Questions: Epistasis and Complementary Gene Interactions

Question 1

Two true-breeding white-flowered plant varieties (AABB and aabb) are crossed. All F1 plants have purple flowers. F1 × F1 crosses give F2 offspring in a 9:7 ratio of purple:white. What does this pattern tell you about the relationship between the two genes?

• A. Gene A is epistatic to Gene B — a dominant A allele prevents Gene B from being expressed, producing a 9:7 modified ratio
• B. Both genes must each contribute a functional product to produce purple pigment; losing either one results in white, regardless of the other gene's genotype
• C. The two white-flowered parents carried the same recessive mutation, and the F1 plants are heterozygous at both loci
• D. The 9:7 ratio indicates incomplete dominance at both loci, producing three phenotypic classes instead of four

Answer: (B)

A 9:7 ratio is the fingerprint of complementary gene action: the 9 A_B_ class has both functional gene products (purple), while the 3 A_bb, 3 aaB_, and 1 aabb classes each lack at least one functional product and are all white. The key insight is that both pathways must contribute — think of two subunits of the same enzyme complex. Crucially, the two original white parents carried mutations in DIFFERENT genes in the same pathway; crossing them produces plants that are heterozygous at both loci and can make both products. This is the complementation test in action.

Question 2

In dominant epistasis producing a 12:3:1 ratio, what is the biochemical logic that generates the 12-class phenotype?

• A. The 12 class includes all individuals with at least one dominant A allele, because A overrides Gene B by directly suppressing its transcription
• B. The 12 class includes A_B_ (9) plus A_bb (3) — all individuals where Gene A is functional, regardless of Gene B, because Gene A supplies a substrate that only Gene B converts further
• C. The 12 class arises because dominant alleles at either locus produce the same phenotype through redundant pathways
• D. Gene A is hypostatic to Gene B — it requires a functional B product before it can act

Answer: (B)

In dominant epistasis, Gene A acts earlier in a pathway and Gene B acts on Gene A's product. Individuals with at least one functional A allele (A_) produce the intermediate substrate, regardless of whether Gene B is functional. If Gene B is also functional (A_B_), the substrate is converted to a further product. If Gene B is non-functional (A_bb), the substrate accumulates, giving a different phenotype — but still not the baseline white of the 'aa' class. The 4 aaB_ and aabb individuals have no substrate at all because Gene A is broken, so Gene B is irrelevant; they all show the epistatic phenotype.

Question 3

Modified dihybrid ratios like 12:3:1 or 9:7 result from violations of Mendel's law of segregation — alleles at epistatic loci do not segregate in the expected 1:2:1 ratios.

• T. True
• F. False

Answer: False

This is the most important misconception to correct. Modified ratios are not violations of segregation. The alleles at each locus still segregate in perfect 1:2:1 ratios, and genes on different chromosomes still assort independently. What changes is only the phenotypic outcome of different genotypic combinations — because gene products interact in biochemical pathways. If you scored each locus separately in a 9:7 cross, each would show exactly the Mendelian 3:1 ratio. Epistasis modifies the phenotypic ratios, not the genetic ones.

Question 4

A 9:7 phenotypic ratio from a dihybrid F2 cross and a 9:3:4 ratio both sum to 16 and both involve two genes that interact, but they indicate different types of gene interaction.

• T. True
• F. False

Answer: True

All modified dihybrid ratios sum to 16 because they are rearrangements of the same 16 genotypic classes (from 4 × 4 gamete combinations). The 9:7 ratio signals complementary gene action (both products required for phenotype); the 9:3:4 ratio signals recessive epistasis (homozygous recessive at one locus masks expression of the other). Each ratio is a diagnostic fingerprint of a distinct type of genetic interaction. The fact that they all sum to 16 is a direct consequence of independent assortment, which is not being violated.

Question 5

Why do modified dihybrid ratios (like 9:7 or 12:3:1) always sum to 16, and what does this tell us about what Mendel's laws are — and are not — being violated?

Model answer: All modified ratios sum to 16 because a standard dihybrid cross (AaBb × AaBb) always produces exactly 16 equally probable genotypic classes: 1 AABB + 2 AABb + 1 AAbb + 2 AaBB + 4 AaBb + 2 Aabb + 1 aaBB + 2 aaBb + 1 aabb. These classes arise from independent assortment, which is not violated. What changes in epistatic crosses is how many of these genotypic classes produce the same phenotype. In 9:7 complementary action, 7 of the 16 classes all share a 'white' phenotype. The laws of segregation and assortment describe how alleles move through generations — they still hold perfectly. The gene interaction affects phenotypic expression, not allele distribution.

This question forces students to distinguish genotype ratios (which obey Mendel's laws) from phenotype ratios (which reflect gene interactions). Once students grasp that all 16 classes are still present in any dihybrid F2 cross, they can decode any modified ratio by asking: which genotypic classes have been lumped together into the same phenotypic category, and why?