Opening your learning space
Starting the app …
Opening your learning space
Starting the app …
In a monohybrid cross between two heterozygous parents, independent combinations of equally likely gametes produce a 1:2:1 genotype ratio.
Free to browse · flashcards or a multiple-choice quiz · import from Essential
Formula
P and all probabilities are dimensionless; proportions range from 0 to 1 and percentages are multiplied by 100.
| Symbol | Meaning | Unit |
|---|---|---|
| A, a | Two alleles at the autosomal locus under consideration | alleles |
| Probability of a homozygous recessive offspring | dimensionless | |
| Probability of a heterozygous offspring | dimensionless | |
| Probability of a homozygous dominant offspring | dimensionless |
P(aa\mid Aa\times Aa)=\frac12\cdot\frac12=\frac14Under Mendelian segregation, each heterozygous parent contributes A or a with probability 1/2. With random fertilization, gamete probabilities multiply: aa occurs with 1/2·1/2=1/4; the two routes to Aa sum to 1/2; and AA also has probability 1/4. The 1:2:1 genotype ratio is not the same as a phenotype ratio: under complete dominance, AA and Aa look alike, giving a 3:1 phenotype ratio. The calculation assumes one autosomal locus, Mendelian segregation, random fertilization and equal survival. It is not an individual medical risk prediction.
Under Mendelian segregation, each heterozygous parent contributes A or a with probability 1/2. With random fertilization, gamete probabilities multiply: aa occurs with 1/2·1/2=1/4; the two routes to Aa sum to 1/2; and AA also has probability 1/4. The 1:2:1 genotype ratio is not the same as a phenotype ratio: under complete dominance, AA and Aa look alike, giving a 3:1 phenotype ratio. The calculation assumes one autosomal locus, Mendelian segregation, random fertilization and equal survival. It is not an individual medical risk prediction.
In a monohybrid cross between two heterozygous parents, independent combinations of equally likely gametes produce a 1:2:1 genotype ratio.
Solve for a quantity
This rearrangement assumes the other quantities are known and the denominator is nonzero.
Two heterozygous parents Aa×Aa have a child. What are the genotype probabilities?
P(AA)=1/4, P(Aa)=1/2 and P(aa)=1/4.
With complete dominance, the phenotype is 3/4 dominant and 1/4 recessive.
Interpreting a metric as universal without its reference, validity limits or consistent units.
Under Mendelian segregation, each heterozygous parent contributes A or a with probability 1/2. With random fertilization, gamete probabilities multiply: aa occurs with 1/2·1/2=1/4; the two routes to Aa sum to 1/2; and AA also has probability 1/4. The 1:2:1 genotype ratio is not the same as a phenotype ratio: under complete dominance, AA and Aa look alike, giving a 3:1 phenotype ratio. The calculation assumes one autosomal locus, Mendelian segregation, random fertilization and equal survival. It is not an individual medical risk prediction.
Formula cluster
The formula structures the calculation; biological interpretation must also account for study design, population and measurement assumptions.
For Aa×Aa, each parent transmits A or a with probability 1/2. Thus P(AA)=1/4, P(Aa)=1/4+1/4=1/2 and P(aa)=1/4. Check: 1/4+1/2+1/4=1. With complete dominance, 3/4 of offspring show the dominant phenotype and 1/4 the recessive phenotype.
Analyze simple Mendelian crosses and distinguish genotype from phenotype probabilities.
Here is how to work through a typical Probability in Monohybrid Crosses (Aa × Aa: ¼ AA, ½ Aa, ¼ aa) task step by step:
Task
Two heterozygous parents Aa×Aa have a child. What are the genotype probabilities?
Solution path
P(AA)=1/4, P(Aa)=1/2 and P(aa)=1/4.
With complete dominance, the phenotype is 3/4 dominant and 1/4 recessive.
Under Mendelian segregation, each heterozygous parent produces A and a gametes with probability 1/2. With random fertilization, the combinations AA, Aa, aA and aa each have probability 1/4. The two heterozygous combinations Aa and aA describe the same genotype and are combined. This gives the genotype ratio 1 AA : 2 Aa : 1 aa. The probabilities sum to one. A Punnett square displays the same four combinations, while the calculation makes the probability factors and grouping explicit. Each combination has the same probability.
With complete dominance, both AA and Aa show the dominant phenotype; only aa shows the recessive phenotype. The 1:2:1 genotype ratio therefore becomes three dominant combinations and one recessive, giving 3:1. This result assumes a fully dominant allele. With incomplete dominance or codominance, heterozygous offspring can look different and the phenotype ratio changes. Genotype ratio is a separate statement about allele combinations and should be reported separately from the observable trait. Group the four Punnett-square cells by phenotype to obtain that ratio.
The simple cross describes one autosomal locus with two alleles, heterozygous parents, Mendelian segregation, random fertilization and equally likely gametes. It also assumes genotypes survive equally before observation and no other influence changes the trait. Sex-linked inheritance, linkage, segregation distortion, selection or variable penetrance require another model. State the assumptions alongside the ratio in coursework or a report. This makes clear which conditions support the 1:2:1 result. The model tracks alleles at the specified locus. Verify the inheritance pattern at the stated locus.
The 1/4 is the probability for each conception under the model. For four offspring, one aa child is the expected count, but the observed number can be zero, one, two, three or four. Under the simplified assumptions, each conception is a separate random event; earlier children do not change the probability for the next conception. Across many offspring, the observed proportion often approaches the model value while remaining subject to chance variation. A particular small family therefore need not show an exact 1:2:1 ratio.
Aa denotes a heterozygous genotype with one A and one a allele. For a fully recessive trait, aa shows the recessive phenotype; in the simple model, Aa is a carrier without that recessive appearance. Thus the recessive genotype and phenotype probabilities for Aa×Aa are each 1/4 when the dominance assumption applies. Real genetic traits may involve several genes, environmental factors or variable expression. This school-level equation explains a Mendelian model and is not an individual medical risk assessment. Genotype names the alleles; phenotype describes the observable trait.
Quanta exam set
Curated exam set for "Probability in Monohybrid Crosses":
Question (front)
Which formula describes Probability in Monohybrid Crosses?
Answer in your set
Question (front)
How do you rearrange Probability in Monohybrid Crosses for Solve for a quantity?
Answer in your set
Question (front)
Which common mistake happens with Probability in Monohybrid Crosses?
Answer in your set
+ 7 more cards: units, variables, derivation, example, exam task
The formula set contains 10 cards. After importing, review them and start practicing; your ratings determine later review intervals.
This list names subject references and further reading. A page or section number appears when it is part of the reference; these are not verbatim quotations.
Create a curated FSRS exam set for Aa × Aa: ¼ AA, ½ Aa, ¼ aa: formula recall, variables, derivation, rearrangement, worked example, common mistakes and exam context.
Curated formula set · LaTeX · FSRS spaced repetition · import from Essential