Mendelian inheritance: calculate from the cross and its assumptions

MCAT content library / Biology & biochemistry / Mendelian inheritance: calculate from the cross and its assumptions

Make the parental genotypes, phenotype rules and independence assumptions explicit before using a familiar ratio.

About 15 minutes, including the worked examples and checks.

Translate the phenotype into an inheritance model

A genetic probability question becomes manageable when its assumptions are explicit. Identify the loci, possible alleles, parental genotypes, the phenotype assigned to each genotype, and whether the loci assort independently. Then calculate. A familiar-looking trait description does not justify importing a dominance rule that the passage never supplied. The AAMC includes these distinctions and testcross reasoning in its inheritance scope.¹

A locus is a genomic location; alleles are alternative versions at that location. In a typical diploid model, an individual has two alleles at an autosomal locus, while a gamete carries one. A homozygote has matching alleles; a heterozygote has different alleles. Genotype describes the inherited combination, whereas phenotype describes the measured or observed characteristic.²

Under complete dominance, AA and Aa share the dominant phenotype while aa has the recessive phenotype. The recessive allele still segregates from Aa; it has not been erased. A phenotype written A_ therefore leaves two possible genotypes. Dominance describes the heterozygote’s phenotype, not whether an allele is common, beneficial or physically stronger.

Build the gametes before combining them

With ordinary equal segregation, Aa produces A and a gametes with probabilities 1/2 each. If A/a and B/b assort independently, AaBb produces AB, Ab, aB and ab with probabilities 1/4 each. Independence can be supplied directly or justified by the genetic context. Nearby loci on the same chromosome may be linked, so multiplication across loci is not an automatic rule for every cross.³

For Aa × Aa, the distinct fertilization routes are AA, Aa, aA and aa. The middle two routes represent the same unordered genotype, so the genotype probabilities are 1/4 AA, 1/2 Aa and 1/4 aa. Complete dominance groups AA and Aa into a 3/4 dominant phenotype probability. Changing the phenotype rule changes that grouping without changing allele segregation.

Multiply probabilities when the required events are independent. Add probabilities for mutually exclusive routes to the same requested outcome. If an “or” question includes overlapping events, subtract the overlap rather than counting it twice. For offspring counts, expected number equals sample size multiplied by the relevant probability; it is not a promise that the observed count will be exact.⁴

Worked testcross: infer what the parent can transmit

Consider an original model-plant experiment. C gives colored tissue and c gives uncolored tissue; L gives a smooth leaf edge and l gives a serrated edge. Dominance is complete at both loci. Assume no new mutations, reliable phenotype scoring, equal survival of offspring genotypes, random fertilization and independent assortment. A colored, smooth parent C_L_ is crossed with an uncolored, serrated tester ccll.

The tester supplies only cl gametes. Therefore, each offspring phenotype reveals the unknown parent’s transmitted alleles. The existence of uncolored offspring shows that the parent can supply c; colored offspring show that it can supply C. Smooth and serrated offspring similarly demonstrate transmission of both L and l. Under the stated model, the parent is CcLl. The approximately equal four counts are consistent with independent assortment and a 1:1:1:1 expectation.

These are two different inferences. Observing both alleles’ recessive outcomes identifies heterozygosity given the assumptions. Nearly equal frequencies support the independence model but do not prove exact equality of biological probabilities. A larger dataset, a statistical test or knowledge of chromosome positions can address departures from that model.

A testcross uses a recessive homozygous tester because its contribution is predictable. A backcross mates offspring to a parental genotype; it is a testcross when that parental genotype is the appropriate recessive tester. Labels such as parental, F1 and F2 specify generations, not a fixed genotype or a universal ratio.²⁴

Change the cross, then calculate again

Now mate the inferred CcLl parent with a known Ccll plant. The first locus is Cc × Cc, so P(colored) = 3/4. The second is Ll × ll, so P(serrated) = 1/2. With independence, P(colored and serrated) = (3/4)(1/2) = 3/8. In 400 offspring, the expected count is 400 × 3/8 = 150.

The other phenotype probabilities are 3/8 colored/smooth, 1/8 uncolored/smooth and 1/8 uncolored/serrated. They sum with 3/8 to one. This quick check catches omitted outcomes. It also shows why importing the familiar 9:3:3:1 ratio would fail: the second parent is ll at the leaf locus, not Ll.

Conditioning changes the denominator. Among the colored offspring at the C locus, CC contributes 1/4 of all offspring and Cc contributes 1/2. Once cc individuals are excluded, P(Cc | colored) = (1/2)/(3/4) = 2/3. The answer is neither the original 1/2 heterozygote probability nor a 1/2 choice between two genotype names. The two remaining genotypes were never equally likely.

Separate allele transmission from phenotype expression

With incomplete dominance, a heterozygote has an intermediate phenotype for the measured trait. With codominance, both allele-associated features are detectably expressed. Thus, a heterozygote × heterozygote cross can show a 1:2:1 phenotype ratio in either case, even though “intermediate intensity” and “both distinct products present” are different observations.²

For example, suppose a supplied assay detects protein M from allele M and protein N from allele N. MM has only M, NN only N, and MN both bands. MN × MM gives 1/2 MM and 1/2 MN: half the offspring have one M band and half have both M and N. Calling M “dominant” merely because its band appears in every offspring would confuse a cross-specific observation with the heterozygote’s phenotype rule.

Use observed counts without overstating them

A small sample can miss a possible class. Under Aa × aa, the probability that eight independently sampled offspring all show the dominant phenotype is (1/2)⁸ = 1/256. That observation is unusual but possible for a heterozygous parent; it does not logically establish AA. In contrast, an aa offspring demonstrates an a contribution from the other parent under the stated no-mutation, accurate-scoring assumptions.

Before finishing, state whether the answer concerns a gamete, a genotype, a phenotype, an expected count or a conditional probability. Recheck independence and survival assumptions. Those decisions determine the arithmetic and often matter more than drawing a larger Punnett square.

Check your reasoning

Choose an answer before opening its feedback. Each option explains the reasoning behind that choice.

Continue learning

Related lessons

• Meiosis and recombination: track chromosomes, DNA and alleles
• DNA replication and repair: follow strands, fragments and errors
• Mean, median, and spread: describing a dataset

Browse biology & biochemistry · MCAT content library

MCAT preparation and roadmap · Aim High, Score Higher

Sources

1. AAMC: B/B Content Category 1C

2. OpenStax Biology 2e: Characteristics and Traits

3. OpenStax Biology 2e: Laws of Inheritance

4. OpenStax Biology 2e: Mendel’s Experiments and the Laws of Probability