In the study of rare disorders, four general patterns of inheritance are distinguishable by pedigree analysis: autosomal recessive, autosomal dominant, X-linked recessive, and X-linked dominant.<\/p>\n
The affected phenotype of an autosomal recessive disorder is determined by a recessive allele, and the corresponding unaffected phenotype is determined by a dominant allele. For example, the human disease phenylketonuria is inherited in a simple Mendelian manner as a recessive phenotype, with PKU determined by the allele p and the normal condition by P . Therefore, sufferers from this disease are of genotype p \/p , and people who do not have the disease are either P \/P or P \/p . What patterns in a pedigree would reveal such an inheritance? The two key points are that (1) generally the disease appears in the progeny of unaffected parents and (2) the affected progeny include both males and females. When we know that both male and female progeny are affected, we can assume that we are dealing with simple Mendelian inheritance, not sex-linked inheritance. The following typical pedigree illustrates the key point that affected children are born to unaffected parents: <\/p>\n
From this pattern, we can immediately deduce simple Mendelian inheritance of the recessive allele responsible for the exceptional phenotype (indicated in black). Furthermore, we can deduce that the parents are both heterozygotes, say A \/a ; both must have an a allele because each contributed an a allele to each affected child, and both must have an A allele because they are phenotypically normal. We can identify the genotypes of the children (in the order shown) as A \/, a \/a , a \/a , and A \/. Hence, the pedigree can be rewritten as follows: <\/p>\n
Note that this pedigree does not support the hypothesis of X-linked recessive inheritance, because, under that hypothesis, an affected daughter must have a heterozygous mother (possible) and a hemizygous father, which is clearly impossible, because he would have expressed the phenotype of the disorder. <\/p>\n
Notice another interesting feature of pedigree analysis: even though Mendelian rules are at work, Mendelian ratios are rarely observed in families, because the sample size is too small. In the preceding example, we see a 1:1 phenotypic ratio in the progeny of a monohybrid cross. If the couple were to have, say, 20 children, the ratio would be something like 15 unaffected children and 5 with PKU (a 3:1 ratio); but, in a sample of 4 children, any ratio is possible, and all ratios are commonly found. <\/p>\n
The pedigrees of autosomal recessive disorders tend to look rather bare, with few black symbols. A recessive condition shows up in groups of affected siblings, and the people in earlier and later generations tend not to be affected. To understand why this is so, it is important to have some understanding of the genetic structure of populations underlying such rare conditions. By definition, if the condition is rare, most people do not carry the abnormal allele. Furthermore, most of those people who do carry the abnormal allele are heterozygous for it rather than homozygous. The basic reason that heterozygotes are much more common than recessive homozygotes is that, to be a recessive homozygote, both parents must have had the a allele, but, to be a heterozygote, only one parent must carry the a allele. <\/p>\n
Geneticists have a quantitative way of connecting the rareness of an allele with the commonness or rarity of heterozygotes and homozygotes in a population. They obtain the relative frequencies of genotypes in a population by assuming that the population is in Hardy-Weinberg equilibrium, to be fully discussed in Chapter 24 . Under this simplifying assumption, if the relative proportions of two alleles A and a in a population are p and q , respectively, then the frequencies of the three possible genotypes are given by p 2 for A \/A , 2pq for A \/a , and q 2 for a \/a . A numerical example illustrates this concept. If we assume that the frequency q of a recessive, disease-causing allele is 1\/50, then p is 49\/50, the frequency of homozygotes with the disease is q 2 =(1\/50)2 =1\/250, and the frequency of heterozygotes is 2pq =249\/501\/50 , or approximately 1\/25. Hence, for this example, we see that heterozygotes are 100 times as frequent as disease sufferers, and, as this ratio increases, the rarer the allele becomes. The relation between heterozygotes and homozygotes recessive for a rare allele is shown in the following illustration. Note that the allele frequencies p and q can be used as the gamete frequencies in both sexes. <\/p>\n
The formation of an affected person usually depends on the chance union of unrelated heterozygotes. However, inbreeding (mating between relatives) increases the chance that a mating will be between two heterozygotes. An example of a marriage between cousins is shown in . Individuals III-5 and III-6 are first cousins and produce two homozygotes for the rare allele. You can see from that an ancestor who is a heterozygote may produce many descendants who also are heterozygotes. Hence two cousins can carry the same rare recessive allele inherited from a common ancestor. For two unrelated persons to be heterozygous, they would have to inherit the rare allele from both their families. Thus matings between relatives generally run a higher risk of producing abnormal phenotypes caused by homozygosity for recessive alleles than do matings between nonrelatives. For this reason, first-cousin marriages contribute a large proportion of the sufferers of recessive diseases in the population. <\/p>\n
Pedigree of a rare recessive phenotype determined by a recessive allele a . Gene symbols are normally not included in pedigree charts, but genotypes are inserted here for reference. Note that individuals II-1 and II-5 marry into the family; they are assumed (more...) <\/p>\n
What are some examples of human recessive disorders? PKU has already served as an example of pedigree analysis, but what kind of phenotype is it? PKU is a disease of processing of the amino acid phenylalanine, a component of all proteins in the food that we eat. Phenylalanine is normally converted into tyrosine by the enzyme phenylalanine hydroxylase: <\/p>\n
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