Heredity — AP Biology Bio Study Guide

For: AP Biology candidates sitting AP Biology.

Covers: Mendelian genetics, linkage and recombination, sex linkage, pedigree analysis, and chromosomal inheritance as outlined in the AP Biology CED Unit 5.

You should already know: High-school biology, basic chemistry.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Biology style for educational use. They are not reproductions of past College Board papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official College Board mark schemes for grading conventions.

1. What Is Heredity?

Heredity is the biological process by which genetic information is transmitted from parent organisms to their offspring, forming the basis of both shared family traits and genetic variation across generations. It is the core focus of AP Biology Unit 5, and covers both simple Mendelian inheritance patterns for single-gene traits and complex non-Mendelian patterns driven by gene linkage, chromosomal abnormalities, and sex-linked gene expression. Common synonyms for heredity include genetic inheritance and trait transmission, and questions on this topic make up 8-11% of your total AP Biology exam score, per the official CED.

2. Mendelian genetics

Mendelian genetics describes the foundational inheritance patterns first identified by Gregor Mendel in his 19th-century pea plant breeding experiments. Key terms you must memorize include: allele (a variant form of a gene), dominant (an allele that is expressed when present in one copy), recessive (an allele that is only expressed when present in two copies), homozygous (having two identical alleles for a gene), heterozygous (having two different alleles for a gene), genotype (the genetic makeup of an organism), and phenotype (the observable physical trait of an organism).

Mendel’s two core laws govern simple inheritance:

1. Law of Segregation: The two alleles for a single gene separate during gamete formation in meiosis, so each gamete receives only one allele for each gene.
2. Law of Independent Assortment: Alleles of two or more different genes assort independently of one another during gamete formation, as long as the genes are located on different chromosomes or very far apart on the same chromosome.

Worked Example

A monohybrid cross between two heterozygous tall pea plants (genotype $Tt$, where $T$ = dominant tall allele, $t$ = recessive short allele) produces the following offspring:

• Genotypic ratio: $1TT:2Tt:1tt$
• Phenotypic ratio: $3$ tall : $1$ short

For a dihybrid cross between two heterozygous yellow, round pea plants ($YyRr$), unlinked genes produce a 9:3:3:1 phenotypic ratio (9 yellow round, 3 yellow wrinkled, 3 green round, 1 green wrinkled). Examiners often ask you to use the multiplication rule to calculate specific offspring probabilities: the probability of getting a green wrinkled offspring ($yyrr$) from this cross is $\frac{1}{4}\times \frac{1}{4}=\frac{1}{16}$.

3. Linkage and recombination

Linked genes are genes located close together on the same chromosome, which tend to be inherited together and violate the Law of Independent Assortment. The only way linked genes are separated is via crossing over, the exchange of homologous chromosome segments that occurs during prophase I of meiosis. Crossing over produces recombinant gametes, which carry new combinations of alleles not present in either parent.

The recombination frequency (RF) is the percentage of recombinant offspring produced in a genetic cross, calculated as: $$\text{Recombination Frequency (RF)}=\frac{\text{Number of Recombinant Offspring}}{\text{Total Number of Offspring}}\times 100\%$$ A 1% RF is equal to 1 map unit (or centimorgan, cM), the standard unit for measuring relative distance between genes on a chromosome. An RF of 50% indicates genes are either on separate chromosomes or so far apart on the same chromosome that crossing over occurs between them in every meiosis, so they assort independently.

Worked Example

In a cross between a heterozygous wild-type fruit fly (gray body, normal wings: $b^{+}bv{g}^{+}vg$) and a double mutant fly (black body, vestigial wings: $bbvgvg$), you observe 2300 total offspring: 965 wild type, 944 double mutant, 206 gray vestigial, 185 black normal.

• Recombinant offspring = 206 + 185 = 391
• RF = $\frac{391}{2300}\times 100=17\%$
• The two genes are 17 cM apart on the same chromosome.

4. Sex linkage

Sex-linked genes are genes located on the sex chromosomes (X and Y in humans and most mammals). Most sex-linked traits are X-linked, as the Y chromosome is very small and carries only ~50 functional genes, compared to ~1000 on the X chromosome. Males are hemizygous for all X-linked genes, meaning they only have one copy, so they will express any recessive X-linked allele even if only one copy is present. Females need two copies of a recessive X-linked allele to express the associated trait, so X-linked recessive traits are far more common in males than females. Common examples of X-linked recessive traits include red-green color blindness, hemophilia A, and Duchenne muscular dystrophy.

Worked Example

A cross between a female carrier for hemophilia A ($X^H{X}^h$, where $X^h$ = recessive hemophilia allele, $X^H$ = dominant normal allele) and a normal male ($X^{H}Y$) produces the following offspring:

• Female offspring: 50% $X^H{X}^H$ (normal, non-carrier), 50% $X^H{X}^h$ (normal, carrier)
• Male offspring: 50% $X^{H}Y$ (normal), 50% $X^{h}Y$ (has hemophilia) The probability of a male child having hemophilia is 50%, while the probability of a female child having hemophilia is 0%. Examiners frequently mark students down for failing to separate male and female probabilities when calculating X-linked trait inheritance.

5. Pedigree analysis

A pedigree is a standardized family tree diagram that tracks the occurrence of a genetic trait across multiple generations, used to infer the inheritance pattern of the trait. Standard symbols include: circles = female, squares = male, filled symbols = affected individual, horizontal line between two symbols = mating, vertical lines = offspring.

You can identify common inheritance patterns using these rules:

1. Autosomal dominant: Affected individuals have at least one affected parent, the trait appears in every generation, and males and females are equally likely to be affected.
2. Autosomal recessive: Affected individuals often have unaffected carrier parents, the trait skips generations, and males and females are equally likely to be affected.
3. X-linked dominant: Affected males pass the trait to 100% of their daughters and 0% of their sons, the trait does not skip generations.
4. X-linked recessive: More males are affected, the trait skips generations, and there is no father-to-son transmission (fathers pass the Y chromosome to sons, not the X chromosome).

Worked Example

A pedigree shows two unaffected parents who have one affected son and three unaffected daughters. There is no recorded father-to-son transmission of the trait across 3 generations, and 80% of affected individuals are male. The inheritance pattern is X-linked recessive: the mother is a carrier of the recessive allele, which she passed to her affected son, and the father has a normal dominant allele on his X chromosome.

6. Chromosomal inheritance

The chromosomal theory of inheritance, first proposed in the early 1900s, states that genes are located at specific fixed positions (loci) on chromosomes, and the segregation and independent assortment of chromosomes during meiosis directly explain Mendel’s laws of inheritance.

Most large-scale genetic disorders are caused by chromosomal abnormalities, which fall into two categories:

1. Aneuploidy: An abnormal number of a single chromosome, caused by nondisjunction (failure of chromosomes to separate during meiosis I or II). Common examples include Down syndrome (trisomy 21, 3 copies of chromosome 21), Turner syndrome (XO, females with only one X chromosome), and Klinefelter syndrome (XXY, males with an extra X chromosome). Nondisjunction during meiosis I produces 4 abnormal gametes, while nondisjunction during meiosis II produces 2 abnormal and 2 normal gametes.
2. Structural abnormalities: Changes to the structure of individual chromosomes, including deletion (e.g., Cri-du-chat syndrome, deletion of part of the short arm of chromosome 5), duplication, inversion, and translocation (e.g., chronic myelogenous leukemia, translocation between chromosomes 9 and 22).

Worked Example

Nondisjunction of chromosome 21 during meiosis I in a female produces an egg with two copies of chromosome 21. When this egg is fertilized by a normal sperm carrying one copy of chromosome 21, the resulting zygote has 3 copies of chromosome 21, causing Down syndrome.

7. Common Pitfalls (and how to avoid them)

• Wrong move: Assuming all dihybrid crosses produce a 9:3:3:1 phenotypic ratio. Why students do it: They memorize Mendel’s results without understanding the underlying condition (unlinked genes). Correct move: First check if the genes are linked (RF < 50%); if they are, use recombination frequency to calculate expected offspring ratios instead of the standard 9:3:3:1.
• Wrong move: Calculating X-linked trait probabilities without separating male and female offspring. Why students do it: They treat X-linked traits the same as autosomal traits. Correct move: Always split offspring by sex when calculating X-linked probabilities, remembering that males are hemizygous for all X-linked genes.
• Wrong move: Ruling out X-linked recessive inheritance just because a trait skips generations. Why students do it: They confuse autosomal recessive and X-linked recessive patterns. Correct move: First check for sex bias (more males affected = likely X-linked recessive) and father-to-son transmission (if present, the trait cannot be X-linked).
• Wrong move: Stating that linked genes can never assort independently. Why students do it: They forget that crossing over occurs regularly between genes that are far apart on the same chromosome. Correct move: Genes that are more than 50 cM apart on the same chromosome have a 50% recombination frequency, so they assort independently just like genes on separate chromosomes.
• Wrong move: Treating map units (cM) as a measure of absolute physical distance on chromosomes. Why students do it: They assume 1 cM equals a fixed number of DNA base pairs. Correct move: Map units reflect recombination frequency, not physical distance; recombination rates vary across the genome, so 1 cM can correspond to different numbers of base pairs in different chromosomal regions.

8. Practice Questions (AP Biology Style)

Question 1

In pea plants, purple flower color ($P$) is dominant to white ($p$), and tall stem length ($T$) is dominant to short ($t$). A plant heterozygous for both traits is crossed with a white, short plant. The resulting offspring are: 125 purple tall, 121 white short, 27 purple short, 28 white tall. (a) Are these two genes linked? Justify your answer. (b) Calculate the recombination frequency between the two genes. (c) What is the map distance between the two genes in centimorgans?

Solution

(a) Yes, the genes are linked. If the genes were unlinked, we would expect a 1:1:1:1 phenotypic ratio (equal numbers of all four offspring types). The observed offspring have far more parental phenotypes (purple tall, white short) than recombinant phenotypes (purple short, white tall), which is characteristic of linked genes. (b) Total offspring = $125+121+27+28=301$. Recombinant offspring = $27+28=55$. $$RF=\frac{55}{301}\times 100\approx 18.3\%$$ (c) 1% RF = 1 cM, so the map distance between the two genes is ~18.3 cM.

Question 2

Hemophilia A is an X-linked recessive trait. A man with hemophilia A has children with a woman who has no family history of hemophilia. What is the probability that their first daughter is a carrier for hemophilia A? What is the probability that their first son has hemophilia A? Show your work.

Solution

Let $X^h$ = recessive hemophilia allele, $X^H$ = dominant normal allele. The man’s genotype is $X^{h}Y$ (he has hemophilia, so he carries the recessive allele on his X chromosome). The woman has no family history of hemophilia, so her genotype is $X^H{X}^H$ (she is not a carrier).

• Daughters inherit one X chromosome from each parent: all daughters will get $X^h$ from their father and $X^H$ from their mother, so their genotype is $X^H{X}^h$. Probability daughter is a carrier = 100% (or 1).
• Sons inherit the Y chromosome from their father and an X chromosome from their mother: all sons will get $X^H$ from their mother, so their genotype is $X^{H}Y$. Probability son has hemophilia = 0% (or 0).

Question 3

A pedigree for a rare genetic trait shows the following: Affected individuals always have at least one affected parent. Affected males pass the trait to 100% of their daughters, and 0% of their sons. Affected females pass the trait to 50% of their sons and 50% of their daughters. What is the inheritance pattern of this trait? Justify your answer.

Solution

The trait is X-linked dominant. Justification:

1. The trait does not skip generations (all affected individuals have an affected parent), so it is a dominant trait, not recessive.
2. Affected males pass the trait to all daughters and no sons: males pass their X chromosome only to their daughters, and their Y chromosome only to their sons, so the gene for the trait must be located on the X chromosome. If the trait were autosomal dominant, affected males would pass the trait to ~50% of their sons, not 0%.

9. Quick Reference Cheatsheet

10. What's Next

Heredity is a foundational topic that connects directly to later core units in the AP Biology syllabus. A strong understanding of inheritance patterns is required to master gene expression and regulation (Unit 6), where you will learn how environmental factors modify the expression of inherited traits, and natural selection (Unit 7), where you will use knowledge of genetic variation to model how adaptive traits spread through populations over time. For example, the recombination frequencies you use to map genes are the same values used to track the spread of antibiotic resistance alleles in bacterial populations, a common exam question topic for both Units 7 and 8.

To reinforce your mastery of heredity, practice working through pedigree analysis and genetic cross problems regularly, and make sure you can connect every inheritance pattern to the underlying molecular events of meiosis. If you have any questions about specific concepts, practice problems, or exam grading conventions, you can ask Ollie, our AI tutor, at any time for personalized explanations and feedback. You can also find more AP Biology study resources, full-length practice tests, and targeted unit quizzes on the homepage to help you prepare for your exam.