📖 ContentsChapter 3
Genetics · Chapter 3

Sex Linkage and Extranuclear Inheritance

Mendel's peas gave the same results no matter which parent carried which allele, but many real traits refuse to behave that way. A son inherits color blindness from his mother, and a deletion on chromosome 15 causes one disease when it comes from the father and another when it comes from the mother. This chapter explains why the sex of the parent, and the sex of the child, can change the outcome of a cross.

By the end of this chapter you can

  • Compare chromosomal, genic and environmental sex determination systems, including XX/XY, XX/XO, ZZ/ZW, haplodiploidy and temperature dependent sex determination.
  • Explain how SRY directs testis development in mammals and how the X to autosome ratio sets sex in Drosophila.
  • Predict the results of reciprocal crosses for X linked genes and explain why Morgan and Bridges took them as proof that genes lie on chromosomes.
  • Recognize X linked recessive, X linked dominant and Y linked patterns in a pedigree and calculate carrier and recurrence risks.
  • Describe three strategies of dosage compensation and the molecular steps of X inactivation, including the role of XIST.
  • Distinguish sex influenced and sex limited traits from sex linked traits and solve crosses involving them.
  • Explain maternal inheritance of mitochondrial and chloroplast DNA, including heteroplasmy, the bottleneck and the threshold effect.
  • Contrast maternal effect with maternal inheritance using snail shell coiling.
  • Explain genomic imprinting and use it to account for Prader Willi and Angelman syndromes.

Section 1Sex determination systems

In most animals and in a minority of flowering plants, individuals come in two sexes that make different gametes: females make large, nutrient rich eggs and males make small, mobile sperm or pollen. Sex determination is the mechanism that decides which developmental path an embryo takes. It matters here because when sex is chromosomal, the chromosomes that decide sex also carry ordinary genes whose inheritance is tied to sex.

Chromosomal sex determination

In chromosomal systems, the two sexes differ in their sex chromosomes, and every other chromosome is an autosome. The sex that carries two different sex chromosomes makes two kinds of gametes and is called the heterogametic sex; the sex with two matching sex chromosomes makes one kind of gamete and is the homogametic sex.

Haplodiploidy

Bees, wasps and ants (the order Hymenoptera) have no sex chromosomes at all. Fertilized eggs develop into diploid females and unfertilized eggs develop into haploid males, a system called haplodiploidy. A queen honeybee controls whether each egg she lays is fertilized, and so its sex. Because a male is haploid, he makes sperm by mitosis, and every sperm he produces is genetically identical to every other.

In many hymenopterans the real switch is a single locus called complementary sex determiner (csd). Individuals heterozygous at csd become female; individuals that are hemizygous (haploid) become male. Inbreeding produces some eggs homozygous at csd, which develop as sterile diploid males.

Mammals: SRY decides

The study of people with unusual numbers of sex chromosomes showed that in humans it is the presence of a Y chromosome, not the number of X chromosomes, that makes an embryo male. A person with 45,X (one X, no second sex chromosome) has Turner syndrome and develops as a female, usually with short stature and ovaries that fail early. A person with 47,XXY has Klinefelter syndrome and develops as a male, typically tall with small testes and reduced fertility. People with 47,XXX are female and those with 47,XYY are male.

The critical gene on the Y is SRY (sex determining region Y), located on the short arm near the boundary of the pseudoautosomal region. SRY encodes a transcription factor with an HMG box, a DNA binding domain that bends DNA. In the bipotential gonad of a six to seven week embryo, SRY switches on the autosomal gene SOX9. SOX9 makes gonad cells become Sertoli cells, which secrete anti Müllerian hormone to remove the female duct system, while Leydig cells make testosterone to masculinize the rest of the body. Without SRY, the gonad follows the ovarian pathway, supported by genes such as WNT4, RSPO1 and FOXL2.

Two rare conditions confirm SRY's role. In XX males, an abnormal crossover during the father's meiosis has carried SRY from the Y onto the X, so the child has two X chromosomes and testes. In XY females (Swyer syndrome in some cases), the Y is present but SRY is deleted or carries a loss of function mutation, and the gonads develop as nonfunctional streaks.

The pseudoautosomal regions

The human X and Y are very different in size and gene content: the X carries roughly 800 to 900 protein coding genes while the Y carries only a few dozen. Yet in male meiosis the X and Y must pair and segregate. They do so through short regions of shared sequence at their tips, the pseudoautosomal regions (PAR1 at the end of the short arms and PAR2 at the end of the long arms). In male meiosis at least one crossover occurs in PAR1, which is why genes there, such as SHOX, are inherited as though they were autosomal. Outside the PARs, the male specific region of the Y does not recombine and passes intact from father to son.

Drosophila: the X to autosome ratio

Fruit flies have XX females and XY males, but their sex is decided differently. In flies the Y carries no sex determining gene; instead, sex depends on the ratio of X chromosomes to sets of autosomes, the X:A ratio. Calvin Bridges worked this out in the 1920s by studying flies with unusual chromosome sets. A ratio of 1.0 or higher produces a female, 0.5 or lower produces a male, and values in between produce intersexes. Molecularly, a double dose of X linked signal genes switches on Sex lethal (Sxl), which starts a cascade of alternative RNA splicing (Sxl to transformer to doublesex) that produces female development. The Y is still needed in male flies for fertility, because it carries genes for sperm motility, but an XO fly is a sterile male and an XXY fly is a fertile female.

Fly chromosomesX:A ratioSexual phenotype
XX; AA2/2 = 1.0Female
XXY; AA2/2 = 1.0Female (fertile; the Y does not make it male)
XXX; AA3/2 = 1.5Metafemale (weak, usually dies)
XX; AAA2/3 = 0.67Intersex
XY; AA1/2 = 0.5Male
XO; AA1/2 = 0.5Male (sterile, because Y fertility genes are missing)
XY; AAA1/3 = 0.33Metamale (weak, sterile)

Environmental sex determination

In some animals, sex is not decided by genotype at all. Many turtles, all crocodilians and some lizards show temperature dependent sex determination (TSD): the temperature at which an egg incubates during a sensitive window of development decides the sex. In many turtles warm nests produce females and cool nests produce males; in the American alligator the pattern is reversed, with cool incubation giving females. Temperature acts partly through aromatase, the enzyme that converts testosterone to estrogen. Some fish, such as clownfish, even change sex as adults in response to social cues.

XX / XYXX / XOZZ / ZWHaplodiploid mammals, fliesgrasshoppersbirds, mothsbees, ants, wasps XXXY XXXO ZZZW 2nn femalemale (heterogametic) femalemale (heterogametic) malefemale (heterogametic) female (fertilized)male (unfertilized)
Figure 3.1 Four chromosomal ways to make two sexes. In XX/XY and XX/XO systems the male is heterogametic; in the ZZ/ZW system the female is. In haplodiploid insects there are no sex chromosomes: diploid eggs become females and haploid eggs become males.
Study tip. Worked problem: In a haplodiploid ant colony founded by one queen who mated with one male, what fraction of their genes do two worker sisters share by descent? Step 1. Each worker gets half her genome from her father and half from her mother. Step 2. The father is haploid, so every sperm he makes carries his entire genome. The paternal half of the two sisters is therefore identical: they share all of it, contributing 1/2 × 1 = 1/2. Step 3. The mother is diploid, so for any gene the two sisters have a 1/2 chance of receiving the same maternal copy. That contributes 1/2 × 1/2 = 1/4. Step 4. Add the two parts: 1/2 + 1/4 = 3/4. Answer: 3/4, compared with 1/2 for sisters in a diploid species.
Key point

Sex can be set by chromosomes, by ploidy or by the environment. In mammals a single Y gene, SRY, starts testis development, so the Y decides maleness; in flies the X:A ratio decides and the Y matters only for fertility; in birds the female is the heterogametic sex; and in bees haploid eggs become males. Knowing which sex is heterogametic is the first step in solving any cross involving sex chromosomes.

Section 2X linked inheritance and pedigrees

Genes on the X chromosome are X linked. A male has one X, so for X linked genes he is neither homozygous nor heterozygous but hemizygous: whatever allele his single X carries is expressed. A female has two X chromosomes and can be homozygous or heterozygous, just as for autosomal genes. This asymmetry gives X linked traits their signature: results that change when the sexes of the parents are swapped.

Morgan's white eyed fly

In 1910 Thomas Hunt Morgan found a single male fruit fly with white eyes in a bottle of normal, red eyed flies. He crossed it to a red eyed female. All the F1 flies had red eyes, so white was recessive. When he intercrossed the F1, the F2 contained 3 red to 1 white, a Mendelian ratio, but with a twist: every white eyed fly was male. Half the F2 males had white eyes and none of the females did.

Morgan proposed that the gene lies on the X and that the Y has no copy. Using the convention X+ for the wild type (red) allele and Xw for white, the original male was XwY and the female X+X+. The F1 females were X+Xw and the F1 males X+Y. In the F2, daughters received X+ from their father and so were all red eyed, while sons received an X from their heterozygous mother and were X+Y or XwY in equal numbers.

The decisive test was the reciprocal cross: a white eyed female (XwXw) crossed to a red eyed male (X+Y). For an autosomal gene the reciprocal cross gives the same result as the original. Here it did not. All F1 daughters were red eyed (X+Xw) and all F1 sons were white eyed (XwY). This pattern, in which sons resemble their mothers and daughters resemble their fathers, is called crisscross inheritance. Intercrossing these F1 flies gives an F2 of 1 red female : 1 white female : 1 red male : 1 white male.

Cross ACross B (reciprocal) PF1F2 red ♀ X⁺X⁺ × white ♂ XʷY white ♀ XʷXʷ × red ♂ X⁺Y ♀ X⁺Xʷ red♂ X⁺Y red ♀ X⁺Xʷ red♂ XʷY white intercrossintercross 1/4 X⁺X⁺ red ♀1/4 X⁺Xʷ red ♀ 1/4 X⁺Y red ♂1/4 XʷY white ♂ 1/4 X⁺Xʷ red ♀1/4 XʷXʷ white ♀ 1/4 X⁺Y red ♂1/4 XʷY white ♂ 3 red : 1 whiteall white flies are males 1 : 1 : 1 : 1F1 shows crisscross pattern
Figure 3.2 Morgan's reciprocal crosses with the X linked white eye gene. The two crosses give different F1 and F2 results, the hallmark of a sex linked gene. In cross B the F1 sons have their mother's phenotype and the daughters their father's.

Bridges and nondisjunction: proof of the chromosome theory

Morgan's student Calvin Bridges pushed the argument to its conclusion. When he crossed white eyed females to red eyed males, about 1 in 2,000 offspring broke the crisscross rule: a few daughters were white eyed and a few sons were red eyed. Bridges predicted that these exceptions came from nondisjunction, the failure of the two X chromosomes to separate in the mother's meiosis. An egg carrying both Xw chromosomes, fertilized by a Y bearing sperm, would give an XwXwY fly: female (X:A = 1.0) and white eyed. An egg with no X, fertilized by an X+ sperm, would give an X+O fly: a red eyed, sterile male. Under the microscope, the exceptional females had a Y and the exceptional males lacked one, exactly as predicted: the most convincing proof yet that genes are carried on chromosomes.

Reading X linked recessive pedigrees

In human families we cannot set up crosses, so we read pedigrees. An X linked recessive trait leaves a characteristic set of fingerprints:

IIIIII affectedgrandfather unaffected soncarrierspousespousecarrier affected grandsoncarrier risk 1/2affected affected carrier
Figure 3.3 An X linked recessive pedigree. The affected man in generation I passes his X to both daughters, who are obligate carriers (dotted circles), and his Y to his son, who is unaffected. The trait reappears in grandsons. Square = male, circle = female, filled = affected.

X linked dominant and Y linked patterns

An X linked dominant trait appears in anyone with at least one copy of the allele. Its pedigree features are different: an affected father passes the trait to all of his daughters and none of his sons, and an affected heterozygous mother passes it to half of her children of either sex. Because females have two chances to inherit the allele, affected females outnumber affected males, often about two to one. Examples include X linked hypophosphatemic rickets, in which a defect in the PHEX gene causes the kidney to waste phosphate, and incontinentia pigmenti and Rett syndrome, in which the mutation is usually lethal or far more severe in hemizygous males, so most patients are female.

Y linked (holandric) traits pass only from father to son, to every son, and never to daughters. Nearly all known Y linked traits concern male fertility, such as deletions that block sperm production.

PatternWho is affectedTransmission cluesExamples
Autosomal recessiveBoth sexes equallyAffected children of unaffected parents; consanguinity raises riskCystic fibrosis, sickle cell disease
Autosomal dominantBoth sexes equallyEvery affected person has an affected parent; male to male transmission occursHuntington disease, achondroplasia
X linked recessiveMostly malesNo male to male transmission; all daughters of affected males are carriers; skips generationsHemophilia A and B, Duchenne muscular dystrophy, red green color blindness, G6PD deficiency
X linked dominantFemales about twice as often as malesAffected father to all daughters, no sons; affected mother to half of all childrenHypophosphatemic rickets, incontinentia pigmenti
Y linkedMales onlyFather to all sons; never to daughtersY microdeletions causing azoospermia
Connection. Duchenne muscular dystrophy is caused by loss of function mutations in the dystrophin gene, which spans more than 2 million base pairs of the X and is one of the largest genes known. Its size makes it a large target, so roughly one third of cases arise from a new mutation; an isolated affected boy does not prove that his mother is a carrier, which changes the risk quoted to his sisters.
Study tip. Worked problem: A woman with normal color vision, whose father was red green color blind, marries a man who is color blind. What fraction of their children will be color blind, and what is the chance that their first child is a color blind daughter? Step 1. Assign genotypes. The woman received her father's X, which carries the color blind allele (Xc). She has normal vision, so she is XCXc. Her husband is XcY. Step 2. Daughters get Xc from the father and either XC or Xc from the mother: 1/2 XCXc (carrier, normal vision) and 1/2 XcXc (color blind). Step 3. Sons get Y from the father and either maternal X: 1/2 XCY (normal) and 1/2 XcY (color blind). Step 4. So half of all children, of either sex, are expected to be color blind. Step 5. The chance that the first child is a girl is 1/2, and the chance a girl is color blind is 1/2: 1/2 × 1/2 = 1/4. Answer: 1/2 of the children overall; 1/4 for a color blind daughter. Keep joint and conditional probabilities separate: a color blind daughter is 1/4 of all births, but 1/2 of daughters.
Study tip. Worked problem: Anna's brother has hemophilia A. Anna, her parents and her husband are unaffected. What is the probability that Anna's first child will be a son with hemophilia? Step 1. The brother's X came from their mother, who is unaffected, so the mother must be a carrier (assuming no new mutation). Step 2. Anna received one of her mother's two X chromosomes, so her chance of being a carrier is 1/2. Step 3. If Anna is a carrier, each child has a 1/2 chance of being a son and a son has a 1/2 chance of receiving the mutant X: 1/2 × 1/2 = 1/4. Step 4. Multiply by the chance she is a carrier: 1/2 × 1/4 = 1/8. Answer: 1/8. If the question had asked about a child already known to be a son, the answer would be 1/2 × 1/2 = 1/4.
Key point

Males are hemizygous for X linked genes, so reciprocal crosses give different results and recessive X linked traits show up mainly in males. In pedigrees, look for the absence of male to male transmission and for carrier daughters of affected men; an X linked dominant trait goes from an affected father to every daughter and no son. Bridges' nondisjunction exceptions tied these patterns to visible chromosomes and proved that genes ride on them.

Section 3Dosage compensation and X inactivation

Gene dosage matters. Extra or missing copies of an autosome are usually lethal or cause serious disease because hundreds of genes are suddenly expressed at the wrong level. Yet in XX/XY species, females have two copies of every X linked gene and males have one, with no harm to either sex. The reason is dosage compensation, a set of mechanisms that equalize the expression of X linked genes between the sexes. Remarkably, three well studied animals solve the problem in three different ways.

OrganismStrategyMolecular machinery
MammalsSilence one X in every female cellXIST RNA coats the future inactive X; Polycomb complexes add H3K27 methylation; DNA methylation locks it in
DrosophilaDouble transcription of the single X in malesMale specific lethal (MSL) complex with roX RNAs; MOF acetylates histone H4 at lysine 16, opening chromatin
C. elegansHalve transcription of each X in hermaphroditesDosage compensation complex related to condensin binds both X chromosomes and compacts them

Barr bodies and the Lyon hypothesis

In 1949, Murray Barr and Ewart Bertram noticed a small, darkly staining mass at the edge of the nucleus in nerve cells of female cats that was missing in males. This Barr body turned out to be a condensed, transcriptionally silent X chromosome, a form of facultative heterochromatin. In 1961 Mary Lyon proposed what is now called the Lyon hypothesis:

  1. In each somatic cell of a female mammal, one X chromosome is inactivated and becomes a Barr body.
  2. Inactivation happens early in embryonic development.
  3. In the embryo proper, the choice of which X to silence, maternal or paternal, is random and is made independently in each cell.
  4. Once made, the choice is stable and clonally inherited: all descendants of a cell keep the same X inactive.

A female mammal is therefore a mosaic of two cell populations, one expressing her maternal X and one expressing her paternal X. The rule extends to people with extra X chromosomes: cells keep exactly one X active and inactivate all the others, so the number of Barr bodies equals the number of X chromosomes minus one (the n minus 1 rule).

How an X is silenced

X inactivation is controlled by a region of the X called the X inactivation center (XIC). Its key gene is XIST (X inactive specific transcript), which encodes a long noncoding RNA of about 17,000 nucleotides that is never translated. The process can be pictured in steps:

  1. Counting. The cell senses how many X chromosomes it has relative to its autosome sets. If there is more than one X, inactivation proceeds.
  2. Choice. One X is chosen to stay active. On the future inactive X, XIST is switched on; on the future active X, XIST is kept off. In mice, an antisense transcript called Tsix, made from the opposite DNA strand across Xist, helps repress Xist on the active X.
  3. Spreading. XIST RNA coats the chromosome that makes it, working only in cis (on its own chromosome). It spreads outward from the XIC, using repetitive sequences such as LINE elements as way stations.
  4. Silencing and maintenance. XIST recruits proteins that remove active histone marks and add repressive ones, especially trimethylation of histone H3 at lysine 27 by Polycomb repressive complex 2. The inactive X gains the histone variant macroH2A, its CpG island promoters become methylated, and it replicates late in S phase. DNA methylation makes the silent state heritable through cell division.

In the germ line the process is reversed. In the developing oocyte, the inactive X is reactivated so that each egg carries a working X. In mouse extraembryonic tissues and in all tissues of marsupials, inactivation is imprinted: the paternal X is always the one silenced.

1. Early embryo2. Random choice3. Clonal patches both X activematernal (red)paternal (blue) paternal X silencedmaternal X silenced XIST RNA coat(gold) = Barr body red: maternal X activeblue: paternal X active Barr bodies = number of X chromosomes minus 1 46,XY: 046,XX: 147,XXY: 147,XXX: 245,X: 0
Figure 3.4 Random X inactivation. Each cell of the early female embryo silences one X, chosen at random, by coating it with XIST RNA. The choice is inherited by every daughter cell, so adult tissues are mosaics of patches in which either the maternal or the paternal X is active.

Escape from inactivation

Inactivation is not complete. In humans about 15 percent of X linked genes escape inactivation and are expressed from both X chromosomes, and another 10 percent or so escape in some individuals or tissues. The escapees include genes in the pseudoautosomal regions, which have partners on the Y and so are present in two active copies in both sexes. Escape explains a puzzle: if one X is inactivated anyway, why are 45,X and 47,XXY not harmless? The answer is that people with Turner syndrome have only one copy of the escape genes instead of two, and people with Klinefelter syndrome have an extra copy. Loss of one copy of the pseudoautosomal gene SHOX, for example, accounts for much of the short stature in Turner syndrome.

Visible mosaics

The most famous mosaic is the calico or tortoiseshell cat. An X linked gene, Orange, has an allele XO that makes orange fur and an allele Xo that allows black fur. A heterozygous female (XOXo) has patches of orange where the Xo chromosome was inactivated and patches of black where XO was inactivated. White spotting from an unrelated autosomal gene adds the white areas of a calico. Because the patches depend on two different X chromosomes, calico cats are nearly always female. The rare male calico is usually XXY, and like an XXY man, he is sterile.

Human females show the same mosaicism, though usually less visibly. Women heterozygous for X linked anhidrotic ectodermal dysplasia have patches of skin without sweat glands, mapped by spraying the skin with a starch iodine indicator. Women heterozygous for G6PD deficiency have two populations of red cells, one normal and one deficient. Because inactivation is random, the proportion of cells using each X varies from woman to woman. When most cells happen to inactivate the normal X, called skewed X inactivation, a carrier can show symptoms. Such women are called manifesting heterozygotes; for example, some carriers of Duchenne muscular dystrophy develop muscle weakness or heart muscle disease.

Connection. Cloning a calico cat does not produce a copy of her coat. The first cloned cat, made from the cell of a calico, had a different pattern of patches because the inactivation choices in the embryo are made anew at random, even though her genome was identical. It is a striking demonstration that an identical genotype does not guarantee an identical phenotype, and that epigenetic events made during development can be as decisive as the DNA sequence.
Study tip. Worked problem: A tortoiseshell female cat (XOXo) mates with a black male (XoY). Predict the coat colors of the kittens by sex, ignoring white spotting. Step 1. Female gametes: 1/2 XO, 1/2 Xo. Male gametes: 1/2 Xo, 1/2 Y. Step 2. Daughters receive Xo from the father: 1/2 XOXo (tortoiseshell) and 1/2 XoXo (black). Step 3. Sons receive Y from the father and one maternal X: 1/2 XOY (orange) and 1/2 XoY (black). Step 4. Over all kittens: 1/4 tortoiseshell females, 1/4 black females, 1/4 orange males, 1/4 black males. Answer: No male kitten can be tortoiseshell unless he has two X chromosomes. Quick check on Barr bodies while you are here: 47,XXY has 2 − 1 = 1, 48,XXXY has 3 − 1 = 2, and 45,X has 0.
Key point

Dosage compensation equalizes X linked gene expression between the sexes: mammals silence one X in females, flies double the male X, and worms halve each X in hermaphrodites. In mammals, XIST RNA coats the chosen X in cis and recruits repressive chromatin marks; the choice is random and clonally inherited, making every female a mosaic. About 15 percent of human X genes escape inactivation, which is why X aneuploidies still have effects.

Section 4Sex influenced and sex limited traits

Not every trait that differs between males and females is caused by a gene on a sex chromosome. Many autosomal genes are expressed differently in the two sexes because males and females have different hormonal environments. Two categories are worth distinguishing, because they produce patterns that can be mistaken for sex linkage.

Sex influenced traits

In a sex influenced trait, both sexes can express the trait, but the heterozygote's phenotype depends on its sex: the allele behaves as dominant in one sex and recessive in the other. Homozygotes have the same phenotype in both sexes.

The standard example is horns in certain sheep breeds. Let H be the allele for horns and h the allele for no horns. HH sheep of both sexes are horned and hh sheep of both sexes are hornless. Hh rams are horned, while Hh ewes are hornless. In effect H is dominant in males and recessive in females, a difference driven by testosterone. Beards in goats follow the same logic. Pattern baldness is often taught the same way, dominant in men and recessive in women. In reality many genes contribute, including variants near the X linked androgen receptor gene, so treat the one gene model as a practice simplification.

Sex limited traits

In a sex limited trait, the phenotype is expressed in only one sex, although both sexes carry and transmit the genes. Milk yield in cattle is the classic example: bulls carry genes that influence milk production and pass them to their daughters, but bulls themselves give no milk. Breeders judge a bull by his daughters' yields. Other examples include genes affecting egg production in hens, the timing of puberty in boys with familial male limited precocious puberty (an autosomal dominant activating mutation in the luteinizing hormone receptor that has no effect in carrier females), and the cock feathering trait of chickens.

In chickens, males usually have long, curved, pointed tail and neck feathers (cock feathering) and females have short, rounded feathers (hen feathering). An autosomal allele H produces hen feathering in both sexes. Females are hen feathered no matter what their genotype is, so the trait of cock feathering is limited to hh males.

GenotypeSheep horns, maleSheep horns, femaleChicken plumage, maleChicken plumage, female
HHHornedHornedHen featheredHen feathered
HhHornedHornlessHen featheredHen feathered
hhHornlessHornlessCock featheredHen feathered
TypeSex influenced: heterozygote differs by sexSex limited: one phenotype appears only in males
Sex influenced (horns)Sex limited (cock feathers) malefemalemalefemale HHHhhh hornedhorned hornedhornless hornlesshornless henhen henhen cockhen Only the heterozygote differsbetween the sexes The cock phenotype occursonly in males (hh)
Figure 3.5 Sex influenced versus sex limited inheritance of autosomal genes. Dashed outlines mark the cells where sex changes the outcome. Both genes are on autosomes, so fathers can pass them to sons.

Telling them apart from sex linkage

Sex influenced traits can mimic X linkage in a single generation. Cross a hornless ram (hh) to a horned ewe (HH): every lamb is Hh, so sons are horned like their mother and daughters hornless like their father, which looks like crisscross inheritance. Two tests reveal the truth. An autosomal gene can pass from father to son, which an X linked gene cannot, and the F2 of a sex influenced gene shows 3:1 in one sex and 1:3 in the other.

Study tip. Worked problem: Two Hh sheep are mated and produce many lambs. What fraction of rams and of ewes will be horned, and what fraction of all lambs? Step 1. Hh × Hh gives 1/4 HH : 1/2 Hh : 1/4 hh, the same in each sex because the gene is autosomal. Step 2. Rams: HH and Hh are horned, so 1/4 + 1/2 = 3/4 horned and 1/4 hornless. Step 3. Ewes: only HH is horned, so 1/4 horned and 3/4 hornless. Step 4. With equal numbers of each sex, all lambs: (1/2 × 3/4) + (1/2 × 1/4) = 3/8 + 1/8 = 1/2 horned. Answer: 3/4 of rams, 1/4 of ewes, 1/2 overall. A good check: the ram and ewe ratios are mirror images (3:1 and 1:3), the signature of a sex influenced gene.
Key point

Sex influenced and sex limited traits are controlled by autosomal genes whose expression depends on the sex of the individual, usually through hormones. In sex influenced traits the heterozygote differs between the sexes; in sex limited traits one phenotype appears in only one sex. Father to son transmission and sex specific F2 ratios of 3:1 and 1:3 distinguish them from X linkage.

Section 5Mitochondrial and chloroplast inheritance, maternal effect and genomic imprinting

This section covers situations in which the two parents do not contribute equally. In extranuclear inheritance, genes in mitochondria or chloroplasts come mostly from one parent. In maternal effect, the mother's nuclear genotype, not the offspring's, decides the offspring's phenotype. In genomic imprinting, a gene inherited from one parent is silenced. They are easily confused, so learn to tell them apart.

The mitochondrial genome

Mitochondria descend from bacteria that were engulfed by an ancestral cell more than a billion and a half years ago, and they still carry a small genome of their own. The human mitochondrial DNA (mtDNA) is a circular molecule of 16,569 base pairs containing 37 genes: 13 encode polypeptides of the oxidative phosphorylation system (subunits of complexes I, III, IV and ATP synthase, but none of complex II), 22 encode transfer RNAs and 2 encode ribosomal RNAs. The genome is compact, with almost no noncoding DNA except the D loop control region, and it uses a slightly different genetic code (for example, UGA codes for tryptophan instead of stop). The other 1,000 or more proteins that a mitochondrion needs are encoded in the nucleus, made in the cytosol and imported. For that reason, many mitochondrial diseases are inherited in ordinary Mendelian patterns.

A cell holds hundreds to thousands of copies of mtDNA. Its mutation rate is much higher than that of nuclear DNA, and with no recombination this makes mtDNA useful for tracing maternal ancestry.

Maternal inheritance, heteroplasmy and the threshold

In humans and most animals, mtDNA is inherited from the mother. An egg contains roughly 100,000 or more copies of mtDNA, while the few mitochondria in the sperm midpiece are tagged with ubiquitin and destroyed in the early embryo. A mother with an mtDNA mutation therefore can pass it to all of her children, sons and daughters alike, but an affected father passes it to none.

When all mtDNA copies in a cell are identical, the cell is homoplasmic; when a cell has a mixture of normal and mutant copies, it is heteroplasmic. Three ideas explain why mitochondrial diseases are so variable, even among siblings:

mother: oocyteprecursor, 50% mutant bottleneckfew mtDNAsrepopulate egg threshold (80%) 15%: healthy55%: healthy90%:disease normal mtDNAmutant mtDNA (red bar = mutant fraction in child)
Figure 3.6 The mitochondrial bottleneck. Because only a small sample of a mother's mtDNA molecules populates each egg, her children can inherit very different mutant loads. Disease appears only in tissues where the mutant fraction exceeds a threshold, shown here at an illustrative 80 percent.
DisorderTypical mtDNA changeMain features
Leber hereditary optic neuropathy (LHON)Point mutations in complex I genes, most often m.11778G>A in MT-ND4; usually homoplasmicSudden painless loss of central vision in young adults; incomplete penetrance and more males affected
MELASm.3243A>G in the tRNA for leucine; heteroplasmicMitochondrial encephalomyopathy, lactic acidosis and strokelike episodes
MERRFm.8344A>G in the tRNA for lysine; heteroplasmicMyoclonic epilepsy with ragged red fibers in muscle biopsy
Connection. Mitochondrial replacement therapy aims to let a woman carrying a severe mtDNA mutation have a genetically related child without passing on the disease. The nucleus of her egg or early zygote is transferred into a donor egg that has had its own nucleus removed but keeps healthy mitochondria. The child carries nuclear DNA from both parents and mtDNA from the donor. Because a little of the mother's mtDNA can be carried over, the method reduces risk rather than removing it.

Chloroplast inheritance

Chloroplasts also carry their own DNA, typically a circle of 120,000 to 160,000 base pairs with roughly 100 to 130 genes for photosynthesis proteins, ribosomal RNAs and transfer RNAs. In 1909 Carl Correns studied the four o'clock plant Mirabilis jalapa, whose branches can be green, white (lacking chlorophyll) or variegated. He found that the phenotype of seedlings depended only on the branch that bore the seed, the female parent, and not at all on the pollen source. Seeds from green branches gave green seedlings, seeds from white branches gave white seedlings (which die once their seed reserves run out), and seeds from variegated branches gave a mix of green, white and variegated seedlings. The variegated branches carried a mixture of normal and defective plastids, and the egg cells received different mixtures, the plant equivalent of heteroplasmy.

Maternal effect

A maternal effect is not the same as maternal inheritance. In a maternal effect, the gene is an ordinary nuclear gene that follows Mendel's laws, but the offspring's phenotype is decided by the mother's genotype, because the mother loads the egg with mRNA and proteins made from her own genes before fertilization. Those products direct the earliest stages of development, before the embryo's own genes are switched on.

The classic example is the direction of shell coiling in the pond snail Lymnaea peregra. The allele D for right handed (dextral) coiling is dominant to d for left handed (sinistral) coiling, but the direction is set by the orientation of the spindle in the first cleavage divisions, which depends on products the mother deposited in the egg. Snails are hermaphrodites, so a single snail can be crossed or self fertilized. The key results are:

PF1F2F3 ♀ dd (sinistral) × ♂ DD (dextral) all Dd, all SINISTRAL (mother was dd) self 1 DD : 2 Dd : 1 dd, all DEXTRAL (mother was Dd) DD selfsDd selfsdd selfs dextraldextralsinistral 3 dextral broods : 1 sinistral brood, one generation late
Figure 3.7 Maternal effect inheritance of shell coiling. Each snail's coiling matches its mother's genotype, not its own, so the expected 3:1 Mendelian ratio appears only when the F2 snails' broods are scored.

Maternal effect genes are central to early development. In Drosophila, the mother deposits bicoid mRNA at the front end of the egg. Its protein gradient specifies head and thorax, and a homozygous mutant mother lays eggs that develop without heads even when the embryo carries a normal paternal allele.

Study tip. Worked problem: A Dd snail that is itself sinistral is used as the female in a cross to a dd male. What is the coiling of her offspring, and what was her mother's genotype? Step 1. The offspring's phenotype follows the mother's genotype. She is Dd, so every offspring is dextral, regardless of whether it is Dd or dd. Step 2. Her own sinistral phenotype was set by her mother's genotype, so her mother must have been dd. Step 3. Check: a dd mother could have produced a Dd daughter only if the father contributed D, which is consistent. Answer: all offspring dextral; her mother was dd. The rule to remember: a snail's phenotype tells you its mother's genotype, and its genotype tells you its offspring's phenotype.

Genomic imprinting

For most genes, the copy from the mother and the copy from the father are expressed equally. For a few hundred mammalian genes (estimates for humans run from about 100 to more than 200), expression depends on the parent of origin: one copy is silenced according to which parent it came from. This is genomic imprinting. An imprinted gene is "maternally imprinted" if the maternal copy is silenced (so only the paternal copy is expressed) and "paternally imprinted" if the paternal copy is silenced. Because these terms are easy to flip, "paternally expressed" or "maternally expressed" is clearer.

Imprints are set by DNA methylation at imprinting control regions (ICRs) during gamete formation. In the primordial germ cells of the embryo, old imprints are erased, and new ones are written according to the sex of the individual: a male writes paternal imprints on every chromosome he will pass on, and a female writes maternal imprints. After fertilization, imprinted methylation is protected from the wave of demethylation that resets most of the genome, and it is maintained in somatic cells for life.

The Igf2 and H19 genes on mouse chromosome 7 (human 11p15) show how an imprint works. Igf2 encodes insulin like growth factor 2, a fetal growth factor, and is expressed only from the paternal chromosome. H19, a noncoding RNA gene, is expressed only from the maternal chromosome. Between them lies an ICR. On the maternal chromosome the ICR is unmethylated and binds the insulator protein CTCF, which blocks shared downstream enhancers from reaching Igf2, so the enhancers activate H19 instead. On the paternal chromosome the ICR is methylated, CTCF cannot bind, the enhancers reach Igf2, and methylation spreading into the H19 promoter silences it. A mouse that inherits a disrupted Igf2 gene from its father is small, but one that inherits the same disrupted gene from its mother is normal, because the maternal copy was silent anyway.

Maternal chromosomePaternal chromosome Igf2 offIgf2 ON ICRICR H19 ONH19 off enhancersenhancers C CTCF bound: insulatormethylated (dots): no CTCF
Figure 3.8 Imprinting at the Igf2 and H19 locus. The methylation state of the imprinting control region (ICR) decides whether CTCF binds and therefore which gene the shared enhancers activate: H19 on the maternal chromosome and Igf2 on the paternal chromosome.

Prader Willi and Angelman syndromes

The clearest human example of imprinting is a cluster of genes on chromosome 15q11 to q13. Some genes in the region, including the small nucleolar RNA cluster SNORD116, are expressed only from the paternal chromosome; one gene, UBE3A, is expressed only from the maternal chromosome in neurons. The same deletion of about 5 million base pairs therefore causes two different disorders depending on its parent of origin:

Imprinting defects at 11p15 cause Beckwith Wiedemann syndrome (overgrowth and embryonal tumor risk, often from excess IGF2) and its mirror image, Silver Russell syndrome (growth restriction). The parental conflict hypothesis proposes that imprinting evolved as a tug of war over fetal growth: paternally expressed genes such as Igf2 tend to promote it and maternally expressed genes tend to restrain it.

PatternWhere the gene isWhat decides the phenotypeSignature in crosses
Maternal (organelle) inheritancemtDNA or cpDNAThe organelle genomes in the eggAffected mother passes trait to all children; affected father to none; variable because of heteroplasmy
Maternal effectNucleus, normal Mendelian geneThe mother's nuclear genotypeReciprocal crosses differ; Mendelian ratios appear one generation late
Genomic imprintingNucleusWhich parent passed on the active copySame mutation causes disease only when inherited from one parent; can be silent for generations
X linked recessiveX chromosomeThe offspring's own genotype and sexMostly males affected; no father to son transmission
Study tip. Worked problem: A gene is expressed only from the paternal copy. A man is heterozygous for a loss of function mutation in it. Is he necessarily affected, and what is the risk to his children? Compare a heterozygous woman. Step 1. The man's phenotype depends on which parent gave him the mutant copy. If he got it from his mother, it was silent in him and he is unaffected; if from his father, he is affected. Step 2. Either way, the copy he passes to his children will be expressed in them, because it is now paternal. Step 3. Each child has a 1/2 chance of receiving the mutant copy, so each child has a 1/2 risk of being affected. Step 4. If instead a heterozygous woman passes the mutant copy, it is silenced in her children: risk 0, although half her children will be silent carriers. Answer: 1/2 of a carrier father's children are affected; none of a carrier mother's children are. A pedigree of such a gene shows the disorder appearing and disappearing depending on which parent transmitted it.
Key point

Organelle genomes are usually inherited from the mother, and heteroplasmy, the bottleneck and the threshold effect make mitochondrial disease highly variable. A maternal effect gene is Mendelian, but its products in the egg set the offspring's phenotype, so ratios appear one generation late. Imprinting silences one parental copy through germ line DNA methylation, which is why a single deletion of 15q11 to q13 causes Prader Willi syndrome when paternal and Angelman syndrome when maternal.

The chapter in one breath

  • Sex can be determined by sex chromosomes (XX/XY, XX/XO, ZZ/ZW), by ploidy in haplodiploid insects, or by the environment, as in temperature dependent sex determination.
  • In mammals the Y gene SRY starts testis development; in Drosophila the X:A ratio sets sex.
  • Males are hemizygous for X linked genes, so reciprocal crosses differ and recessive X linked traits appear mainly in males.
  • X linked recessive pedigrees show no father to son transmission and carrier daughters of affected men.
  • Dosage compensation silences one X in female mammals, doubles the X in male flies and halves both X chromosomes in worm hermaphrodites.
  • XIST RNA coats the future inactive X in cis; random, clonally stable inactivation makes every female mammal a mosaic, and the number of Barr bodies is the number of X chromosomes minus one.
  • Sex influenced traits differ between the sexes only in heterozygotes; sex limited traits appear in only one sex; both are autosomal.
  • mtDNA (16,569 bp, 37 genes) is inherited maternally; heteroplasmy, the bottleneck and the threshold effect make mitochondrial disease variable.
  • In a maternal effect the mother's nuclear genotype sets the offspring's phenotype, so Mendelian ratios show up a generation late, as in snail coiling.
  • Genomic imprinting silences one parental copy by germ line methylation; loss of paternal 15q11 to q13 causes Prader Willi syndrome and loss of maternal UBE3A causes Angelman syndrome.

Words worth knowing

TermWhat it means
Heterogametic sexThe sex with two different sex chromosomes (XY males in mammals, ZW females in birds), which makes two kinds of gametes.
Temperature dependent sex determinationSex decided by the incubation temperature of the egg rather than by genotype, as in many turtles and all crocodilians.
SRYThe Y chromosome gene encoding an HMG box transcription factor that starts testis development in mammals.
X:A ratioThe ratio of X chromosomes to sets of autosomes, which determines sex in Drosophila.
Pseudoautosomal regionA region at the tips of the X and Y that is shared by both, pairs and crosses over in male meiosis, and is inherited like an autosome.
HemizygousHaving only one copy of a gene, as males do for most X linked genes.
Y linked (holandric)Describes a gene on the male specific part of the Y, passed from father to every son.
Crisscross inheritanceThe pattern in which sons show the mother's phenotype and daughters the father's, typical of X linked recessive genes.
NondisjunctionFailure of homologous chromosomes or sister chromatids to separate during cell division, producing gametes with an extra or missing chromosome.
Dosage compensationAny mechanism that equalizes the expression of X linked genes between the sexes.
Barr bodyThe condensed, inactive X chromosome seen at the edge of the nucleus in cells of female mammals.
Lyon hypothesisThe proposal that one X in each female somatic cell is inactivated early, at random and permanently for that cell lineage.
XISTA long noncoding RNA made by the future inactive X that coats that chromosome and recruits silencing proteins.
Manifesting heterozygoteA female carrier of an X linked recessive allele who shows symptoms, often because of skewed X inactivation.
Sex influenced traitAn autosomal trait whose heterozygote phenotype differs between the sexes, as with horns in some sheep breeds.
Sex limited traitAn autosomal trait expressed in only one sex, such as milk production or cock feathering.
Maternal inheritanceInheritance of organelle genes through the egg cytoplasm, so traits pass from mother to all offspring.
HeteroplasmyThe presence of more than one type of mitochondrial or chloroplast genome in the same cell or individual.
Maternal effectDetermination of an offspring's phenotype by the mother's nuclear genotype through products she places in the egg.
Genomic imprintingParent of origin specific silencing of a gene, set by DNA methylation in the germ line.
Uniparental disomyInheritance of both copies of a chromosome, or part of one, from a single parent.

Check yourself

A person has a 46,XX karyotype but has testes. What is the most likely genetic explanation?
Answer. The SRY gene was moved onto one of the X chromosomes by an abnormal crossover between the X and Y during the father's meiosis. SRY alone is enough to start testis development, so the rest of the Y is not required for male sex determination, although missing Y fertility genes make such men infertile.
In birds, a breeder wants to sort chicks by sex at hatching using a Z linked color gene. Should the dominant allele be carried by the mother or the father?
Answer. By the mother. Cross a ZW female carrying the dominant allele to a ZZ male homozygous recessive. Sons receive the father's Z and the mother's Z, so they show the dominant phenotype; daughters receive only the father's Z (plus the W), so they show the recessive phenotype.
A white eyed female fly is crossed to a red eyed male. Among thousands of offspring, a few red eyed sons appear. What chromosome constitution do they have, and are they fertile?
Answer. They are X+O. Nondisjunction in the mother produced an egg with no X, and the father's X+ sperm fertilized it. The X:A ratio is 0.5, so the fly is male, but without a Y he lacks sperm motility genes and is sterile.
A man with hemophilia A and a woman who is not a carrier have children. What is the risk for their sons and daughters, and for the sons of their daughters?
Answer. Their sons are unaffected, because they receive the father's Y. All daughters are carriers, because they receive his X. Each son of a carrier daughter has a 1/2 chance of having hemophilia, so the condition skips a generation.
How many Barr bodies would you expect in cells of a person who is 48,XXXY, and what sex would the person be?
Answer. Two Barr bodies (three X chromosomes minus one). The presence of the Y with SRY makes the person male.
If one X is inactivated in every female cell, why does 45,X cause Turner syndrome rather than a normal female phenotype?
Answer. About 15 percent of X genes, including pseudoautosomal genes such as SHOX, escape inactivation and are normally expressed from both X chromosomes. A 45,X person has only one copy of these genes, and the reduced dose causes features such as short stature. A single X in the germ line also leads to early loss of oocytes.
A hornless ram is mated with a horned ewe, and all sons are horned and all daughters hornless. A student concludes the gene is X linked. What cross would you do to test this, and what would you expect if the gene is sex influenced?
Answer. Intercross the F1 (Hh rams with Hh ewes). For a sex influenced autosomal gene, the F2 rams would be 3/4 horned and the F2 ewes 1/4 horned. Father to son transmission would also rule out X linkage.
A woman with MELAS has a son and a daughter. Why might the son be severely affected while the daughter has only mild symptoms, and can the son pass the condition to his children?
Answer. The mother is heteroplasmic, and the bottleneck in oogenesis gave each egg a different fraction of mutant mtDNA. The son likely received a fraction above the threshold in brain and muscle and the daughter one below it. The son cannot pass the mutation on, because mtDNA is inherited from the mother; the daughter can.
In snails, a cross between a dextral female and a sinistral male produces all sinistral offspring. What is the female's genotype?
Answer. dd. Offspring coiling reflects the mother's genotype, so a mother whose offspring are all sinistral must be dd. She herself is dextral because her own mother carried at least one D allele.
A child has a deletion of 15q11 to q13 and severe intellectual disability, absent speech and seizures. From which parent did the deletion most likely come, and which gene's loss explains the neurological features?
Answer. From the mother. These are features of Angelman syndrome, caused by loss of the maternally expressed gene UBE3A in neurons; the paternal copy is silenced there by imprinting, so nothing replaces the missing maternal copy. The same deletion from the father would cause Prader Willi syndrome.