Questions: Test Crosses: Determining Unknown Genotypes

Question 1

An organism showing the dominant phenotype for a single trait is crossed with a homozygous recessive (aa) organism. All 60 offspring display the dominant phenotype. What conclusion is best supported?

• A. The unknown parent is definitely Aa — the 1:1 ratio just wasn't realized by chance in this small sample
• B. No conclusion is possible; a test cross requires at least 200 offspring to be informative
• C. The unknown parent is most likely AA, because if it were Aa, we would expect roughly 30 recessive offspring — an outcome with near-zero probability
• D. The unknown parent could be either AA or Aa; only molecular sequencing can determine the genotype

Answer: (C)

The test cross is a probabilistic tool. If the unknown parent were Aa, each offspring would have a 50% chance of being recessive (aa). The probability of getting 60 dominant offspring in a row from an Aa × aa cross is (0.5)^60 ≈ 10^−18 — essentially impossible. Observing all dominant offspring strongly supports the hypothesis that the parent is AA. This is not certainty (the parent could theoretically be Aa with extraordinary luck), but 60/60 dominant offspring is overwhelmingly consistent with AA and incompatible in practice with Aa.

Question 2

A dihybrid test cross (suspected AaBb × aabb) produces 41 AaBb, 39 Aabb, 42 aaBb, and 43 aabb offspring. What do these results indicate about the genes?

• A. The two genes are linked — the equal proportions indicate suppressed recombination
• B. The two genes assort independently — the 1:1:1:1 ratio matches the expectation under independent assortment
• C. The recombination frequency is approximately 50%, indicating the genes are on different chromosomes but very close together
• D. The AaBb parent must have been homozygous recessive at one of the two loci

Answer: (B)

With 165 total offspring and roughly equal numbers in all four classes (~41 each), the observed ratio is approximately 1:1:1:1, which is exactly what independent assortment predicts. The four gamete types (AB, Ab, aB, ab) are being produced in equal frequencies by the heterozygous parent, confirming that the two genes are on different chromosomes (or far apart on the same chromosome). Linkage would produce an excess of parental-type classes and a deficit of recombinant types.

Question 3

In a test cross, the primary role of the homozygous recessive parent is to ensure that all offspring phenotypes directly reflect the gamete types produced by the unknown parent.

• T. True
• F. False

Answer: True

Because the aa (or aabb) parent can only contribute recessive alleles, it acts as a genetic 'blank' — it adds no dominant alleles that could mask the gametes coming from the other parent. Whatever allele combination each offspring shows in its phenotype came from the unknown parent's gamete. This is why the test cross is a 'direct readout' of gamete frequencies, and why the choice of a homozygous recessive tester is essential to the method.

Question 4

A deviation from the expected 1:1:1:1 ratio in a dihybrid test cross generally indicates an error in crossing technique or scoring.

• T. True
• F. False

Answer: False

Deviations from 1:1:1:1 are the primary evidence for genetic linkage. When two genes are physically located on the same chromosome, they tend to be inherited together rather than assorting independently. In a test cross, linked genes produce an excess of parental-type offspring (matching the original parent's chromosome arrangements) and a deficit of recombinant types (new combinations produced by crossing over). The magnitude of the deviation estimates recombination frequency and genetic distance between the loci.

Question 5

Explain why recombination frequency measured in a dihybrid test cross can be used to estimate genetic distance between two genes.

Model answer: In a test cross, the aabb parent contributes only recessive alleles to every offspring, so each offspring class directly represents one gamete type produced by the heterozygous parent. For linked genes, most gametes are parental-type (original chromosome arrangements) and recombinant gametes arise only when crossing over occurs between the two loci. Crossing over occurs more frequently between genes that are far apart (more physical space for crossover events) and less frequently between genes that are close together. The recombination frequency — recombinants divided by total offspring — therefore estimates how often the chromosomal segment between the two genes undergoes crossing over, which correlates with physical distance.

This is why 1 map unit (1 centimorgan) is defined as a 1% recombination frequency. The linear relationship between recombination frequency and genetic distance (up to about 50 cM, beyond which the genes behave as if unlinked) is what allows test cross data to build genetic maps and order genes along chromosomes.