Dihybrid crosses and test crosses

Explanation

For AaBb with independently assorting loci, gametes AB, Ab, aB and ab each have probability 1/4. In AaBb x AaBb, combining the gametes gives 16 equally likely cells in a Punnett square. Group genotypes into phenotypes only after applying the stated dominance relationships.

With complete dominance at both loci and no gene interaction affecting the classes, A_B_, A_bb, aaB_ and aabb occur in expected proportions 9/16, 3/16, 3/16 and 1/16. The underscore means either allele can occupy the second position, not an unknown extra gene.

A test cross pairs an individual showing dominant phenotypes with a homozygous recessive tester. For AaBb x aabb under independent assortment, the offspring directly reveal the four gamete types from the heterozygote and are expected in a 1:1:1:1 ratio. An all-dominant small sample does not prove homozygosity with certainty.

Use multiplication for independent events: P(aa and bb) = 1/4 x 1/4 = 1/16 in the heterozygote cross. Linkage or selection among offspring can invalidate simple independence assumptions. Expected ratios describe probabilities; observed small samples need not match exactly.

A genetic diagram earns marks row by row, so label every row: parental phenotypes, parental genotypes, gametes (each circled), offspring genotypes, and offspring phenotypes with their ratio. Write the phenotype in words as well as the genotype.

To test whether observed offspring fit 9:3:3:1, complete a chi-squared table with one row per phenotype class and columns for O, E, O - E, (O - E) squared and (O - E) squared / E. Chi-squared is the sum of the last column. With four classes there are 3 degrees of freedom, and the critical value at p = 0.05 is 7.82. A smaller value means the difference is not significant, so the results are consistent with independent assortment; a larger value suggests the loci are linked or interact.