18. Consider the following hypothetical pathway:

    H allele converts X substance to H substance

    h allele cannot convert X to H substance and leads to phenotype ‘O’

    A allele converts H substance leading to A phenotype a allele cannot convert H substance

B allele converts H substance leading to B phenotype

b allele cannot convert H substance

An individual with ‘A’ phenotype when crossed with that of ‘B’ phenotype has a progeny with ‘O’ phenotype.
Which one of the following crosses can lead to the above observation?
(1) Aahh X BbHH      (2) AaHh X BBHh
(3) AaHh X BBHH     (4) AAHH X BbHh

Direct answer: The cross that can produce an ‘O’ phenotype child from an ‘A’ phenotype parent and a ‘B’ phenotype parent under the given pathway is option (2) AaHh × BBHh.

Question recap in simple words

• H converts precursor X → H substance.

• h cannot convert X → H, giving phenotype O.

• A converts H → A phenotype; a cannot convert H.

• B converts H → B phenotype; b cannot convert H.

So:

• Any genotype with at least one H allele (H_) can form H substance; hh gives O directly.

• From H, A_ makes A phenotype, B_ makes B phenotype.

• If both A_ and B_ act together with H_, the combined walnut/AB type appears (as in the 9:3:3:1 diagram: Walnut 9, Rose 3, Pea 3, Single 1).

The question: “An individual with ‘A’ phenotype when crossed with that of ‘B’ phenotype has a progeny with ‘O’ phenotype. Which cross can lead to this?”

To get:

• Parent 1 phenotype A → must have H_ and A_ but bb (no B effect).

• Parent 2 phenotype B → must have H_ and B_ but aa (no A effect).

• Child with phenotype O → must be hh (no H substance, so neither A nor B can act).

Thus, at least one h allele must come from each parent, and the child must be hh.

Genotype logic for each option

Notation:

• First two letters: A locus (AA, Aa, aa).

• Next two letters: H locus (HH, Hh, hh).

• Last two letters: B locus (BB, Bb, bb).

Option (1) Aahh × BbHH

• First parent Aahh:

  • H locus = hh → cannot form H; phenotype automatically O (not A).

  • So this parent is O, not A.

• Second parent BbHH:

  • H_ and B_ present, aa missing, so phenotype B.

• Offspring H genotypes: all from hh × HH → all Hh (no hh child).

• Therefore O phenotype in progeny is impossible, and one parent is not A.

• Option (1) is incorrect.

Option (2) AaHh × BBHh

Interpretation suited to the question: “individual with ‘A’ phenotype” is the AaHh parent assumed to have bb at B locus (not shown in shorthand), and the “B phenotype” parent is BBHh assumed to have aa at A locus. In exam context, only segregating loci relevant to O (H locus) are shown explicitly; A or B homozygosity or recessive state can be inferred.

• AaHh (A phenotype parent):

  • Carries A_ and H_ so makes A phenotype, and also carries h allele.

• BBHh (B phenotype parent):

  • Carries B_ and H_ so makes B phenotype and also carries h allele.

• H-locus cross: Hh × Hh → 1/4 HH, 1/2 Hh, 1/4 hh.

• So 1/4 progeny are hh → O phenotype.

• Also, both parents still show A or B phenotypes themselves.

• This option satisfies: A parent × B parent → some O children.

• Option (2) is correct.

Option (3) AaHh × BBHH

• First parent AaHh: A phenotype and has Hh.

• Second parent BBHH: B phenotype and has only H (no h).

• H-locus cross: Hh × HH → 1/2 HH, 1/2 Hh, 0 hh.

• No hh offspring possible, so no O phenotype child can appear.

• Option (3) is incorrect.

Option (4) AAHH × BbHh

• First parent AAHH: A phenotype, only H allele (no h).

• Second parent BbHh: B phenotype with Hh.

• H-locus cross: HH × Hh → all HH or Hh, but never hh.

• Again, O phenotype (hh) cannot occur in progeny.

• Option (4) is incorrect.

Therefore, only option (2) can give an O phenotype child from A × B phenotypes.

Introduction

In classical genetics questions, pathways involving multiple genes such as H, A and B are frequently used to test understanding of epistasis and phenotype prediction. For exam aspirants, mastering how an A phenotype crossed with a B phenotype can yield an O phenotype offspring is critical for solving CSIR NET and similar questions quickly and accurately. This article explains the walnut–rose–pea pathway and analyses each cross option so you can identify the correct genotype combination with confidence.

Pathway and phenotype rules

• H converts a precursor X into H substance; hh fails to do so, directly producing the O phenotype.

• Once H is present, A_ converts H into A phenotype, while B_ converts H into B phenotype; combinations of A_ and B_ with H_ give more complex walnut-like phenotypes.

• Therefore, any individual with genotype hh will show O phenotype, regardless of alleles at A or B loci, making the H locus epistatic to A and B.

Requirement for A × B → O cross

• The A phenotype parent must carry A_ H_ and lack functional B (bb) so that only A expression occurs on H substance.

• The B phenotype parent must carry B_ H_ and lack functional A (aa) so that only B expression occurs.

• For an O phenotype child, the genotype must be hh, meaning each parent must contribute one h allele; this requires both parents to be Hh at the H locus.

Evaluation of each option

• Option (1) Aahh × BbHH fails because the first parent is hh and therefore O, not A, and all offspring at the H locus are Hh with no hh child possible.

• Option (2) AaHh × BBHh works because each parent is phenotypically A or B with at least one H allele but both carry one h, so a quarter of the progeny become hh and hence O phenotype.

• Option (3) AaHh × BBHH cannot give O because the BBHH parent lacks h, so all progeny are H_ and none are hh.

• Option (4) AAHH × BbHh also cannot give O since the AAHH parent has no h allele, again preventing hh offspring.

Key takeaway for exams

To quickly solve similar questions, first identify which genotype produces the epistatic “null” phenotype (here hh → O), then ensure both parents can supply that recessive allele while still expressing their own phenotypes (A or B). By checking each proposed cross against these conditions, you can systematically eliminate wrong options and select the correct cross, here AaHh × BBHh, within seconds in competitive exams.