D3.2 Inheritance
SL & HL+ AHL sectionsSL and HL, with AHL content · Organisms · Theme D
Big questions
- Why do offspring resemble their parents, yet never exactly?
- How do the movements of chromosomes explain the ratios we see in genetic crosses?
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Gametes, zygotes and genotypes
Offspring resemble their parents because they inherit genes from them, and in every eukaryote with a sexual life cycle the route is the same:
- Body cells of the parents are diploid (2n): they have two sets of chromosomes, so they carry two copies of each autosomal gene, one on each chromosome of a homologous pair.
- Meiosis halves the chromosome number to make haploid (n) gametes, each with only one copy of each gene (see D2.1 Cell and nuclear division).
- At fertilization a male and a female gamete fuse to form a zygote, which is diploid again. Every body cell of the new organism is made from it by mitosis, so they all have the zygote’s alleles.
Because both gametes carry the same number of chromosomes, the two parents make an equal contribution to the nuclear genes of their offspring.
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Text description of this animation
Two diploid body cells, one from each parent, each contain a homologous pair of chromosomes carrying one gene. The mother is heterozygous, Aa, and the father is homozygous, AA. Meiosis halves the chromosome number, so each parent makes haploid gametes carrying just one allele of the gene: here an egg with a and a sperm with A. At fertilization the sperm and egg fuse to form a diploid zygote with genotype Aa, one allele from each parent. Because A is dominant, the phenotype of Aa is the same as that of AA. A final panel shows phenotypic plasticity: skin with the same genotype makes more melanin and darkens after more exposure to sunlight, then becomes paler again when exposure falls, without any change in genotype.
- Diploid parents. Body cells of both parents are diploid (2n). Each has two copies of every autosomal gene, one on each chromosome of a homologous pair. The mother here is Aa; the father is AA.
- Meiosis makes haploid gametes. Meiosis halves the chromosome number, so each gamete is haploid (n) and carries just one allele of each gene. Half of this mother's eggs carry A and half carry a; all of the father's sperm carry A.
- Fertilization. The sperm and egg fuse; their chromosomes combine in the zygote, so it is diploid again. It has one allele from each parent, so its genotype is Aa. Mitosis copies this genotype into every body cell.
- Genotype and phenotype. Aa is heterozygous. A is dominant, because one copy makes enough of the functional protein, so the phenotype of Aa is the same as that of AA. The recessive allele a has no effect in a heterozygote.
- Phenotypic plasticity. Phenotype also depends on the environment. More UV light changes gene expression in skin so that more melanin is made and the skin darkens. When exposure falls the skin becomes paler again. The genotype never changed.
Genes, alleles and genotypes
A gene is a heritable factor at a particular locus (position) on a chromosome. Alleles are different versions of the same gene. A diploid organism has two alleles of each autosomal gene, and the combination it has is its genotype.
| Genotype | Term | Gametes made |
|---|---|---|
| AA | homozygous (dominant) | all A |
| aa | homozygous (recessive) | all a |
| Aa | heterozygous | ½ A, ½ a |
Separation of the two alleles of a gene into different gametes is called segregation. A heterozygous parent therefore passes on each allele with a probability of ½.
Carrying out crosses in flowering plants
Patterns of inheritance were worked out by crossing plants with different characteristics, as Gregor Mendel did with pea plants (Pisum sativum). In a flowering plant, pollen grains made in the anthers contain the male gametes, and the female gametes are in ovules inside the ovary. Pollination (transfer of pollen to a stigma) is therefore needed before fertilization can happen.
Peas carry anthers and an ovary in the same flower, so they self-pollinate and therefore self-fertilize unless prevented. To cross two chosen plants, the breeder controls which pollen reaches each stigma.
1. Remove the anthers of the flower chosen as the female parent while it is still in bud, before any pollen is shed. This stops self-pollination.
2. Cover the flower with a bag so that wind and insects cannot bring pollen from any other plant.
3. Transfer pollen (which contains the male gametes) from anthers of the chosen male parent onto the stigma when it is ripe, then re-cover the flower.
4. Collect the seeds. A pollen tube carries a male gamete down the style to the female gamete in an ovule. Each fertilized ovule becomes a seed; growing the seeds gives the F1 generation.
Naming the generations
- P generation: the parents, usually two pure-breeding (homozygous) varieties.
- F1 generation: the offspring of the P cross, grown from the seeds produced.
- F2 generation: offspring of the F1, often obtained by letting F1 plants self-pollinate.
The results of a cross are predicted with a Punnett grid: the possible gametes of one parent go along the top and those of the other down the side, and each box shows one equally likely offspring genotype.
The same technique of controlled cross-pollination is used by plant breeders to combine useful characteristics and so produce new crop varieties (for example higher yield or disease resistance) and new ornamental flower varieties.
Phenotype, dominance and plasticity
The phenotype is the set of observable characteristics of an organism: structural features such as seed shape and functional ones such as an enzyme’s activity. Phenotype depends on the genotype and on the environment.
| Cause | Human examples |
|---|---|
| Genotype only | ABO blood group, haemophilia, cystic fibrosis |
| Environment only | scars, tattoos, the language a person speaks |
| Genotype and environment | height, body mass, skin colour |
Dominant and recessive alleles
Mendel crossed pure-breeding round-seeded and wrinkled-seeded peas. All the F1 were round, and when the F1 self-pollinated about 3 round : 1 wrinkled appeared in the F2. The allele for round (smooth) seeds, S, is dominant and the allele for wrinkled seeds, s, is recessive.
Why do SS and Ss plants look the same? The S allele codes for a working starch-branching enzyme; the s allele codes for a non-functional version. A heterozygote has one working allele, which makes enough enzyme for normal starch synthesis, so its phenotype is identical to SS. Only ss seeds lack the enzyme, store more sugar, take up more water and wrinkle as they dry. This is a common reason for recessiveness: the recessive allele codes for a non-functional product and one functional allele is enough.
Pea plant (Pisum sativum), seed shape
S = allele for smooth (round) seeds, dominant; s = allele for wrinkled seeds, recessive.
| Female parent ↓ · Male parent → | S | s |
|---|---|---|
| S | SSsmooth seeds | Sssmooth seeds |
| s | Sssmooth seeds | sswrinkled seeds |
Genotype ratio 1 : 2 : 1
- 1/4 SS
- 2/4 Ss
- 1/4 ss
Phenotype ratio 3 : 1
- 3/4 smooth seeds
- 1/4 wrinkled seeds
Ss × Ss: 3 in 4 smooth seeds, 1 in 4 wrinkled seeds. Phenotype ratio 3 : 1. Ss plants have smooth seeds, just like SS plants: one S allele makes enough of the starch-branching enzyme, so the s allele has no effect on the phenotype.
Ratios are probabilities for each offspring. Real families and small samples rarely match them exactly; large numbers of offspring (as in pea crosses) come close.
Text description of this simulation
A Punnett grid builder for monohybrid crosses. Choose one of four patterns: dominant and recessive alleles (pea seed shape, S and s), incomplete dominance (four o'clock flower, Mirabilis jalapa, with red, pink and white flowers), multiple alleles with codominance (human ABO blood groups, alleles IA, IB and i) or sex linkage (haemophilia, with alleles written as superscripts on X). Pick each parent's genotype from a list. The gametes of each parent are placed along the edges of a 2 × 2 grid, each box shows an equally likely offspring genotype and phenotype, and the genotype and phenotype ratios are listed, for example 1 : 2 : 1 and 3 : 1 for two heterozygotes with complete dominance.
Phenotypic plasticity
Phenotypic plasticity is the capacity of an organism to develop traits that suit the environment it experiences, by changing patterns of gene expression. The genotype does not change, and many plastic changes are reversible during the organism’s lifetime:
- Tanning: more ultraviolet light on the skin increases expression of genes for melanin synthesis, so skin darkens; it becomes paler when exposure falls.
- Altitude: people who move to high altitude make more red blood cells, which falls again when they return to sea level.
Plasticity is useful where the environment varies over the lifetime of an individual.
Phenylketonuria: a recessive genetic disease
Phenylketonuria (PKU) is caused by a recessive allele of an autosomal gene, PAH, that codes for the enzyme phenylalanine hydroxylase. This liver enzyme converts the amino acid phenylalanine into tyrosine.
- People with one or two normal alleles (AA or Aa) make enough working enzyme, so heterozygotes are unaffected carriers.
- People with two PKU alleles (aa) make no working enzyme. Phenylalanine from protein in food builds up in the blood and damages the developing brain, causing intellectual disability if untreated. Less tyrosine is made, and tyrosine is needed for melanin, so untreated patients often have lighter skin and hair.
Most babies are tested for PKU a few days after birth using a heel-prick blood sample. An affected child who follows a diet low in phenylalanine develops normally. PKU is therefore a clear example of a phenotype that depends on genotype and environment together: the aa genotype causes harm only if the diet contains normal amounts of phenylalanine.
Many single-gene disorders in humans are caused by recessive alleles, because the usual mutation destroys the function of a protein and a carrier’s other allele still works.
SNPs, multiple alleles and codominance
Single-nucleotide polymorphisms and multiple alleles
New alleles arise by mutation (see D1.3 Mutation and gene editing). Most alleles of a gene differ from each other at only one or a few positions in the base sequence. A position where different bases are found in different members of a species is a single-nucleotide polymorphism (SNP). One gene can contain many SNPs, so the gene pool of a species (all the alleles of all its genes) can contain many alleles of a gene: multiple alleles. Any number of alleles can exist in the population, but a diploid individual has only two.
ABO blood groups
The ABO blood groups are controlled by one gene with three common alleles, written IA, IB and i. IA and IB code for slightly different enzymes that add different sugars to a glycoprotein on red blood cells, giving the A antigen or the B antigen. The i allele codes for a non-functional enzyme, so neither antigen is added.
| Genotype | Blood group |
|---|---|
| IAIA or IAi | A |
| IBIB or IBi | B |
| IAIB | AB |
| ii | O |
Six genotypes give four phenotypes. Parents of groups A and B who are both heterozygous (IAi × IBi) can have children of all four groups; try this in the Punnett grid generator above.
Codominance and incomplete dominance
Not every pair of alleles shows simple dominance. The difference lies in the phenotype of the heterozygote:
- Codominance: both alleles are fully expressed, so the heterozygote has a dual phenotype. IAIB red cells carry both A and B antigens: group AB is not halfway between A and B, it is both.
- Incomplete dominance: the heterozygote has an intermediate phenotype. In the four o’clock flower (Mirabilis jalapa, also called marvel of Peru), red CRCR × white CWCW gives pink CRCW F1 plants, because one CR allele makes less red pigment than two. Pink × pink gives 1 red : 2 pink : 1 white.
The heterozygote looks the same as the dominant homozygote.
The heterozygote is intermediate: a blend of the two homozygous phenotypes.
The heterozygote shows both phenotypes at once: A and B antigens on the same red cell.
Sex determination and haemophilia
Sex determination in humans
Humans have 22 pairs of autosomes and one pair of sex chromosomes. Females are usually XX, so every egg contains an X. Males are usually XY, so half their sperm carry an X and half a Y. The sex chromosome in the sperm therefore decides the sex of the zygote, and about equal numbers of each are expected.
The Y chromosome carries the SRY gene. Early in development, its product causes the embryonic gonads to develop as testes, which secrete testosterone; this hormone causes male-typical features to develop. In an embryo without a Y, the gonads develop as ovaries and female-typical features develop.
The X chromosome is much larger and carries far more genes than the Y: several hundred, many of them needed by both sexes, against only a few dozen on the Y.
Haemophilia: a sex-linked disorder
A gene on the X chromosome shows sex linkage. Males have only one allele of each X-linked gene (they are hemizygous), so a recessive allele on their X is always expressed. Females have two, so they are affected only if both are recessive and are otherwise carriers.
Haemophilia is caused by a recessive allele of a gene on the X chromosome for a blood-clotting protein (factor VIII in the commonest form). Without it, blood clots slowly and bleeding is prolonged. Alleles are written as superscripts on X: XH (normal clotting) and Xh (haemophilia).
| Cross | Daughters | Sons |
|---|---|---|
| carrier XHXh × XHY | ½ XHXH, ½ carriers | ½ unaffected, ½ haemophilia |
| XHXH × affected XhY | all carriers | all unaffected |
Haemophilia is much commoner in males, sons inherit it from their mothers, and an affected father cannot pass it to his sons.
Pedigree charts
A pedigree chart records the phenotypes of members of a family over several generations:
- squares are males and circles are females; affected individuals are shaded
- a horizontal line joins two parents, and a vertical line leads down to their children
- generations are numbered I, II, III… and individuals 1, 2, 3… from left to right
Clues to the pattern of inheritance
| Pattern | Key evidence |
|---|---|
| Autosomal recessive | Unaffected parents have an affected child; can skip generations; affects both sexes |
| Autosomal dominant | Every affected person has an affected parent; two affected parents can have an unaffected child |
| X-linked recessive | Mostly males affected, through unaffected carrier mothers; no father-to-son transmission; affected daughters only if the father is affected |
- unaffected male
- unaffected female
- affected male
- affected female
- carrier (shown after you reveal genotypes)
- marriage between relatives
1. Which pattern of inheritance does this family show?
Look for two key clues: unaffected parents with an affected child, and whether affected daughters have affected fathers.
2. Deduce genotypes
Use A for the allele that makes the working enzyme and a for the PKU allele.
What is the genotype of III-1?
What can be deduced about the genotype of IV-2?
III-1 and III-3 have another child. What is the probability that this child has PKU?
The genotypes shown are worked out by testing every possible combination against the chart. Where two genotypes are shown (for example AA / Aa) the chart does not contain enough information to decide.
Text description of this simulation
Three pedigree charts drawn with standard symbols: squares for males, circles for females, filled symbols for affected people, horizontal lines joining parents and vertical lines to their children, and a double line for a marriage between relatives. Family A shows phenylketonuria (autosomal recessive), with an affected daughter born to unaffected first cousins. Family B shows Huntington's disease (autosomal dominant), with an unaffected son born to two affected parents. Family C shows haemophilia (X-linked recessive), with only males affected and carrier mothers. For each chart, choose the pattern of inheritance and answer questions about the genotypes of individuals; feedback explains the reasoning. A button reveals every genotype that can be deduced, with carriers marked by a dot.
Inductive and deductive reasoning
Working with pedigrees uses both kinds of reasoning. Inductive reasoning draws a general conclusion from a limited set of observations: noticing that only males are affected in a family leads to the hypothesis that the condition is X-linked recessive. Deductive reasoning applies a general rule to a particular case: if the condition is X-linked recessive, the unaffected mother of an affected son must be XHXh. Inductive conclusions can be wrong, which is why they are tested against further cases.
Marriage between close relatives
Rare recessive alleles are usually hidden in carriers. Close relatives share many alleles inherited from common ancestors, so if they have children together, both parents are much more likely to carry the same recessive allele, and the chance of a homozygous recessive child with a genetic disease rises. This raised risk helps explain why many cultures and legal systems forbid close relatives to marry. Family A in the analyser shows this: the PKU allele reaches both of the cousins’ children from a shared grandparent.
Continuous variation and box-and-whisker plots
Discrete and continuous variation
- Discrete (discontinuous) variation: individuals fall into a few distinct categories with nothing in between, such as ABO blood group. It is usually controlled by one or a few genes, with little effect of the environment.
- Continuous variation: any value within a range is possible, as for height or skin colour. It results from polygenic inheritance (many genes, each with a small additive effect), from the environment, or from both.
Human skin colour is the classic example. The amount of the pigment melanin in the skin is influenced by alleles of many genes, some increasing and some decreasing melanin, so many different combinations are possible. Exposure to sunlight also increases melanin production. The combined effects give a continuous range from very pale to very dark skin.
For a continuous variable, a frequency distribution is often roughly bell-shaped (normal). Three measures of central tendency summarise it:
- Mean: sum of all values ÷ number of values.
- Median: the middle value when the data are placed in order (the mean of the two middle values if there is an even number). It is less affected by outliers than the mean.
- Mode: the most frequent value or category; the only one that can be used for discrete categories such as blood groups.
Box-and-whisker plots
A box-and-whisker plot shows the spread of continuous data, such as the heights of students in a class, using six features: the minimum, the lower quartile (Q1), the median, the upper quartile (Q3), the maximum and any outliers. The box spans the interquartile range (IQR = Q3 − Q1), which contains the middle half of the data.
Any value lying further than 1.5 × IQR below Q1, or further than 1.5 × IQR above Q3, counts as an outlier.
Dots: each individual value (repeats stack). Box: Q1 to Q3, split at the median. Whiskers: smallest and largest values that are not outliers. Thin vertical lines: the outlier limits, 1.5 × IQR beyond the quartiles.
- n
- 21
- Mean
- 166.5
- Median
- 166
- Mode
- 165
- Q1 / Q3
- 161.5 / 171.5
- IQR = Q3 − Q1
- 10
- Outlier limits
- below 146.5 or above 186.5
- Outliers
- none
21 values. Median 166 cm, Q1 161.5, Q3 171.5, IQR 10. 1.5 × IQR = 15, so outliers are below 146.5 or above 186.5: there are none. Whiskers run from 152 to 181. Mean 166.5.
Illustrative data. Quartiles here are the medians of the lower and upper halves of the ordered data; calculators and spreadsheets use slightly different rules, so their quartiles may differ a little.
Text description of this simulation
A box-and-whisker plot drawn above an axis, with every data value shown as a dot. Choose a data set (student heights, heights with one very tall student, or skin melanin index) or type your own values. The box runs from the lower quartile (Q1) to the upper quartile (Q3) and is split at the median. The outlier limits are 1.5 × IQR below Q1 and above Q3; values beyond them are plotted as separate outlier points, and the whiskers end at the smallest and largest values that are not outliers. The mean, median, mode, quartiles, interquartile range and outliers are listed. A slider changes the largest value so you can watch it become an outlier.
Segregation, independent assortment and dihybrid crosses
A dihybrid cross follows two genes at once. The outcome depends on how chromosomes move in meiosis (D2.1 Cell and nuclear division):
- Segregation: in anaphase I, the homologous chromosomes of each bivalent move to opposite poles, so the two alleles of each gene end up in different gametes.
- Independent assortment: if two genes are on different chromosomes (unlinked), the orientation of one bivalent at the equator in metaphase I is random and does not affect the orientation of the other. A double heterozygote AaBb therefore makes four types of gamete, AB, Ab, aB and ab, in equal proportions (25% each).
Predicting dihybrid ratios
Mendel crossed pure-breeding round yellow peas (RRYY) with wrinkled green peas (rryy). The F1 were all RrYy, round and yellow. Each F1 plant makes four types of gamete, so the F2 Punnett grid has 4 × 4 = 16 equally likely boxes, giving phenotypes in the ratio 9 round yellow : 3 round green : 3 wrinkled yellow : 1 wrinkled green. This is (3 : 1) × (3 : 1): each gene still gives 3 : 1 on its own.
A test cross of the double heterozygote with a double recessive (RrYy × rryy) gives 1 : 1 : 1 : 1, because the rryy parent contributes only ry and the four gamete types of the heterozygote appear directly in the offspring.
F1 × F1: RrYy (round, yellow) × RrYy (round, yellow), genes on different chromosomes
R = round seed, r = wrinkled; Y = yellow seed, y = green (pea, Pisum sativum).
G = grey body, g = black body; L = long wings, l = vestigial wings (fruit fly, Drosophila melanogaster, both genes on chromosome 2).
0% = loci so close that crossing over between them almost never happens; 50% = so far apart (or on different chromosomes) that they assort independently. These two Drosophila genes give about 17% recombinants. The heterozygote must be female, because male fruit flies do not undergo crossing over.
| gametes | RY¼ | Ry¼ | rY¼ | ry¼ |
|---|---|---|---|---|
| RY25% | RRYYround, yellow | RRYyround, yellow | RrYYround, yellow | RrYyround, yellow |
| Ry25% | RRYyround, yellow | RRyyround, green | RrYyround, yellow | Rryyround, green |
| rY25% | RrYYround, yellow | RrYyround, yellow | rrYYwrinkled, yellow | rrYywrinkled, yellow |
| ry25% | RrYyround, yellow | Rryyround, green | rrYywrinkled, yellow | rryywrinkled, green |
Expected phenotype ratio 9 : 3 : 3 : 1. Each of the 16 boxes is equally likely; 9 contain at least one R and one Y.
Chi-squared test on the offspring
| Phenotype | O | E | (O − E)² ÷ E |
|---|---|---|---|
| round, yellow | 312.75 | 0.02 | |
| round, green | 104.25 | 0.13 | |
| wrinkled, yellow | 104.25 | 0.10 | |
| wrinkled, green | 34.75 | 0.22 | |
| Total | 556 | 556 | χ² = 0.47 |
H₀: the genes assort independently, so the offspring fit a 9 : 3 : 3 : 1 ratio. χ² = 0.47; degrees of freedom = 4 − 1 = 3; critical value at p = 0.05 = 7.815. 0.47 ≤ 7.815, so H₀ is not rejected: the differences from 9 : 3 : 3 : 1 can be explained by chance.
Samples are simulated with a seeded random generator from the true probabilities of the cross chosen (for linked genes, using the recombination frequency set above). Chance alone gives a significant result in about 1 sample in 20 when H₀ is true.
Text description of this simulation
Three dihybrid crosses. In pea plants (round R / wrinkled r, yellow Y / green y, genes on different chromosomes) a double heterozygote self-cross gives a 16-box grid and a 9 : 3 : 3 : 1 phenotype ratio, and a test cross with a double recessive gives 1 : 1 : 1 : 1. In fruit flies (grey G / black g body, long L / vestigial l wings, genes linked on chromosome 2) a test cross gives many more parental types than recombinants; a slider sets the recombination frequency from 0 to 50%, and a chromosome diagram shows how crossing over makes the recombinant gametes. Below, a sample of offspring can be bred (or typed in, or Mendel's own F2 data loaded) and tested with chi-squared: expected values from the Mendelian ratio, (O − E)² ÷ E for each class, degrees of freedom 3 and a critical value of 7.815 at p = 0.05.
These ratios rest on what is often 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 between them happens in nearly every meiosis (recombination near 50%). Like many biological “laws”, it has exceptions; linked genes are the main one.
Loci of human genes and their products
The human genome has roughly 20 000 protein-coding genes. Each has a locus, a fixed position on one of the 22 autosomes or on the X or Y chromosome. Loci are written as chromosome, arm (p short, q long) and band: 11p15.4 is band 15.4 on the short arm of chromosome 11.
Online databases such as NCBI Gene and Ensembl let you search for a gene by name and find its locus, its base sequence, its alleles and the polypeptide it codes for. Some examples:
| Gene | Polypeptide product | Locus |
|---|---|---|
| HBB | β-globin (subunit of haemoglobin) | 11p15.4 |
| HBD | δ-globin (subunit of a minor adult haemoglobin) | 11p15.4 |
| INS | insulin (made as preproinsulin) | 11p15.5 |
| PAH | phenylalanine hydroxylase | 12q23.2 |
| CFTR | CFTR chloride channel protein | 7q31.2 |
| ABO | glycosyltransferase that adds A or B sugar | 9q34.2 |
| F8 | blood-clotting factor VIII | Xq28 |
| OPN1LW | red-sensitive (long-wavelength) cone opsin | Xq28 |
| SRY | sex-determining region Y protein | Yp11.2 |
- Different chromosomes: PAH (chromosome 12) and CFTR (chromosome 7) are unlinked and assort independently.
- Close together on one chromosome: HBB and HBD sit next to each other in the β-globin gene cluster on chromosome 11, and F8 and OPN1LW are both in band Xq28. Alleles of such neighbours are rarely separated by crossing over, so they are strongly linked.
Gene linkage and recombinants
Linked genes have loci on the same chromosome. In most cases this is an autosome, so the pattern is autosomal gene linkage (not to be confused with sex linkage). Alleles of linked genes are carried together on one chromosome, so they tend to be inherited together and do not assort independently.
In crosses with linked genes, show the alleles beside vertical lines that represent the homologous chromosomes, so that it is clear which alleles are on the same chromosome.
Recombinants
Linkage is rarely complete. In prophase I, crossing over between non-sister chromatids can exchange the alleles between the two loci, making chromatids with new combinations. Recombinants are gametes, offspring genotypes or offspring phenotypes with a combination of alleles or characters that differs from those of both original (P generation) parents.
The clearest way to detect recombinants is a test cross between a double heterozygote and a double homozygous recessive:
- Unlinked genes: four offspring phenotypes in a 1 : 1 : 1 : 1 ratio; the two recombinant classes make up 50%.
- Linked genes: the two parental classes are much commoner and the two recombinant classes are rarer. In fruit flies (Drosophila melanogaster), GgLl females (grey body, long wings, with G and L on the same chromosome) crossed with ggll males give mostly grey long and black vestigial offspring, with a smaller number of grey vestigial and black long recombinants.
Recombination frequency = number of recombinants ÷ total offspring. It is higher for loci further apart, because a crossover is more likely to fall between them, but it cannot exceed 50%.
The chi-squared test for dihybrid crosses
Real results never match a predicted ratio exactly, because which gametes fuse is a matter of chance. The chi-squared (χ²) test decides whether the difference between observed (O) and expected (E) results is small enough to be due to chance. The F2 (or test-cross) offspring are a sample that stands for all the offspring the cross could produce.
- State the null hypothesis (H₀): there is no significant difference between observed and expected results; the genes assort independently (9 : 3 : 3 : 1 or 1 : 1 : 1 : 1). The alternative hypothesis (H₁): there is a significant difference (for example because the genes are linked).
- Calculate expected numbers from the ratio and the total, e.g. 9/16 × total.
- Calculate χ² = Σ (O − E)² ÷ E.
- Degrees of freedom = number of classes − 1 = 4 − 1 = 3.
- Compare with the critical value at the p = 0.05 significance level: 7.815 for df = 3.
- If χ² > 7.815, reject H₀: there is a less than 5% probability that a difference this large arose by chance. Otherwise, H₀ is not rejected.
Worked example: linked genes
A test cross of 1000 fruit flies gave 415 grey long, 88 grey vestigial, 92 black long and 405 black vestigial. H₀ predicts 1 : 1 : 1 : 1, so E = 250 for each class.
χ² = (165² + 162² + 158² + 155²) ÷ 250 = 409.8
409.8 is far above 7.815, so H₀ is rejected (p < 0.05): the genes do not assort independently. Parental types are commoner than recombinants, so the genes are linked. Recombination frequency = (88 + 92) ÷ 1000 = 18%.
By contrast, Mendel’s F2 data (315 : 108 : 101 : 32 from 556 seeds) give χ² ≈ 0.47, well below 7.815, so they fit 9 : 3 : 3 : 1. The dihybrid simulation above runs this test on any data you enter or breed.
Must-know definitions
- Gene and allele
- A gene is a heritable factor: a length of DNA at a particular locus that influences a characteristic. Alleles are different versions of the same gene, differing slightly in base sequence (often by single nucleotides).
- Genotype and phenotype
- Genotype is the combination of alleles of a gene that an organism has. Phenotype is its observable characteristics, which result from the genotype and its interaction with the environment.
- Homozygous and heterozygous
- Homozygous: two identical alleles of a gene (e.g. AA or aa), so all gametes get the same allele. Heterozygous: two different alleles (Aa), so half the gametes get each.
- Dominant and recessive alleles
- A dominant allele determines the phenotype even when only one copy is present (in a heterozygote). A recessive allele affects the phenotype only when there is no dominant allele, i.e. in the homozygous recessive.
- Codominance and incomplete dominance
- In codominance both alleles are expressed and the heterozygote shows both phenotypes at once (IAIB, blood group AB). In incomplete dominance the heterozygote has an intermediate phenotype (pink CRCW Mirabilis jalapa).
- Sex linkage
- The pattern of inheritance of a gene on a sex chromosome, usually the X. Males have one X, so a recessive X-linked allele such as the haemophilia allele Xh is always expressed in males; females can be carriers (XHXh).
- Phenotypic plasticity
- The capacity of an organism to develop different traits in different environments by changing patterns of gene expression, without any change in genotype. The changes may be reversible, as in tanning of skin.
- Polygenic inheritance
- Control of one characteristic by many genes, each with a small additive effect. With environmental effects added, it produces continuous variation, as in human skin colour.
- Independent assortmentAHL
- Alleles of genes on different chromosomes are distributed to gametes independently of each other, because each bivalent orients randomly in metaphase I. A double heterozygote makes four gamete types in equal proportions.
- Linked genes and recombinantsAHL
- Linked genes have loci on the same chromosome, so their alleles tend to be inherited together. Recombinants have a new combination of alleles, made by crossing over between the loci in prophase I.
- Chi-squared testAHL
- A statistical test of whether observed counts differ significantly from those expected under a null hypothesis: χ² = Σ (O − E)² ÷ E. For four phenotype classes, df = 3 and the critical value at p = 0.05 is 7.815.
Common exam mistakes
Loses marks Using “gene” and “allele” as if they mean the same thing (“he has the blue-eye gene”).
Write instead Everyone has the gene; people differ in which alleles of it they carry. An allele is one version of a gene.
Loses marks A dominant allele is the most common allele in the population, or it “overpowers” the recessive allele.
Write instead Dominance only describes which phenotype a heterozygote shows. A dominant allele can be rare (the Huntington’s allele is), and the recessive allele is still present and can be passed on.
Loses marks Writing haemophilia genotypes as Hh or Xh, or leaving out the Y in males.
Write instead Write X-linked alleles as superscripts on a capital X, and show the Y: XHXh for a carrier female, XhY for an affected male.
Loses marks If two carriers have had one child with PKU, their next three children will be unaffected.
Write instead Each child is a separate fertilization, so each has the same 1 in 4 chance of being aa. A 3 : 1 ratio is a probability, not a schedule.
Loses marks The AB blood group is an example of incomplete dominance because it is “in between” A and B.
Write instead AB is codominance: red cells carry both the A and the B antigens. Incomplete dominance gives an intermediate phenotype, such as pink flowers from red and white parents.
Loses marks A haemophiliac father can pass haemophilia to his son.
Write instead A father gives his son a Y chromosome, not his X, so sons never inherit an X-linked allele from their father. All of his daughters inherit his Xh and are carriers.
Loses marks Phenotypic plasticity means the alleles change to suit the environment.
Write instead Plasticity involves changes in which genes are expressed and how much; the genotype (base sequence) stays the same, and the change can often be reversed.
Loses marks Drawing whiskers to the outliers, or calling any extreme value an outlier.
Write instead Only values further than 1.5 × IQR beyond Q1 or Q3 are outliers. Plot it as a separate point; the whisker stops at the most extreme value that is not an outlier.
Loses marks Linked genes are the same as sex-linked genes.
Write instead Linked genes are any genes with loci close together on the same chromosome (usually an autosome). Sex-linked genes are on a sex chromosome.
AHLLoses marks A χ² value below the critical value proves the genes are unlinked.
Write instead It means the null hypothesis is not rejected: the differences could be due to chance. Statistics cannot prove a hypothesis, and genes far apart on one chromosome also give a 1 : 1 : 1 : 1 ratio.
AHL
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Multiple-choice questions
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Why does a diploid organism have two copies of each autosomal gene, but its gametes only one?
Body cells contain two sets of chromosomes, one from each parent, so each autosomal gene is present on both chromosomes of a homologous pair. Meiosis halves the chromosome number, so each haploid gamete receives one chromosome of each pair and one allele of each gene.
Homozygous vs heterozygous
Homozygous — two identical alleles (AA or aa), so all gametes carry the same allele. Heterozygous — two different alleles (Aa), so half the gametes carry A and half carry a.
How do you prevent self-pollination when crossing two pea plants?
Remove the anthers of the flower chosen as the female parent before they shed pollen, then cover it with a bag. When the stigma is ripe, transfer pollen from the chosen male parent and re-cover it.
Why do AA and Aa individuals have the same phenotype?
The dominant allele codes for a functional product (often an enzyme). One copy makes enough of it for the normal phenotype, so the recessive allele, which codes for a non-functional product, has no visible effect.
Give a human trait due to genotype only, one due to environment only and one due to both.
Genotype only — ABO blood group or haemophilia. Environment only — a scar or a tattoo. Both — height, body mass or skin colour.
What is phenotypic plasticity? Give an example.
The ability of one genotype to produce different phenotypes, suited to its surroundings, by changing gene expression, with no change in genotype; often reversible. Example — skin darkens with more sunlight because more melanin is made, then pales again.
What causes phenylketonuria?
Two copies of a recessive allele of an autosomal gene (PAH) for phenylalanine hydroxylase. Without the working enzyme, phenylalanine is not converted to tyrosine; it accumulates and harms brain development unless a low-phenylalanine diet is followed.
What is a SNP, and how does it relate to multiple alleles?
A single-nucleotide polymorphism is a position in a gene where different bases are found in different individuals. One gene can contain many SNPs, so many alleles can exist in the gene pool, although each diploid individual carries only two.
Codominance vs incomplete dominance
Codominance — the heterozygote shows both phenotypes (IᴬIᴮ, blood group AB, with A and B antigens). Incomplete dominance — the heterozygote is intermediate (CᴿCᵂ pink four o'clock flowers, Mirabilis jalapa).
What determines the sex of a human baby?
The sex chromosome in the sperm. All eggs carry X; an X sperm gives XX (female), a Y sperm gives XY (male). SRY on the Y makes the gonads develop into testes, which secrete testosterone.
Why is haemophilia more common in males than in females?
The gene is on the X chromosome. Males have one X, so one recessive Xʰ allele causes haemophilia (XʰY). Females need two copies (XʰXʰ); with one copy they are unaffected carriers.
Why do many societies forbid marriage between close relatives?
Relatives share alleles from common ancestors, so both partners are more likely to carry the same rare recessive allele, which raises the chance of a homozygous recessive child with a genetic disease.
How is an outlier defined for a box-and-whisker plot?
A value more than 1.5 × IQR (IQR = Q3 − Q1) below the lower quartile or above the upper quartile. It is plotted as a separate point beyond the whisker.
Why does skin colour show continuous variation?
It is polygenic — alleles of many genes each add a little to melanin production — and sunlight also increases melanin. Many combinations of alleles plus environmental effects give a continuous range of skin colours.
Explain the 9 : 3 : 3 : 1 ratio.
For two unlinked genes, independent assortment in meiosis lets the double heterozygote (AaBb) make AB, Ab, aB and ab gametes equally. Random fusion of these gametes gives 16 equally likely combinations — 9 A_B_ : 3 A_bb : 3 aaB_ : 1 aabb.
Why do linked genes not assort independently?
Their loci are on the same chromosome, so their alleles move together into gametes. Only crossing over between the two loci in prophase I separates them, producing recombinants.
Expected test-cross results for unlinked and linked genes
Double heterozygote × double recessive. Unlinked — 1 : 1 : 1 : 1 (50% recombinants). Linked — mainly the two parental types, with fewer recombinants; recombination frequency = recombinants ÷ total, at most 50%.
Chi-squared on a dihybrid cross — df, critical value, decision?
df = 4 classes − 1 = 3. Critical value at p = 0.05 is 7.815. If χ² > 7.815, reject H₀ (results differ significantly from the Mendelian ratio, e.g. because of linkage); otherwise H₀ is not rejected.
Last reviewed 19 October 2026. Spotted a mistake? Report it.