Hartl, Jones - Genetics. Principlers and analysis - 1998 (522927), страница 24
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. . .[Wehave found that a] 500 kb segment is the most likely site of the genetic defect. [The abbreviation kb stands forkilobase pairs; 1 kb equals 1000 base pairs.] Within this region, we have identified a large gene, spanningapproximately 210 kb, that encodes a previously undescribed protein. The reading frame contains a polymorphic(CAG)n trinucleotide repeat with at least 17 alleles in the normal population, varying from 11 to 34 CAG copies.On HD chromosomes, the length of the trinucleotide repeat is substantially increased.
. . . Elongation of atrinucleotide repeat sequence has been implicated previously as the cause of three quite different humandisorders, the fragile-X syndrome, myotonic dystrophy, and spino-bulbar muscular atrophy. . . . It can beexpected that the capacity to monitor directly the size of the trinucleotide repeat in individuals "at risk" for HDwill revolutionize testing for the disorder. . . . We consider it of the utmost importance that the currentinternationally accepted guidelines and counseling protocols for testing people at risk continue to be observed,and that samples from unaffected relatives should not be tested inadvertently or without full consent. .
. . With themystery of the genetic basis of HD apparently solved, [it opens] the next challenges in the effort to understandand to treat this devastating disorder.Source: Cell 72:971–983affected parent are affected. These are characteristic features of simple Mendelian dominance. The dominant allele,HD, that causes Huntington disease is very rare. All affected persons in the pedigree have the heterozygousgenotype HD hd, whereas nonaffected persons have the homozygous normal genotype hd hd.A pedigree pattern for a trait due to a homozygous recessive allele is shown in Figure 2.18.
The trait is albinism,absence of pigment in the skin, hair, and iris of the eyes. Both sexes can be affected, but the affected individualsneed not have affected parents. The nonaffected parents are called carriers because they are heterozygous for therecessive allele; in a mating between carriers (Aa × Aa), each offspring has a 1/4 chance of being affected. Thepedigree also illustrates another feature found withPage 54Figure 2.18Pedigree of albinism.
With recessive inheritance, affected persons (filledsymbols) often have unaffected parents. The double horizontal lineindicates a mating between relatives—in this case, first cousins.recessive traits, particularly rare traits, which is that the parents of affected individuals are often related.
A matingbetween relatives, in this case first cousins, is indicated with a double line connecting the partners.Matings between relatives are important for observing rare recessive alleles, because when a recessive allele israre, it is more likely to become homozygous through inheritance from a common ancestor than from parents whoare completely unrelated. The reason is that the carrier of a rare allele may have many descendants who arecarriers. If two of these carriers should mate (for example, in a first-cousin mating) the recessive allele can becomehomozygous with a probability of 1/4. Mating between relatives constitutes inbreeding, and the consequences ofinbreeding are discussed further in Chapter 15.2.5—Genetic AnalysisWhen a geneticist makes the statement that a single gene with alleles P and p determines whether the color of theflowers on a pea plant will be purple or white, the statement does not imply that this gene is the only oneresponsible for flower color.
The statement means only that this particular gene affecting flower color has beenidentified owing to the discovery of the recessive p mutation that, when homozygous, changes the color frompurple to white. Many genes beside P are also necessary for purple flower coloration. Among these are genes thatencode enzymes in the biochemical pathway for the synthesis of the purple pigment, anthocyanin. A geneticistinterested in understanding the genetic basis of flower color would rarely be satisfied in having identified only theP gene necessary for purple coloration.
The ultimate goal of a genetic analysis of flower color would be to isolateat least one mutation in every gene necessary for purple coloration and then, through further study of the mutantphenotypes, determine the normal function of each of the genes that affect the trait.The Complementation Test in Gene IdentificationIn a genetic analysis of flower color, a geneticist would begin by isolating many new mutants with white flowers.Although mutations are usually very rare, their fre-Page 55quency can be increased by treatment with radiation or certain chemicals.
The isolation of a set of mutants, all ofwhich show the same type of defect in phenotype, is called a mutant screen. Among the mutants that are isolated,some will contain mutations in genes already identified. For example, a genetic analysis of flower color in peasmight yield one or more new mutations that changed the wildtype P allele into a defective p allele that prevents theformation of the purple pigment. Each of the Pp mutations might differ in DNA sequence, but all of the newlyisolated p alleles would be defective forms of P that prevent formation of the purple pigment. On the other hand, amutant screen should also yield mutations in genes not previously identified. Each of the new genes might be alsorepresented by several recurrences of mutation, analogous to the multiple Pp mutations.In a mutant screen for flower color, all of the new mutations are identified in plants with white flowers.
Most of thenew mutant alleles will encode an inactive protein needed for the formation of the purple pigment. The mutantalleles will be recessive, because in the homozygous recessive genotype, neither of the mutant alleles can producethe wildtype protein needed for pigmentation, and so the flowers will be white. In the heterozygous genotype,which carries one copy of the mutant allele along with one copy of the wildtype allele, the flowers will be purplebecause the wildtype allele codes for a functional protein that compensates for the defective protein encoded by themutant.Because white flowers may be caused by mutations in any of several genes, any two genotypes with white flowersmay be homozygous recessive for alleles of the same gene or for alleles of different genes. After a mutant screen,how can the geneticist determine which pairs of mutations are alleles and which pairs of mutations are not alleles?The issue is illustrated for three particular white-flower mutations in Figure 2.19.
Each of the varieties ishomozygous for a recessive mutation that causes the flowers to be white. On the one hand, the varieties might carryseparate occurrences of a mutation in the same gene; the mutations would be alleles. On the other hand, eachmutation might be in a different gene; they would not be alleles.The issue of possible allelism of the mutations is resolved by observing the phenotype of the progeny producedfrom a cross between the varieties.
As indicated in Figure 2.19. there are two possible outcomes of the cross. TheF1 progeny have either the wildtype phenotype (Figure 2.19A, purple flowers) or the mutant phenotype (Figure2.19B, white flowers). If the progeny have purple flowers, it means that the mutations in the parental plants are indifferent genes; this result is called complementation. When complementation is observed, it implicates twodifferent genes needed for purple flowers. In this example, mutant strain 1 is homozygous pp for the recessive pallele. The other parent is homozygous for a mutant allele in a different gene, designated cc.
The completegenotype of the parental strains should therefore be written pp CC for mutant strain 1 and PP cc for mutant strain2. The cross yields the F1 genotype Pp Cc, which is heterozygous Pp and heterozygous Cc. Because p and c areboth recessive, the phenotype of the F1 progeny is purple flowers.The other possible outcome of the cross is shown in Figure 2.19B. In this case, the F1 progeny have white flowers,which is the mutant phenotype. This outcome is called noncomplementation. The lack of complementationindicates that both parental strains have a mutation in the same gene, because neither mutant strain can provide thegenetic function missing in the other.
In this example, mutant strain 1 is known to have the genotype pp. Mutantstrain 3 is homozygous for a different mutation (possibly a recurrence of p). Because the F1 has white flowers, thegenotype of the F1 must be pp. But the only source of the second p allele must be mutant strain 3, which means thatmutant strain 3 is also homozygous pp. In the figure, the particular allele in mutant strain 3 is designated p* toindicate that this mutation arose independently of the original p allele.The kind of cross illustrated in Figure 2.19 is a complementation test.
As we have seen, it is used to determinewhether recessive mutations in each of two different strains are alleles of the same gene.Page 56Figure 2.19The complementation test reveals whether two recessive mutations are alleles of the same gene.In the complementation test, homozygous recessive genotypes are crossed. If the phenotype ofthe F1 progeny is nonmutant (A), it means that the mutations in the parental strains are allelesof different genes. If the phenotype of the F1 progeny is mutant (B), it means that the mutationsin the parental strains are alleles of the same gene.Because the result indicates the presence or absence of allelism, the complementation test is one of the keyexperimental operations in genetics.
To illustrate the application of the test in practice, suppose a mutant screenwere carried out to isolate new mutations for white flowers in peas. Starting with a true-breeding strain with purpleflowers, we treat pollen with x rays and use the irradiated pollen to fertilize ovules to obtain seeds. The F1 seeds aregrown and the resulting plants allowed to self-fertilize, after which the F2 plants are grown. A few of the F1 seedsmay contain a new mutation for white flowers, but because the white phenotype is recessive, the flower will bepurple. However, the resulting F1 plant will be heterozygous for the new white mutation, so selffertilization willresult in the formation of F2 plants with a 3 : 1 ratio of purple : white flowers. Because mutations resulting in aparticular phenotype are quite rare, even when induced by radiation, only a few among many thousands of selffertilized plants, will be found to have a new white-Page 57flower mutation.
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