Hartl, Jones - Genetics. Principlers and analysis - 1998 (522927), страница 30
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Codominance is often observed for proteins when eachalternative allele codes for a different amino acid replacement, because it may be possible to distinguish thealternative forms of the protein by chemical or physical means. Genes are not always expressed to the same extentin different organisms; this phenomenon is called variable expressivity. A genotype that is not expressed at all insome organisms is said to have incomplete penetrance.Page 73Key Termsaddition ruleheterozygousPunnett SquarealbinismhomozygousrecessivealleleHuntington diseasereciprocal crossamorphhybridsegregationantibodyhypermorphsibantigenhypomorphsiblingantimorphincomplete dominancesibshipbackcrossincomplete penetrancetestcrosscarrierindependent assortmenttrue breedingcodominanceMendelian geneticsvariable expressivitycomplementation groupmonohybridzygotecomplementation testmonohybriddihybridmultiplication ruledominantmutant screenectopic expressionneomorphF1generationP1generationF2generationpartial dominancegamatepedigreegenepenetrancegenotypephenotypeReview the Basics• What is segregation? How would the segregation of a pair of alleles be exhibited in the progeny of a testcross?• Explain the following statement: "Among the F2 progeny of a dihybrid cross, the ratio of genotypes is 1: 2 : 1, butamong the progeny that express the dominant phenotype, the ratio of genotypes is 1 : 2."• What is a mutant screen and how is it used in genetic analysis?• What is a complementation test? How does this test enable a geneticist to determine whether two differentmutations are or are not mutations in the same gene?• What do we mean by a "modified dihybrid F2 ratio"? Give two examples of a modified dihybrid F2 ratio andexplain the gene interactions that result in the modified ratio.• What is the distinction between incomplete dominance and codominance? Give an example of each.Guide to Problem SolvingProblem 1: In tomatoes, the shape of the fruit is inherited, and both round fruit and elongate fruit are true breeding.The cross round × elongate produces F1 progeny with round fruit, and the cross F1 × F1 produces 3/4progeny withround fruit and 1/4 progeny with elongate fruit.
What kind of genetic hypothesis can explain these data?Answer: In this kind of problem, a good strategy is to look for some indication of Mendelian segregation. The 3 : 1ratio in the F2 generation is characteristic of Mendelian segregation when there is dominance. This observationsuggests the genetic hypothesis of a dominant gene R for round fruit and a recessive allele r for elongate fruit. Ifthe hypothesis were correct, then the true-breeding round and elongate genotypes would be RR and rr, respectively.The F1 progeny of the cross RR (round) × rr (elongate) would be Rr, which has round fruit, as observed.
The F1 ×F1 cross (Rr × Rr) yields 1/4 RR, 1/2 Rr, and 1/4 rr. Because both RR and Rr have round fruit, the expected F2 ratioof round : elongate phenotypes is 3 : 1, as expected from the single-gene hypothesis.Problem 2: In Shorthorn cattle, both red coat color and white coat color are true breeding. Crosses of red × whiteproduce progeny that are uniformly reddish brown but thickly sprinkled with white hairs; this type of coat color iscalled roan. Crosses of roan × roan produce 1/4 : 1/2 roan : 1/4 white.
What kind of genetic hypothesis can explainthese data?Answer: In this case, the 1:2:1 ratio of phenotypes in the cross roan × roan suggests Mendelian segregation,because this is the ratio expected from a mating betweenPageChapter 2 GeNETics on the webGeNETics on the web will introduce you to some of the most important sites for finding genetic information on theInternet. 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 will bepresented 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 that makesuse of the information available at the site. This report, or an alternative, may be assigned by your instructor.1. Mendel's paper is one of the few nineteenth-century scientific papers that reads almost as clearly as though it hadbeen written today. It is important reading for every aspiring geneticist. You can access a conveniently annotatedtext by using the keyword Mendel. Although modern geneticists make a clear distinction between genotype andphenotype, Mendel made no clear distinction between these concepts. If assigned to do so, make a list of threespecific instances in Mendel's paper, each supported by a quotation, in which the concepts of genotype andphenotype are not clearly separated; rewrite each quotation in a way that makes the distinction clear.2.
Although the incidence of Huntingon disease is only 30 to 70 per million people in most Western countries, ithas received great attention in genetics because of its late age of onset and autosomal dominant inheritance. Use thekeyword to learn more about this condition. Under the heading History there is an Editor's Note quoting the blindseer Tiresias confronting Oedipus with the following paradox: "It is sorrow to be wise when wisdom profits not." Ifassigned to do so, write a 250-word essay explaining what this means in reference to Huntington disease and whyDNA-based diagnosis is regarded as an ethical dilemma.3.
The red and purple colors of flowers, as well as of autumn leaves, result from members of a class of pigmentscalled anthocyanins. The biochemical pathway for anthocyanin synthesis(text box continued to next page)heterozygotes when dominance is incomplete. Supposing that roan is heterozygous (say, Rr). Then the cross roan ×roan (Rr × Rr) is expected to produce 1/4 RR, 1/2 Rr, and 1/2 rr genotypes. the observed result, that 1/2 of the progeare roan (Rr), fits this hypothesis, which implies that the RR and rr genotypes correspond to red and white.
Theproblem states that red and white are true breeding, which is consistent with their being homozygous genotypes.Additional confirmation comes from the cross RR × rr, which yields Rr (roan) progeny, as expected. Note that the gensymbols R and r are assigned to red and white arbitrarily, swo it does not matter whether RR stands for red and rr fowhite, or the other way around.Problem 3: The tailess trait in the mouse results from an allele of a gene in chromosome 17. The cross tailless ×tailless produces tailless and wildtype progency in a ratio of 2 tailless: 1 wildtype.
All tailless progency from this crowhen mated with wildtype, produce a 1 : 1 ratio of tailless to wildtype progeny.(a) Is the allele for the tailless trait dominant or recessive?(b) What genetic hypothesis can account for the 2 : 1 ratio of tailless: wildtype and the results of the crosses betweenthe tailless animlas?Answer: (a) If the tailles phenotype were homozygous recessive, then the cross tailless × tailless should produce onltailless progeny. This is not the case, so the tailless phenotype must result from a dominanat allele, say T.
(b) Becausthe cross tailless × tailless produces both tailless and wildtype progeny, both parents must be heterozygous Tt. Theexpected ration of genotypes among the zygotes is 1/4 TT, 1/2 Tt, and 1/4 tt. Because T is dominant, the Tt animals artailless and the tt animals are wildtype. The 2 : 1 ration can be explained if the TT zygotes do not survive (that is, theTT genotype is lethal). Because all surviving tailless animals must be Tt, this genetic hypothesis would also explain wall of the tailless animals from the cross, when mated with tt, give a 1 : 1 ration of tailless (Tt) to wildtype (tt).(Developmental studies confirm that about 25 percent of the embryous do not survive.)Problem 4: The accompanying illustration shows four alternative types of combs in chickens; they are called rose,pea, single, and walnut. The following data summarize the results of crosses.
The rose and pea strains used in crosse1, 2, and 5 are true breeding.1. rose × singlerose2. pea × singlepea3. (rose × single) F1 × (rose × single) F13 rose : 1 singlePage(text box continued from previous page)in the snapdragon, Antirrhinum majus, can be found at this keyword site. The enzyme responsible for the first step ithe pathway limits the amount of pigment formed, which explains why red and white flowers in Antirrhinum showincomplete dominance. If assigned to do so, identify the enzyme responsible for the first step in the pathway, andgive the molecular structures of the substrate (or substrates) and product.
Also, examine all of the intermediates inthe anthocyanin pathway, and identify which atom that is so prominent in the purine and pyrimidine bases is notfound in anthocyanin.MUTABLE SITE EXERCISESThe Mutable Site Exercise changes frequently. Each new update includes a different exercise that makes use ofgenetics resources avilable on the World Wide Web. Select the Mutable Site for Chapter 2, and you will be linkedto the current exercise that relates to the material presented in this chapter.PIC SITEThe Pic Site showcases some of the most visually appealing genetics sites on the World Wide Web.
To visit theshowcase genetics site, select the Pic Site for Chapter 2.4. (pea × single) F1 × (pea × single) F15. rose × pea3 pea : 1 singlewalnut6. (rose × pea) F1 × (rose × pea) F19 walnut : 3 rose : 3 pea : 1 single(a) What genetic hypothesis can explain these results?(b) What are the genotypes of parents and progeny in each of the crosses?(c) What are the genotypes of true-breeding strains of rose, pea, single, and walnut?Answer:(a) Cross 6 gives the Mendelian ratios expected when two genes are segregating, so a genetic hypothesis with twogenes is necessary.
Crosses 1 and 3 give the results expected if rose comb were due to a dominant allele (say, R).Crosses 2 and 4 give the results expected if pea comb were due to a dominant allele (say, P). Cross 5 indicates thatwalnut comb results from the interaction of R and P. The segregation in cross 6 means that R and P are not alleles ofthe same gene.(b) 1.
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