5.3Heredity
Mendelian Genetics — AP BIOLOGY Revision Notes & Topical Questions
5 min read 11 parts Lesson 3 of 5 1 quiz
Lesson summary
5.3 Mendelian Genetics — Notes — AP Biology notes: Apply Mendel's laws, probability rules, chi-square analysis and pedigrees to inheritance.Gene = a section of DNA coding for a trait; alleles are alternative versions of a gene at the same locus.Locus — the position of a gene on a…
- Law of segregation
- Law of independent assortment
- Test cross
- Chi-square test
- Null hypothesis
- Locus — the position of a gene on a chromosome
Section 1 of 11Basic terms
1
Gene = a section of DNA coding for a trait; alleles are alternative versions of a gene at the same locus.
2
Locus — the position of a gene on a chromosome.
3
Genotype — the allele combination (homozygous AA or aa; heterozygous Aa).
4
Phenotype — the observable trait.
5
P generation — the true-breeding parents; F1 is their offspring and F2 the offspring of crossing F1 individuals.
6
Mendel — bred large numbers of pea plants; F1 plants all resembled one parent and F2 showed both parental traits in a 3:1 ratio, showing inheritance depends on discrete factors rather than blending.
Section 2 of 11Mendel's laws
| Aspect | Law of segregation | Law of independent assortment |
|---|---|---|
| Statement | The two alleles of a gene separate randomly during gamete formation | Genes on different chromosomes (or far apart) assort independently |
| Alleles in gametes | One allele per gene | Random allele combinations |
| Meiosis stage | Anaphase I (homologues separate) | Metaphase I (random orientation of pairs) |
| Evidence | F1 all dominant, F2 3 : 1 | Dihybrid cross gives 9 : 3 : 3 : 1 |
Mendel's laws
Fertilisation restores two alleles for each gene.
The alleles of one gene segregate randomly and do not affect the assortment of another gene.
Section 3 of 11Working a cross
1
Choose letters for the alleles (capital for dominant) and write the parental genotypes and phenotypes.
2
List the possible gametes of each parent (one allele of each gene per gamete).
3
Place the gametes on the axes of a Punnett grid.
4
Combine gametes in the boxes to give the F1 genotypes.
5
List the genotype and phenotype ratios and state the phenotypes, not only the genotypes.
Working a cross
Dihybrid — a double heterozygote makes four equally likely gametes (YG, Yg, yG, yg), so the grid is 4 × 4.
Separate-gene method — treat each gene as its own monohybrid cross and multiply the results, which is quicker than a large grid.
Test cross — crossing with a homozygous recessive reveals an unknown genotype: all dominant offspring means the parent is AA and 1:1 means Aa.
Section 4 of 11Mendelian crosses compared
| Cross | Genotype ratio | Phenotype ratio | Use |
|---|---|---|---|
| Aa × Aa | 1 AA : 2 Aa : 1 aa | 3 : 1 | Monohybrid |
| Aa × aa | 1 Aa : 1 aa | 1 : 1 | Test cross |
| AaBb × AaBb | 9 genotype classes | 9 : 3 : 3 : 1 | Dihybrid |
| AaBb × aabb | 1 : 1 : 1 : 1 | 1 : 1 : 1 : 1 | Dihybrid test cross |
Section 5 of 11Probability
1
Randomness — allele combinations in offspring are random because of random assortment in meiosis and random fertilisation.
2
Product (multiplication) rule — for independent events P(A and B) = P(A) × P(B).
- Gg × GgP(gg) = ½ × ½ = ¼.
3
Sum (addition) rule — for mutually exclusive events P(A or B) = P(A) + P(B).
- Gg × GgP(heterozygote) = ¼ + ¼ = ½, as G from father with g from mother and the reverse are separate routes.
4
Gamete types — an organism heterozygous for n unlinked genes makes 2n gamete types.
5
Multi-gene cross — multiply the probability for each gene, e.g. aabbcc from AaBbCc × AaBbCc = ¼ × ¼ × ¼ = 1/64.
6
Human sex — females are XX and males XY, the father determines the sex, and each child has a 0.5 probability of being male.
7
Limit — the rules apply accurately only to unlinked genes.
Section 6 of 11Chi-square test
Chi-square test
Purpose — decides whether a difference between observed and expected results (including those predicted by Mendel's laws) is small enough to be due to chance; the data must be categorical.
Null hypothesis — no significant difference between observed and expected.
1
Calculate the expected numbers (E) from the total and the predicted ratio.
2
Calculate O − E for each class; the sign is irrelevant because it is squared.
3
Divide each squared difference by E and sum over all classes: χ² = Σ(O − E)² ÷ E (no units).
4
Find degrees of freedom = number of classes − 1.
5
Compare χ² with the critical value at p = 0.05.
1
Critical values at p = 0.05 — df 1 = 3.84, df 2 = 5.99, df 3 = 7.82.
2
If χ² ≥ critical value — the difference is significant, something other than chance acts (e.g. linkage) and the null hypothesis is rejected.
3
If χ² is below the critical value — the difference is due to chance and the null hypothesis is not rejected.
4
9:3:3:1 data — four classes give df = 3, so a small χ² such as 0.56 is well below 7.82 and the data fit the expected ratio.
Section 7 of 11Pedigrees
1
Symbols — squares are males, circles females, shading shows affected individuals, Roman numerals label generations, horizontal lines join parents and vertical lines join parent to child.
2
Purpose — pedigrees trace patterns of inheritance, so the mode of inheritance and probabilities can be predicted.
3
Information given — dominant or recessive allele, sex bias, genotypes and the probability of inheritance.
4
Autosomal recessive — can skip generations, and two unaffected parents can have an affected child.
5
Autosomal dominant — appears in every generation, and an affected child has at least one affected parent.
6
Method — write the homozygous recessive genotype onto affected individuals first, then deduce the rest.
7
Carriers — two unaffected parents with an affected child must both be heterozygous (Aa), and their unaffected children could be AA or Aa.
Section 8 of 11Pedigree patterns and human disorders
| Pattern | Key clues |
|---|---|
| Autosomal recessive | Can skip generations; affected child from unaffected parents; both sexes equally |
| Autosomal dominant | Appears every generation; unaffected parents never have affected children |
| X-linked recessive | Mostly males; passed from carrier mothers to sons; never father to son |
| X-linked dominant | Affected father passes it to all daughters and no sons |
| Mitochondrial | Affected mother passes it to all children; affected father to none |
- Check sex bias and father-to-son transmission, which rules X-linkage in or out.
| Disorder | Inheritance | Gene | What to state |
|---|---|---|---|
| Sickle-cell anaemia | Autosomal recessive | HBB | Abnormal β-globin; half-moon cells block capillaries |
| Sickle-cell carrier | Heterozygous | HBB | Makes both haemoglobins; may have no symptoms |
| Tay-Sachs disease | Autosomal recessive | HEXA | Hexosaminidase A deficient; symptoms from 3–6 months |
| Cystic fibrosis | Autosomal recessive | CFTR | Skips generations; carriers unaffected |
| Huntington's disease | Autosomal dominant | HTT | Faulty huntingtin; late onset, fatal; 50% risk per child of an affected parent |
Section 9 of 11More from the specification
1Phenotypic ratios across observation scales
- Phenotypic ratios across observation scales evaluating large sample sizes of offspring across macroscopic scales confirms predicted 3:1 monohybrid Mendelian proportions.
2Scale
- The ratio between the size of a drawing and the real object.
3Genetic variation
- Differences in alleles within a population.
4Pedigree diagram
- A family tree showing how a characteristic is inherited.
Section 10 of 11Key terms
| Term | Meaning |
|---|---|
| Allele | version of a gene |
| Genotype | allele combination |
| Phenotype | observable trait |
| Law of segregation | the two alleles separate into different gametes |
| Law of independent assortment | genes on different chromosomes sort independently |
| Test cross | cross with a homozygous recessive to reveal a genotype |
| Chi-square test | statistical test comparing observed with expected results |
| Null hypothesis | no significant difference between observed and expected |
| Pedigree | family tree showing inheritance of a trait |
| Carrier | heterozygote for a recessive allele |
Section 11 of 11Exam technique
1
Using the product rule for linked genes — Mendel's probability rules apply only to unlinked genes.
2
Accepting or rejecting the null hypothesis the wrong way — a χ² below the critical value means no significant difference.
3
Using the wrong degrees of freedom — df is the number of classes minus one.
4
Writing the wrong gametes in a dihybrid cross — each gamete has one allele of each gene.
5
Calling unaffected children of carriers AA — they could be AA or Aa.
Marked practiceQuizzes for this lesson
Last updated: