Even if a character shows multiple allelism, an individual will only have two alleles for that character. Why?

NCERT Exemplar · Q32

Q. Even if a character shows multiple allelism, an individual will only have two alleles for that character. Why?

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Concept understanding — Dihybrid Cross Ratio

Let’s begin with something you already know from everyday life. Think about a family where the parents have two different traits — say, one parent has curly hair and brown eyes, the other has straight hair and blue eyes. Their children might inherit any combination: curly hair with brown eyes, straight hair with blue eyes, curly hair with blue eyes, or straight hair with brown eyes. You can see that traits don’t always travel together; they can mix and match.

That mixing is exactly what a dihybrid cross is about. In biology, a dihybrid cross is a breeding experiment that tracks two different traits at the same time — for example, seed shape (round vs wrinkled) and seed colour (yellow vs green) in pea plants. The “dihybrid cross ratio” is the predictable pattern in which these two traits appear in the offspring when both parents are hybrid (carrying one dominant and one recessive version) for both traits.

The classic result, as stated in the NCERT textbook, is a 9:3:3:1 ratio in the second generation. That means:

9 out of 16 offspring show both dominant traits (e.g., round and yellow)
3 out of 16 show the first dominant trait and the second recessive trait (e.g., round and green)
3 out of 16 show the first recessive trait and the second dominant trait (e.g., wrinkled and yellow)
1 out of 16 shows both recessive traits (e.g., wrinkled and green)

Important

The 9:3:3:1 ratio is not a random outcome. It is the direct consequence of independent assortment — the principle that genes for different traits are inherited independently of one another. This is one of Mendel’s key laws, and the ratio is its visible proof.

Why does this matter for a commerce or humanities student? Because this ratio is a classic example of probability in action. It shows how combinations of independent events produce predictable patterns — the same logic that underlies risk assessment in insurance, portfolio diversification in finance, or even the likelihood of certain combinations in a game of cards. You don’t need to calculate anything; you just need to see that nature follows rules, and those rules can be expressed as simple proportions. …

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An individual diploid organism carries only two alleles for any gene because it inherits one chromosome (and thus one allele) from each parent, regardless of how many different allelic forms exist in the population.

The concept of multiple allelism often confuses students when they first encounter it, because it seems to suggest that a single organism might carry three, four, or even more versions of the same gene. But that's not what multiple allelism means at all. It refers to the existence of more than two allelic forms of a gene in the population as a whole, not within a single individual.

Consider the classic example of human ABO blood groups. The gene for blood type has three alleles: $I^A$, $I^B$, and $i$. These three versions circulate in the human population, and different people carry different combinations. But here's the key: you, as an individual, can only possess two of these three alleles. You might be $I^A I^A$, $I^A i$, $I^B i$, $I^A I^B$, $I^B I^B$, or $ii$, but you cannot be $I^A I^B i$ or carry all three simultaneously.

The reason is rooted in the chromosomal basis of inheritance. Humans are diploid organisms, meaning every somatic cell contains two sets of chromosomes—one inherited from the mother, one from the father. Each chromosome in a pair carries genes at corresponding locations (loci). So for any given gene, you have exactly two copies: one on the maternal chromosome and one on the paternal chromosome. These two copies are your two alleles for that character.

Note

The term "multiple alleles" describes genetic diversity at the population level, not the genetic makeup of an individual. A population may harbor dozens of allelic variants, but the diploid condition restricts each organism to just two. …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

AP EAPCET 2025Set ap-2025-05-19-FN1 markMCQ
$Q. O-group child cannot have parents of following blood groups. (A) B and B (B) A and B (C) O and O (D) AB and O$
›Reveal solutionSolution
An AB parent never carries the recessive 'i' allele, so AB × O parents can never produce an O-group (ii) child. Answer: (D).

Concept and Intuition

ABO blood group inheritance follows multiple allele genetics with codominance: the alleles are Iᴬ, Iᴮ (both dominant over i, and codominant with each other), and i (recessive). Genotypes: AA/Ai → Group A; BB/Bi → Group B; IᴬIᴮ → Group AB; ii → Group O.

For a child to be blood group O (genotype ii), it must inherit one 'i' allele from EACH parent. This means each parent must be capable of passing on an 'i' allele — i.e., each parent's genotype must include at least one 'i' (so a parent of blood group A must be Ai, not AA; a parent of blood group B must be Bi, not BB; and a parent of blood group AB, genotype IᴬIᴮ, can NEVER pass on an 'i' allele, since it has none).

Step-by-Step Solution

Recall that O-group child requires genotype ii, needing an 'i' allele from both parents.

Check (A) B and B: If both are heterozygous Bi × Bi, a quarter of offspring would be ii (O group) — POSSIBLE.

Check (B) A and B: If A is Ai and B is Bi, a quarter of offspring would be ii (O group) — POSSIBLE.

Check (C) O and O: ii × ii → all offspring are ii (O group) — not just possible, but guaranteed. …
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AP EAPCET 2025Set ap-2025-05-20-FN1 markMCQ
$F_2$
›Reveal solutionSolution
Summing the relevant phenotype classes from the 9:3:3:1 dihybrid ratio gives wrinkled seeds = 4/16 and yellow seeds = 12/16, so the ratio wrinkled : yellow simplifies to 1:3 — option (A).

Concept and Intuition

In a dihybrid cross of two independently assorting traits (seed shape: round R dominant over wrinkled r; seed colour: yellow Y dominant over green y), the F2 generation shows the classic 9:3:3:1 ratio across the four combined phenotype classes. To find the ratio for a single trait (ignoring the other), you sum across all classes sharing that trait's phenotype.

Step-by-Step Solution

Cross: RRYY (round, yellow) × rryy (wrinkled, green) → F1 all RrYy (round, yellow).

F2 dihybrid ratio (16 parts total): 9 R_Y_ (round yellow) : 3 R_yy (round green) : 3 rrY_ (wrinkled yellow) : 1 rryy (wrinkled green). …
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AP EAPCET 2025Set ap-2025-05-20-FN1 markMCQ
$F_2$
›Reveal solutionSolution
Recombinants make up 6 parts out of the 16-part dihybrid F2 ratio (the two 3/16 classes); 1280 × 6/16 = 480 — option (C).

Concept and Intuition

In a dihybrid F2 (9:3:3:1), the 9-part class (both dominant, matching one parent) and the 1-part class (both recessive, matching the other parent) are the parental (non-recombinant) types, while the two 3-part classes (each combining one dominant and one recessive trait in a way neither original parent showed) are the recombinant types.

Step-by-Step Solution

F2 ratio: 9 (parental) : 3 (recombinant) : 3 (recombinant) : 1 (parental), total 16 parts.

Recombinant fraction = 3 + 3 = 6 parts out of 16 = 6/16 = 3/8. …
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AP EAPCET 2024Set ap-2024-05-16-AN1 markMCQ
$Q. In the progeny of dihybrid cross of pea plant, the phenotype ratio between the proportions of recombinants and parent plants respectively is (A) 2:4 (B) 4:2 (C) 3:5 (D) 5:3$
›Reveal solutionSolution
Out of the F2 9:3:3:1 dihybrid ratio, the 9+1=10 classes reproduce the original parental phenotype combinations, and the 3+3=6 classes are new recombinants, giving a recombinant:parental ratio of 3:5.

Concept and Intuition

When a dihybrid cross starts with two true-breeding parents differing in two independently assorting genes (e.g., YYRR x yyrr), the F1 is heterozygous for both (YyRr), and selfing the F1 gives the classic F2 phenotypic ratio 9:3:3:1. Two of these four phenotype classes reproduce the exact combinations seen in the original parents (the double-dominant class 9, resembling one parent, and the double-recessive class 1, resembling the other parent); these are called parental types. The other two classes (each ratio 3) represent new combinations of traits not seen in either original parent, arising from independent assortment during meiosis; these are the recombinant types.

Step-by-Step Solution

Recall the F2 dihybrid phenotype ratio: 9 (Y_R_) : 3 (Y_rr) : 3 (yyR_) : 1 (yyrr).

Identify parental-type classes: the 9 (both dominant traits together, as in one parent) and the 1 (both recessive traits together, as in the other parent), total = 9+1 = 10. …
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AP EAPCET 2024Set ap-2024-05-16-AN1 markMCQ
$F_2$
›Reveal solutionSolution
Using the standard 9-class F2 dihybrid genotype distribution (1:2:1:2:4:2:1:2:1 out of 16), the four requested genotypes YYrr, yyRR, YyRR, Yyrr occur in the ratio 1:1:2:2.

Concept and Intuition

For two independently assorting gene pairs, each behaves according to Mendel's law of segregation, giving a 1:2:1 genotypic ratio on its own (1 homozygous dominant : 2 heterozygous : 1 homozygous recessive). Because the two genes assort independently, the combined genotype ratio for both genes together is simply the product of each gene's individual 1:2:1 ratio, expanded as a 3x3 grid, yielding nine distinct genotype classes out of 16 total offspring, with frequencies 1, 2, 1, 2, 4, 2, 1, 2, 1 for the various combinations.

Step-by-Step Solution

Write the individual monohybrid ratios: Y-gene -> 1 YY : 2 Yy : 1 yy; R-gene -> 1 RR : 2 Rr : 1 rr.

Multiply to get combined genotype frequencies (out of 16): YYRR=1x1=1; YYRr=1x2=2; YYrr=1x1=1; YyRR=2x1=2; YyRr=2x2=4; Yyrr=2x1=2; yyRR=1x1=1; yyRr=1x2=2; yyrr=1x1=1. …
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AP EAPCET 2024Set ap-2024-05-16-FN1 markMCQ
$Q. Choose correct pair from the following I RrYy 4/16 II Rryy 1/16 III RrYY 2/16 IV rrYy 1/16 (A) I, III (B) II, IV (C) I, II (D) III, IV$
›Reveal solutionSolution
Checking each genotype's true F2 dihybrid frequency out of 16 against the frequencies stated: RrYy=4/16 and RrYY=2/16 are correctly stated; Rryy and rrYy are actually 2/16 each, not 1/16 as claimed.

Concept and Intuition

For a dihybrid cross RrYy × RrYy, the F2 genotypic ratio (out of 16) expands from independent 1:2:1 monohybrid ratios for each gene multiplied together:

Homozygous-homozygous combinations (RRYY, RRyy, rrYY, rryy) each occur 1/16.

Heterozygous-homozygous combinations (RRYy, RrYY, Rryy, rrYy) each occur 2/16.

Doubly heterozygous (RrYy) occurs 4/16.

Step-by-Step Solution

RrYy (both loci heterozygous) = 2×2 = 4/16 → statement I (4/16) is correct.

Rryy (Rr heterozygous × yy homozygous) = 2×1 = 2/16 → statement II claims 1/16, which is wrong. …
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AP EAPCET 2024Set ap-2024-05-16-FN1 markMCQ
$F_2$
›Reveal solutionSolution
Recombinant phenotypes make up 6/16 of an F2 dihybrid population; for 1280 progeny that's 480.

Concept and Intuition

In Mendel's classic dihybrid cross (e.g. round-yellow × wrinkled-green pea, RRYY × rryy), the F2 generation shows a 9:3:3:1 phenotypic ratio:

9/16 Round Yellow and 1/16 Wrinkled Green are the parental-type combinations (matching the original parents) — together 10/16.

3/16 Round Green and 3/16 Wrinkled Yellow are the recombinant combinations (new combinations not seen in either parent), arising from independent assortment — together 6/16.

Step-by-Step Solution

Recombinant fraction of F2 = 3/16 + 3/16 = 6/16 = 3/8. …
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AP EAPCET 2024Set ap-2024-05-17-AN1 markMCQ
$Q. If one parent has AB blood group and the other parent has A blood group (homozygous), these blood groups are not expected in their children. (A) B and O (B) A and AB (C) A and B (D) B and AB$
›Reveal solutionSolution
This tests ABO blood group genetics via a simple cross; AB × homozygous-A parents can only produce A or AB children, never B or O.

Concept and Intuition

The ABO blood group is controlled by three alleles at one locus: IA and IB (both dominant/codominant to each other) and i (recessive). A homozygous "A" parent has genotype IAIA (contributes only IA gametes). An "AB" parent has genotype IAIB (contributes either IA or IB gametes, each with 50% probability).

Step-by-Step Solution

Homozygous A parent: gametes = IA, IA (always IA).

AB parent: gametes = IA, IB (each 50%).

Cross the gametes (Punnett square): IA(from A-parent) × IA(from AB-parent) → IAIA (blood group A); IA(from A-parent) × IB(from AB-parent) → IAIB (blood group AB). …
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AP EAPCET 2023Set ap-2023-05-22-AN1 markMCQ
$Q. The number of phenotypes and their genotypic ratio found when heterozygous round and yellow seeded pea plant crossed with other pea plant showing seeds with wrinkled and heterozygous yellow (A) Eight and 3:1:3:1 (B) Eight and 1:2:1:1:2:1 (C) Six and 1:2:1:2 (D) four and 1:1:1:1$
›Reveal solutionSolution
This cross (RrYy × rrYy) is a testcross for seed-shape combined with a monohybrid heterozygous cross for seed-colour; the full Punnett square has 8 cells and the resulting genotypic ratio is 1:2:1:1:2:1.

Concept and Intuition

When two genes are tracked simultaneously but the parents differ in "zygosity" at each locus (one parent heterozygous at both loci, the other homozygous recessive at one locus and heterozygous at the other), the overall genotype ratio is the product of the ratios from each locus considered independently (assuming independent assortment), and the total number of Punnett-square cells is the product of the numbers of distinct gamete types each parent produces.

Step-by-Step Solution

Parent 1 = RrYy (round, yellow, heterozygous both genes) → produces 4 gamete types: RY, Ry, rY, ry (each 1/4).

Parent 2 = rrYy (wrinkled, heterozygous yellow) → produces 2 gamete types: rY, ry (each 1/2).

Total distinct gamete combinations in the Punnett square = 4 × 2 = 8.

For the R locus: Rr × rr → 1 Rr : 1 rr (a straightforward testcross, 1:1).

For the Y locus: Yy × Yy → 1 YY : 2 Yy : 1 yy (1:2:1). …
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AP EAPCET 2021Set ap-2021-09-03-AN1 markMCQ
$Q. When homozygous 'A' blood group Man marries homozygous 'B' group Woman in 'F1' generation and 'AB' group of springs are born. When these 'F1' individuals marry each other, in 'F2' generation individuals with how many blood groups are born? (A) 1 (B) 2 (C) 3 (D) 4$
›Reveal solutionSolution
This tests ABO multiple-allele inheritance; since both original parents are homozygous for I^A and I^B (no 'i' allele present anywhere in the cross), the O blood group can never appear, leaving only 3 possible groups in F2.

Concept and Intuition

The ABO blood group is controlled by a single gene with three alleles: I^A and I^B (co-dominant to each other) and i (recessive to both). Blood group O requires the genotype ii. If neither parent ever carries an i allele, no descendant in any generation can be ii, so blood group O is structurally impossible in this pedigree, regardless of how many generations you cross.

Step-by-Step Solution

P generation: I^A I^A (blood group A) × I^B I^B (blood group B).

F1: all offspring are I^A I^B → blood group AB (co-dominance), 100% AB — consistent with the question's given F1.

F1 × F1 cross: I^A I^B × I^A I^B → Punnett square gives I^A I^A : I^A I^B : I^A I^B : I^B I^B = 1 : 2 : 1. …
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Q.Even if a character shows multiple allelism, an individual will only have two alleles for that character. Why?

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