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Spec Mapping — OCR H420 Module 6.1.2 — Patterns of inheritance, content statements covering dihybrid crosses, the expected 9:3:3:1 ratio under independent assortment, and the use of Punnett squares (refer to the official OCR H420 specification document for exact wording). This lesson formalises the dihybrid-cross framework that brings together monohybrid principles, linkage and epistasis into a single quantitative tool — the cornerstone of A-Level genetics problem-solving.
The historical scholarship is rich. Gregor Mendel (1866, paraphrased) explicitly performed dihybrid crosses on pea plants (round/wrinkled × yellow/green) and reported the 9:3:3:1 F2 ratio. The arithmetic was the foundation of his law of independent assortment. The rediscoverers — William Bateson, Edith Saunders and Reginald Punnett (1900s, paraphrased) — confirmed and extended Mendel's dihybrid work in chickens, sweet peas and other species. Punnett's diagrammatic square (a 4×4 grid for dihybrid) is the universal tool for predicting cross outcomes.
A dihybrid cross follows the inheritance of two genes simultaneously. When the two genes are on different chromosomes and assort independently, the classic Mendelian ratio of 9:3:3:1 appears in the F2 generation. But as we have seen in the previous lessons, this ratio can be modified by linkage (genes on the same chromosome) and epistasis (interactions between genes). Being able to predict and interpret dihybrid ratios is one of the key skills tested in OCR A-Level Biology A, covered by specification module 6.1.2(b) and 6.1.2(f).
Key Definitions:
- Dihybrid cross — a genetic cross examining the inheritance of two genes at once.
- Independent assortment — Mendel's second law: alleles of different genes segregate independently during gamete formation (provided the genes are on different chromosomes).
- Phenotypic ratio — the ratio of different visible traits in the offspring.
- Genotypic ratio — the ratio of different allele combinations in the offspring.
Mendel's pea-plant experiments are the textbook example. He crossed plants that were homozygous for two traits:
Round, yellow (RRYY) × wrinkled, green (rryy). All F1 offspring are RrYy, round and yellow.
Cross RrYy × RrYy. Each parent makes four gamete types in equal proportions: RY, Ry, rY, ry. A 4 × 4 Punnett square shows all 16 equally likely combinations:
| RY | Ry | rY | ry | |
|---|---|---|---|---|
| RY | RRYY | RRYy | RrYY | RrYy |
| Ry | RRYy | RRyy | RrYy | Rryy |
| rY | RrYY | RrYy | rrYY | rrYy |
| ry | RrYy | Rryy | rrYy | rryy |
Grouping by phenotype:
This is the 9:3:3:1 ratio — the signature of a dihybrid cross between heterozygotes when the two genes assort independently.
The 9:3:3:1 ratio is simply the product of two independent 3:1 ratios:
Total: 9 + 3 + 3 + 1 = 16. The fact that you can simply multiply the two one-gene probabilities is a statement of independence — it only works when the genes assort independently.
A test cross uses a homozygous recessive partner to reveal the genotype of a dihybrid. For example, RrYy × rryy:
| RY | Ry | rY | ry | |
|---|---|---|---|---|
| ry | RrYy | Rryy | rrYy | rryy |
Offspring: 1 round yellow : 1 round green : 1 wrinkled yellow : 1 wrinkled green. The 1:1:1:1 test-cross ratio is a strong test for independence — if you see this, the genes are almost certainly on different chromosomes and the parent is heterozygous at both loci.
Real-world dihybrid ratios are often not 9:3:3:1. The two main reasons are linkage and epistasis.
If A and B are on the same chromosome with the dominant alleles coming from one parent (AB on one chromosome, ab on the other), then crossing over produces few recombinants and many parentals. In a dihybrid test cross (AaBb × aabb) you expect 1:1:1:1 under independence; under linkage you see mostly AB and ab offspring and few Ab or aB. In a dihybrid intercross (AaBb × AaBb) the 9:3:3:1 is skewed — you see an excess of the parental combinations and a deficit of recombinants, although every class is still present because of crossing over.
Epistasis merges some of the four phenotype classes:
| Epistasis type | Ratio |
|---|---|
| None (independent genes) | 9 : 3 : 3 : 1 |
| Recessive epistasis (e.g. agouti mice) | 9 : 3 : 4 |
| Dominant epistasis (e.g. squash colour) | 12 : 3 : 1 |
| Complementary (sweet pea) | 9 : 7 |
Note that all the ratios still sum to 16. The classes are merged, not lost.
Cross grey-bodied, long-winged flies BBVV with black-bodied, vestigial-winged flies bbvv. F1 are all BbVv (grey, long) as expected.
Observed data (Morgan's classic experiment):
Total = 2300. These numbers are nothing like 1:1:1:1. The two parental classes (grey long and black vestigial) dominate, while the two recombinant classes (grey vestigial and black long) are rare. This tells us B and V are linked on the same chromosome.
Cross-over value = (206 + 185)/2300 × 100 ≈ 17%.
Because dihybrid Punnett squares are 4 × 4, they take a little care. Follow this procedure:
flowchart TB
A[Step 1: Write each parent's genotype]
A --> B[Step 2: List all possible gametes]
B --> C[Step 3: Draw a 4x4 grid]
C --> D[Step 4: Fill each square with the combined genotype]
D --> E[Step 5: Group by phenotype and count]
E --> F[Step 6: Simplify to a ratio]
Parents: RrYy × Rryy
| Ry | ry | |
|---|---|---|
| RY | RRYy (round yellow) | RrYy (round yellow) |
| Ry | RRyy (round green) | Rryy (round green) |
| rY | RrYy (round yellow) | rrYy (wrinkled yellow) |
| ry | Rryy (round green) | rryy (wrinkled green) |
Counting phenotypes: 3 round yellow : 3 round green : 1 wrinkled yellow : 1 wrinkled green. A 3:3:1:1 ratio.
| Cross | Expected (independent) |
|---|---|
| AaBb × AaBb | 9 : 3 : 3 : 1 |
| AaBb × aabb | 1 : 1 : 1 : 1 |
| AaBb × AABB | All A_B_ |
| AaBb × Aabb | 3 : 3 : 1 : 1 |
| AABb × aaBb | 3 A_B_ : 1 A_bb |
When asked to work out a dihybrid cross, always show the gametes explicitly before filling in the Punnett square. Then write each offspring genotype and its phenotype in the cell (or in a separate table), so the examiner can follow your reasoning. Finish by writing the simplified phenotypic ratio clearly — e.g. "phenotypic ratio 9 round yellow : 3 round green : 3 wrinkled yellow : 1 wrinkled green".
The ABO blood group (with three alleles Iᴬ, Iᴮ, Iᴼ) and the Rhesus system (Rh+ dominant, Rh− recessive) are inherited independently because they are on different chromosomes. A genetic counsellor might be asked the probability of specific combinations in a couple's children.
Consider a father of genotype IᴬIᴼ Rr (Group A, Rh+) and a mother of genotype IᴮIᴼ Rr (Group B, Rh+). What are the possible blood-group phenotypes of their children and their probabilities?
Each gene is inherited independently, so we can apply the product rule:
Multiply the two for each combined phenotype:
| ABO | Rh+ | Rh− |
|---|---|---|
| AB (1/4) | 3/16 | 1/16 |
| A (1/4) | 3/16 | 1/16 |
| B (1/4) | 3/16 | 1/16 |
| O (1/4) | 3/16 | 1/16 |
Eight possible combined phenotypes with a 3:3:3:3:1:1:1:1 ratio — which simplifies to 3:1 Rh+ : Rh− within each ABO class. This is a dihybrid cross problem solved by multiplying two monohybrid results.
Trihybrid crosses (three genes at once) are an extension of the same logic but become tedious with a Punnett square (8 × 8 = 64 squares). Use the branching "forked-line" method instead: handle one gene at a time.
For AaBbCc × AaBbCc with all three genes unlinked:
Multiply for each of the eight phenotype classes:
Trihybrid phenotypic ratio: 27:9:9:9:3:3:3:1 (sums to 64 = 4³). You do not need to memorise this for OCR, but being able to derive it shows you understand the underlying probability structure of Mendelian ratios.
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