Hartl, Jones - Genetics. Principlers and analysis - 1998 (522927), страница 42
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In birds, moths, butterflies, and some reptiles, the situation is the reverse:Females are the heterogametic sex (WZ) and males the homogametic sex (WW). The Y chromosome in manyspecies contains only a few genes. In human beings and other mammals, the Y chromosome includes a maledetermining factor. In Drosophila, sex is determined by a male-specific or female-specific pattern of geneexpression that is regulated by the ratio of the number of X chromosomes to the number of sets of autosomes. Inmost organisms, the X chromosome contains many genes unrelated to sexual differentiation.
These X-linked genesshow a characteristic pattern of inheritance that is due to their location in the X chromosome.The progeny of genetic crosses often conform to the theoretical predictions of the binomial probability formula.The degree to which the observed numbers of different genetic classes of progeny fit theoretically expectednumbers is usually found with a chi-square (X2) test. On the basis of the criterion of the X2 test, Mendel's data fitthe expectations somewhat more closely than chance would dictate. However, the bias in the data is relativelysmall and is unlikely to be due to anything more than recounting or repeating certain experiments whose resultswere regarded as unsatisfactory.Key Termsanaphaseheterogameticsporeanaphase Ihighly significantsporeanaphase IIhomogameticstatistical significanceautosomeshomologoussynapsisbivalentinterphasetelophasecell cyclekinetochoretelophase Icentromereleptotenetelophase IIchiasmaM periodtetradchi-squaremeiocyteX chromosomechromatidmeiosisX-linkedchromomeremetaphaseY chromosomechromosomemetaphase platechromosome complementmetaphase Icrossing-overmetaphase IIcyclinsmitosiscytokinesismitotic spindledegrees of freedomnondisjunctiondiakinesisnucleolusdiploidpachytenediplotenePascal's triangleequational divisionprophasefirst meiotic divisionprophase IG1 periodprophase IIG2 periodreductional divisiongametophyteS phasegerm cellsecond meiotic divisiongoodness of fitsister chromatidshaploidsex chromosomehempphilia Asomatic cellzygotenePage 117Review the Basics• Explain the following statement: "Independent alignment of nonhomologous chromosomes at metaphase I ofmeiosis is the physical basis of independent assortment of genes on different chromosomes."• Draw a diagram of a bivalent, and label the following parts: centromere, sister chromatids, nonsister chromatids,homologous chromosomes, chiasma.• What is the genetic consequence of the formation of a chiasma in a bivalent?• T.
H. Morgan discovered X-linkage by following up his observation that reciprocal crosses in which one parentwas wildtype for eye color and the other had white eyes yielded different types of progeny. Diagram the reciprocalcrosses, indicating the X and Y chromosomal genotypes of each parent and each class of offspring.• What does it mean to say that two outcomes of a cross are mutually exclusive? What does it mean to say that twooutcomes of a cross are independent?• In what way does the chi-square value indicate "goodness of fit"?• What are the conventional P values for "significant" and "highly significant" and what do these numbers mean?• If Mendel did discard the results of some experiments because he considered them excessively deviant as a resultof to pollen contamination or some other factor, do you consider this a form of "cheating"? Why or why not?Guide to Problem SolvingProblem 1: The black and yellow pigments in the fur of cats are determined by an X-linked pair of alleles, cb(black) and cy (yellow).
Males are black (cb) or yellow (cy), and females are either homozygous black cbcb,homozygous yellow (cycy), or heterozygous (cbcy). The phenotype of the heterozygous female has patches of blackand patches of yellow, a pattern knows as calico. (The white spotting usually also present in domestic short-haircats is caused by a separate gene.)(a) What genotypes and phenotypes would be expected among the offspring of a cross between a black female anda yellow male?(b) In a litter of eight kittens, there are two calico females, one yellow female, two black males, and three yellowmales. What are the genotypes and phenotypes of the parents?(c) Rare calico males are the result of nondisjunction.
What are their sex-chromosome constitution and theirgenotype?Answer: (a) The black female has genotype cbcb and the yellow male has genotype cy. The female offspring receivean X chromosome from each parent, so their genotype is cbcy and their phenotype is calico.
The male offspringreceive an X chromosome from their mother, so their genotype is cb and their phenotype is black. (b) The maleoffspring provide information about the X chromosomes in the mother. Because some males are black (cb) andsome yellow (cy), the mother must have the genotype cbcy and have a calico coat. The fact that one of the offspringis a yellow female means that both parents carry an X chromosome with the cy allele, so the father must have thegenotype cy and have a yellow coat.
The occurrence of the calico female offspring is also consistent with theseparental genotypes. (c) Nondisjunction in a female of genotype cbcy can produce an XX egg with the alleles cb andcy. When the XX egg is fertilized by a Y-bearing sperm, the result is an XXY male of genotype cbcy, which is thegenotype of a male calico cat. (The XXY males are sterile but otherwise are similar to normal males.)Problem 2: Certain breeds of chickens have reddish gold feathers because of a recessive allele, g, on the Wchromosome (the avian equivalent of the X chromosome); presence of the dominant allele results in silverplumage. An autosomal recessive gene, s, results in feathers called silkie that remain soft like chick down. Whatgenotypes and phenotypes of each sex would be expected from a cross of a red rooster heterozygous for silkie anda silkie hen with silver plumage? (Remember that, in birds, females are the heterogametic sex, WZ, and males thehomogametic sex, WW.)Answer: In this problem, you need to keep track of both the W-linked and the autosomal inheritance, and thesituation has the additional complication that males are WW and females WZ for the sex chromosomes.
Theparental red rooster that is heterozygous for silkie has the genotype gg for the W chromosome and Ss for therelevant autosome, and the silver, silkie hen has the genotype G for the W chromosome and ss for the autosome.The female (WZ) offspring from the cross receive their W chromosome from their father (the reverse of thesituation in most animals), so they have genotype g (reddish gold feathers) and half of them are Ss (wildtype) andhalf ss (silkie). The male (WW) offspring have genotype Gg (silver feathers), and again, half are Ss (wildtype) andhalf ss (silkie).Page 118Chapter 3 GeNETics on the webGeNETics on the web will introduce you to some of the most important sites for finding genetic information onthe Internet.
To complete the exercises below, visit the Jones and Bartlett home page athttp://www.jbpub.com/geneticsSelect the link to Genetics: Principles and Analysis and then choose the link to GeNETics on the web. You willbe presented with a chapter-by-chapter list of highlighted keywords.GeNETics EXERCISESSelect the highlighted keyword in any of the exercises below, and you will be linked to a web site containing thegenetic information necessary to complete the exercise. Each exercise suggests a specific, written report thatmakes use of the information available at the site.
This report, or an alternative, may be assigned by yourinstructor.1. See meiosis in action by using this keyword. Then locate the still photographs of chromosomes in variousstages of prophase I. How many bivalents are formed in this organism? What is the diploid chromosome numberof the organism? If assigned to do so, identify each of the prophase I stages depicted as leptotene, zygotene,pachytene, diplotene, or diakinesis.2. Some of the main characteristics of X-linked inheritance can be examined at this site. Mate wildtype femaleswith white males and observe the results.
Then cross the F1 females with their white-eye fathers and observe theresults. Make a diagram of the crosses, giving the genotypes and phenotypes of all the flies and the numbersobserved in each class of offspring. If assigned to do so, prepare a similar report of an initial mating of white-eyedfemales with wildtype males, followed by mating of the F1 female progeny with their wildtype fathers.3. FlyBase is the main internet repository of information about Drosophila genetics.
Using the keyword white,you can learn about the metabolic defect in the mutation that Thomas Hunt Morgan originally discovered. Enterthe keyword into the search engine, and(text box continued on next page)Problem 3: Ranch mink with the dark gray coat color known as aleutian are homozygous recessive, aa. GenotypesAA and Aa have the standard deep brown color. A mating of Aa × Aa produces eight pups.(a) What is the probability that none of them has the aleutian coat color?(b) What is the probability of a perfect 3 : 1 distribution of standard to aleutian?(c) What is the probability of a 1 : 1 distribution in this particular litter?Answer: This kind of problem demonstrates the effect of chance variation in segregation ratios in small sibships.
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