AQA A-Level Biology: Sex Linkage, Autosomal Linkage and Epistasis
A clear revision guide for AQA A-Level Biology to sex linkage, autosomal linkage and epistasis: why males express recessive X-linked alleles, how linked genes are inherited together, and how one gene can mask another.
The tidy ratios of a dihybrid cross assume that genes are inherited independently. Often they are not. This guide explains three situations that change the pattern of inheritance: sex linkage, where a gene sits on a sex chromosome; autosomal linkage, where two genes travel together on the same chromosome; and epistasis, where one gene overrides another. Each explains why real offspring ratios can depart from the ones you would predict.
Sex linkage
A sex-linked gene is one whose locus is on a sex chromosome, normally the X chromosome. This matters because the two sexes carry different sex chromosomes.
In humans, females are XX and males are XY. The Y chromosome is much shorter than the X and carries few genes, so for a gene on the X chromosome:
- Females (XX) have two alleles. A female only shows a recessive X-linked condition if she is homozygous recessive; with one recessive allele she is an unaffected carrier.
- Males (XY) have only one allele of the gene, inherited on the single X chromosome from their mother. There is no second allele on the Y to mask it, so a recessive allele is always expressed in a male.
This is why recessive X-linked conditions, such as red-green colour blindness and haemophilia, are far more common in males. Pedigree diagrams can reveal sex linkage: if a condition appears mainly in males, it suggests the gene is on the X chromosome, and an unaffected mother with an affected child must be a carrier. You can also rule sex linkage out. If an affected father has an unaffected daughter, the gene cannot be a recessive one on the X chromosome, because he would have passed his only X-linked allele to her and she would be affected or a carrier accordingly.
(In some organisms, such as birds, the system is reversed, with males XX and females XY. The same reasoning then applies with the sexes swapped.)
Worked example: a sex-linked cross
Red-green colour blindness in humans is caused by a recessive allele of a gene on the X chromosome. The normal allele is XN and the colour-blind allele is Xn.
A woman with normal colour vision whose father was colour blind has children with a man with normal colour vision.
Draw a genetic diagram to show the probability that a son of theirs is colour blind.
| Parental phenotypes | Normal vision (carrier) | Normal vision | |||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|
| Parental genotypes | XNXn | XNY | |||||||||
| Gamete genotypes | XN and Xn | XN and Y | |||||||||
| Genetic diagram |
| ||||||||||
| Offspring genotypes | XNXN, XNY, XNXn, XnY | ||||||||||
| Offspring phenotypes | Normal daughter, normal son, carrier daughter, colour-blind son | ||||||||||
| Ratio | 1 in 2 sons colour blind, no daughters affected | ||||||||||
The father's column is the one to look at, because it is what makes a sex-linked square different from an ordinary one. He contributes either XN or Y, so every daughter gets his XN and is protected, while every son gets his Y and has nothing to mask whatever came from his mother.
So the answer is 1 in 2 of the sons, not 1 in 4 of the children. A question that asks for the probability of a colour-blind son wants you to look only at the two male boxes, and a question that asks for the probability of a colour-blind child wants all four. Read which one is being asked.
How the woman is described matters too. She has normal vision but her father was colour blind, so he could only have passed her his Xn: she must be a carrier, XNXn. Working that out from the family information is usually the first mark.
Worked example: a sex-linked cross where the mother is affected
A single gene on the X chromosome controls eye colour in a species of mouse. The dominant allele XE gives black eyes and the recessive allele Xe gives red eyes.
A red-eyed female is crossed with a black-eyed male.
Complete the genetic diagram and give the phenotypes of the offspring.
| Parental phenotypes | Red-eyed female | Black-eyed male | |||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|
| Parental genotypes | XeXe | XEY | |||||||||
| Gamete genotypes | Xe only | XE and Y | |||||||||
| Genetic diagram |
| ||||||||||
| Offspring genotypes | XEXe, XeY, XEXe, XeY | ||||||||||
| Offspring phenotypes | All daughters black-eyed carriers, all sons red-eyed | ||||||||||
| Ratio | 1 black-eyed female : 1 red-eyed male | ||||||||||
This cross is worth learning alongside the first one, because it comes out the other way round. The mother is homozygous recessive, so the only allele she has to give is Xe. Every son gets that allele and a Y, with nothing to mask it, so every son is red-eyed, not one in two. Every daughter gets her father's XE as well, so all the daughters are black-eyed carriers.
That is the pattern to recognise: a sex-linked recessive allele passes from an affected mother to all of her sons. The phrase the mark scheme wants is that a son inherits his only X chromosome from his mother, so the father's genotype cannot protect him.
Worked example: sex linkage with codominance
In a species of fish, scale colour is controlled by a gene on the X chromosome that has two codominant alleles: XR gives red scales and XY gives yellow scales.
A heterozygous female has patches of both colours and is described as mottled.
A mottled female is crossed with a red male. Complete the genetic diagram and give the ratio of phenotypes expected in the offspring.
| Parental phenotypes | Mottled female | Red male | |||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|
| Parental genotypes | XRXY | XRY | |||||||||
| Gamete genotypes | XR and XY | XR and Y | |||||||||
| Genetic diagram |
| ||||||||||
| Offspring genotypes | XRXR, XRY, XRXY, XYY | ||||||||||
| Offspring phenotypes | Red female, red male, mottled female, yellow male | ||||||||||
| Ratio | 1 : 1 : 1 : 1 | ||||||||||
Codominance and sex linkage are being tested together here, so both sets of rules apply at once. In the females, neither allele is masked, so a female with one of each is mottled rather than red or yellow. In the males, there is only ever one allele of the gene, so a male is red or yellow and can never be mottled.
That last point is the one questions are usually built around. A mottled fish must be female, because being mottled requires two different alleles, and only an individual with two X chromosomes can have them. The same reasoning gives the answer whenever a question asks why one phenotype is only ever seen in one sex.
Worked example: a dihybrid cross with sex linkage
In a species of beetle, a gene on an autosome controls the wing cases. The allele for spotted wing cases (S) is dominant to the allele for plain wing cases (s).
A second gene, on the X chromosome, controls antenna length. The allele for long antennae (XA) is dominant to the allele for short antennae (Xa).
A spotted female with long antennae, heterozygous for both genes, is crossed with a plain male with short antennae.
Complete the genetic diagram and give the ratio of phenotypes expected in the offspring.
| Parental phenotypes | Spotted, long antennae | Plain, short antennae | |||||||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| Parental genotypes | SsXAXa | ssXaY | |||||||||||||||
| Gamete genotypes | SXA, SXa, sXA and sXa | sXa and sY | |||||||||||||||
| Genetic diagram |
| ||||||||||||||||
| Offspring phenotypes | Spotted with long antennae, spotted with short antennae, plain with long antennae, and plain with short antennae, each in both sexes | ||||||||||||||||
| Ratio | 1 : 1 : 1 : 1 : 1 : 1 : 1 : 1 | ||||||||||||||||
Write the gamete genotypes carefully, because that is where sex linkage changes the method. The autosomal allele and the sex chromosome go into the same gamete, so each one has to be written as a pair, and the father's two gametes differ in their sex chromosome rather than in the gene: he gives sXa or sY.
That has a consequence worth stating in an answer. Every son receives his father's Y, so a son's antenna length depends only on which X his mother passed on. The father is short-antennaed, but none of his sons inherit that from him, because the allele he carries is on the X chromosome he only ever passes to his daughters.
The reason all eight classes come out equally likely is that the male is recessive for both genes, which makes this a test cross. Each of the mother's four gamete types is expressed directly in the offspring, and the father's X or Y then splits each of those four classes evenly between daughters and sons.
One mark here is purely about notation. The alleles of the sex-linked gene must be written on the X, as XA and Xa, not as a bare A and a. Writing them without the X loses the mark, because the symbol is what shows the examiner that the gene is on the sex chromosome.
Autosomal linkage
Autosomal linkage is when two genes lie on the same autosome, a chromosome that is not a sex chromosome. Because the genes are on the same chromosome, their alleles travel together into a gamete rather than being sorted independently.
- During meiosis, the two linked alleles stay together through the independent segregation of homologous chromosomes, so they are usually inherited as a unit.
- The exception is crossing over, which can swap sections between homologous chromosomes and produce new combinations of the linked alleles.
- The closer together two genes are on the chromosome, the less likely crossing over is to separate them, so the more tightly they are inherited together.
Autosomal linkage means a dihybrid cross does not give the expected 9 : 3 : 3 : 1 ratio. Instead, the parental combinations of alleles appear far more often than expected, with only a small number of recombinant offspring produced by crossing over.
Worked example: autosomal linkage. In a species of snail, shell colour and shell pattern are controlled by two genes on the same pair of homologous chromosomes. Brown (B) is dominant to yellow (b), and unbanded (N) is dominant to banded (n). Snails heterozygous for both genes, BbNn, were crossed; in each parent, B and N lie on one chromosome of the pair and b and n on the other. Explain the results in the table, and give the ratio of phenotypes that would have been expected if the two genes had been on different pairs of homologous chromosomes.
| Phenotype of offspring | Number of offspring |
|---|---|
| Brown, unbanded | 1518 |
| Brown, banded | 47 |
| Yellow, unbanded | 39 |
| Yellow, banded | 504 |
The shape of the data is the answer. Two phenotypes are common and two are rare, and the two common ones are the parental combinations: brown with unbanded, and yellow with banded. The genes are linked, so the alleles that started on the same chromosome usually stay together, and most gametes are BN or bn. Those two gamete types alone give a ratio close to 3 : 1, which 1518 to 504 very nearly is.
The 47 and the 39 are the recombinants. They can only be produced by crossing over between the two loci during meiosis, which swaps sections between the homologous chromosomes and puts B with n, or b with N, in the same gamete. Crossing over is rare between loci this close together, which is why these classes are small.
The second half of the question wants a number. On different pairs of homologous chromosomes the two genes would segregate independently, so the cross would be an ordinary dihybrid and the expected ratio would be 9 : 3 : 3 : 1: 9 brown unbanded, 3 brown banded, 3 yellow unbanded, 1 yellow banded. Saying how far the observed numbers depart from that is the point of the question, and comparing them properly is what a chi-squared test is for.
Epistasis
Epistasis is the interaction of two non-linked genes where one gene masks or suppresses the expression of the other. In other words, whether the second gene shows up at all depends on the alleles present at the first.
A common example is a gene that controls whether a pigment is produced at all, and a second gene that controls the colour of that pigment. If the first gene prevents any pigment being made, the alleles of the second gene make no difference to the phenotype, because there is no pigment for them to colour. Because one gene overrides the other, epistasis also distorts the expected offspring ratios of a dihybrid cross.
Worked example: a dihybrid cross with epistasis
Petal colour in a species of orchid is controlled by two genes that are not linked. A colourless precursor is converted into a pink pigment by enzyme 1, and the pink pigment is then converted into a purple pigment by enzyme 2.
Gene C codes for enzyme 1. The dominant allele C gives a functional enzyme, but a cc plant makes no functional enzyme 1, so the precursor is never converted and the petals stay white.
Gene D codes for enzyme 2. The dominant allele D gives a functional enzyme, but a dd plant makes no functional enzyme 2, so the pathway stops at the pink pigment.
Two plants, each heterozygous for both genes, are crossed. Complete the genetic diagram and give the ratio of phenotypes expected in the offspring.
| Parental phenotypes | Purple | Purple | |||||||||||||||||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|---|
| Parental genotypes | CcDd | CcDd | |||||||||||||||||||||||||
| Gamete genotypes | CD, Cd, cD and cd | CD, Cd, cD and cd | |||||||||||||||||||||||||
| Genetic diagram |
| ||||||||||||||||||||||||||
| Offspring phenotypes | 9 purple, 3 pink, 4 white | ||||||||||||||||||||||||||
| Ratio | 9 : 3 : 4 | ||||||||||||||||||||||||||
The square is filled in exactly as any other dihybrid square. What epistasis changes is how the sixteen boxes are grouped into phenotypes, so work through the pathway for each genotype rather than reaching for a remembered ratio.
Nine boxes have at least one C and at least one D, so both enzymes work and the pigment is taken all the way to purple. Three boxes have at least one C but are dd, so the pathway stops at the pink pigment. The remaining four boxes are cc, and all four are white, whatever their alleles of gene D, because with no functional enzyme 1 there is no pink pigment for enzyme 2 to act on. That last clause is the mark.
So the usual 9 : 3 : 3 : 1 becomes 9 : 3 : 4: the 3 and the 1 of the old ratio both contain cc, so they merge into one phenotype. Gene C is epistatic to gene D, and gene D is hypostatic. A ratio of 9 : 3 : 4 in a set of results is itself evidence of epistasis, and spotting that is often what a question is really asking.
How this fits together
Sex linkage, autosomal linkage and epistasis are the three main reasons real inheritance departs from the simple predictions of a dihybrid cross, and they are exactly the sort of thing a chi-squared test can flag up. Inheritance so far has been about individuals and families; the same alleles, considered across a whole population, are the subject of the populations and the Hardy-Weinberg principle guide.