Mendelian Genetics

Begin with a model, not a ratio

In a fictional plant, let R and r be two alleles at one locus. Assume RR and Rr plants have violet flowers and rr plants have white flowers: complete dominance for this measured character. RR and rr are homozygous genotypes; Rr is heterozygous. Violet and white are phenotypes. Dominant does not mean common, stronger, healthier, or more likely to enter a gamete.

A true-breeding RR line crossed with an rr line produces Rr offspring because each gamete contributes one allele. If those first-generation offspring are crossed, r can meet r again. This is the logic behind using parental, F₁, and F₂ generations: the F₁ appearance alone does not reveal whether an allele was lost.

For the examples that follow, assume a diploid organism, unbiased segregation, random fertilization, reliable phenotype scoring, and no genotype-dependent survival differences. We will add independent assortment only when analyzing loci for which that assumption is justified.

Segregation gives the gametes; fertilization gives the offspring

An Rr individual makes gametes carrying R or r, each with probability \(1/2\) under this model. This reflects separation of homologous chromosomes during meiosis. The gamete does not normally carry “half an R” and “half an r,” and dominance does not change the segregation probabilities.

For \(Rr\times Rr\), combine one gamete from each parent. A Punnett square records possible unions; it is not a prediction that every four siblings must occupy its four boxes once.

The RR probability is \((1/2)(1/2)=1/4\). There are two mutually exclusive routes to Rr, so its probability is \(1/4+1/4=1/2\). The rr probability is \(1/4\). Thus the genotype ratio is 1:2:1, whereas grouping RR and Rr together produces the phenotype ratio 3:1. One does not replace the other.

A visible phenotype can hide two genotypes

Suppose an offspring from this cross is known to be violet. The rr possibility is now excluded, but RR and Rr remain. Among the three equally weighted violet boxes, two are Rr.

This conditional probability is not \(1/2\). It uses information about the offspring's phenotype to restrict the sample space. The result applies to offspring of this particular cross under complete dominance, not to every violet individual in a population.

A test cross asks the unknown parent to reveal its gametes

Cross an individual with a dominant phenotype, written R_, to rr. The recessive parent always contributes r, so offspring differences reveal whether the unknown parent contributes R only or both R and r. RR predicts all Rr; Rr predicts Rr and rr in a 1:1 ratio.

One white offspring supports Rr for the unknown parent within this model. But several violet offspring do not logically prove RR: an Rr parent can also produce a run of violet offspring by chance.

Build a two-locus prediction from two one-locus predictions

Add an independently assorting locus S/s for a fictional seed character, again with complete dominance. For \(RrSs\times RrSs\), each parent makes RS, Rs, rS, and rs gametes with probability \(1/4\). At each locus the dominant phenotype has probability \(3/4\), and the recessive phenotype \(1/4\).

The 9:3:3:1 ratio follows from the specified cross and assumptions. It is not a universal rule for every cross involving two genes. Nearby linked loci may not assort independently; gene interactions can change how genotypes map to phenotypes even when segregation is normal.

Choose addition, multiplication, or a complement deliberately

Multiply for independent events that must both occur. Add for mutually exclusive routes to the same outcome. When routes overlap, subtract the overlap rather than counting it twice. For example, in the two-locus heterozygote cross, the probability of rr or ss is \(1/4+1/4-1/16=7/16\).

Earlier offspring do not make the next one “due” to be white. With parental genotypes fixed and independent fertilizations, its white probability remains 1/4. Earlier observations can, however, change what we infer about an initially unknown parental genotype.

Test a predicted ratio without demanding an exact match

Imagine an Rr × Rr plant experiment produces 200 scored offspring: 142 violet and 58 white. These are invented teaching data. A 3:1 model predicts 150 and 50. The difference between observed and expected counts is not automatically evidence against the model; random sampling produces variation.

Set the null hypothesis before testing: the phenotype probabilities are 3/4 and 1/4 under the stated cross. For each category, square the observed-minus-expected difference and divide by its expected count, then add:

Here O and E are observed and expected counts. With two categories and no fitted probability parameter, the degrees of freedom are \(2-1=1\). At a preselected significance level of 0.05, the critical value for one degree of freedom is approximately 3.84. Because 1.707 is smaller, fail to reject the 3:1 model.

That does not prove the model true, or assign it a 95 percent probability of being true. The discrepancy is not large enough to reject it at this threshold. Use counts rather than percentages, ensure independent observations and adequate expected counts, and check that scoring or survival has not biased the sample. If a result rejects the model, the test alone cannot tell whether linkage, phenotype assumptions, contamination, or another cause explains the mismatch.

Read a pedigree as constraints, not a diagnosis

Consider a hypothetical pedigree in which two individuals without a marked phenotype have an offspring with that phenotype. Under an autosomal recessive model with complete penetrance and no new mutation, the offspring is aa. Each parent must have contributed a, while each parent's unmarked phenotype requires an A. Both parents are therefore Aa.

An unmarked sibling from that same cross has a 2/3 carrier probability, using the same conditional reasoning as the violet-flower example. But a small pedigree can fit more than one biological model. Sex-linked inheritance, incomplete penetrance, new mutation, and missing information require separate consideration. A teaching pedigree is not sufficient for a medical conclusion about a real family.

Make the biological assumptions testable

To test a single-locus complete-dominance model in a fictional plant, begin with established parental lines, control pollination, record F₁ phenotypes, and generate F₂ and test-cross offspring. Score at a consistent developmental stage and include every eligible offspring rather than selecting examples that resemble the expected ratio. Use independent families to check whether one family's result generalizes.

The model predicts different outcomes for different crosses: one F₁ phenotype from opposite homozygous parents, approximately 3:1 in an F₂ heterozygote cross, and approximately 1:1 in a heterozygote test cross. Agreement across these different predictions is more informative than forcing one dataset to look like a familiar ratio.

The chromosome explanation connects these observations to the previous lessons: homolog separation accounts for segregation, independent pair orientation accounts for independent assortment, and fertilization restores allele pairs. Shared chromosome-based inheritance machinery across organisms also supports common ancestry; it does not imply that all organisms have the same chromosome number or identical inheritance patterns.

Before choosing a familiar ratio

Write the parental genotypes, list their gametes and probabilities, and specify how each offspring genotype produces a phenotype. If those steps are sound, the ratio follows. If the phenotype model or independence assumption is missing, a neatly completed Punnett square can still give the wrong biological prediction.