Genetics — IB Biology HL HL Study Guide

For: IB Biology HL candidates sitting IB Biology HL.

Covers: Mendelian genetics, Punnett squares, sex linkage, autosomal inheritance, codominance, multiple alleles, mutations, genetic engineering basics, polygenic traits, and HL chi-squared statistical analysis.

You should already know: IGCSE 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 IB Biology HL style for educational use. They are not reproductions of past IBO papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official IBO mark schemes for grading conventions.

1. What Is Genetics?

Genetics is the branch of biology that studies the inheritance of phenotypic traits from parent to offspring, the molecular structure and function of genes, and patterns of genetic variation across populations and species. In the IB DP Biology HL syllabus, this topic bridges foundational molecular biology (Topic 2) and evolution (Topic 10), with ~15% of total HL paper marks allocated to genetics questions across Papers 1, 2, and 3. It draws on prior knowledge of DNA structure, alleles, and meiosis, and is assessed through multiple choice, short answer, long answer, and data analysis questions.

2. Mendelian genetics and Punnett squares

Gregor Mendel’s 19th-century pea plant experiments established three core laws of inheritance that form the basis of classical genetics:

1. Law of Segregation: Two alleles for a single trait separate during gamete formation, so each gamete inherits only one allele per gene.
2. Law of Independent Assortment: Alleles of different unlinked genes assort independently of one another during meiosis.
3. Law of Dominance: Dominant alleles mask the expression of recessive alleles in heterozygous individuals.

A Punnett square is a graphical tool used to predict the genotypic (genetic makeup) and phenotypic (observable trait) ratios of offspring from a given parental cross. Key terms to define first:

• Homozygous: Two identical alleles for a gene (e.g. $TT$ for tall pea plants, $tt$ for short)
• Heterozygous: Two different alleles for a gene (e.g. $Tt$ for tall pea plants, where $T$ is dominant)

Worked example: Monohybrid cross

You cross two heterozygous tall pea plants ($Tt\times Tt$), where $T$ (tall) is dominant over $t$ (short). Each parent produces gametes with $T$ and $t$ alleles at 50% probability each. The Punnett square gives the following offspring:

Genotypic ratio: 1 $TT$ : 2 $Tt$ : 1 $tt$. Phenotypic ratio: 3 tall : 1 short. For dihybrid crosses of two unlinked heterozygous genes, the expected phenotypic ratio is 9:3:3:1. Examiners frequently ask you to explicitly label both genotypic and phenotypic ratios, so never omit these labels in your answers.

3. Sex linkage and autosomal inheritance

Autosomal inheritance

Autosomal genes are located on non-sex chromosomes (pairs 1-22 in humans), so inheritance patterns are equal across male and female offspring. There are two common autosomal inheritance patterns:

• Autosomal dominant: Traits appear in every generation, affected individuals have at least one affected parent (e.g. Huntington’s disease)
• Autosomal recessive: Traits often skip generations, affected individuals are usually born to unaffected carrier parents (e.g. cystic fibrosis, sickle cell anemia)

Sex linkage

Sex-linked genes are located on the X or Y sex chromosomes. ~99% of sex-linked traits are X-linked, as the Y chromosome is far smaller and carries very few functional genes. X-linked recessive traits are far more common in males, because males only have one X chromosome (XY) so there is no second allele to mask a recessive mutation. Common examples include red-green color blindness and hemophilia A.

Worked example: X-linked recessive cross

A carrier female for hemophilia ($X^H{X}^h$, where $X^h$ is the recessive hemophilia allele) has children with a normal male ($X^{H}Y$). The Punnett square gives the following offspring:

0% of female offspring are affected, while 50% of male offspring are affected. A key exam tip for pedigree analysis: if you observe male-to-male transmission of a trait, it cannot be X-linked, as fathers only pass Y chromosomes to their sons, not X chromosomes.

4. Codominance and multiple alleles

In standard complete dominance, the dominant allele fully masks the recessive allele in heterozygotes. In codominance, both alleles are fully and separately expressed in the heterozygous phenotype, with no blending of traits. Many genes also demonstrate multiple alleles: more than two alleles for a single gene exist in a population, even though each individual only inherits two copies.

The most common IB example of both codominance and multiple alleles is the human ABO blood group system:

• Three alleles exist: $I^A$, $I^B$, and $i$
• $I^A$ and $I^B$ are codominant, while $i$ is recessive to both
• Genotype to phenotype mapping:
• $I^A{I}^A$ / $I^{A}i$ = Blood group A
• $I^B{I}^B$ / $I^{B}i$ = Blood group B
• $I^A{I}^B$ = Blood group AB (both A and B antigens expressed)
• $ii$ = Blood group O

Worked example: ABO blood group cross

A parent with blood group AB ($I^A{I}^B$) has children with a parent with blood group O ($ii$). Possible offspring genotypes are $I^{A}i$ (50%, blood group A) and $I^{B}i$ (50%, blood group B). No AB or O offspring are possible from this cross. Examiners deduct marks for describing codominance as "blending", so always use the phrasing "both alleles are fully expressed" in your answers.

5. Mutations and genetic engineering basics

Mutations

Mutations are permanent, random changes in the nucleotide sequence of an organism’s genome, and are the ultimate source of all genetic variation. There are two broad categories:

1. Point mutations: Single base pair changes (substitution, insertion, deletion) that may alter a single protein product (e.g. the point mutation that causes sickle cell anemia)
2. Chromosomal mutations: Changes to chromosome number or structure (e.g. non-disjunction during meiosis leading to trisomy 21, or Down syndrome) Most mutations are neutral, some are harmful, and a small minority are beneficial and act as raw material for evolution.

Genetic engineering

Genetic engineering is the direct manipulation of an organism’s genome using biotechnology to introduce new desirable traits. The most frequently assessed IB example is the production of human insulin using recombinant E. coli bacteria:

1. Extract the human insulin gene using restriction endonuclease enzymes, which cut DNA at specific recognition sites, leaving complementary "sticky ends"
2. Cut a bacterial plasmid (small circular DNA vector) with the same restriction enzyme to match the sticky ends of the insulin gene
3. Use DNA ligase enzyme to join the insulin gene to the plasmid, creating a recombinant plasmid
4. Insert the recombinant plasmid into E. coli host cells, which are cultured in large fermenters
5. The E. coli express the human insulin gene, producing insulin that is extracted, purified, and used to treat diabetic patients

Always mention both restriction endonuclease and DNA ligase in your answers to genetic engineering questions, as these are standard mark scheme points.

6. Polygenic and chi-squared tests (HL)

Polygenic inheritance

Polygenic traits are controlled by multiple independent genes, each contributing a small additive effect to the final phenotype. Unlike monogenic (single-gene) traits that show discrete variation (e.g. pea plant height is either tall or short), polygenic traits show continuous variation that forms a normal bell-shaped distribution curve in a population. Common examples include human height, skin color, and seed mass in plants. Environmental factors often also influence polygenic traits, further broadening the distribution curve.

Chi-squared (${\chi }^2$) test

The chi-squared test is a statistical tool used to determine if the difference between observed phenotypic ratios from a genetic cross and expected Mendelian ratios is statistically significant, or due to random chance. The formula is: $${\chi }^2=\sum \frac{(O-E{)}^2}{E}$$ Where $O$ = observed count for each phenotypic category, $E$ = expected count for each category, and the sum is taken across all categories.

Steps to complete a chi-squared test:

1. State the null hypothesis: There is no significant difference between observed and expected values, any difference is due to chance.
2. Calculate expected counts for each category using Mendelian ratios.
3. Calculate the ${\chi }^2$ value using the formula above.
4. Calculate degrees of freedom ($df$) = number of phenotypic categories - 1.
5. Compare your calculated ${\chi }^2$ value to the critical value given for $p=0.05$ (the standard 5% significance level used in biology):

• If ${\chi }^2<$ critical value: Accept the null hypothesis, differences are due to chance.
• If ${\chi }^2>$ critical value: Reject the null hypothesis, differences are statistically significant.

Worked example: Chi-squared test

A dihybrid cross of pea plants is expected to produce a 9:3:3:1 phenotypic ratio. Observed counts: 315 round yellow, 108 round green, 101 wrinkled yellow, 32 wrinkled green (total 556 offspring).

• Expected counts: $\frac{9}{16}\times 556=312.75$, $\frac{3}{16}\times 556=104.25$, $\frac{3}{16}\times 556=104.25$, $\frac{1}{16}\times 556=34.75$
• ${\chi }^2=\frac{(315-312.75{)}^2}{312.75}+\frac{(108-104.25{)}^2}{104.25}+\frac{(101-104.25{)}^2}{104.25}+\frac{(32-34.75{)}^2}{34.75}=0.47$
• $df=4-1=3$, critical value at $p=0.05$ is 7.815
• Since 0.47 < 7.815, accept the null hypothesis: the cross follows the expected 9:3:3:1 ratio.

7. Common Pitfalls (and how to avoid them)

• Wrong move: Mixing up genotypic and phenotypic ratios in Punnett square answers. Why students do it: Rushing through questions without reading the prompt carefully. Correct move: Always label both ratios explicitly, and double-check which ratio the question asks for before writing your answer.
• Wrong move: Stating that males can be carriers of X-linked recessive traits. Why students do it: Confusing autosomal and X-linked inheritance patterns. Correct move: Remember that males have only one X chromosome, so they either express the trait or are normal; only females can be carriers of X-linked recessive traits.
• Wrong move: Describing codominance as "blending of traits". Why students do it: Confusing codominance with incomplete dominance (not tested in core IB HL Biology). Correct move: Always describe codominance as both alleles being fully and separately expressed in the heterozygous phenotype.
• Wrong move: Forgetting to subtract 1 from the number of categories to calculate degrees of freedom in chi-squared tests. Why students do it: Memorizing the formula but not understanding its logic. Correct move: Once you know $n-1$ category counts, the last count is fixed, so $df=\text{number of categories}-1$.
• Wrong move: Omitting key enzymes (restriction endonuclease, DNA ligase) when describing genetic engineering steps. Why students do it: Focusing on the big picture and forgetting small molecular details required by the mark scheme. Correct move: Make a flashcard of the 4 key steps of bacterial gene transfer including both enzymes to recall them easily.

8. Practice Questions (IB Biology HL Style)

Question 1

A couple is planning to have children. The father has hemophilia A, an X-linked recessive condition. The mother has no family history of hemophilia and is not a carrier. Calculate the probability that their first child will be a son with hemophilia, and explain your answer with a Punnett square. [3 marks]

Worked Solution

Let $X^h$ = recessive hemophilia allele, $X^H$ = dominant normal allele. Father’s genotype: $X^{h}Y$ (affected male). Mother’s genotype: $X^H{X}^H$ (homozygous normal, non-carrier).

Question 2

A man with blood group A, whose father was blood group O, has children with a woman with blood group B, whose mother was blood group O. What are the possible blood groups of their children, and the probability of each? [3 marks]

Worked Solution

The man has blood group A, his father was O ($ii$), so he inherited an $i$ allele from his father, giving him a genotype of $I^{A}i$. The woman has blood group B, her mother was O ($ii$), so she inherited an $i$ allele from her mother, giving her a genotype of $I^{B}i$.

Question 3 (HL only)

A genetic cross between two heterozygous sweet pea plants for flower color and pollen shape was performed, with an expected 9:3:3:1 dihybrid ratio. Observed offspring counts: 296 purple long, 19 purple round, 27 red long, 85 red round (total 427 offspring). Calculate the chi-squared value, state the degrees of freedom, and state whether the null hypothesis is accepted or rejected (critical value for $p=0.05$ and $df=3$ is 7.815). [4 marks]

Worked Solution

1. Expected counts: $\frac{9}{16}\times 427=240.19$, $\frac{3}{16}\times 427=80.06$, $\frac{3}{16}\times 427=80.06$, $\frac{1}{16}\times 427=26.69$
2. ${\chi }^2=\frac{(296-240.19{)}^2}{240.19}+\frac{(19-80.06{)}^2}{80.06}+\frac{(27-80.06{)}^2}{80.06}+\frac{(85-26.69{)}^2}{26.69}=13.0+46.6+35.2+127.4=222.2$
3. Degrees of freedom = 4 categories - 1 = 3
4. Calculated ${\chi }^2$ (222.2) > critical value (7.815), so reject the null hypothesis. The difference between observed and expected values is statistically significant, so the genes are linked and do not assort independently. Mark scheme: 1 mark for correct expected values, 1 mark for correct ${\chi }^2$ calculation, 1 mark for correct degrees of freedom, 1 mark for correct conclusion.

9. Quick Reference Cheatsheet

10. What's Next

This genetics topic forms the foundation for two higher-level IB Biology units that follow: Topic 10 (Genetics and Evolution), which covers advanced meiosis, gene linkage, and population genetics, and Topic 11 (Animal Physiology), which includes immune system genetics and monoclonal antibody production. Understanding inheritance patterns and chi-squared analysis is also critical for your Internal Assessment (IA) if you choose to investigate genetic variation in plant or animal populations, as well as for Paper 3 Option B (Biotechnology and Bioinformatics) if you select that option for your exam.

If you are struggling with any part of this guide, from interpreting pedigree charts to calculating chi-squared values, you can ask Ollie, our AI tutor, for personalized explanations, extra practice questions, or feedback on your written answers. You can also browse more IB Biology HL study guides and past paper practice resources on the homepage to build your exam confidence and maximize your final score.