Question 1
A researcher crossed a tomato plant heterozygous for both fruit colour (Aa) and plant height (Bb) with a plant homozygous recessive for both traits. The offspring were distributed as follows:
Phenotype Count Red fruit, tall 248 Red fruit, dwarf 252 Yellow fruit, tall 251 Yellow fruit, dwarf 249 What is the most reasonable conclusion from these data?
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Method #1Approach 1Step 1: Identify the cross type
This is a test cross: an AaBb individual is crossed with aabb (homozygous recessive for both traits). If the two genes assort independently, we expect a 1:1:1:1 ratio of the four phenotypes.
Step 2: Calculate expected vs. observed ratios
Total offspring = . Each of the four phenotype classes is approximately 250, giving a ratio very close to 1:1:1:1. This is exactly what independent assortment predicts.
Step 3: Interpret the result
If the genes were linked, parental combinations (e.g., red tall and yellow dwarf) would dominate, and recombinant types would be rare — the ratio would deviate significantly from 1:1:1:1. Here, all four classes are equally frequent, ruling out linkage.
Step 4: Select the correct conclusion
The near-perfect 1:1:1:1 ratio strongly supports independent assortment of the two genes, meaning they are located on different chromosomes (or far apart on the same chromosome).
Method #2Approach 2Step 1: Identify what is being asked
We need to interpret the phenotypic distribution from a test cross (AaBb × aabb) and select the most valid biological conclusion.
Step 2: Eliminate 'same chromosome and complete linkage'
Complete linkage would produce only two phenotype classes (parental types) in equal proportions — not all four classes equally. This option is eliminated.
Step 3: Eliminate 'codominance between A and a'
Codominance would produce a blended or intermediate phenotype in heterozygotes, not an equal distribution of four distinct phenotype classes. This option does not apply here.
Step 4: Eliminate 'mutation during meiosis'
A mutation is an unpredictable event and would not reliably produce a consistent 1:1:1:1 ratio across 1000 offspring. This explanation is not supported by the data.
Step 5: Select the correct answer
The equal distribution of all four phenotype classes (~250 each) is precisely what independent assortment predicts for a dihybrid test cross. This is the correct conclusion.
Question 2
In a certain species of rabbit, the allele C causes spotted coat pattern and the allele c causes plain coat. Genotype CC results in embryonic death, so only surviving offspring are counted. A breeder crosses two spotted rabbits and records the following surviving offspring:
Phenotype Count Spotted 60 Plain 28 What are the most likely genotypes of both parent rabbits?
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Method #1Approach 1Step 1: Identify the lethal genotype constraint
Because CC is lethal, offspring from a Cc × Cc cross that survive are: CC (die) + Cc (spotted) + cc (plain). Among survivors, the ratio simplifies to 2 spotted : 1 plain.
Step 2: Check observed ratio
Observed: 60 spotted : 28 plain 2.14 : 1, which is close to the expected 2:1 ratio when a lethal homozygous dominant genotype removes one class of offspring.
Step 3: Match to parental genotypes
Both parents are spotted, so neither can be cc. Since CC is lethal, surviving spotted parents must be Cc. A Cc × Cc cross produces the 2:1 surviving ratio observed.
Step 4: Confirm the answer
The genotype of each parent is Cc, making the cross Cc × Cc. This is the only cross that explains both the parental phenotype (spotted) and the 2:1 offspring ratio.
Method #2Approach 2Step 1: Identify what is being asked
We must determine parental genotypes based on the offspring ratio and the fact that CC is lethal.
Step 2: Eliminate CC × CC
CC × CC would produce only CC offspring, all of which would die — no surviving offspring would be observed. This is impossible given the data.
Step 3: Eliminate cc × cc
cc × cc would produce only plain-coated offspring. But both parents are spotted, so neither parent can be cc. Eliminated.
Step 4: Eliminate CC × Cc
CC × Cc would produce 50% CC (all die) and 50% Cc (all spotted). No plain offspring would survive. The data includes ~28 plain offspring, so this cross is ruled out.
Step 5: Select Cc × Cc
Cc × Cc produces CC (lethal), Cc (spotted), and cc (plain). Among survivors, the ratio is 2:1 (spotted:plain), matching the observed ~60:28. This is the correct answer.