DP Biology HL D3.2 Inheritance

DP Biology HL D3.2 Inheritance - Full notes

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DP Biology HL | D3.2 Inheritance

DP BIOLOGY · THEME D: CONTINUITY AND CHANGE

D3.2

Inheritance

Full notes for HL · IB-style

Contents

1	Genes, alleles and genotypes
2	Mendel and monohybrid crosses
3	Punnett grids and predicting ratios
4	Codominance and multiple alleles: ABO blood groups
5	Sex-linked inheritance
6	Pedigree charts and genetic diseases
7	Continuous variation and polygenic traits
8	Dihybrid crosses and independent assortment (HL only)
9	Gene linkage and the chi-squared test (HL only)
10	Key genetic terms
11	Monohybrid crosses
12	Test crosses
13	Codominance
14	Multiple alleles: ABO blood groups
15	Sex determination
16	Sex-linked inheritance
17	Pedigree charts
18	Genetic diseases
19	Continuous variation and polygenic inheritance
20	Environmental effects on phenotype
21	Mixed review: worked problems
22	Additional exam-style practice
23	Dihybrid crosses and linkage (HL)
24	Practical: modelling inheritance
25	Common misconceptions and links
26	Exam-style question with markscheme
27	Check your understanding
28	Summary and topic map

These are full teaching notes. Read each section in order, try the Think about it question before you read on, and do every Quick check before moving to the next section.

Before you start

Rate yourself on each statement now, and again after you've worked through the notes. Shade R (not yet), A (getting there) or G (confident).

I can…	R	A	G
define gene, allele, genotype, phenotype, dominant, recessive, homozygous, heterozygous
explain Mendel's experiments and monohybrid crosses
use Punnett grids to predict genotype and phenotype ratios
explain codominance and multiple alleles using ABO blood groups
explain sex-linked inheritance using haemophilia and colour blindness
deduce genotypes from pedigree charts
explain continuous variation and polygenic inheritance
predict ratios from dihybrid crosses
explain gene linkage and recombinants
use the chi-squared test to compare observed and expected ratios

Sections marked HL only (Sections 8 and 9) are Additional Higher Level content. Every other section is common to SL and HL.

Guiding questions

How are characteristics passed from parents to offspring? How can we predict the outcomes of genetic crosses?

How this topic is examined

D3.2 is one of the most heavily examined subtopics: genetic crosses with Punnett grids, pedigree analysis, blood group and sex-linkage problems. HL questions include dihybrid crosses, linkage and chi-squared tests.

Always define symbols, show parental genotypes, gametes, the Punnett grid, and offspring genotypes, phenotypes and ratios.

Command term	What the examiner wants
State / Identify	a brief answer with no explanation needed
Outline	a brief account or summary
Describe	a detailed account, including trends or features, without reasons
Explain	a detailed account including causes, reasons or mechanisms
Distinguish / Compare and contrast	give the differences (and similarities), point by point
Calculate	a numerical answer with working and units
Deduce / Suggest	reach a conclusion from data; propose a reasonable explanation
Evaluate / Discuss	weigh strengths and limitations and reach a judgement

How to use these notes

Each section teaches one part of the syllabus in full. You'll see these boxes throughout; each has one job:

Box	What it's for
Think about it	a question to get you thinking before a section; the answer comes in the section
Key point	the one idea you must not forget
Common mistake	an error examiners see every year, and how to avoid it
In the real world	where the biology shows up outside the classroom
Try this	something quick you can do yourself to test the idea
Worked example	full solutions with IB-style marks (M1 method, A1 answer, R1 reasoning)
Quick check	short questions to test yourself at the end of each section

1 Genes, alleles and genotypes

The vocabulary of genetics.

Think about it

Two brown-eyed parents have a blue-eyed child. How is that possible?

Term	Meaning
Gene	a heritable factor: a section of DNA that influences a characteristic
Locus	the position of a gene on a chromosome
Allele	one of the alternative forms of a gene
Genotype	the combination of alleles an organism has (e.g. Bb)
Phenotype	the observable characteristics, resulting from genotype and environment
Dominant allele	expressed in the phenotype when present in one copy (heterozygote)
Recessive allele	expressed only when two copies are present (homozygous)
Homozygous	two identical alleles (BB or bb)
Heterozygous	two different alleles (Bb)
Carrier	heterozygous for a recessive disease allele; not affected

Diploid organisms have two copies of each gene, one on each of a pair of homologous chromosomes, one inherited from each parent. The two copies may be the same allele or different alleles. Gametes are haploid and carry one allele of each gene, because homologous chromosomes separate in meiosis.

2 Mendel and monohybrid crosses

Peas revealed the rules of inheritance.

Gregor Mendel, working with garden peas in the 1860s, crossed pure-breeding plants differing in one characteristic, such as tall and dwarf. The first generation (F₁) were all tall. When F₁ plants were crossed with each other, the F₂ generation contained tall and dwarf plants in a ratio of about 3 : 1 (787 tall to 277 dwarf).

Mendel concluded that characteristics are controlled by pairs of factors (now called alleles), one from each parent; that one factor can be dominant over the other; and that the two factors segregate (separate) into different gametes, so each gamete carries only one. This is Mendel's law of segregation, explained by the separation of homologous chromosomes in meiosis.

A monohybrid cross follows the inheritance of one gene.

Quick check

1. What ratio did Mendel find in the F₂?

2. What does segregation mean?

Answers: 1. 3 : 1 2. the two alleles of a gene separate into different gametes

3 Punnett grids and predicting ratios

A grid shows every combination of gametes.

A Punnett grid shows all the possible combinations of gametes from two parents and the expected proportions of offspring genotypes. The steps: (1) choose symbols, using a capital letter for the dominant allele and the same lower-case letter for the recessive; (2) state the parental genotypes and phenotypes; (3) list the possible gametes; (4) draw the grid; (5) state offspring genotypes, phenotypes and ratios.

	T	t
T	TT (tall)	Tt (tall)
t	Tt (tall)	tt (dwarf)

WORKED EXAMPLE 1 A monohybrid cross

In pea plants, tall (T) is dominant to dwarf (t). Two heterozygous tall plants are crossed. Predict the genotypes and phenotypes of the offspring.

Solution

Parents: Tt × Tt; gametes: T, t from eachM1

Grid gives TT : Tt : tt in the ratio 1 : 2 : 1A1

Phenotypes: 3 tall : 1 dwarfA1

A test cross determines whether an organism showing the dominant phenotype is homozygous or heterozygous: it is crossed with a homozygous recessive individual. If any offspring show the recessive phenotype, the unknown parent must be heterozygous.

Common mistake

Writing ratios as percentages of actual offspring. Ratios are probabilities: a 3 : 1 ratio means each offspring has a 3/4 chance of being tall, not that exactly three of every four will be.

Quick check

1. What genotype ratio does Tt × Tt give?

2. What is a test cross?

Answers: 1. 1 TT : 2 Tt : 1 tt 2. cross with homozygous recessive to reveal the unknown genotype

4 Codominance and multiple alleles: ABO blood groups

When neither allele is fully dominant.

In codominance, both alleles in a heterozygote are fully expressed in the phenotype. Some genes have more than two alleles in the population (multiple alleles), though each individual still has only two.

The ABO blood group gene has three alleles: IA (antigen A on red cells), IB (antigen B) and i (no antigen). IA and IB are codominant; both are dominant to i.

Genotype	Blood group
IAIA or IAi	A
IBIB or IBi	B
IAIB	AB (codominance)
ii	O

WORKED EXAMPLE 2 Blood group cross

A man with blood group A (IAi) and a woman with blood group B (IBi) have children. What blood groups are possible, and in what ratio?

Solution

Gametes: IA, i and IB, iM1

Offspring: IAIB, IAi, IBi, iiM1

Blood groups AB : A : B : O = 1 : 1 : 1 : 1A1

Quick check

1. Define codominance.

2. Which genotype gives blood group AB?

Answers: 1. both alleles expressed in the heterozygote 2. IAIB

5 Sex-linked inheritance

Genes on the X chromosome show different patterns in males and females.

Humans have 22 pairs of autosomes and one pair of sex chromosomes: XX in females, XY in males. The X chromosome is much larger and carries many genes that are absent from the Y. Genes carried on the X chromosome are sex-linked.

Because males have only one X chromosome, a male with a recessive allele on his X will show the recessive phenotype; there is no second X to carry a dominant allele. So sex-linked recessive conditions are much more common in males. Females can be carriers (heterozygous).

Haemophilia (a blood clotting disorder) and red-green colour blindness are sex-linked recessive conditions. Alleles are written as superscripts on X: XH (normal clotting) and Xh (haemophilia).

A son inherits his X chromosome from his mother, so sex-linked conditions are passed from carrier mothers to sons. Affected fathers pass the allele to all their daughters (who become carriers) but to none of their sons.

WORKED EXAMPLE 3 Sex-linked cross

A carrier woman (XHXh) and an unaffected man (XHY) have children. Predict the outcomes.

Solution

Gametes: XH, Xh and XH, YM1

Offspring: XHXH, XHXh, XHY, XhYM1

Daughters: all unaffected (half carriers); sons: 1 in 2 chance of haemophiliaA1

Quick check

1. Why are sex-linked recessive conditions more common in males?

2. Can an affected father pass haemophilia to his son?

Answers: 1. males have one X, so one recessive allele is expressed 2. no: sons receive his Y chromosome

6 Pedigree charts and genetic diseases

Family trees reveal patterns of inheritance.

A pedigree chart shows the inheritance of a characteristic in a family over several generations. Squares are males, circles are females; shaded symbols are affected individuals; horizontal lines join parents, and vertical lines lead to children.

Recessive conditions can 'skip' generations: two unaffected parents (both carriers) can have an affected child. Dominant conditions appear in every generation: every affected person has at least one affected parent. Sex-linked recessive conditions mainly affect males, and pass through carrier females.

Examples of genetic diseases: cystic fibrosis (autosomal recessive, affecting a chloride channel protein, causing thick mucus); Huntington's disease (autosomal dominant, causing degeneration of the nervous system in mid-life); phenylketonuria (autosomal recessive); haemophilia (sex-linked recessive).

WORKED EXAMPLE 4 Deducing a pedigree

Two unaffected parents have an affected daughter. Deduce whether the condition is dominant or recessive, and give the parents' genotypes.

Solution

Unaffected parents with an affected child: the allele is recessiveR1

Both parents must be carriers (Aa)A1

Not sex-linked recessive: an affected daughter would need an affected fatherR1

Quick check

1. How can you recognize a recessive condition in a pedigree?

2. Give an example of an autosomal dominant condition.

Answers: 1. unaffected parents can have affected children 2. Huntington's disease

7 Continuous variation and polygenic traits

Many genes, many phenotypes.

Characteristics controlled by a single gene often show discontinuous variation: distinct categories (blood groups, tall or dwarf peas). Many human characteristics, such as height, skin colour and body mass, show continuous variation: a range of values, often with a normal distribution.

Continuous variation results from polygenic inheritance (many genes each with a small additive effect) combined with environmental factors. Skin colour depends on several genes affecting melanin production, and on exposure to sunlight. Height depends on hundreds of genes and on nutrition.

Figure 1. Continuous variation in height produces a normal distribution.

8 Dihybrid crosses and independent assortment (HL only)

Two genes at once.

HL only

This section is Additional Higher Level content. SL students do not need it.

A dihybrid cross follows two genes at the same time. If the genes are on different chromosomes, they are inherited independently: the alleles of one gene can combine with either allele of the other in the gametes, because homologous pairs orient randomly in meiosis (independent assortment, D2.1).

A cross between two individuals heterozygous for both genes (AaBb × AaBb) produces four types of gamete from each (AB, Ab, aB, ab), giving a 16-square Punnett grid and a phenotype ratio of 9 : 3 : 3 : 1.

WORKED EXAMPLE 5 A dihybrid cross

In peas, round seed (R) is dominant to wrinkled (r), and yellow (Y) to green (y). Predict the phenotype ratio from RrYy × RrYy.

Solution

Gametes from each parent: RY, Ry, rY, ryM1

16-square gridM1

9 round yellow : 3 round green : 3 wrinkled yellow : 1 wrinkled greenA1

9 Gene linkage and the chi-squared test (HL only)

Genes on the same chromosome tend to be inherited together.

HL only

This section is Additional Higher Level content. SL students do not need it.

Genes on the same chromosome are linked: they tend to be inherited together and do not assort independently, so dihybrid crosses do not give the 9 : 3 : 3 : 1 ratio. Combinations of alleles from the parents (parental types) are more common than expected. Recombinants with new combinations arise only if crossing over occurs between the two loci in prophase I. The further apart the genes, the more often crossing over separates them.

The chi-squared test compares observed results with those expected from a hypothesis (such as independent assortment). Null hypothesis: there is no significant difference between observed and expected. χ² = Σ (O − E)² ÷ E. Degrees of freedom = number of categories − 1. If χ² exceeds the critical value at p = 0.05, the difference is significant, suggesting the genes may be linked.

Phenotype	Observed	Expected (9:3:3:1 of 160)	(O − E)² ÷ E
dominant both	102	90	1.60
dominant / recessive	18	30	4.80
recessive / dominant	16	30	6.53
recessive both	24	10	19.60
Total	160	160	χ² = 32.53

WORKED EXAMPLE 6 Chi-squared for a dihybrid cross

Using the table, test whether the results fit a 9 : 3 : 3 : 1 ratio (critical value for 3 degrees of freedom at p = 0.05 is 7.81).

Solution

χ² = 1.60 + 4.80 + 6.53 + 19.60 = 32.53M1 A1

32.53 > 7.81: reject the null hypothesisR1

Parental types are over-represented: the genes are probably linkedR1

Quick check

1. What ratio does a dihybrid cross of two double heterozygotes give?

2. Why do linked genes not give this ratio?

Answers: 1. 9 : 3 : 3 : 1 2. they are on the same chromosome; only crossing over separates them

10 Key genetic terms

Precise vocabulary.

Term	Meaning
gene	a heritable factor; a sequence of DNA coding for a polypeptide
allele	an alternative form of a gene
genotype	the alleles an organism has (e.g. Bb)
phenotype	observable characteristics
homozygous	two identical alleles
heterozygous	two different alleles
dominant	expressed in heterozygotes
recessive	expressed only when homozygous

11 Monohybrid crosses

Punnett grids.

Write the parental genotypes, the possible gametes, a Punnett grid, then the offspring genotype and phenotype ratios. A cross between two heterozygotes (Aa × Aa) gives 1 AA : 2 Aa : 1 aa, a 3 : 1 phenotype ratio.

WORKED EXAMPLE 7 Heterozygous cross (standard)

In pea plants, tall (T) is dominant to dwarf (t). Predict the offspring of Tt × Tt.

Solution

Gametes T, t from eachM1

1 TT : 2 Tt : 1 tt; 3 tall : 1 dwarfA1

Quick check

1. Predict the offspring of Tt × tt.

2. Two brown-eyed parents have a blue-eyed child. What are the parents' genotypes?

Answers: 1. 1 tall : 1 dwarf 2. both heterozygous (Bb)

12 Test crosses

Revealing unknown genotypes.

A test cross crosses an individual with the dominant phenotype with a homozygous recessive. If all offspring show the dominant phenotype, the individual is probably homozygous; if half show the recessive phenotype, it is heterozygous.

WORKED EXAMPLE 8 Test cross (exam level)

A black guinea pig crossed with a white one (bb) gives 6 black and 5 white offspring. Deduce the black parent's genotype.

Solution

White offspring must have received b from the black parentR1

Black parent is BbA1

13 Codominance

Both alleles expressed.

In codominance, both alleles are expressed in heterozygotes. Use superscripts: CRCW gives roan cattle (red and white hairs), or in ABO blood groups, IAIB gives group AB.

WORKED EXAMPLE 9 Flowers (standard)

Red (CRCR) × white (CWCW) four-o'clock flowers give pink offspring. Predict F2 ratio.

Solution

CRCW × CRCWM1

1 red : 2 pink : 1 whiteA1

14 Multiple alleles: ABO blood groups

Three alleles.

The ABO gene has three alleles: IA and IB (codominant) and i (recessive). Genotypes: IAIA or IAi (group A), IBIB or IBi (group B), IAIB (AB), ii (O).

WORKED EXAMPLE 10 Blood groups (exam level)

Can parents with groups A and B have a child with group O? Explain.

Solution

Yes, if both are heterozygous (IAi × IBi)R1

Probability 1/4 for iiA1

Quick check

1. Give the possible blood groups of children of IAIB × ii.

Answers: 1. A or B (1 : 1)

15 Sex determination

XX and XY.

In humans, females are XX and males XY. The Y chromosome carries the SRY gene, which triggers male development. Each child has a 50% chance of being male, because half of sperm carry X and half Y.

Quick check

1. Which parent determines the sex of the child?

Answers: 1. the father (X or Y sperm)

16 Sex-linked inheritance

Genes on the X chromosome.

Genes on the X chromosome show sex-linked patterns. Males have only one X, so recessive X-linked conditions (red–green colour blindness, haemophilia) are more common in males. Females can be carriers (XHXh).

WORKED EXAMPLE 11 Haemophilia (exam level)

A carrier female (XHXh) has children with a normal male (XHY). Predict the outcomes.

Solution

Daughters: 1/2 XHXH, 1/2 carriersA1

Sons: 1/2 normal, 1/2 haemophiliacA1

Quick check

1. Can a colour-blind man pass the allele to his son?

Answers: 1. no: sons receive his Y chromosome

17 Pedigree charts

Tracing inheritance in families.

Pedigrees use squares (males) and circles (females); shaded symbols show affected individuals. Clues: two unaffected parents with an affected child indicates a recessive condition; an affected child must have at least one affected parent for a dominant condition; more affected males suggests X-linked recessive.

WORKED EXAMPLE 12 Interpreting (exam level)

Two unaffected parents have an affected daughter. Deduce the pattern of inheritance.

Solution

Recessive (parents are carriers)R1

Autosomal, because an affected daughter of an unaffected father rules out X-linked recessiveR1

18 Genetic diseases

Examples.

Condition	Inheritance	Cause
cystic fibrosis	autosomal recessive	faulty CFTR chloride channel
Huntington's disease	autosomal dominant	CAG repeat expansion
haemophilia	X-linked recessive	missing clotting factor
sickle cell anaemia	autosomal codominant/recessive	base substitution in β-globin

19 Continuous variation and polygenic inheritance

Many genes, many phenotypes.

Characteristics such as height and skin colour are controlled by many genes (polygenic) plus environmental factors, producing continuous variation with a normal distribution.

Quick check

1. Why does height show continuous variation?

Answers: 1. controlled by many genes and influenced by environment (diet)

20 Environmental effects on phenotype

Genotype + environment.

Phenotype results from genotype and environment. Hydrangea flower colour depends on soil pH; human height depends on nutrition; identical twins can differ in weight.

Quick check

1. Give an example of the environment affecting phenotype.

Answers: 1. hydrangea colour and soil pH

21 Mixed review: worked problems

Bringing the ideas together.

WORKED EXAMPLE 13 Mixed problem (exam level)

A woman with group A blood whose father was group O has a child with a man of group AB. List the possible blood groups of the child and their probabilities.

Solution

Woman IAi; man IAIBM1

Offspring: IAIA, IAIB, IAi, IBiA1

A 1/2, AB 1/4, B 1/4A1

22 Additional exam-style practice

Paper 1 and Paper 2.

WORKED EXAMPLE 14 Paper 1 question (standard)

What ratio is expected from Aa × aa? A. 3 : 1 B. 1 : 1 C. 1 : 2 : 1 D. all dominant

Solution

BA1

WORKED EXAMPLE 15 Paper 2 question (exam level)

Explain why red–green colour blindness is more common in males.

Solution

Gene on X chromosome; males have one XR1

One recessive allele is expressed in males; females need twoR1

23 Dihybrid crosses and linkage (HL)

Two genes at once.

For two unlinked genes, AaBb × AaBb gives a 9 : 3 : 3 : 1 ratio (independent assortment). Linked genes on the same chromosome are inherited together, giving more parental-type offspring; recombinants arise from crossing over. The chi-squared test compares observed and expected ratios.

WORKED EXAMPLE 16 Chi-squared (exam level)

A dihybrid cross gives 315, 108, 101, 32 (total 556). Expected 9 : 3 : 3 : 1. Calculate χ2 (critical value 7.81).

Solution

Expected 312.75, 104.25, 104.25, 34.75M1

χ2 = 0.016 + 0.135 + 0.101 + 0.218 = 0.47A1

0.47 < 7.81: fits 9 : 3 : 3 : 1R1

24 Practical: modelling inheritance

Coins and beads.

Tossing two coins (each representing a heterozygous parent's gametes) many times models a monohybrid cross. Large samples get closer to the expected 1 : 2 : 1 ratio.

WORKED EXAMPLE 17 Sample size (standard)

Why do large samples give ratios closer to expected?

Solution

Random variation has less effect on large samplesR1

Try this

Toss two coins 100 times, recording HH, HT and TT. Compare your results with the expected 25 : 50 : 25.

25 Common misconceptions and links

Students often think…	Actually…
dominant alleles are more common	dominance is about expression, not frequency
ratios predict exact numbers	ratios are probabilities
males can be carriers of sex-linked recessive conditions	males have one X, so they are affected or unaffected
all dihybrid crosses give 9:3:3:1	linked genes do not

How D3.2 connects to the rest of the course

Topic	Connection
D2.1 Meiosis	segregation and independent assortment
D1.3 Mutations	origin of alleles
D4.1 Natural selection	allele frequencies
B1.1 Glycoproteins	ABO antigens
C3.2 Defence	blood transfusions

26 Exam-style question with markscheme

Try this under exam conditions (about 18 minutes) before reading the markscheme.

Text A. In a family pedigree, two unaffected parents have three children: an unaffected son, an affected daughter and an unaffected daughter.

(a) Deduce whether the condition in the pedigree is dominant or recessive. [2]

(b) Two carriers of cystic fibrosis have a child. Calculate the probability the child has cystic fibrosis. [2]

(d) A man with blood group AB and a woman with blood group O have a child. State the possible blood groups. [2]

(e) Explain why the dihybrid results differ significantly from 9:3:3:1. [2]

Markscheme

Part	Answer	Marks
(a)	recessive	A1
	unaffected parents have an affected child	R1
(b)	Aa × Aa; Punnett grid	M1
	1/4 (25%)	A1
(c)	gene on X chromosome	R1
	males have one X	R1
	one recessive allele expressed; females need two	R1
(d)	IAi and IBi	M1
	A or B (1 : 1)	A1
(e)	genes linked on same chromosome	R1
	parental combinations inherited together; recombinants from crossing over	R1

Examiner's comments. In (d), weaker answers included AB or O; the child must receive IA or IB from the father and i from the mother.

27 Check your understanding

Section A: multiple choice

1. The phenotype ratio from Tt × Tt is

A. 3 : 1 B. 1 : 2 : 1 C. 1 : 1 D. 9 : 3 : 3 : 1

2. A heterozygous genotype is

A. BB B. bb C. Bb D. B

3. Blood group O has genotype

A. IAi B. ii C. IAIB D. IBi

4. Haemophilia is

A. autosomal dominant B. codominant C. polygenic D. sex-linked recessive

5. Height shows

A. discontinuous variation B. continuous variation C. codominance D. sex linkage

6. A test cross uses a

A. homozygous recessive individual B. heterozygote C. homozygous dominant D. carrier

7. AaBb × AaBb (unlinked) gives the ratio

A. 3 : 1 B. 1 : 1 : 1 : 1 C. 1 : 2 : 1 D. 9 : 3 : 3 : 1

8. Recombinants of linked genes arise by

A. independent assortment B. mutation C. crossing over D. non-disjunction

Section B: structured

9. Explain Mendel's law of segregation. [3]

10. Explain codominance using ABO blood groups. [3]

11. A colour-blind man and a carrier woman have children. Predict outcomes. [4]

12. Explain how linkage affects dihybrid ratios. [3]

Answers

1. A 2. C 3. B 4. D 5. B 6. A 7. D 8. C

9. Two alleles per gene; separate into different gametes in meiosis; each gamete has one allele.

10. IA and IB both expressed in AB; both dominant to i.

11. XbY × XBXb: daughters 1/2 colour-blind, 1/2 carriers; sons 1/2 colour-blind.

12. Linked genes inherited together; fewer recombinants; ratio deviates from 9:3:3:1; chi-squared test.

28 Summary and topic map

Idea	Remember
Terms	gene, allele, genotype, phenotype, dominant, recessive
Mendel	3 : 1; segregation
Punnett grid	gametes, grid, ratios; test cross
Codominance	ABO: IA, IB codominant, i recessive
Sex linkage	X chromosome; haemophilia; males affected more
Pedigrees	recessive skips generations; dominant does not
Continuous variation	polygenic + environment
HL	dihybrid 9:3:3:1; linkage; chi-squared

Before the exam, make sure you…

•define symbols and show every step of crosses
•use X and Y superscripts for sex linkage
•deduce pedigree patterns with reasons
•treat ratios as probabilities

Inheritance on one page

Cover this page with a sheet of paper and try to rebuild it from memory. Whatever you can't recall is what to revise next.

Figure 2. Topic map for Inheritance.

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