The unusual phenotype of a recessive disorder is determined by homozygosity for a recessive allele, and the unaffected phenotype is determined by the corresponding dominant allele. In Chapter 3 we saw that phenylketonuria (PKU) is a recessive phenotype. PKU is determined by an allele that we can call p, and the normal condition by P. Therefore, sufferers of this disease are of genotype p/p, and unaffected people are either P/P or P/p. What patterns in a pedigree would reveal such an inheritance? Two key points are that generally the disease appears in the progeny of unaffected parents and that the affected progeny include both males and females equally. When we know that both male and female phenotypic proportions are equal, we can assume that we are dealing with autosomal inheritance, not X-linked inheritance. The following typical pedigree illustrates the key point that affected children are born to unaffected parents:

From this pattern we can immediately deduce autosomal inheritance, with the recessive allele responsible for the exceptional phenotype (indicated by shading). Furthermore, we can deduce that the parents must both be heterozygotes, P/p. (Both must have a p allele because each contributed one to each affected child, and both must have a P allele because the people are phenotypically normal.) We can identify the genotypes of the children (in the order shown) as P/–, p/p, p/p, and P/–. Hence, the pedigree can be rewritten

Notice another interesting feature of pedigree analysis: even though Mendelian rules are at work, Mendelian ratios are rarely observed in single families because the sample sizes are too small. In the above example, we see a 1:1 phenotypic ratio in the progeny of what is clearly a monohybrid cross, in which we might expect a 3:1 ratio. If the couple were to have, say, 20 children, the ratio would undoubtedly be something like 15 unaffected children and 5 with PKU (the expected monohybrid 3:1 ratio), but in a sample of four any ratio is possible and all ratios are commonly found.

In the case of a rare recessive allele, in the population most of these alleles will be found in heterozygotes, not in homozygotes. The reason is a matter of probability: to conceive a recessive homozygote, both parents must have had the p allele, but to conceive a heterozygote all that is necessary is one parent with the allele. The formation of an affected individual usually depends on the chance union of unrelated heterozygotes, and for this reason the pedigrees of autosomal recessives look rather bare, generally with only siblings of one cross affected.

Inbreeding (mating between relatives) increases the chance that a mating will be between two heterozygotes. An example of a cousin marriage is shown in Figure 4-18. Individuals III-5 and III-6 are first cousins and produce two children. You can see from the figure that an ancestor who is a heterozygote may produce many descendants who are also heterozygotes. Matings between relatives thus run a higher risk of producing abnormal homozygous recessives than do matings between nonrelatives. It is for this reason that first cousin marriages are responsible for a large portion of recessive diseases in human populations.

Figure 4-18

Pedigree of a rare recessive phenotype determined by a recessive allelea. Gene symbols normally are not included in pedigree charts, but genotypes are inserted here for reference. Note that individuals II-1 andII-5 marry into the family; (more...)

Albinism (Figure 4-19) is another rare condition that is inherited in a Mendelian manner as an autosomal recessive phenotype in many animals, including humans. The striking “white” phenotype is caused by a defect in an enzyme that synthesizes melanin, the pigment responsible for most black and brown coloration of animals. In humans, such coloration is most evident in hair, skin, and retina, and its absence in albinos (who have the homozygous recessive genotype a/a) leads to white hair, white skin, and eye pupils that are pink because of the unmasking of the red hemoglobin pigment in blood vessels in the retina. The inheritance and molecular genetics of albinism are integrated in Figure 4-20.

Figure 4-19

An albino. The phenotype is caused by homo-zygosity for a recessive allele, say,a/a. The dominant alleleA determines one step in the chemical synthesis of the dark pigment melanin in the cells of skin, hair, and eye retinas. In (more...)

Figure 4-20

Genetics and the molecular biology of albinism. In the pedigree, parents heterozygous for the recessive albinism allele produce three A/– progeny, who have melanin in their cells, and one a/a male, who is albino. The three panels (more...)

In autosomal dominant disorders, the normal allele is recessive and the abnormal allele is dominant. It might seem paradoxical that a rare disorder can be dominant, but remember that dominance and recessiveness are simply reflections of how alleles act and are not defined in terms of predominance in the population. An example of a rare autosomal dominant phenotype is achondroplasia, a type of dwarfism (see Figure 4-21). In this case, people with normal stature are genotypically d/d, and the dwarf phenotype in principle could be D/d or D/D. However, it is believed that in D/D individuals the two “doses” of the D allele produce such a severe effect that this genotype is lethal. If true, all achondroplastics are heterozygotes.

Figure 4-21

The human achondroplasia pheno-type, illustrated by a family of five sisters and two brothers. The pheno-type is determined by a dominant allele, which we can call D, that interferes with bone growth during development. Most members of (more...)

In pedigree analysis, the main clues for identifying an autosomal dominant disorder are that the phenotype tends to appear in every generation of the pedigree and that affected fathers and mothers transmit the phenotype to both sons and daughters. Again, the representation of both sexes among the affected offspring argues against X-linked inheritance. The phenotype appears in every generation because generally the abnormal allele carried by an individual must have come from a parent in the previous generation. (Abnormal alleles can arise de novo by mutation. This is relatively rare, but must be kept in mind as a possibility.) A typical pedigree for a dominant disorder is shown in Figure 4-22. Once again, notice that Mendelian ratios are not necessarily observed in families. As with recessive disorders, individuals bearing one copy of the rare allele (A/a) are much more common than those bearing two copies (A/A), so most affected people are heterozygotes, and virtually all matings involving dominant disorders are A/a × a/a. Therefore, when the progeny of such matings are totaled, a 1:1 ratio is expected of unaffected (a/a) to affected individuals (A/a).

Figure 4-22

Pedigree of a dominant phenotype determined by a dominant allele A. In this pedigree, all the genotypes have been deduced.

Huntington’s disease is an example of an autosomal dominant disorder. The phenotype is one of neural degeneration, leading to convulsions and premature death. However, it is a late-onset disease, the symptoms generally not appearing until after the person has begun to have children. Each child of a carrier of the abnormal allele stands a 50 percent chance of inheriting the allele and the associated disease. This tragic pattern has led to a drive to find ways of identifying people who carry the abnormal allele before they experience the onset of the disease. The discovery of the molecular nature of the mutant allele, and of neutral DNA mutations that act as “markers” close to the affected allele on the chromosome, has revolutionized this sort of diagnosis.

In human populations there are many examples of polymorphisms (generally dimorphisms) in which the alternative phenotypes of the character are determined by alleles of a single gene, for example, the dimorphisms for chin dimple versus none, attached earlobes versus unattached, widow’s peak versus none, and so on. The interpretation of pedigrees for dimorphisms is somewhat different from those for rare disorders, because by definition the morphs in a dimorphism are common. Let’s look at a pedigree for an interesting human dimorphism. Most human populations are dimorphic for the ability to taste the chemical phenylthiocarbamide (PTC): people can either detect it as a foul, bitter taste or—to the great surprise and disbelief of tasters—cannot taste it at all. From the pedigree in Figure 4-23, we can see that two tasters sometimes produce nontaster children. This makes it clear that the allele for ability to taste is dominant and that the allele for nontasting is recessive. Notice, however, that almost all people who marry into this family carry the recessive allele either in heterozygous or in homozygous condition. Such a pedigree thus differs from those of rare recessive disorders, for which it is conventional to assume that all who marry into a family are homozygous normal. As both PTC alleles are common, it is not surprising that all but one of the family members in this pedigree married individuals with at least one copy of the recessive allele.

Figure 4-23

Pedigree for the ability to taste the chemical PTC.

Few phenotypes determined by alleles on the differential region of the X chromosome are related to sex determination. Phenotypes with X-linked recessive inheritance typically show the following patterns in pedigrees:

1.

Many more males than females show the phenotype under study. This is because a female showing the phenotype can result only from a mating in which both the mother and the father bear the allele (for example, X ^A /X ^a  × X ^a /Y), whereas a male with the phenotype can be produced when only the mother carries the allele. If the recessive allele is very rare, almost all individuals showing the phenotype are males.

2.

None of the offspring of an affected male are affected, but all his daughters must be heterozygous “carriers” because females must receive one of their X chromosomes from their fathers. Half the sons born to these carrier daughters are affected (Figure 4-24).

Perhaps the best-known example is hemophilia, a malady in which a person’s blood fails to clot. Many proteins must interact in sequence to make blood clot. The most common type of hemophilia is caused by the absence or malfunction of one of these proteins, called factor VIII. The most famous cases of hemophilia are found in the pedigree of the interrelated royal families of Europe (Figure 4-25). The original hemophilia allele in the pedigree arose spontaneously (as a mutation) in the reproductive cells of Queen Victoria’s parents or of Queen Victoria herself. Alexis, the son of the last czar of Russia, inherited the allele ultimately from Queen Victoria, who was the grandmother of his mother Alexandra. Nowadays, hemophilia can be treated, but it was formerly a potentially fatal condition. It is interesting to note that in the Jewish Talmud there are rules about exemptions to male circumcision which show clearly that the mode of transmission of the disease through unaffected carrier females was well understood in ancient times. For example, one exemption was for the sons of women whose sisters’ sons had bled profusely when they were circumcised.

Duchenne muscular dystrophy is a fatal X-linked recessive disease. The phenotype is a wasting and atrophy of muscles. Generally the onset is before the age of 6, with confinement to a wheelchair by 12 and death by 20. The gene for Duchenne muscular dystrophy has now been isolated and shown to encode a muscle protein, dystrophin. Such insight holds out hope for a better understanding of the physiology of this condition and, ultimately, a therapy.

A rare X-linked recessive phenotype that is interesting from the point of view of sexual differentiation is a condition called testicular feminization syndrome, which has a frequency of about 1 in 65,000 male births. People afflicted with this syndrome are chromosomally males, 44A + XY, but they develop as females (Figure 4-26). They have female external genitalia, a blind vagina, and no uterus. Testes may be present either in the labia or in the abdomen. Although many such people are happily married, they are, of course, sterile. The condition is not reversed by treatment with male hormone (androgen), so it is sometimes called androgen insensitivity syndrome. The reason for the insensitivity is that the causative allele codes for a malfunctioning androgen receptor protein, so male hormone can have no effect on the target organs that are involved in maleness. In humans, femaleness results when the male-determining system is not functional.

Figure 4-24

Pedigree showing that X-linked recessive alleles expressed in males are then carried unexpressed by their daughters in the next generation, to be expressed again in their sons. Note that III-3 and III-4 cannot be distinguished phenotypically. (more...)

Figure 4-25

The inheritance of the X-linked recessive condition hemophilia in the royal families of Europe. A recessive allele causing hemophilia (failure of blood clotting) arose in the reproductive cells of Queen Victoria, or one of her parents, through (more...)

Figure 4-26

Four siblings with testicular feminization syndrome (congenital insensitivity to androgens). All four subjects in this photograph have 44 autosomes plus an X and a Y, but they have inherited the recessive X-linked allele conferring insensitivity (more...)

Pedigrees of rare X-linked dominant phenotypes show the following characteristics:

1.

Affected males pass the condition on to all their daughters but to none of their sons (Figure 4-27).

2.

Females married to unaffected males pass the condition on to half their sons and daughters.

Figure 4-27

Pedigree of an X-linked dominant disorder.

There are few examples of X-linked dominant phenotypes in humans. One is hypophosphatemia, a type of vitamin D–resistant rickets.

The mechanisms of X-linked dominance and recessiveness in humans are somewhat complicated by the phenomenon of X chromosome inactivation found in mammals. This topic will be covered in Chapter 16.

When a disease allele is known to be present in a family, knowledge of simple gene transmission patterns can be used to calculate the probability of prospective parents’ having a child with the disorder. For example, a married couple finds out that each had an uncle with Tay-Sachs disease (a severe autosomal recessive disease). The pedigree is as follows:

The probability of their having a child with Tay-Sachs can be calculated in the following way. The question is whether or not the man and woman are heterozygotes (it is clear that they do not have the disease) because if they are both heterozygotes then they stand a chance of having an affected child.

1.

The man’s grandparents must have both been heterozygotes T/t because they produced a t/t child (the uncle). Therefore, they effectively constituted a monohybrid cross. The man’s father could be T/T or T/t, but we know that the relative probabilities of these genotypes must be 1/4 and 1/2, respectively (the expected progeny ratio in a monohybrid cross is 1/4 T/T, 1/2 T/t, and 1/4 t/t). Therefore, there is a 2/3 probability that the father is a heterozygote [calculated as 1/2 divided by (+ 1/4+1/2)].

2.

The man’s mother must be assumed to be T/T, since she married into the family and disease alleles generally are rare. Thus if the father is T/t, then the mating to the mother was a cross T/t × T/T and the expected progeny proportions are 1/2 T/T and 1/2 T/t.

3.

The overall probability of the man’s being a heterozygote must be calculated using a statistical rule called the product rule, which states that the probability of two independent events both occurring is the product of their individual probabilities. Hence the probability of the man’s being a heterozygote is the probability of his father’s being a heterozygote times the probability of the father having a heterozygous son, which is 2/3 × 1/2 = 1/3.

4.

Likewise the probability of the man’s wife being heterozygous is also 1/3.

5.

If they are both heterozygous (T/t), then the probability of their having a t/t child is 1/4, so overall the probability of the couple having an affected child is 1/3 × 1/3 × 1/4 = 1/36; in other words, a 1 in 36 chance.