cl+gl+/clgl (F1 hybrids) x clgl/clgl (tester plants)

If the chlorina and glabrous loci are assorting independently, then the cross above would be expected to produce F2 offspring with what phenotypic ratio?

Question

cl+gl+/clgl (F1 hybrids) x clgl/clgl (tester plants)

If the chlorina and glabrous loci are assorting independently, then the cross above would be expected to produce F2 offspring with what phenotypic ratio?

BlackTom AI · Accepted Answer

We start by identifying the cross and what it implies for inheritance across two loci.
The F1 hybrid is cl+ gl+ / cl gl, meaning it is a dihybrid for two genes with coupling phase (the + alleles on one chromosome and the recessive alleles on the other).
The tester plant is clgl / clgl, homozygous recessive at both loci. This setup is a standard dihybrid testcross, so the F1 makes four possible gametes in equal frequencies if the two loci assort independently: AB, ab, Ab, aB, each at 1/4.
Here, AB corresponds to cl+ gl+ (wildtype), ab to cl gl (double mutant), Ab to cl+ gl (single mutant for gl), and aB to cl (single mutant for cl).
When these gametes combine with the tester’s ab genotype, the resulting phenotypes reflect the following zygotes:
- AB × ab yields AB/ab, which has both dominant phenotypes, i.e., wildtype.
- Ab × ab yields Ab/ab, which has the cl+ and gl recessive combination, i.e., single mutant for gl.
- aB × ab yields aB/ab, which has the cl recessive and gl+ dominant combination, i.e., single mutant for chlorina.
- ab × ab yields ab/ab, which is double recessive for both traits, i.e., double mutant.
Since each F1 gamete type is 1/4, the four resulting phenotypic classes should occur with equal frequency: 1/4 wildtype, 1/4 single mutant (chlorina), 1/4 single mutant (glabrous), and 1/4 double mutant.
Now evaluate each answer option:
- Option: 9 wildtype : 3 single mutant : 3 single mutant : 1 double mutant
This is the classic 9:3:3:1 ratio from a dihybrid self-cross, not a testcross. The question describes a testcross, so this ratio is incorrect in this context.
- Option: 1 wildtype : 1 single mutant : 1 single mutant : 1 double mutant
This matches the expected 1:1:1:1 distribution for a dihybrid testcross with complete independence between loci. It correctly assigns one class to wildtype, one class to single mutant for gl, one class to single mutant for cl, and one class to double mutant.
- Option: 1 homozygous dominant : 2 heterozygous : 1 homozygous recessive
This is the typical phenotype/genotype ratio for a single-gene monohybrid cross (dominant vs recessive) or a Punnett square for a two-allele locus without considering a second gene; it does not apply to a dihybrid testcross involving two loci.
- Option: 3 wildtype : 1 mutant
This resembles a monohybrid expectation skewed toward wildtype and a single mutant phenotype, not matching the two-locus, independent-assortment scenario described here.
In summary, the correct interpretation for a dihybrid F1 crossed to a double-recessive tester with independent assortment yields a 1:1:1:1 phenotypic ratio, which aligns with the second option above.

AGRI10051_2025_SM2 Practical 4 Post-prac Test

cl+gl+/clgl (F1 hybrids) x clgl/clgl (tester plants) If the chlorina and glabrous loci are assorting independently, then the cross above would be expected to produce F2 offspring with what phenotypic ratio?

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