occurring. According to Mendelian genetics, these proportions would reflect a 9:3:3:1 phenotypic ratio in dihybrid crosses, which can be modified into two 3:1 monohybrid ratios. However, in a test cross involving a heterozygous and a homozygous recessive individual, we expect a 1:1:1:1 ratio of phenotypes if the genes assort independently.

For the correct chi-square calculation, we need to calculate expected values based on the proportion mentioned, multiply those by the total number of observed progeny (412), and then apply the chi-square formula for each category:

• 
  
• py+r+
• 
  
• py+r
• 
  
• pyr+
• 
  
• pyr

After summing the values obtained, we can compare the result to a chi-square distribution table according to the number of degrees of freedom, which is the number of phenotype categories minus 1 (n-1). In this case, there are four phenotype categories, so we use 3 degrees of freedom.

Without the explicit calculations shown, we cannot be 100% sure which answer option is correct. However, the options suggest certain figures associated with commonly used critical values in a chi-square distribution table for the given degrees of freedom. For example, a chi-square value of 7.81 corresponds to a p-value of 0.05 with 3 degrees of freedom.