Educerie
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Educerie · IB Diploma · Biology

Theme D Continuity and change · D3.2 Inheritance

Level
SL and HL. Sections 11 to 15 are HL only. If you are SL, skip them; nothing in your papers tests them.
Themes (key concepts)
continuity and change, at the level of organisms. Continuity, because every sexually reproducing eukaryote passes its genes on the same way, one set from each parent; change, because each child receives a new combination of alleles and so is never a copy of either parent.
The question this unit answers
what patterns of inheritance exist in plants and animals, and what is the molecular basis of those patterns?
Where it is examined
Paper 1A multiple choice, where a cross or a pedigree is set as a four-option puzzle; Paper 1B, where you read a box-and-whisker plot, a table of cross results or a chi-squared calculation; Paper 2 Section A, where "use a Punnett grid to determine…" (3–4 marks) and "deduce the mode of inheritance from the pedigree" (2–3 marks) are regular questions; and Paper 2 Section B, where "explain the inheritance of haemophilia" or, at HL, "explain how linkage affects the outcome of a dihybrid cross" are extended-response parts worth 4 to 7 marks.

What you must be able to do

You must be able toLevelWhat it looks like in the exam
Explain how haploid gametes fuse to form a diploid zygote, and why a diploid cell has two copies of each autosomal geneSL, HL"Outline how genes pass from parents to offspring" (3 marks)
Describe how a cross is carried out in a flowering plant, and use P, F1, F2 and Punnett grid correctlySL, HL"Outline the method for crossing two pea plants" (3 marks)
Use genotype, homozygous, heterozygous, gene and allele accuratelySL, HLDefinitions in multiple choice; "distinguish between gene and allele" (2 marks)
Explain phenotype as the result of genotype and environment, with examples of each kind of traitSL, HL"Suggest one human trait due to environment only" (1 mark)
Explain why AA and Aa have the same phenotypeSL, HL"Explain why a heterozygous individual does not have PKU" (2 marks)
Explain phenotypic plasticity, and say why it is not a change in genotypeSL, HL"Distinguish between phenotypic plasticity and mutation" (2 marks)
Explain PKU as an autosomal recessive conditionSL, HLA cross between two carriers (3 marks)
Explain SNPs and multiple alleles; use Iᴬ, Iᴮ and i for ABO blood groupsSL, HL"Determine the possible blood groups of the children" (3–4 marks)
Distinguish codominance from incomplete dominance, using AB blood and four o'clock flowersSL, HL"Distinguish between…" (2–3 marks)
Explain sex determination and the inheritance of sex-linked genes, using haemophilia and Xᴴ / Xʰ notationSL, HLA carrier cross, then "explain why more males than females…" (4 marks)
Deduce patterns of inheritance and genotypes from pedigree charts; distinguish inductive from deductive reasoningSL, HLPaper 1B or 2: "deduce, with a reason, whether the allele is dominant or recessive" (2 marks)
Explain continuous variation by polygenic inheritance and environment; use mean, median and modeSL, HL"Explain why human skin colour shows continuous variation" (3 marks)
Draw and read a box-and-whisker plot, including the 1.5 × IQR outlier ruleSL, HLPaper 1B: "determine whether the value is an outlier" (2 marks)
Link chromosome movements in meiosis to segregation and independent assortmentHL only"Explain how meiosis produces four types of gamete" (3 marks)
Use a Punnett grid for a dihybrid cross and derive 9:3:3:1 and 1:1:1:1HL only"Predict the phenotypic ratio" (3–4 marks)
Find gene loci and their polypeptides in a databaseHL only"Suggest why these two genes are inherited together" (2 marks)
Explain autosomal linkage and identify recombinants in gametes, genotypes and phenotypesHL onlyA test cross with linked genes (4–5 marks)
Carry out and interpret a chi-squared test on dihybrid cross dataHL onlyPaper 1B or 2: calculate χ², compare with the critical value, conclude (4–5 marks)

Before you start

You need D2.1 (meiosis halves the chromosome number and pairs homologous chromosomes), D3.1 (fertilisation and pollination), and A1.2 (a gene is a stretch of DNA with a base sequence). Everything here is those three ideas put together: meiosis deals out one copy of each chromosome, fertilisation brings two copies back together, and the base sequence of each copy decides what protein is made.


1The idea in one paragraph

Each parent makes gametes by meiosis, and each gamete carries one copy of every chromosome, so one version of every gene. Fertilisation joins two gametes, so the child has two versions of each gene, one from each parent. Those versions are alleles. What the child looks like, its phenotype, depends on which two alleles it holds and on the environment it grows up in. Sometimes one allele hides the other (dominance), sometimes both show (codominance), sometimes the result is a blend (incomplete dominance). Genes on the X chromosome follow a different pattern in males and females. Many traits are controlled by several genes at once and vary continuously. At HL you follow two genes at a time, see how their position on the chromosomes changes the result, and test the result with statistics.

2Gametes, zygotes and two copies of every gene

Every eukaryote that reproduces sexually runs the same cycle. Body cells are diploid (2n): they have two sets of chromosomes, one inherited from each parent. Meiosis makes haploid (n) gametes with one set. Two haploid gametes fuse at fertilisation and form a diploid zygote. The cycle then repeats in the next generation. Figure 1 follows one gene round it.

Figure 1 · One gene, from two parents to one zygote Figure 1 · One gene, from two parents to one zygote diploid 2n haploid n diploid 2n A a Parent 1 · Aa A a Parent 2 · Aa meiosis meiosis A a A a gametes: one allele each fertilisation a A Zygote · Aa (one chromosome, one allele, from each parent) Meiosis puts one allele of the gene into each gamete; fertilisation brings two together again.
Figure 1 · One gene, from two parents to one zygote

The consequence for genes is simple and it drives everything else in this page. Each autosome (any chromosome other than the sex chromosomes) comes in a matching pair, so a diploid cell holds two copies of each autosomal gene. The two copies sit at the same position, the same locus, on the two homologous chromosomes. They may be identical, or they may differ slightly in base sequence. Each gamete receives only one of the two, chosen by which chromosome of the pair goes into it at meiosis.

Meiosis halves: one allele of each gene goes into each gamete. Fertilisation doubles: two alleles of each gene come together in the zygote.

3Genes, alleles, genotype and phenotype

Four words, and the marks depend on keeping them apart.

  • A gene is a length of DNA at a particular locus that codes for a polypeptide (or an RNA). The gene for flower colour in a pea is one gene.
  • An allele is one version of a gene, differing from other versions in its base sequence, often by only one or a few bases. The purple and the white version are two alleles of the same gene.
  • The genotype is the combination of alleles an organism has inherited for a gene, written with letters: PP, Pp or pp.
  • The phenotype is the observable trait: purple flowers, blood group A, a height of 170 cm.

If the two alleles are the same (PP or pp) the organism is homozygous for that gene. If they differ (Pp) it is heterozygous.

Why a heterozygote can look exactly like a homozygote. A dominant allele has its effect whether one copy or two copies are present. A recessive allele shows only when there is no dominant allele, so only in the homozygous recessive genotype. The reason is usually molecular. The dominant allele codes for a working protein, often an enzyme. The recessive allele codes for a protein that does not work, or for no protein at all. One working copy makes enough enzyme to produce the normal phenotype, so PP and Pp look the same: both make the purple pigment. Only pp, with no working copy, makes no pigment and has white flowers. Write it in an answer as: one functional allele is enough to produce sufficient protein.

Phenotype is genotype plus environment. The guide expects you to suggest examples of three kinds.

Kind of traitExamplesWhy
Due to genotype onlyABO blood group; red–green colour blindnessNo normal environment changes them
Due to environment onlyA scar; a tattoo; the language you speakNo allele codes for them
Due to genotype and environment togetherHeight (alleles set the range, childhood nutrition decides where in it); skin colour (alleles plus sun exposure); the symptoms of PKU (section 6)Both have to be known to predict the phenotype

4Phenotypic plasticity

Phenotypic plasticity is the capacity of an organism to develop traits suited to the environment it experiences, by changing which genes are expressed and how strongly. The genotype does not change. The same alleles are there; the environment switches some of them on or up, others off or down. Because it is a matter of expression, the change can often be reversed within the individual's lifetime.

  • Tanning. Ultraviolet light increases the expression of melanin-producing genes in skin cells, and the darker skin protects deeper cells. Out of the sun, the tan fades.
  • Living at altitude. Low oxygen raises the rate of red blood cell production, so the blood carries more haemoglobin. At sea level the count falls back.
  • Water crowfoot. This aquatic buttercup grows thread-like leaves under water and broad, flat leaves on the surface, on the same plant, from the same genes.

The exam trap is to call these mutations or adaptations inherited by the next generation. They are neither. The child of a tanned parent is born with the same skin as the child of an untanned one.

5How a cross is done in a flowering plant

In a flowering plant the male gametes are inside pollen grains, made in the anthers, and the female gametes are inside ovules, in the ovary. So a cross needs pollination: pollen from one plant has to reach the stigma of the other. Many plants, such as peas, carry both anthers and ovaries in the same flower, and pea flowers normally stay closed, so they pollinate themselves. That is self-pollination and leads to self-fertilisation. It is useful, because a line bred by self-fertilisation for generations becomes homozygous, or pure-breeding. It is also a problem, because a cross between two chosen plants must stop self-pollination first. Figure 2 sets out the method.

Figure 2 · Crossing two plants by hand Figure 2 · Crossing two plants by hand 1 · Choose parents two pure-breeding plants with contrasting traits (P generation) 2 · Remove anthers from the seed parent's flower before pollen is released 3 · Transfer pollen from the other parent to the stigma with a small brush 4 · Bag the flower so no other pollen can reach the stigma 5 · Sow the seeds the plants that grow are the F1 generation 6 · Self the F1 let F1 plants self-pollinate; their offspring are the F2 Removing the anthers and bagging the flower make sure the seeds come from the chosen cross.
Figure 2 · Crossing two plants by hand

The two chosen parents are the P generation. Removing the anthers from the seed parent stops it fertilising itself; the bag keeps out stray pollen. The seeds grow into the F1 generation (first filial), and the offspring of F1 plants selfed or crossed with each other are the F2 generation.

A Punnett grid predicts the result: the gametes of one parent along the top, the gametes of the other down the side, and each box is one possible zygote. Figure 3 follows tall and dwarf pea plants, where the tall allele T is dominant over the dwarf allele t.

Figure 3 · A monohybrid cross from P to F2 Figure 3 · A monohybrid cross from P to F2 P generation TT tall × tt dwarf gametes T t F1 generation Tt all tall F1 × F1: Tt × Tt F2 generation gametes of one F1 parent across, the other down T t T t TT tall Tt tall Tt tall tt dwarf genotypes 1 TT : 2 Tt : 1 tt phenotypes 3 tall : 1 dwarf The F1 is all tall; the F2 is 1 TT : 2 Tt : 1 tt, which is 3 tall : 1 dwarf.
Figure 3 · A monohybrid cross from P to F2

Every F1 plant is Tt and tall. The F2 has the genotypes TT, Tt and tt in the ratio 1 : 2 : 1, so the phenotypes are tall and dwarf in the ratio 3 : 1. Suppose a grower counts 306 tall and 106 dwarf F2 plants. That is 306 ÷ 106 ≈ 2.9 : 1, close to 3 : 1. A ratio from real counts is never exact, because fertilisation is random; section 15 shows how to test whether a gap is too big to be chance.

A ratio is a probability, not a promise. "3 : 1" means each F2 seed has a 1 in 4 chance of being dwarf. In a family of four children, all four can be affected, or none.

Crosses like this are how plant breeders produce new varieties of crops and garden flowers: they cross two plants with useful traits, select the offspring that combine them, and self those until the new line breeds true.

6Phenylketonuria: a disease due to a recessive allele

Phenylketonuria (PKU) is a recessive genetic condition. The gene involved is autosomal and codes for the enzyme phenylalanine hydroxylase, which converts the amino acid phenylalanine into another amino acid, tyrosine. The mutant allele codes for an enzyme that does not work.

Use A for the normal allele and a for the PKU allele.

  • AA: two working copies. No PKU.
  • Aa: one working copy makes enough enzyme. No PKU. This person is a carrier.
  • aa: no working enzyme. Phenylalanine from the diet builds up in the blood, and at high concentration it damages the developing brain.

Two carriers have a 1 in 4 chance, each time, of a child with PKU:

parents: Aa × Aa
gametes: A or a from each parent, each with probability 1/2
children: AA : Aa : aa = 1 : 2 : 1
chance of PKU (aa) = 1/2 × 1/2 = 1/4 = 25%the same for every pregnancy

PKU is also the clearest example of genotype and environment acting together. Newborn babies in many countries have a heel-prick blood test for it. A child with genotype aa who is fed a diet very low in phenylalanine grows up with normal brain development. The genotype is the same; the environment, the diet, changes the phenotype.

7Gene pools, SNPs, multiple alleles, and dominance that is not complete

A single-nucleotide polymorphism (SNP) is a position in the genome where the base varies between individuals of a species: some people have A, others G, at the same place. A SNP inside a gene gives two different alleles of that gene. A gene may contain many SNPs, and each combination is a different allele, so a species can hold many alleles of one gene in its gene pool, all the alleles of all the genes in a population. Keep one limit in mind: however many alleles exist in the gene pool, one diploid individual inherits only two of them, one from each parent, and they may be the same.

ABO blood groups: multiple alleles. The ABO gene has three common alleles. The guide's symbols are Iᴬ, Iᴮ and i. Iᴬ and Iᴮ code for enzymes that add different sugars to a glycoprotein on the surface of red blood cells, making the A antigen or the B antigen. The i allele codes for an enzyme that does not work, so neither antigen is made.

GenotypeBlood groupWhy
IᴬIᴬ or IᴬiAIᴬ is dominant over i
IᴮIᴮ or IᴮiBIᴮ is dominant over i
IᴬIᴮABIᴬ and Iᴮ are codominant: both antigens are made
iiONeither antigen

Six genotypes, four phenotypes.

Codominance and incomplete dominance both describe a heterozygote that does not look like either homozygote, but in different ways. Figure 4 puts them side by side.

Figure 4 · Two ways a heterozygote can differ from both parents Figure 4 · Two ways a heterozygote can differ from both parents (a) Codominance · ABO blood groups (b) Incomplete dominance · four o'clock flower IᴬIᴬ group A IᴬIᴮ group AB IᴮIᴮ group B A antigen B antigen The heterozygote makes both antigens: a dual phenotype. CᴿCᴿ red CᴿCᵂ pink CᵂCᵂ white F1 pink × F1 pink → F2: 1 red : 2 pink : 1 white The heterozygote makes about half as much pigment: an intermediate phenotype. Codominance: both alleles show in full. Incomplete dominance: the heterozygote is in between.
Figure 4 · Two ways a heterozygote can differ from both parents
  • In codominance, the heterozygote has a dual phenotype: both alleles are fully expressed and both effects show. An IᴬIᴮ person has A antigens and B antigens on every red blood cell.
  • In incomplete dominance, the heterozygote has an intermediate phenotype. In the four o'clock flower (Mirabilis jalapa, also called marvel of Peru), a plant with two red-flower alleles (CᴿCᴿ) has red flowers, CᵂCᵂ has white flowers, and CᴿCᵂ has pink flowers: one working allele makes only about half as much red pigment. Cross two pink plants and the F2 is 1 red : 2 pink : 1 white. Here, unlike Figure 3, the phenotype ratio equals the genotype ratio, because every genotype looks different.

Notice the notation: both alleles take the same base letter with different superscripts, because neither is dominant.

8Sex determination and sex-linked genes

Humans have 22 pairs of autosomes and one pair of sex chromosomes. Females are XX and males XY. Every egg carries one X. Half the sperm carry an X and half a Y, so the sperm decides: an X sperm gives an XX zygote, a Y sperm an XY zygote, in equal numbers on average. A gene on the Y chromosome (called SRY) switches on the development of testes in the embryo, and the hormones the testes release then produce male-typical physical characteristics. Without it, female-typical characteristics develop.

The X and Y are very different in size. The X carries several hundred genes; the Y carries only a few dozen, mostly concerned with male development and sperm production. So most genes on the X have no partner on the Y. A male has only one copy of each of those genes, and whatever allele it is, it shows. That is sex linkage: a pattern of inheritance that differs between the sexes because the gene is on a sex chromosome.

Haemophilia is the guide's example. Blood clotting needs a cascade of proteins, and one of them, factor VIII, is coded by a gene on the X chromosome. A recessive allele codes for a faulty factor VIII, and blood does not clot properly, so small injuries bleed for a long time. It is treated by injections of the missing factor.

Write the alleles as superscripts on the X, as the guide requires: Xᴴ is the normal allele, Xʰ the haemophilia allele. The Y is written as plain Y, with no allele, because the gene is not on it.

GenotypePhenotype
XᴴXᴴfemale, normal
XᴴXʰfemale, normal, carrier
XʰXʰfemale, haemophilia (very rare)
XᴴYmale, normal
XʰYmale, haemophilia

Figure 5 crosses a carrier woman with a man who does not have haemophilia.

Figure 5 · A carrier mother and an unaffected father Figure 5 · A carrier mother and an unaffected father Mother XᴴXʰ (carrier) × Father XᴴY (unaffected) father's sperm mother's eggs Xᴴ Y Xᴴ Xʰ XᴴXᴴ unaffected daughter XᴴY unaffected son XᴴXʰ carrier daughter XʰY son with haemophilia Xᴴ = normal allele · Xʰ = haemophilia allele · Y carries no copy Each son: 1/2 chance of haemophilia. Each daughter: 1/2 chance of being a carrier. Half the sons are expected to have haemophilia; half the daughters are expected to be carriers.
Figure 5 · A carrier mother and an unaffected father

Each son has a 1 in 2 chance of haemophilia. Each daughter has a 1 in 2 chance of being a carrier and no chance of being affected, because she receives her father's Xᴴ. Three rules follow, and they are the answers to most sex-linkage questions.

  1. A male with haemophilia got his Xʰ from his mother: his father gave him a Y.
  2. An affected father passes his Xʰ to every daughter, who are all carriers at least, and to no son.
  3. Haemophilia is far commoner in males because one recessive allele is enough in a male, while a female needs two, one from each parent.

9Pedigree charts

A pedigree chart is a family tree that records who shows a trait. Squares are males and circles are females. A horizontal line between a square and a circle is a mating; a vertical line down from it leads to the children, shown in birth order from left to right. A filled symbol means the person is affected. Figure 6 is an invented family with a rare disorder.

Figure 6 · An invented pedigree, and what it lets you deduce Figure 6 · An invented pedigree, and what it lets you deduce I II III I-1 I-2 II-1 II-2 II-3 II-4 III-1 III-2 III-3 Key unaffected male unaffected female affected What you can deduce III-2 is affected, parents are not: recessive. Her father II-3 is unaffected: not X-linked, so autosomal. III-2 is dd; II-3 and II-4 are Dd; III-1 and III-3 are DD or Dd. Unaffected parents II-3 and II-4 have an affected daughter: the allele is recessive and autosomal.
Figure 6 · An invented pedigree, and what it lets you deduce

Work through it the way a Paper 2 question wants.

Is the allele dominant or recessive? Look for two unaffected parents with an affected child. Individuals II-3 and II-4 are unaffected, yet their daughter III-2 is affected. She must have inherited the allele from them, so they carry it without showing it. The allele is recessive.

Is it autosomal or sex-linked? For an X-linked recessive, an affected daughter needs an Xʰ from her father, so her father would be affected. III-2's father, II-3, is unaffected. So the allele is not X-linked: it is autosomal recessive.

Genotypes. Call the alleles D and d. III-2 is dd. Both her parents are therefore Dd. The chart agrees: II-3's father, I-1, is affected (dd), so II-3 must have received a d from him. Her unaffected brother III-1 is DD or Dd, and from the chart alone you cannot say which. Write "DD or Dd" and give the reason; a single guessed genotype loses the mark.

Nature of science: inductive and deductive reasoning. Inductive reasoning moves from particular observations to a general conclusion: from several families in which unaffected parents have affected children, a geneticist concludes that the condition is recessive. The conclusion goes beyond the evidence, so it could be overturned by a new case. Deductive reasoning moves from a general rule to a particular conclusion that must follow: if the allele is autosomal recessive, then III-2 is dd, and therefore each parent is Dd. On a pedigree you use both, in that order: induce the pattern from part of the chart, then deduce individual genotypes from the pattern.

Why many societies prohibit marriage between close relatives. Most harmful alleles are recessive and rare, so an unrelated couple seldom both carry the same one. Close relatives inherited many of their alleles from the same recent ancestors. First cousins, for example, share on average one eighth of their genes by that route. If one carries a rare recessive allele, the other is much more likely to carry it too, and their children have a much higher chance of being homozygous recessive and affected. The rule long predates genetics, but genetics explains why it reduces harm.

10Continuous variation, and how to display it

Blood group is a discrete variable: each person falls into one of four categories, and there is nothing in between. Height, mass and skin colour are continuous variables: they can take any value within a range, and a large sample produces a smooth, usually bell-shaped distribution.

Two things produce continuous variation.

Polygenic inheritance. The trait is controlled by several genes, each with a small additive effect. Imagine skin colour controlled by three genes, each with an allele that adds pigment and one that does not. A person can carry anything from zero to six pigment-adding alleles, giving seven classes. In a cross between two people who are heterozygous for all three genes, the middle classes are far more likely than the extremes, because there are many more ways to get three pigment-adding alleles than to get six. Figure 7 shows the prediction.

Figure 7 · Three genes, seven classes (a simplified model) Figure 7 · Three genes, seven classes (a simplified model) Offspring (out of 64) Number of pigment-adding alleles 1 0 6 1 15 2 20 3 15 4 6 5 1 6 AaBbCc × AaBbCc 0 = lightest, 6 = darkest with more genes and environment added Many genes with small additive effects, plus environment, give a smooth continuous curve.
Figure 7 · Three genes, seven classes (a simplified model)

Real human skin colour involves more genes than this, and then environment smooths the steps away: sun exposure increases melanin production (section 4), so people with the same genotype differ in shade. More genes plus environment gives a continuous curve.

Summarising a continuous variable. The mean is the sum of the values divided by their number. The median is the middle value when they are in order. The mode is the most frequent value. Take these twenty invented heights of sixteen-year-olds, in centimetres, already in order:

152, 158, 161, 163, 164, 166, 167, 168, 170, 170, 170, 172, 173, 174, 175, 177, 178, 180, 183, 199

mean = 3420 ÷ 20 = 171 cm
median = mean of 10th and 11th values = (170 + 170) ÷ 2 = 170 cm
mode = 170 cmit appears three times

The mean sits above the median because of the one very tall student. That is a clue that the data contain an outlier.

Box-and-whisker plots. A box-and-whisker plot shows six things: the minimum, the first quartile (Q1), the median, the third quartile (Q3), the maximum, and any outliers. The quartiles are the medians of the lower and upper halves of the data. The interquartile range (IQR) is Q3 − Q1, the spread of the middle half. A value is an outlier if it lies more than 1.5 × IQR above Q3 or below Q1.

lower half: 152 … 170 → Q1 = (164 + 166) ÷ 2 = 165 cm
upper half: 170 … 199 → Q3 = (175 + 177) ÷ 2 = 176 cm
IQR = 176 − 165 = 11 cm
1.5 × IQR = 16.5 cm
upper limit = 176 + 16.5 = 192.5 cm199 > 192.5, so 199 is an outlier
lower limit = 165 − 16.5 = 148.5 cm152 > 148.5, so no low outlier

Figure 8 draws the result. The whisker stops at the largest value that is not an outlier (183 cm), and the outlier is plotted on its own as a point.

Figure 8 · A box-and-whisker plot of twenty student heights Figure 8 · A box-and-whisker plot of twenty student heights 135 145 155 165 175 185 195 205 Height (cm) lower limit Q1 − 1.5 × IQR = 148.5 upper limit Q3 + 1.5 × IQR = 192.5 minimum 152 Q1 165 median 170 Q3 176 maximum 183 outlier 199 IQR = 176 − 165 = 11 cm The whiskers stop at 152 and 183 cm; 199 cm lies more than 1.5 × IQR above Q3, so it is an outlier.
Figure 8 · A box-and-whisker plot of twenty student heights

11HLSegregation and independent assortment in meiosis

SL students can skip to section 16.

Everything so far followed one gene. HL follows two, and the key is what meiosis does with two pairs of chromosomes.

Segregation. In anaphase I the two homologous chromosomes of each pair move to opposite poles. So the two alleles of a gene separate, and each gamete receives one of them. This is why an Aa parent makes A gametes and a gametes in equal numbers.

Independent assortment. In metaphase I each pair of homologous chromosomes lines up on the equator, and which way round it lies is random. It is also independent of how every other pair lies. Take a plant that is AaBb, with gene A on one chromosome pair and gene B on a different pair. Figure 9 shows the two ways the pairs can line up.

Figure 9 · Two chromosome pairs, two orientations, four kinds of gamete (HL) Figure 9 · Two chromosome pairs, two orientations, four kinds of gamete (HL) Orientation 1 A a B b pole pole metaphase I Orientation 2 A a b B pole pole metaphase I AB ab Ab aB gametes after meiosis: AB : Ab : aB : ab = 1 : 1 : 1 : 1 Each orientation is equally likely, so an AaBb cell line makes AB, Ab, aB and ab in equal numbers.
Figure 9 · Two chromosome pairs, two orientations, four kinds of gamete (HL)

One orientation puts A with B and a with b; the other puts A with b and a with B. Each happens in half the cells, so the plant makes AB, Ab, aB and ab gametes in equal proportions, 1 : 1 : 1 : 1. This holds only for unlinked genes, genes on different chromosomes. Section 14 is what happens when they are on the same one.

12HLDihybrid crosses and where 9:3:3:1 comes from

A dihybrid cross follows two genes at once. Take peas where Y (yellow seeds) is dominant over y (green) and R (round seeds) is dominant over r (wrinkled), on different chromosomes.

Cross pure-breeding yellow round (YYRR) with green wrinkled (yyrr). Every F1 plant is YyRr, yellow and round. Self the F1. Each parent makes four gamete types in equal numbers, so the Punnett grid is 4 × 4, with 16 equally likely boxes. Figure 10 fills it in.

Figure 10 · The F2 of a dihybrid cross (HL) Figure 10 · The F2 of a dihybrid cross (HL) YR Yr yR yr YR Yr yR yr YYRR YYRr YyRR YyRr YYRr YYrr YyRr Yyrr YyRR YyRr yyRR yyRr YyRr Yyrr yyRr yyrr gametes of one F1 parent gametes of the other 9 yellow round 3 yellow wrinkled 3 green round 1 green wrinkled Y yellow is dominant to y green · R round is dominant to r wrinkled YyRr × YyRr: sixteen equally likely boxes, which sort into 9 : 3 : 3 : 1.
Figure 10 · The F2 of a dihybrid cross (HL)

Count the phenotypes:

PhenotypeGenotypes that give itBoxes
yellow roundat least one Y and at least one R9
yellow wrinkledat least one Y, and rr3
green roundyy, and at least one R3
green wrinkledyyrr1

That is the 9 : 3 : 3 : 1 ratio. A quicker route is to multiply the two monohybrid ratios: yellow : green is 3 : 1, round : wrinkled is 3 : 1, and (3 : 1) × (3 : 1) = 9 : 3 : 3 : 1. The multiplication only works because the two genes assort independently.

The second ratio to know comes from a test cross: the heterozygote YyRr crossed with the double homozygous recessive yyrr. The recessive parent can only make yr gametes, so each offspring's phenotype shows exactly which gamete it received from the heterozygote. Four gamete types in equal numbers give four phenotypes in the ratio 1 : 1 : 1 : 1.

Nature of science: laws with exceptions. These ratios rest on what has been called Mendel's second law, the law of independent assortment. It only holds if the genes are on different chromosomes, or so far apart on one chromosome that crossing over separates them in half of all meioses. Biology has few laws without exceptions; a "law" is a pattern that holds under stated conditions, and the conditions matter.

13HLWhere genes are: loci and their products

The locus of a gene is its position on a particular chromosome, written as chromosome number, arm (p short, q long) and band. Online databases, such as the gene databases run by the NCBI and Ensembl, let you look up any human gene and see its locus and the polypeptide it codes for. The guide asks you to explore them and to find pairs of genes on different chromosomes and pairs close together on the same chromosome. A few you can check yourself:

GeneLocusPolypeptide product
PAHchromosome 12phenylalanine hydroxylase (section 6)
ABOchromosome 9the enzyme that adds the A or B sugar (section 7)
F8X chromosome, long armclotting factor VIII (section 8)
HBB and HBDside by side on chromosome 11the β-globin and δ-globin chains of haemoglobin
OPN1LW and OPN1MWside by side on the X, long armthe red-sensitive and green-sensitive opsins of cone cells

PAH and ABO are on different chromosomes, so they assort independently. HBB and HBD sit next to each other and are almost always inherited together. So are the two opsin genes; because they lie on the X, red–green colour blindness is X-linked, like haemophilia.

14HLLinkage and recombinants

Linked genes are genes whose loci are on the same chromosome. They tend to be inherited together, because in anaphase I the whole chromosome goes to one pole, carrying all its alleles with it. So linked genes do not assort independently.

For linked genes the guide wants a special notation: draw the two homologous chromosomes as vertical lines and write the alleles alongside. A plant with A and B on one chromosome and a and b on its homologue is written as AB over ab, the two alleles of each chromosome on the same side of the lines. Figure 11 uses it.

Figure 11 · Linked genes: parental and recombinant gametes (HL) Figure 11 · Linked genes: parental and recombinant gametes (HL) The heterozygote A B a b written AB/ab: A and B on one chromosome crossed with ab/ab (test cross) No crossover between the loci A B a b parental gametes: the majority A crossover between the loci A b a B recombinant gametes: the minority Test cross offspring (invented, 1,000 plants) A and B 440 a and b 430 A and b 68 a and B 62 recombinants 130 ÷ 1000 = 13% Crossing over between the loci makes recombinants, so they are always the smaller classes.
Figure 11 · Linked genes: parental and recombinant gametes (HL)

Without crossing over, this plant would make only two kinds of gamete, AB and ab, the combinations it inherited. These are the parental combinations. But in prophase I, homologous chromosomes can exchange segments by crossing over. If a crossover happens between the two loci, the chromatids involved come out as Ab and aB. These new combinations are recombinants. Because a crossover falls between two particular loci in only some meioses, recombinant gametes are always fewer than parental ones, and the closer the loci, the rarer they are.

Reading a test cross with linked genes. Cross the AB/ab plant with ab/ab. The invented results for 1,000 offspring:

Offspring phenotypeGamete from the heterozygoteNumberType
A and BAB440parental
a and bab430parental
A and bAb68recombinant
a and BaB62recombinant
recombinants = 68 + 62 = 130
recombinant frequency = 130 ÷ 1000 × 100 = 13%

If the genes were unlinked, the four classes would be about 250 each (1 : 1 : 1 : 1), and the recombinants would make up 50%. Here there are two large classes and two small ones: the signature of linkage. You can identify recombinants three ways, and the guide asks for all three: as gametes (Ab and aB, combinations not present on either parental chromosome), as offspring genotypes (Aabb and aaBb, each made by a recombinant gamete), and as offspring phenotypes (the two small classes).

15HLThe chi-squared test

Real crosses never give exact ratios. The chi-squared (χ²) test decides whether the difference between observed and expected numbers is small enough to be chance, or too large for chance to explain.

  1. State the null hypothesis (H₀): there is no significant difference between observed and expected; for a dihybrid cross, the genes assort independently and the ratio is 9 : 3 : 3 : 1. The alternative hypothesis (H₁): there is a significant difference, for example because the genes are linked.
  2. Calculate the expected number in each class from the total and the ratio.
  3. Calculate χ² = Σ (O − E)² ÷ E.
  4. Degrees of freedom = number of classes − 1.
  5. Compare χ² with the critical value at the p = 0.05 significance level, from a table. If χ² is greater than the critical value, the difference would happen by chance less than 5% of the time: reject H₀. If χ² is smaller, the difference is not significant: H₀ is not rejected.

Critical values at p = 0.05: 3.841 (1 degree of freedom), 5.991 (2), 7.815 (3).

Worked example. An F2 from a dihybrid cross gives 191, 55, 52 and 22 plants in the four phenotype classes (invented data), 320 in all.

expected: 320 × 9/16 = 180, 320 × 3/16 = 60, 60, 320 × 1/16 = 20
(191 − 180)2 ÷ 180 = 121 ÷ 180 = 0.672
(55 − 60)2 ÷ 60 = 25 ÷ 60 = 0.417
(52 − 60)2 ÷ 60 = 64 ÷ 60 = 1.067
(22 − 20)2 ÷ 20 = 4 ÷ 20 = 0.200
χ2 = 0.672 + 0.417 + 1.067 + 0.200 = 2.36
degrees of freedom = 4 − 1 = 3; critical value = 7.815
2.36 < 7.815 → not significant: do not reject H0the results fit 9:3:3:1

Say what the test does and does not show. A χ² below the critical value means the data are consistent with independent assortment; it does not prove it. And say what was sampled.

Nature of science: a sample stands for a population. The F2 you counted is a sample of all the F2 offspring the cross could produce, and the test asks how likely your sample would be if H₀ were true of that population. In other experiments the sample is your set of replicates or repeated measurements.

16Linking questions

What are the principles of effective sampling in biological research? A sample must be random and large enough for chance deviations to average out: an F2 of 20 plants can miss a 1-in-16 class altogether, an F2 of 500 will not.

What biological processes involve doubling and halving? Meiosis halves the chromosome number and fertilisation doubles it (Figure 1); DNA replication doubles the DNA before either division.

17Where marks are lost

Using "gene" and "allele" as if they were the same. Everyone has the gene for ABO blood group; people differ in which alleles of it they have. "She has the gene for blood group A" should be "she has the Iᴬ allele".

Saying a dominant allele is commoner or stronger. Dominance describes what shows in a heterozygote, nothing more. A dominant allele can be rare in a population; the allele for Huntington's disease is dominant and uncommon.

Treating a ratio as a guarantee. A 1 in 4 chance of PKU applies to every child separately. Having one affected child does not make the next three unaffected.

Confusing codominance with incomplete dominance. Codominance shows both phenotypes at once (A and B antigens); incomplete dominance shows a blend (pink). Pink flowers are never codominance.

Giving a male two alleles of a sex-linked gene. A male is XᴴY or XʰY, never XᴴXʰY or "Hh". Always write the alleles as superscripts on X.

Saying a son inherits haemophilia from his father. The father gives a son his Y, not his X. An affected son's allele came from his mother.

Calling phenotypic plasticity a mutation or an adaptation passed on to offspring. The genotype does not change; only gene expression does.

HL: confusing linked genes with sex-linked genes. Linkage means two genes on the same chromosome, any chromosome. Sex linkage means one gene on a sex chromosome. And a χ² value below the critical value does not "prove the hypothesis"; it means the difference is not significant.

18Draw it right

  1. Punnett grid: gametes of one parent across the top, of the other down the side, each gamete circled or labelled as a gamete, and one allele per gene in each gamete (A, not Aa).
  2. Show the parental genotypes and phenotypes, the gametes, the offspring genotypes and the offspring phenotypes with the ratio. Each of those is usually a separate mark.
  3. Define your symbols in a key, using the same letter for alleles of one gene (T and t, not T and d). For codominance and incomplete dominance use one base letter with superscripts (Cᴿ, Cᵂ; Iᴬ, Iᴮ).
  4. Sex-linked: superscripts on an uppercase X (Xᴴ, Xʰ) and a bare Y.
  5. Pedigree: squares for males, circles for females, filled for affected; roman numerals for generations and numbers for individuals, so you can name them in your answer.
  6. Box-and-whisker plot: a labelled scale with units; whiskers to the furthest values that are not outliers; outliers as separate points.
  7. HL, linked genes: two vertical lines for the homologous chromosomes with the alleles written alongside, the alleles of each chromosome on the same side.
  8. HL, χ²: show H₀, the expected numbers, each (O − E)² ÷ E term, the total, the degrees of freedom, the critical value at p = 0.05, and a conclusion in words.

19Try it

Marks in brackets. Answers and marker's notes are at the end.

Q1. Distinguish between genotype and phenotype, and give one human trait that is influenced by both genotype and environment. 3 marks

Q2. A woman with blood group A, whose father was blood group O, has children with a man of blood group B who is heterozygous. Using a Punnett grid, determine the possible blood groups of their children and the probability of each. 4 marks

Q3. A woman who is a carrier of haemophilia has children with a man who does not have haemophilia.

(a) State the probability that their first son will have haemophilia. 1 mark

(b) Explain why haemophilia is much more common in males than in females. 3 marks

Q4. A box-and-whisker plot of the mass of 60 newborn lambs (invented data) has Q1 = 4.2 kg, median = 4.8 kg and Q3 = 5.4 kg. The heaviest lamb has a mass of 7.3 kg.

(a) Calculate the interquartile range. 1 mark

(b) Determine whether the 7.3 kg lamb is an outlier. Show your working. 2 marks

Q5 (HL). Two pure-breeding pea lines, one yellow round and one green wrinkled, are crossed. The F1 is self-pollinated. The F2 contains 150 yellow round, 34 yellow wrinkled, 31 green round and 25 green wrinkled plants (invented data).

(a) Calculate the value of χ² for the hypothesis that the ratio is 9 : 3 : 3 : 1. 3 marks

(b) The critical value at p = 0.05 with 3 degrees of freedom is 7.815. Evaluate the hypothesis, and suggest a reason for the result. 2 marks

Q6 (HL). Explain why linked genes do not show a 1 : 1 : 1 : 1 ratio in a test cross, and how recombinants can still appear. 4 marks

20In one breath

Meiosis puts one allele of each gene into each haploid gamete and fertilisation brings two together, so a diploid zygote has two copies of each autosomal gene. Genotype is the pair of alleles; phenotype is what shows, set by genotype and environment. AA and Aa look alike because one working allele makes enough protein; PKU shows only in aa, with no phenylalanine hydroxylase. Plasticity changes gene expression, not genes. A plant cross needs anthers removed, pollen transferred and the flower bagged; a monohybrid F2 is 3 : 1. A gene pool can hold many alleles, often differing by SNPs, but an individual holds two: ABO has Iᴬ, Iᴮ and i, with Iᴬ and Iᴮ codominant, while four o'clock flowers show incomplete dominance. Sperm decide sex; X-linked recessives such as haemophilia show in XʰY males, who inherit them from carrier mothers. Pedigrees are read inductively for the pattern and deductively for genotypes. Polygenic traits plus environment give continuous variation, shown in box plots with outliers beyond 1.5 × IQR. HL: independent assortment at metaphase I gives 9 : 3 : 3 : 1 and 1 : 1 : 1 : 1; linked genes stay together unless crossing over makes the minority recombinants; χ² above the critical value at p = 0.05 rejects the null hypothesis.


Answers

Q1. The genotype is the combination of alleles an organism has inherited for a gene; the phenotype is the observable characteristic, which results from the genotype and the environment. Height is influenced by both: alleles set the potential, and nutrition during childhood affects how much of it is reached. 1 for genotype as the alleles, 1 for phenotype as the observable trait, 1 for a valid example with both influences named. Skin colour, body mass or PKU symptoms also score. Blood group scores 0 for the example, because it is set by genotype alone.

Q2. The woman's father was ii, so he gave her i: she is Iᴬi. The man is Iᴮi. Gametes: Iᴬ or i from the mother; Iᴮ or i from the father. The grid gives IᴬIᴮ, Iᴬi, Iᴮi and ii, one of each. So the children can be AB, A, B or O, each with a probability of 1/4 (25%). 1 for the mother's genotype Iᴬi with the reason, 1 for correct gametes, 1 for the four offspring genotypes in a grid, 1 for the four blood groups each 1/4. Using letters other than Iᴬ, Iᴮ and i loses the first mark only if the key is missing.

Q3. (a) 1/2 (0.5 or 50%). [1] (b) The gene for factor VIII is on the X chromosome, and the Y chromosome carries no copy of it; a male has only one X, so a single recessive allele (XʰY) produces haemophilia; a female must inherit the recessive allele from both parents (XʰXʰ) to be affected, and with one normal allele she is a carrier who clots normally; since the allele is rare, inheriting two copies is far less likely than inheriting one. 1 for the gene on the X and not the Y, 1 for one allele being enough in males, 1 for females needing two, or heterozygous females being carriers. "Males are weaker" or "it is on the Y chromosome" scores 0.

Q4. (a) IQR = 5.4 − 4.2 = 1.2 kg. [1] (b) 1.5 × IQR = 1.5 × 1.2 = 1.8 kg; upper limit = Q3 + 1.8 = 5.4 + 1.8 = 7.2 kg; 7.3 kg is greater than 7.2 kg, so the lamb is an outlier. 1 for the limit of 7.2 kg with working, 1 for the conclusion stated from the comparison. A conclusion with no calculation scores 0.

Q5 (HL). (a) Total = 240. Expected: 135, 45, 45, 15. Terms: (150 − 135)² ÷ 135 = 1.667; (34 − 45)² ÷ 45 = 2.689; (31 − 45)² ÷ 45 = 4.356; (25 − 15)² ÷ 15 = 6.667. χ² = 15.4. M1 for all four expected values, M1 for the (O − E)² ÷ E terms, A1 for 15.4 (accept 15.38). (b) 15.4 is greater than 7.815, so the difference between observed and expected is significant at p = 0.05; the null hypothesis of a 9 : 3 : 3 : 1 ratio is rejected. There are more yellow round and green wrinkled plants (the parental combinations) than expected, and fewer recombinant classes, which suggests the two genes are linked on the same chromosome. 1 for the comparison and rejection of H₀, 1 for linkage with reference to the excess of parental types. "The hypothesis is proved wrong" loses the first mark; say rejected at the 5% level.

Q6 (HL). Linked genes are on the same chromosome, so their alleles move together to the same pole in anaphase I and do not assort independently; a heterozygote therefore produces mostly gametes with the parental combinations (for example AB and ab), and the parental phenotypes form the two largest classes in a test cross; crossing over between the two loci during prophase I exchanges segments between non-sister chromatids, producing chromatids with new combinations (Ab and aB); these recombinant gametes give the two small classes of offspring, which are always fewer than the parental classes because a crossover falls between the loci in only some meioses. 1 for same chromosome and moving together, 1 for parental gametes predominating, 1 for crossing over in prophase I between non-sister chromatids, 1 for recombinants being the minority with a reason. An answer about sex linkage scores 0.


Educerie · written from the published IB Diploma Programme Biology guide, first assessment 2025, section D3.2 Inheritance. Original text, examples and questions. Diagrams drawn by Educerie. Last reviewed 25 September 2026.

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