Hartl, Jones - Genetics. Principlers and analysis - 1998 (522927), страница 56
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These genes are separated by about 12 map units.Plants of genotype B- have purple flowers, and plants of the genotype bb have red flowers. Plants of genotype Ehave elongate pollen, and those of genotype ee have round pollen. Determine what genotypes and phenotypesamong the progeny would be expected from crosses of F1 hybrids × red, round when the F1 hybrids are obtained asfollows(a) purple, elongate × red, roundpurple, elongate F1(b) purple, round × red, elongatepurple, elongate F1Answer: The red, round parents have genotype b e/b e.
You must deduce the genotype of the F1 parent. Althoughboth F1 hybrids have the dominant purple, elongate phenotype, they are different kinds of double heterozygotes. (a)In this mating, both the b and e alleles come from the red, round parent, so the F1 genotype is the cis, or coupling,heterozygote B E/b e. (b) In this mating, the b allele comes from one parent and the e allele from the other, so the F1genotype is the trans, or repulsion, heterozygote B e/b E.
To calculate the types of progeny expected, use the factthat 1 map unit corresponds to 1 percent recombination: Because the genes are 12 map units apart, 12 percent of thegametes will be recombinant (6 percent of each of the reciprocal classes) and 88 percent will be nonrecombinant (44percent of each of the reciprocal classes).
Therefore, the two types of heterozygotes produce gametes with thefollowing expected percentages:GametesBE/beBe/bEBE446Be644bE644be446The answers to the specific questions are as follows.(a) BE/be (purple, elongate) 44 percent; Be/be (purple, round) 6 percent; bE/be (red, elongate) 6 percent; be/be (red,round) 44 percent.Page 166(b) BE/be (purple, elongate) 6 percent; Be/be (purple, round) 44 percent; bE/be (red, elongate) 44 percent; be/be(red, round) 6 percent.Problem 3: In Drosophila, the genes ct (cut wing margin), y (yellow body), and v (vermilion eye color) are Xlinked. Females heterozygous for all three markers were mated with wildtype males, and the following maleprogeny were obtained.
As is conventional in Drosophila genetics, the wildtype allele of each gene is designatedby a + sign in the appropriate column.ctyv4cty+93ct+v54ct++349+yv331+y+66++v97+++6Total1000(a) What was the genotype of the female parents?(b) What is the order of the genes?(c) Calculate the recombination frequencies and the interference between the genes.(d) Draw a genetic map of the genes.Answer: This is a conventional kind of three-point linkage problem.
The way to solve it is to start by deducing thegenotype of the triply heterozygous parent. This is done by noting that the most common classes of offspring arethe nonrecombinants, in this case ct+ + and +y v. Then deduce the order of the genes. This is done by noting thatthe least common classes of offspring are the double recombinants, in this case ct y v and + + +. Moreover, eachclass of double recombinants will be identical with one of the nonrecombinants except for the marker that is in themiddle. In this case, comparing ct y v with + y v and + + + with ct + + shows that ct is in the middle.
(a) Therefore,the answer to part (a) is that the mothers' genotype was + ct +/y + v. (b) The answer to part (b) is that ct is in themiddle. (It is not possible to determine from these data whether the gene order is y ct v or v ct y.)The next step is to rearrange the data with the genes in the correct order, with reciprocal chromosomes adjacent andwith nonrecombinants at the top, single recombinants for region 1 and 2 in the middle, and nonrecombinants at thebottom.+a+349 Nonrecombinanty+v331 Nonrecombinantya+93 Region 1 crossing-over++v97 Region 1 crossing-over+av54 Region 2 crossing-overy++66 Region 2 crossing-overyav4 Region 1 + Region 2 crossing-over+++6 Region 1 + Region 2 crossing-over(c) Now, to answer part (c), calculate the recombination frequencies.
The most common mistake is to ignore thedouble recombinants. In fact, the double recombinants are recombinant in both regions and so must be countedtwice—once in calculating the recombination frequency in region 1 and again in calculating the recombinationfrequency in region 2. The frequency of recombination in region 1 (y-ct) is (93 + 97 + 4 + 6)/1000 = 0.20, or 20percent.
The frequency of recombination in region 2 (ct-v) is (54 + 66 + 4 + 6)/1000 = 0.13, or 13 percent. Tocalculate the interference, note that the expected number of double recombinants is 0.20 × 0.13 × 1000 = 26,whereas the actual number is 10. The coincidence is the ratio 10/26 = 0.385, and therefore the interference acrossthe region is 1 - 0.385 = 0.615. (d) For part (d), the genetic map based on these data is shown below.Problem 4: In the yeast Saccharomyces cerevisiae, the gene MET14 codes for an enzyme used in synthesis of theamino acid methionine and is very close to the centromere of chromosome XI.
For simplicity, denote MET14 withthe symbol a, and suppose that a is used in mapping a new mutation—say, b—whose location in the genome isunknown. Diploids are made by mating a B × A b, in which the uppercase symbols denote the wildtype alleles, andasci are examined. In this mating, what kinds of asci correspond to parental ditype (PD), nonparental ditype (NPD)and tetratype (TT)? What relative proportions of PD, NPD and TT asci would be expected in each of the followingcases:(a) b closely linked with a?(b) b on a different chromosome from the one a is on but close to the centromere?(c) b on a different chromosome from the one a is on and far from the centromere?Answer: The genotype of the diploid is the trans, or repulsion, heterozygote a B/A b, so the parental ditype (PD)tetrads contain two a B spores and two A b spores; the nonparental ditype (NPD) tetrads contain two a b spores andtwo A B spores; and the tetratype (TT) tetrads contain one spore each of a B, A b, a b, and A B.
In this problem, thespecific questions illustrate the principle that any marker tightly linked to its centromere in a fungus withunordered tetrads serves to identify first-division and second-division segregation of every other gene in theorganism. This principle makes possible the mapping of other genes with respect to their centromeres. Reasoningthrough problems of tetrad analysis is best approached by considering the possible chromosome configurationsduring the meiotic divisions. The various possibilities are illustrated in the adjoining figure, in which parts Athrough C correspond to questions (a) through (c).Page 167(a) Part A shows the genes closely linked to each other (and therefore to the same centromere).
They are shown in different arms,but they could just as well be in the same arm. With no crossing-over between either gene and the centromere, all tetrads are PD.(b) In part B, the genes are closely linked to different centromeres. With no crossing-over between either gene and its centromere, halfthe tetrads are PD and half NPD. (c) In part C, the b gene is not centromere-linked.
Again PD = NPD (indicating that A and B arein different chromosomes), but crossing-over between the gene and the centromere yields TT tetrads. The frequency of TT tetradsincreases with the recombination frequency between the gene and the centromere, to a maximum of 2/3 when the gene andcentromere undergo free recombination.Analysis and Applications4.1 Each of two recessive mutations, m1 and m2, can produce albino guinea pigs. The genes are close together inone chromosome and may be allelic. The cross m1m1 × m2m2 produces progeny with wildtype pigmentation. Are themutations in the same gene?4.2 What gametes are produced by an individual of genotype A b/a B if the genes are(a) in different chromosomes?(b) in the same chromosome with no recombination between them?4.3 What gametes are produced by an individual of genotype A b/a B if recombination can occur between thegenes? Which gametes are the most frequent?4.4 In the absence of recombination, what genotypes of progeny are expected from the cross A B/a b × A b/a B? IfA and B are dominant, what ratio of phenotypes is expected?4.5 Which of the following arrays represents the cis configuration of the alleles A and B: a b/A B or a B/A b?4.6 What gametes, and in what ratios, are produced by male and female Drosophila of genotype A B/a b when thegenes are present in the same chromosome and the frequency of recombination between them is 5 percent?4.7 If a large number of genes were mapped, how many linkage groups would be found in(a) a haploid organism with 17 chromosomes per somatic cell?Page 168(b) a bacterial cell with a single, circular DNA molecule?(c) the rat, with 42 chromosomes per diploid somatic cell?4.8 If the recombination frequency between genes A and B is 6.2 percent, what is the distance between the genes inmap units in the linkage map?4.9 Consider an organism heterozygous for two genes located in the same chromosome, A B/a b.
If a singlecrossing-over occurs between the genes in every cell undergoing meiosis, and no multiple crossing-overs occurbetween the genes, what is the recombination frequency between the genes? What effect would multiple crossingover have on the recombination frequency?4.10 Normal plant height in wheat requires the presence of either or both of two dominant alleles, A and B. Plantshomozygous for both recessive alleles (a b/a b) are dwarfed but otherwise normal.
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