IB BIOLOGY · THEME D · CONTINUITY AND CHANGE

How traits are handed down

Half of your genes came from each parent, and which half is a matter of chance. Follow the alleles from the gamete to the grid to the family tree, learn why some traits hide for a generation, and at higher level test whether two genes really travel apart.

IB Biology D3.2SL and HL5 hours SL, 8 HL15 chaptersdry lab

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P p P p PP Pp pp 3 : 1 Pp P p I II A pea flower Two gametes meet A Punnett grid: 3 to 1 The same pattern in a family
Inheritance in one loop: a pea flower's gametes meet, a Punnett grid fills in the offspring of two Pp parents, and a family tree shows the same pattern in people. Build the crosses yourself in the guide.

01Learn

Fifteen chapters, the last three for higher level. Read one, play its figure or tap its parts, and make each prediction before you open the answer.

  1. 1Halve, then fuse
  2. 2How to make a cross
  3. 3Genes, alleles and the traits you see
  4. 4Dominant and recessive alleles
  5. 5Same genes, different outcomes
  6. 6One faulty enzyme: PKU
  7. 7Many alleles, not just two
  8. 8Neither wins, or both do
  9. 9X, Y and sex
  10. 10Reading a family tree
  11. 11A spread, not a few types
  12. 12Showing a spread with a box plot
  13. 13Two genes at once HL
  14. 14Genes that travel together HL
  15. 15Is it chance? The chi-squared test HL

D3.2.1 Halving and doubling

Halve, then fuse

Your body cells hold two copies of every chromosome, one set from each parent. A cell like that is diploid. A human body cell has 46 chromosomes, written 2n = 46. Most of them are autosomes, so a diploid cell has two copies of each autosomal gene: one on each chromosome of a pair.

If a child simply received all the chromosomes of both parents, the number would double every generation. Life avoids this by halving first. Special cells divide by meiosis to make gametes, and each gamete gets one chromosome from every pair. A cell with a single set is haploid. In humans, n = 23.

Then fertilisation joins two gametes. Their nuclei fuse to make a zygote, which is diploid again, with one set from each parent. Halve, then double: that is the loop that carries genes from one generation to the next.

This pattern is common to all eukaryotes with a sexual life cycle. You, a pea plant, a mushroom and a fruit fly all make haploid gametes and fuse them to form a diploid zygote. The details of the life cycle differ from group to group, but the loop is the same.

Linking question What biological processes involve doubling and halving? Here it is the chromosome number: meiosis halves it and fertilisation doubles it. The same idea returns in D2.1 (cell and nuclear division), where cells divide, and in D1.1 (DNA replication), where the DNA doubles before they do.
Watch out A gamete is not "half a cell". It is a whole cell with half the number of chromosomes. And a parent does not pass both of its copies of a gene to a gamete: it passes one.

A human body cell has 46 chromosomes. How many chromosomes are in a sperm, and how many in the zygote it helps to form?

Halve, then fuseMother2n = 46Father2n = 46meiosismeiosisEggn = 23Spermn = 23fusionZygote, 2n = 46same gene, one copyfrom each parent
  1. Two parents, each with diploid body cells (2n = 46). Three of the 23 pairs are drawn.
  2. Meiosis makes gametes with one chromosome from each pair: an egg and a sperm, each haploid (n = 23).
  3. Fusion of the two gametes makes a zygote. It is diploid again, with one set from each parent.
  4. In the zygote, a pair of chromosomes carries the same gene twice, one copy from each parent.
Follow the chromosomes through meiosis to four haploid gametes, then watch two gametes fuse. Try one, two or three pairs of chromosomes.

D3.2.2 Genetic crosses in flowering plants

How to make a cross

To study inheritance you need to know the parents, so you choose them and then count the offspring. That is a genetic cross. Flowering plants are good for it: you can control which plants breed, and the seeds are ready within weeks.

The male gametes are in pollen, which forms in the anthers. The female gametes are in the ovules, inside the ovary at the base of the flower. A pollen grain has to reach the stigma before it can fertilise an ovule, so a cross needs pollination.

Pea plants add a twist. One flower makes both pollen and ovules, so it can pollinate itself: self-fertilisation. To make a cross on purpose you remove the anthers from the flower that will receive pollen, before it can pollinate itself, and carry pollen from the other plant to its stigma.

Crosses are described in generations. The two parents you choose are the P generation. Parents that are true-breeding give the F1 generation (F is for filial, meaning of a son or daughter). Let the F1 plants pollinate each other, or themselves, and you get the F2 generation. A Punnett grid predicts what the F2 should look like. The letters in the grid are versions of a gene, which chapter 3 explains.

Remember it The generations of a cross

  • ParentsP generationthe plants you choose to cross
  • FirstF1 generationtheir offspring
  • FollowsF2 generationthe offspring of two F1 plants
Example Crosses are not only for laboratories. Breeders cross plants to join the useful traits of two varieties, such as the disease resistance of one with the yield of another, and to make new varieties of ornamental flowers.
Watch out A Punnett grid gives chances, not promises. It predicts the proportions you should expect in a large number of offspring, not exactly which seeds you will get.

A student wants to cross a purple-flowered pea plant with a white-flowered one, but forgets to remove the anthers from the white plant. What is likely to happen?

Making a cross in a pea plantStigmaAnther: pollen grainshold the male gametesOvary: ovuleshold the female gametesA pea flower pollinatesitself: pollen lands on itsown stigmaPollen donorReceiving floweranthers removedpollen carried acrossP generationYY×yyyellow seedsgreen seedscrossF1 generationYyall yellow seedsself-pollinateF2 generationpollen from the F1 plantYyYyYYYyYyyy3 yellow : 1 green1 YY : 2 Yy : 1 yy
  1. A pea flower makes both kinds of gamete. Pollen forms in the anthers and holds the male gametes. The ovules in the ovary hold the female gametes.
  2. Left alone, a pea flower pollinates itself: its pollen reaches its own stigma.
  3. For a cross, the anthers are removed from the receiving flower, and pollen is carried from the donor to its stigma.
  4. Cross two true-breeding parents (the P generation). Every seed of the F1 generation has one allele from each parent.
  5. Let the F1 plants self-pollinate to make the F2 generation. The Punnett grid shows every way their gametes can combine: 3 yellow seeds for every 1 green.

D3.2.3 D3.2.4 Genotype and phenotype

Genes, alleles and the traits you see

A gene is a length of DNA at one place on a chromosome. Different versions of the same gene are alleles. The seed colour gene in peas is a single place on one chromosome, and its alleles are one for yellow seeds (Y) and one for green (y). Keep the two words apart: a gene is the place, an allele is the version.

You are diploid, so you have two alleles of every autosomal gene, one on each chromosome of a pair. The pair of alleles is your genotype for that gene. If the two alleles are the same, you are homozygous (YY or yy). If they differ, you are heterozygous (Yy).

Your phenotype is what can be observed: the colour of a seed, your blood group, your height, even your behaviour. It comes from the genotype and the environment. Some traits depend on the genotype only, some on the environment only, and many on both.

Example In humans: genotype only, ABO blood group (diet and exercise cannot change it). Environment only, a scar from a cut, or a language you have learned. Both, adult height (genes set a range, and nutrition and health decide where you land in it) and skin colour (which also depends on how much sun you get).
Watch out There is no "gene for yellow seeds". There is a gene for seed colour, and the Y allele of it makes the seed yellow. A gene has alleles; a trait is made by a genotype.

Two plants have the same genotype for every gene, but one grows much taller because it was given more water and fertiliser. What does this show?

A pea plant has the genotype Yy. Which statement is correct?

Genes, alleles and genotypesYyallele Yallele yThe gene is the place; the alleles are its versions.YYYYhomozygousboth alleles the sameYyYyheterozygoustwo different allelesyyyyhomozygousboth alleles the sameGenotype: the alleles an organism has for a gene.ABO blood groupa scar from a cutadult heightgenotypeenvironmentBar lengths show the idea, not measurements.
  1. A pair of homologous chromosomes. The gene is one place on the chromosome. Each chromosome carries one allele of it, here Y and y.
  2. The two alleles make the genotype. Two the same is homozygous. Two different is heterozygous.
  3. The phenotype comes from the genotype, the environment, or both: blood group, a scar and height.

D3.2.5 Dominant and recessive alleles

Dominant and recessive alleles

In Mendel's peas, a cross between a yellow-seeded plant and a green-seeded one gave an F1 in which every seed was yellow. The green trait had not vanished: about a quarter of the F2 seeds were green. Something was hiding it.

A dominant allele shows its effect when it is present, in one copy or in two. A recessive allele shows only when there is no dominant allele alongside it. So YY and Yy have the same phenotype, yellow, and only yy is green. By convention, a capital letter is the dominant allele and a small letter the recessive one.

Why is one copy enough? An allele works through the protein its gene makes. For many genes, a recessive allele is a version that does not make a working protein. A heterozygote has one working copy, and for many genes one working copy makes enough protein to give the trait. A homozygous recessive has no working copy, and the trait changes. There is no single mechanism for all genes, but this is the usual reason.

Watch out Dominant does not mean common, strong or better. Some dominant alleles are rare, and some cause disease. Dominant only describes what shows in a heterozygote. Recessive alleles do not fade away either: in a heterozygote they are still there, and they can reappear in a later generation.

Two yellow-seeded pea plants are crossed, and some of their offspring have green seeds. What must be true of both parents?

Two Yy plants are crossed. What fraction of their offspring have green seeds?

Three genotypes of one geneYYYYYyYyyyyyplentyenoughtoo littleworking protein made; dashed line = enough to show the traityellowyellowgreen
  1. Three genotypes of the seed colour gene: two homozygous and one heterozygous.
  2. Each gene makes a protein. A Y allele makes a working one, and one working copy makes enough to give the trait. With no working copy, the trait changes.
  3. The phenotypes: YY and Yy are both yellow, so the dominant allele shows in a heterozygote. Only yy is green.
P generationfemale parentYYyellow seedsmale parentyygreen seedsxF1 offspringmale gametes, in pollenfemale gametesyyYYyellowgreenP generationfemale parentYYyellow seedsmale parentyygreen seedsxF1 offspringmale gametes, in pollenfemale gametesyyYYYyYyYyYyyellowgreen
  1. Each parent makes gametes by meiosis, one allele of each gene in each. The female parent's go down the side of the Punnett grid, the male parent's along the top.
  2. Each cell joins one gamete from each parent: a possible offspring, every cell as likely as the next. The F1: genotypes all Yy; phenotypes all yellow.
Pick the two parents' genotypes, predict the offspring, then breed 30 and see how the sample scatters around the expectation.

D3.2.6 Phenotypic plasticity

Same genes, different outcomes

Genotype is not destiny. One individual, with one set of alleles, can develop different traits in different conditions, and the change often suits the conditions. That is phenotypic plasticity.

Take two cuttings from one plant. They have identical alleles. Grow one in full sun and the other in shade. A shaded plant tends to grow taller, with longer gaps between its leaves and broader leaves, reaching for light. The sunlit plant stays shorter and sturdier. Neither plant changed its genotype.

The difference is in gene expression: which genes are switched on, where, and how strongly. Light conditions change the pattern, so the same DNA builds a different body. The base sequence has not changed, which means plasticity is not a mutation.

Because the alleles are the same, the change can be reversible during the life of the individual: move the shaded plant into the sun, and its new growth comes in short and sturdy, like the sun plant's. Growth that has already happened stays as it was.

Example Your skin tans in strong sun and fades again afterwards, with no change in your DNA. An Arctic hare's coat is white in winter and darker in summer, so the same animal is camouflaged in each season.
Watch out A plastic change is not passed to the next generation through the alleles: the shaded plant's seeds do not inherit its stretching. The individual does not choose to change either. The capacity to respond is itself in the genotype, and the environment triggers it.

Two clones grow differently, one in sun and one in shade. What explains the difference?

Which of these is an example of phenotypic plasticity?

Same alleles, different plantsparent plantcutting Acutting Bsame allelesin both cuttingsfull sunshadegrown in sungrown in shadesunsungrown in sunmoved into the sunnew growthis like thesun plant
  1. One parent plant, with its alleles.
  2. Two cuttings from it carry exactly the same alleles.
  3. One cutting grows in sun and one in shade. Same alleles, but different gene expression makes a short, sturdy plant and a tall one with broad leaves.
  4. Move the shaded plant into the sun, and its new growth comes in like the sun plant’s. The change is reversible, and the alleles never changed.

D3.2.7 Phenylketonuria: a recessive disease

One faulty enzyme: PKU

Phenylketonuria (PKU) is a human disease caused by a recessive allele of an autosomal gene. The gene codes for an enzyme, phenylalanine hydroxylase, which converts the amino acid phenylalanine into another amino acid, tyrosine.

A person with at least one working allele makes enough enzyme. A person who is homozygous for the faulty allele makes none that works. Phenylalanine from food cannot be converted, so it builds up, and in a baby the high levels can damage the developing brain.

Because the allele is recessive, a person with one working and one faulty allele has no symptoms. That person is a carrier. Two carrier parents have a 1 in 4 chance of a child with PKU each time they have a baby, and most carriers do not know that they are one.

PKU shows genotype and environment working together. The genotype cannot be changed, but the environment can. Babies are screened soon after birth, and a child found to have PKU is given a diet low in phenylalanine, which protects the developing brain. The phenotype depends on the diet as well as on the alleles.

Example Two carriers, Nn × Nn. Each makes N and n gametes in equal numbers. The offspring are NN (1 in 4), Nn (2 in 4) and nn (1 in 4). Only nn has PKU. Of the children who do not have PKU, two in three are carriers.
Watch out The chance is 1 in 4 for each child, every time. One child with PKU does not make the next three safe.

Two parents do not have PKU, but their first child does. What is the chance that their second child has PKU?

A baby with PKU is put on a diet low in phenylalanine. Has its genotype changed?

One step in the pathway failsPhenylalanineOH addedTyrosineenzymephenylalanine hydroxylasefaulty enzymelittle tyrosine madePhenylalanine builds upLess in the diet, less builds upTwo carrier parentsNn×NnNnNnNNNnNnnnNN: not a carrierNn: carrier, no PKUnn: has PKUEach child has a 1 in 4 chance of PKU.
  1. In a person with a working allele, the enzyme converts phenylalanine to tyrosine, which has one extra OH group.
  2. In a person with two faulty alleles the enzyme does not work. Little tyrosine is made and phenylalanine builds up.
  3. A diet low in phenylalanine gives the body less to build up. The genotype is the same, and the phenotype changes.
  4. Two carrier parents (Nn): each child has a 1 in 4 chance of nn, so of PKU.

D3.2.8 D3.2.9 SNPs, multiple alleles and ABO blood groups

Many alleles, not just two

So far every gene had two alleles. In a population, a gene can have many. New alleles arise by mutation, and alleles of one gene usually differ by only one or a few bases. A place in the genome where people differ by a single base, and where each version is common enough to count (often taken as more than 1 in 100), is a single-nucleotide polymorphism, or SNP ("snip").

All the alleles of all the genes in a population make up its gene pool. A gene can have any number of alleles in the gene pool, but you are diploid, so you inherit only two of them: one from each parent. Different people carry different pairs.

The ABO blood group is the textbook example. One gene has three alleles: IA, IB and i. IA and IB each lead to a different marker (an antigen) on the surface of red blood cells, and i leads to neither. Both IA and IB are dominant to i. Three alleles make six genotypes and four blood groups; what happens when IA meets IB is the subject of chapter 8.

Remember it The four blood groups

  • AppleGroup AIAIA or IAi
  • BananaGroup BIBIB or IBi
  • BothGroup ABIAIB, with both markers
  • OrangeGroup Oii, with neither marker
Example A parent of group A (IAi) and a parent of group B (IBi) can have children of any of the four groups: IAIB (AB), IAi (A), IBi (B) or ii (O).
Watch out A person never has three ABO alleles. The gene pool has three; each person has two. And group O is not "no gene": the i allele is a real allele, and a person with ii has two copies of it.

A man with blood group AB (IAIB) and a woman with group O (ii) have children. Which groups are possible?

How many alleles of the ABO gene does one person have?

SNPs, alleles and the gene poolPerson 1ATGGCACTGACTPerson 2ATGGCACTGACTPerson 3ATGGCATTGACTPerson 4ATGGCACTGACTA SNP: one position where people differ by one baseGene pool112314213212134one gene, four alleles in the population13One personinherits just twogenotypeblood groupIAIAgroup AIAigroup AIBIBgroup BIBigroup BIAIBgroup ABiigroup OThree alleles in the pool, so six genotypes and four groups.
  1. DNA from four people at one place. They differ at a single base: a SNP.
  2. Across a whole population a gene can have many alleles. The gene pool holds them all.
  3. Each person inherits just two of them.
  4. ABO: three alleles in the pool give six genotypes and four blood groups.

D3.2.10 Incomplete dominance and codominance

Neither wins, or both do

Dominance is not the only pattern. Sometimes a heterozygote looks like neither homozygote. Two patterns are worth telling apart, and the difference is in the phenotype.

In incomplete dominance the heterozygote has an intermediate phenotype. The standard example is the four o'clock plant, Mirabilis jalapa, also called the marvel of Peru. The alleles are written CR (red) and CW (white). CRCR plants are red, CWCW white and CRCW pink. Cross two pink plants and you get red, pink and white in the ratio 1 : 2 : 1, so here the phenotype ratio matches the genotype ratio.

In codominance both alleles are expressed fully, and the heterozygote has a dual phenotype. The AB blood group is the example: a person with IAIB makes both the A marker and the B marker on every red blood cell. Neither allele hides the other, and nothing is in between: both are there.

  • Complete dominance: the heterozygote looks like one homozygote.
  • Incomplete dominance: the heterozygote is in between.
  • Codominance: the heterozygote shows both at once.
Watch out Incomplete dominance is not blending. If the red and white alleles mixed like paints, pink plants could never give red or white again. They do, in the F2, because each allele stays whole and separates in the gametes.

A red four o’clock plant is crossed with a white one. What colour are the F1 flowers?

What blood group does a person with the genotype IAIB have, and why?

When neither allele wins outrightCRCRredhomozygousCRCWpinkheterozygousCWCWwhitehomozygousFour o’clock flowers (Mirabilis jalapa)pink × pinkCRCWCRCWCRCRCRCWCRCWCWCW1 red : 2 pink : 1 whiteThe phenotype ratio matches the genotype ratio.IAIAgroup Aonly A markersIAIBgroup ABboth A and B markersIBIBgroup Bonly B markersA markerB marker
  1. Incomplete dominance in four o’clocks: red (CRCR), pink (CRCW) and white (CWCW). The heterozygote is intermediate.
  2. Cross two pink plants and the offspring are 1 red : 2 pink : 1 white. The phenotype ratio matches the genotype ratio, and red and white reappear.
  3. Codominance in blood: a red blood cell of group AB carries both markers, so the heterozygote has a dual phenotype.

D3.2.11 D3.2.12 Sex determination and sex linkage

X, Y and sex

Most of your 23 pairs of chromosomes are autosomes, matching pairs. The last pair is different. Human females have two X chromosomes (XX). Males have one X and a smaller Y (XY). Every egg carries an X. A sperm carries an X or a Y, so the sperm decides: an X sperm makes an XX zygote, which develops female-typical characteristics, and a Y sperm makes an XY zygote, which develops male-typical ones. Half the sperm carry each, so about half of children are girls. The Y chromosome carries genes that start the development of male characteristics.

The X and the Y are very unequal. The X carries far more genes than the Y, and most of them have nothing to do with sex. A sex-linked gene is one that sits on a sex chromosome, and in practice nearly all of them are on the X. A male has only one X, so a recessive allele on it always shows: there is no second allele on the Y to mask it. A female has two Xs and can be a carrier.

Haemophilia is the standard example. A recessive allele on the X means a protein needed to clot blood is missing or faulty, so even small cuts can bleed for a long time. Write the alleles as superscripts on an uppercase X: XH (normal clotting) and Xh (haemophilia). A girl XHXh is a carrier. A boy XhY has haemophilia, because he has no other X. The Y gets no letter, because it has no allele of this gene.

Example Red-green colour blindness is another X-linked recessive condition, so it is more common in males than in females.
Watch out A father cannot pass an X-linked allele to his son. A son gets his Y from his father and his X from his mother. A father does pass his X to every daughter.

A man with haemophilia (XhY) and a woman who is not a carrier (XHXH) have children. What are the children like?

Why are X-linked recessive conditions more common in males than in females?

X, Y and sexXYmany genesfar fewer genesTwo X chromosomes: female characteristics develop.One X and one Y: male characteristics develop.mother XX × father XYXYspermXXegggirlboygirlboyHalf the sperm carry X and half carry Y: the sperm decides.carrier mother × unaffected fatherXHYXHXhXHXHgirl, not affectedXHYboy, not affectedXHXhgirl, carrierXhYboy, has haemophiliaMother: row of eggs. Father: column of sperm.
  1. The X is big and carries many genes. The Y is small and carries far fewer. Two Xs develop female, an X and a Y develop male.
  2. Every egg carries an X. A sperm carries X or Y, so the sperm decides the sex: half girls, half boys.
  3. Haemophilia: a carrier mother (XHXh) and an unaffected father (XHY). Half the sons have haemophilia and half the daughters are carriers. No daughter is affected.

D3.2.13 Pedigree charts

Reading a family tree

A pedigree chart follows a trait through a family. Squares are males and circles are females. A filled shape means the person has the condition, and a line across joins a couple, with their children below. Generations are numbered I, II, III from the top.

You cannot cross people to see what happens, so you read the pattern instead. The aim is to work out whether the condition is dominant or recessive, and whether the gene is autosomal or on the X. Some patterns settle it. Others leave more than one possibility.

Remember it Four clues for reading a pedigree

  • QuietUnaffected parents, affected childrecessive: both parents carry the allele
  • LoudAffected parents, unaffected childdominant: both parents carry a recessive allele
  • DadAffected daughter, unaffected fathernot X-linked recessive
  • BoysMostly sons, skipping generationssuggests X-linked recessive

Chart reading uses two kinds of reasoning. Looking over the family and deciding "this pattern fits a recessive trait" is inductive reasoning: you generalise from the cases you can see. Once you have the pattern, you use deductive reasoning to find the genotype of a particular person: if the trait is recessive, an affected woman must be nn, and her unaffected parents must both be Nn.

Pedigrees also explain a rule found in many societies, that close relatives may not marry. Relatives share alleles they inherited from common ancestors. If a harmful recessive allele runs in a family, two relatives are more likely to carry the same one, so their children are more likely to be homozygous recessive and have the condition. That is the genetic basis only; the rule has other roots too.

Watch out A pattern that fits is not proof. A small family may fit more than one inheritance pattern, so say "consistent with" unless the chart rules the others out.

In a pedigree, two unaffected parents have an affected daughter. Which statement is correct?

You decide, from several families, that a trait is probably recessive. What kind of reasoning is this?

Nature of science Scientists draw general conclusions by inductive reasoning: they base a theory on some of the cases, not all. A pattern deduced from part of a pedigree is a theory, and it then lets you deduce the genotypes of particular people. Keep the two kinds of reasoning apart in your answers.
A family tree for one conditionmalefemaleaffectedopen = unaffectedIIIIIItwo unaffectedparentsaffected daughterNnNnnnN_N_affected daughter
  1. The key: squares are males, circles are females, filled symbols are affected. A line across joins a couple, and lines down lead to their children.
  2. Two unaffected parents (generation I) have an affected daughter (generation II). The condition was hidden in both parents.
  3. Deducing the genotypes: a recessive condition needs nn in the daughter and one n from each parent, so both parents are Nn.
IIIIII121234512IIIIII121234512malefemaleaffectedcousins married
  1. A condition no one in the family had heard of appears in a daughter. Her parents are first cousins. Squares are men, circles women; a filled symbol is affected. Roman numerals number the generations.
  2. Look first at I-1, I-2 and II-4, ringed: they rule out at least one way of inheriting. Which patterns could give this tree, and which are ruled out?
Mark who is affected, then test four ideas (autosomal dominant, autosomal recessive, X-linked recessive, X-linked dominant) and see which the family rules out, and why.

D3.2.14 Continuous variation

A spread, not a few types

Some traits come in a few clear types. ABO blood group is A, B, AB or O, and nobody is halfway. That is discrete variation, and it usually comes from one gene with a few alleles. Other traits, such as height, skin colour or the mass of a seed, form a smooth range with every value in between. That is continuous variation.

What makes the difference? Two things, often together. A trait can be polygenic: controlled by several genes whose small effects add up. With one gene you get a few types. With many genes you get many combinations of alleles and a smooth spread. And the environment blurs the steps further, because it nudges every genotype a little up or down.

Human skin colour is the standard example. The amount of the pigment melanin depends on several genes, and also on sun exposure. The figure uses a very simple model: six alleles, each making the skin a little lighter or a little darker, like six coins. Most combinations land near the middle, and the extremes are rare. The real picture has more genes with unequal effects, but the pattern is the same: a smooth spread with most people near the middle.

You describe the middle of such data with three numbers. The mean is the sum divided by the number of values. The median is the middle value when they are in order. The mode is the most frequent value.

Example Five heights in centimetres: 150, 158, 160, 160 and 182. The mean is 810 ÷ 5 = 162, the median is 160 and the mode is 160. One tall value pulls the mean up but leaves the median where it was.
Watch out Continuous does not mean "not genetic". Height is strongly affected by genes. And polygenic does not mean "no dominance": each of the genes can still have dominant and recessive alleles.

Which of these shows discrete variation?

Discrete or continuous?0510Blood group of 30 students (made-up numbers)number of students11A4B2AB13Oblood group01020Six alleles, each lighter or darker (a simple model)combinations out of 6410611522031546516number of darker allelesAdd sun, diet and other environmental effectslighterdarkerskin colour (many shades, no steps)
  1. Discrete variation: the bars are separate, because nobody is halfway between two blood groups.
  2. A simple model of a polygenic trait: six alleles, each lighter or darker. Combinations with three darker alleles are commonest, and the extremes are rare.
  3. Add the environment and the steps blur into a smooth, continuous spread.

D3.2.15 Box-and-whisker plots

Showing a spread with a box plot

A graph of continuous data should show the middle and the spread. A box-and-whisker plot does it with six aspects of the data: the minimum, the first quartile, the median, the third quartile, the maximum and any outliers.

Put the values in order. The median splits them in half. The quartiles split each half again: a quarter of the values are below Q1 and a quarter are above Q3. The box runs from Q1 to Q3 with the median marked inside, so half of all the data are in the box. The length of the box, Q3 − Q1, is the interquartile range (IQR). Calculators and spreadsheets work out quartiles by slightly different rules, so use the rule you are given.

A value is an outlier if it is more than 1.5 × IQR above Q3 or below Q1. In the figure the IQR is 172 − 160 = 12, so the fences are at 160 − 18 = 142 and 172 + 18 = 190. The whiskers run to the lowest and highest values inside the fences, 150 and 181, and the 194 cm student is plotted on their own.

Example For the 25 made-up heights: median 166 cm, mean 166.8 cm (to one decimal place), mode 160 cm and range 44 cm (150 to 194). The one outlier pulls the mean a little above the median.
Linking question What are the principles of effective sampling in biological research? Twenty-five students tell you about all students only if they were chosen fairly (at random, not just the basketball team), if there are enough of them, and if each was measured the same way. A box plot of a poor sample is a neat picture of the wrong thing.

In a box plot, what fraction of the data lie inside the box?

A student is 194 cm tall. Q3 is 172 cm and the IQR is 12 cm. Is 194 an outlier?

25 student heights (made-up data)each dot is one student140150160170180190200height (cm)Q1160median166Q3172IQR = 12lower fence 142upper fence 190outlier
  1. 25 student heights, one dot each (made up). Where is the middle, and how spread out are they?
  2. The box runs from Q1 to Q3 with the median marked. The whiskers run to the smallest and largest values that are not outliers.
  3. IQR = 172 − 160 = 12. The fences lie 1.5 × IQR beyond the box, at 142 and 190.
  4. A value beyond a fence is an outlier: 194 is above 190, so it is plotted on its own.
130140150160170180190200210height / cm130140150160170180190200210height / cmlower fence 142median 166upper fence 190150Q1 160Q3 172181194 outlier
  1. Heights of 25 students in one class (made-up data). Height varies continuously: each dot is one student, and the values spread out with most near the middle.
  2. The box runs from Q1 (160) to Q3 (172) with the median (166) inside; the whiskers reach the furthest values within 1.5 x IQR of the box (the dashed fences, 142 and 190). Beyond them, 194 is an outlier.
Enter or drag 25 made-up heights and watch the median, the quartiles and the outlier fences move.

D3.2.16 D3.2.17 Segregation, independent assortment and dihybrid crosses

Two genes at once

A dihybrid cross follows two genes. The question is whether the alleles of the two genes travel together or separately. Take Mendel's two seed genes: colour (Y yellow, y green) and shape (R round, r wrinkled).

First, one gene. In meiosis I the two chromosomes of a homologous pair go to opposite poles, so the two alleles of a gene end up in different gametes: each gamete gets one. That is segregation.

Now two genes on different chromosomes. In meiosis I each pair lines up on the equator and faces the poles one way or the other, independently of every other pair. A YyRr parent can send Y with R and y with r, or Y with r and y with R. Both are equally likely, so it makes four kinds of gamete, YR, Yr, yR and yr, in equal numbers. This is independent assortment.

Cross two YyRr plants. Each makes four kinds of gamete, so a 4 × 4 Punnett grid gives 16 equally likely combinations. Count the phenotypes: 9 yellow round, 3 yellow wrinkled, 3 green round and 1 green wrinkled. That is 9 : 3 : 3 : 1. It is the product of two 3 : 1 ratios: (3 yellow : 1 green) × (3 round : 1 wrinkled). Cross a double heterozygote YyRr with a double recessive yyrr and the recessive parent adds only yr, so the offspring show the heterozygote's four gametes: 1 : 1 : 1 : 1.

Steps (1) Write both parents' genotypes. (2) Write each parent's gametes: one allele of each gene, so a YyRr parent gives YR, Yr, yR and yr. (3) Put one parent's gametes along the top and the other's down the side, and fill in the grid. (4) Count the phenotypes, remembering that capitals show.
Nature of science These ratios are based on what has been called Mendel's second law. It only applies if the genes are on different chromosomes, or far enough apart on one chromosome that recombination reaches 50%. Chapter 14 shows genes that break it. There are exceptions to all biological "laws" under some conditions.

A plant YyRr is crossed with a plant yyrr. What proportion of the offspring are yellow and round?

What must be true for the 9 : 3 : 3 : 1 ratio to appear?

Two genes, two pairs of chromosomeschromosomes line up in pairsYyRrEach pair lines upat the equator.Which partner faceswhich pole is chance.one wayor the other wayYyrRgametes: YR and yrgametes: Yr and yRBoth ways are equally likely: four kinds of gamete, in equal numbers.YyRr × YyRrYRYRYrYryRyRyryrYYRRYYRrYyRRYyRrYYRrYYrrYyRrYyrrYyRRYyRryyRRyyRrYyRrYyrryyRryyrr9 yellow, round3 yellow, wrinkled3 green, round1 green, wrinkled9 : 3 : 3 : 1
  1. In meiosis I each pair of homologous chromosomes lines up at the equator, and the partners go to opposite poles.
  2. The pairs line up either way round, independently. Two equally likely ways give four kinds of gamete: YR, yr, Yr and yR.
  3. A YyRr × YyRr cross: put each parent’s four kinds of gamete along the sides of a 4 × 4 grid.
  4. Fill in the 16 genotypes, each equally likely.
  5. Read off the phenotypes: 9 : 3 : 3 : 1.
Watch two pairs of chromosomes line up, each its own way, and see which four kinds of gamete result.
P generationfemale parentYYRRyellow, roundmale parentyyrrgreen, wrinkledxF1 offspringmale gametes, in pollenfemale gametesyrYRyellow roundyellow wrinkledgreen roundgreen wrinkledP generationfemale parentYYRRyellow, roundmale parentyyrrgreen, wrinkledxF1 offspringmale gametes, in pollenfemale gametesyrYRYyRryellow roundyellow wrinkledgreen roundgreen wrinkled
  1. Each parent makes gametes by meiosis, one allele of each gene in each. The female parent's go down the side of the Punnett grid, the male parent's along the top.
  2. Each cell joins one gamete from each parent: a possible offspring, every cell as likely as the next. The F1: genotypes all YyRr; phenotypes all yellow round.
Choose two parents for the seed colour and shape genes. Predict the ratio, then see the 16-box grid fill and breed a sample.

D3.2.18 D3.2.19 D3.2.20 Gene loci, linkage and recombinants

Genes that travel together

Independent assortment works only if the genes are on different chromosomes. A chromosome carries hundreds or thousands of genes, so most genes share a chromosome with others. Genes on the same chromosome are linked: they travel together in meiosis because they sit on the same piece of DNA.

A gene database lists each human gene with its locus (the chromosome and the position on it) and the polypeptide it codes for, so you can look pairs up. Try the genes for the beta chain of haemoglobin (HBB) and the alpha chain (HBA1 and HBA2): they sit on different chromosomes, so they assort independently. Then look at the genes next to HBB, such as HBD for the delta chain: they are close neighbours on the same chromosome, so they are linked. Your teacher will name a database.

Linked alleles are shown with a vertical line for each homologous chromosome and the alleles beside it: AB over ab means A and B are on one chromosome and a and b on its partner. With no crossing over, only two kinds of gamete form, AB and ab: the combinations that were already on the chromosomes, called parental types.

But chromosomes do swap pieces. In prophase I, homologous chromosomes pair up and non-sister chromatids can cross over. If a crossover falls between the two genes, some gametes get Ab or aB. A gamete, a genotype or a phenotype with a combination of alleles that differs from the combinations on the parent's chromosomes is a recombinant. The further apart two genes are, the more often a crossover falls between them, so the more recombinants. Genes far enough apart to give 50% recombinants behave as if unlinked.

The test cross shows this. Cross an AaBb individual with an aabb one. The aabb parent adds only ab, so each offspring's phenotype shows the gamete it got from the heterozygote. Unlinked genes give 1 : 1 : 1 : 1, half of them recombinants. Linked genes give two large parental classes and two small recombinant classes. The recombinant frequency is the recombinants divided by all the offspring. You can find recombinants in gametes, in the genotypes of offspring and in their phenotypes.

Try it yourself in the dry lab: choose unlinked genes or genes linked at different distances, set up a test cross and count the offspring. Or try the linked or not? mystery.

Watch out Linked genes are not "always inherited together": crossing over separates them sometimes. And a recombinant is not a mutant. No allele has changed; only the combination of alleles is new.

A test cross AaBb × aabb gives 48 AB, 47 ab, 3 Ab and 2 aB. What does this suggest?

Two genes lie far apart at opposite ends of a long chromosome. What will a test cross look like?

D3.2.21 The chi-squared test

Is it chance? The chi-squared test

Real offspring never match the expected ratio exactly. From 160 plants from a 9 : 3 : 3 : 1 cross you would not expect exactly 90, 30, 30 and 10, even when the genes are unlinked, because chance decides which gametes meet. So when is a difference too big to blame on chance? A chi-squared test answers that.

Begin with two hypotheses. The null hypothesis says there is no real difference between observed and expected: the genes are unlinked, and chance explains the scatter. The alternative hypothesis says there is a real difference, for example that the genes are linked.

The test gives χ² = the sum of (O − E)² ÷ E over every class, where O is observed and E is expected. The bigger the differences, the bigger χ². Compare it with a critical value from a table, chosen by the degrees of freedom (the number of classes minus 1, so 3 for four phenotype classes) and the significance level, p = 0.05. If χ² is bigger than the critical value, the difference is statistically significant: a difference this large would arise by chance alone less than 5% of the time, so you reject the null hypothesis. If χ² is smaller, chance could explain the difference and you do not reject it.

The dry lab does the arithmetic with you: breed a brood and test it, then decide whether to reject the null hypothesis.

Remember it The five steps of a chi-squared test

  • HungryHypothesesstate the null and the alternative
  • ElvesExpectedwork out the expected counts
  • CookCalculateχ² = sum of (O − E)² ÷ E
  • DeliciousDegrees of freedomnumber of classes minus 1
  • CakesComparewith the critical value at p = 0.05
Critical values at p = 0.05 1 degree of freedom: 3.84. 2: 5.99. 3: 7.81. 4: 9.49.
Nature of science Statistical testing often uses a sample to represent a population. Here the sample is the F2 generation: 160 plants standing for every plant this cross could ever produce. In many experiments the sample is the set of repeated measurements. A bigger sample gives a more reliable test, so geneticists breed many offspring and repeat crosses. That is effective sampling, the linking question of this subtopic.
Watch out A result that is not significant does not prove the null hypothesis: it means these data are not enough to reject it. And a significant result says the data differ from the expected ratio, not why. Linkage is the explanation to test next. The 5% level is a convention: about 1 test in 20 on unlinked genes would look significant by chance alone.

A chi-squared value of 2.1 with 3 degrees of freedom is compared with a critical value of 7.81. What do you conclude?

You keep every percentage the same but double the number of offspring counted. What happens to chi-squared?

Observed against expectedexpected from 9 : 3 : 3 : 1observed in 160 plants9084yellowround3036yellowwrinkled3028greenround1012greenwrinkledclassOE(O − E)² / Eyellow round84900.40yellow wrinkled36301.20green round28300.13green wrinkled12100.40sum = chi-squared2.13024681012chi-squared valuecritical value 7.81do not rejectthe null hypothesisreject thenull hypothesisdf = 4 classes − 1 = 3, at p = 0.052.132.13127.5
  1. 160 F2 plants: the expected counts from 9 : 3 : 3 : 1 (90, 30, 30 and 10) against observed counts (made up).
  2. χ² is the sum of (O − E)² ÷ E over the four classes: 2.13.
  3. With 3 degrees of freedom and p = 0.05 the critical value is 7.81. 2.13 is below it, so you do not reject the null hypothesis: chance can explain the difference.
  4. A test cross of linked genes (made-up counts 112, 21, 18 and 109 from 260 offspring) gives χ² = 127.5, far past 7.81. Reject the null: the genes are probably linked.

02Dry lab: Fly crosses: linked or not?

Cross flies that carry two invented genes, count each brood and use a chi-squared test to decide whether the genes assort independently or travel together. Then try the mystery pairs. This bench is for higher level.

Your questionDo two genes on the same chromosome sort independently, as genes on different chromosomes do? Start with one pair of genes and one cross, work out the ratio you expect if they are unlinked, then see what a real brood gives you.
What you can changeWhether the two genes sit on different chromosomes or are linked at 40, 20 or 5 percent recombination, the cross (a test cross or F1 by F1), and the size of the brood (100, 200, 500 or 1000). The genes are invented and are not real fruit fly data. Change one thing at a time.
What to recordFor each brood: the settings, the count of each of the four phenotypes, the percentage of recombinants, the expected counts, your chi-squared value, the degrees of freedom, and your decision to reject or not reject the null hypothesis. Repeat each set-up at least three times: broods scatter, as real ones do. The coach under the bench helps you plan, process and evaluate.
F1 parentheterozygous for both genesWwRrTest-cross parentdouble recessivewwrr×W straight, w curled wings; R red, r plum eyesinvented genes, not real fly dataGametes from the F1 parent200 offspring a broodW Rparental · 40 %W rrecombinant · 10 %w Rrecombinant · 10 %w rparental · 40 %straight wingsred eyesstraight wingsplum eyescurled wingsred eyescurled wingsplum eyesThe settingsGenes: wing shape and eye colour (invented, not real fruit fly genes).Where they sit: the same chromosome, nearer (20 % recombinants).Cross: test cross, F1 heterozygote x double recessive. 200 offspring a brood.Record each brood: the number of offspring of each phenotype, then test it.Chi-squared, one broodcritical value at p = 0.05, 3 degrees of freedom: 7.81PhenotypeObserved OExpected EO − E(O − E)² ÷ Estraight wings, red eyesstraight wings, plum eyescurled wings, red eyescurled wings, plum eyesTotal, and χ²
Fly crosses. Cross a fly heterozygous for two invented genes, breed a brood, count each phenotype and test the counts with chi-squared against the 1 : 1 : 1 : 1 that independent assortment predicts. The genes and flies are made up for practice: not real fruit fly data.

IA coach

Write it up as an investigation

Your results from this dry lab can be written up the way an IB scientific investigation is: four criteria, each worth 6 marks. Tick each line as your write-up does it. The coach on the page checks these as you go; it never gives a mark.

1 Research design

A focused question set in its biology; your choices of what to measure, how, over what range and how often, explained rather than listed; what you kept the same and how; safety, ethics and the environment; a method someone else could follow and repeat.

  • My question names what I changed.
  • It gives the range I used, with units.
  • It names what I measured and how.
  • It names the organism or system.
  • My background explains, with the biology, why the two should be linked.
  • I cite my sources, and each reference can be traced.
  • If I give a hypothesis, it predicts and explains.
  • Every variable has its How, Why and Units.
  • For each thing kept the same, I say how I kept it so.
  • I list at least three things I kept the same.
  • I name the apparatus that set what I changed, and the instrument that measured.
  • I test enough values.
  • My values spread wide enough.
  • I run at least three trials at every value.
  • I run a control.
  • I explain why these values, this many and this often.
  • Someone else could repeat my method.
  • I deal with safety, ethics and the environment.

2 Data analysis

Raw data recorded clearly, with titles, units, uncertainties and the same decimal places down a column; what you saw as well as what you measured; processing that suits the question, shown with a worked example; the spread shown; a graph a reader can trust.

  • Each table has a numbered title that says what it shows.
  • Every heading has its quantity and its unit.
  • Every raw data heading has its uncertainty.
  • Each uncertainty is realistic for the instrument.
  • Each uncertainty has the same decimal places as its readings.
  • I record what I saw as well as what I measured.
  • I show one worked example of my processing, and it is right.
  • I say why I left out any trial.
  • My error bars show a spread my repeats can support.
  • My graph has a numbered title and labelled axes with units.
  • I say what the error bars show.
  • A straight line, if I draw one, has its R².
  • A statistical test, if I run one, has its hypotheses and suits the data.

3 Conclusion

Your question answered from the processed data, with numbers quoted and the spread taken into account; the biology that explains the pattern; a comparison with published work; no claim bigger than the data.

  • I answer my question in a sentence, naming both variables.
  • I quote my processed numbers.
  • I say how sure the spread lets me be.
  • I say whether my hypothesis is supported.
  • I explain the pattern with biology.
  • I compare my result with published work, and cite it.
  • I claim no more than my data show.

4 Evaluation

Weaknesses and limitations of this method (not slips in carrying it out), with evidence from your own data, the effect each had on the results, which mattered most and why, and a realistic improvement for each.

  • I give at least three weaknesses or limitations.
  • Every row is filled in.
  • No slips in carrying it out: weaknesses of the method itself.
  • Each improvement is specific: which instrument, how many, what changes.
  • Each effect says what it did to my data.
  • At least two rows point to evidence in my own data.
  • At least one limitation: what this experiment cannot tell me.
  • I say which mattered most, and why.
  • I deal with what stands out in my data.
  • My improvements improve this investigation, not a new one.

Variables

VariableHow?Why?Units
Independent: position of the two genes (where their loci sit)
Dependent: percentage of recombinant offspring
Kept the same: the cross: F1 heterozygote x double recessive (a test cross)
Kept the same: number of offspring counted in each brood
Kept the same: the two genes studied (wing shape and eye colour)
Kept the same: how the alleles sit in the F1 parent (both dominant alleles from one pure-breeding parent)
Kept the same: temperature the flies are raised at
Kept the same: food medium and vial size
Kept the same: when the offspring are counted (after all have emerged)

Evaluation

Weakness or limitationWhich?Evidence in my dataEffect on my resultsImprovement

Data generated with the fly cross dry lab, biologybybradford.com (a simulation of invented genes in an invented fly; not real fruit fly data). These checks catch what is missing. They cannot judge how good it is: that is what your teacher does.

Find out

  • What do unlinked genes give? Set the two genes on different chromosomes, make a test cross and breed 200 offspring. Write down the ratio you expect before you breed. How close is the brood to it, and how far apart are the four classes?
  • What does linkage do to the ratio? Repeat with the genes linked at 40, then 20, then 5 percent recombination. Which two classes grow, which two shrink, and which classes are the recombinants?
  • Does the size of the brood matter? Use 100, then 500, then 1000 offspring for the same linked pair. At what size does the chi-squared test start to reject the unlinked expectation every time?
  • Can you tell by looking? Open the chi-squared working for one brood. Say which hypothesis you are testing, work out the expected counts, then let the bench check your value and the critical value for 3 degrees of freedom.
  • Linked or not? Switch to the mystery. You get hidden pairs of genes. Breed as many broods as you need, give your verdict, and see how often you are right.

Record

  • For every brood: the two genes, how they are placed (different chromosomes or linked, and how far), the cross and the number of offspring.
  • The count of each of the four phenotypes, the percentage that are recombinants, and the expected count of each class if the genes are unlinked.
  • Your chi-squared value, the degrees of freedom, the critical value at p = 0.05, and your decision to reject or not reject the null hypothesis.
  • Change one setting at a time. Repeat each set-up at least three times: no two broods match, because chance decides which gametes meet.
  • Use the live graph to compare observed and expected bars, or download the CSV and plot your own. The genes here are invented, so no result is a fact about real flies.

You make a test cross with two genes on different chromosomes and breed 200 offspring. What do you expect for the four phenotypes?

Now the two genes are linked, with 20 percent recombination, and you breed 200 offspring from a test cross. Which result fits best?

A brood from a test cross gives a chi-squared value above the critical value for the unlinked expectation. What should you conclude?

03Check

A fresh paper from the exam bank on this guide's topics each time: past-paper style questions, marked as you go, and saved to your Exam practice progress when you are signed in. It opens on biologybybradford.com, so it needs the internet.

Exam practice

Exam questions on D3.2, marked

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04Downloads

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Saved from biologybybradford.com/guides/ib-biology-d3-2-inheritance, where the guide is kept up to date with the notes, the revision guide and, for the teacher, the lesson plans and slides.

05Teacher resources

The guide as five one-hour lessons, and three more at higher level. Each has a plan to print (the timings, what to project and where to stop, the activities with what they need, what students get wrong) and slides to teach it from, in the same order, with the teacher's moves in the notes.

LessonCoversFocusChaptersLesson planSlides
1Halving, crossing and the genotype D3.2.1 to D3.2.460 min Meiosis as halving and fertilisation as doubling, how a cross in flowering plants is made and described, and the difference between genotype and phenotype 1, 2, 3 Lesson planPDF 721 KB SlidesPPTX 3.9 MB
2Dominance, plasticity and PKU D3.2.5 to D3.2.760 min Why a heterozygote can look like one homozygote, how the environment changes which genes are expressed, and phenylketonuria as a disease of a recessive allele 4, 5, 6 Lesson planPDF 710 KB SlidesPPTX 856 KB
3Many alleles, codominance and sex linkage D3.2.8 to D3.2.1260 min Single-nucleotide polymorphisms and multiple alleles, the ABO blood groups, incomplete dominance and codominance, and the inheritance of genes on the sex chromosomes 7, 8, 9 Lesson planPDF 718 KB SlidesPPTX 605 KB
4Reading a family tree D3.2.1360 min Using a pedigree chart to deduce whether a condition is dominant or recessive and autosomal or sex-linked, and reasoning from some cases to a general pattern 10 Lesson planPDF 628 KB SlidesPPTX 438 KB
5A spread, not a few types D3.2.14 and D3.2.1560 min Continuous variation from polygenic inheritance and the environment, and drawing and reading box-and-whisker plots 11, 12 Lesson planPDF 652 KB SlidesPPTX 533 KB
6Two genes at once HL D3.2.16 and D3.2.1760 min Segregation and independent assortment of unlinked genes, and the Punnett grid for a dihybrid cross 13 Lesson planPDF 668 KB SlidesPPTX 4.1 MB
7Genes that travel together HL D3.2.18 to D3.2.2060 min The loci of human genes, linkage on a chromosome, recombinants made by crossing over, and reading a test cross 14 Lesson planPDF 634 KB SlidesPPTX 435 KB
8Is it chance? Chi-squared and the dry lab HL D3.2.2160 min Using the chi-squared test to decide whether a dihybrid cross matches the expected ratio, and applying it in a simulated experiment 15, dry lab Lesson planPDF 649 KB SlidesPPTX 326 KB
PDFEvery lesson planAll eight plans in one PDF, with this overview at the front. 1.8 MB KEYNote template, filled inThe class’s note template with every gap filled. 1.4 MB

The note template, the revision guide and the offline copy are in Downloads, for you and the class.

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How this was made. The course structure, the activities and the teaching choices come from my own classroom, my presentations and my years of teaching IB and AP Biology. I use Claude, an AI, to help write and build the interactive guides, lesson plans, slides and practice questions, and to draw the diagrams and code the interactives. I direct the work and review it against the IB and AP specifications before it goes live.