Mendelian Genetics and Inheritance Patterns
General Biology · Section 2.5 · 18 study cards
Segregation, independent assortment, crosses and ratios, then the extensions: codominance, epistasis, sex linkage, linkage and pedigrees.
Practice this set → Spaced repetition, card by card. No account needed.
Method
Decide what kind of problem you have
Before drawing anything, classify the question. Are you given the parents and asked for the offspring, or given the offspring and asked for the parents? Forward problems want a Punnett square or a probability product. Backward problems, including every test cross and every pedigree, want you to reason from the ratio in the offspring to the genotypes that could have produced it.
The procedure for setting up and reading a cross
- Assign a letter to each gene, capital for the dominant allele and the same letter in lower case for the recessive. State in words what each symbol means before you use it.
- Write the genotype of each parent. If a genotype is unknown, write it with a blank, as in A_, and let the offspring resolve it.
- List the gametes each parent can make. A heterozygote at n independently assorting genes makes 2n gamete types, so AaBb makes four: AB, Ab, aB, ab.
- Combine the gametes, either in a grid or by multiplying probabilities gene by gene. For more than two genes, always multiply — the grid becomes unusable.
- Collect the offspring genotypes into phenotype classes, applying the dominance relationship you stated in step one.
- Reduce to a ratio and compare it with the standard ones: 3:1, 1:2:1, 9:3:3:1, 1:1:1:1.
Read the ratio backwards
Ratios are diagnostic. A 1:1 ratio from a single gene means one parent was heterozygous and the other homozygous recessive. A 1:1:1:1 ratio in a two-gene cross means a dihybrid test cross with independent assortment. A 1:2:1 phenotypic ratio means the heterozygote is distinguishable, so incomplete dominance or codominance. Any total of sixteen with fewer than four phenotype classes means epistasis. A large excess of the two parental classes means linkage, and the minority classes are the recombinants.
Check the sex and the generation
For every pedigree and every cross involving a disorder, ask two questions before answering: does the trait affect the sexes unequally, and does it skip generations. Unequal sexes points to the X chromosome; skipping points to a recessive allele. Answer those two first and the genotypes usually follow without any further work.
Definitions and theorems
Worked example
In peas, tall (T) is dominant to dwarf (t) and purple flowers (P) are dominant to white (p). A tall, purple-flowered plant of unknown genotype is crossed with a dwarf, white-flowered plant. The offspring are 48 tall purple, 47 tall white, 45 dwarf purple and 44 dwarf white. Determine the genotype of the unknown parent and state what the result shows about the two genes.
Identify the cross. The second parent is dwarf and white, so it must be homozygous recessive for both genes, ttpp. A cross against a double homozygous recessive is a dihybrid test cross, so the offspring phenotypes report the gametes of the unknown parent directly.
Reduce the counts to a ratio. The four classes are 48, 47, 45 and 44 out of 184 offspring, which is close to equal and rounds to 1:1:1:1. Small deviations from exactly 46 each are expected sampling noise.
Read the gametes. Four equally frequent offspring classes mean the unknown parent produced four equally frequent gamete types: TP, Tp, tP and tp. Producing more than one allele of a gene means being heterozygous for it.
Assign the genotype. Heterozygous at both loci gives TtPp. Check it against the phenotype given: Tt is tall and Pp is purple, which matches the tall purple parent described.
Interpret the ratio. Equal numbers of all four classes mean the parental combinations (tall purple and dwarf white) are no more common than the new combinations (tall white and dwarf purple). The alleles therefore entered gametes independently.
State the conclusion. The unknown parent is TtPp, and the genes for height and flower colour assort independently, consistent with their lying on different chromosomes or very far apart on the same one. Had they been linked, the two parental classes would have greatly outnumbered the other two.
Common mistakes
- Treating dominant as common or strong. Students assume the dominant phenotype must be the frequent one in a population, and then misread pedigrees. Dominance describes only what the heterozygote looks like; allele frequency is a separate matter entirely.
- Drawing a Punnett square for three or more genes. A trihybrid cross has 64 boxes and almost guarantees an arithmetic slip under time pressure. Multiply the single-gene probabilities instead: the chance of a particular three-gene genotype is the product of three simple fractions.
- Forgetting to condition on information already given. Asked for the chance that an unaffected child of two carriers is a carrier, students answer 1/2 from the 1:2:1 ratio. The word unaffected eliminates the homozygous recessive class, so the denominator is three, not four, and the answer is 2/3.
- Confusing codominance with incomplete dominance. Both give a 1:2:1 ratio, so the ratio cannot distinguish them. Look at the heterozygote: a blend or intermediate is incomplete dominance, while both products appearing side by side is codominance.
- Expecting 9:3:3:1 from every dihybrid cross. That ratio requires two independently assorting genes with simple dominance and no interaction. Epistasis collapses classes into ratios such as 9:3:4 or 12:3:1, and linkage skews the counts towards the parental types.
Practice it
Reading the method is not the same as being able to recall it under pressure. This set drills 18 cards one at a time and schedules each card separately, so the ones you keep missing come back sooner.