Biology · Meiosis & Patterns of Inheritance
Mendelian patterns, segregation and independent assortment
Pp describes two alleles at one locus. It does not by itself tell us whether a mark is visible. First follow the alleles into gametes; then apply a separately stated trait rule. Doing the two steps in that order makes a Mendelian ratio understandable rather than something to memorise.
Begin with the cell-division lesson and Heredity, Variation & Evolution: recognise homologous chromosomes, sister chromatids, a locus and an allele. DNA, Replication & Repair explains how copying makes two sister DNA molecules. Gene Expression helps with the later distinction between transmitting a sequence and expressing a trait. Here, fractions and conditional probabilities are explained alongside the cases. All chromosome-count examples use a specified diploid model; the final lessons introduce extensions rather than requiring advanced genetics beforehand.
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Derive one-locus and two-locus predictions from chromosome behaviour, and state the additional conditions needed to turn genotype probabilities into phenotype ratios.
Follow one locus into gametes
Mendel's plant crosses helped establish that hereditary factors can retain their identities across generations rather than permanently blending. The chromosome account now gives a physical route for allele segregation: in a standard diploid reproductive pathway, each gamete receives one of the two alleles at an ordinary autosomal locus.
For our Pp parent, assume P and p each occur in half the gametes. The two parents' contributions unite independently. A Pp × Pp cross therefore has four equally probable union paths: PP, Pp, pP and pp. The middle two describe the same unordered diploid genotype, so the genotype probabilities are 1/4 PP, 1/2 Pp and 1/4 pp.
Equal transmission is a condition, not a promise for every locus and organism. Segregation distortion, new sequence changes or unequal survival can alter the record. Specify whether probabilities concern gametes, fertilised zygotes or offspring at a later scoring stage.
Add a trait rule only when it is supplied
Assign our hypothetical mark to PP and Pp, and its absence to pp, under the stated conditions. This is complete dominance of P for that scored trait. Adding the first three union paths gives probability 3/4 for the mark and 1/4 for its absence. The genotype ratio remains 1:2:1 while the phenotype ratio is 3:1.
Dominance compares the heterozygote's phenotype with the two homozygotes for a particular measurement. It does not say that one DNA sequence destroys the other, that a dominant allele is more common, or that it improves reproductive success. A Pp individual still carries p and can transmit it.
Gene products and biological pathways help explain dominance relationships. For instance, in a specified pathway one functional copy might supply enough product to meet a scoring threshold. That is a possible mechanism to investigate; an uppercase letter alone proves neither that mechanism nor complete dominance for every trait of the organism.
Four union paths, three marked paths
Read the diagram in words
PP has probability 1/4, Pp 1/2 and pp 1/4. The supplied complete-dominance rule gives a mark to PP and Pp, so mark present has probability 3/4. The grid's four cells are possible paths rather than a fixed set of four offspring.
Pp × Pp
| Gamete from parent 1 | Gamete from parent 2: P | Gamete from parent 2: p |
|---|---|---|
| Gamete from parent 1: P | PP; Mark present; 1/4 | Pp; Mark present; 1/4 |
| Gamete from parent 1: p | Pp; Mark present; 1/4 | pp; Mark absent; 1/4 |
| PP | 1/4 |
|---|---|
| Pp | 2/4 |
| pp | 1/4 |
| Mark present | 3/4 |
|---|---|
| Mark absent | 1/4 |
| PP | Mark present |
|---|---|
| Pp | Mark present |
| pp | Mark absent |
- Four equally likely union paths
- Genotype grouping: 1:2:1
- Assigned phenotype grouping: 3:1
Ask whether two loci travel independently
Put A/a on the long chromosome pair and B/b on the short pair. In the no-crossover marker drawing, one relative orientation sends A with B and a with b. The other sends A with b and a with B. Across many meioses with equal independent orientation and equal sampling of products, AB, Ab, aB and ab each have probability 1/4.
A single meiosis in this stripped-down drawing produces two repeated complementary types, not necessarily all four types. Four equally likely gamete types describe the distribution across meioses. Crossing over can further change combinations, but it is not needed for this between-pair assortment result.
Now use independently assorting P/p and Q/q in a PpQq × PpQq cross. If P separately determines a mark and Q a ridge by complete dominance, with full penetrance, no trait interaction and no differential loss, the phenotype probabilities are 9/16 for both, 3/16 for mark only, 3/16 for ridge only and 1/16 for neither.
The 9:3:3:1 pattern depends on those transmission and expression conditions. Nearby loci on the same chromosome can travel together through linkage. Interacting gene products can change phenotype categories even when the alleles assort independently. The next lessons explore both possibilities.
Two relative orientations give four types across meioses
Read the diagram in words
One relative orientation gives AB, AB, ab, ab; the alternative gives Ab, Ab, aB, aB. With each orientation occurring half the time and all products equally sampled, each of the four gamete types has probability 1/4.
| Long pair | A / a |
|---|---|
| Short pair | B / b |
A travels with B
1/2
| 1 | A, B |
|---|---|
| 2 | a, b |
| 1 | AB |
|---|---|
| 2 | AB |
| 3 | ab |
| 4 | ab |
A travels with b
1/2
| 1 | A, b |
|---|---|
| 2 | a, B |
| 1 | Ab |
|---|---|
| 2 | Ab |
| 3 | aB |
| 4 | aB |
| AB | 1/4 |
|---|---|
| Ab | 1/4 |
| aB | 1/4 |
| ab | 1/4 |
- Meiosis I pole contents
- Four final products in this one orientation
- The two relative orientations are alternatives
- Across independent meioses and equally sampled products
- No crossover in the depicted marker paths
Worked example
Build 9/16 from two smaller predictions
For the independently assorting PpQq cross, P(P_) = 3/4 and P(Q_) = 3/4; an underscore means either allele can occupy the second position. Independence lets us multiply: P(P_Q_) = (3/4)(3/4) = 9/16. For mark only, P(P_qq) = (3/4)(1/4) = 3/16. The four stated class probabilities add to 16/16.
In 160 independently scored offspring, the expected class counts are 90, 30, 30 and 10. An actual group can differ through sampling. A ratio describes the model's probabilities; deciding whether observed departures support linkage or another explanation requires data quality and an appropriate statistical comparison.
Check your understanding
Why does Pp × Pp give a 1:2:1 genotype ratio but a 3:1 mark ratio in our example?
Show reasoning
Two union paths give Pp. The assigned complete-dominance rule groups PP and Pp into one marked phenotype. Without that expression rule, the mark ratio would not follow.
Must one meiosis of our no-crossover A/a, B/b model produce AB, Ab, aB and ab once each?
Show reasoning
No. One relative orientation gives AB, AB, ab, ab; the alternative gives Ab, Ab, aB, aB. Equal probabilities of the four types arise across independently oriented meioses and equally sampled products.
Could a cross have independent allele assortment yet fail to show four phenotype classes in a 9:3:3:1 ratio?
Show reasoning
Yes. Dominance, penetrance, gene interactions or survival may differ from the assigned conditions. Independent transmission alone does not fix the phenotype grouping.
What to carry forward
Derive transmission first, then apply a stated expression rule and scoring stage.
References
Resources supporting this lesson’s explanations and examples.
- Biology 2e — Inheritance and probability (2018)Mary Ann Clark, Matthew Douglas and Jung Choi · OpenStaxOpen textbook sectionsBiology 2e — Inheritance and probability (2018)Characteristics, traits and dominanceSegregation and independent assortment
- Introduction to Genetics — Chromosomes, crosses and mapping (2023 adaptation of Open Genetics Lectures)Natasha Ramroop Singh · Thompson Rivers UniversityOpen university textbookIntroduction to Genetics — Chromosomes, crosses and mapping (2023 adaptation of Open Genetics Lectures)DNA amount and chromosome numberAssumptions of a dihybrid ratioConditional pedigree probabilitiesAllele phaseLinkage and recombinant fractionsGenetic maps and their limitsTwo-point and three-point crossesTypes of dominanceBiochemical basis of dominanceGene interactions and phenotype groupingPseudoautosomal regions