Dihybrid Inheritance

Dihybrid inheritance extends Mendelian analysis to two genes simultaneously. Where monohybrid crosses establish how single genes segregate, dihybrid crosses reveal how the segregation of one gene relates to that of another. Mendel's experiments with two-trait combinations in pea plants (seed shape × seed colour; pod colour × stem length) led him to the law of independent assortment: provided the two genes are located on different chromosomes, their alleles assort independently of one another during gamete formation. The result is the iconic 9:3:3:1 phenotypic ratio from a cross between two double heterozygotes — a ratio that is the signature of unlinked genes in dihybrid inheritance and one of the most heavily examined topics at A-Level.

Spec mapping: This lesson sits in AQA 7402 Section 3.7.1 — Inheritance, building directly on the monohybrid genetics of the previous lesson. The relevant content covers Mendel's second law (independent assortment), the use of dihybrid crosses to predict offspring ratios, the relationship between the 9:3:3:1 ratio and the underlying chromosomal mechanisms, and the chi-squared test for assessing whether observed offspring numbers fit the expected ratio. (Refer to the official AQA specification document for exact wording.)

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Mendel's Second Law — The Law of Independent Assortment

Mendel's second law states that the alleles of different genes assort independently of one another during gamete formation, provided the genes are located on different chromosomes (i.e. they are unlinked).

The chromosomal basis is direct:

• During meiosis I, homologous chromosomes pair as bivalents and line up at the metaphase plate.
• The orientation of one bivalent is random and independent of the orientation of any other bivalent — maternal homologue can face either pole regardless of how other bivalents are oriented.
• For two genes on different chromosomes, the four possible bivalent orientations are equally likely, generating four gamete types in equal proportions.
• The allele inherited for one gene therefore does not influence which allele is inherited for another (when the genes are on different chromosomes).

Mendel formulated this law without knowing about chromosomes; the cytological explanation came with Sutton and Boveri's chromosome theory of inheritance (1902–1903).

Key Point: Independent assortment only applies to genes on different chromosomes. Genes on the same chromosome are linked and tend to be inherited together. Recombination between linked genes occurs only by crossing over, the frequency of which depends on the distance between the loci (covered in the next lesson). Mendel's law is therefore not universal — it is true for unlinked genes only.

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The Dihybrid Cross

Worked Example 1 — Pea Plant Seed Characteristics

In pea plants, Mendel's original two-gene system involves:

• Seed shape: Round (R) is dominant over wrinkled (r). The wrinkled phenotype is due to a defective starch-branching enzyme — wrinkled seeds accumulate sucrose, take up water, then shrink as they dry, producing the wrinkled appearance.
• Seed colour: Yellow (Y) is dominant over green (y). The yellow phenotype is due to enzymatic conversion of chlorophyll to carotenoid pigments during seed development; the green phenotype reflects retained chlorophyll.

These two genes are on different chromosomes, so they assort independently — Mendel's second law holds.

Cross: RrYy × RrYy (both parents are heterozygous for both genes)

Gametes from each parent: RY, Ry, rY, ry (four types of gamete, produced in equal proportions)

	RY	Ry	rY	ry
RY	RRYY	RRYy	RrYY	RrYy
Ry	RRYy	RRyy	RrYy	Rryy
rY	RrYY	RrYy	rrYY	rrYy
ry	RrYy	Rryy	rrYy	rryy

Counting the Phenotypes

With complete dominance at both loci, the 16 offspring genotypes fall into four phenotypic classes determined by whether at least one dominant allele is present at each locus. The underscore notation (e.g. R_) means "either R or r" — the second allele can be anything without affecting the phenotype when the dominant allele is present.

From the 16 possible offspring:

Phenotype	Genotypes	Count
Round, yellow	R_Y_	9
Round, green	R_yy	3
Wrinkled, yellow	rrY_	3
Wrinkled, green	rryy	1

Expected phenotypic ratio: 9:3:3:1

This 9:3:3:1 ratio is the signature of a dihybrid cross with complete dominance at both loci and independent assortment.

Key Definition: The underscore notation (e.g., R_) means "either homozygous dominant or heterozygous" — the second allele can be either R or r without affecting the phenotype when the dominant allele is present.

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Predicting Offspring Ratios

Shortcut Method — The Branch (Forked-Line) Diagram

Rather than drawing a full 4×4 Punnett square, the branch (forked-line) method uses the multiplication rule of independent probabilities. Because the two genes assort independently, the probability of any particular combined phenotype is the product of the individual probabilities.

1. Consider each gene separately as a monohybrid cross.
2. Multiply the probabilities from each gene to find the combined probability for each phenotype.

For Rr × Rr: 3/4 round, 1/4 wrinkled For Yy × Yy: 3/4 yellow, 1/4 green

Combined probabilities:

• Round and yellow: 3/4 × 3/4 = 9/16
• Round and green: 3/4 × 1/4 = 3/16
• Wrinkled and yellow: 1/4 × 3/4 = 3/16
• Wrinkled and green: 1/4 × 1/4 = 1/16

This gives the 9:3:3:1 ratio without the need to enumerate 16 cells in a Punnett square. The branch method generalises easily: for a trihybrid cross (AaBbCc × AaBbCc), each gene independently gives 3/4 : 1/4 for its dominant/recessive phenotype, and the eight phenotypic classes have probabilities (3/4)³ : (3/4)²(1/4) : … through to (1/4)³, giving the 27:9:9:9:3:3:3:1 ratio.

Practical note: When the cross involves any homozygous parent at one locus (e.g. AaBb × AABb), the dihybrid is reduced — that locus produces only one gamete type from one parent — and the Punnett square collapses to a 2 × 4 or 4 × 1 grid. Calculate the expected ratio for each locus first, then combine.

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Worked Example 2 — Guinea Pig Coat

In guinea pigs (a useful A-Level example because the two genes are on different chromosomes, so they assort independently):

• Black coat (B) is dominant over brown (b). The B allele encodes a functional tyrosinase enzyme producing the black pigment eumelanin; bb genotypes produce a variant that defaults to brown.
• Short hair (S) is dominant over long hair (s). The S allele encodes an active FGF5 protein that regulates hair-cycle length; ss genotypes have the long-hair phenotype seen in Peruvian guinea pigs.

Cross: BbSs × BbSs (both parents heterozygous at both loci)

Using the branch method:

• Bb × Bb → 3/4 black : 1/4 brown
• Ss × Ss → 3/4 short : 1/4 long

Phenotype	Probability	Ratio out of 16
Black, short	3/4 × 3/4 = 9/16	9
Black, long	3/4 × 1/4 = 3/16	3
Brown, short	1/4 × 3/4 = 3/16	3
Brown, long	1/4 × 1/4 = 1/16	1

This is the 9:3:3:1 phenotypic ratio. In a litter of 32 offspring (twice the ratio total) we would expect 18 black short, 6 black long, 6 brown short, 2 brown long — but small samples will fluctuate substantially.

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Worked Example 3 — Test Cross with Two Genes

A test cross for a dihybrid follows the same logic as a monohybrid test cross: cross the unknown individual with one whose genotype is fully recessive at both loci. The recessive parent provides only one type of gamete; the unknown parent's gametes are revealed directly in the offspring phenotypes.

The test cross involves crossing a double heterozygote (RrYy) with a double homozygous recessive (rryy).

Cross: RrYy × rryy

Gametes from RrYy: RY, Ry, rY, ry (equal proportions) Gametes from rryy: ry (only one type)

	ry
RY	RrYy (round, yellow)
Ry	Rryy (round, green)
rY	rrYy (wrinkled, yellow)
ry	rryy (wrinkled, green)

Expected phenotypic ratio: 1:1:1:1 (equal numbers of all four phenotypes)

This 1:1:1:1 ratio is the expected result when a dihybrid heterozygote is test-crossed with a homozygous recessive. Any deviation from this ratio suggests that the genes may be linked (on the same chromosome).

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Worked Example 4 — Parental Genotypes from Offspring Data

A cross between two tomato plants produces the following offspring:

• 315 tall, red fruit
• 108 tall, yellow fruit
• 101 short, red fruit
• 32 short, yellow fruit

Step 1: Calculate the ratio. 315 : 108 : 101 : 32 ≈ 9 : 3 : 3 : 1 (dividing through by the smallest class: 315/32 ≈ 9.8, 108/32 ≈ 3.4, 101/32 ≈ 3.2, 32/32 = 1)

Step 2: A 9:3:3:1 ratio is the signature of a dihybrid cross between two heterozygous parents with independent assortment and complete dominance at both loci.

Step 3: Determine dominance.

• Tall (T) is dominant over short (t) — because the larger class (315 + 108 = 423 tall) shows the tall phenotype, consistent with 3/4 of offspring being tall in a heterozygous monohybrid.
• Red fruit (R) is dominant over yellow (r) — because the larger class (315 + 101 = 416 red) shows the red phenotype.

Step 4: Both parents must be TtRr (heterozygous for both genes). Their gametes are TR, Tr, tR, tr in equal proportions, and the Punnett square or branch method gives the 9:3:3:1 ratio.

Step 5 (optional verification): Apply the chi-squared test. Expected from a total of 556: 9/16 × 556 = 312.75; 3/16 × 556 = 104.25 (twice); 1/16 × 556 = 34.75. χ² = (315−312.75)²/312.75 + (108−104.25)²/104.25 + (101−104.25)²/104.25 + (32−34.75)²/34.75 ≈ 0.02 + 0.13 + 0.10 + 0.22 = 0.47. At 3 df, p = 0.05 critical value = 7.815. Since 0.47 ≪ 7.815, the deviation is far from statistically significant — the data strongly support the 9:3:3:1 hypothesis and therefore the model of two unlinked, completely-dominant genes.

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Deviations from the 9:3:3:1 Ratio

The 9:3:3:1 ratio is the signature of unlinked, completely-dominant dihybrid inheritance. It depends on a checklist of assumptions:

• Both genes show complete dominance (heterozygotes are indistinguishable from homozygous dominants).
• The two genes are on different chromosomes (unlinked) — or so far apart on the same chromosome that the recombination frequency reaches 50%.
• There is no epistasis — neither gene modifies the expression of the other.
• There is no selection acting differentially on the offspring genotypes (no lethal combinations; no fertility differences).
• The sample size is large enough for the ratio to be statistically distinguishable from random fluctuations.

If any of these conditions is violated, the observed ratio departs from 9:3:3:1 in characteristic ways:

Cause	Typical observed ratio	Interpretation
Linkage (genes on same chromosome)	Parental phenotypes over-represented; recombinants under-represented	Recombination frequency < 50%
Recessive epistasis (complementary)	9:7	Both dominant alleles needed for one phenotype
Recessive epistasis (with intermediate)	9:3:4	Homozygous recessive at one locus masks the other
Dominant epistasis	12:3:1	Dominant allele at one locus masks the other
Duplicate dominant epistasis	15:1	Dominant allele at either locus produces the same phenotype
Codominance at one locus	6:3:3:2:1:1 or similar	Heterozygote phenotypically distinct from homozygotes
Selection / inviability	Class missing entirely	Lethal genotype eliminated